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Physical Chemistry
Understanding our Chemical World

Physical Chemistry
Understanding our Chemical World
Paul Monk
Manchester Metropolitan University, UK

Copyright  2004

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Library of Congress Cataloging-in-Publication Data
Monk, Paul M. S.
Physical chemistry : understanding our chemical world / Paul Monk.
p. cm.
Includes bibliographical references and index.
ISBN 0-471-49180-2 (acid-free paper) – ISBN 0-471-49181-0 (pbk. : acid-free paper)
1. Chemistry, Physical and theoretical. I. Title.
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541 – dc22
2004004224
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0-471-49181-0 paperback
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Contents
Preface

xv

Etymological introduction

xix

List of symbols

xxiii

Powers of ten: standard prefixes

1 Introduction to physical chemistry
1.1 What is physical chemistry: variables, relationships and laws
Why do we warm ourselves by a radiator?
Why does water get hot in a kettle?
Are these two colours complementary?
Does my radio get louder if I vary the volume control?
Why does the mercury in a barometer go up when the air pressure increases? Why does a radiator feel hot to the touch when ‘on’, and cold when ‘off’?

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2
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5
7

1.2 The practice of thermodynamic measurement

9

What is temperature?
How long is a piece of string?
How fast is ‘greased lightning’?
Why is the SI unit of mass the kilogram?
Why is ‘the material of action so variable’?

9
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15
17
18

1.3 Properties of gases and the gas laws
Why
Why
How
Why
Why
How

do we see eddy patterns above a radiator? does a hot-air balloon float? was the absolute zero of temperature determined? pressurize the contents of a gas canister? does thunder accompany lightning? does a bubble-jet printer work?

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vi

CONTENTS

What causes pressure?
Why is it unwise to incinerate an empty can of air freshener?

1.4 Further thoughts on energy
Why is the room warm?
What do we mean by ‘room temperature’?
Why do we get warmed-through in front of a fire, rather than just our skins? 2 Introducing interactions and bonds
2.1 Physical and molecular interactions
What is ‘dry ice’?
How is ammonia liquefied?
Why does steam condense in a cold bathroom?
How does a liquid-crystal display work?
Why does dew form on a cool morning?
How is the three-dimensional structure maintained within the DNA double helix? How do we make liquid nitrogen?
Why is petrol a liquid at room temperature but butane is a gas?

2.2 Quantifying the interactions and their influence
How does mist form?
How do we liquefy petroleum gas?
Why is the molar volume of a gas not zero at 0 K?

2.3 Creating formal chemical bonds
Why
Why
How
Why
Why
Why
Why
Why
Why
Why
Why
Why
Why

is chlorine gas lethal yet sodium chloride is vital for life? does a bicycle tyre get hot when inflated? does a fridge cooler work? does steam warm up a cappuccino coffee? does land become more fertile after a thunderstorm? does a satellite need an inert coating? does water have the formula H2 O? is petroleum gel so soft? does salt form when sodium and chlorine react? heat a neon lamp before it will generate light? does lightning conduct through air? is argon gas inert? is silver iodide yellow?

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3 Energy and the first law of thermodynamics

77

3.1 Introduction to thermodynamics: internal energy

77

Why
Why
Why
Why

does the mouth get cold when eating ice cream? is skin scalded by steam? do we sweat? do we still feel hot while sweating on a humid beach?

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81
83

CONTENTS

Why is the water at the top of a waterfall cooler than the water at its base?
Why is it such hard work pumping up a bicycle tyre?
Why does a sausage become warm when placed in an oven?
Why, when letting down a bicycle tyre, is the expelled air so cold?
Why does a tyre get hot during inflation?
Can a tyre be inflated without a rise in temperature?
How fast does the air in an oven warm up?
Why does water boil more quickly in a kettle than in a pan on a stove?
Why does a match emit heat when lit?
Why does it always take 4 min to boil an egg properly?
Why does a watched pot always take so long to boil?

3.2 Enthalpy
How does a whistling kettle work?
How much energy do we require during a distillation?
Why does the enthalpy of melting ice decrease as the temperature decreases? Why does water take longer to heat in a pressure cooker than in an open pan? Why does the temperature change during a reaction?
Are diamonds forever?
Why do we burn fuel when cold?
Why does butane burn with a hotter flame than methane?

3.3 Indirect measurement of enthalpy
How
How
Why
Why
How

do we make ‘industrial alcohol’? does an ‘anti-smoking pipe’ work? does dissolving a salt in water liberate heat? does our mouth feel cold after eating peppermint? does a camper’s ‘emergency heat stick’ work?

4 Reaction spontaneity and the direction of thermodynamic change
4.1 The direction of physicochemical change: entropy
Why does the colour spread when placing a drop of dye in a saucer of clean water?
When we spill a bowl of sugar, why do the grains go everywhere and cause such a mess?
Why, when one end of the bath is hot and the other cold, do the temperatures equalize?
Why does a room containing oranges acquire their aroma?
Why do damp clothes become dry when hung outside?
Why does crystallization of a solute occur?

4.2 The temperature dependence of entropy
Why do dust particles move more quickly by Brownian motion in warm water? Why does the jam of a jam tart burn more than does the pastry?

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CONTENTS

4.3 Introducing the Gibbs function
Why is burning hydrogen gas in air (to form liquid water) a spontaneous reaction? How does a reflux condenser work?

4.4 The effect of pressure on thermodynamic variables
How
Why
Why
How

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148

much energy is needed? does a vacuum ‘suck’? do we sneeze? does a laboratory water pump work?

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4.5 Thermodynamics and the extent of reaction

156

Why
Why
Why
Why
Why

is a ‘weak’ acid weak? does the pH of the weak acid remain constant? does the voltage of a battery decrease to zero? does the concentration of product stop changing? do chicken eggs have thinner shells in the summer?

4.6 The effect of temperature on thermodynamic variables
Why does egg white denature when cooked but remain liquid at room temperature? At what temperature will the egg start to denature?
Why does recrystallization work?

5 Phase equilibria
5.1 Energetic introduction to phase equilibria
Why does an ice cube melt in the mouth?
Why does water placed in a freezer become ice?
Why was Napoleon’s Russian campaign such a disaster?

5.2 Pressure and temperature changes with a single-component system: qualitative discussion
How
How
How
How

is the ‘Smoke’ in horror films made? does freeze-drying work? does a rotary evaporator work? is coffee decaffeinated?

5.3 Quantitative effects of pressure and temperature change for a single-component system
Why is ice so slippery?
What is ‘black ice’?
Why does deflating the tyres on a car improve its road-holding on ice?
Why does a pressure cooker work?

5.4 Phase equilibria involving two-component systems: partition
Why
How
Why
How
Why

does a fizzy drink lose its fizz and go flat? does a separating funnel work? is an ice cube only misty at its centre? does recrystallization work? are some eggshells brown and some white?

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CONTENTS

5.5 Phase equilibria and colligative properties
Why
How
Why
Why

does a mixed-melting-point determination work? did the Victorians make ice cream? boil vegetables in salted water? does the ice on a path melt when sprinkled with salt?

5.6 Phase equilibria involving vapour pressure
Why does petrol sometimes have a strong smell and sometimes not?
How do anaesthetics work?
How do carbon monoxide sensors work?
Why does green petrol smell different from leaded petrol?
Why do some brands of ‘green’ petrol smell different from others?
Why does a cup of hot coffee yield more steam than above a cup of boiling water at the same temperature?
How are essential oils for aromatherapy extracted from plants?

6 Acids and Bases
6.1 Properties of Lowry–Brønsted acids and bases
Why does vinegar taste sour?
Why is it dangerous to allow water near an electrical appliance, if water is an insulator?
Why is bottled water ‘neutral’?
What is ‘acid rain’?
Why does cutting an onion make us cry?
Why does splashing the hands with sodium hydroxide solution make them feel ‘soapy’?
Why is aqueous ammonia alkaline?
Why is there no vinegar in crisps of salt and vinegar flavour?
How did soldiers avoid chlorine gas poisoning at the Second Battle of
Ypres?
How is sherbet made?
Why do steps made of limestone sometimes feel slippery?
Why is the acid in a car battery more corrosive than vinegar?
Why do equimolar solutions of sulphuric acid and nitric acid have different pHs?
What is the pH of a ‘neutral’ solution?
What do we mean when we say blood plasma has a ‘pH of 7.4’?

6.2 ‘Strong’ and ‘weak’ acids and bases
Why
Why
Why
Why

is a nettle sting more painful than a burn from ethanoic acid? is ‘carbolic acid’ not in fact an acid? does carbonic acid behave as a mono-protic acid? is an organic acid such as trichloroethanoic acid so strong?

6.3 Titration analyses
Why does a dock leaf bring relief after a nettle sting?
How do indigestion tablets work?

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CONTENTS

6.4 pH buffers
Why does the pH of blood not alter after eating pickle?
Why are some lakes more acidic than others?
How do we make a ‘constant-pH solution’?

6.5 Acid–base indicators
What is ‘the litmus test’?
Why do some hydrangea bushes look red and others blue?
Why does phenolphthalein indicator not turn red until pH 8.2?

7 Electrochemistry
7.1 Introduction to cells: terminology and background
Why does putting aluminium foil in the mouth cause pain?
Why does an electric cattle prod cause pain?
What is the simplest way to clean a tarnished silver spoon?
How does ‘electrolysis’ stop hair growth?
Why power a car with a heavy-duty battery yet use a small battery in a watch? How is coloured (‘anodized’) aluminium produced?
How do we prevent the corrosion of an oil rig?
What is a battery?
Why do hydrogen fuel cells sometimes ‘dry up’?
Why bother to draw cells?
Why do digital watches lose time in the winter?
Why is a battery’s potential not constant?
What is a ‘standard cell’?
Why aren’t electrodes made from wood?
Why is electricity more dangerous in wet weather?

7.2 Introducing half-cells and electrode potentials
Why are the voltages of watch and car batteries different?
How do ‘electrochromic’ car mirrors work?
Why does a potential form at an electrode?

7.3 Activity
Why does the smell of brandy decrease after dissolving table salt in it?
Why does the smell of gravy become less intense after adding salt to it?
Why add alcohol to eau de Cologne?
Why does the cell emf alter after adding LiCl?
Why does adding NaCl to a cell alter the emf, but adding tonic water doesn’t? Why does MgCl2 cause a greater decrease in perceived concentration than
KCl?
Why is calcium better than table salt at stopping soap lathering?
Why does the solubility of AgCl change after adding MgSO4 ?

7.4 Half-cells and the Nernst equation
Why does sodium react with water yet copper doesn’t?

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CONTENTS

Why does a torch battery eventually ‘go flat’?
Why does EAgCl,Ag change after immersing an SSCE in a solution of salt?
Why ‘earth’ a plug?

7.5 Concentration cells
Why does steel rust fast while iron is more passive?
How do pH electrodes work?

7.6 Transport phenomena
How do nerve cells work?
What is a ‘salt bridge’?

7.7 Batteries
How does an electric eel produce a current?
What is the earliest known battery?

8 Chemical kinetics
8.1 Kinetic definitions
Why does a ‘strong’ bleach clean faster than a weaker one does?
Why does the bleaching reaction eventually stop?
Why does bleach work faster on some greases than on others?
Why do copper ions amminate so slowly?
How fast is the reaction that depletes the ozone layer?
Why is it more difficult to breathe when up a mountain than at ground level? 8.2 Qualitative discussion of concentration changes
Why does a full tank of petrol allow a car to travel over a constant distance? Why do we add a drop of bromine water to a solution of an alkene?
When magnesium dissolves in aqueous acid, why does the amount of fizzing decrease with time?

8.3 Quantitative concentration changes: integrated rate equations
Why do some photographs develop so slowly?
Why do we often refer to a ‘half-life’ when speaking about radioactivity?
How was the Turin Shroud ‘carbon dated’?
¨
How old is Otzi the iceman?
Why does the metabolism of a hormone not cause a large chemical change in the body?
Why do we not see radicals forming in the skin while sunbathing?

8.4 Kinetic treatment of complicated reactions
Why is arsenic poisonous?
Why is the extent of Walden inversion smaller when a secondary alkyl halide reacts than with a primary halide?
Why does ‘standing’ a bottle of wine cause it to smell and taste better?
Why fit a catalytic converter to a car exhaust?
Why do some people not burn when sunbathing?
How do Reactolite sunglasses work?

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xii

CONTENTS

8.5 Thermodynamic considerations: activation energy, absolute reaction rates and catalysis
Why prepare a cup of tea with boiling water?
Why store food in a fridge?
Why do the chemical reactions involved in cooking require heating?
Why does a reaction speed up at higher temperature?
Why does the body become hotter when ill, and get ‘a temperature’ ?
Why are the rates of some reactions insensitive to temperature?
What are catalytic converters?

9 Physical chemistry involving light: spectroscopy and photochemistry 9.1 Introduction to photochemistry
Why is ink coloured?
Why do neon streetlights glow?
Why do we get hot when lying in the sun?
Why is red wine so red?
Why are some paints red, some blue and others black?
Why can’t we see infrared light with our eyes?
How does a dimmer switch work?
Why does UV-b cause sunburn yet UV-a does not?
How does a suntan protect against sunlight?
How does sun cream block sunlight?
Why does tea have a darker colour if brewed for longer?
Why does a glass of apple juice appear darker when viewed against a white card?
Why are some paints darker than others?
What is ink?

9.2 Photon absorptions and the effect of wavelength
Why
Why
Why
Why
Why
Why

do radical reactions usually require UV light? does photolysis require a powerful lamp? are spectroscopic bands not sharp? does hydrogen look pink in a glow discharge? do surfaces exposed to the sun get so dusty? is microwave radiation invisible to the eye?

9.3 Photochemical and spectroscopic selection rules
Why is the permanganate ion so intensely coloured?
Why is chlorophyll green?
Why does adding salt remove a blood stain?
What is gold-free gold paint made of?
What causes the blue colour of sapphire?
Why do we get hot while lying in the sun?
What is an infrared spectrum?
Why does food get hot in a microwave oven?
Are mobile phones a risk to health?

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CONTENTS

9.4 Photophysics: emission and loss processes
How are X-rays made?
Why does metal glow when hot?
How does a light bulb work?
Why is a quartz–halogen bulb so bright?
What is ‘limelight’?
Why do TV screens emit light?
Why do some rotting fish glow in the dark?
How do ‘see in the dark’ watch hands work?
How do neon lights work?
How does a sodium lamp work?
How do ‘fluorescent strip lights’ work?

9.5 Other optical effects
Why is the mediterranean sea blue?
Do old-master paintings have a ‘fingerprint’?

10 Adsorption and surfaces, colloids and micelles
10.1 Adsorption and definitions
Why is steam formed when ironing a line-dried shirt?
Why does the intensity of a curry stain vary so much?
Why is it difficult to remove a curry stain?
Why is iron the catalyst in the Haber process?
Why is it easier to remove a layer of curry sauce than to remove a curry stain? How does water condense onto glass?
How does bleach remove a dye stain?
How much beetroot juice does the stain on the plate contain?
Why do we see a ‘cloud’ of steam when ironing a shirt?

10.2 Colloids and interfacial science
Why is milk cloudy?
What is an ‘aerosol’ spray?
What is ‘emulsion paint’?
Why does oil not mix with water?

xiii

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10.3 Colloid stability

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How
How
How
Why
Why
How

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513

are cream and butter made? is chicken soup ‘clarified’ by adding eggshells? is ‘clarified butter’ made? does hand cream lose its milky appearance during hand rubbing? does orange juice cause milk to curdle? are colloidal particles removed from waste water?

10.4 Association colloids: micelles
Why does soapy water sometimes look milky?
What is soap?
Why do soaps dissolve grease?

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518

xiv

CONTENTS

Why is old washing-up water oily when cold but not when hot?
Why does soap generate bubbles?
Why does detergent form bubbles?

519
521
522

Answers to SAQs

525

Bibliography

533

Index

565

Preface
This book
Some people make physical chemistry sound more confusing than it really is. One of their best tricks is to define it inaccurately, saying it is ‘the physics of chemicals’. This definition is sometimes quite good, since it suggests we look at a chemical system and ascertain how it follows the laws of nature. This is true, but it suggests that chemistry is merely a sub-branch of physics; and the notoriously mathematical nature of physics impels us to avoid this otherwise useful way of looking at physical chemistry.
An alternative and more user-friendly definition tells us that physical chemistry supplies ‘the laws of chemistry’, and is an addition to the making of chemicals. This is a superior lens through which to view our topic because we avoid the bitter aftertaste of pure physics, and start to look more closely at physical chemistry as an applied science: we do not look at the topic merely for the sake of looking, but because there are real-life situations requiring a scientific explanation. Nevertheless, most practitioners adopting this approach are still overly mathematical in their treatments, and can make it sound as though the science is fascinating in its own right, but will sometimes condescend to suggest an application of the theory they so clearly relish.
But the definition we will employ here is altogether simpler, and also broader: we merely ask ‘why does it happen?’ as we Now published as Revfocus on the behaviour of each chemical system. Every example elations of Divine Love, we encounter in our everyday walk can be whittled down into by Mother Julian of small segments of thought, each so simple that a small child can Norwich. understand. As a famous mystic of the 14th century once said, ‘I saw a small hazelnut and I marvelled that everything that exists could be contained within it’. And in a sense she was right: a hazelnut looks brown because of the way light interacts with its outer shell – the topic of spectroscopy (Chapter 9); the hazelnut is hard and solid – the topic of bonding theory (Chapter 2) and phase equilibria
(Chapter 5); and the nut is good to eat – we say it is readily metabolized, so we think

xvi

PREFACE

of kinetics (Chapter 8); and the energetics of chemical reactions (Chapters 2–4). The sensations of taste and sight are ultimately detected within the brain as electrical impulses, which we explain from within the rapidly growing field of electrochemistry
(Chapter 7). Even the way a nut sticks to our teeth is readily explained by adsorption science (Chapter 10). Truly, the whole of physical chemistry can be encompassed within a few everyday examples.
So the approach taken here is the opposite to that in most other books of physical chemistry: each small section starts with an example from everyday life, i.e. both the world around us and also those elementary observations that a chemist can be certain to have pondered upon while attending a laboratory class. We then work backwards from the experiences of our hands and eyes toward the cause of why our world is the way it is.
Nevertheless, we need to be aware that physical chemistry is not a closed book in the same way of perhaps classical Latin or Greek. Physical chemistry is a growing discipline, and new experimental techniques and ideas are continually improving the data and theories with which our understanding must ultimately derive.
Inevitably, some of the explanations here have been over-simplified because physical chemistry is growing at an alarming rate, and additional sophistications in theory and experiment have yet to be devised. But a more profound reason for caution is in ourselves: it is all too easy, as scientists, to say ‘Now I understand!’ when in fact we mean that all the facts before us can be answered by the theory. Nevertheless, if the facts were to alter slightly – perhaps we look at another kind of nut – the theory, as far as we presently understand it, would need to change ever so slightly. Our understanding can never be complete.
So, we need a word about humility. It is said, probably too often, that science is not an emotional discipline, nor is there a place for any kind of reflection on the human side of its application. This view is deeply mistaken, because scientists limit themselves if they blind themselves to any contradictory evidence when sure they are right. The laws of physical chemistry can only grow when we have the humility to acknowledge how incomplete is our knowledge, and that our explanation might need to change. For this reason, a simple argument is not necessary the right one; but neither is a complicated one. The examples in this book were chosen to show how the world around us manifests Physical Chemistry. The explanation of a seemingly simple observation may be fiendishly complicated, but it may be beautifully simple. It must be admitted that the chemical examples are occasionally artificial. The concept of activity, for example, is widely misunderstood, probably because it presupposes knowledge from so many overlapping branches of physical chemistry. The examples chosen to explain it may be quite absurd to many experienced teachers, but, as an aid to simplification, they can be made to work. Occasionally the science has been simplified to the point where some experienced teachers will maintain that it is technically wrong. But we must start from the beginning if we are to be wise, and only then can we progress via the middle . . . and physical chemistry is still a rapidly growing subject, so we don’t yet know where it will end.
While this book could be read as an almanac of explanations, it provides students in further and higher education with a unified approach to physical chemistry. As a

PREFACE

xvii

teacher of physical chemistry, I have found the approaches and examples here to be effective with students of HND and the early years of BSc and MChem courses. It has been written for students having the basic chemical and mathematical skills generally expected of university entrants, such as rearrangement of elementary algebra and a little calculus. It will augment the skills of other, more advanced, students.
To reiterate, this book supplies no more than an introduction to physical chemistry, and is not an attempt to cover the whole topic. Those students who have learned some physical chemistry are invited to expand their vision by reading more specialized works. The inconsistencies and simplifications wrought by lack of space and style in this text will be readily overcome by copious background reading. A comprehensive bibliography is therefore included at the end of the book. Copies of the figures and bibliography, as well as live links can be found on the book’s website at http://www.wileyeurope.com/go/monkphysical. Acknowledgements
One of the more pleasing aspects of writing a text such as this is the opportunity to thank so many people for their help. It is a genuine pleasure to thank Professor S´ amus e Higson of Cranfield University, Dr Roger Mortimer of Loughborough University, and Dr Michele Edge, Dr David Johnson, Dr Chris Rego and Dr Brian Wardle from my own department, each of whom read all or part of the manuscript, and whose comments have been so helpful.
A particular ‘thank you’ to Mrs Eleanor Riches, formerly a high-school teacher, who read the entire manuscript and made many perceptive and helpful comments.
I would like to thank the many students from my department who not only saw much of this material, originally in the form of handouts, but whose comments helped shape the material into its present form.
Please allow me to thank Michael Kaufman of The Campaign for a Hydrogen Economy (formerly the Hydrogen Association of UK and Ireland ) for helpful discussions to clarify the arguments in Chapter 7, and the Tin Research Council for their help in constructing some of the arguments early in Chapter 5.
Concerning permission to reproduce figures, I am indebted to The Royal Society of
Chemistry for Figures 1.8 and 8.26, the Open University Press for Figure 7.10, Elsevier Science for Figures 4.7 and 10.3, and John Wiley & Sons for Figures 7.19, 10.11 and 10.14. Professor Robin Clarke FRS of University College London has graciously allowed the reproduction of Figure 9.28.
Finally, please allow me to thank Dr Andy Slade, Commissioning Editor of Wiley, and the copy and production editors Rachael Ballard and Robert Hambrook. A special thank you, too, to Pete Lewis.
Paul Monk
Department of Chemistry & Materials
Manchester Metropolitan University
Manchester

Etymological introduction The hero in The Name of the Rose is a medieval English monk. He acts as sleuth, and is heard to note at one point in the story how, ‘The study of words is the whole of knowledge’. While we might wish he had gone a little further to mention chemicals, we would have to agree that many ‘‘Etymology’’ means of our technical words can be traced back to Latin or Greek roots. the derivation of a
The remainder of them originate from the principal scientists who word’s meaning. pioneered a particular field of study, known as etymology.
Etymology is our name for the science of words, and describes the sometimestortuous route by which we inherit them from our ancestors. In fact, most words change and shift their meaning with the years. A classic example describes how King
George III, when first he saw the rebuilt St Paul’s Cathedral in London, described it as ‘amusing, artificial and awful’, by which he meant, respectively, it ‘pleased him’, was ‘an artifice’ (i.e. grand) and was ‘awesome’ (i.e. breathtaking).
Any reader will soon discover the way this text has an unusual etymological emphasis: the etymologies are included in the belief that taking a word apart actually helps us to understand it and related concepts. For example, as soon as we know the Greek for
‘green’ is chloros, we understand better the meanings of the proper nouns chlorophyll and chlorine, both of which are green. Incidentally, phyll comes from the Greek for
‘leaf’, and ine indicates a substance.
Again, the etymology of the word oxygen incorporates much historical information: oxys is the Greek for ‘sharp’, in the sense of an extreme sensory experience, such as the taste of acidic vinegar, and the ending gen comes from gignesthaw (pronounced ‘gin-es-thaw’), meaning ‘to be produced’. The classical roots of ‘oxygen’ reveal how the French scientists who first made the gas thought they had isolated the distinguishing chemical component of acids.
The following tables are meant to indicate the power of this approach. There are several dozen further examples in the text. The bibliography on p. 533 will enable the interested reader to find more examples.

charge, electricity capacitance, electrochemistry electricity, potential

Coulomb, Charles
Faraday, Michael

Voltaire, Allesandro

hundred times ten times

giant thousand, thousands bed, couch, coverlet

under, beneath, directly under above, on, over, on top of

Centi(n)s
Decie(n)s

Giga(n)s
Milli
Stratum

Sub

Super

Original Latin meaning

Word or root

Words and roots from Latin

above, bigger than

below, less than

subterfuge, subterranean sub-standard, subset superstar, superlative

century, cent (= $/100) decimate (i.e. to kill 1 in 10) decimal gigantic millipede, millennium stratify (many layered)

decimetre (= metre ÷ 10) giga (symbol G) = factor of 109 milli (symbol m) = factor of 10−3 strata of rock (in geology) substrate (chemistry and physics) substrate (i.e. underlying strata) subscript (in typesetting) superscript (in typesetting)

centi (symbol c) = factor of 102 deci (symbol d) factor of 10

Scientific examples

amp (unit of current); amperometry; amperometric; voltammetry coulomb (unit of charge); coulombic faraday (molar electronic charge); faradaic; farad (unit of capacitance) volt (unit of potential); voltaic; voltammetry

Derived wordsa

Modern examples

potential

charge passed electricity current

Present meaning(s)

very large thousand, thousandth something beneath

hundred ten Modern meaning

Note that derived words do not start with a capital letter.

current, electricity

Amp` re, Andr´ e e

a

Field of study

Scientist

Words derived from a scientist’s name

small

one

straight

near, beyond, contrary

four, four times

energy, temperature

Mono (µoνoς )

Ortho (oρθ oς )

Para (π αρα-)

Tetra (τ ετ τ αρεoς )

Thermo (θ ερµo)

Mega (µεγ α-)

meter

great, large

Kilo (κιλoς )

Micro (µιχρoς )

lots of, many

Iso (ισ o)

Meter (µετρητ ης )

level, equality

Di (δις )

middle, mid

two, twice

Cyclo (κυκλoς )

afterwards

circle, circular

Cathode (καθ oδoς )

Mesos (µεσ oς )

descent

Baro (βαρoς )

Meta (µετ α-)

ascent

weigh down, heaviness

Anode (ανoδoς )

Original Greek meaning

Word or root

Words and roots from Greek

Heat

to do with four

opposite

straight, right

one, alone

tiny, small

a meter, to meter

after, beyond

mid, intermediate

very large

factor of a thousand

same

to do with two

cycle, circle

negative electrode

to do with atmosphere

positive electrode

Modern meaning

to do with two (coordination chemistry)

cyclotron, cyclization

cathode, cathodic, cathodize

barometer, isobar, bar (unit of pressure)

anode, anodic, anodize

Scientific examples

thermos, thermometer

tetrahedron

paranormal (beyond normal) paradox (contrary to standard)

orthodox (i.e. to the standard) othopædic (straightening bones)

monorail, monologue monotonous (i.e. on one note)

microscope, micrometer

gas meter, metrical

metaphor (i.e. beyond the real meaning) mezzanine (mid floor)

megabyte

Kilometre

thermos, thermometer

to do with four (in coodination chemistry)

position opposite the primary carbon

adjacent position on a ring

monomer, e.g. mono-substituted

micro (symbolµ) = factor of 10−6

barometer (i.e. measures pressure)

position beyond ortho on a ring metathesis (i.e. product of mixing)

mesophase (i.e. phase between two extremes)

mega (symbol M) = factor of 106

kilo (symbol k) = factor of 103

e.g. isobutane, isomer

dilemma (two options)
Isomer

dimer, di-stereoisomer (i.e. one of two images)

dihedral

bicycle, cylinder, cyclone

cathode

barometer, barometric

anode

Modern examples

List of Symbols
Symbols for variables a a
A
A b B c c c C
C
Cp
CV
E
E
Ea
E(ea)
Ej
E(load)
emf

Activity van der Waals constant optical absorbance
Area
van der Waals constant virial coefficient concentration the intercept on the y-axis of a graph constant of proportionality virial coefficient heat capacity heat capacity determined at constant pressure heat capacity determined at constant volume energy potential activation energy electron affinity junction potential potential of a battery or cell when passing a current potential of a cell, determined at zero current EO,R
O
EO,R

E(LHS)
E(RHS)
f f F
G
G
GO
G‡ h H
H
H(ads)
HO
HBE
Hc
Hf
H‡
I
I

electrode potential for the couple
O + ne− = R standard electrode potential for the couple O + ne− = R electrode potential of the negative electrode in a cell electrode potential of the positive electrode in a cell fugacity frequency force Gibbs function change in Gibbs function standard Gibbs function
Gibbs function of activation height enthalpy change in enthalpy enthalpy of adsorption standard enthalpy bond enthalpy enthalpy of combustion enthalpy of formation enthalpy of activation electrical current intensity of light following absorption

xxiv

Io
I
I
J

J k k k k kn k−n k(n) ka kd kH
K
K
Ka
Ka(n)
Kb
Kc
Kp
Ks

Kw
K‡
l m LIST OF SYMBOLS

intensity of incident light beam ionic strength ionization energy rotational quantum number; rotational quantum number of an excited state rotational quantum number of ground state force constant of a bond proportionality constant rate constant pseudo rate constant rate constant of an nth-order reaction rate constant for the back reaction of an nth-order reaction rate constant of the nth process in a multi-step reaction rate constant of adsorption rate constant of desorption
Henry’s law constant equilibrium constant correction constant of an ion-selective electrode acidity constant (‘acid dissociation’ constant) acidity constant for the nth dissociation reaction basicity constant equilibrium constant formulated in terms of concentration equilibrium constant formulated in terms of pressure equilibrium constant of solubility
(sometimes called ‘solubility product’ or ‘solubility constant’) autoprotolysis constant of water equilibrium constant of forming a transition state ‘complex’ length gradient of a graph

m
M
n n nm
N
p p(i) O p(i) pO q Q
Q
r r r r ro
R
s s S
S
S
S‡
t t1 O

mass relative molar mass number of moles number of electrons in a redox reaction amount of material in an adsorbed monolayer number pressure partial pressure of component i vapour pressure of pure i standard pressure of 105 Pa heat energy charge reaction quotient separation between ions radius of a circle or sphere bond length bond length in an optically excited species equilibrium bond length electrical resistance solubility stoichiometric ratio entropy change in entropy standard entropy entropy of activation time half life

2

T
T
To
TK
U
U
v v temperature optical transmittance optical transmittance without a sample
Krafft temperature internal energy change in internal energy, e.g. during reaction quantum-number of vibration quantum-number of vibration in an excited-state species

LIST OF SYMBOLS

v
V
V
V
Vm w x x xi y z
Z
γ γ± γ γ δ


ε εr εo θ θ κ λ λ(max) µ µi µiO ν quantum-number of vibration in a ground-state species volume voltage, e.g. of a power pack
Coulomb potential energy molar volume work controlled variable on the horizontal axis of a graph deviation of a bond from its equilibrium length mole fraction of i observed variable on the vertical of a graph charge on ion (so z+ for a cation and z− for an anion) compressibility activity coefficient mean ionic activity coefficient fugacity coefficient surface tension small increment partial differential change in a variable (so
X = X(final form) − X(initial form) ) extinction coefficient (‘molar decadic absorptivity’) relative permittivity permittivity of free space adsorption isotherm angle ionic conductivity wavelength the wavelength of a peak in a spectrum reduced mass chemical potential of i standard chemical potential of i stoichiometric constant

ν ν νo ξ ρ σ σ φ φ φ Φ χ ω

xxv

velocity frequency (the reciprocal of the period of an event) frequency following transmission
(in Raman spectroscopy) extent of reaction density electrical conductivity standard deviation electric field strength (electrostatic interaction) work function of a metal primary quantum yield quantum yield of a reaction electronegativity wavenumber of a vibration
(determined as ω = λ ÷ c)

Symbols for constants
A
c cO e f F kB L
NA
g h R

Debye–H¨ ckel ‘A’ factor u the speed of light in vacuo standard concentration charge on an electron, of value
1.6 × 10−19 C mathematical operator (‘function of’)
Faraday constant, of value
96 485 C mol−1
Boltzmann constant, of value
1.38 × 10−23
Avogadro constant, of value
6.022 × 1023 mol−1
Avogadro number, of value
6.022 × 1023 mol−1 acceleration due to gravity, of value
9.81 m s−2
Planck constant, of value
6.626 × 10−34 J s gas constant, of value
8.314 J K−1 mol−1

xxvi

LIST OF SYMBOLS

Symbols for units
A
˚
A
bar
C

C
g
Hz
J
K
kg m mmHg mol N
Pa
s
S
V
W
yr

amp` re e angstr¨ m, length of value 10−10 m o ˚
(non-IUPAC)
standard pressure of 105 Pa
(non-SI unit) coulomb centigrade (non-SI) gram hertz joule kelvin kilogram metre millimetre of mercury (non-SI unit of pressure) mole newton pascal second (SI unit) siemen volt watt year ohm Acronyms and abbreviations
CT
d
HOMO
IQ
IR
IUPAC

charge transfer differential operator (which never appears on its own) highest occupied molecular orbital intelligence quotient infrared International Union of Pure and
Applied Chemistry

IVF
LCD
LHS
LUMO

in vitro fertilization liquid crystal display left-hand side lowest unoccupied molecular orbital MLCT metal-to-ligand charge transfer
MRI
magnetic resonance imaging
NIR
near-infra red
NMR
nuclear magnetic resonance
O
general oxidized form of a redox couple p mathematical operator,
− log10 [variable], so pH = − log10 [H+ ]
PEM
proton exchange membrane
R
general reduced form of a redox couple RHS right-hand side
s.t.p.
standard temperature and pressure SAQ self-assessment question
SCE
saturated calomel electrode
SCUBA self-contained underwater breathing apparatus SHE standard hydrogen electrode
SHM
simple harmonic motion
SI
Syst` me Internationale e SN 1 unimolecular nucleophilic substitution process SN 2 bimolecular nucleophilic substitution process SSCE silver–silver chloride electrode
TS
transition state
TV
television
UPS
UV-photoelectron spectroscopy
UV
ultraviolet
UV–vis ultraviolet and visible
XPS
X-ray photoelectron spectroscopy

LIST OF SYMBOLS

Standard subscripts (other than those where a word or phrase is spelt in full) ads aq c eq f g l LHS m p
Pt
r
RHS
s sat’d t
V
0


adsorption; adsorbed aqueous combustion at equilibrium formation gas liquid left-hand side of a cell molar at constant pressure platinum (usually, as an electrode) reaction right-hand side of cell solid saturated at time t (i.e. after a reaction or process has commenced) at constant volume initially (i.e. at time t = 0) measurement taken after an infinite length of time

Standard superscripts (other than those where a word or phrase is spelt in full)


+
*

activated quantity anion cation excited state


+•
O

xxvii

radical radical cation standard state

Chemicals and materials
A
Bu
CFC
DMF
DMSO
DNA e− EDTA
HA
LPG
M
MB
MV
O
PC
Ph
R
R
SDS
TFA α β γ general anion butyl chlorofluorocarbon
N,N-dimethylformamide
dimethylsulphoxide deoxyribonucleic acid electron ethylenediamine tetra-acetic acid general Lowry–Brønsted acid liquid petroleum gas general cation methylene blue methyl viologen
(1,1 -dimethyl-4,4 -bipyridilium) general oxidized form of a redox couple propylene carbonate phenyl substituent general alkyl substituent general reduced form of a redox couple sodium dodecyl sulphate tetrafluoroacetic acid particle emitted during radioactive disintegration of nucleus particle emitted during radioactive disintegration of nucleus high-energy photon (gamma ray)

a atto f femto

p pico n nano µ micro

m milli

c centi

d deci

1

k kilo

M mega

G giga

T tera
P peta

10−18
10−15

10−12
10−9
10−6

10−3

10−2

10−1

100= 1

103

106

109

1012
1015

1, 000, 000, 000, 000
1, 000, 000, 000, 000, 000

1, 000, 000, 000

1, 000, 000

1, 000

1

0.1

0.01

0.001

0.000, 000, 000, 001
0.000, 000, 001
0.000, 001

0.000, 000, 000, 000, 000, 001
0.000, 000, 000, 000, 001

Powers of ten: energy in joules
1 aj
1 fJ = energy of a single high-energy photon (γ -ray of λ = 10−10 m)
1 pJ = energy consumption of a single nerve impulse
1 nJ = energy per beat of a fly’s wing
1µJ = energy released per second by a single phosphor on a TV screen
1 mJ = energy consumption per second of an LCD watch display
1 cJ = energy per mole of low-energy photons (radio wave of λ = 10 m)
1 dJ = energy released by passing 1 mA across 1 V for 100 s (energy = V it)
1 J = half the kinetic energy of 0.1 kg (1 N) travelling at a velocity of 1 m s−1
1 kJ = half the energy of room temperature (E at RT of 298 K = 2.3 kJ)
1 MJ = energy of burning 1 /3 th mole of glucose
(about a quarter a Mars bar)
1 GJ = energy of 1 mole of γ -ray photons
(λ = 10−10 m)
1 TJ = energy (via E = mc2 ) held in a mass of 100 g
1 PJ = energy released by detonating a very small nuclear bomb

Standard prefixes powers of ten

k kilo

M mega
G giga

T tera

P peta

106
109

1012

1015

µ micro

10−6

m milli c centi d deci

p pico n nano

10−12
10−9

10−3
10−2
10−1
100= 1
103

a atto f femto

10−18
10−15

1,000,000,000,000,000

1,000,000,000,000

1,000,000
1,000,000,000

0.001
0.01
0.1
1
1,000

0.000,001

0.000,000,000,001
0.000,000,001

0.000,000,000,000,000,001
0.000,000,000,000,001

Powers of ten: time in seconds
1 as = the very fastest laser flash
1 fs = 10 × the time required for photon absorption 1 ps = time required for bond rearrangement
1 ns = time required for solvent rearrangement e.g. after redox change
1 µs = time for 166 calculations in a Pentium microprocessor chip
1 ms = time for a nerve impulse
1 cs = fastest human reflexes
1 ds = time required to blink fast
1 s = time required to sneeze
1 ks = average time required to walk 1 mile (ca.
16 minutes)
1 Ms = life expectancy of a housefly (11.5 days)
1 Gs = average life span of an adult in the third world (31.7 years)
1 Ts = age of fossil remains of Neanderthal man
(31,700 years)
1 Ps = time since dinosaurs were last in their prime (31.7 million years ago)

xxx
STANDARD PREFIXES POWERS OF TEN

a atto

f femto

p pico

n nano µ micro

m milli c centi d deci
1
k kilo

M mega

G giga

T tera

P peta

10−18

10−15

10−12

10−9
10−6

10−3
10−2
10−1
100= 1
103

106

109

1012

1015
1,000,000,000,000,000

1,000,000,000,000

1,000,000,000

1,000,000

0.001
0.01
0.1
1
1,000

0.000,000,001
0.000,001

0.000,000,000,001

0.000,000,000,000,001

0.000,000,000,000,000,001

Powers of ten: length in metres
1 am = diameter of a γ -ray photon (if it is taken to be a particle)
1 fm = diameter of an electron (if it is taken to be a particle)
1 pm = ten times the diameter of an atomic nucleus
1 nm = ten bond lengths
1 µm = wavelength of near infra-red light
(i.e. heat)
1 mm = diameter of a grain of sand
1 cm = width of a human finger
1 dm = length of the human tongue
1 m = half the height of an adult human
1 km = distance walked in about 12 minutes 1 Mm = distance between London and
Edinburgh
1 Gm = distance travelled by light in 3 minutes 1 Tm = ninth of the distance between the sun and the earth
1 Pm = tenth of a light year (distance travelled by a photon in 1 year)

STANDARD PREFIXES POWERS OF TEN

xxxi

a atto

f femto

p pico

n nano

µ micro m milli c centi

d deci

1 k kilo
M mega
G giga
T tera
P peta

10−18

10−15

10−12

10−9

10−6
10−3
10−2

10−1

100= 1
103
106
109
1012
1015
1
1,000
1,000,000
1,000,000,000
1,000,000,000,000
1,000,000,000,000,000

0.1

0.000,001
0.001
0.01

0.000,000,001

0.000,000,000,001

0.000,000,000,000,001

0.000,000,000,000,000,001

Powers of ten: mass in grams
1 ag = mass of 1 molecule of polystyrene
(of molar mass 106 g mol−1 )
1 fg = mass of 1 molecule of DNA (of molar mass 109 g mol−1 )
1 pg = mass of phosphorous used to dope
1 g of silicon (for a microchip)
1 ng = mass of a single ferrite particle for use in a computer floppy disc
1µg = mass of ink in a full stop
1 mg = mass of ink on a banknote
1 cg = mass of nicotine absorbed from a single cigarette
1 dg = mass of the active component within 1 aspirin tablet
1 g = mass of a single lentil
1 kg = mass of a grapefruit
1 Mg = mass of a baby elephant
1 Gg = mass of a large crane (1000 tonnes)
1 Tg = mass of a mountain
1 Pg = mass of a large island such as the
Isle of White (109 tonnes)

xxxii
STANDARD PREFIXES POWERS OF TEN

1

Introduction to physical chemistry Introduction
In this, our introductory chapter, we start by looking at the terminology of physical chemistry. Having decided what physical chemistry actually is, we discuss the nature of variables and relationships. This discussion introduces the way relationships underlying physical chemistry are formulated.
We also introduce the fundamental (base) units of the Syst` me Internationale (SI), e and discuss the way these units are employed in practice.
We look at the simple gas laws to explore the behaviour of systems with no interactions, to understand the way macroscopic variables relate to microscopic, molecular properties. Finally, we introduce the statistical nature underlying much of the physical chemistry in this book when we look at the Maxwell–Boltzmann relationship.

1.1

What is physical chemistry: variables, relationships and laws
Why do we warm ourselves by a radiator?
Cause and effect
We turn on the radiator if we feel cold and warm ourselves in front of it. We become warm because heat travels from the radiator to us, and we absorb its heat energy, causing our own energy content to rise. At root, this explains why we feel more comfortable. While this example is elementary in the extreme, its importance lies in the way it illustrates the concept of cause and effect. We would not feel warmer if the radiator

2

INTRODUCTION TO PHYSICAL CHEMISTRY

was at the same temperature as we were. We feel warmer firstly because the radiator is warmer than us, and secondly because some of the heat energy leaves the radiator and we absorb it. A transfer of energy occurs and, therefore, a change. Without the cause, the effect of feeling warmer could not have followed.
We are always at the mercy of events as they occur around us. The physical chemist could do nothing if nothing happened;
A variable is an experchemists look at changes. We say a physical chemist alters variimental parameter we can change or ‘tweak’. ables, such as pressure or temperature. Typically, a chemist causes one variable to change and looks at the resultant response, if any.
Even a lack of a response is a form of result, for it shows us what is and what is not a variable.

Why does water get hot in a kettle?
The fearsome-looking word ‘physicochemical’ means ‘relating to physical chemistry’.

Physicochemical relationships

Putting water into an electric kettle does not cause the water to get hot. The water stays cold until we turn on the power to the kettle element, which converts electrical energy from the mains into heat energy. The heat energy from the kettle element is then absorbed by the water, which gets hot as a direct consequence.
The temperature of the water does not increase much if a small amount of electrical energy is consumed; conversely, the water gets hotter if a greater amount of energy is consumed and thereafter passed to the water. A physical chemist says a ‘relationship’ exists (in this case) between heat
In words, the symbols
T = f (energy) means ‘T input and temperature, i.e. the temperature of the water depends on is a function of energy’. the amount of energy consumed.
Note how variables are
Mathematically, we demonstrate the existence of a relationship usually printed in italic by writing T = f (energy), where T is temperature and the small type. f means ‘is a function of’.
So the concept of variables is more powerful than just changing parameters; nor do physical chemists merely vary one parameter and see what happens to others. They search for ‘physicochemical’ relationships between the variables.

Are these two colours complementary?
Qualitative and quantitative measurements
We often hear this question, either at the clothes shop or at a paint merchant. Either someone wants to know if the pink colour of a sweatshirt matches the mauve of a skirt, or perhaps a decorator needs to know if two shades of green will match when painted on opposing bedroom walls.

WHAT IS PHYSICAL CHEMISTRY: VARIABLES, RELATIONSHIPS AND LAWS

3

But while asking questions concerning whether a series of colours are complementary, we are in fact asking two questions Complementary means at once: we ask about the colour in relation to how dark or light ‘to make complete’. it is (‘What is the brightness of the colour?’); but we also ask a more subjective question, saying ‘Is the pink more red or more white: what kind of pink is it?’ We are looking for two types of relationship.
In any investigation, we first look for a qualitative relationship. In effect, we ask questions like, ‘If I change the variable x, is there is a response in a different variable y?’ We look at what kind of response we can cause – a scientist wants to know about the qualities of the response, hence QUAL-itative. An obvious question relating to qualitative relationships is, ‘If I mix solutions of A and B, does a reaction occur?’
Only after we know whether or not there is a response (and of what general kind) does a physical chemist ask the next question, seeking a quantitative assessment. He asks, ‘How much of the response is caused?’ In effect, physical chemists want to know if the magnitude (or quantity) of a response is big, small or intermediate. We say we look for a QUANT-itative aspect of the relationship. An obvious question relating to quantitative relationships is, ‘I now know that a reaction occurs when I mix solutions of A and B, but to what extent does the reaction occur; what is the chemical yield ?’

Does my radio get louder if I vary the volume control?
Observed and controlled variables
We want to turn up the radio because it’s noisy outside, and we want to hear what is broadcast. We therefore turn the volume knob toward ‘LOUD’. At its most basic, the volume control is a variable resistor, across which we pass a current from the battery, acting much like a kettle element. If we turn up the volume control then a larger current is allowed to flow, causing more energy to be produced by the resistor. As a listener, we hear a response because the sound from the speakers becomes louder.
The speakers work harder.
But we must be careful about the way we state these relationships. We do not ‘turn up the volume’ (although in practice we might say these exact words and think in these terms). Rather, we vary the volume control and, as a response, our ears experience an increase in the decibels coming through the radio’s speakers. The listener controls the magnitude of the noise by deciding how far the volume-control knob needs to be turned. Only then will the volume change. The process does not occur in reverse: we do not change the magnitude of the noise and see how it changes the position of the volume-control knob.
We consciously, careWhile the magnitude of the noise and the position of the volume fully, vary the magniknob are both variables, they represent different types, with one tude of the controlled depending on the other. The volume control is a controlled variable variable and look at because the listener dictates its position. The amount of noise is the the response of the observed variable because it only changes in response to variations observed variable. in the controlled variable, and not before.

4

INTRODUCTION TO PHYSICAL CHEMISTRY

Relationships and graphs
Physical chemists often depict relationships between variables by drawing graphs. The controlled variable is always drawn along the x-axis, and the observed variable is drawn up the y-axis.
Figure 1.1 shows several graphs, each demonstrating a different kind of relationship. Graph (a) is straight line passing through the origin. This graph says: when we vary the controlled variable x, the observed variable y changes in direct proportion. An obvious example in such a case is the colour intensity in a glass of blackcurrant cordial: the intensity increases in linear proportion to the concentration of the cordial, according to the Beer–Lambert law
(see Chapter 9). Graph (a) in Figure 1.1 goes through the origin because there is no purple colour when there is no cordial (its concentration is zero).
Graph (b) in Figure 1.1 also demonstrates the existence of a relationship between the variables x and y, although in this case not a linear relationship. In effect, the graph tells us that the observed variable y increases at a faster rate than does the controlled variable x. A simple example is the distance travelled by a ball as a function of time t as it accelerates while rolling down a hill. Although the graph is not straight, we still say there is a relationship, and still draw the controlled variable along the x-axis.
Observed variable y

Observed variable y

The x-axis (horizontal) is sometimes called the abscissa and the y-axis (vertical) is the ordinate. A simple way to remember which axis is which is to say,
‘an eXpanse of road goes horizontally along the x-axis’, and ‘a YoYo goes up and down the y-axis’.

(b)
Observed variable y

Controlled variable x

(a)
Observed variable y

Controlled variable x

Controlled variable x

Controlled variable x

(c)

(d)

Figure 1.1 Graphs of observed variable (along the y-axis) against controlled variable (along the x-axis). (a) A simple linear proportionality, so y = constant × x; (b) a graph showing how y is not a simple function of x, although there is a clear relationship; (c) a graph of the case where variable y is independent of variable x; (d) a graph of the situation in which there is no relationship between y and x, although y does vary

WHAT IS PHYSICAL CHEMISTRY: VARIABLES, RELATIONSHIPS AND LAWS

5

Graph (c) in Figure 1.1 is a straight-line graph, but is horizontal. In other words, whatever we do to the controlled variable x, the observed variable y will not change.
In this case, the variable y is not a function of x because changing x will not change
y. A simple example would be the position of a book on a shelf as a function of time.
In the absence of other forces and variables, the book will not move just because it becomes evening.
Graph (d) in Figure 1.1 shows another situation, this time the data do not demonstrate a straightforward relationship; it might Data is plural; the sindemonstrate there is no relationship at all. The magnitude of the gular is datum. controlled variable x does not have any bearing on the observed variable y. We say the observed variable y is independent of the When two variables are controlled variable x. Nevertheless, there is a range of results for multiplied together, we y as x varies. Perhaps x is a compound variable, and we are being call them a compound simplistic in our analysis: an everyday example might be a stu- variable. dent’s IQ as x and his exam performance as y, suggesting that, while IQ is important, there must be another variable controlling the magnitude of the exam result, such as effort and commitment. Conversely, the value of y might be completely random (so repeating the graphs with the same values of x would generate a different value of y – we say it is irreproducible). An example of this latter situation would be the number of people walking along a main road as a function of time.

Why does the mercury in a barometer go up when the air pressure increases?
Relationships between variables
The pressure p of the air above any point on the Earth’s surface relates ultimately to the amount of air above it. If we are standing high up, for example on the top of a tall mountain, there is less air between us and space for gravity to act upon.
Conversely, if we stand at the bottom of the Grand Canyon (one of the lowest places on Earth) then more air separates us from space, causing the air pressure p to be much greater.
A barometer is an instrument designed to measure air pressure p. It consists of a pool of liquid mercury in a trough. A long, thin glass tube (sealed at one end) is placed in the centre of the trough with its open-side beneath the surface of the liquid; see Figure 1.2. The pressure of the air acts as a force on the surface of the mercury, forcing it up and into the capillary within the tube. If the air pressure is great, then the force of the air on the mercury is also great, causing much mercury up the tube. A lower pressure is seen as a shorter length h of mercury in the tube.
By performing experiments at different pressures, it is easy to prove the existence of a relationship between the air pressure p and the height h of the mercury column

6

INTRODUCTION TO PHYSICAL CHEMISTRY
Vacuum

Thick-walled glass tube

h

Trough of mercury

Figure 1.2 A barometer is a device for measuring pressures. A vacuum-filled glass tube (sealed at one end) is placed in a trough of mercury with its open end beneath the surface of the liquid metal. When the tube is erected, the pressure of the external air presses on the surface and forces mercury up the tube. The height of the mercury column h is directly proportional to the external pressure p
In fact, the value of the constant c in
Equation (1.1) comprises several natural constants, including the acceleration due gravity g and the density ρ of the mercury.

in the tube. This relationship follows Equation (1.1): h=c×p (1.1)

where c is merely a proportionality constant.
In practice, a barometer is merely an instrument on which we look at the length of the column of mercury h and, via
Equation (1.1), calculate the air pressure p. The magnitude of h is in direct relation to the pressure p. We ascertain the magnitude of h if we need to know the air pressure p.
While physical chemistry can appear to be horribly mathematical, in fact the mathematics we employ are simply one way (of many) to describe the relationships between variables. Often, we do not know the exact nature of the function until a later stage of our investigation, so the complete form of the relationship has to be discerned in several stages. For example, perhaps we first determine the existence of a linear equation, like Equation (1.1), and only then do we seek to measure an accurate value of the constant c.
We might see this
But we do know a relationship holds, because there is a response. situation written mathWe would say there was no relationship if there was no response. ematically as, h = f (p),
For example, imagine we had constructed a poor-quality barometer where the ‘=’ means
(meaning it does not follow Equation (1.1)) and gave it a test run. If
‘is not equal to’. In we could independently verify that the pressure p had been varied other words, h is not a function of p in a poor over a wide range of values yet the length of the mercury h in barometer. the barometer did not change, then we would say no relationship existed between p and h.

WHAT IS PHYSICAL CHEMISTRY: VARIABLES, RELATIONSHIPS AND LAWS

7

Why does a radiator feel hot to the touch when ‘on’, and cold when ‘off’?
Laws and the minus-oneth law of thermodynamics
Feeling the temperature of a radiator is one of the simplest of experiments. No one has ever sat in front of a hot radiator and A ‘law’ in physical felt colder. As a qualitative statement, we begin with the excellent chemistry relates to a wide range of situageneralization, ‘heat always travels from the hotter to the colder tions. environment’. We call this observation a law because it is universal.
Note how such a law is not concerned with magnitudes of change but simply relays information about a universal phenomenon: energy in the form of heat will travel from a hotter location or system to a place which is colder. Heat energy never travels in the opposite direction.
We can also notice how, by saying ‘hotter’ and ‘colder’ rather than just ‘hot’ and
‘cold’, we can make the law wider in scope. The temperature of a radiator in a living room or lecture theatre is typically about 60 ◦ C, whereas a human body has an ideal temperature of about 37 ◦ C. The radiator is hotter than we are, so heat travels to us from the radiator. It is this heat emitted by the radiator which we absorb in order to feel warmer.
Conversely, now consider placing your hands on a colder radiator having a temperature of 20 ◦ C (perhaps it is broken or has not been switched on). In this second example, although our hands still have the same temperature of 37 ◦ C, this time the heat energy travels to the radiator from our hands as soon as we touch it. The direction of heat flow has been reversed in response to the reversal of the relative difference between the two temperatures. The direction in which the heat energy is transferred is one aspect of why the radiator feels cold. We see how the movement of energy not only has a magnitude but also a direction.
Such statements concerning the direction of heat transfer are sometimes called the minus-oneth law of thermodynamics, which The ‘minus-oneth law sounds rather daunting. In fact, the word ‘thermodynamics’ here of thermodynamics’ may be taken apart piecemeal to translate it into everyday English. says, ‘heat always
First the simple bit: ‘dynamic’ comes from the Greek word travels from hot to dunamikos, which means movement. We obtain the conventional cold’.
English word ‘dynamic’ from the same root; and a cyclist’s
‘dynamo’ generates electrical energy from the spinning of a bicycle wheel, i.e. from a moving object. Secondly, thermo is another commonly encountered Greek root, and means energy or temperature. We encounter the root thermo incorporated into such everyday words as ‘thermometer’, ‘thermal’ and ‘thermos flask’. A ‘thermodynamic’ property, therefore, relates to events or processes in which there are ‘changes in heat or energy’.

8

INTRODUCTION TO PHYSICAL CHEMISTRY

Aside
We need to explain the bizarre name of this law, which is really an accident of history.
Soon after the first law of thermodynamics was postulated in the mid nineteenth century, it was realized how the law presupposed a more elementary law, which we now call the zeroth law (see below). We call it the ‘zeroth’ because zero comes before one. But scientists soon realized how even the zeroth law was too advanced, since it presupposed a yet more elementary law, which explains why the minus-oneth law had to be formulated.

How does a thermometer work?
Thermal equilibrium and the zeroth law of thermodynamics
A fever is often the first visible sign of someone developing an illness. The body’s temperature rises – sometimes dramatically – above its preferred value of 37 ◦ C. As a good generalization, the temperature is hotter when the fever is worse, so it is wise to monitor the temperature of the sick person and thereby check the progress of the illness. A thermometer is the ideal instrument for this purpose.
When measuring a temperature with a thermometer, we place the mercury-containing end into the patient’s mouth or armpit and allow the reading to settle. The mercury is encased within a thinwalled glass tube, which itself is placed in contact with the patient.
A ‘reading’ is possible because the mercury expands with increasing temperature: we take the length l of the mercury in the tube to be an accurate function of its temperature T . We read the patient’s temperature from the thermometer scale only when the length of the mercury has stopped changing.
But how does the thermometer work in a thermodynamic sense, since at no time can the toxic mercury be allowed to touch the patient?
Consider the flow of heat: heat energy first flows from the patient to the glass, and thence flows through the glass into the mercury.
Bodies together at the
Only when all three – mercury, glass and patient – are at the same same temperature are said to be in ‘thermal temperature can the thermometer reading become steady. We say equilibrium’. we have thermal equilibrium when these three have the same temperature; see Figure 1.3.
Although in some respects a trivial example, a thermometer helps us see a profound truth: only when both (i) the mercury and the glass, and (ii) the glass and the patient are at thermal equilibrium can the patient and the mercury truly be said to be at the same temperature. By this means, we have measured the temperature of the patient by

The word ‘thermometer’ has two roots: meter denotes a device to measure something, and thermo means
‘energy’ or ‘temperature’. Thus, a ‘thermometer’ is a device for measuring energy as a function of temperature.

THE PRACTICE OF THERMODYNAMIC MEASUREMENT

A

Mouth

B

Glass

9

C

Mercury

Patient's tongue

Mercury-in-glass thermometer

Figure 1.3 The zeroth law states, ‘Imagine three bodies, A, B and C. If A and B are in thermal equilibrium, and B and C are in thermal equilibrium, then A and C are also in thermal equilibrium’
(see inset). A medic would rephrase the law, ‘If mercury is in thermal equilibrium with the glass of a thermometer, and the glass of a thermometer is in thermal equilibrium with a patient, then the mercury and the patient are also in thermal equilibrium’

utilizing a temperature-dependent property of the liquid metal inside the thermometer, yet at no time do we need to expose the patient to the toxic mercury.
We begin to understand the power of thermodynamics when we realize how often this situation arises: in effect, we have made The zeroth law of theran indirect measurement – a frequent occurrence – so we need to modynamics says: formulate another law of thermodynamics, which we call the zeroth imagine three bodies, law. Imagine three bodies, A, B and C. If A and B are in thermal A, B and C. If A and B are equilibrium, and B and C are in thermal equilibrium, then A and in thermal equilibrium, and B and C are also
C are also in thermal equilibrium.
While sounding overly technical, we have in fact employed the in thermal equilibrium, then A and C will be in zeroth law with the example of a thermometer. Let us rephrase thermal equilibrium. the definition of the zeroth law and say, ‘If mercury is in thermal equilibrium with the glass of a thermometer, and the glass of a thermometer is in thermal equilibrium with a patient, then the mercury and the patient are also in thermal equilibrium’. A medic could not easily determine the temperature of a patient without this, the zeroth law.
From now on we will assume the zeroth law is obeyed each time we use the phrase
‘thermal equilibrium’.

1.2

The practice of thermodynamic measurement What is temperature?
Scientific measurement
Although the answer to the simple question ‘what is temperature?’ seems obvious at first sight, it is surprisingly difficult to answer to everyone’s satisfaction. In fact, it is

10

INTRODUCTION TO PHYSICAL CHEMISTRY

generally easier to state the corollary, ‘a body has a higher temperature if it has more energy, and a lower temperature if it has less energy’. We have been rather glib so far when using words such as
‘heat’ and ‘temperature’, and will be more careful in future. Heat is merely one way by which we experience energy. Everything contains energy in various amounts, although the exact quantity of the energy is not only unknown but unknowable.
Much of the time, we, as physical chemists, will be thinking about energy and the way energetic changes accompany chemical
The word ‘thermochanges (i.e. atoms, ions or whole groups of atoms combine, or chemistry’ has two add, or are being lost, from molecules). While the total energy roots: thermo, meancannot be known, we can readily determine the changes that occur ing ‘temperature or in tandem with chemical changes. We sometimes give the name energy’, and chemthermochemistry to this aspect of physical chemistry. istry, the science of the combination of
In practice, the concept of temperature is most useful when deterchemicals. We see mining whether two bodies are in thermal equilibrium. Firstly, we how ‘thermochemistry’ need to appreciate how these equilibrium processes are always studies the energy and dynamic, which, stated another way, indicates that a body simultatemperature changes neously emits and absorbs energy, with these respective amounts accompanying chemiof energy being equal and opposite. Furthermore, if two bodies cal changes. participate in a thermal equilibrium then we say that the energy emitted by the first body is absorbed by the second; and the first body also absorbs a similar amount of energy to that emitted by
A body in ‘dynamic equithe second body. librium’ with another
Temperature is most conveniently visualized in terms of the exchanges energy with senses: we say something is hotter or is colder. The first therit, yet without any net mometer for studying changes in temperature was devised in 1631 change. by the Frenchman Jean Rey, and comprised a length of water in a glass tube, much like our current-day mercury-in-glass thermometers but on a much bigger scale. The controlled variable in this thermometer was temperature T , and the observed variable was the length l of the water in the glass tube.
Rey’s thermometer was not particularly effective because the density of water is so low, meaning that the volume of the tube had to be large. And the tube size caused an additional problem. While the water expanded with temperature (as required for the thermometer to be effective), so did the glass encapsulating it. In consequence of both water and glass expanding, although the water expanded in a straightforward way with increasing temperature, the visible magnitude of the expansion was not in direct proportion to the temperature rise.
Although we could suggest that a relationship existed between the length l and the temperature T (saying one is a function of the
Scientists use the word other), we could not straightforwardly ascertain the exact nature
‘ideal’ to mean obeying of the function. In an ideal thermometer, we write the mathematthe laws of science. ical relationship, l = f (T ). Because Rey’s thermometer contained
‘Corollary’ means a deduction following on from another, related, fact or series of facts.

THE PRACTICE OF THERMODYNAMIC MEASUREMENT

11

Observed variable (length of mercury in a thermometer, l )

water, Rey was not able to observe a linear dependence of l on T for his thermometer, so he could not write l = aT + b (where a and b are constants).
For a more dense liquid, such as mercury, the relationship between l and T is linear – at least over a relatively narrow range ‘Narrow’ in this case of temperatures – so a viable mercury-in-glass thermometer may means 50–70 ◦ C at be constructed. But, because the temperature response is only lin- most. ear over a narrow range of temperatures, we need to exercise caution. If we assume the existence of a linear response for such a thermometer, then the thermometer is ‘calibrated’ by correlating the To ‘calibrate’ an instrureadings of length l using the known properties of the standard, as ment such as a therfollows. First, the thermometer is placed in a trough of pure ice at mometer, we correlate its melting temperature, and the end of the mercury bead marked a physico-chemical as 0 ◦ C. The same thermometer is then placed in water at its boil- property (such as the ing point and the end of the mercury bead marked as 100 ◦ C. The length l of the mercury) physical distance between these two extremes is subdivided into using the temperaturedependent properties
100 equal portions, each representing a temperature increment of of a known standard.
1 ◦ C. This centigrade scale is satisfactory for most purposes. (The same scale is sometimes called Celsius after a Swedish physicist who championed its use.)
This formulation of the centigrade scale presupposed a linear relationship between length l and temperature T (i.e. the straight The ‘centigrade’ scale line (a) on the graph in Figure 1.4), but we must be aware the was first proposed in relationship might only be approximately linear (e.g. the curved 1694 by Renaldi. Centi line (b) on the graph in Figure 1.4). The straight and the curved is a Latin prefix meanlines only agree at the two temperatures 0 ◦ C and 100 ◦ C merely ing ‘hundred’. because they were defined that way.

(a) Ideal response

(b) More realistic response

0°C

100 °C
Controlled variable (temperature, T )

Figure 1.4 In using a thermometer, we assume the existence of a linear response between the length l of the mercury and the controlled variable temperature T . Trace (a) shows such a relationship, and trace (b) shows a more likely situation, in which there is a close approximation to a linear relationship between length l and temperature T

12

INTRODUCTION TO PHYSICAL CHEMISTRY

This last paragraph inevitably leads to the questions, ‘So how do we know what the exact temperature is?’ and ‘How do I know if my thermometer follows profile
(a) or profile (b) in Figure 1.4?’ Usually, we do not know the answer. If we had a single thermometer whose temperature was always accurate then we could use it as a primary standard, and would simply prepare a calibrated thermometer against which all others are calibrated.
But there are no ideal (perfect) thermometers in the real world. In practice, we generally experiment a bit until we find a thermometer for which a property X is as close to being a linear function of temperature as possible, and call it a standard thermometer (or ‘ideal thermometer’). We then calibrate other thermometers in relation to this, the standard. There are several good approximations to a standard thermometers available today: the temperature-dependent (observed) variable in a gas thermometer is the volume of a gas V . Provided the pressure of the gas is quite low (say, onehundredth of atmospheric pressure, i.e. 100 Pa) then the volume V and temperature
T do indeed follow a fairly good linear relationship.
A second, popular, standard is the platinum-resistance thermometer. Here, the electrical resistance R of a long wire of platinum increases with increased temperature, again with an essentially linear relationship.
Worked Example 1.1 A platinum resistance thermometer has a resistance R of 3.0 ×
10−4 at 0 ◦ C and 9.0 × 10−4 at 100 ◦ C. What is the temperature if the resistance R is measured and found to be 4.3 × 10−4 ?
We first work out the exact relationship between resistance R and temperature T . We must assume a linear relationship between the two to do so.
The change per degree centigrade is obtained as ‘net change in resistance ÷ net change in temperature’. The resistance R increases
These discussions are by 6.0 × 10−4 while the temperature is increased over the 100 ◦ C expressed in terms of range; therefore, the increase in resistance per degree centigrade is centigrade, although given by the expression absolute temperatures are often employed – see next section.



R per C =

6.0 × 10−4
100 ◦ C

= 6 × 10−6



C−1

Next, we determine by how much the resistance has increased in going to the new
(as yet unknown) temperature. We see how the resistance increases by an amount (4.3 −
3.0) × 10−4 = 1.3 × 10−4 .
The increase in temperature is then the rise in resistance divided by the change in resistance increase per degree centigrade.
We obtain
1.3 × 10−4
6 × 10−6 ◦ C−1 so the new temperature is 21.7 ◦ C.

THE PRACTICE OF THERMODYNAMIC MEASUREMENT

13

We should note, before proceeding, firstly how the units of on both top and bottom of this fraction cancel; and secondly, how ◦ C−1 is in the denominator of the fraction. As a consequence of it being on the bottom of the fraction, it is inverted and so becomes ◦ C. In summary, we see how a simple analysis of the units in this sum automatically allows the eventual answer to be expressed in terms of ◦ C. We are therefore delighted, because the answer we want is a temperature, and the units tell us it is indeed a temperature.

Aside
This manipulation of units is sometimes called dimensional analysis. Strictly speaking, though, dimensional analysis is independent of the units used. For example, the units of speed may be in metres per second, miles per hour, etc., but the dimensions of speed are always a length [L] divided by a time [T]:
[speed] = [L] ÷ [T]
Dimensional analysis is useful in two respects. (1) It can be used to determine the units of a variable in an equation. (2) Using the normal rules of algebra, it can be used to determine whether an equation is dimensionally correct, i.e. the units should balance on either side of the equation. All equations in any science discipline are dimensionally correct or they are wrong!
A related concept to dimensional analysis is quantity calculus, a method we find particularly useful when it comes to setting out table header rows and graph axes.
Quantity calculus is the handling of physical quantities and their units using the normal rules of algebra. A physical quantity is defined by a numerical value and a unit: physical quantity = number × unit
e.g.
H = 40.7 kJ mol−1 which rearranges to

H /kJ mol−1 = 40.7

SAQ 1.1 A temperature is measured with the same platinum-resistance thermometer used in Worked Example 1.1, and a resistance R = 11.4 ×
10−4 determined. What is the temperature?

14

INTRODUCTION TO PHYSICAL CHEMISTRY

The word ‘philosophical’ comes from the
Greek words philos meaning ‘love’ and sophia meaning ‘wisdom’. Philosophy is therefore the love of wisdom. This same usage of ‘wisdom’ is seen with the initials
PhD, which means a
‘philosophy doctorate’.

Some people might argue that none of the discussion above actually answers the philosophical question, ‘What is temperature?’
We will never come to a completely satisfactory answer; but we can suppose a body has a higher temperature if it contains more energy, and that it has a lower temperature if it has less energy.
More importantly, a body will show a rise in temperature if its energy content rises, and it will show a lower temperature if its energy content drops. This is why we sit in front of a fire: we want to absorb energy, which we experience as a higher temperature.

How long is a piece of string?
This definition of ‘more energy means hotter’ needs to be handled with care: consider two identical weights at the same temperature. The higher weight has a greater potential energy. The SI unit of length

A common problem in Anglo Saxon England, as well as much of contemporary Europe, was the way cloth merchants could so easily cheat the common people. At a market, it was all too easy to ask for a yard of cloth, to see it measured against the merchant’s yardstick, and pay for the cloth only to get home to learn just how short the merchant’s stick was. Paying for 10 yards and coming home with only 9 yards was common, it seems; and the problem was not restricted to just cloth, but also to leather and timber.
According to legend, the far-sighted English King Edgar (AD 959–975) solved the problem of how to stop such cheating by standardizing the length. He took 100 foot soldiers and measured the length of the right foot of each, one after the other, as they stood in line along the floor of his threshing hall. This overall length was then subdivided into 100 equal parts to yield the standard length, the foot. The foot is still commonly employed as a unit of length in Britain to this day. Three of these feet made up 1 yard. The king was said to keep in his treasury a rod of gold measuring exactly 1 yard in length. This is one theory of how the phrase ‘yardstick’ originated.
Any merchant accused of cheating was required to bring his yardstick and to compare its length against that of the king. Therefore, a merchant whose stick was shorter was a cheat and paid the consequences. A merchant whose stick was longer was an idiot.
While feet and yards are still used in Britain and other countries, the usual length is now the metre. At the time of the French RevSI units are selfolution in the 18th century and soon after, the French Academy consistent, with all of Sciences sought to systemize the measurement of all scientific units being defined in quantities. This work led eventually to the concept of the Syst` me e terms a basis of seven
Internationale, or SI for short. Within this system, all units and fundamental units. The definitions are self-consistent. The SI unit of length is the metre.
SI unit of length l is the
The original metre rule was kept in the International Bureau metre (m). of Weights and Measures in S` vres, near Paris, and was a rod of e THE PRACTICE OF THERMODYNAMIC MEASUREMENT

platinum–iridium alloy on which two deep marks were scratched
1 m apart. It was used in exactly the same way as King Edgar’s yardstick 10 centuries earlier.
Unfortunately, platinum–iridium alloy was a poor choice, for it has the unusual property of shrinking (albeit microscopically) with time. This SI metre rule is now about 0.3 per cent too short. King
Edgar’s yardstick, being made of gold, would still be the same length today as when it was made, but gold is too ductile, and could have been stretched, bent or re-scored.
In 1960, the SI unit of length was redefined. While keeping the metre as the unit of length, it is now defined as 1 650 763.73 wavelengths of the light emitted in vacuo by krypton-86. This is a sensible standard, because it can be reproduced in any laboratory in the world.

15

‘Ductile’ means the ability of a metal to be drawn to form a wire, or to be worked.
Ductile is the opposite of ‘brittle’.
In vacuo is Latin for
‘in a vacuum’. Many properties are measured in a vacuum to avoid the complication of interference effects.

How fast is ‘greased lightning’?
Other SI standards
In comic books of the 1950s, one of the favourite phrases of super-heroes such as
Superman was ‘greased lightning!’ The idea is one of extreme speed. The lightning we see, greased or otherwise, is a form of light and travels very, very fast. For example, it travels through a vacuum at 3 × 108 m s−1 , which we denote with the symbol c.
But while the speed c is constant, the actual speed of light may not be: in fact, it alters very slightly depending on the medium through which it travels. We see how a definition of time involving the speed of light is inherently risky, explaining why we now choose to define time in terms of the duration (or fractions and multiples thereof) between static events. And by ‘static’ we mean unchanging.

SI ‘base units’
Time is one of the so-called ‘base units’ within the SI system, and so is length.
Whereas volume can be expressed in terms of a length (for example, a cube has a volume l 3 and side of area l 2 ), we cannot define length in terms of something simpler. Similarly, whereas a velocity is a length per unit There are seven base time, we cannot express time in terms of something simpler. In fact, SI units: length, time, just as compounds are made up of elements, so all scientific units mass, temperature, are made up from seven base units: length, time, mass, temperature, current, luminous intensity and amount current, amount of material and luminous intensity.
Table 1.1 summarizes the seven base (or ‘fundamental’) SI phys- of material. ical quantities and their units. The last unit, luminous intensity, will not require our attention any further.
The SI unit of ‘time’ t is
The SI unit of ‘time’ t is the second. The second was originally the second (s). defined as 1/86 400th part of a mean solar day. This definition is

16

INTRODUCTION TO PHYSICAL CHEMISTRY

Table 1.1

The seven fundamental SI physical quantities and their units

Physical quantity
Length
Mass
Time
Electrical current
Thermodynamic temperature
Amount of substance
Luminous intensity

Symbola

SI unit

Abbreviation

l m t
I
T n Iv

metre kilogram second amp` re e kelvin mole candela

m kg s
A
K mol cd

a
Notice how the abbreviation for each quantity, being a variable, is always italicized, whereas the abbreviation for the unit, which is not a variable, is printed with an upright typeface. None of these unit names starts with a capital.

again quite sensible because it can be reproduced in any laboratory in the world.
While slight changes in the length of a solar year do occur, the word ‘mean’ in our definition obviates any need to consider them. Nevertheless, it was felt necessary to redefine the second; so, in the 1960s, the second was redefined as 9 192 631 770 periods of the radiation corresponding to the transition between two of the hyperfine levels in the ground state of the caesium-133 atom. Without discussion, we note how the heart of a so-called ‘atomic clock’ contains some caesium-133.
In a similar way, the Syst` me Internationale has ‘defined’ other e The SI unit of ‘temperacommon physicochemical variables. The SI unit of ‘temperature’ ture’ T is the kelvin (K).
T is the kelvin. We define the kelvin as 1/273.16th part of the thermodynamic temperature difference between absolute zero (see
Section 1.4) and the triple point of water, i.e. the temperature at which liquid water is at equilibrium with solid water (ice) and gaseous water (steam) provided that the pressure is 610 Pa.
The SI unit of ‘current’ I is the amp` re (A). An amp` re was e e
The SI unit of ‘current’ first defined as the current flowing when a charge of 1 C (coulomb)
I is the amp`re (A). e passed per second through a perfect (i.e. resistance-free) conductor.
The SI definition is more rigorous: ‘the amp` re is that constant e current which, if maintained in two parallel conductors (each of negligible resistance) and placed in vacuo 1 m apart, produces a force between of exactly 2 × 10−7 N per metre of length’. We will not employ this latter definition.
The SI unit of the ‘amount of substance’ n is the mole. CuriThe SI unit of ‘amount ously, the SI General Conference on Weights and Measures only of substance’ n is the decided in 1971 to incorporate the mole into its basic set of fundamole (mol). mental parameters, thereby filling an embarrassing loophole. The mole is the amount of substance in a system that contains as many elementary entities as does 0.012 kg (12 g) of carbon-12. The amount of substance must be stated in terms of the elementary entities chosen, be they photons, electrons, protons, atoms, ions or molecules.
The number of elementary entities in 1 mol is an experimentally determined quantity, and is called the ‘Avogadro constant’ L, which has the value 6.022 × 1023 mol−1 .
The Avogadro constant is also (incorrectly) called the ‘Avogadro number’. It is

THE PRACTICE OF THERMODYNAMIC MEASUREMENT

17

Table 1.2 Several of the more common units that are not members of the Syst` me e Internationale
Quantity

Non-SI unit

Abbreviation

Conversion from non-SI to SI

Energy
Length
Pressure
Pressure
Volume

calorie angstr¨ m o ˚ atmosphere bar litre cal
˚
A atm bar dm3 1
1
1
1
1

cal = 4.184 J
˚
A = 10−10 m atm = 101 325 Pa bar = 105 Pa dm3 = 10−3 m3

increasingly common to see the Avogadro constant given a different symbol than
L. The most popular alternative symbol at present seems to be NA .

Non-SI units
It is important to be consistent with units when we start a calculation. An enormously expensive spacecraft crashed on the surface of the planet Mars in 1999 because a distance was calculated by a NASA scientist in terms of inches rather than centimetres.
Several non-SI units persist in modern usage, the most common being listed in
Table 1.2. A calculation performed wholly in terms of SI units will be self-consistent.
Provided we know a suitable way to interchange between the SI and non-SI units, we can still employ our old non-SI favourites.

Aside
In addition to the thermodynamic temperature T there is also the Celsius temperature t, defined as t = T − T0 where T0 = 273.15 K.
Sometimes, to avoid confusion with the use of t as the symbol for time, the Greek symbol θ (theta) is substituted for the Celsius temperature t instead.
Throughout this book we adopt T to mean temperature. The context will make clear whether T is required to be in degrees Celsius or kelvin. Beware, though, that most formulae require the use of temperature in kelvin.

Why is the SI unit of mass the kilogram?
Multiples and the SI unit of mass m
The definition of mass in the Syst` me Internationale scheme departs from the stated e aim of formulating a rigorous, self-consistent set of standards. The SI unit of ‘mass’

18

INTRODUCTION TO PHYSICAL CHEMISTRY

Table 1.3 Selection of a few physicochemical parameters that comprise combinations of the seven SI fundamental quantities
Quantity
Acceleration
Area
Density
Force
Pressure
Velocity
Volume

The SI unit of ‘mass’ m is the kilogram (kg).

In the SI system, 1 g is defined as the mass of 5.02 × 1022 atoms of carbon-12. This number comes from L/12, where L is the Avogadro number.

Symbol a A ρ F p v
V

SI units m s−2 m2 kg m−3 kg m s−2 kg m−1 s−2 m s−1 m3 m is the ‘kilogram’. Similar to the metre, the original SI standard of mass was a block of platinum metal in S` vres, near Paris, which e weighted exactly 1 kg. The current SI definition is more complicated: because 12.000 g in the SI system represents exactly 1 mol of carbon-12, then 1 g is one-twelfth of a mole of carbon-12.
The problem with the SI base unit being a kilogram is the ‘kilo’ part. The philosophical idea behind the SI system says any parameter (physical, chemical, mechanical, etc.) can be derived from a suitable combination of the others. For example, the SI unit of velocity is metres per second (m s−1 ), which is made up of the two
SI fundamental units of length (the metre) and time (the second).
A few of these combinations are cited in Table 1.3.

Why is ‘the material of action so variable’?
Writing variables and phrases
The classical author Epictetus (ca 50–ca 138 AD) once said, ‘The materials of action are variable, but the use we make of them should be constant’. How wise.
When we build a house, we only require a certain number of building materials: say, bricks, tubes and window panes. The quanWe give the name tity surveyor in charge of the building project decides which materi‘compound unit’ to als are needed, and writes a quantity beside each on his order form: several units written
10 000 bricks, 20 window panes, etc. Similarly, when we have a together. We leave a velocity, we have the units of ‘m’ and ‘s−1 ’, and then quantify space between each it, saying something like, ‘The man ran fast, covering a distance constituent unit when of 10 metres per second’. By this means, any parameter is defined we write such a comboth qualitatively (in terms of its units) and quantitatively (in terms pound unit. of a number). With symbols, we would write v = 10 m s−1 .

THE PRACTICE OF THERMODYNAMIC MEASUREMENT

19

A variable (mass, length, velocity, etc.) is written in a standard format, according to Equation (1.2):
Variable or physicochemical quantity = number × units
We sometimes call it a ‘phrase’. Because some numbers are huge and others tiny, the SI system allows us a simple and convenient shorthand. We do not need to write out all the zeros, saying the velocity of light c is 300 000 000 m s−1 : we can write it as c = 3 ×
108 m s−1 or as 0.3 Gm s−1 , where the capital ‘G’ is a shorthand for ‘giga’, or 1 000 000 000. The symbol G (for giga) in this context is called a ‘factor’. In effect, we are saying 300 000 000 m s−1 =
0.3 Gm s−1 . The standard factors are listed on pp. xxviii–xxxi.
Most people find that writing 300 000 000 m s−1 is a bit long winded. Some people do not like writing simple factors such as G for giga, and prefer so-called scientific notation. In this style, we write a number followed by a factor expressed as ten raised to an appropriate power. The number above would be 3.0 × 108 m s−1 .
Worked Example 1.2 Identify the variable, number, factor and unit in the phrase, ‘energy = 12 kJ mol−1 ’.
Energy

Variable

= 12

Number

k

(1.2)

‘Giga’ comes from the
Latin gigas, meaning
‘giant’ or ‘huge’. We also get the everyday words ‘giant’ and
‘gigantic’ from this root. In physical chemistry, a ‘factor’ is a number by which we multiply the numerical value of a variable. Factors are usually employed with a shorthand notation.

J mol−1

Factor

Compound unit Reasoning
Variable – in simple mathematical ‘phrases’ such as this, we almost always write the variable on the left. A variable is a quantity whose value can be altered.
Number – the easy part! It will be made up of numbers 1, 2, 3, . . . , 0.
Factor – if we need a factor, it will always be written between the number and the units (compound or single). A comprehensive list of the simple factors is given on pp. xxviii–xxxi.
Units – the units are always written on the right of a phrase such as this. There are two units here, joules (J) and moles (as ‘mol−1 ’, in this case). We should leave a space between them.
A factor is simply shorthand, and is dispensable. We could have dispensed with the factor and written the number differently, saying energy = 12 000 J mol−1 . This same energy in scientific notation would be 12 × 103 J mol−1 . But units are not dispensable.

20

INTRODUCTION TO PHYSICAL CHEMISTRY

SAQ 1.2 Identify the variable, number, factor and unit in the phrase,
‘length = 3.2 km’.

1.3

Properties of gases and the gas laws
Why do we see eddy patterns above a radiator?
The effects of temperature on density
The air around a hot radiator soon acquires heat. We explain this observation from the ‘minus oneth law of thermodynamics’ (see Section 1.1), since heat travels from hot to cold.
The density of a gas depends quite strongly on its temperature, so hot air has a smaller density than does cold air; colder air is more dense than hot air. From everyday experience, we know that something is dense if it tries to drop, which is why a stone drops to the bottom of a pond and a coin sinks to the bottom of a pan of water. This relative motion occurs because both the stone and the coin have higher densities than does water, so they drop. Similarly, we are more dense than air and will drop if we fall off a roof.
Just like the coin in water, cold air sinks because it is denser than warmer air.
We sometimes see this situation stated as warm air ‘displaces’ the cold air, which subsequently takes its place. Alternatively, we say ‘warm air rises’, which explains why we place our clothes above a radiator to dry them, rather than below it.
Light entering the room above the radiator passes through these pockets of warm air as they rise through colder air, and therefore passes through regions of different density. The rays of light bend in transit as they pass from region to region, much in the same way as light twists when it passes through a glass of water. We say the light is refracted. The eye responds to light, and interprets these refractions and twists as different intensities.
So we see swirling eddy (or ‘convective’) patterns above a radiator because the density of air is a function of temperature. If all the air had the same temperature, then no such difference in density would exist, and hence we would see no refraction and no eddy currents – which is the case in the summer when the radiator is switched off.
Then again, we can sometimes see a ‘heat haze’ above a hot road, which is caused by exactly the same phenomenon.

Why does a hot-air balloon float?
The effect of temperature on gas volume
A hot-air balloon is one of the more graceful sights of summer. A vast floating ball, powered only by a small propane burner, seems to defy gravity as it floats effortlessly above the ground. But what is it causing the balloon to fly, despite its considerable weight?

PROPERTIES OF GASES AND THE GAS LAWS

The small burner at the heart of the balloon heats the air within the canvas hood of the balloon. The densities of all materials – solid, liquid or gas – alter with temperature. Almost universally, we find the density ρ increases with cooling. Density ρ is defined as the ratio of mass m to volume V , according to density ρ =

mass, m volume, V

21

‘Density’ ρ is defined as mass per unit volume.

(1.3)

It is not reasonable to suppose the mass m of a gas changes by heating or cooling it
(in the absence of chemical reactions, that is), so the changes in ρ caused by heating must have been caused by changes in volume V . On the other hand, if the volume were to decrease on heating, then the density would increase.
So the reason why the balloon floats is because the air inside its voluminous hood has a lower density than the air outside. The exterior air, therefore, sinks lower than the less-dense air inside. And the sinking of the cold air and the rising of the warm air is effectively the same thing: it is movement of the one relative to the other, so the balloon floats above the ground. Conversely, the balloon descends back to earth when the air it contains cools to the same temperature as the air outside the hood.

How was the absolute zero of temperature determined? Charles’s law
J. A. C. Charles (1746–1823) was an aristocratic amateur scientist of the 18th century.
He already knew that the volume V of a gas increased with increasing temperature T , and was determined to find a relationship between these variables.
The law that now bears his name can be stated as, ‘The ratio of
According to ‘Charles’s volume and temperature for a fixed mass of gas remains constant’, law’, a linear relationprovided the external pressure is not altered. ship exists between
Stated mathematically, Charles demonstrated
V and T (at constant
V
= constant
T

pressure p).

(1.4)

where the value of the constant depends on both the amount and the identity of the gas. It also depends on the pressure, so the data are obtained at constant pressure p.
This is one form of ‘Charles’s law’. (Charles’s law is also called
‘Gay–Lussac’s law’.) Alternatively, we could have multiplied both sides of Equation (1.4) by T , and rewritten it as
V = constant × T

(1.5)

A ‘straight line’ will always have an equation of the type y = mx + c, where m is the gradient and c is the intercept on the y-axis
(i.e. when the value of x = 0).

22

INTRODUCTION TO PHYSICAL CHEMISTRY

Lord Kelvin (1824–1907) was a great thermodynamicist whom we shall meet quite often in these pages. He noticed how the relationship in Equation (1.5) resembles the equation of a straight line, i.e. takes the form y = mx + c

(1.6)

observed gradient controlled constant variable variable

except without an intercept, i.e. c = 0. Kelvin obtained good-quality data for the volume of a variety of gases as a function of temperature, and plotted graphs of volume V (as y) against temperature T (as x) for each; curiously, however, he was unable to draw a graph with a zero intercept for any of them.
Kelvin then replotted his data, this time extrapolating each graph till the volume of the gas was zero, which he found to occur at a
Note: degrees in the temperature of −273.15 ◦ C; see Figure 1.5. He then devised a new
Kelvin scale do not have the degree symtemperature scale in which this, the coldest of temperatures, was bol. The units have a the zero. He called it absolute zero, and each subsequent degree capital K, but the noun was equal to 1 ◦ C. This new scale of temperature is now called
‘kelvin’ has a small the thermodynamic (or absolute) scale of temperature, and is also letter. sometimes called the Kelvin scale.
The relationship between temperatures T on the centigrade and the absolute temperature scales is given by


T in C = T in K − 273.15

(1.7)

Equation (1.7) demonstrates how 1 ◦ C = 1 K.
0.04

Volume/m3

0.03

0.02

0.01

0
0

100

200
300
Temperature T / K

400

Figure 1.5 A graph of the volume V of a gas (as y) against temperature T (as x) is linear.
Extrapolating the gas’s volume suggests its volume will be zero if the temperature is −273.15 ◦ C
(which we call 0 K, or absolute zero)

PROPERTIES OF GASES AND THE GAS LAWS

SAQ 1.3 What is the temperature T expressed in kelvin if the temperature is 30 ◦ C?
SAQ 1.4 What is the centigrade temperature corresponding to 287.2 K?
SAQ 1.5 The data in the table below relate to gaseous helium. Demonstrate the linear relationship between the volume V and the temperature T .
Temperature T /K
Volume V /m3

280
0.023

300
0.025

320
0.027

340
0.028

360
0.030

380
0.032

400
0.033

Charles’s law is often expressed in a slightly different form than
Equation (1.4), as
V1
V2
=
(1.8)
T1
T2 which is generally regarded as superior to Equation (1.4) because we do not need to know the value of the constant.
Equation (1.8) is also preferred in situations where the volume of a fixed amount of gas changes in response to temperature changes
(but at constant pressure). The subscripts refer to the two situations; so, for example, the volume at temperature T1 is V1 and the volume at temperature T2 is V2 .
SAQ 1.6 The gas inside a balloon has a volume V1 of
1 dm3 at 298 K. It is warmed to 350 K. What is the volume following warming? Assume the pressure remained constant. 23

We divide each temperature, both kelvin and centigrade, by its respective unit to obtain a number, rather than the temperature.

420
0.035

440
0.037

Note how we write the controlled variable along the top row of a table, with the observed following. (If the table is vertical, we write the controlled variable on the far left.)

The subscripts written to the right of a variable are called ‘descriptors’.
They are always written as a subscript, because a superscripted number means a power, i.e. V 2 means V × V.

Why pressurize the contents of a gas canister?
The effect of pressure on gas volume: Boyle’s law
It is easy to buy canisters of gas of many sizes, e.g. as fuel when we wish to camp in the country, or for a portable welding kit. The gas will be n-butane if the gas is for heating purposes, but might be oxygen or acetylene if the gas is to achieve the higher temperatures needed for welding.
Typically, the components within the can are gaseous at most temperatures. The typical volume of an aerosol can is about 0.3 dm3 (3 × 10−4 m3 ), so it could contain very little gas if stored at normal pressure. But if we purchase a canister of gas and release its entire contents at once, the gas would occupy a volume similar that of

24

INTRODUCTION TO PHYSICAL CHEMISTRY

Care: a small p indicates pressure, yet a big P is the symbol for the element phosphorus. Similarly, a big V indicates volume and a small v is the symbol for velocity.

an entire living room. To ensure the (small) can contains this (large) amount of gas, we pressurize it to increase its capacity. We see how volume and pressure are interrelated in a reciprocal way: the volume decreases as the pressure increases.
Robert Boyle was the first to formulate a relationship between p and V . Boyle was a contemporary of the greatest scientist the world has ever seen, the 17th-century physicist Sir Isaac Newton.
Boyle’s law was discovered in 1660, and states pV = constant

(1.9)

where the numerical value of the constant on the right-hand side of the equation depends on both the identity and amount of the gas, as well as its temperature T .
Figure 1.6 shows a graph of pressure p (as y) against volume V
An ‘isotherm’ is a line
(as x) for 1 mol of neon gas. There are several curves, each repreon a graph representsenting data obtained at a different temperature. The temperature ing values of a variable per curve was constant, so we call each curve an isotherm. The obtained at constant word isotherm has two Greek roots: iso means ‘same’ and thermo temperature. means temperature or energy. An isotherm therefore means at the same energy.
The actual shape of the curves in Figure 1.6 are those of reciprocals. We can prove this mathematical form if we divide both sides
‘Reciprocal’ means to of Equation (1.9) by V , which yields

turn a fraction upside down. X can be thought of as ‘X ÷ 1’, so its reciprocal is 1/X (i.e.
1 ÷ X).

p=

1
× constant
V

(1.10)

Figure 1.7 shows a graph of volume p (as y) against 1/volume
V (as x), and has been constructed with the same data as used for
100 000

Pressure p /Pa

80 000
60 000
40 000
20 000
0
0

1

2

3

4

Volume V /m3

Figure 1.6 Graph of pressure p (as y) against volume V (as x) for 1 mol of an ideal gas as a
) 800 K function of temperature: (· · · · ·) 200 K; (– · – · –) 400 K; (– – –) 600 K; (

PROPERTIES OF GASES AND THE GAS LAWS

25

7.00E+05
6.00E+05
800 K
Pressure p / Pa

5.00E+05
4.00E+05

600 K

3.00E+05
400 K
2.00E+05
1.00E+05

200 K

0.00E+00
0

20

40

60
1/(V/m3)

80

100

Figure 1.7 Graph of pressure p (as y) against reciprocal volume 1 ÷ V (as x) for 1 mol of an ideal gas as a function of temperature. The data are the same as those from Figure 1.6. The temperatures are indicated. We need to appreciate how plotting the same data on a different set of axes yields a linear graph, thereby allowing us to formulate a relationship between p and 1 ÷ V

Figure 1.6. Each of the lines on the graph is now linear. Again, we find these data are temperature dependent, so each has the gradient of the respective value of ‘constant’.
At constant temperature T , an increase in pressure (so p2 > p1 ) causes a decrease in volume (so V2 < V1 ). This observation At constant temperexplains why the graph for the gas at the higher temperatures has ature T , an increase in pressure (p2 > p1 ) a smaller value for the constant. causes a decrease in
An alternative way of writing Equation (1.9) is p1 V1 = p2 V2

volume (V2 < V1 ).

(1.11)

SAQ 1.7 The usual choice of propellant within an aerosol of air freshener is propane gas. What is the volume of propane following compression, if
1 dm3 of gaseous propane is compressed from a pressure of 1 atm to a pressure of 2.5 atm? Assume the temperature is kept constant during the compression. Why does thunder accompany lightning?
Effect of changing both temperature and pressure on gas volume Lightning is one of the most impressive and yet frightening manifestations of nature.
It reminds us just how powerful nature can be.

26

INTRODUCTION TO PHYSICAL CHEMISTRY

Lightning is quite a simple phenomenon. Just before a storm breaks, perhaps following a period of hot, fine weather, we often note how the air feels ‘tense’. In fact, we are expressing an experiential truth: the air contains a great number of ions – charged particles. The existence of a large charge on the Earth is mirrored by a large charge in the upper atmosphere. The only difference between these two charges is that the Earth bears a positive charge and the atmosphere bears a negative charge.
Accumulation of a charge difference between the Earth and the upper atmosphere cannot proceed indefinitely. The charges must eventually equalize somehow: in practice, negative charge in the upper atmosphere passes through the air to neutralize the positive charge on the Earth. The way we see this charge conducted between the Earth and the sky is lightning: in effect, air is ionized to make it a conductor, allowing electrons in the clouds and upper atmosphere to conduct through the air to the Earth’s surface. This movement of electrical charge is a current, which we see as lightning. Incidentally, ionized air emits light, which explains why we see lightning (see Chapter 9). Lightning comprises a massive amount of energy, so the local air through which it conducts tends to heat up to as much as a few thousand degrees centigrade.
And we have already seen how air expands when warmed, e.g. as described mathematically by Charles’s law (Equation (1.6)). In fact, the air through which the lightning passes increases in volume to an almost unbelievable extent because of its rise in temperature. And the expansion is very rapid.

‘Experiential’ means the way we notice something exists following an experience or sensation.

SAQ 1.8 Show, using the version of Charles’s law in Equation (1.8), how a rise in temperature from 330 K to 3300 K is accompanied by a tenfold increase in volume.

We hear the sensation of sound when the ear drum is moved by compression waves travelling through the air; we hear people because their speech is propagated by subtle pressure changes in the surrounding air. In a similar way, the huge increase in air volume is caused by huge changes in air pressure, itself manifested as sound: we hear the thunder caused by the air expanding, itself in response to lightning.
And the reason why we see the lightning first and hear the thunder later is because light travels faster than sound. The reason why thunder accompanies lightning, then, is because pressure p, volume V and temperature T are interrelated.

How does a bubble-jet printer work?
The ideal-gas equation
A bubble-jet printer is one of the more useful and versatile inventions of the last decade. The active component of the printer is the ‘head’ through which liquid ink passes before striking the page. The head moves from side to side over the page. When

PROPERTIES OF GASES AND THE GAS LAWS

27

the ‘head’ is positioned above a part of the page to which an image is required, the computer tells the head to eject a tiny bubble of ink. This jet of ink strikes the page to leave an indelible image. We have printing.
The head is commonly about an inch wide, and consists of a row of hundreds of tiny pores (or ‘capillaries’), each connecting the ink reservoir (the cartridge) and the page. The signals from the computer are different for each pore, allowing different parts of the page to receive ink at different times. By this method, images or letters are formed by the printer.
The pores are the really clever bit of the head. Half-way along each pore is a minute heater surrounded by a small pocket of air. In front of the heater is a small bubble of ink, and behind it is the circuitry of the printer, ultimately connecting the heater to the computer. One such capillary is shown schematically in Figure 1.8.
Just before the computer instructs the printer to eject a bubble of ink, the heater is activated, causing the air pocket to increase in temperature T at quite a rapid rate. The temperature increase causes the air to expand to a greater volume V . This greater volume increases the pressure p within the air pocket. The enhanced air pressure p is sufficient to eject the ink bubble from the pore and onto the page. This pressure-activated ejection is similar to spitting.
This ejection of ink from a bubble-jet printer ingeniously utilizes the interconnectedness of pressure p, volume V and temperature T . Experiments with simple gases show how p, T and V are related by the relation pV = constant
T

(1.12)

which should remind us of both Boyle’s law and Charles’s law.

Ink jet nozzle

Resistor

t > 5 µs

t ∼ 10 µs

t ∼ 20 µs

Figure 1.8 Schematic diagram of a capillary (one of hundreds) within the printing ‘head’ of a bubble-jet printer. The resistor heats a small portion of solution, which boils thereby increasing the pressure. Bubbles form within 5 µs of resistance heating; after 10 µs the micro-bubbles coalesce to force liquid from the aperture; and a bubble is ejected a further 10 µs later. The ejected bubble impinges on the paper moments afterwards to form a written image. Reproduced by permission of
Avecia

28

INTRODUCTION TO PHYSICAL CHEMISTRY

If there is exactly 1 mol of gas, the pressure is expressed in pascals (Pa), the temperature is in kelvin and the volume is in cubic metres (both SI units), then the value of the constant is 8.314 J K−1 mol−1 . We call it the gas constant and give it the symbol R. (Some old books may call R the ‘universal gas constant’, ‘molar gas constant’ or just ‘the gas constant’. You will find a discussion about R on p. 54)
More generally, Equation (1.12) is rewritten as
Equation (1.13) tells us the constant in Boyle’s law is ‘nRT ’ and the
(different) constant in
Charles’s law is
‘nR ÷ p’.

pV = nRT

(1.13)

where n is the number of moles of gas. Equation (1.13) is called the ideal-gas equation (or, sometimes, in older books the ‘universal gas equation’). The word ‘ideal’ here usually suggests that the gas in question obeys Equation (1.13).

Worked Example 1.3 What is the volume of 1 mol of gas at a room temperature of
25 ◦ C at an atmospheric pressure of 105 Pa?
First, we convert the data into the correct SI units. In this example, only the temperature needs to be converted. From Equation (1.7), the temperature is 298 K.
Secondly, we rearrange Equation (1.13) to make V the subject, by dividing both sides by p: nRT V = p and then insert values:
V =

1 mol × 8.314 J K−1 mol−1 × 298 K
105 Pa

So the volume V = 0.0248 m3 .
If we remember how there are 1000 dm3 in 1 m3 , we see how 1 mol of gas at room temperature and standard pressure has a volume of 24.8 dm3 .
SAQ 1.9 2 mol of gas occupy a volume V = 0.4 m3 at a temperature
T = 330 K. What is the pressure p of the gas?

An alternative form of Equation (1.13) is given as p 2 V2 p 1 V1
=
T1
T2

(1.14)

and is used when we have to start with a constant number of moles of gas n housed in a volume V1 . Its initial pressure is p1 when the temperature is T1 . Changing one variable causes at least one of the two to change. We say the new temperature is T2 , the new

PROPERTIES OF GASES AND THE GAS LAWS

29

pressure is p2 and the new volume is V2 . Equation (1.14) then holds provided the number of moles does not vary.
Worked Example 1.4 Nitrogen gas is housed in a sealed, hollow cylinder at a pressure of 105 Pa. Its temperature is 300 K and its volume is 30 dm3 . The volume within the cylinder is increased to 45 dm3 , and the temperature is increased at the same time to
310 K. What is the new pressure, p2 ?
We first rearrange Equation (1.14) to make the unknown volume p2 the subject, writing p2 =

p 1 V1 T 2
T 1 V2

We insert values into the rearranged equation: p2 =

105 Pa × 30 dm3 × 310 K
300 K × 45 dm3

Note how the units of volume cancel, meaning we can employ any unit of volume provided the units of V1 and V2 are the same.

so p2 = 0.69 × 105 Pa. The answer demonstrates how the pressure drops by about a third on expansion.
SAQ 1.10 The pressure of some oxygen gas is doubled from 1.2 × 105 Pa to 2.4 × 105 Pa. At the same time, the volume of the gas is decreased from 34 dm3 to 29 dm3 . What is the new temperature T2 if the initial temperature T1 was 298 K?

Justification Box 1.1
We start with n moles of gas at a temperature T1 , housed in a volume V1 at a pressure of p1 . Without changing the amount of material, we change the volume and temperature to V2 and T2 respectively, therefore causing the pressure to change to p2 .
The number of moles remains unaltered, so we rearrange Equation (1.13) to make n the subject: n = p1 V1 ÷ RT1
Similarly, the same number of moles n under the second set of conditions is n = p2 V2 ÷ RT2
Again, although we changed the physical conditions, the number of moles n remains constant, so these two equations must be the same. We say p 2 V2 p 1 V1
=n=
RT1
RT2

30

INTRODUCTION TO PHYSICAL CHEMISTRY

As the value of R (the gas constant) does not vary, we can simplify the equation by multiplying both sides by R, to obtain p 2 V2 p 1 V1
=
T1
T2
which is Equation (1.14).

What causes pressure?
Motion of particles in the gas phase
The question, ‘What is pressure?’ is another odd question, but is not too difficult to answer.
The constituent particles of a substance each have energy. In practice, the energy is manifested as kinetic energy – the energy of movement – and explains why all molecules and atoms move continually as an expression of that kinetic energy. This energy decreases as the temperature decreases. The particles only stop moving when cooled to a temperature of absolute zero: 0 K or −273.15 ◦ C.
The particles are not free to move throughout a solid substance, but can vibrate about their mean position. The frequency and amplitude of such vibration increases as the temperature rises. In a liquid, lateral motion of the particles is possible, with the motion becoming faster as the temperature increases. We call this energy translational energy. Furthermore, as the particles acquire energy with increased temperature, so the interactions (see Chapter 2) between the particles become comparatively smaller, thereby decreasing the viscosity of the liquid and further facilitating rapid motion of the particles. When the interactions become negligible (comparatively), the particles can break free and become gaseous.
And each particle in the gaseous state can move at amazingly high speeds; indeed, they are often supersonic. For example, an average atom of helium travels at a mean speed of 1204 m s−1 at 273.15 K. Table 1.4 lists the mean speeds of a few other gas molecules at 273.15 K. Notice how heavier molecules travel more slowly, so carbon dioxide has a mean speed of 363 m s−1 at the same temperature. This high speed of atomic and molecular gases as they move is a manifestation of
The gas particles are their enormous kinetic energy. It would not be possible to travel widely separated. so fast in a liquid or solid because they are so much denser – we call them condensed phases.
The separation between each particle in gas is immense, and
Particles of gas travel usually thousands of times greater than the diameter of a single fast and in straight gas particle. In fact, more than 99 per cent of a gas’s volume lines, unless they colis empty space. The simple calculation in Worked Example 1.5 lide. demonstrates this truth.

PROPERTIES OF GASES AND THE GAS LAWS

31

Table 1.4 The average speeds of gas molecules at 273.15 K, given in order of increasing molecular mass. The speeds c are in fact root-mean-square speeds, obtained by squaring each velocity, taking their mean and then taking the square root of the sum
Gas
Monatomic gases
Helium
Argon
Mercury
Diatomic gases
Hydrogen
Deuterium
Nitrogen
Oxygen
Carbon monoxide
Chlorine
Polyatomic gases
Methane
Ammonia
Water
Carbon dioxide
Benzene

Speed c/m s−1

1204.0
380.8
170.0
1692.0
1196.0
454.2
425.1
454.5
285.6
600.6
582.7
566.5
362.5
272.8

Worked Example 1.5 What is the molar volume of neon, assuming it to be a straightforward solid?
We must first note how the neon must be extremely cold if it is to be
The ‘molar volume’ is a solid – probably no colder than about 20 K. the name we give to
We know that the radius of a neon atom from tables of X-ray crystalthe volume ‘per mole’.
−10
lographic data is about 10 m, so the volume of one atom (from the
4
3
−30
3 m . If we assume the equation of a sphere, V = 3 πr ) is 4.2 × 10 neon to be a simple solid, then 1 mol of neon would occupy a volume of 4.2 × 10−30 m3 per atom × 6.022 × 1023 atoms per mole = 2.5 × 10−6 m3 mol−1 . This volume represents
2.5 cm3 mol−1 .

A volume of 2.5 cm3 mol−1 is clearly much smaller than the value we calculated earlier in Worked Example 1.3 with the ideal-gas equation, Equation (1.13). It is also smaller than the volume of solid neon made in a cryostat, suggesting the atoms in a solid are also separated by much empty By corollary, if the gas particles move fast and space, albeit not so widely separated as in a gas.
In summary, we realize how each particle of gas has enor- the gas is ideal, the gas mous kinetic energy and are separated widely. Yet, like popcorn particles must travel in straight lines between in a popcorn maker, these particles cannot be classed as wholly collisions. independent, one from another, because they collide. They collide

32

INTRODUCTION TO PHYSICAL CHEMISTRY

Newton’s first law states that every action has an equal but opposite reaction. His second law relates the force acting on an object to the product of its mass multiplied by its acceleration.

The pressure of a gas is a ‘macroscopic’ manifestation of the
‘microscopic’ gas particles colliding with the internal walls of the container. The surface area inside a cylinder of radius r and height h is 2π rh.
Don’t forget to include the areas of the two ends, each of which is π r2.

firstly with each other, and secondly with the internal walls of the container they occupy.
Just like the walls in a squash court, against which squash balls continually bounce, the walls of the gas container experience a force each time a gas particle collides with them. From Newton’s laws of motion, the force acting on the wall due to this incessant collision of gas particles is equal and opposite to the force applied to it. If it were not so, then the gas particles would not bounce following a collision, but instead would go through the wall.
We see how each collision between a gas particle and the internal walls of the container causes the same result as if we had applied a force to it. If we call the area of the container wall A and give the symbol F to the sum of the forces of all the particles in the gas, then the pressure p exerted by the gas-particle collisions is given by pressure, p =

force, F area, A

(1.15)

In summary, the pressure caused by a container housing a gas is simply a manifestation of the particles moving fast and colliding with the container walls.
SAQ 1.11 A cylindrical can contains gas. Its height is
30 cm and its internal diameter is 3 cm. It contains gas at a pressure of 5 × 105 Pa. First calculate the area of the cylinder walls (you will need to know that 1 m = 100 cm, so 1 m2 = 104 cm2 ), and then calculate the force necessary to generate this pressure.

Aside
A popular misconception says a molecule in the gas phase travels faster than when in a liquid. In fact, the molecular velocities will be the same in the gas and liquid phases if the temperatures are the same. Molecules only appear to travel slower in a liquid because of the large number of collisions between its particles, causing the overall distance travelled per unit time to be quite short.

Why is it unwise to incinerate an empty can of air freshener? The molecular basis of the gas laws
The writing printed on the side of a can of air freshener contains much information.
Firstly, it cites the usual sort of advertising prose, probably saying it’s a better product

FURTHER THOUGHTS ON ENERGY

33

than anyone else’s, and smells nicer. Few people seem to bother reading these bits.
But in most countries, the law says the label on the can should also gives details of the can’s contents, both in terms of the net mass of air freshener it contains and also perhaps a few details concerning its chemical composition. Finally, a few words of instruction say how to dispose safely of the can. In this context, the usual phrase printed on the can is, ‘Do not incinerate, even when CFC stands for chloempty’. But why? rofluorocarbon. Most
It is common for the can to contain a propellant in addition to the CFCs have now been actual components of the air freshener mixture. Commonly, butane banned because of or propane are chosen for this purpose, although CFCs were the their ability to damage the ozone layer in favoured choice in the recent past.
Such a can is thrown away when it contains no more air fresh- the upper atmosphere. ener, although it certainly still contains much propellant. Incineration of the can leads to an increase in the kinetic energy of the remaining propellant molecules, causing them to move faster and Pressure increases with faster. And as their kinetic energy increases, so the frequency with increasing temperature which they strike the internal walls of the can increases. The force because the collisions of each collision also increases. In fact, we rediscover the ideal gas between the gas partiequation, Equation (1.13), and say that the pressure of the gas (in cles and the container a constant-volume system) increases in proportion to any increase wall are more enerin its temperature. In consequence, we should not incinerate an old getic and occur more can of air freshener because the internal pressure of any residual frequently. propellant increases hugely and the can explodes. Also note the additional scope for injury afforded by propane’s flammability.

1.4

Further thoughts on energy
Why is the room warm?
The energy of room temperature
Imagine coming into a nice, warm room after walking outside in the snow. We instantly feel warmer, because the room is warmer. But what exactly is the energy content of the room? Stated another way, how much energy do we get from the air in the room by virtue of it being at its own particular temperature?
For simplicity, we will consider only the molecules of gas. Each molecule of gas will have kinetic energy (the energy of movement) unless the temperature is absolute zero. This energy may be transferred through inelastic molecules collisions. But how much kinetic energy does the gas have?
At a temperature T , 1 mol of gas has a kinetic energy of 3 RT , where T is the ther2 modynamic temperature and R is the gas constant. This energy is directly proportional to the thermodynamic temperature, explaining why we occasionally call the kinetic energy ‘thermal motion energy’. This simple relationship says that temperature is merely a measure of the average kinetic energy of gas molecules moving chaotically.

34

INTRODUCTION TO PHYSICAL CHEMISTRY

It is important to appreciate that this energy relates to the average energy of 1 mol of gas molecules. The concept of temperature has no meaning when considering a single molecule or atom. For example, the velocity (and hence the kinetic energy) of a single particle changes with time, so in principle its temperature also changes.
Temperature only acquires any thermodynamic meaning when we consider average velocities for a large number of particles.
Provided we know the temperature of the gas, we know its energy – the energy it has simply by existing at the temperature T .
Worked Example 1.6 What is the energy of 1 mol of gas in a warm room at 310 K?

The ‘room energy’ 3 RT
2
derives from the kinetic
(movement) energy of a gas or material.

The energy per mole is 3 × R × T ; so, inserting values, energy =
2
3
× 8.314 J K−1 mol−1 × 310 K.
2
Energy = 3866 J mol−1 ≈ 3.9 kJ mol−1

The molar energy of these molecules is about 4 kJ mol−1 , which is extremely slight compared with the energy of the bonds connecting the respective atoms within a molecule (see Chapters 2 and 3). There is little chance of this room energy causing bonds to break or form.

SAQ 1.12 What is the room energy per mole on a cold winter’s day, at
−8 ◦ C (265 K)?

What do we mean by ‘room temperature’?
Standard temperature and pressure
Suppose two scientists work on the same research project, but one resides in the far north of the Arctic Circle and the other lives near the equator. Even if everything else is the same – such as the air pressure, the source of the chemicals and the manufacturers of the equipment – the difference between the temperatures in the two laboratories will cause their results to differ widely. For example, the ‘room energy’
RT will differ. One scientist will not be able to repeat the experiments of the other, which is always bad science.
An experiment should always be performed at known temperature. Furthermore, the temperature should be constant throughout
An experiment should the course of the experiment, and should be noted in the laboraalways be performed at a known, fixed tempertory notebook. ature. But to enable complete consistency, we devise what is called a set of standard conditions. ‘Standard pressure’ is given the symbol p O , and has a value of 105 Pa. We sometimes call it ‘1 bar’. Atmospheric pressure has a value of 101 325 Pa, so it is larger than p O . We often give atmospheric pressure the symbol ‘atm’.

FURTHER THOUGHTS ON ENERGY

‘Standard temperature’ has the value of 298 K exactly, which equates to just below 25 ◦ C. If both the pressure and the temperature are maintained at these standard conditions, then we say the measurement was performed at ‘standard temperature and pressure’, which is universally abbreviated to ‘s.t.p.’ If the scientists at the equator and the Arctic Circle perform their work in thermostatically controlled rooms, both at s.t.p., then the results of their experiments will be identical.

35

A ‘thermostat’ is a device for maintaining a temperature.
Thermo is Greek for
‘energy’ or ‘temperature’, and ‘stat’ derives from the Greek root statikos, meaning ‘to stand’, i.e. not move or alter. Why do we get warmed-through in front of a fire, rather than just our skins?
The Maxwell–Boltzmann distribution of energies
If no heat was distributed, then our faces and those parts closest to the fire would quickly become unbearably hot, while the remainder of our flesh would continue to feel cold. Heat conducts through the body principally by the fire warming the blood on the surface of the skin, which is then pumped to other parts of the body through the circulatory system. The energy in the warmed blood is distributed within We often see this relationship called merely cooler, internal tissues. the ‘Boltzmann disIt is important to note how the heat energy is distributed around tribution’, after the the body, i.e. shared and equalized. Nature does not like diversity Austrian Physicist Ludin terms of energetic content, and provides many mechanisms by wig Boltzmann (1844– which the energy can be ‘shared’. We shall discuss this aspect of 1906), who played a thermochemistry in depth within Chapter 4. pivotal role in marrying
We can be certain that molecules do not each have the same thermodynamics with energy, but a distribution of energies. The graph in Figure 1.9 statistical and molecuconcerns the energies in a body. The x-axis gives the range of lar physics. energies possible, and the y-axis represents the number of particles in the body (molecules, atoms, etc.) having that energy. The graph clearly shows how few particles possess a large energy and how a few particles have a tiny energy, but the majority have lesser The thermodynamic energies. We call this spread of energies the ‘Maxwell–Boltzmann temperature is the sole variable required to distribution’. All speeds are found at all temperatures, but more molecules define the Maxwell–
Boltzmann distribution: travel at faster speeds at the higher temperatures. raising the temperature
The distribution law depicted in Figure 1.9 may be modelled increases the spread of mathematically, to describe the proportions of molecules of molar energies. mass M with energies E in the range E to E + dE that exist in

36

INTRODUCTION TO PHYSICAL CHEMISTRY

Energy distribution

100 K

200 K
300 K
500 K
700 K
1000 K

500

1000

1500

Velocity v /m s−1

Figure 1.9 Molecular energies follow the Maxwell–Boltzmann distribution: energy distribution of nitrogen molecules (as y) as a function of the kinetic energy, expressed as a molecular velocity
(as x). Note the effect of raising the temperature, with the curve becoming flatter and the maximum shifting to a higher energy

thermal equilibrium with each other at a temperature T : f (E) = 4π

M
2π RT

3/2

E 2 exp −

Ms 2
2RT

(1.16)

where f on the far left indicates the ‘function’ that is to be applied to the variable E: the mathematical nature of this function is given by the right-hand side of the equation.
So, in summary, we feel warmer in front of a fire because energy is distributed between those parts facing the flames and the more hidden tissues within.

2

Introducing interactions and bonds
Introduction
We look first at deviations from the ideal-gas equation, caused by inter-particle interactions. Having described induced dipoles (and hydrogen bonds) the interaction strengths are quantified in terms of the van der Waals and virial equations of state.
Next, formal bonds are described, both covalent (with electrons shared between participating atoms) and ionic (in which electrons are swapped to form charged ions; these ions subsequently associate in response to electrostatic forces). Several underlying factors are expounded, such as ionization energy I and electron affinity E(ea) .
The energy changes occurring while forming these interactions are alluded to, but are treated properly in Chapter 3.

2.1

Physical and molecular interactions
What is ‘dry ice’?
Deviations from the ideal-gas equation
We call solid carbon dioxide (CO2 ) ‘dry ice’. To the eye, it looks just like normal ice, although it sometimes appears to ‘smoke’; Substances sublime if they pass directly from see below. Carbon dioxide is a gas at room temperature and only a solid to form a gas solidifies (at atmospheric pressure) if the temperature drops to about without being a liquid

−78 C or less, so we make dry ice by cooling gaseous CO2 below as an intermediate its freezing temperature. We call it dry ice because, unlike normal phase; see Chapter 5. ice made with water, warming it above its melting temperature leaves no puddle of liquid, because the CO2 converts directly to a gas. We say it sublimes.
Gases become denser as we lower their temperature. If CO2 was still a gas at
−90 ◦ C, then its molar volume would be 15 200 cm3 . In fact, the molar volume of

38

INTRODUCING INTERACTIONS AND BONDS

solid CO2 at this temperature is about 30 cm3 . We deduce that CO2 does not obey the ideal-gas equation (Equation (1.13)) below its freezing temperature, for the very obvious reason that it is no longer a gas.
SAQ 2.1 Show that the volume of 1 mol of CO2 would be 15 200 cm3 at p O and −90 ◦ C (183 K). [Hint: use the ideal-gas equation. To express this answer in cubic metres, you will need to remember that 1 m3 = 103 dm3 and 106 cm3 .]

Although solidifying CO2 is an extreme example, it does show how deviations from the ideal-gas equation occur.

How is ammonia liquefied?
Intermolecular forces
Compressing ammonia gas under high pressure forces the molecules into close proximity. In a normal gas, the separation between each molecule is generally large – approximately 1000 molecular diameters is a good generalization. By contrast, the separation between the molecules in
We sometimes call a solid or a liquid a a condensed phase (solid or liquid) is more likely to be one to two
‘condensed phase’. molecular diameters, thereby explaining why the molar volume of a solid or liquid is so much smaller than the molar volume of a gas.
As a direct consequence of the large intermolecular separations, we can safely say no interactions form between the molecules in
‘Intermolecular’ means ammonia gas. The molecules are simply too far apart. We saw
‘between molecules’. in the previous chapter how the property known as pressure is a macroscopic manifestation of the microscopic collisions occurring between gas particles and, say, a solid object such as a container’s walls. But the gas particles can also strike each other on the same microscopic scale: we say the resultant interactions between molecules are intermolecular.
Intermolecular interactions only operate over relatively short distances, so we assume that, under normal conditions, each molecule
A ‘formal bond’ involves in a gas is wholly unaffected by all the others. By contrast, when the permanent involvement of electrons in the gas is compressed and the particles come to within two or three covalent or ionic bonds; molecular diameters of each other, they start to ‘notice’ each other. see p. 64. Interactions
We say the outer-shell electrons on an atom are perturbed by the between molecules charges of the electrons on adjacent atoms, causing an interaction. in a compressed gas
We call these interactions bonds, even though they may be too are temporary. weak to be formal bonds such as those permanently connecting the atoms or ions in a molecule.
The intermolecular interactions between molecules of gas are generally attractive; so, by way of response, we find that, once atoms are close enough to interact, they prefer to remain close – indeed, once a tentative interaction forms, the atoms or

PHYSICAL AND MOLECULAR INTERACTIONS

39

molecules generally draw themselves closer, which itself makes the interaction stronger. We see a simple analogy with everyday Translational motion magnets: once two magnets are brought close enough to induce an is movement through space, rather than a interaction, we feel the attractive force dragging them closer still. vibration about a mean
As soon as the particles of a gas attract, the inertia of the point or a rotation aggregate species increases, thereby slowing down all translational about an axis. motion. And slower particles, such as these aggregates, are an easier target for further collisions than fast-moving gas atoms and
Formation of an agmolecules. The same principle explains why it is impossible to gregate facilitates furcatch someone who is running very fast during a playground game ther coalescence (evenof ‘tig’. Only when the runners tire and slow down can they tually forming a conbe caught. In practice, as soon as an aggregate forms, we find densed phase). We say that other gas particles soon adhere to it, causing eventual coa- ‘nucleation’ occurs. lescence and the formation of a droplet of liquid. We say that nucleation occurs.
With the same reasoning as that above, we can force the molecules of ammonia still closer together by applying a yet larger pressure, to form a denser state such as a solid.

Why does steam condense in a cold bathroom?
Elastic and inelastic collisions
In the previous example, we looked at the interactions induced when changing the external pressure, forcing the molecules into close proximity. We look here at the effects of changing the temperature.
A bathroom mirror is usually colder than the temperature of the steam rising from a hot bath. Each molecule of steam (gaseous water) has an enormous energy, which comes ultimately from the boiler that heats the water. The particles of steam would remain as liquid if they had less energy. In practice, particles evaporate from the bath to form energetic molecules of steam. We see this energy as kinetic energy, so the particles move fast (see p. 30). The typical speeds at which gas particles move make it inevitable that steam molecules will collide with the mirror.
We say such a collision is elastic if no energy transfers during the No energy is exchangcollision between the gas particle and the mirror; but if energy does ed during an ‘elastic’ collision, but energy is transfer – and it usually does – we say the collision is inelastic. exchanged during an
The energy transferred during an inelastic collision passes from
‘inelastic’ collision. the hot molecule of steam to the cooler mirror. This energy flows in this direction because the steam initially possessed more energy per molecule than the mirror as a consequence of its higher temperature. It is merely a manifestation of the minus-oneth law of thermodynamics, as discussed in Chapter 1.
But there are consequences to the collisions being inelastic: the molecules of steam have less energy following the collision because some of their energy has transferred.

40

INTRODUCING INTERACTIONS AND BONDS

Energy is never lost or gained, only transferred or converted; see Chapter 3.

We generally assume that all particles in an ideal gas do not interact, meaning that the gas obeys the ideal-gas equation. This assumption is sometimes poor.

We perceive this lower energy as a cooler temperature, meaning that the water vapour in a steam-filled bathroom will cool down; conversely, the mirror (and walls) become warmer as they receive the energy that was previously possessed by the steam. These changes in the temperatures of gas and mirror occur in a complementary sense, so no energy is gained or lost.
These changes in temperature represent a macroscopic proof that microscopic processes do occur. Indeed, it is difficult to envisage a transfer of energy between the gas particles with the cold mirror without these microscopic interactions.
We spent quite a lot of time looking at the concept of an ideal gas in Chapter 1. The simplest definition of an ideal gas is that it obeys the ideal-gas equation (Equation (1.13)). Most gases can be considered as ideal most of the time. The most common cause of a gas disobeying the ideal-gas equation is the formation of interactions, and the results of intermolecular collisions.

How does a liquid-crystal display work?
Electronegativity and electropositivity
Liquid crystals are organic compounds that exhibit properties somewhere between those of a solid crystal and a liquid. Compounds I and II in Figure 2.1 both form liquid crystals at room temperature.
We observe that liquid crystals can flow like any other viscous liquid, but they also possess some of the properties of crystalline solids, such as physical order, rather than random chaos. Unlike most other liquids, liquid crystals have some properties

C4H9

N
O
CH3
(I)

O
O

N

C10H21
O

(II)

Figure 2.1

Compounds that form room-temperature liquid crystals

PHYSICAL AND MOLECULAR INTERACTIONS

41

Crossed light polarizers

Light transmitted

Lower polarizer blocks the transmission of light

No voltage applied

Voltage applied

Figure 2.2 The transparent electrodes in an LCD are coated with crossed polarizers. The liquid crystals (depicted as slender lozenges) form helices, thereby ‘guiding’ polarized light from the upper electrode through the LCD, enabling transmission through to the lower polarizer. This is why the display has no colour. The helical structure is destroyed when a voltage is applied, because the polar liquid crystals align with the electrodes’ field. No light can transmit, so the display looks black

that depend on the direction of measurement, because of the alignment of their long, rod-like structures.
In a liquid-crystal display (LCD) device, the two electrodes are parallel and separated by a thin layer of liquid crystal (see A physicist would say
Figure 2.2). The liquid crystals in this layer naturally adopt a heli- the liquid crystal adopted a twisted cal structure.
Light can be represented as a transverse electromagnetic wave nematic structure. made up of fluctuating electric and magnetic fields, moving in mutually perpendicular directions (see Chapter 9). Ordinary light is made up of waves that fluctuate at all possible angles, which normally cannot be separated. A polarizer is a material that allows only light with a specific angle of vibration to transmit.
We place a light polarizer on one side of either transparent electrode in the LCD, each similar to one lens in a pair of polaroid sunglasses. The helix of the liquid crystal twists the polarized light as it transmits through the LCD, guiding it from the upper polarizer and allowing it unhindered passage through the ‘sandwich’ and lower polarizer. The transmitting state of an LCD (at zero voltage) is thus ‘clear’.
Applying a voltage to a pixel within the cell causes the molecules to move, aligning themselves parallel with the electric field ‘Pixel’ is short for ‘picimparted by the electrodes. This realignment destroys the helical ture element’. An LCD structure, precluding the unhindered transmission of light, and the image comprises many thousands of pixels. display appears black.
Molecules of this type are influenced by an external electric field because they possess a dipole: one end of the molecule is electron withdrawing while the other is electron attracting, with the result that one end possesses a higher electron density than the other. As a result, the molecule behaves much like a miniature bar magnet. Applying a voltage between the two

42

INTRODUCING INTERACTIONS AND BONDS

electrodes of the LCD causes the ‘magnet’ to reorientate in just the same way as a magnet moves when another magnet is brought close to it.
These dipoles form because of the way parts of the molecule attract electrons to differing extents. The power of an element
Atoms or groups are
(when part of a compound) to attract electrons is termed its ‘elec‘electronegative’ if they tend to acquire negtronegativity’ χ . Highly electron-attracting atoms tend to exert ative charge at the control over the outer, valence electrons of adjacent atoms. The expense of juxtaposed most electronegative elements are those placed near the right-hand atoms or groups. Groups side of the periodic table, such as oxygen and sulphur in Group acquiring a positive
VI(b) or the halogens in Group VII(b). charge are ‘electroposThere have been a large number of attempts to quantify elecitive’. tronegativities χ , either theoretically or semi-empirically, but none has been wholly successful. All the better methods rely on bond strengths or the physical dimensions of atoms.
Similar to the concept of electronegativity is the electropositivity of an element, which is the power of its atoms (when part of a compound) to lose an electron. The most electropositive elements are the metals on the far-left of the periodic table, particularly
Groups I(a) and II(a), which prefer to exist as cations. Being the opposite concept to electronegativity, electropositivity is not employed often. Rather, we tend to say that an atom such as sodium has a tiny electronegativity instead of being very electropositive.

Why does dew form on a cool morning?
Van der Waals forces
Many people love cool autumn mornings, with the scent of the cool air and a rich dew underfoot on the grass and paths. The dew forms when molecules of water from the air coalesce, because of the cool temperature, to form minute aggregates that subsequently nucleate to form visible drops of water. These water drops form a stable colloid (see Chapter 10).
Real gases are never wholly ideal: there will always be some extent of non-ideality.
At one extreme are the monatomic rare gases such as argon and neon, which are nonpolar. Hydrocarbons, like propane, are also relatively non-polar, thereby precluding stronger molecular interactions. Water, at the opposite extreme, is very polar because some parts of the molecule are more electron withdrawing than
The symbol δ means ‘a others. The central oxygen is relatively electronegative and the two small amount of . . .’, so hydrogen atoms are electropositive, with the result that the oxygen
‘δ − ’ is a small amount is more negative than either of the hydrogen atoms. We say it has of negative charge. a slight excess charge, which we write as δ − . Similar reasoning shows how the hydrogen atoms are more positive than the oxygen, with excess charges of δ + .
These excess charges form in consequence of the molecule incorporating a variety of atoms. For example, the magnitude of δ − on the chlorine of H–Cl is larger than the excess charges in the F–Cl molecule, because the difference in electronegativity

PHYSICAL AND MOLECULAR INTERACTIONS

Table 2.1

43

Values of electronegativity χ for some main-group elements

H
2.1
Li
1.0

Be
1.5

B
2.0

C
2.5

N
3.0

O
3.5

F
4.0

Na
0.9

Mg
1.2

Al
1.5

Si
1.8

P
2.1

S
2.5

Cl
3.0

K
0.8

Ca
1.0

Ga
1.6

Ge
1.8

As
2.0

Se
2.4

Br
2.8

Rb
0.8

Sr
1.0

I
2.5

χ between H and Cl is greater than the difference between F and Cl. There will be no excess charge in the two molecules H–H or Cl–Cl because the atoms in both are the same – we say they are homonuclear. Table 2.1 contains a few electronegativities.
SAQ 2.2 By looking at the electronegativities in Table 2.1, suggest whether the bonds in the following molecules will be polar or non-polar: (a) hydrogen bromide, HBr; (b) silicon carbide, SiC; (c) sulphur dioxide, O=S=O; and
(d) sodium iodide, NaI.

The actual magnitude of the excess charge is generally unknown, although we do know they are small. Whereas some calculations suggest that δ is perhaps as much as 0.1 of a full, formal charge, others suggest about 0.01 or even less.
While debate persists concerning the magnitudes of each excess charge within a molecule, it is certain that the overall charge on the molecule is zero, meaning that the two positive charges in water cancel out the central negative charge on the oxygen. We reason this by saying that water is a neutral molecule.
Figure 2.3 shows the ‘V’ shape of the water molecule. The top of the molecule (as drawn) has a negative excess charge and the bottom is positive. The δ + and δ − charges are separated spatially, which we call a dipole. Such dipoles are crucial when explaining why water vapour so readily forms a liquid: those parts of the molecule bearing a slight positive charge (δ + ) attract those parts of adjacent molecules that bear a slight negative charge (δ − ).
The interaction is electrostatic, and forms in much a similar manner to the north pole of a magnet attracting the south pole of another magnet.
Electrostatic interactions of this type are called ‘dipole–dipole interactions’, or ‘van der Waals forces’ after the Dutch physicist
Johannes Diderik van der Waals (1837–1923) who first postulated their existence. A van der Waals force operates over a relatively

Water is a neutral molecule, so the central negative charge in the water molecule counteracts the two positive charges.

A ‘dipole’ forms when equal and opposite charges are separated by a short distance. ‘Di’ means two, and ‘pole’ indicates the two ends of a magnet.

‘Van der Waals forces’ are electrostatic interactions between dipoles. (Note how we pronounce ‘Waals’ as
‘vahls’.)

44

INTRODUCING INTERACTIONS AND BONDS

d−

d+

O

d+
H

H

Figure 2.3 The water molecule has a ‘V’ shape. Experiments show
˚
that gaseous water has an O–H length of 0.957 18 A; the H–O–H angle is 104.474◦ . Water is polar because the central oxygen is electronegative and the two hydrogen atoms are electropositive. The vertical arrow indicates the resultant dipole, with its head pointing toward the more negative end of the molecule
H
H

H

O

H

O
H

H

O

O
O

H
H

H

H

H

O

O
H

H
O
H

H

H
O

H

H

Figure 2.4 Water would be a gas rather than a liquid at room temperature if no van der Waals forces were present to ‘glue’ them together, as indicated with dotted lines in this two-dimensional representation. In fact, water coalesces as a direct consequence of this three-dimensional network of dipole–dipole interactions. Note how all the O–H · · · O bonds are linear

short distance because the influence of a dipole is not large. In practice, we find that the oxygen atoms can interact with hydrogen atoms on an adjacent molecule of water, but no further.
The interactions between the two molecules helps to ‘glue’ them together. It is a sobering thought that water would be a gas rather than a liquid if hydrogen bonds
(which are merely a particularly strong form of van der Waals forces) did not promote the coalescence of water. The Earth would be uninhabitable without them. Figure 2.4 shows the way that liquid water possesses a three-dimensional network, held together with van der Waals interactions.
Each H2 O molecule in liquid water undergoes at least one interaction with another molecule of H2 O (sometimes two). Nevertheless, the interactions are not particularly strong – perhaps as much as 20 kJ mol−1 .
Whereas the dipoles themselves are permanent, van der Waals interactions are not. They are sufficiently weak that they continually break and re-form as part of a dynamic process.

How is the three-dimensional structure maintained within the DNA double helix?
Hydrogen bonds
DNA is a natural polymer. It was first isolated in 1869 by Meischer, but its role in determining heredity remained unrecognized until 1944, by which time it was

PHYSICAL AND MOLECULAR INTERACTIONS

appreciated that it is the chromosomes within a cell nucleus that dictate hereditary traits. And such chromosomes consist of DNA and protein. In 1944, the American bacteriologist Oswald Avery showed how it was the DNA that carried genetic information, not the protein.
The next breakthrough came in 1952, when Francis Crick and
Donald Watson applied X-ray diffraction techniques to DNA and elucidated its structure, as shown schematically in Figure 2.5. They showed how its now famous ‘double helix’ is held together via a series of unusually strong dipole–dipole interactions between precisely positioned organic bases situated along the DNA polymer’s backbone.
There are four bases in DNA: guanine, thymine, cytosine and adenine. Each has a ketone C=O group in which the oxygen is quite electronegative and bears an excess negative charge δ − , and an amine in which the electropositive hydrogen atoms bear an excess

45

The word ‘theory’ comes from the Greek theoreo, meaning ‘I look at’. A theory is something we look at, pending acceptance or rejection.

The rules of ‘base pairing’ (or nucleotide pairing) in DNA are: adenine (A) always pairs with thymine (T); cytosine (C) always pairs with guanine (G).

1 nm

3.4 nm

(a)
H
H

O
C

H

N

N

H

C
C

O

H

C

N

C

DNA backbone

N
C

C

C
N

H

H
C
N
DNA backbone

N

N
H

Cytosine

Guanine

H
H3C

O
C

H

DNA backbone

N

N
C

C

C

N
N

H
H

N

C

C
C

O

C
N

H
C
N
DNA backbone

H

Thymine

Adenine
(b)

Figure 2.5 (a) The structure of the ‘double helix’ at the heart of DNA. The slender ‘rods’ represent the hydrogen bonds that form between the organic bases situated on opposing strands of the helix.
(b) Hydrogen bonds (the dotted lines) link adenine with thymine, and guanine with cytosine

46

INTRODUCING INTERACTIONS AND BONDS

Table 2.2

The energies of hydrogen bonds

Atoms in H-bond

Typical energy/kJ mol−1

H and N
H and O
H and F

20
25
40

positive charge δ + . Since the hydrogen atom is so small and so electropositive, its excess charge leads to the formation of an unusually strong dipole, itself leading to a strong van der Waals bond. The bond is usually permanent (unlike a typical dipole–dipole interaction), thereby ‘locking’ the structure of DNA into its pair of parallel helices, much like the interleaving teeth of a zip binding together two pieces of cloth.
We call these extra-strong dipole–dipole bonds ‘hydrogen bonds’, and these are defined by the IUPAC as ‘a form of association between
IUPAC is the Internaan electronegative atom and a hydrogen atom attached to a second, tional Union of Pure relatively electronegative atom’. All hydrogen bonds involve two and Applied Chemistry.
It defines terms, quandipoles: one always comprises a bond ending with hydrogen; the tities and concepts in other terminates with an unusually electronegative atom. It is best chemistry. considered as an electrostatic interaction, heightened by the small size of hydrogen, which permits proximity of the interacting dipoles or charges. Table 2.2 contains typical energies for a few hydrogen bonds. Both electronegative atoms are usually (but not necessarily)
Strictly, the dipole– from the first row of the periodic table, i.e. N, O or F. Hydrogen dipole interactions disbonds may be intermolecular or intramolecular. cussed on p. 42 are
Finally, as a simple illustration of how weak these forces are, also hydrogen bonds, note how the energy required to break the hydrogen bonds in since we discussed the interactions arising liquid hydrogen chloride (i.e. the energy required to vaporize it) between H–O-bonded is 16 kJ mol−1 , yet the energy needed break the chemical bond species. between atoms of hydrogen and chlorine in H–Cl is almost 30 times stronger, at 431 kJ mol−1 .

Aside
The ancient Greeks recognized that organisms often pass on traits to their offspring, but it was the experimental work of the Austrian monk Gregor Mendel (1822–1884) that led to a modern hereditary ‘theory’. He entered the Augustinian monastery at Br¨ nn u (now Brno in the Czech Republic) and taught in its technical school.
He cultivated and tested the plants at the monastery garden for 7 years. Starting in
1856, painstakingly analysing seven pairs of seed and plant characteristics for at least
28 000 pea plants. These tedious experiments resulted in two generalizations, which were later called Mendel’s laws of heredity. Mendel published his work in 1866, but it remained almost unnoticed until 1900 when the Dutch botanist Hugo Marie de Vries referred to it. The full significance of Mendel’s work was only realized in the 1930s.

PHYSICAL AND MOLECULAR INTERACTIONS

47

His observations also led him to coin two terms that still persist in present-day genetics: dominance for a trait that shows up in an offspring, and recessiveness for a trait masked by a dominant gene.

How do we make liquid nitrogen?
London dispersion forces
Liquid nitrogen is widely employed when freezing sperm or eggs when preparing for the in vitro fertilization (IVF). It is also essential A ‘homonuclear’ molefor maintaining the cold temperature of the superconducting magnet cule comprises atoms at the heart of the NMR spectrometers used for structure elucidation from only one element; homo is Greek for or magnetic resonance imaging (MRI).
‘same’. Most molecules

Nitrogen condenses to form a liquid at −196 C (77 K), which are ‘heteronuclear’ and is so much lower than the temperature of 373.15 K at which water comprise atoms from condenses that we suspect a different physicochemical process is in several elements; hetevidence. Below −196 ◦ C, molecules of nitrogen interact, causing ero is Greek for ‘other’ condensation. That there is any interaction at all should surprise us, or ‘different’. because the dipoles above were a feature of heteronuclear bonds, but the di-nitrogen molecule (N≡N) is homonuclear, meaning both atoms are the same.
We need to invoke a new type of interaction. The triple bond Dipoles are usually a between the two nitrogen atoms in the di-nitrogen molecule incor- feature of heteronuporates a huge amount of electron density. These electrons are clear bonds, although a never still, but move continually; so, at any instant in time, one fuller treatment needs end of a molecule might be slightly more negative than the other. to consider the elecA fraction of a second later and the imbalance departs. But a tiny tronic environment dipole forms during the instant while the charges are unbalanced: of atoms and groups beyond the bond of we call this an induced dipole; see Figure 2.6.
The electron density changes continually, so induced dipoles interest. never last more than about 10−11 s. Nevertheless, they last sufficiently long for an interaction to form with the induced dipole of another nitrogen molecule nearby. We call this new interaction the London dispersion force after Fritz
London, who first postulated their existence in 1930.
London dispersion forces form between all molecules, whether polar or non-polar.
In a large atom or molecule, the separation between the nucleus and valence electrons is quite large; conversely, the nucleus–electron separation in a lighter atom or molecule is smaller, implying that the electrons are more tightly held. The tighter binding precludes the ready formation of an induced dipole. For this reason, larger
(and therefore heavier) atoms and molecules generally exhibit stronger dispersion forces than those that are smaller and lighter.

48

INTRODUCING INTERACTIONS AND BONDS

d−

d+

Electrons

d−

d+

Atomic nucleus Figure 2.6 Schematic diagram to show how an induced dipole forms when polarizable electrons move within their orbitals and cause a localized imbalance of charge (an ‘induced dipole’ in which the negative electrons on one atom attract the positive nucleus on another). The dotted line represents the electrostatic dipole interaction

Aside
The existence of an attractive force between non-polar molecules was first recognized by van der Waals, who published his classic work in 1873. The origin of these forces was not understood until 1930 when Fritz London (1900–1954) published his quantummechanical discussion of the interaction between fluctuating dipoles. He showed how these temporary dipoles arose from the motions of the outer electrons on the two molecules. We often use the term ‘dispersion force’ to describe these attractions. Some texts prefer the term ‘London–van der Waals’ forces.

Polarizability
The ease with which the electron distribution around an atom or molecule can be distorted is called its
‘polarizability’.

The weakest of all the intermolecular forces in nature are always London dispersion forces.

The electrons in a molecule’s outer orbitals are relatively free to move. If we could compare ‘snapshots’ of the molecule at two different instants in time then we would see slight differences in the charge distributions, reflecting the changing positions of the electrons in their orbitals. The ease with which the electrons can move with time depends on the molecule’s polarizability, which itself measures how easily the electrons can move within their orbitals. In general, polarizability increases as the orbital increases in size: negative electrons orbit the positive nucleus at a greater distance in such atoms, and consequently experience a weaker electrostatic interaction. For this reason, London dispersion forces tend to be stronger between molecules that are easily polarized, and weaker between molecules that are not easily polarized.

PHYSICAL AND MOLECULAR INTERACTIONS

49

Why is petrol a liquid at room temperature but butane is a gas?
The magnitude of London dispersion forces
The major component of the petrol that fuels a car is octane, present as a mixture of various isomers. It is liquid at room temperature because its boiling point temperature
T(boil) is about 125.7 ◦ C. The methane gas powering an oven has a T(boil) = −162.5 ◦ C and the butane propellant in a can of air freshener has T(boil) = −0.5 ◦ C. Octadecane is a gel and paraffin wax is a solid. Figure 2.7 shows the trend in T(boil) for a series of straight-chain alkanes Each hydrocarbon experiences exactly the same intermolecular forces, so what causes the difference in T(boil) ?
Interactions always form between molecules because London forces cannot be eradicated. Bigger molecules experience greater intermolecular forces. These dispersion forces are weak, having a magnitude of between 0.001 and 0.1 per cent of the strength of a typical covalent bond binding the two atoms in diatomic molecule, H2 . These forces, therefore, are so small that they may be ignored within molecules held together by stronger forces, such as network covalent bonds or large permanent dipoles or ions.
The strength of the London dispersion forces becomes stronger with increased polarizability, so larger molecule (or atoms) form stronger bonds. This observation helps explain the trends in physical state of the Group VII(b) halogens: I2 is a solid,
Br2 is a liquid, and Cl2 and F2 are gases.
But the overall dispersion force strength also depends on the total number of electrons in the atom or molecule. It is a cumulative effect. Butane contains 14 atoms and
58 electrons, whereas octane has 26 atoms and 114 electrons. The greater number of electrons increases the total number of interactions possible and, since both melting
150
Boiling point of n-hydrocarbons/°C

C8H18
100

C7H16
C6H14

50

C5H12

0
0

20

60 C4H10 80
C3H8

40

−50

−200

120

140

C2H6

−100
−150

100

CH4
Molecular mass / g mol−1

Figure 2.7 The boiling temperature of simple linear hydrocarbons increases as a function of molecular mass, as a consequence of a greater number of induced dipoles

50

INTRODUCING INTERACTIONS AND BONDS

and boiling points depend directly on the strength of intermolecular bonds, the overall strength of the London forces varies as the molecule becomes larger.

Aside
When is a dispersion force sufficiently strong that we can safely call it a hydrogen bond?
Hydrogen bonds are much stronger than London dispersion forces for two principal reasons:
(1)

The induced dipole is permanent, so the bond is permanent.

(2)

The molecule incorporates a formal H–X covalent bond in which X is a relatively electronegative element (see p. 42).

We call an interaction ‘a hydrogen bond’ when it fulfils both criteria.

2.2

Quantifying the interactions and their influence
How does mist form?
Condensation and the critical state

Why is it that no dew forms if the air pressure is low, however cool the air temperature?
To understand this question, we must first appreciate how molecules come closer together when applying a pressure. The Irish physical chemist Thomas Andrews
(1813–1885) was one of the first to study the behaviour of gases as they liquefy: most of his data refer to CO2 . In his most famous experiments, he observed liquid CO2 at constant pressure, while gradually raising its temperature. He readily discerned a clear meniscus between condensed and gaseous phases in his tube at low temperatures, but the boundary between the phases vanished at temperatures of about 31 ◦ C. Above this temperature, no amount of pressure could bring about liquefaction of the gas.
Andrews suggested that each gas has a certain ‘critical’ temperature, above which condensation is impossible, implying that
The ‘critical temperno liquid will form by changes in pressure alone. He called this ature’ T(critical) is that temperature above temperature the ‘critical temperature’ T(critical) . which it is impossible
Figure 2.8 shows a Boyle’s-law plot of pressure p (as y) against to liquefy a gas. volume V (as x) for carbon dioxide. The figure is drawn as a function of temperature. Each line on the graph represents data obtained at a single, constant temperature, and helps explain why we call each line an isotherm. The uppermost isotherm represents data collected at 31.5 ◦ C. Its shape is essentially straightforward, although it clearly shows distortion. The middle trace (at

QUANTIFYING THE INTERACTIONS AND THEIR INFLUENCE

51

75

Applied pressure /pO

74

31.523°
73

31.013°

72

30.409°

71

29.929°

32

36

40

44

48

52

56

60

64

Molar volume/cm3 mol−1

Figure 2.8 Isotherms of carbon dioxide near the critical point of 31.013 ◦ C. The shaded parabolic region indicates those pressures and volumes at which it is possible to condense carbon dioxide

the cooler temperature of 30.4 ◦ C) is more distorted – it even shows a small region where the pressure experienced is independent of temperature, which explains why the graph is linear and horizontal. And in the bottom trace the isotherm at 29.9 ◦ C is even more distorted still.
The area enclosed by the parabola at the bottom of Figure 2.8 represents those values of pressure and temperature at which CO2 will condense. If the temperature is higher than that at the apex of the parabola (i.e. warmer than
31.013 ◦ C) then, whatever the pressure, CO2 will not liquefy. For Critical fluids are discussed in Chapter 5, this reason, we call 31.013 ◦ C the critical temperature T(critical) of where values of T
(critical)
CO2 . Similarly, the gas will not liquefy unless the pressure is above are listed. a minimum that we call the ‘critical pressure’ p(critical) .

52

INTRODUCING INTERACTIONS AND BONDS

As the temperature rises above the critical temperature and the pressure drops below the critical pressure, so the gas approximates increasingly to an ideal gas,
i.e. one in which there are no interactions and which obeys the ideal-gas equation
(Equation (1.13)).

How do we liquefy petroleum gas?
Quantifying the non-ideality
Liquefied petroleum gas (LPG) is increasingly employed as a fuel. We produce it by applying a huge pressure (10–20 × p O ) to the petroleum gas obtained from oil fields.
Above a certain, critical, pressure the hydrocarbon gas condenses; we say it reaches the dew point, when droplets of liquid first form. The proportion of the gas liquefying increases with increased pressure until, eventually, all of it has liquefied.
Increasing the pressure forces the molecules closer together, and the intermolecular interactions become more pronounced. Such interactions are not particularly strong because petroleum gas is a non-polar hydrocarbon, explaining why it is a gas at room temperature and pressure. We discuss other ramifications later.
The particles of an ideal gas (whether atoms or molecules) do
We often call a gas that not interact, so the gas obeys the ideal-gas equation all the time. is non-ideal, a real gas.
As soon as interactions form, the gas is said to be non-ideal, with the result that we lose ideality, and the ideal-gas equation
(Equation (1.13)) breaks down. We find that pV = nRT .
When steam (gaseous water) is cooled below a certain temThe physical chemperature, the molecules have insufficient energy to maintain their istry underlying the high-speed motion and they slow down. At these slower speeds, liquefaction of a gas they attract one another, thereby decreasing the molar volume. is surprisingly compliFigure 2.9 shows a graph of the quotient pV ÷ nRT (as y) cated, so we shall not against pressure p (as x). We sometime call such a graph an return to the question
Andrews plot. It is clear from the ideal-gas equation (Equa until Chapter 5. tion (1.13)) that if pV = nRT then pV ÷ nRT should always equal to one: the horizontal line drawn through y = 1, therefore,
We form a ‘quotient’ indicates the behaviour of an ideal gas. when dividing one thing
But the plots in Figure 2.9 are not horizontal. The deviation of a by another. We meet trace from the y = 1 line quantifies the extent to which a gas devithe word frequently ates from the ideal-gas equation; the magnitude of the deviation when discussing a depends on pressure. The deviations for ammonia and ethene are person’s IQ, their ‘intelclearly greater than for nitrogen or methane: we say that ammoligence quotient’, which nia deviates from ideality more than does nitrogen. Notice how we define as: (a perthe deviations are worse at high pressure, leading to the empirison’s score in an intelcal observation that a real gas behaves more like an ideal gas at ligence test ÷ the averlower pressures. age score) × 100.
Figure 2.10 shows a similar graph, and displays Andrews plots for methane as a function of temperature. The graph clearly

QUANTIFYING THE INTERACTIONS AND THEIR INFLUENCE

53

2.0

Compressibility factor Z

H2
1.5

1.0

N2
CH4

NH3

C2H4

0.5

0

0

200

400

600

800

Applied pressure / p

1000

1200

O

Figure 2.9 An Andrews plot of P V ÷ nRT (as y) against pressure p (as x) for a series of real gases, showing ideal behaviour only at low pressures. The function on the y-axis is sometimes called the compressibility Z

Compressibility factor Z

2.5

200 K

2.0

300 K
400 K
600 K

1.5
1.0
0.5
0.0

0

200

400

600

800

1000

Applied pressure /p O

Figure 2.10 An Andrews plot of P V ÷ nRT (as y) against pressure p (as x) for methane gas as a function of temperature. Methane behaves more like an ideal gas at elevated temperatures

demonstrates how deviations from ideality become less severe with increasing temperature. In fact, we should expect the deviations to decrease as the temperature increases, because a higher temperature tells us how the particles have more energy, decreasing the likelihood of interparticle interactions being permanent.
Drawing graphs such as Figure 2.10 for other gases suggests a
Gases behave more like second empirical law, that gases behave more like ideal gases as ideal gases at higher the temperature rises. The ideal-gas equation (Equation (1.13)) is temperatures. so useful that we do not want to lose it. Accordingly, we adapt it

54

INTRODUCING INTERACTIONS AND BONDS

somewhat, writing it as pV = Z × (nRT )
We sometimes call the function ‘pV ÷ nRT ’ the ‘compressibility’ or ‘compressibility factor’ Z.

(2.1)

where Z is the compressibility or compressibility factor. The value of Z will always be one for an ideal gas, but Z rarely has a value of one for a real gas, except at very low pressures. As soon as p increases, the gas molecules approach close enough to interact and pV = nRT . The value of Z tells us a lot about the interactions between gas particles.

Aside
The gas constant R is generally given the value 8.314 J K−1 mol−1 , but in fact this numerical value only holds if each unit is the SI standard, i.e. pressure expressed in pascals, temperature in kelvin and volume in cubic metres.
The value of R changes if we express the ideal-gas equation (Equation (1.13)) with different units. Table 2.3 gives values of R in various other units. We must note an important philosophical truth here: the value of the gas constant is truly constant, but the actual numerical value we cite will depend on the units with which we express it.
We met a similar argument before on p. 19, when we saw how a standard prefix (such as deca, milli or mega) will change the appearance of a number, so V = 1 dm3 = 103 cm3 .
In reality, the number remains unaltered.
We extend this concept here by showing how the units themselves alter the numerical value of a constant.
Table 2.3 Values of the gas constant R expressed with various unitsa
8.3145 J K−1 mol−1
2 cal K−1 mol−1
0.083 145 dm3 bar mol−1 K−1
83.145 cm3 bar mol−1 K−1
0.082 058 dm3 atm mol−1 K−1
82.058 cm3 atm mol−1 K−1
1 bar = p O = 105 Pa. 1 atm = 1.013 25 × 105
Pa. The ‘calorie’ is a wholly non-SI unit of energy;
1 cal = 4.157 J. a Why is the molar volume of a gas not zero at 0 K?
The van der Waals equation
In Chapter 1 we recalled how Lord Kelvin devised his temperature scale after cooling gases and observing their volumes. If the simplistic graph in Figure 1.5 was obeyed,

QUANTIFYING THE INTERACTIONS AND THEIR INFLUENCE

55

then a gas would have a zero volume at −273.15 ◦ C. In fact, the molar volume of a gas is always significant, even at temperatures close to absolute zero. Why the deviation from Kelvin’s concept?
Every gas consists of particles, whether as atoms (such as neon) or as molecules
(such as methane). To a relatively good first approximation, any atom can be regarded as a small, incompressible sphere. The reason why we can compress a gas relates to the large separation between the gas particles. The first effect of compressing a gas is to decrease these interparticle distances.
Particles attract whenever they approach to within a minimum distance. Whatever the magnitude of the interparticle attraction, energetic molecules will separate and continue moving after their encounter; but, conversely, molecules of lower energy do not separate after the collision because the attraction force is enough to overwhelm the momentum that would cause the particles to bounce apart. The process of coalescence has begun.
Compressing a gas brings the particles into close proximity, thereby increasing the probability of interparticle collisions, and magnifying the number of interactions. At this point, we need to consider two physicochemical effects that operate in opposing directions. Firstly, interparticle interactions are usually attractive, encouraging the particles to get closer, with the result that the gas has a smaller molar volume than expected. Secondly, since the particles have their own intrinsic volume, the molar volume of a gas is described not only by the separations between particles but also by the particles themselves. We need to account for these two factors when we describe the physical properties of a real gas.
The Dutch scientist van der Waals was well aware that the idealgas equation was simplistic, and suggested an adaptation, which The a term reflects the strength of the we now call the van der Waals equation of state: p+ n2 a
(V − nb) = nRT
V2

(2.2)

interaction between gas particles, and the b term reflects the particle’s size.

where the constants a and b are called the ‘van der Waals constants’, the values of which depend on the gas and which are best obtained experimentally. Table 2.4 contains a few sample values. The constant a reflects the strength of the interaction between gas molecules; so, a value of 18.9 for benzene suggests a strong interaction whereas 0.03 for helium represents a negligible interaction. Incidentally, this latter value reinforces the idea that inert gases are truly inert. The magnitude of the constant b reflects the physical size of the gas particles, and are again seen to follow a predictable trend. The magnitudes of a and b dictate
Equation (2.2) simplithe extent to which the gases deviate from ideality. fies to become the
Note how Equation (2.2) simplifies to become the ideal-gas equa- ideal-gas equation tion (Equation (1.13)) if the volume V is large. We expect this (Equation (1.13)) result, because a large volume not only implies a low pressure, whenever the volume but also yields the best conditions for minimizing all instances of V is large. interparticle collisions.

56

INTRODUCING INTERACTIONS AND BONDS

Table 2.4

Van der Waals constants for various gases a/(mol dm−3 )−2 bar

b/(mol dm−3 )−1

Monatomic gases
Helium
Neon
Argon
Krypton

0.034 589
0.216 66
1.3483
2.2836

0.023 733
0.017 383
0.031 830
0.038 650

Diatomic gases
Hydrogen
Nitrogen
Oxygen
Carbon monoxide

0.246 46
1.3661
1.3820
1.4734

0.026 665
0.038 577
0.031 860
0.039 523

Gas

Polyatomic gases
Ammonia
Methane
Ethane
Propane
Butane
Benzene

The value of p calculated with the ideal-gas equation is 3.63 ×
105 Pa, or 3.63 bar.

4.3044
2.3026
5.5818
9.3919
13.888
18.876

0.037 847
0.043 067
0.065 144
0.090 494
0.116 41
0.119 74

Worked Example 2.1 0.04 mol of methane gas is enclosed within a flask of volume 0.25 dm3 . The temperature is 0 ◦ C. From Table 2.4, a = 2.3026 dm6 bar mol−2 and b = 0.043 067 dm3 mol−1 . What is the pressure p exerted?
We first rearrange Equation (2.2), starting by dividing both sides by the term (V − nb), to yield p+ n2 a nRT =
2
V
(V − nb)

We then subtract (n2 a) ÷ V 2 from both sides: p= Calculations with the van der Waals equation are complicated because of the need to convert the units to accommodate the SI system. The value of R comes from Table 2.3.

n nRT −
(V − nb)
V

2

a

Next, we insert values and convert to SI units, i.e. 0 ◦ C is expressed as 273.15 K. p= 0.04 mol×(8.314×10−2 dm3 bar K−1 mol−1 )×273.15 K
(0.25 dm3 − 0.04 mol×0.043 067 dm3 mol−1 )


0.04 mol
0.25 dm3

2

× 2.3026 (dm3 mol−1 )2 bar

QUANTIFYING THE INTERACTIONS AND THEIR INFLUENCE

57

p = 3.65887 bar − 0.05895 bar p = 3.59992 bar
The pressure calculated with the ideal-gas equation (Equation (1.13)) is 3.63 bar, so the value we calculate with the van der Waals equation (Equation (2.2)) is 1 per cent smaller. The experimental value is 3.11 bar, so the result with the van der Waals equation is superior.

The lower pressure causes coalescence of gas particles, which decreases their kinetic energy. Accordingly, the impact between the aggregate particle and the container’s walls is less violent, which lowers the observed pressure.

The virial equation
An alternative approach to quantifying the interactions and deviations from the idealgas equation is to write Equation (1.13) in terms of ‘virial coefficients’: p V n = RT (1 + B p + C p2 + . . .)

(2.3)

where the V ÷ n term is often rewritten as Vm and called the molar volume.
Equation (2.3) is clearly similar to the ideal-gas equation,
Equation (1.13), except that we introduce additional terms, each The word ‘virial’ comes expressed as powers of pressure. We call the constants, B , C etc., from the Latin for force
‘virial coefficients’, and we determine them experimentally. We or powerful. call B the second virial coefficient, C the third, and so on.
Equation (2.3) becomes the ideal-gas equation if both B and C are tiny. In fact, these successive terms are often regarded as effectively ‘fine-tuning’ the values of p or
Vm . The C coefficient is often so small that we can ignore it; and D is so minuscule that it is extremely unlikely that we will ever include a fourth virial coefficient in any calculation. Unfortunately, we must exercise care, because B constants are themselves a function of temperature.
Worked Example 2.2 What is the molar volume Vm of oxygen gas at 273 K and p O ? Ignore the third and subsequent virial terms, and take B = −4.626 × 10−2 bar−1 .
From Equation (2.3)
Vm =

RT
V
=
× (1 + B p) n p

Care: the odd-looking units of B’ require us to cite the gas constant R in SI units with prefixes. 58

INTRODUCING INTERACTIONS AND BONDS

Inserting numbers (and taking care how we cite the value of R) yields
Vm =
The value of Vm calculated with the idealgas equation (Equation
(1.13)) is 4.4 per cent higher. 83.145 cm3 bar mol−1 K−1 × 273 K
1 bar
× (1 − 4.626 × 10−2 bar−1 × 1 bar)

Vm = 22 697 × 0.954 cm3
Vm = 21 647 cm3
1 cm3 represents a volume of 1 × 10−6 m3 , so expressing this value of Vm in SI units yields 21.6 × 10−3 m3 .

SAQ 2.3 Calculate the temperature at which the molar volume of oxygen is 24 dm3 . [Hint: you will need some of the data from Worked Example 2.2.
Assume that B has not changed, and be careful with the units, i.e.
Vm = 24 000 cm3 .]
The relationship between B and B is B =
(B ÷ RT ).
A positive virial coefficient indicates repulsive interactions between the particles.
The magnitude of B indicates the strength of these interaction.

An alternative form of the virial equation is expressed in terms of molar volume Vm rather than pressure:
P Vm = RT

1+

B
C
+
+ ...
Vm Vm

(2.4)

Note that the constants in Equation (2.4) are distinguishable from those in Equation (2.3) because they lack the prime symbol. For both Equations (2.3) and (2.4), the terms in brackets represents the molar compressibility Z. Table 2.5 lists a few virial coefficients.

SAQ 2.4 Calculate the pressure of 1 mol of gaseous argon housed within
2.3 dm3 at 600 K. Take B = 11.9 cm3 mol−1 , and ignore the third virial term, C . [Hint: take care with all units; e.g. remember to convert the volume to m3 .]
Table 2.5 Virial coefficients B for real gases as a function of temperature, and expressed in units of cm3 mol−1
Gas
Argon
Hydrogen
Helium
Nitrogen
Neon
Oxygen

100 K

273 K

373 K

−187.0
−2.0
11.4
−160.0
−6.0
−197.5

−21.7
13.7
12.0
−10.5
10.4
−22.0

−4.2
15.6
11.3
6.2
12.3
−3.7

CREATING FORMAL CHEMICAL BONDS

2.3

59

Creating formal chemical bonds
Why is chlorine gas lethal yet sodium chloride is vital for life?
The interaction requires electrons
Chlorine gas is very reactive, and causes horrific burns to the eyes and throat; see p. 243. The two atoms are held together by means of The word ‘chlorine’ a single, non-polar covalent bond. Cl2 has a yellow–green colour derives from the Greek chloros, meaning and, for a gas, is relatively dense at s.t.p. Conversely, table salt
‘green’.
(sodium chloride) is an ionic solid comprising Na+ and Cl− ions, held together in a three-dimensional array. What is the reason for their differences in behaviour?
The outer shell of each ‘atom’ in Cl2 possesses a full octet of electrons: seven electrons of its own (which explains why it belongs to Group VII(b) of the periodic table) and an extra electron from covalent ‘sharing’ with the other atom in the Cl2 molecule. The only other simple interactions in molecular chlorine are the inevitable induced dipolar forces, which are too weak at room temperature to allow for the liquefying of Cl2(g) .
Each chloride ion in NaCl also has eight electrons: again, seven electrons come from the element prior to formation of a chloride ion, but the extra eighth electron comes from ionizing the sodium counter ion. This extra electron resides entirely on the chloride ion, so no electrons are shared. The interactions in solid NaCl are wholly ionic in nature. Induced dipoles will also exist within each ion, but their magnitude is utterly negligible when compared with the strength of the formal charges on the
Na+ and Cl− ions. We are wise to treat them as absent.
So, in summary, the principal differences between Cl2(g) and NaCl(s) lie in the location and the interactions of electrons in the atoms’ outer shells. We say these electrons reside in an atom’s frontier orbitals, meaning that we can ignore the inner electrons, which are tightly bound to the nucleus.

Why does a bicycle tyre get hot when inflated?
Bonds and interactions involve energy changes
A bicycle tyre gets quite hot during its inflation. The work of inflating the tyre explains in part why the temperature increases, but careful calculations (e.g. see pp. 86 and 89) show that additional factors are responsible for the rise in temperature.
On a macroscopic level, we say we compress the gas into the confined space within the tyre; on a microscopic level, interparticle interactions form as soon as the gas particles come into close proximity. We look on p. 86 at the effect of performing
‘work’ while inflating a bicycle tyre, and the way work impinges on the internal energy of the gas.

60

INTRODUCING INTERACTIONS AND BONDS

All matter seeks to minimize its energy and entropy; see Chapter 4. This concept explains, for example, why a ball rolls down a hill, and only stops when it reaches its position of lowest potential energy. These interparticle interactions form for a similar reason.
When we say that two atoms interact, we mean that the outer electrons on the two atoms ‘respond’ to each other. The electrons within the inner orbitals are buried too deeply within the atom to be available for interactions or bonding. We indicate this situation by saying the electrons that interact reside within the ‘frontier’ orbitals.
And this interaction always occurs in such as way as to minimize the energy. We could describe the interaction schematically by
A + B − → product + energy


(2.5)

where A and B are particles of gas which interact when their frontier orbitals are sufficiently close to form a ‘product’ of some kind; the product is generally a molecule or association complex. (A less na¨ve view should also accommodate changes in ı entropy; see Chapter 4.)
We saw earlier (on p. 33) that measuring the temperature is the simplest macroscopic test for an increased energy content. ThereEnergy is liberated fore, we understand that the tyre becomes warmer during inflation when bonds and interactions form. because interactions form between the particles with the concurrent release of energy (Equation (2.5)).

How does a fridge cooler work?
Introduction to the energetics of bond formation
At the heart of a fridge’s cooling mechanism is a large flask containing volatile organic liquids, such as alkanes that have been partially fluorinated and/or chlorinated, which are often known as halons or chlorofluorocarbons (CFCs). We place this flask behind the fridge cabinet, and connect it to the fridge interior with a thin-walled pipe.
The CFCs circulate continually between the fridge interior and the rear, through a heat exchanger.
Now imagine placing a chunk of cheese in the refrigerator. We need to cool the cheese from its original temperature to, say, 5 ◦ C. Because the cheese is warmer than the fridge interior, energy in the form of heat transfers from the cheese to the fridge, as a consequence of the zeroth law of
Converting the liquid thermodynamics (see p. 8). This energy passes ultimately to the
CFC to a gas (i.e. boilvolatile CFCs in the cooling system. ing) is analogous to
The CFC is initially a liquid because of intermolecular interacputting energy into a tions (of the London dispersion type). Imagine that the interactions kettle, and watching involves 4 kJ of energy but cooling the cheese to 5 ◦ C we liberate the water boil off as about 6 kJ of energy: it should be clear that more energy is libersteam. ated than is needed to overcome the induced dipoles. We say that

CREATING FORMAL CHEMICAL BONDS

61

absorption of the energy from the cheese ‘overcomes’ the interactions – i.e. breaks them – and enables the CFC to convert from its liquid form to form a gas:

CFC(l) + energy − → CFC(g)

(2.6)

The fridge pump circulates the CFC, so the hotter (gaseous) CFC is removed from the fridge interior and replaced with cooler CFC (liquid). We increase the pressure of the gaseous CFC with a pump. The higher pressure causes the CFC to condense back to a liquid. The heat is removed from the fridge through a so-called heat exchanger.
Incidentally, this emission of heat also explains why the rear of a fridge is generally warm – the heat emitted is the energy liberated Forming bonds and when the cheese cooled. interactions liberates
In summary, interactions form with the liberation of energy, but energy; breaking bonds adding an equal or greater amount of energy to the system can and interactions rebreak the interactions. Stated another way, forming bonds and inter- quires the addition actions liberates energy; breaking bonds and interactions requires of energy. the addition of energy.
A similar mechanism operates at the heart of an air-conditioning mechanism in a car or office.

Why does steam warm up a cappuccino coffee?
Forming a bond releases energy: introducing calorimetry
To make a cappuccino coffee, pass high-pressure steam through a cup of cold milk to make it hot, then pour coffee through the milk froth. The necessary steam comes from a kettle or boiler.
A kettle or boiler heats water to its boiling point to effect the process:

H2 O(l) + heat energy − → H2 O(g)

(2.7)

Water is a liquid at room temperature because cohesive forces bind together the molecules; the bonds in this case are hydrogen bonds – see p. 44. To effect the phase transition, liquid → gas, we overcome the hydrogen bonds, which explains why we must put energy into liquid water to generate gaseous steam. Stated another way, steam is a high-energy form of water.
Much of the steam condenses as it passes through the cool milk. This condensation occurs in tandem with forming the hydrogen bonds responsible for the water being a liquid. These bonds form concurrently with the The word ‘calorimetry’ liberation of energy. This energy transfers to the milk, explaining comes from the Latin calor, which means why its temperature increases.
Calorimetry is the measurement of energy changes accompany- heat. We get the word ing chemical or physical changes. We usually want to know how ‘calorie’ from the same root. much energy is liberated or consumed per unit mass or mole of

62

INTRODUCING INTERACTIONS AND BONDS

substance undergoing the process. Most chemists prefer data to be presented in the form of energy per mole. In practice, we measure accurately the amount of heat energy liberated or consumed by a known amount of steam while it condenses.
A physical chemist reading from a data book learns that 40.7 kJ mol−1 of energy are liberated when 1 mol of water condenses and will ‘translate’ this information to say that when 1 mol (18 g) of steam condenses to form liquid water, bonds form concurrently with the liberation of 40 700 J of energy.
As 40.7 kJ mol−1 is the molar energy (the energy per mole), we can readily calculate the energy necessary, whatever the amount of water involved.
In fact, every time the experiment is performed, the same amount of energy will be liberated when 18 g condense.

Strictly, this amount of energy is liberated only when the temperature remains at 100 ◦ C during the condensation process.
Any changes in temperature need to be considered separately.

Worked Example 2.3 How much energy is liberated when 128 g of water condenses?
Note the way the units of ‘g’ cancel, to leave n expressed in the units of moles.

Firstly, we calculate the amount of material n involved using amount of material n =

mass in grams molar mass in grams per mole

(2.8)

so, as 1 mol has a mass of 18 g mol−1 n= 128 g
18 g mol−1

n = 7.11 mol
Secondly, the energy liberated per mole is 40.7 kJ mol−1 , so the overall amount of energy given out is 40.7 kJ mol−1 × 7.11 mol = 289 kJ.
Cappuccino coffee is named after Marco d’Aviano, a ‘Capuchin’ monk who was recently made a saint. He entered a looted Turkish army camp, and found sacks of roasted coffee beans. He mixed it with milk and honey to moderate its bitter flavour.

SAQ 2.5 How much energy will be liberated when 21 g of water condense?

A physical chemist will go one stage further, and say that this energy of 40.7 kJ mol−1 relates directly to processes occurring during the condensation process. In this case, the energy relates to the formation of hydrogen bonds.
As each water molecule forms two hydrogen bonds, so 1 mol of water generates 2 mol of hydrogen bonds. The energy per hydrogen bond is therefore (40.7 kJ mol−1 ÷ 2); so the energy of forming a hydrogen bond is 20.35 kJ mol−1 .
In summary, the macroscopic changes in energy measured in an experiment such as this are a direct reflection of microscopic energy changes occurring on the molecular level. The milk of a cappuccino coffee is warmed when steam passes through it because the steam

CREATING FORMAL CHEMICAL BONDS

63

condenses to form liquid water; and the water is a liquid because of the formation of intermolecular forces in the form of hydrogen bonds.

Why does land become more fertile after a thunderstorm? Breaking bonds requires an input of energy
A plant accumulates nutrients from the soil as it grows. Such accumulation depletes the amount of nutrient remaining in the soil; so, harvesting an arable crop, such as maize, barley or corn, removes nutrients from the field. A farmer needs to replenish the nutrients continually if the land is not to The reaction of elemenbecome ‘exhausted’ after a few seasons. tal nitrogen to form
In the context here, ‘nutrients’ principally comprise compounds compounds that can of nitrogen, most of which come from bacteria that employ natu- be readily metabolized rally occurring catalysts (enzymes) which feed on elemental nitro- by a plant is termed gen – a process known as fixing. An example is the bacterium ‘fixing’. All the principal
Rhizobium which lives on beans and peas. The bacteria convert means of fixing nitroatmospheric nitrogen into ammonia, which is subsequently avail- gen involve bacteria. able for important biological molecules such as amino acids, proteins, vitamins and nucleic acids.
Other than natural fixing, the principal sources of nutrients are the man-made fertilizers applied artificially by the farmer, the most common being inorganic ammonium nitrate (NH4 NO3 ), which is unusually rich in nitrogen.
But lightning is also an efficient fertilizer. The mixture of gases we breathe comprises nitrogen (78 per cent), oxygen (21 per cent) Notice the difference and argon (1 per cent) as its principal components. The nitro- between the two words gen atoms in the N2 molecule are bound together tightly via a ‘princiPAL’ (meaning triple bond, which is so strong that most reactions occurring dur- ‘best’, ‘top’ or ‘most ing plant growth (photosynthesis) cannot cleave it: N2 is inert. But important’) and ‘printhe incredible energies unleashed by atmospheric lightning are able ciPLE’ (meaning ‘idea’,
‘thought’ or ‘concept’). to overcome the N≡N bond.
The actual mechanism by which the N≡N molecule cleaves is very complicated, and is not fully understood yet. It is nevertheless clear that much nitrogen is oxidized to form nitrous oxide, NO. This NO dissolves in the water that inevitably accompanies lightning and forms water-soluble nitrous acid
HNO2 , which further oxidizes during the storm to form nitric acid, HNO3 . Nitric acid functions as a high-quality fertilizer. It has been estimated that a thunderstorm can yield many tonnes of fertilizer per acre of land.
To summarize, the N≡N bond in the nitrogen molecule is very strong and cannot be cleaved unless a large amount of energy is We require energy to available to overcome it. Whereas bacteria can fix nitrogen, the cleave bonds: bond energies are discussed biological processes within crops, such as corn and maize, canon p. 114. not provide sufficient energy. But the energy unleashed during a

64

INTRODUCING INTERACTIONS AND BONDS

thunderstorm easily overcomes the N≡N bond energy, fixing the nitrogen without recourse to a catalyst.

Aside
In the high mountains of Pashawa in Pakistan, near the border with Afghanistan, thunderstorms are so common that the soil is saturated with nitrates deriving from the nitric acid formed by lightning. The soil is naturally rich in potassium compounds.
Ion-exchange processes occur between the nitric acid and potassium ions to form large amounts of potassium nitrate, KNO3 , which forms a thick crust of white crystals on the ground, sometimes lending the appearance of fresh snow.
High concentrations of KNO3 are relatively toxic to plant growth because the ratio of K+ to Na+ is too high, and so the soil is not fertile.

Why does a satellite need an inert coating?
Covalency and bond formation
A satellite, e.g. for radio or TV communication, needs to be robust to withstand its environment in space. In particular, it needs to be protected from the tremendous gravitational forces exerted during take off, from the deep vacuum of space, and from atoms in space.
Being a deep vacuum, there is a negligible ‘atmosphere’ surrounding a satellite as it orbits in space. All matter will exist solely
The hydrogen atoms in as unattached atoms (most of them are hydrogen). These atoms space form a ‘hydride’ with the materials on impinge on the satellite’s outer surface as it orbits. On Earth, hydrothe surface of the satelgen atoms always seek to form a single bond. The hydrogen atoms lite. in space interact similarly, but with the satellite’s tough outer skin.
Such interactions are much stronger than the permanent hydrogen bonds or the weaker, temporary induced dipoles we met in Section 2.1. They form a stronger interaction, which we call a covalent bond.
The great American scientist G. N. Lewis coined the word covalent, early in the
20th century. He wanted to express the way that a bond formed by means of electron sharing. Each covalent bond comprises a pair of electrons. This pairing is permanent, so we sometimes say a covalent bond is a formal bond, to distinguish it from weak and temporary interactions such as induced dipoles.
The extreme strength of the covalent bond derives from the way electrons accumulate between the two atoms. The space occupied by the electrons as they accumulate is not random; rather, the two electrons occupy a molecular orbital that is orientated spatially in such a way that the highest probability of finding the electronic charges is directly between the two atomic nuclei.
As we learn about the distribution of electrons within a covalent bond, we start with a popular representation known as a Lewis structure. Figure 2.11 depicts the

CREATING FORMAL CHEMICAL BONDS

65

Atomic nucleus
Electrons

Figure 2.11 Lewis structure of the covalent hydrogen molecule in which electrons are shared

Lewis structure of the hydrogen molecule, in which each atom of hydrogen (atomic number 1) provides a single electron. The resultant molecule may be defined as two atoms held together by means of sharing electrons. Incidentally, we note that the glue holding the two atoms together (the ‘bond’) involves two electrons. This result is common: each covalent bond requires two electrons.

We often call these
Lewis structures ‘dot– cross diagrams’.

Aside
Why call it a ‘molecule’?
The word ‘molecule’ has a long history. The word itself comes from the old French mol´ cule, itself derived from the Latin molecula, the diminutive of moles, meaning e ‘mass’.
One of the earliest cited uses of the word dates from 1794, when Adams wrote,
‘Fermentation disengages a great quantity of air, that is disseminated among the fluid molecules’; and in 1799, Kirwen said, ‘The molecules of solid abraded and carried from some spots are often annually recruited by vegetation’. In modern parlance, both
Kirwen and Adams meant ‘very small particle’.
Later, by 1840, Kirwen’s small particle meant ‘microColloids are discussed scopic particle’. For example, the great Sir Michael Farain Chapter 10. day described a colloidal suspension of gold, known then as Purple of Cassius, as comprising molecules which were ‘small particles’. The surgeon William Wilkinson said in 1851, ‘Molecules are merely indistinct granules; but under a higher magnifying power, molecules become
[distinct] granules’.
Only in 1859 did the modern definition come into being, when the Italian scientist
Stanislao Cannizarro (1826–1910) defined a molecule as ‘the smallest fundamental unit comprising a group of atoms of a chemical compound’. This statement arose while
Cannizarro publicized the earlier work of his compatriot, the chemist and physicist
Amedeo Avogadro (1776–1856).
This definition of a molecule soon gained popularity. Before modern theories of bonding were developed, Tyndall had clearly assimilated Cannizarro’s definition of a molecule when he described the way atoms assemble, when he said, ‘A molecule is a group of atoms drawn and held together by what chemists term affinity’.

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INTRODUCING INTERACTIONS AND BONDS

Why does water have the formula H2 O?
Covalent bonds and valence
The water molecule always has a composition in which two hydrogen atoms combine with one oxygen. Why?
The Lewis structure in Figure 2.11 represents water, H2 O. Oxygen is element number eight in the periodic table, and each oxygen atom possesses six electrons in its outer shell. Being a member of the second row of the periodic table, each oxygen atom seeks to have an outer shell of eight electrons – we call this trend the ‘octet rule’. Each oxygen atom, therefore, has a deficiency of two electrons. As we saw immediately above, an atom of hydrogen has a single electron and, being a row 1 element, requires just one more to complete its outer shell.
The Lewis structure in Figure 2.12 shows the simplest way in which nature satisfies the valence requirements of each element:
The concept of the full each hydrogen shares its single electron with the oxygen, meaning outer shell is crucial if we wish to understand that the oxygen atom now has eight electrons (six of its own – the covalent bonds. crosses in the figure) and two from the hydrogen atoms. Looking now at each hydrogen atom (the two are identical), we see how each now has two electrons: its own original electron (the dot in the diagram) together with one extra electron each from the oxygen (depicted as crosses).
We have not increased the number of electrons at all. All we have done is shared them between the two elements, thereby enabling each atom to have a full outer shell.
This approach is known as the electron-pair theory.

Valence bond theory
The valence bond theory was developed by Linus Pauling (1901–1992) and others in the 1930s to amalgamate the existing electron-pair bonding theory of G. N. Lewis and new data concerning molecular geometry. Pauling wanted a single, unifying theory.
He produced a conceptual framework to explain molecular bonding, but in practice it could not explain the shapes of many molecules.

Atomic nucleus
Electrons

Figure 2.12 Lewis structure of the covalent water molecule. The inner shell of the oxygen atom has been omitted for clarity

CREATING FORMAL CHEMICAL BONDS

67

Nevertheless, even today, we often discuss the bonding of organic compounds in terms of Lewis structures and valence bond theory.

Why is petroleum gel so soft?
Properties of covalent compounds
Clear petroleum gel is a common product, comprising a mixture of simple hydrocarbons, principally n-octadecane (III). It is not quite a solid at room temperature; neither is it really a liquid, because it is very viscous. We call it a gel. Its principal applications are to lubricate (in a car) or to act as a water-impermeable barrier (e.g. between a baby and its nappy, or on chapped hands).

(III)

We saw on p. 52 how methane is a gas unless condensed by compression at high pressure or frozen to low temperatures. But octadecane is neither a solid nor a gas. Why?
There are several, separate types of interaction in III: both covalent bonds and dipoles. Induced dipoles involve a partial charge, which we called δ + or δ − , but, by contrast, covalent bonds involve whole numbers of electrons. A normal covalent bond, such as that between a hydrogen atom and one of the carbon atoms in the backbone of III, requires two electrons. A ‘double bond’ consists simply of two covalent bonds, so four electrons are shared. Six electrons are incorporated in each of the rare instances of a covalent ‘triple bond’. A few quadruple bonds occur in organometallic chemistry, but we will ignore them here.
Most covalent bonds are relatively non-polar. Some are completely non-polar: the diatomic hydrogen molecule is held together with two electrons located equidistantly from the two hydrogen nuclei. Each of the two atoms has an equal ‘claim’ on the electrons, with the consequence that there is no partial charge on the atoms: each is wholly neutral. Only homonuclear molecules such as H2 , F2 , O2 or N2 are wholly non-polar, implying that the majority of covalent bonds do possess a slight polarity, arising from an unequal sharing of the electrons bound up within the bond.
We see the possibility of a substance having several types of bond. Consider water for example. Formal covalent bonds hold together the hydrogen and oxygen atoms, but the individual water molecules cohere by means of hydrogen bonds. Conversely, paraffin wax (n-C15 H32 ) is a solid. Each carbon is bonded covalently

Molecules made of only one element are called
‘homonuclear’, since homo is Greek for
‘same’. Examples of homonuclear molecules are H2 , N2 , S8 and fullerene C60 .

Even a covalent bond can possess a permanent induced dipole.

Covalent compounds tend to be gases or liquids. Even when solid, they tend to be soft.
But many covalent compounds are only solid at lower temperatures and/or higher pressures, i.e. by maximizing the incidence of induced dipoles.

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INTRODUCING INTERACTIONS AND BONDS

Table 2.6 Typical properties covalent compounds
Property

Example

Low melting point
Low boiling point
Physically soft
Malleable, not brittle
Low electrical conductivity
Dissolve in non-polar solvents
Insoluble in polar solvents a b

Ice melts in the mouth
Molecular nitrogen is a gas at room temperature We use petroleum jelly as a lubricant
Butter is easily spread on a piece of bread
We insulate electrical cables with plastica
We remove grease with methylated spiritb
Polyurethane paint protects the window frame from rain

The polythene coating on an electrical wire comprises a long-chain alkane.
‘Methylated spirit’ is the industrial name for a mixture of ethanol and methanol.

Figure 2.13 Diamond has a giant macroscopic structure in which each atom is held in a rigid three-dimensional array. Other covalent solids include silica and other p-block oxides such as Al2 O3

to one or two others to form a linear chain; the hydrogen atoms are bound to this backbone, again with covalent bonds. But the wax is a solid because dispersion forces ‘glue’ together the molecules. Table 2.6 lists some of the common properties of covalent compounds.
Finally, macromolecular covalent solids are unusual in comprising atoms held together in a gigantic three-dimensional array of bonds. Diamond and silica are the simplest examples; see Figure 2.13. Giant macroscopic structures are always solid. Aside
The word covalent was coined in 1919 when the great American Chemist Irving Langmuir said, ‘it is proposed to define valence as the number of pairs of electrons which a given atom shares with others. In view of the fact . . . that ‘valence’ is very often used to express something quite different, it is recommended that the word covalence be used to denote valence defined as above.’ He added, ‘In [ionic] sodium chloride, the covalence of both sodium and chlorine is zero’.
The modern definition from IUPAC says, ‘A covalent bond is a region of relatively high electron density between nuclei which arises (at least partially) from sharing of electrons, and gives rise to an attractive force and characteristic inter-nuclear distance’. CREATING FORMAL CHEMICAL BONDS

69

Why does salt form when sodium and chlorine react?
Bond formation with ions
Ionic interactions are electrostatic by nature, and occur between ions of opposite charge.
The overwhelming majority of ionic compounds are solids, although a few biological exceptions do occur. Table 2.7 lists a few typical properties of ionic compounds.
It is generally unwise to think of ionic compounds as holding together with physical bonds; it is better to think of an array of point charges, held together by the balance of their mutual electrostatic interactions. (By ‘mutual’ here, we imply equal numbers of positive and negative ions, which therefore impart an overall charge of zero to the solid.)
Ionic compounds generally form following the reaction of metallic elements; non-metals rarely have sufficient energy to provide Care: chlorINE is an elemental gas; chlothe necessary energy needed to form ions (see p. 123).
The structure in Figure 2.14 shows the result of an ionic reac- rIDE is a negatively charged anion. tion: sodium metal has reacted with chlorine gas to yield white crystalline sodium chloride, NaCl. Each Na atom has lost an electron to form an Na+ cation and each chlorine atom has gained an electron and is hence a Cl− anion. In practice, the new electron The chloride ion has a negative charge possessed by the chloride came from the sodium atom. because, following ionThe electron has transferred and in no way is it shared. Sodium ization, it possesses chloride is a compound held together with an ionic bond, the more electrons than strength of the bond coming from an electrostatic interaction bet- protons. ween the positive and negative charges on the ions.

Why heat a neon lamp before it will generate light?
Ionization energy
Neon lamps generate a pleasant pink–red glow. Gaseous neon within the tube (at low pressure) is subjected to a strong electric discharge. One electron per neon atom
Table 2.7 Typical properties of ionic compounds
Property
High melting point
High boiling point
Physically hard
Often physically brittle
High electrical conductivity in solution Dissolve in polar solvents
Insoluble in non-polar solvents

Example
We need a blast furnace to melt metals
A lightning strike is needed to volatilize some substances
Ceramics (e.g. plates) can bear heavy weights
Table salt can be crushed to form a powder
Using a hair dryer in the bath risks electrocution
Table salt dissolves in water
We dry an organic solvent by adding solid CaCl2

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INTRODUCING INTERACTIONS AND BONDS
Sodium cation, Na+

Chloride anion, Cl−

Atomic nucleus
Electrons

Figure 2.14 Lewis structure of ionic sodium chloride. Note how the outer shell of the sodium ion is empty, so the next (inner) shell is full

is lost, forming positively charged Ne+ ions:
Ne(g) − → Ne+ (g) + e− (g)

Generally, the flask holding the neon gas contains a small amount of sodium to catalyse
(‘kick start’) the ionization process – see
p. 481.

‘Monatomic’ is an abbreviation for ‘monoatomic’, meaning the
‘molecule’ contains only one atom. The word generally applies to the Group VIII(a) rare gases.

(2.9)

We ‘ionized’ the neon atoms to form Ne+ cations, i.e. each bears a positive charge. (On p. 480 we discuss in detail the photochemical processes occurring at the heart of the neon lamp.)
We generally need quite a lot of energy to ionize an atom or molecule. For example, 2080 kJ of energy are required to ionize
1 mol of monatomic neon gas. This energy is large and explains the need to heat the neon strongly via a strong electric discharge.
We call this energy the ionization energy, and give it the symbol
I (some people symbolize it as Ie ). Ionization energy is defined formally as the minimum energy required to ionize 1 mol of an element, generating 1 mol of electrons and 1 mol of positively charged cations.
The energy required will vary slightly depending on the conditions employed, so we need to systematize our terminology. While the definition of I is simple enough for neon gas, we need to be more careful for elements that are not normally gaseous. For example, consider the process of ionizing the sodium catalyst at the heart of the neon lamp. In fact, there are two energetically distinct processes:

(1)

Vaporization of the sodium, to form a gas of sodium atoms: Na(s) → Na(g) .

(2)

Ionization of gaseous atoms to form ions: Na(g) → Na+ (g) + e− (g) .

To remove any possible confusion, we further refine the definition of ionization energy, and say that I is the minimum energy required to ionize 1 mol of a gaseous element. The ionization energy I relates to process (2); process (1) is additional.

CREATING FORMAL CHEMICAL BONDS

71

Table 2.8 Ionization energies I(n) . For convenience, the figures in the table are given in MJ mol−1 rather than the more usual kJ mol−1 to emphasize their magnitudes
Element
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon

I(1)

I(2)

1.318
2.379
0.526
0.906
0.807
1.093
1.407
1.320
1.687
2.097

I(3)

I(4)

I(5)

I(6)

I(7)

I(8)

I(9)

I(10)

5.257
7.305 11.822
1.763 14.855 21.013
2.433 3.666 25.033 32.834
2.359 4.627 6.229 37.838 47.285
2.862 4.585 7.482 9.452 53.274 64.368
3.395 5.307 7.476 10.996 13.333 71.343 84.086
3.381 6.057 8.414 11.029 15.171 17.874 92.047 106.443
3.959 6.128 9.376 12.184 15.245 20.006 23.076 115.389 131.442

First ionization energy I / kJ mol −1

2500

2000

1500

1000

500

0
0

20

40

60

80

100

Atomic number

Figure 2.15 The first ionization energies I of the first 105 elements (as y) against atomic number
(as x)

Table 2.8 lists several ionization energies: notice that all of them are positive.
Figure 2.15 depicts the first ionization energies I(1) (as y) for the elements hydrogen to nobelium (elements 1–102) drawn as a function of atomic number (as x).
It is clear from Figure 2.15 that the rare gases in Group VIII(b) have the highest values of I , which is best accounted for by noting that they each have a full outer shell of electrons and, therefore, are unlikely to benefit energetically from being ionized.
Similarly, the halogens in Group VII(b) have high values of I because their natural tendency is to accept electrons and become anions X− , rather than to lose electrons.
The alkali metals in Group I(a) have the lowest ionization energies, which is again expected since they always form cations with a +1 valence. There is little variation in I across the d-block and f-block elements, with a slight increase in I as the atomic number increases.

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INTRODUCING INTERACTIONS AND BONDS

Why does lightning conduct through air?
Electron affinity
Lightning is one of the more impressive manifestations of the power in nature: the sky lights up with a brilliant flash of light, as huge amounts of electrical energy pass through the air.
As an excellent generalization, gases may be thought of as electrical insulators, so why do we see the lightning travel through the air? How does it conduct? Applying a huge voltage across a sample of gas generates an electric discharge, which is apparent by the appearance of light. In fact, the colour of the light depends on the nature of the gas, so neon gives a red colour, krypton gives a green colour and helium is invisible to the eye, but emits ultraviolet light.
The source of the light seen with an electric discharge is the plasma formed by the electricity, which is a mixture of ions and electrons, and unionized atoms. If, for example, we look solely at nitrogen, which represents 78 per cent of the air, an electric discharge would form a plasma comprising N2 + , N+ , electrons e− , nitrogen radicals N• , as well as unreacted N2 . Incidentally, the formation of these ions explains how air may conduct electricity.
Very soon after the electric discharge, most of the electrons and nitrogen cations reassemble to form uncharged nitrogen, N2 . The recombination produces so much energy that we see it as visible light – lightning. Some of the electrons combine with nitrogen atoms to form nitrogen anions N− , via the reaction

N(g) + e− (g) − → N− (g)

(2.10)

and, finally, some, N2 + cations react with water or oxygen in the air to form ammonium or hydroxylamine species.
The energy exchanged during the reaction (in Equation (2.10)) is called the ‘electron affinity’ E(ea) . This energy (also called the electron attachment energy) is defined as the change in the internal energy that occurs when 1 mol of atoms in the gas phase are converted by electron attachment to form 1 mol of gaseous ions.
The negative ions formed in Equation (2.10) are called anions.
Most elements are sufficiently electronegative that their electron
The negative value of affinities are negative, implying that energy is given out during
E(ea) illustrates how the energy of the species the electron attachment. For example, the first electron affinity
X− (g) is lower than that of nitrogen is only 7 kJ mol−1 , but for chlorine E(ea) = of its precursor, X(g) .
−364 kJ mol−1 . Table 2.9 lists the electron affinities of gaseous halogens, and Figure 2.16 depicts the electron affinities for the first
20 elements (hydrogen–calcium).
The electron affinity measures the attractive force between the incoming electron and the nucleus: the stronger the attraction, the more energy there is released. The factors that affect this attraction are exactly the same as those relating to ionization
Care: do not confuse the symbols for electron affinity E(ea) and activation energy
Ea from kinetics (see
Chapter 8).

CREATING FORMAL CHEMICAL BONDS

73

Table 2.9 The first electron affinities of the Group VII(a) elements
Gas

E(ea) /kJ mol−1

F2
Cl2
Br2
I2

−348
−364
−342
−314

Note: there is much disagreement in the literature about the exact values of electron affinity. These values are taken from the Chemistry Data Book by Stark and
Wallace. If we use a different data source, we may find slightly different numbers.
The trends will be the same, whichever source we consult.

100

Electron affinity / kJ mol−1

He

Ar

Ne

0
0

5

10

15

20

25

−100

−200

−300
F

CI

−400
Atomic number

Figure 2.16 Graph of the electron affinities E(ea) of the first 30 elements (as y) against atomic number (as x)

energies – nuclear charge, the distance between the nucleus and the electron, as well as the number of electrons residing between the nucleus and the outer, valence electrons.

Aside
We must be careful with the definition above: in many older textbooks, the electron affinity is defined as the energy released when an electron attaches to a neutral atom.
This different definition causes E(ea) to change its sign.

74

INTRODUCING INTERACTIONS AND BONDS

Why is argon gas inert?
First electron affinity and reactivity
Gases such as helium, neon and argon are so unreactive that we call them the inert gases. They form no chemical compounds, and their only interactions are of the London dispersion force type.
They cannot form hydrogen bonds, since they are not able to bond with hydrogen and are not electronegative.
The outer shell of the helium atom is full and complete: the shell can only accept two electrons and, indeed, is occupied by two electrons. Similarly, argon has a complete octet of electrons in its outer shell. Further reaction would increase the number of electrons if argon were to undergo a covalent bond or become an anion, or would decrease the number of electrons below the ‘perfect’ eight if a cation were to form.
There is no impetus for reaction because the monatomic argon is already at its position of lowest energy, and we recall that bonds form in order to decrease the energy.
Sodium atoms always seek to lose a single electron to form the Na+ monocation, because the outer valence shell contains only one electron – that is why we assign sodium to Group I(a) of the periodic table. This single electron helps us explain why it is so favourable, energetically, to form the Na+ cation: loss of the electron empties the outer shell, to reveal a complete inner shell, much like removing the partial skin of an onion to expose a perfectly formed inner layer. So, again, removal of sodium’s single outer electron occurs in order to generate a full shell of electrons.
But if we look at an element like magnesium, there are several ionization processes possible:

Krypton, xenon and radon will form a very limited number of compounds, e.g. with fluorine, but only under quite exceptional conditions.

(1)

Formation of a monocation: Mg(g) → Mg+ (g) + e− (g) .

(2)

Formation of a dication: Mg+ (g) → Mg2+ (g) + e− (g) .

The energy change in reaction (1) is called the first ionization energy and the energy associated with reaction (2) is the second ionization energy. We symbolize the two processes as I(1) and I(2) respectively. The second ionization energy is always larger than the first, because we are removing a negative electron from a positively charged cation, so we need to overcome the attractive force between them. The value of I(1) for a magnesium atom is 734 kJ mol−1 , but I(2) for removing an electron from the Mg+ monocation is 1451 kJ mol−1 . Both ionization energies are huge, but I(2) is clearly much the larger. Table 2.8 contains many other ionization energies for elements 1–10.
It is clear from Table 2.8 that each ionization energy is larger than the one before.
Also note that the last two ionization energies of an element are always larger than the others. The sudden rise follows because the last two energies represent the removal of the two 1 s electrons: removal of electrons from the 2s and 2p orbitals is easier.

Care: do not confuse the symbols for molecular iodine I2 and the second ionization energy I(2) . Hint: note carefully the use of italic type.

CREATING FORMAL CHEMICAL BONDS

75

Why is silver iodide yellow?
Mixed bonding
Silver chloride is white; silver bromide is pale yellow; and silver iodide has a rich yellow colour. We might first think that the change in colour was due to AgI incorporating the iodide anion, yet NaI or HI are both colourless, so the colour does not come from the iodide ions on their own. We need to find a different explanation.
Silver iodide also has other anomalous properties: it is physically soft – it can even be beaten into a sheet, unlike the overwhelming majority of ionic compounds.
More unusual still, it is slightly soluble in ethanol. Clearly, silver iodide is not a straightforward ionic compound. In fact, its properties appear to overlap between covalent (see Table 2.6) and ionic (see Table 2.7).
Silver iodide is neither wholly covalent nor wholly ionic; its bonding shows contributions from both. In fact, most formal chemical bonds comprise a contribution from both covalent and ionic forces. The only exceptions to this general rule are homonuclear molecules such as hydrogen or chlorine, in which the bonding is 100 per cent covalent. The extent of covalency in compounds we prefer to think of as ionic will usually be quite small: less than 0.1 per A ‘trimer’ is a species cent in NaCl. For example, each C–H bond in methane is about 4% comprising three comionic, but the bonding can be quite unusual in compounds compris- ponents (the Latin ing elements from the p- and d-blocks of the periodic table. For tri means ‘three’). example, aluminium chloride, AlCl3 , has a high vapour pressure The W3 O9 trimer has a triangle structure,
(see p. 221); tungsten trioxide will sublime under reduced preswith a WO3 unit at sures to form covalent W3 O9 trimers; sulphur trioxide is a gas but each vertex. will dissolve in water. Each, therefore, demonstrates a mixture of ionic and covalent bonding.
In other words, the valence bonds approach is suitable for compounds showing purely ionic or purely covalent behaviour; we require molecular orbitals for a more mature description of the We require ‘molecbonding in such materials. So the yellow colour of silver iodide ular orbitals’ for a reflects the way the bonding is neither ionic nor covalent. We find, more mature descripin fact, that the charge clouds of the silver and iodide ions overlap tion of the bonding in to some extent, allowing change to transfer between them. We will such materials. look at charge transfer in more detail on p. 459.

Oxidation numbers
Valency is the number of electrons lost, borrowed or shared in a chemical bond. Formal charges are indicated with Arabic numerals, so the formal charge on a copper cation is expressed as Cu2+ , meaning each copper cation has a deficiency of two electrons. In this system of thought, the charge on the central carbon of methane is zero.

Numbers written as
1,2,3,. . ., etc. are called
‘Arabic
numerals’.

76

INTRODUCING INTERACTIONS AND BONDS

Table 2.10 Rule for assigning oxidation numbers
1. In a binary compound, the metal has a positive oxidation number and, if a non-metal, it has a negative oxidation number.
2. The oxidation number of a free ion equals the charge on the ion, e.g. in
Na+ the sodium has a +I oxidation number and chlorine in the Cl− ion has an oxidation number of −I. The oxidation number of the MnO4 − ion is −I, oxide O2− is −II and the sulphate SO4 2− ion is −II.
3. The sum of the oxidation numbers in a polyatomic ion equals the oxidation number of the ions incorporated: e.g. consider MnO4 − ion.
Overall, its oxidation number is −I (because the ion’s charge is −1). Each oxide contributes −II to this sum, so the oxidation number of the central manganese must be +VII.
4. The oxidation number of a neutral compound is zero. The oxidation number of an uncombined element is zero.
5. Variable oxidation numbers:
H = +I (except in the case of hydrides)
Cl = −I (except in compounds and ions containing oxygen)
O = −II (except in peroxides and superoxides)

Unfortunately, many compounds contain bonds that are a mixture of ionic and covalent. In such a case, a formal charge as written is unlikely to represent the actual number of charges gained or lost. For example, the complex ferrocyanide anion
[Fe(CN)6 ]4− is prepared from aqueous Fe2+ , but the central iron atom in the complex definitely does not bear a +2 charge (in fact, the charge is likely to be nearer +1.5). Therefore, we employ the concept of oxidation
Numbers written as I, number. Oxidation numbers are cited with Roman numbers, so the
II, III, . . . etc. are called oxidation number of the iron atom in the ferrocyanide complex
‘Roman numerals’. is +II. The IUPAC name for the complex requires the oxidation number: we call it hexacyanoferrate (II).
Considering the changes in oxidation number during a reaction can dramatically simplify the concept of oxidation and reduction: oxidation is an increase in oxidation number and reduction is a decrease in oxidation number (see Chapter 7). Be aware, though, oxidation numbers rarely correlate with the charge on an ion. For example, consider the sulphate anion SO2− (IV).
4
O
O



S

O

O
(IV)

The central sulphur has eight bonds. The ion has an overall charge of −2. The oxidation number of the sulphur is therefore 8 − 2 = +6. We generally indicate oxidation numbers with roman numerals, though, so we write S(VI). Table 2.10 lists the rules required to assign an oxidation number.

3

Energy and the first law of thermodynamics
Introduction
In this chapter we look at the way energy may be converted from one form to another, by breaking and forming bonds and interactions. We also look at ways of measuring these energy changes.
While the change in internal energy U is relatively easy to visualize, chemists generally concentrate on the net energy H , where H is the enthalpy. H relates to changes in U after adjusting for pressure–volume expansion work, e.g. against the atmosphere and after transfer of energy q into and out from the reaction environment.
Finally, we look at indirect ways of measuring these energies. Both internal energy and enthalpy are state functions, so energy cycles may be constructed according to
Hess’s law; we look also at Born–Haber cycles for systems in which ionization processes occur.

3.1

Introduction to thermodynamics: internal energy
Why does the mouth get cold when eating ice cream?
Energy
Eating ice cream soon causes the mouth to get cold, possibly to the extent of making it feel quite uncomfortable. The mouth of a normal, healthy adult has a temperature of about 37 ◦ C, and the ice cream has a maximum temperature of 0 ◦ C, although it is likely to be in the range −5 to −10 ◦ C if it recently came from the freezer.
A large difference in temperature exists, so energy transfers from the mouth to the ice cream, causing it to melt.

Ice cream melts as it warms in the mouth and surpasses its normal melting temperature; see Chapter 5.

78

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

The evidence for such a transfer of energy between the mouth and the ice cream is the change in temperature, itself a response to the minus-oneth law of thermodynamics
(p. 7), which says heat travels from hot to cold. Furthermore, the zeroth law (p. 8) tells us energy will continue to transfer from the mouth (the hotter object) to the ice cream (the colder) until they are at the same temperature, i.e. when they are in thermal equilibrium.

Internal energy U
Absolutely everything possesses energy. We cannot ‘see’ this energy directly, nor do we experience it except under certain conditions. It appears to be invisible because it is effectively ‘locked’ within a species. We call the energy possessed by the object the ‘internal energy’, and give it the symbol U .
The internal energy U is defined as the total energy of a body’s components. Unfortunately, there is no way of telling how much
We cannot know how energy is locked away. In consequence, the experimentalist can much energy a body or system has ‘locked’ only look at changes in U . within it. ExperimenThe energy is ‘locked up’ within a body or species in three printally, we can only study cipal ways (or ‘modes’). First, energy is locked within the atomic changes in the internal nuclei. The only way to release it is to split the nucleus, as hapenergy, U. pens in atomic weapons and nuclear power stations to yield nuclear energy. The changes in energy caused by splitting nuclei are massive. We will briefly mention nuclear energy in Chapter 8, but the
The energy E locked topic will not be discussed otherwise. It is too rare for most physical into the atomic nucleus chemists to consider further. is related to its mass
This second way in which energy is locked away is within chemm and the speed of ical bonds. We call this form of energy the chemical energy, which light c, according to is the subject of this chapter. Chemical energies are smaller than the Einstein equation, nuclear energies.
E = mc2 .
And third, energy is possessed by virtue of the potential energy, and the translational, vibrational, rotational energy states of the atoms and bonds within the substance, be it atomic, molecular or ionic. The energy within each of these states is quantized, and will be discussed in greater detail in
Chapter 9 within the subject of spectroscopy. These energies are normally much smaller than the energies of chemical bonds.
As thermodynamicists, we generally study the second of these modes of energy change, following the breaking and formation of
Strictly, the bonds bonds (which are held together with electrons), although we occaare held together with
‘outer-shell’ electrons. sionally consider potential energy. The magnitude of the chemical energy will change during a reaction, i.e. while altering the number and/or nature of the bonds in a chemical. We give the name calorimetry to the study of energy changes occurring during bond changes.

79

INTRODUCTION TO THERMODYNAMICS: INTERNAL ENERGY

Chemists need to understand the physical chemistry underlying these changes in chemical energy. We generally prefer to write in shorthand, so we don’t say ‘changes in internal energy’ nor the shorter phrase ‘changes in U ’, but say instead ‘ U ’. But we need to be careful: the symbol does not just mean ‘change in’. We define it more precisely with Equation (3.1):
U(overall) = U(final

state)

− U(initial

state)

(3.1)

where the phrases ‘initial state’ and ‘final state’ can refer to a single chemical or to a mixture of chemicals as they react. This way, U has both a magnitude and a sign.

Placing the Greek letter
(Delta) before the symbol for a parameter such as U indicates the change in U while passing from an initial to a final state. We define the change in a parameter X as X =
X(final state) − X(initial state) .

Aside
In some texts, Equation (3.1) is assumed rather than defined, so we have to work out which are the final and initial states each time, and remember which comes first in expressions like Equation (3.1). In other texts, the final state is written as a subscript and the initial state as a superscript. The value of U for melting ice cream would
(final
(melted ice cream) be written as U(initial state) , i.e. U(ice cream before melting) . It may even be abbreviated to state) Usl , where s = solid and l = liquid.
To further complicate matters, other books employ yet another notation. They retain the sub- and super-scripts, but place them before the variable, so the last expression in the previous paragraph would be written as ls U .
We will not use any of these notation styles in this book.

Why is skin scalded by steam?
Exothermic reactions
Water in the form of a gas is called ‘steam’. Two things happen concurrently when human skin comes into close contact with steam – it could happen, for example, when we get too close to a boiling kettle. Firstly, the flesh in contact with the steam gets burnt and hurts. Secondly, steam converts from its gaseous form to become liquid water. We say it condenses. We summarize the condensation reaction thus:

H2 O(g) − → H2 O(l)

The ‘condensation’ reaction is one of the simplest forms of a
‘phase change’, which we discuss in greater depth in Chapter 5.

(3.2)

80

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

From Equation (3.1), which defines the changes to internal energy,
U for the process in Equation (3.2) is U(condensation) = U(water, l)
− U(water, g) .
As we saw in Chapter 2, the simplest way of telling whether something gains energy is to ascertain whether its temperature goes up.
The temperature of the skin does increase greatly (so it feels hot); its energy increases following the condensation reaction. Conversely, the temperature of the water decreases – indeed, its temperature decreases to below its boiling temperature, so it condenses. The water has lost energy. In summary, we see how energy is transferred, with energy passing from the steam to the skin.
When energy passes from one body to another, we say the process is thermodynamic. The condensation of water is a therThe word ‘exothermodynamic process, with the energy of the water being lower folmic’ comes from two
Greek roots: thermo, lowing condensation. Stated another way, the precursor steam had meaning ‘energy’ or more energy than the liquid water product, so U(final) is lower than
‘temperature’, and exo
U(initial) . Figure 3.1 represents this situation visually, and clearly meaning ‘outside’ or shows how the change in internal energy U during steam con‘beyond’. An exotherdensation is negative. We say the change in U is exothermic. mic process therefore
The energy lost by the steam passes to the skin, which therefore gives out energy. gains energy. We experience this excess energy as burning: with the skin being an insulator, the energy from the steam remains within the skin and causes damaging thermal processes. Nerve endings in the skin report the damage to the brain, which leads to the experience of pain.
But none of the energy is lost during condensation, so exactly the same amount of energy is given out by the steam as is given to the skin. (In saying this, we assume no other thermodynamic processes occur, such as warming of the surrounding air.
Even if other thermodynamic processes do occur, we can still say confidently that no energy is lost. It’s just more difficult to act as an ‘energy auditor’, and thereby follow where it goes.)

Internal energy

We will use the word
‘process’ here to mean any physical chemistry requiring a change in energy. Initially
(before reaction)

Finally
(after reaction)

Figure 3.1 In an exothermic process, the final product has less energy than the initial starting materials. Energy has been given out

INTRODUCTION TO THERMODYNAMICS: INTERNAL ENERGY

Worked Example 3.1 Use Equation (3.1) to demonstrate that U is negative for the condensation of steam if, say, U(final) = 12 J and
U(initial) = 25 J.
Inserting values into Equation (3.1):
U = U(final) − U(initial)
U = (12 − 25) J
U = −13 J
So we calculate the value of U as −13 J. The change in U is negative and, therefore, exothermic, as expected.
We see that U is negative. We could have reasoned this result by saying U(final) < U(initial) , and subtracting a larger energy from a smaller one generates a deficit.

81

Important: although we have assigned numerical values to
U(final) and U(initial) , it is, in fact, impossible to know their values.
In reality, we only know the difference between them.

The symbol ‘J’ here means joule, which is the SI unit of energy.

Why do we sweat?
Endothermic reactions
We all sweat at some time or other, e.g. after running hard, living in a hot climate or perhaps during an illness when our temperature is raised due to an infection (which is why we sometimes say, we have ‘got a temperature’).
Producing sweat is one of the body’s natural ways of cooling itself, and it operates as follows. Sweat is an aqueous solution of salt and natural oils, and is secreted by glands just below the surface of the skin. The glands generate this mixture whenever the body feels too hot. Every time air moves over a sweaty limb, from a mechanical fan or natural breeze, the skin feels cooler following evaporation of water from the sweat.
When we say the water evaporates when a breeze blows, we mean it undergoes a phase transition from liquid to vapour, i.e. a phase transition proceeding in the opposite direction to that in the previous example, so Equation (3.2) occurs backwards. When we consider the internal energy changes, we see U(final) = U(water, g) and U(initial) = U(water, l) , so the final state of the water here is more energetic than was its initial state. Figure 3.2 shows a schematic representation of the energy change involved.

We need the salt in sweat to decrease the water’s surface tension in order to speed up the evaporation process
(we feel cooler more quickly). The oils in sweat prevents the skin from drying out, which would make it susceptible to sunburn.

Evaporation is also called ‘vaporization’.
It is a thermodynamic process, because energy is transferred.

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

Internal energy

82

Initially
(before reaction)

Finally
(after reaction)

Figure 3.2 In an endothermic process, the final product has more energy than the initial starting materials. Energy has been taken in

A process is endothermic if the final state has more energy than does the initial state. The word derives from the Greek roots thermo (meaning ‘energy’ or ‘temperature’) and endo
(meaning ‘inside’ or
‘within’). An endothermic process takes in energy. Worked Example 3.2 What is the change in internal energy during sweating?
The definition of U in Equation (3.1) is U = U(final) − U(initial) , so the value of U(evaporation) is obtained as U(water, g) − U(water, l) . We already know that the final state of the water is more energetic than its initial state, so the value of U is positive. We say such a process is endothermic.

We feel cooler when sweating because the skin loses energy by transferring it to the water on its surface, which then evaporates. This process of water evaporation (sweating) is endothermic because energy passes from the skin to the water, and a body containing less energy has a lower temperature, which is why we feel cooler.

Aside
Heat is absorbed from the surroundings while a liquid evaporates. This heat does not change the temperature of the liquid because the energy absorbed equates exactly to the energy needed to break intermolecular forces in the liquid (see Chapter 2). Without these forces the liquid would, in fact, be a gas.
At constant temperature, the heat absorbed during evaporation is often called the latent heat of evaporation. This choice of words arises from the way evaporation occurs without heating of the liquid; ‘latent’ is Latin for ‘hidden’, since the energy added to is not ‘seen’ as a temperature rise.

INTRODUCTION TO THERMODYNAMICS: INTERNAL ENERGY

83

Why do we still feel hot while sweating on a humid beach?
State functions
Sometimes we feel hot even when sweating, particularly in a humid environment like a beach by the sea on a hot day. Two processes occur in tandem on the skin: evaporation (liquid water → gaseous water) and condensation (gaseous water → liquid water). It is quite possible that the same water condenses on our face as evaporated earlier. In effect, then, a cycle of ‘liquid → gas → liquid’ occurs. The two halves of this cycle operate in opposite senses, since both exo- and endo-thermic processes occur simultaneously. The net change in energy is, therefore, negligible, and we feel no cooler.
These two examples of energy change involve water. The only difference between them is the direction of change, and hence the sign of U . But these two factors are related. If we were to condense exactly 1 mol of steam then the amount of energy released into the skin would be 40 700 J. The change in internal energy U (ignoring volume changes) is negative because energy is given out during the condensation process, so U = −40 700 J.
Conversely, if we were to vaporize exactly 1 mol of water from the skin of a sweaty body, the change in internal energy would be +40 700 J. In other words, the magnitude of the change is identical, but the sign is different.
While the chemical substance involved dictates the magnitude of U (i.e. the amount of it), its sign derives from the direction of the thermodynamic process. We can go further: if the same mass of substance is converted from state A to state B, then the change in internal energy is equal and opposite to the same process occurring in the reverse direction, from B to A. This essential truth is depicted schematically in Figure 3.3.
The value of U when condensing exactly 1 mol of water is termed the molar change in internal energy. We will call it Um (condensation) , where the small ‘m’ indicates that a mole is involved in the thermodynamic process. Similarly, the molar
∆U(reaction)

B

A
−∆U(reaction)

Figure 3.3 The change in internal energy when converting a material from state A to state B is equal and opposite to the change in U obtained when performing the same process in reverse, from
B to A

84

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

We often omit the small ‘m’; so, from now on, we assume that changes in internal energy are molar quantities. change in energy during vaporization can be symbolized as
Um (vaporization) . If we compare the molar energies for these two similar processes, we see the following relation:
Um

(condensation)

= − Um

(vaporization)

(3.3)

The two energies are equal and opposite because one process occurs in the opposite direction to the other, yet the same amount of
Internal energy U is material (and hence the same amount of energy) is involved in both. a ‘state function’ beFollowing from Equation (3.3), we say that internal energy is a cause: (1) it is a therstate function. A more formal definition of state function is, ‘A thermodynamic property; modynamic property (such as internal energy) that depends only on and (2) its value the present state of the system, and is independent of its previous depends only on the present state of the history’. In other words, a ‘state function’ depends only on those system, i.e. is indepenvariables that define the current state of the system, such as how dent of the previous much material is present, whether it is a solid, liquid or gas, etc. history. The concept of a state function can be quite difficult, so let us consider a simple example from outside chemistry. Geographical position has analogies to a thermodynamic state function, insofar as it does not matter whether we have travelled from London to New York via Athens or flew direct.
The net difference in position is identical in either case. Figure 3.4 shows this truth diagrammatically. In a similar way, the value of U for the process A → C is the same as the overall change for the process A → B → C. We shall look further at the consequences of U being a state function on p. 98.

New York

Direct

London

Indirect

Athens

Figure 3.4 If geographical position were a thermodynamic variable, it would be a state function because it would not matter if we travelled from London to New York via Athens or simply flew direct. The net difference in position would be identical. Similarly, internal energy, enthalpy, entropy and the Gibbs function (see Chapter 4) are all state functions

INTRODUCTION TO THERMODYNAMICS: INTERNAL ENERGY

85

Furthermore, because internal energy is a state function, the overThe ‘first law of therall change in U is zero following a series of changes described by modynamics’ says a closed loop. As an example, imagine three processes: a change energy can neither be from A → B, then B → C and finally from C → A. The only rea- created nor destroyed, son why the net value of U for this cycle is zero is because we only converted from have neither lost nor picked up any energy over the cycle. We can one form to another. summarize this aspect of physical chemistry by saying, ‘energy cannot be created or destroyed, only converted’ – a vital truth called the first law of thermodynamics.
If we measure U over a thermodynamic cycle and obtain a non-zero value, straightaway we know the cycle is either incomplete (with one or more processes not accounted for) or we employed a sloppy technique while measuring U .

Aside
William Rankine was the first to propose the first law of thermodynamics explicitly, in
1853 (he was famous for his work on steam engines). The law was already implicit in the work of other, earlier, thermodynamicists, such as Kelvin, Helmholtz and Clausius. None of these scientists sought to prove their theories experimentally; only Joule published experimental proof of the first law.

Why is the water at the top of a waterfall cooler than the water at its base?
The mechanical equivalence of work and energy
Two of the architects of modern thermodynamics were William Thompson (better known as Lord Kelvin) and his friend James Prescott Joule – a scientist of great vision, and a master of accurate thermodynamic measurement, as well as being something of an English eccentric. For example, while on a holiday in Switzerland in 1847,
Thompson met Joule. Let Thompson describe what he saw:
I was walking down from Chamonix to commence a tour of Mont Blanc, and whom should I meet walking up but Joule, with a long thermometer in his hand and a carriage with a lady in, not far off. He told me that he had been married since we parted in
Oxford [two weeks earlier] and that he was going to try for the elevation of temperature in waterfalls.

Despite it being his honeymoon, Joule possessed a gigantic thermometer fully 4 to
5 feet in length (the reports vary). He spent much of his spare time during his honeymoon in making painstaking measurements of the temperature at the top and bottom of elongated Swiss waterfalls. He determined the temperature difference between the

86

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

water at the bottom and top of the waterfall, finding it to be about 1 ◦ F warmer at the bottom. In Joule’s own words, ‘A [water]fall of 817 feet [249 m] will generate one degree [Fahrenheit] of temperature’. This result is not attributable to colder air at the top of the waterfall, nor due to friction or viscous drag, or other effects occurring during the water’s descent, but is wholly due to a change in internal energy. The water was simply changing its altitude.
The potential energy of a raised object is given by the expression potential energy = mgh

(3.4)

where m is the mass, g is the acceleration due to gravity and h the height by which it is raised. The potential energy of the water decreases during descent because its height decreases. This energy is liberated; and, as we have noted several times already, the simplest way to tell if the internal energy has increased is to determine its temperature.
Joule showed the temperature of the water of the waterfalls had indeed increased.
We could summarize by saying that thermodynamic work w is energetically equivalent to the lowering or raising of a weight (like the water of the waterfall, above), as discussed below.

Why is it such hard work pumping up a bicycle tyre?
Thermodynamic work
No one who has pumped up a bicycle tyre says it’s easy. Pumping a car tyre is harder still. It requires a lot of energy, and we really have to work at it.
We saw in Chapter 1 how increasing the amount of a gas causes its volume to increase. This increase in volume is needed to oppose
The pressure inside a any increases in pressure. It also explains why blowing into a party party balloon is higher than the external, balloon causes it to get bigger. By contrast, a car tyre cannot expand atmospheric pressure, greatly during pumping, so increasing the amount of gas it contains as evidenced by the will increase its internal pressure. In a fully inflated car tyre, the way it whizzes around a internal pressure is about 10 times greater than ‘standard pressure’ room when punctured. p O , where p O has a value of 105 Pa.
The first law of thermodynamics states that energy may be converted between forms, but cannot be created or destroyed. Joule
Work is a form of enerwas a superb experimentalist, and performed various types of work, gy. The word ‘energy’ each time generating energy in the form of heat. In one set of expercomes from the Greek iments, for example, he rotated small paddles immersed in a water en ergon, meaning
‘from work’. trough and noted the rise in temperature. This experiment was apparently performed publicly in St Anne’s Square, Manchester.
Joule discerned a relationship between energy and work (symbol w). We have to perform thermodynamic work to increase the pressure within the tyre. Such work is performed every time a system alters its volume against an opposing pressure or force, or alters the pressure of a system housed within a constant volume.

INTRODUCTION TO THERMODYNAMICS: INTERNAL ENERGY

Work and energy can be considered as interchangeable: we perform work whenever energy powers a physical process, e.g. to propel a car or raise a spoon to the mouth. The work done on a system increases its energy, so the value of U increases, itself causing U to be positive). Work done by a system corresponds to a negative value of U .

87

Work done on a system increases its energy, so U is positive. Work done by a system corresponds to a negative value of U.

Why does a sausage become warm when placed in an oven?
Isothermal changes in heat and work
At first sight, the answer to our title question is obvious: from the minus-oneth law of thermodynamics, heat travels from the hot oven to the cold(er) item of food we place in it. Also, from the zeroth law, thermal equilibrium is attained only when the sausage and the oven are at the same temperature. So the simplest answer to why a sausage gets hot is to say the energy content of the sausage (in the form of heat) increases, causing its internal energy to rise. And, yet again, we see how the simplest test of an increasing internal energy is an increased temperature.
We can express this truth by saying the sausage gets warmer as the magnitude of its internal energy increases; so, from Equation Care: in the past,
U = U(after heating) − Equation (3.5) was
(3.1),
U = U(final) − U(initial) , hence
U(before heating) . We see how the value of U is positive since often written as U = q − w, where the minus
U(after heating) > U(before heating) .
But we can now be more specific. The internal energy U changes sign is intended to in response to two variables, work w and heat energy q, as defi- show how the internal energy decreases folned by lowing work done by
U =q +w
(3.5) a system. We will use
We have already met the first law of thermodynamics. Equation (3.5) here is the definitive statement of this law, and is expressed in terms of the transfer of energy between a system and its environment. In other words, the magnitude of U is the sum of the changes in the heat q added (or extracted) from a system, and the work w performed by (or done to) it.
The internal energy can increase or decrease even if one or other of the two variables, q and w, remains fixed. Although the sausage does no work w in the oven, the magnitude of U increases because the food receives heat energy q from the oven.
Worked Example 3.3 What is the energy of the sausage after heating, if its original energy is 4000 J, and 20 000 J is added to it?
No work is done, so w = 0.

Equation (3.5), which is the more usual form.

Care: The symbol of the joule is J. A small ‘j’ does not mean joules; it represents another variable from a completely different branch of physical chemistry.

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ENERGY AND THE FIRST LAW OF THERMODYNAMICS

The definition of

U is given by Equation (3.1):
U = U(final) − U(initial)

Care: this is a highly artificial calculation and is intended for illustrative purposes only.
In practice, we never know values of U, only changes in U, i.e. U.

Rearranging to make U(final) the subject, we obtain
U(final) = U(initial) +
Equation (3.5) is another expression for
Equation (3.1) allows us to say

U
U . Substituting for

U in

U(final) = U(initial) + q + w
Inserting values into this equation, we obtain

The word ‘isothermal’ can be understood by looking at its Greek roots. Iso means ‘same’ and thermo means
‘energy’ or ‘temperature’, so a measurement is isothermal when performed at a constant temperature.

U(final) = 4000 J + 20 000 J = 24 000 J

The example above illustrates how energy flows in response to the minus-oneth law of thermodynamics, to achieve thermal equilibrium. The impetus for energy flow is the equalization of temperature (via the zeroth law), so we say that the measurement is isothermal.
We often want to perform thermodynamic studies isothermally because, that way, we need no subsequent corrections for inequalities in temperature; isothermal measurements generally simplify our calculations.

Why, when letting down a bicycle tyre, is the expelled air so cold?
Thermodynamic work
When a fully inflated car tyre is allowed to deflate, the air streaming through the nozzle is cold to the touch. The pressure of the air within the tyre is fairly high, so opening the tyre valve allows it to leave the tyre rapidly – the air movement may even cause a breeze. We could feel a jet of cold air on our face if we were close enough.
As it leaves the tyre, this jet of air pushes away atmospheric air, which requires an effort. We say that work is performed. (It is a form of pressure–volume work, and will be discussed in more depth later, in
Energy added to, or
Section 3.2.) work done on, a sysThe internal energy of the gas must change if work is performed, tem is positive. Energy because U = q + w. It is unlikely that any energy is exchanged removed from, or work so, in this simplistic example, we assume that q = 0. done by, a system is
Energy is consumed because work w is performed by the gas, negative. causing the energy of the gas to decrease, and the change in internal

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energy is negative. If U is negative and q = 0, then w is also negative. By corollary, the value of w in Equation (3.5) is negative whenever the gas performs work.
From Chapter 2, we remember again that the simplest way to tell whether the internal energy decreases is to check whether the temperature also decreases. We see that the gas coming from the tyre is cold because it performs work, which decreases its internal energy.

Why does a tyre get hot during inflation?
Adiabatic changes
Anyone inflating a tyre with a hand pump will agree that much
The temperature of hard work is needed. A car or bicycle tyre usually gets hot during a tyre also increases inflation. In the previous example, the released gas did thermody- when inflated, and is namic work and the value of w was negative. In this example, work caused by interparticle is done to the gas in the tyre, so the value of w is positive. Again, interactions forming; we assume that no energy is transferred, which again allows us to see p. 59. take q as zero.
Looking again at Equation (3.5), U = q + w, we see that if q = 0 and w increases (w is positive), then U increases. This increase in U explains why the temperature of the gas in the tyre increases.
Let us return to the assertion that q is zero, which implies that the system is energetically closed, i.e. that no energy can enter or leave the tyre. This statement is not wholly true because the temperature of the gas within the tyre will equilibrate eventually with the rubber of the tyre, and hence with the outside air, so the tyre becomes cooler in accordance with the minus-oneth and zeroth laws of thermodynamics. But the rubber with which tyre is made is a fairly good thermal insulator, and equilibration is slow. We then make the good approximation that the system is closed, energetically. We say the change in energy is adiabatic.
Energetic changes are adiabatic if they can be envisaged to occur while contained within a boundary across which no energy can A thermodynamic propass. In other words, the energy content within the system stays cess is adiabatic if fixed. For this reason, there may be a steep temperature jump in it occurs within a going from inside the sealed system to its surroundings – the gas (conceptual) boundary across which no in the tyre is hot, but the surrounding air is cooler.
In fact, a truly adiabatic system cannot be attained, since even energy can flow. the most insulatory materials will slowly conduct heat. The best approximations are devices such as a Dewar flask (sometimes called a ‘vacuum flask’). Can a tyre be inflated without a rise in temperature?
Thermodynamic reversibility
A tyre can indeed be inflated without a rise in temperature, most simply by filling it with a pre-cooled gas, although some might regard this ‘adaptation’ as cheating!

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Alternatively, we could consider inflating the tyre with a series of, say, 100 short steps – each separated by a short pause. The difference in pressure before and after each of these small steps would be so slight that the gas within the tyre would be allowed to reach equilibrium with its surroundings after adding each increment, and before the next. Stated a different way, the difference between the pressure of the gas in the hand pump and in the tyre will always be slight.
We can take this idea further. We need to realize that if there is
‘Infinitesimal’ is the no real difference in pressure between the hand pump and the gas reciprocal of infinite, within the tyre, then no work would be needed to inflate because
i.e. incredibly small. there would never be the need to pump against a pressure. Alternatively, if the inflation were accomplished at a rate so slow that it was infinitesimally slow, then there would never be a difference in pressure, ensuring w was always zero. And if w was zero, then U would stay constant per increment. (We need to be aware that this argument requires us to perform the process isothermally.)
It should be clear that inflating a tyre under such conditions is never going to occur in practice, because we would not have the time, and the inflation would never be complete. But as a conceptual experiment, we see that working at an infinitesimally slow rate does not constitute work in the thermodynamic sense.
It is often useful to perform thought experiments of this type, changing a thermodynamic variable at an infinitely slow rate: we
A thermodynamic prosay we perform the change reversibly. (If we perform a process in a cess is reversible if an infinitesimal change in non-reversible manner then we say it is ‘irreversible’.) As a simple an external variable definition, a process is said to be reversible if the change occurs
(e.g. pressure) can at an infinitesimal rate, and if an infinitesimal change in an exterchange the direction nal variable (such as pressure) could change the direction of the in which the process thermodynamic process. It is seen that a change is only reversible occurs. if it occurs with the system and surroundings in equilibrium at all times. In practice this condition is never attained, but we can sometimes come quite close.
Reversibility can be a fairly difficult concept to grasp, but it
The amount of work is invaluable. In fact, the amount of work that can be performed that can be performed during a thermodynamic process is maximized when performing during a thermodynamic process is maxiit reversibly. mized by performing it
The discussion here has focused on work done by changes in reversibly. pressure, but we could equally have discussed it in terms of volume changes, electrical work (see Chapter 7) or chemical changes (see
Chapter 4).

How fast does the air in an oven warm up?
Absorbing energy
The air inside an oven begins to get warm as soon as we switch it on. We can regard the interior of the oven as a fixed system, so the internal energy U of the

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gases increases as soon as energy is added, since U = q + w (Equation (3.5)).
For simplicity, in this argument we ignore the expansion of the atmosphere inside the oven.
But what is the temperature inside the oven? And by how much does the temperature increase? To understand the relationship between the temperature and the amount of heat entering the system, we must first appreciate that all energies are quantized. The macroscopic phenomenon of temperature rise reflects the microscopic absorption of energy. During absorption, quanta of energy enter a substance at the lowest energy level possible, and only enter higher quantal states when the lower energy states are filled. We see the same principle at work when we fill a jar with marbles: the first marbles fall to the jar bottom (the position of lowest potential energy); and we only see marbles at the top of the jar when all the lower energy levels are filled. Continuing the analogy, a wide jar fills more slowly than does a narrower jar, even when we add marbles at a constant rate.
On a macroscopic level, the rate at which the quantal states are filled as a body absorbs energy is reflected by its heat capacity C. We can tell how quickly the quantum states are occupied because the temperature of a body is in direct proportion to the proportion of states filled. A body having a large number of quantum states requires a large number of energy quanta for the temperature to increase, whereas a body having fewer quantum states fills more quickly, and becomes hot faster.

Why does water boil more quickly in a kettle than in a pan on a stove?
Heat capacity
Most modern kettles contain a powerful element (the salesman’s word for ‘heater’), operating at a power of 1000 W or more. A heater emits 1 W if it gives out 1 J s−1 ; so, a heater rated at 1000 W emits 1000 J s−1 . We may see this power expressed as 1 kW (remember that a small ‘k’ is shorthand The SI unit of power is for kilo, meaning 1000). By contrast, an electrical ring on the stove the watt (W). A heater will probably operate between 600 and 800 W, so it emits a smaller rated at a power of 1 W
−1
amount of heat per second. Because the water absorbs less heat emits 1 J s . energy per unit time on a stove, its temperature rises more slowly.
The amount of energy a material or body must absorb for its temperature to increase is termed its ‘heat capacity’ C. A fixed The heat capacity C of a material or body amount of water will, therefore, get warmer at a slower rate if the relates the amount of amount of heat energy absorbed is smaller per unit time. energy absorbed when
Equation (3.6) expresses the heat capacity C in a mathematical raising its temperature. form: dU
(3.6)
CV = dT V

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The expression in Equation (3.6) is really a partial differential: the value of U depends on both T and V , the values of which are connected via Equation (1.13). Accordingly, we need to keep one variable constant if we are unambiguously to attribute changes in
CV to the other. The two subscript ‘V ’ terms tell us C is measured while maintaining the volume constant. When the derivative is a partial derivative, it is usual to write the ‘d’ as ‘∂’.
We call CV ‘the heat capacity at constant volume’. With the volume constant, we measure CV without performing any work
We also call CV the
(so w = 0), so we can write Equation (3.6) differently with dq isochoric heat capacity. rather than dU .
Unfortunately, the value of CV changes slightly with temperature; so, in reality, a value of CV is obtained as the tangent to the graph of internal energy (as y) against temperature (as x); see
A tangent is a straight line that meets a curve
Figure 3.5. at a point, but not
If the change in temperature is small, then we can usually assume does cross it. If the that CV has no temperature dependence, and write an approximate heat capacity changes form of Equation (3.6), saying

slightly with temperature, then we obtain the value of CV as the gradient of the tangent to a curve of U (as y) against T (as x).

CV =

U
T

(3.7)

Analysing Equations (3.6) and (3.7) helps us remember how the
SI unit of heat capacity CV is J K−1 . Chemists usually cite a heat capacity after dividing it by the amount of material, calling it the specific heat capacity, either in terms of J K−1 mol−1 or J K−1 g−1 . As an example, the heat capacity of water is 4.18 J K−1 g−1 , which means that the temperature of 1 g of water increases by 1 K for every 4.18 J of energy absorbed.

SAQ 3.1 Show that the molar heat capacity of water is 75.24 J K−1 mol−1 if CV = 4.18 J K−1 g−1 . [Hint: first calculate the molar mass of H2 O.]

U

Tangential gradient = heat capacity, CV

T

Figure 3.5 The value of the heat capacity at constant volume CV changes slightly with temperature, so its value is best obtained as the gradient of a graph of internal energy (as y) against temperature (as x)

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93

Worked Example 3.4 An electrical heater warms 12 g of water. Its initial temperature is 35.0 ◦ C. The heater emits 15 W for 1 min. What is the new temperature of the water?

Answer strategy. Firstly, we will calculate the energy produced by the heater, in joules. Secondly, knowing the heat capacity of the water C, we divide this energy by C to obtain the temperature rise.
(1) To calculate the energy produced by the heater. Remember that
1 W = 1 J s−1 , so a wattage of 15 W means 15 J s−1 . The heater operates for 1 min (i.e. 60 s), so the energy produced is 15 J s−1 × 60 s =
900 J.
This amount of energy is absorbed by 12 g of water, so the energy absorbed per gram is
900 J
= 75 J g−1
12 g

The word ‘strategy’ comes from the Greek stratos meaning ‘army’.
Strategy originally concerned military manoeuvres. (2) To calculate the temperature rise. The change in temperature T is sufficiently small that we are justified in assuming that the value of CV is independent of temperature. This assumption allows us to employ the approximate equation, Equation (3.7). We rearrange it to make T the subject:
U
T =
CV
Inserting values:
T =

75 J g−1
4.18 J K−1 g−1

yielding
T = 17.9 K
As 1 ◦ C = 1 K, the final temperature of the water is (25.0 + 17.9) ◦ C = 42.9 ◦ C.
SAQ 3.2 How much energy must be added to 1.35 kg of water in a pan if it is to be warmed from 20 ◦ C to its boiling temperature of 100 ◦ C? Assume
CV does not vary from 4.18 J K−1 mol−1 .

The heat capacity CV is an extensive quantity, so its value depends on how much of a material we want to warm up. As chemists, we usually want a value of CV expressed per mole of material. A molar heat capacity is an intensive quantity.

Aside
Another heat capacity is Cp , the heat capacity measured at constant pressure (which is also called the isobaric heat capacity). The values of Cp and CV will differ, by perhaps as much as 5–10 per cent. We will look at Cp in more depth in the next section.

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Why does a match emit heat when lit?
Reintroducing calorimetry
‘Lighting’ a match means initiating a simple combustion reaction. Carbohydrates in the wood combine chemically with oxygen in the air to form water and carbon dioxide. The amount of heat liberated is so great that it catches fire (causing the water to form as steam rather than liquid water).
Heat is evolved because the internal energy of the system changes during the combustion reaction. Previously, the oxygen was a gaseous element characterized by
O=O bonds, and the wood was a solid characterized by C–C, C–H and C–O bonds.
The burning reaction completely changes the number and type of bonds, so the internal energies of the oxygen and the wood alter. This explains the change in U .
We know from Equation (3.5) that U = q + w. Because U changes, one or both of q and w must change. It is certain that much energy is liberated because we feel the heat, so the value of q is negative. Perhaps work w is also performed because gases are produced by the combustion reaction, causing movement of the atmosphere around the match (i.e. w is positive).
The simplest way to measure the change in internal energy U is to perform a reaction in a vessel of constant volume and to look at the amount of heat evolved.
We perform a reaction in a sealed vessel of constant volume called a calorimeter. In practice, we perform the reaction and look at the rise in temperature. The calorimeter is completely immersed in a large reservoir of water (see Figure 3.6) and its temperature is monitored closely before, during, and after the reaction. If we know the heat
Water stirrer

Oxygen inlet Electrical contacts Resistance thermometer Water

Bomb

Sample

Figure 3.6 Schematic representation of the bomb calorimeter for measuring the changes in internal energy that occur during combustion. The whole apparatus approximates to an adiabatic chamber, so we enclose it within a vacuum jacket (like a Dewar flask)

INTRODUCTION TO THERMODYNAMICS: INTERNAL ENERGY

95

capacity C of both the calorimeter itself and the surrounding water, then we can readily calculate the change in energy U accompanying the reaction.

Why does it always take 4 min to boil an egg properly?
Thermochemistry
Most people prefer their eggs to be lightly boiled, with the yellow yolk still liquid and the albumen solid and white. We say the egg white has been ‘denatured’. The variation in egg size is not great. An average egg contains essentially a constant amount of yolk and albumen, so the energy necessary to heat both the yolk and albumen (and to denature the albumen) is, more or less, the same for any egg.
If the energy required to cook an egg is the same per egg, then the simplest way to cook the egg perfectly every time is to ensure carefully that the same amount of energy is absorbed. Most people find that the simplest way to do this is to immerse an egg in boiling water (so the amount of energy entering the egg per unit time is constant), and then to say, ‘total energy = energy per second × number of seconds’.
In practice, it seems that most people prefer an egg immersed in boiling water for about 240 s, or 4 min.
This simple example introduces the topic of thermochemistry.
In a physical chemist’s laboratory, we generally perform a similar Thermochemistry is type of experiment but in reverse, placing a sample in the calorime- the branch of thermoter and measuring the energy released rather than absorbed. The dynamics concerned most commonly performed calorimetry experiment is combustion with the way energy inside a bomb calorimeter (Figure 3.6). We place the sample in the is transferred, released calorimeter and surround it with oxygen gas at high pressure, then or consumed during a seal the calorimeter securely to prevent its internal contents leaking chemical reaction. away, i.e. we maintain a constant volume. An electrical spark then ignites the sample, burning it completely. A fearsome amount of energy is liberated in consequence of the ignition, which is why we call this calorimeter a ‘bomb’.
The overall heat capacity of the calorimeter is a simple function of the amount of steel the bomb comprises and the amount of water surrounding it. If the mass is m and the heat capacity is C, then the overall heat capacity is expressed by
C(overall) = (m(steel) × Cm

(steel) )

+ (m(water) × Cm

(water) )

If the amount of compound burnt in the calorimeter is n, and remembering that no work is done, then a combination of Equations (3.7) and (3.8) suggests that the change in internal energy occurring during combustion is given by
Um

(combustion)

=−

C(overall) T n (3.8)
C(overall) is the heat capacity of the reaction mixture and the calorimeter. (3.9)

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where the ‘m’ means ‘molar’. The negative sign arises from the conventions above, since heat is given out if the temperature goes up, as shown by T being positive.
It is wise first to calibrate the calorimeter by determining an accurate value of C(overall) . This is achieved by burning a comU(combustion) for benpound for which the change in internal energy during combustion zoic acid is −3.2231 MJ
−1
is known, and then accurately warming the bomb and its reservoir mol at 298 K. with an electrical heater. Benzoic acid (I) is the usual standard of choice when calibrating a bomb calorimeter.
O
OH
(I)

The electrical energy passed is q, defined by q =V ×I ×t

(3.10)

where V is the voltage and I the current of the heater, which operates for a time of t seconds.
SAQ 3.3 A voltage of 10 V produces a current of 1.2 A when applied across a heater coil. The heater is operated for 2 min and 40 s. Show that the energy produced by the heater is 1920 J.
We can assume Cp is constant only if T is small. For this reason, we immerse the ‘bomb’ in a large volume of water. This explains why we need to operate the heater for a long time.

Worked Example 3.5 A sample of glucose (10.58 g) is burnt completely in a bomb calorimeter. What is the change in internal energy
U if the temperature rises by 1.224 K? The same heater as that in
SAQ 3.3 is operated for 11 240 s to achieve a rise in temperature of
1.00 K.
Firstly, we calculate the energy evolved by the reaction. From
Equation (3.10), the energy given out by the heater is q = 10 V ×
1.2 A × 11 240 s = 134 880 J.
Secondly, we determine the value of C(overall) for the calorimeter, saying from Equation (3.9)
C=

so

134 880 J energy released
=
change in temperature
1.00 K
C = 134 880 J K−1 .

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Thirdly, we determine the amount of glucose consumed n. We obtain the value of n as ‘amount = mass ÷ molar mass’. The molar mass of glucose is 180 g mol−1 , so the number of moles is 5.88 × 10−2 mol.
Finally, we calculate the value of U from Equation (3.9). Inserting values:
The minus sign is a
U(combustion)
so

134 880 J K−1 × 1.224 K
=−
0.0588 mol

U(combustion) = −2.808 MJ mol−1

Notice how this value of U is negative. As a good generalization, the change in internal energy U liberated during combustion is negative, which helps explain why so many fires are self-sustaining (although see Chapter 4).
The value of U(combustion) for glucose is huge, but most values of
U are smaller, and are expressed in kilo joules per mole.
SAQ 3.4 A sample of anthracene (C14 H10 , II) was burnt in a bomb calorimeter. A voltage of 10 V and a current of 1.2 A were passed for exactly 15 min to achieve the same rise in temperature as that caused by the burning of 0.40 g. Calculate the molar energy liberated by the anthracene. consequence of the way Equation (3.9) is written. An energy change of MJ mol−1 is exceptional.
Most changes in U are smaller, of the order of kJ mol−1 .

We often calculate a volume of U but cite the answer after adjusting for pressure–volume work; see p. 102.

(II)

Aside
The large value of U in Worked Example 3.5 helps explain why sweets, meals and drinks containing sugar are so fattening. If we say a single spoonful of sugar comprises
5 g of glucose, then the energy released by metabolizing it is the same as that needed to raise a 3.5 kg weight from the ground to waist level 1000 times.
(We calculate the energy per lift with Equation (3.4), saying E = m × g × h, where m is the mass, g is the acceleration due to gravity and h is the height through which the weight is lifted.)

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Why does a watched pot always take so long to boil?
Introduction to Hess’s law
We sometimes say, ‘A watched pot never boils’. This empirical observation – that we get bored waiting a long time for the pot to boil – follows because we need to put a lot of energy (heat) in order for the water to boil. The amount of energy we can put into the water per unit time was always low
The popular saying
‘A watched pot never in the days of coal and wood fires. Accordingly, a long time was boils’ arose when most required to boil the water, hence the long wait. fires were wood or
Imagine we want to convert 1 mol of water starting at a room coal, neither of which temperature of, say, 25 ◦ C to steam. In fact we must consider generates heat as fast two separate thermodynamic processes: we first consider the heat as, say, a modern 1 kW needed to warm the water from 25 ◦ C to its boiling temperature kettle. of 100 ◦ C. The water remains liquid during this heating process.
Next, we convert 1 mol of the liquid water at 100 ◦ C to gaseous
This argument relies water (i.e. we boil it), but without altering the temperature. only on words. In realWe will at the moment ignore once more the problems caused ity, the situation is by volume changes. The change in internal energy U(overall) for somewhat more comthe overall process H2 O(l) at 25 ◦ C → H2 O(g) at 100 ◦ C can be plicated because water separated into two components: expands slightly on heating, and greatly on boiling. Energy

U1 relates to the process




H2 O(l) at 25 C − → H2 O(l) at 100 C

Energy U2 relates to the process




H2 O(l) at 100 C − → H2 O(g) at 100 C

so U1 relates to warming the water until it reaches the boiling temperature, and
U2 relates to the actual boiling process itself.
We can obtain U(overall) algebraically, according to
U(overall) =
We can obtain the answer in several different ways because internal energy is a
‘state function’.

Hess’s law states that the value of an energy obtained is independent of the number of intermediate reaction steps taken.

U1 +

U2

(3.11)

In practice, we could have measured U(overall) directly in the laboratory. Alternatively, we could have measured U1 or U2 in the laboratory and found the U values we did not know in a book of tables. Either way, we will get the same answer from these two calculation routes.
Equation (3.11) follows directly from U being a state function, and is an expression of Hess’s law. The great German thermodynamicist Hess observed in 1840 that, ‘If a reaction is performed in more than one stage, the overall enthalpy change is a sum of the enthalpy changes involved in the separate stages’.
We shall see shortly how the addition of energies in this way provides the physical chemist with an extremely powerful tool.

ENTHALPY

99

Hess’s law is a restatement if the first law of thermodynamics. We do not need to measure an energy change directly but can, in practice, divide the reaction into several constituent parts. These parts need not be realizable, so we can actually calculate the energy change for a reaction that is impossible to perform in the laboratory. The only stipulation is for all chemical reactions to balance.
The importance of Hess’s law lies in its ability to access information about a reaction that may be difficult (or impossible) to obtain experimentally, by looking at a series of other, related reactions.

3.2

Enthalpy
How does a whistling kettle work?
Pressure–volume work
The word ‘work’ in the question above could confuse. In common parlance, we say a kettle works or does not work, meaning it either functions as a kettle or is useless.
But following the example in the previous section, we now realize how the word
‘work’ has a carefully defined thermodynamic meaning. ‘Operate’ would be a better choice in this context. In fact, a kettle does not perform any work at all, since it has no moving parts and does not itself move.
In a modern, automatic kettle, an electric heater warms the water inside the kettle – we call it the ‘element’. The electric circuit stops when the water reaches 100 ◦ C because a temperature-sensing bimetallic strip is triggered. But the energy for a more old-fashioned, whistling kettle comes from a gas or a coal hob. The water boils on heating and converts to form copious amounts of gas (steam), which passes through a small valve in the kettle lid to form a shrill note, much like in a football referee’s whistle.
The whistle functions because boiling is accompanied by a change in volume, so the steam has to leave the kettle. And the volume change is large: the volume per mole of liquid water is 18 cm3 (about the size of a small plum) but the volume of a mole of gaseous water (steam) is huge.
SAQ 3.5 Assuming steam to be an ideal gas, use the ideal-gas equation
(Equation (1.13)) to prove that 1 mol of steam at 100 ◦ C (373 K) and standard pressure (p O = 105 Pa) has a volume of is 0.031 m3 .

A volume of 0.031 m3 corresponds to 31 dm3 , so the water increases its volume by a factor of almost 2000 when boiled to generate steam. This staggering result helps us realize just how great the increase in pressure is inside the kettle when water boils.
The volume inside a typical kettle is no more than 2 dm3 . To avoid a rapid build up of pressure within the kettle (which could cause an explosion), the steam seeks to leave the kettle, exiting through the small aperture in the whistle. All the vapour passes through this valve just like a referee blowing ‘time’ after a game. And the

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ENERGY AND THE FIRST LAW OF THERMODYNAMICS

large volume ensures a rapid exit of steam, so the kettle produces an intense, shrill whistle sound.
The steam expelled from the kettle must exert a pressure against the air as it leaves the kettle, pushing it aside. Unless stated otherwise, the pressure of the air surrounding the kettle will be 105 Pa, which we call ‘standard pressure’ p O . The value of p O is
105 Pa. The steam must push against this pressure when leaving the kettle. If it does not do so, then it will not move, and will remain trapped within the kettle. This pushing against the air represents work. Specifically, we call it pressure–volume work, because the volume can only increase by exerting work against an external pressure.
The magnitude of this pressure–volume work is w, and is expressed by w = − (pV )

(3.12)

where (pV ) means a change in the product of p × V . Work is done to the gas when it is compressed at constant pressure, i.e. the minus sign is needed to make w positive.
Equation (3.12) could have been written as p × V if both the pressure and the volume changed at the same time (an example would be the pushing of a piston in a car engine, to cause the volume to decrease at the same time as the pressure increases). In most of the physicochemical processes we will consider here, either p or V will be constant so, in practice, there is only one variable. And with one variable, Equation (3.12) becomes either w = p × V or w = V × p, depending on whether we hold p or V constant.
Most chemists perform experiments in which the contents of our beaker, flask or apparatus are open to the air – obvious examples
For most purposes, a include titrations and refluxes, as well as the kinetic and electrochemist can say w = p × V. chemical systems we consider in later chapters. The pressure is the air pressure (usually p O ), which does not change, so any pressure–volume work is the work necessary to push back the atmosphere. For most purposes, we can say w = p V .
It should be obvious that the variable held constant – whether p or V – cannot be negative, so the sign of w depends on which of the variables we change, so the sign of w in Equation (3.12) depends on the sign(s) of p or V . The sign of w will be negative if we decrease the volume or pressure while performing work.
Worked Example 3.6 We generate 1 mol of water vapour in a kettle by boiling liquid water. What is the work w performed by expansion of the resultant steam?
We have already seen how 1 mol of water vapour occupies a volume of 0.031 m3 (see
SAQ 3.5). This volume of air must be pushed back if the steam is to leave the kettle. The external pressure is p O , i.e. 105 Pa.
The change in volume
V = V(final) − V(initial) so V = V(water,

g)

− V(water,

l)

ENTHALPY

Inserting numbers yields

∆V =

(

0.031

volume per mole of steam



18 × 10−6

)

101

Each aliquot of 1 cm3 represents a volume of
1 × 10−6 m3 .

m3 ≈ 0.031 m3

volume per mole of liquid water

Inserting values into Equation (3.12), w = p V : w = 105 Pa × 0.031 m3 w = 3.1 × 103 J so the pressure–volume work is 3100 J.

We often encounter energies of the order of thousands of joules.
As a shorthand, we often want to abbreviate, so we rewrite the answer to Worked Example 3.6, and say w = 3.1 × 1000 J. Next, we substitute for the factor of 1000 with an abbreviation, generally choosing a small letter ‘k’. We rewrite, saying w = 3.1 kJ.
SAQ 3.6 What is the work done when the gas from a party balloon is released? Assume the inflated balloon has a volume of 2 dm3 , and a volume of 10 cm3 when deflated. Assume there is no pressure change, so p = p O .
[Hint: 1 l = 1 × 10−3 m3 ].

An energy expressed with a letter ‘k’ in the answer means ‘thousands of joules’. We say kilojoules. The choice of ‘k’ comes from the Greek word for thousand, which is kilo. Worked Example 3.7 What is the work performed when inflating a car tyre from p O to 6 × p O . Assume the volume inside the tyre stays constant at 0.3 m3 .
Firstly, we calculate the change in pressure, from an equation like Equation (3.1), p(final) − p(initial) , so p = (6 − 1) × p O , i.e. p = 5 × pO p = 5 × 105 Pa
Then, inserting values into Equation (3.5), w = p × V : w = 5 × 105 Pa × 0.3 m3 yielding w = 15 000 J = 15 kJ

p=

102

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

How much energy do we require during a distillation?
The effect of work on

U: introducing enthalpy H

Performing a simple distillation experiment is every chemist’s delight. We gently warm a mixture of liquids, allowing each component to boil off at its own characteristic temperature (the ‘boiling temperature’ T(boil) ). Each gaseous component cools and condenses to allow collection. Purification and separation are thereby effected.
Although we have looked already at boiling and condensation, until now we have always assumed that no work was done. We now see how invalid this assumption was. A heater located within the distillation apparatus, such as an isomantle, supplies heat energy q to molecules of the liquid. Heating the flask increases the internal energy U of the liquids sufficiently for it to vaporize and thence become a gas.
But not all of the heater’s energy q goes into raising U . We need some of it to perform pressure–volume work, since the vapour formed on boiling works to push back the external atmosphere. The difference between the internal energy U and the available energy (the enthalpy) is given by
H =

U +p V

(3.13)

H is a state function since p, V and U are each state functions. As a state function, the enthalpy is convenient for dealing with systems in which the pressure is constant but the volume is free to change. This way, an enthalpy can be equated with the energy supplied as heat, so q = H .
Worked Example 3.8 A mole of water vaporizes. What is the change in enthalpy,
Take pressure as p O .

H?

We have already seen in the previous section that U = +40.7 kJ per mole of water, and from SAQ 3.5 the volume of 1 mol of water vapour is 0.031 m3 per mole of water.
Inserting values into Equation (3.13):

∆H =

40700 J mol−1

∆U

so and +

(

105 Pa

×

p

0.031 m3 mol−1

)

∆V

H = 40 700 J mol−1 + (3100 J mol−1 )
H = 45.8 kJ mol−1

In this example, the difference between U and H is about 11 per cent.
The magnitude of the difference will increase as the values of H and U get smaller. ENTHALPY

103

Justification Box 3.1
We saw above how the work w performed by a gas is pV. Because performing work will decrease the internal energy, we say w = −pV

(3.14)

Substitution of this simple relationship into the definition of internal energy in Equation
(3.5) yields
U = q − pV
(3.15)
and rearranging Equation (3.15) yields q = U + pV .
This combination of variables occurs so often in physical chemistry, that we give it a name: we call it the enthalpy, and give it the symbol H . Accordingly, we rewrite Equation (3.15) as:
H = U + pV

We often call a collection of variables a
‘compound variable’.

(3.16)

The change in enthalpy H during a thermodynamic process is defined in terms of internal energy and pressure–volume work by
H =

U+

(pV )

(3.17)

Because it is usual to perform a chemical experiment with the top of the beaker open to the open air, the pressure p during most chemical reactions and thermodynamic processes is the atmospheric pressure p O . Furthermore, this pressure will not vary. In other words, we usually simplify (pV ) saying p V because only the volume changes.
Accordingly, Equation (3.17) becomes
H =

U +p V

(3.18)

The equation in the form of Equation (3.18) is the usual form we use. Changes in U are not equal to the energy supplied as heat (at constant pressure p) because the system employs some of its energy to push back the surroundings as they expand. The pV term for work is, therefore, a correction for the loss of energy as work. Because many, if not most, physicochemical measurements occur under conditions of constant pressure, changes in enthalpy are vitally important because it automatically corrects for the loss of energy to the surroundings.

Aside
One of the most common mistakes we make during calculations of this kind is forgetting the way ‘k’ stands for ‘1000’. Think of it this way: a job advertisement offers a salary of £14 k. We would be very upset if, at the end of the first year, we were given just
£14 and the employer said he ‘forgot’ the ‘k’ in his advert!

104

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

Why does the enthalpy of melting ice decrease as the temperature decreases?
Temperature dependence of enthalpy
The enthalpy of freezing water is −6.00 kJ mol−1 at its normal freezing temperature of 273.15 K. The value is negative because energy is liberated during freezing. But the freezing temperature of water changes if the external pressure is altered; so, for example, water freezes at the lower temperature
The word ‘normal’ in this context means ‘at of 253 K when a pressure of about 100 × p O is applied. This high
O
a pressure of p ’. pressure is the same as that along the leading edge of an aeroplane wing. At this lower temperature, H(melt) = 5.2 kJ mol−1 .
The principal cause of H(melt) changing is the decreased temperature. The magnitude of an enthalpy depends on the temperature. For this reason, we need to cite the temperature at which an enthalpy is determined. If the conditions are not cited, we assume a temperature of 298 K and a pressure of p O . We recognize these conditions as s.t.p. Values of enthalpy are often written as HrO298 K for this reason.
The temperature dependence of the standard enthalpy is related by Kirchhoff’s law:
HrOT2 =

Reminder: The ‘curly d’ symbols ∂ tells us the bracketed term in the equation is a ‘partial differential’. HrOT1 +

T2
T1

Cp (T ) dT

(3.19)

where Cp is the molar heat capacity at constant pressure of the substance in its standard state at a temperature of T . We define Cp according to
∂H
Cp =
(3.20)
∂T p

The value of Cp is itself a function of temperature (see p. 140), which explains why we integrate Cp (T ) rather than Cp alone.
The Kirchhoff law is a direct consequence of the heat capacity at constant pressure being the derivative of enthalpy with respect to temperature. It is usually sufficient to assume that the heat capacity
Cp is itself independent of temperature over the range of temperatures required, in which case Equation (3.19) simplifies to

Cp (T ) means Cp as a function of thermodynamic temperature.

HrOT2 =

HrOT1 +

Cp (T2 − T1 )

(3.21)

The experimental scientist should ensure the range of temperatures is slight if calculating with Equation (3.21).
Worked Example 3.9 The standard enthalpy of combustion HcO for benzoic acid
(I) is −3223.1 kJ mol−1 at 20 ◦ C. What is HcO298 K ? The change in Cp during the

105

ENTHALPY

reaction is 118.5 J K−1 mol−1 . Assume this value is temperature independent over this small temperature interval.
Inserting data into Equation (3.21):
HrO298

K

= −3223.7 kJ mol−1 + 118.5 J K−1 mol−1 (298 − 293) K

HrO298

K

= −3223.7 kJ mol−1 + 592.5 J mol−1

so
HrO298

K

= −3223.1 kJ mol−1 or − 3.2231 MJ mol−1

SAQ 3.7 Ethane burns completely in oxygen to form carbon dioxide and
O
O water with an enthalpy of Hc = −1558.8 kJ mol−1 at 25 ◦ C. What is Hc

at 80 C? First calculate the change in heat capacity Cp from the data in the following table and Equation (3.22).
Substance
Cp at 80 ◦ C/J K−1 mol−1

C2 H6(g)
52.6

Cp =

O2(g)
29.4

νCp −

CO2(g)
37.1

H2 O(l)
75.3

νCp

(3.22)

reactants

products

where the upper-case Greek letter Sigma means ‘sum of’, and the lower-case Greek letters ν (nu) represent the stoichiometric number of each species, which are the numbers of each reagent in a fully balanced equation. In the convention we adopt here, the values of ν are positive for products and negative for reactants.

Justification Box 3.2
Starting with the definition of heat capacity in Equation (3.20):
∂H
∂T

p

= Cp

This equation represents Cp for a single, pure substance. Separating the variables yields dH = Cp dT
Then we integrate between limits, saying the enthalpy is H1 at T1 and H2 at T2 :
H2
H1

dH =

T2
T1

Cp dT

106

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

Integrating yields
(H2 − H1 ) = Cp (T2 − T1 )

(3.23)

where the term on the left-hand side is H . Equation (3.23) relates to a single, pure substance.
If we consider a chemical reaction in which several chemicals combine, we can write an expression like this for each chemical. Each chemical has a unique value of H and
Cp , but the temperature change (T2 − T1 ) remains the same for each.
We combine each of the H terms to yield HrOT2 (i.e. HrO at T2 ) and HrOT1 .
Combining the Cp terms according to Equation (3.22) yields Cp . Accordingly, Equation (3.23) then becomes Equation (3.21), i.e.:
HrOT2 =

HrOT1 +

Cp (T2 − T1 )

Why does water take longer to heat in a pressure cooker than in an open pan?
The differences between CV and Cp
A pressure cooker is a sealed cooking pan. Being sealed, as soon as boiling occurs, the pressure of steam within the pan increases dramatically, reaching a maximum pressure of about 6 × p O , causing the final boiling temperature to increase (see Fig. 5.12 on p. 200). Unlike other pans, the internal
See p. 199 to see why a pressure cooker can volume is fixed and the pressure can vary; the pressure in most cook faster than a conpans is atmospheric pressure (∼ p O ), but the volume of the steam ventional, open pan. increases continually.
The heat capacity of the contents in a pressure cooker is CV because the internal volume is constant. By contrast, the heat capacity of the food or whatever inside a conventional pan is Cp . The water is a pressure cooker warms slower because the value of Cp is always smaller than CV . And being smaller, the temperature increases faster per unit input of energy.
In fact, the relationship between CV and Cp is given by
It is relatively rare that we need CV values; most reactions are performed at constant pressure, e.g. refluxing a flask at atmospheric pressure. CV − Cp = nR

(3.24)

where we have met all terms previously.
Worked Example 3.10 What is the heat capacity CV of 1 mol of water? Take the value of Cp from SAQ 3.7.
Rearranging Equation (3.24) slightly yields
CV = nR + Cp

107

ENTHALPY

Inserting values:

CV = (1 × 8.3 + 75.3) J K−1 mol−1

so CV = 83.6 J K−1 mol−1 .

The value of CV is 11 per cent higher than Cp , so the water in the pressure cooker will require 11 per cent more energy than if heated in an open pan.

Justification Box 3.3
Starting with the definition of enthalpy in Equation (3.16):
H = U + pV
The pV term can be replaced with ‘nRT ’ via the ideal-gas equation (Equation (1.13)), giving H = U + nRT
The differential for a small change in temperature is dH = dU + nR dT

The values of n and R are constants and do not change.

dividing throughout by dT yields
∂H
∂T

=

∂U
∂T

+ nR

The first bracket equals CV and the second bracket equals Cp , so
CV = Cp + nR which is just Equation (3.24) rearranged. Dividing throughout by n yields the molar heat capacities:
CV = Cp + R

(3.25)

Why does the temperature change during a reaction?
Enthalpies and standard enthalpies of reaction:

Hr and

O
Hr

One of the simplest definitions of a chemical reaction is ‘changes in the bonds’. All reactions proceed with some bonds cleaving concurrently with others forming. Each

108

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

bond requires energy to form, and each bond liberates energy when breaking (see
p. 63 ff). Typically, the amount of energy consumed or liberated is characteristic of the bond involved, so each C–H bond in methane releases about 220 kJ mol−1 of energy. And, as we have consistently reported, the best macroscopic indicator of a microscopic energy change is a change in temperature.
Like internal energy, we can never know the enthalpy of a reagent; only the change in enthalpy during a reaction or process is knowable. Nevertheless, we can think of changes in H . Consider the preparation of ammonia:

N2(g) + 3H2(g) − → 2NH3(g)

(3.26)

We obtain the standard enthalpy change on reaction HrO as a sum of the molar enthalpies of each chemical participating in the reaction:
HrO =

O νHm −

products

O νHm (3.27)

reactants

The values of ν for the reaction in Equation (3.26) are ν(NH3 ) = +2, ν(H2 ) = −3 and ν(N2 ) = −1. We obtain the standard molar enthalpy of forming ammonia after inserting values into Equation (3.27), as
O
O
O
HrO = 2Hm (NH3 ) − [Hm (N2 ) + 3Hm (H2 )]

SAQ 3.8 Write out an expression for
2NO2 in the style of Equation (3.27).

HrO for the reaction 2NO + O2 →

Unfortunately, we do not know the enthalpies of any reagent. All we can know is a change in enthalpy for a reaction or process. But what is the magnitude of this energy change? As a consequence of Hess’s law (see p. 98), the overall change in enthalpy accompanying a reaction follows from the number and nature of the bonds involved.
We call the overall enthalpy change during a reaction the ‘reaction enthalpy’ Hr , and define it as ‘the change in energy occurring when 1 mol of reaction occurs’. In consequence, its units are J mol−1 , although chemists will usually want to express
H in kJ mol−1 .
In practice, we generally prefer to tighten the definition of Hr above, and look at reagents in their standard states. Furthermore, we maintain the temperature T at
298 K, and the pressure p at p O . We call these conditions standard temperature and pressure, or s.t.p. for short. We need to specify the conditions because temperature and pressure can so readily change the physical conditions of the reactants and products.
As a simple example, elemental bromine is a liquid at s.t.p., so we say the standard state of bromine at s.t.p. is Br2(l) . If a reaction required gaseous bromine Br2(g) then we would need to consider an additional energy – the energy of vaporization to effect the process Br2(l) → Br2(g) . Because we restricted ourselves to s.t.p. conditions, we no longer talk of the reaction enthalpy, but the ‘standard reaction enthalpies’ HrO , where we indicate the standard state with the plimsoll sign ‘ O ’.

ENTHALPY

109

In summary, the temperature of a reaction mixture changes because energy is released or liberated. The temperature of the reaction mixture is only ever constant in the unlikely event of HrO being zero. (This argument requires an adiabatic reaction vessel; see p. 89.)
Some standard enthalpies have special names. We consider below some of the more important cases.

Are diamonds forever?
Enthalpies of formation
We often hear it said that ‘diamonds are forever’. There was even a James Bond novel and film with this title. Under most conditions, a diamond will indeed last forever, or as near ‘for ever’ as makes no difference. But is it an absolute statement of fact?
Diamond is one of the naturally occurring allotropes of carbon, the other common allotrope being graphite. (Other, less common, Some elements exist in allotropes include buckminster fullerine.) If we could observe a several different crysdiamond over an extremely long time scale – in this case, several tallographic forms. The billions of years – we would observe a slow conversion from bril- differing crystal forms liant, clear diamond into grey, opaque graphite. The conversion are called allotropes. occurs because diamond is slightly less stable, thermodynamically, than graphite.
Heating graphite at the same time as compressing it under enormous pressure will yield diamond. The energy needed to convert 1 mol of graphite to diamond is
2.4 kJ mol−1 . We say the ‘enthalpy of formation’ Hf for the diamond is +2.4 kJ mol−1 because graphite is the standard state of carbon.
We define the ‘standard enthalpy of formation’ HfO as the enthalpy change involved in forming 1 mol of a compound from The ‘standard enthalpy
O
its elements, each element existing in its standard form. Both T of formation’ Hf is and p need to be specified, because both variables influence the the enthalpy change magnitude of H . Most books and tables cite HfO at standard involved in forming pressure p O and at a temperature of 298 K. Table 3.1 cites a few 1 mol of a compound or non-stable allotrope representative values of HfO . from its elements, each
It will be immediately clear from Table 3.1 that several val- element being in its ues of Hf are zero. This value arises from the definition we standard form, at s.t.p. chose, above: as Hf relates to forming a compound from its constituent elements, it follows that the enthalpy of forming an element can only be zero, provided it exists in its standard state. Incidentally, it also explains why Hf (Br2 , l) = 0 but Hf (Br2 , g) =
29.5 kJ mol−1 , because the stable form of bromine is liquid at s.t.p. We define the enthalpy
For completeness, we stipulate that the elements must exist in of formation of an eletheir standard states. This sub-clause is necessary, because whereas ment (in its normal most elements exist in a single form at s.t.p. (in which case their state) as zero. enthalpy of formation is zero), some elements, such as carbon

110

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

Table 3.1 Standard enthalpies of formation
Compound

Hf at 298 K
O

Hf /kJ mol−1
O

Organic
Hydrocarbons
methane (CH4 , g) ethane (CH3 CH3 , g) propane (CH3 CH2 CH3 , g) n-butane (C4 H10 , g) ethane (CH2 =CH2 , g) ethyne (CH≡CH, g) cis-2-butene (C4 H8 , g) trans-2-butene (C4 H8 , g) n-hexane (C6 H14 , l) cyclohexane (C6 H12 , l)

−74.8
−84.7
−103.9
−126.2
54.3
226.7
−7.00
−11.2
−198.7
−156

Alcohols methanol (CH3 OH, l) ethanol (C2 H5 OH, l)

−238.7
−277.7

Aromatics benzene (C6 H6 , l) benzene (C6 H6 , g) toluene (CH3 C6 H5 , l)

49.0
82.9
50.0

Sugars α-D-glucose (C6 H12 O6 , s) β-D-glucose(C6 H12 O6 , s) sucrose (C12 H22 O11 , s)
Elements
bromine (Br2 , l) bromine (Br2 , g) chlorine (Cl2 , g) chlorine (Cl, g) copper (Cu, s) copper (Cu, g) fluorine (F2 , g) fluorine (F, g) iodine (I2 , s) iodine (I2 , g) iodine (I, g) nitrogen (N, g) phosphorus (P, white, s) phosphorus (P, red, s) sodium (Na, g) sulphur (S, rhombic, s) sulphur (S, monoclinic, s)
Inorganic
carbon (diamond, s) carbon monoxide (CO, g)

−1274
−1268
−2222
0.00
30.9
0.00
121.7
0.00
338.3
0.00
78.99
0.00
62.4
106.8
472.7
0.00
15.9
107.3
0.00
0.33
2.4
−110.5

ENTHALPY

Table 3.1

111

(continued )

Compound carbon dioxide (CO2 , g) copper oxide (CuO, s) hydrogen oxide (H2 O, l) hydrogen oxide (H2 O, g) hydrogen fluoride (HF, g) hydrogen chloride (HCl, g) nitrogen hydride (NH3 , g) nitrogen hydride (NH3 , aq) nitrogen monoxide (NO, g) nitrogen dioxide (NO2 , g) phosphine (PH3 , g) silicon dioxide (SiO2 , s) sodium hydroxide (NaOH, s) sulphur dioxide (SO2 , g) sulphur trioxide (SO3 , g) sulphuric acid (H2 SO4 , l)

Hf /kJ mol−1
O

−393.0
−157.3
−285.8
−241.8
−271.1
−92.3
−46.1
−80.3
90.3
33.2
5.4
−910.9
−425.6
−296.8
−395.7
−909.3

(above), sulphur or phosphorus, have allotropes. The enthalpy of formation for the stable allotrope is always zero, but the value of Hf for the non-stable allotropes will not be. In fact, the value of Hf for the non-stable allotrope is cited with respect to the stable allotrope. As an example, Hf for white phosphorus is zero by definition
(it is the stable allotrope at s.t.p.), but the value of Hf for forming red phosphorus from white phosphorus is 15.9 kJ mol−1 .
If the value of Hf is determined within these three constraints of standard T , standard p and standard allotropic form, we call the enthalpy a standard enthalpy, which we indicate using the plimsoll symbol ‘ O ’ as HfO .
To conclude: are diamonds forever? No. They convert slowly into graphite, which is the stablest form of carbon. Graphite has the lowest energy for any of the allotropes of carbon, and will not convert to diamond without the addition of energy.

Why do we burn fuel when cold?
Enthalpies of combustion
A common picture in any book describing our Stone Age forebears shows short, hairy people crouched, warming themselves round a flickering fire. In fact, fire was one of the first chemical reactions discovered by our prehistoric ancestors. Primeval fire was needed for warmth. Cooking and warding off dangerous animals with fire was a later
‘discovery’.
But why do they burn wood, say, when cold? The principal reactions occurring when natural materials burn involve chemical oxidation, with carbohydrates combining with elemental oxygen to yield water and carbon dioxide. Nitrogen

112

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

compounds yield nitrogen oxide, and sulphur compounds yield sulphur dioxide, which itself oxidizes to form SO3 .
Let us simplify and look at the combustion of the simplest hydrocarbon, methane.
CH4 reacts with oxygen according to
CH4(g) + 2O2(g) − → CO2(g) + 2H2 O(g)


(3.28)

The reaction is very exothermic, which explains why much of the developed world employs methane as a heating fuel. We can measure the enthalpy change accompanying the reaction inside a calorimeter, or we can calculate a value with thermochemical data.
This enthalpy has a special name: we call it the enthalpy of combustion, and define it as the change in enthalpy accompanying
Most authors abbrethe burning of methane, and symbolize it as H(combustion) or just viate ‘combustion’ to just ‘c’, and symbolize
Hc . In fact, we rarely perform calculations with Hc but with the the enthalpy change standard enthalpy of combustion HcO , where the plimsoll symbol
O
as Hc . Others write
‘ O ’ implies s.t.p. conditions.
O
H(comb) .
Table 3.2 contains values of HcO for a few selected organic compounds. The table shows how all value of HcO are negative, reminding us that energy is given out during a combustion reaction. We say combustion is exothermic, meaning energy is emitted. All exothermic reactions are characterized by a negative value of HcO .
But we do not have to measure each value of HcO : we can calculate them if we know the enthalpies of formation of each
We can use equations chemical, product and reactant, we can adapt the expression in like Equation (3.27) for eq. (3.27), saying: any form of enthalpy, not just combustion.

HcO =

ν HfO − products We could not perform cycles of this type unless enthalpy was a state function.

The word ‘calorific’ means heat containing, and comes from the Latin calor, meaning ‘heat’.

ν HfO

(3.29)

reactants

where each H term on the right-hand side of the equation is a molar enthalpy of formation, which can be obtained from tables.
Worked Example 3.11 The wood mentioned in our title question is a complicated mixture of organic chemicals; so, for simplicity, we update the scene. Rather than prehistoric men sitting around a fire, we consider the calorific value of methane in a modern central-heating system. Calculate the value of Hc for methane at 25 ◦ C using molar enthalpies of formation HfO .
The necessary values of HfO are:
Species (all as gases)
HfO /kJ mol−1

CH4
−74.81

O2
0

CO2
−393.51

H2 O
−285.83

ENTHALPY

113

Table 3.2 Standard enthalpies of combustion Hc for a few organic compounds (all values are at 298 K)
O

Substance

Hc /kJ mol−1
O

Hydrocarbons methane (CH4 , g) ethane (CH3 CH3 , g) propane (CH3 CH2 CH3 , g) n-butane (C4 H10 , g) cyclopropane (C3 H6 , g) propene (C3 H6 , g)
1-butene (C4 H8 , g) cis-2-butene (C4 H8 , g) trans-2-butene (C4 H8 , g)

−890
−1560
−2220
−2878
−2091
−2058
−2717
−2710
−2707

Alcohols methanol (CH3 OH, l) ethanol (C2 H5 OH, l)

−726
−1368

Aromatics benzene (C6 H6 , l) toluene (CH3 C6 H5 , l) naphthalene (C10 H8 , s)

−3268
−3953
−5147

Acids methanoic (HCO2 H, l) ethanoic (CH3 CO2 H, l) oxalic (HCO2 · CO2 H, s) benzoic (C6 H5 · CO2 H, s)

−255
−875
−254
−3227

Sugars α-D-glucose (C6 H12 O6 , s) β-D-glucose (C6 H12 O6 , s) sucrose (C12 H22 O11 , s)

−2808
−2810
−5645

O2 is an element, so its value of HfO is zero. The other values of
Inserting values into Equation (3.29):

HfO are exothermic.

HcO = [(−393.51) + (2 × −285.83)] − [(1 × −74.81) + (2 × 0)] kJ mol−1
HcO = −965.17 − (−74.81) kJ mol−1
HcO = −886.36 kJ mol−1 which is very close to the experimental value of −890 kJ mol−1 in
Table 3.2.
SAQ 3.9 Calculate the standard enthalpy of combustion
O
Hc for burning β-D-glucose, C6 H12 O6 . The required values of HfO may be found in Table 3.1.

The massive value
O
of Hc for glucose explains why athletes consume glucose tablets to provide them with energy.

114

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

We defined the value of HcO during combustion as H(final) − H(initial) , so a negative sign for HcO suggests the final enthalpy is more negative after combustion.
In other words, energy is given out during the reaction. Our Stone Age forebears absorbed this energy by their fires in the night, which is another way of saying ‘they warmed themselves’.

Why does butane burn with a hotter flame than methane?
Bond enthalpies
Methane is easily bottled for transportation because it is a gas. It burns with a clean flame, unlike coal or oil. It is a good fuel. The value of HcO for methane is
−886 kJ mol−1 , but HcO for n-butane is −2878 kJ mol−1 . Burning butane is clearly far more exothermic, explaining why it burns with a hotter flame. In other words, butane is a better fuel.
The overall enthalpy change during combustion is HcO . An alternative way of calculating an enthalpy change during reaction dispenses with enthalpies of formation HfO and looks at the individual numbers of bonds formed and broken. We saw in Chapter 2 how we always need energy to break a bond, and release energy each time a bond forms. Its magnitude depends entirely on the enthalpy change for breaking or making the bonds, and on the respective numbers of each. For example, Equation (3.28) proceeds with six bonds cleaving (four C–H bonds and two O=O bonds) at the same time as six bonds form (two C=O bonds and four H–O bonds). A quick glance at Worked Example 3.11 shows how the energy released during combustion is associated with forming the CO2 and H2 O. If we could generate more CO2 and H2 O, then the overall change in
To simplify the calH would be greater, and hence the fuel would be superior. In culation, we pretend the reaction proceeds fact, many companies prefer butane to methane because it releases with all bonds breakmore energy per mole. ing at once; then, an
We can calculate an enthalpy of reaction with bond enthalpies by instant later, different assuming the reaction consists of two steps: first, bonds break, and bonds form, again all at then different bonds form. This approach can be simplified further once. Such an idea is if we consider the reaction consists only of reactive fragments, mechanistic nonsense and the products form from these fragments. The majority of the but it simplifies the molecule can remain completely unchanged, e.g. we only need to calculation. consider the hydroxyl of the alcohol and the carboxyl of the acid during a simple esterification reaction.
Worked Example 3.12 What fragments do we need to consider during the esterification of 1-butanol with ethanoic acid?

ENTHALPY

We first draw out the reaction in full:

The butyl ethanoate produced by Equation
(3.30) is an ester, and smells of pear drops.

H
OH + O
O

115

CH3

(3.30)
O

CH3 + H

O

H

O

Second, we look for those parts that change and those that remain unchanged. In this example, the bonds that cleave are the O–H bond on the acid and the C–O bond on the alcohol. Such cleavage will require energy. The bonds that form are an O–H bond (to yield water) and a C–O bond in the product ester. All bonds release energy as they form.
In this example, the bonds outside the box do not change and hence do not change their energy content, and can be ignored.
The value of Hr relates to bond changes. In this example, equal numbers of O–H and C–O bonds break as form, so we expect an equal amount of energy to be released as is consumed, leading to an enthalpy change of zero. In fact, the value of Hr is tiny at
−12 kJ mol−1 .

Worked Example 3.12 is somewhat artificial, because most reacO tions proceed with differing numbers of bonds breaking and form- In some texts, HBE is written simply as ‘BE’. ing. A more rigorous approach quantifies the energy per bond – the
O
‘bond enthalpy’ HBE (also called the ‘bond dissociation energy’).
O
O
HBE is the energy needed to cleave 1 mol of bonds. For this reason, values of HBE are always positive, because energy is consumed.
The chemical environment of a given atom in a molecule will influence the magnitude of the bond enthalpy, so tabulated data such as that in Table 3.3 represent average values.
We can calculate a value of HrO with an adapted form of Equation (3.29): products HrO = −

reactants
O
ν HBE −

i

O ν HBE

(3.31)

i

where the subscripted i means those bonds that cleave or form within each reactant or product species during the reaction. We need the minus sign because of the way we defined the bond dissociation enthalpy. All the values in Table 3.3 are positive
O
because HBE relates to bond dissociation.
The stoichiometric numbers ν here can be quite large unless the molecules are small.
The combustion of butane, for example, proceeds with the loss of 10 C–H bonds. A

116

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

Table 3.3 Table of mean bond enthalpies H as a function of bond order and atoms. All
BE
values cited in kJ mol−1 and relate to data obtained (or corrected) to 298 K
O

C

S

O

I

Br

Cl

F

H

414
490
326
272
218
326
803a
1075
289
582
285
515
858
331
590c
812

H–
F–
Cl–
Br–
I–
O–
O=
O≡
S–
S=
N–
N=
N=
C–
C=
C≡

N
389
280
201
163

230
590b



159
473
946

368
343
272
209

423
523

247


464
213
205


142
498


297
280
209
176
151

368
285
218
192

431
255
243

569
159

435

728 if –C=O.
406 if –NO2 ; 368 if –NO3 . c 506 if alternating – and =. a b

moment’s thought suggests an alternative way of writing Equation (3.31), i.e.:
Values of Hr can vary markedly from experimental values if calculated in terms of
O
HBE .
O

bonds formed

HrO = −

bonds broken
O
ν HBE −

i

O ν HBE

(3.32)

i

Note again the minus sign, which we retain for the same reason as for Equation (3.31).
Each of these bond enthalpies is an average enthalpy, measured from a series of
O
similar molecules. Values of HBE for, say, C–H bonds in hydrocarbons are likely to be fairly similar, as shown by the values in Table 3.3. The bond energies of C–H bonds will differ (sometimes quite markedly) in more exceptional molecules, such as
O
those bearing ionic charges, e.g. carbocations. HBE values differ for the OH bond in an alcohol, in a carboxylic acid and in a phenol.
These energies relate to bond rearrangement in gaseous molecules, but calculations are often performed for reactions of condensed phases, by combining the enthalpies of vaporization, sublimation, etc. We can calculate a value without further correction if a crude value of Hr is sufficient, or we do not know the enthalpies of phase changes. ENTHALPY

117

Worked Example 3.13 Use the bond enthalpies in Table 3.3 to calculate the enthalpy of burning methane (Equation (3.28)). Assume all processes occur in the gas phase.
Strategy. We start by writing a list of the bonds that break and form.
Broken: 4 × C–H and 2 × O=O
Formed: 2 × C=O and 4 × O–H so HrO = −[2 ×

O
HBE(C=O) + 4 ×

Inserting values of

O
HBE(O–H) ] − [4 ×

O
HBE(C–H) + 2 ×

O
HBE(O=O) ]

O
HBE from Table 3.3 into Equation (3.32):

Reminder: all HBE values are positive because they relate to dissociation of bonds.
O

HrO = −[(2 × 803) + (4 × 464)]
− [(4 × 414) + (2 × 498)] kJ mol−1
HrO = −(3462 − 2652) kJ mol−1 so HrO = −810 kJ mol−1

which is similar to the value in Worked Example 3.11, but less exothermic.

Aside
O
Calculations with bond enthalpies HBE tend to be relatively inaccurate because each energy is an average. As a simple example, consider the sequential dissociation of ammonia. (1)

NH3 dissociates to form NHž and Hž , and re2 quires an energy of 449 kJ mol−1 .

(2)

NHž dissociates to form NHž and Hž , and re2 quires an energy of 384 kJ mol−1 .

(3)

NHž dissociates to form Nž and Hž , and requires an energy of 339 kJ mol−1 .

The symbol ‘ž’ means a radical species, i.e. with unpaired electron(s).

The variations in HBE are clearly huge, so we usually work with an average bond enthalpy, which is sometimes written as BE or H O . The average bond enthalpy for
BE
the three processes above is 390.9 kJ mol−1 .
O

118

3.3

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

Indirect measurement of enthalpy
How do we make ‘industrial alcohol’?
Enthalpy Cycles from Hess’s Law
Industrial alcohol is an impure form of ethanol made by hydrolysing ethene,
CH2 =CH2 :
H
H

H
H

H2O

HH

H

(3.33)
HH

OH

We pass ethene and water (as a vapour) at high pressure over a suitable catalyst, causing water to add across the double bond of the ethene molecule. The industrial alcohol is somewhat impure because it contains trace quantities of ethylene glycol
(1,2-dihydroxyethane, III), which is toxic to humans. It also contains unreacted water, and some dissolved ethene.
H
H

OH
H
H
We may rephrase
Hess’s law, saying ‘The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which the reaction may be divided’.

OH
(III)

But what is the enthalpy of the hydration reaction in Equation
(3.33)? We first met Hess’s law on p. 98. We now rephrase it by saying ‘The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which the reaction may be divided.’
Accordingly, we can obtain the enthalpy of reaction by drawing a Hess cycle, or we can obtain it algebraically. In this example, we will use the cycle method.

Worked Example 3.14 What is the enthalpy change
(3.33)?
Notice that each of these formation reactions is highly exothermic, explaining why energy is needed to obtain the pure elements.

Hr of the reaction in Equation

We start by looking up the enthalpies of formation Hf for ethene, ethanol and water. Values are readily found in books of data; Table 3.1 contains a suitable selection.
Hf(1) [CH2 =CH2 ] = −52 kJ mol−1
Hf(2) [CH3 CH2 OH] = −235 kJ mol−1
Hf(3) [H2 O] = −286 kJ mol−1

INDIRECT MEASUREMENT OF ENTHALPY

119

(We have numbered these three (1) to (3) simply to avoid the necessity of rewriting the equations.)

To obtain the enthalpy of forming ethanol, we first draw a cycle. It is usual to start by writing the reaction of interest along the top, and the elements parallel, along the bottom. Remember, the value of Hr is our ultimate goal.
DHr

CH2 CH2 + H2O

CH3CH2OH

Reaction of interest

Step 1
1O
2 2

2C + 3H2 +

Elements

The next three stages are inserting the three enthalpies the left-hand side, we insert Hf(1) :
DHr

CH2 CH2 + H2O

Step 2

CH3CH2OH

∆Hf(1)
2C + 3H2 +

1O
2 2

We then put in the enthalpy of forming water,

CH2 CH2 + H2O

Step 3

Hf (1) to (3). Starting on

DHf(1)

D Hr

Hf(3) :

CH3CH2OH

DHf(3)
2C + 3H2 +

1O
2 2

And finally we position the enthalpy of forming the product ethanol,

CH2 CH2 + H2O

Step 4

DHf(1)

D Hr

DHf(3)
2C + 3H2 +

Hf(2) :

CH3CH2OH
DHf(2)

1O
2 2

We can now determine the value of reaction. The rules are as follows:

Hr . Notice how we only draw one arrow per

120

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

We are only allowed to make a choice of route like this because enthalpy is a state function. (3)

(1)

We wish to go from the left-hand side of the reaction to the right-hand side. We can either follow the arrow labelled
Hr , or we pass via the elements (along the bottom line) and thence back up to the ethanol.

(2)

If we go along an arrow in the same direction as the arrow is pointing, then we use the value of H as it is written.

If we have to go along an arrow, but in the opposite direction to the direction in which it points, then we multiply the value of H by ‘−1’.

In the example here, to go from the left-hand side to the right-hand side via the elements, we need to go along two arrows Hf(1) and Hf(3) in the opposite directions to the arrows, so we multiply the respective values of H and multiply each by −1.
We then go along the arrow Hf(2) , but this time we move in the same direction as the arrow, so we leave the sign of the enthalpy unaltered.
And then we tie the threads together and say:
Note there are three arrows, so there are three H terms within
Hr .

Hr = (−1 ×

Hf(1) ) + (−1 ×

Hf(3) ) +

Hf(2)

Inserting values into this equation:

Hr = (−1 × −52 kJ mol−1 ) + (−1 × −286 kJ mol−1 ) + (−235 kJ mol−1 )
Hr = 52 kJ mol−1 + 286 kJ mol−1 + (−235 kJ mol−1 ) so Hr = 103 kJ mol−1

We obtained this value of Hr knowing the other enthalpies in the cycle, and remembering that enthalpy is a state function. Experimentally, the value of Hr =
99 kJ mol−1 , so this indirect measurement with Hess’s law provides relatively good data. Sometimes, these cycles are considerably harder than the example here. In such cases, it is usual to write out a cycle for each reaction, and then use the results from each cycle to compile another, bigger cycle.

How does an ‘anti-smoking pipe’ work?
Hess’s Law Cycles with Enthalpies of Combustion
Smoking causes severe damage to the heart, lungs and respiratory system. The tobacco in a cigarette or cigar is a naturally occurring substance, and principally comprises the elements carbon, oxygen, hydrogen and nitrogen.
Unfortunately, because the tobacco is contained within the bowl of a pipe or a paper wrapper, complete combustion is rare, meaning that the oxidation is incomplete. One

INDIRECT MEASUREMENT OF ENTHALPY

121

of the worse side effects of incomplete combustion during smoking is the formation of carbon monoxide (CO) in relatively large quantities. Gaseous CO is highly toxic, and forms an irreversible complex with haemoglobin in the blood. This complex helps explain why people who smoke are often breathless.
A simple way of overcoming the toxic effects of CO is to oxidize it before the smoker inhales the tobacco smoke. This is where the ‘anti-smoking’ pipe works. (In fact, the name is a misnomer: it does not stop someone smoking, but merely makes the smoke less toxic.) The cigarette is inserted into one end of a long, hollow tube
(see Figure 3.7) and the smoker inhales from the other. Along the tube’s length are a series of small holes. As the smoker inhales, oxygen enters the holes, mixes with the CO and combines chemically with it according to

CO(g) + 1 O2 − → CO2(g)
2

(3.34)

The CO2(g) produced is considerably less toxic than CO(g) , thereby averting at least one aspect of tobacco poisoning.
We might wonder: What is the enthalpy change of forming the CO in Equation
(3.34)? It is relatively easy to make CO in the laboratory (for example by dehydrating formic acid with concentrated sulphuric acid), so the enthalpy of oxidizing CO to CO2 is readily determined. Similarly, it is easy to determine the enthalpy of formation of CO2 , by burning elemental carbon; but it is almost impossible to determined
Hc for the reaction C + 1 O2 → CO, because the pressure of oxygen in a bomb
2
calorimeter is so high that all the carbon is oxidized directly to CO2 rather than CO.
Therefore, we will employ Hess’s law once more, but this time employing enthalpies of combustion Hc .
The enthalpies of combustion of carbon and CO are obtained readily from books of data. We can readily find out the following The enthalpy Hc(1) is huge, and helps explain from such data books or Table 3.2:
Hc(1) [C(s) + O2(g) − → CO2(g) ] = −393.5 kJ mol


−1

Hc(2) [CO(g) + 1 O2(g) − → CO2(g) ] = −283.0 kJ mol−1

2

why we employ coke and coal to warm a house; this reaction occurs when a coal fire burns. Again, we have numbered the enthalpies, to save time.
Once more, we start by drawing a Hess-law cycle with the elements at the bottom of the page. This time, it is not convenient to write the reaction of interest along the

Cigarette

Mouth

Small holes to allow in oxygen

Figure 3.7 An anti-smoking device: the cigarette is inserted into the wider end. Partially oxidized carbon monoxide combines chemically with oxygen inside the device after leaving the end of the cigarette but before entering the smoker’s mouth; the oxygen necessary to effect this oxidation enters the device through the small circular holes positioned along its length

122

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

top, so we have drawn it on the left as
CO +

Step 1

1O
2 2

CO2

Hr .

Compounds

DHr
Constituent elements

C + O2

Next we insert the enthalpies for the reactions of interest; we first insert
CO +

Step 2

1O
2 2

Hc(1) :

CO2

DH r

DHc(1)
C + O2

Hc(2) :

and finally, we insert
CO +

1O
2 2

DHc(2)

CO2

Step 3
DH r

DH c(1)
C + O2

We want to calculate a value of Hr . Employing the same laws that we can either go along the arrow for Hr directly, or along direction as the arrow (so we do not change its sign), then along
Concerning this last arrow, we go in the opposite direction to multiply its value by −1.
The value of Hr is given as
Hr =

Hc(1) + (−1 ×

as before, we see
Hc(1) in the same the arrow Hc(2) . the arrow, so we

Hc(2) )

Inserting values:
Hr = −393.5 kJ mol−1 + (+283 kJ mol−1 ) so Hr = −110.5 kJ mol−1

SAQ 3.10 Calcite and aragonite are both forms of calcium carbonate,
CaCO3 . Calcite converts to form aragonite. If HfO (calcite) = −1206.92 kJ

123

INDIRECT MEASUREMENT OF ENTHALPY

mol−1 and HfO (aragonite) = −1207.13 kJ mol−1 , calculate the value of for the transition process:
CaCO3(s,

calcite)

− − CaCO3(s,
−→

aragonite)

Hr

(3.35)

Why does dissolving a salt in water liberate heat?
Hess’s Law Applied to Ions: Constructing Born–Haber Cycles
Dissolving an ionic salt in water often liberates energy. For example, 32.8 kJ mol−1 of energy are released when 1 mol of potassium nitrate dissolves in water. Energy is released, as experienced by the test tube getting warmer.
Before we dissolved the salt in water, the ions within the crystal were held together by strong electrostatic interactions, which The ‘lattice enthalpy’ is obeyed Coulomb’s law (see p. 313). We call the energetic sum defined as the standard of these interactions the lattice enthalpy (see p. 124). We need change in enthalpy to overcome the lattice enthalpy if the salt is to dissolve. Stated when a solid subanother way, salts like magnesium sulphate are effectively insolu- stance is converted ble in water because water, as a solvent, is unable to overcome the from solid to form gaseous constituent lattice enthalpy.
But what is the magnitude of the lattice enthalpy? We cannot ions. Accordingly, values of H(lattice) are measure it directly experimentally, so we measure it indirectly, with always positive. a Hess’s law energy cycle. The first scientists to determine lattice enthalpies this way were the German scientists Born and Haber: we construct a Born–Haber cycle, which is a form of Hess’slaw cycle.
It is common to see
Before we start, we perform a thought experiment; and, for con- values of H(lattice) venience, we will consider making 1 mol of sodium chloride at called ‘lattice energy’.
25 ◦ C. There are two possible ways to generate 1 mol of gaseous Strictly, this latter term
Na+ and Cl− ions: we could start with 1 mol of solid NaCl and is only correct when the
O
vaporize it: the energy needed is H(lattice) . Alternatively, we could temperature T is 0 K. start with 1 mol of sodium chloride and convert it back to the elements (1 mol of metallic sodium and 0.5 mol of elemental chlorine gas (for which the energy is − HfO ) and, then vaporize the elements one at a time, and ionize each in the gas phase. The energies needed to effect ionization are I for the sodium and
E(ea) for the chlorine.
In practice, we do not perform these two experiments because we can calculate a value of lattice enthalpy H(lattice) with an energy cycle. Next, we appreciate how generating ions from metallic sodium and elemental chlorine involves several processes. If we first consider the sodium, we must: (i) convert it from its solid state to
O
gaseous atoms (for which the energy is H(sublimation) ); (ii) convert the gaseous atoms to gaseous cations (for which the energy is the ionization energy I ). We next consider the chlorine, which is already a gas, so we do not need to volatilize it. But: (i) we
O
must cleave each diatomic molecule to form atoms (for which the energy is HBE );

124

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

(ii) ionize the gaseous atoms of chlorine to form anions (for which the energy is the electron affinity E(ea) ). Finally, we need to account for the way the sodium chloride forms from elemental sodium and chlorine, so the cycle must also include HfO .
Worked Example 3.15 What is the lattice enthalpy
25 ◦ C?

H(lattice) of sodium chloride at

Strategy. (1) We start by compiling data from tables. (2) We construct an energy cycle.
(3) Conceptually, we equate two energies: we say the lattice enthalpy is the same as the sum of a series of enthalpies that describe our converting solid NaCl first to the respective elements and thence the respective gas-phase ions.
(1) We compile the enthalpies: for sodium chloride
These energies are huge. Much of this energy is incorporated into the lattice; otherO wise, the value of Hf would be massive.

HfO = −411.15 kJ mol−1

Na(s) + 1 Cl2(g) − − NaCl(s)
−→
2 for the sodium

O
H(sublimation) = 107.32 kJ mol−1

Na(s) − − Na(g)
−→
Na(g) − − Na+ (g) + e−
−→

I(Na) = 502.04 kJ mol−1

for the chlorine
1
Cl(g)
2

− − Cl·(g)
−→

1
2

Cl·(g) + e− − − Cl− (g)
−→

O
HBE = 121.68 kJ mol−1 (i.e. half of 243.36 kJ mol−1 )

E(ea) = −354.81 kJ mol−1

(2) We construct the appropriate energy cycle; see Figure 3.8. For simplicity, it is usual to draw the cycle with positive enthalpies going up and negative enthalpies going down.
O
(3) We obtain a value of H(lattice) , equating it to the energy needed to convert solid NaCl to its elements and thence the gaseous ions. We
O
We multiply Hf by construct the sum:

‘−1’ because we consider the reverse process to formation. (We travel in the opposite direction to the arrow
O
representing Hf in
Figure 3.8.)

O
H(lattice) = − HfO +

O
H(sublimation) + I(Na) +

1
2

O
HBE + E(ea)

Inserting values:
O
H(lattice) = −(−411.153) + 107.32 + 502.04

+ 121.676 + (−354.81) kJ mol−1 so O
H(lattice) = 787.38 kJ mol−1

SAQ 3.11 Calculate the enthalpy of formation HfO for calcium fluoride.
O
O
Take
H(lattice) = −2600 kJ mol−1 , H(sublimation) = 178 kJ mol−1 ; I1(Ca→Ca+ )

INDIRECT MEASUREMENT OF ENTHALPY

125

1 mol gaseous Na+ and Cl− ions

E(ea)
O

Energy/kJ mol−1

∆H BE

I(Na)

O

∆H(lattice)
1 mol elements (in standard states) O

0

∆H(sublimation)
O

∆Hf

1 mol solid NaCl

Figure 3.8 Born–Haber cycle constructed to obtain the lattice enthalpy H(lattice) of sodium chloride. All arrows pointing up represent endothermic processes and arrows pointing down represent exothermic processes (the figure is not drawn to scale)
O

O
= 596 kJ mol−1 and I2(Ca+ →Ca2+ ) = 1152 kJ mol−1 ; HBE = 157 kJ mol−1
−1
and E(ea) = −334 kJ mol .

Why does our mouth feel cold after eating peppermint? Enthalpy of Solution
Natural peppermint contains several components that, if ingested, lead to a cold sensation in the mouth. The best known and best understood is (−)-menthol (IV), which is the dominant component of the peppermint oil extracted from Mentha piperiia and
M. arvensia.
CH3

OH
CH3 CH3
(IV)

126

ENERGY AND THE FIRST LAW OF THERMODYNAMICS

The cause of the cooling sensation is the unusually positive enthalpy of solution.
O
Most values of H(solution) are positive, particularly for simple inorganic solutes.
Pure IV is a solid at s.t.p. Dissolving IV in the mouth disrupts its molecular structure, especially the breaking of the hydrogen bonds associated with the hydroxyl group. These bonds break concurrently with new hydrogen bonds forming with the water of the saliva. We require energy to break the existing bonds, and liberate energy as new bonds form. Energetically, dissolving (−)-menthol is seen to be endothermic, meaning we require energy. This energy comes from the mouth and, as we saw earlier, the macroscopic manifestation of a lower microscopic energy is a lower temperature.
Our mouth feels cold.
The other substance sometimes added to foodstuffs to cause cooling of the mouth is xylitol (V). It is added as a solid to some sweets, chewing gum, toothpastes and mouth-wash solutions.
CH

CH
CH

CH2
OH
OH

Measuring values of

CH2
OH

OH
(V)

OH

O
H(solution)

O
It is quite difficult to measure an accurate enthalpy of solution H(solution) with a calorimeter, but we can measure it indirectly. Consider the example of sodium chloride, NaCl. The ions in solid NaCl are held together in a tight array by strong ionic bonds. While dissolving in water, the ionic bonds holding the constituent ions of
Na+ and Cl− in place break, and new bonds form between the ions and molecules of water to yield hydrated species. Most simple ions are surrounded with six water molecules, like the [Na(H2 O)6 ]+ ion (VI). Exceptions include the proton with four water molecules (see p. 235) and lanthanide ions with eight.

+

OH2
H2O

OH2
Na

H2O

OH2
OH2

The positive charge does not reside on the central sodium alone. Some charge is distributed over the whole ion.

(VI)

Each hydration bond is partially ionic and partially covalent.
Each oxygen atom (from the water molecules) donates a small amount of charge to the central sodium; hence the ionicity. The orbitals also overlap to impart covalency to the bond.

127

INDIRECT MEASUREMENT OF ENTHALPY

Energy is needed to break the ionic bonds in the solid salt and energy is liberated forming hydration complexes like VI. We also break some of the natural hydrogen bonds in the water. The overall change in enthalpy is termed the enthalpy of soluO tion, H(solution) . Typical values are −207 kJ mol−1 for nitric acid; 34 kJ mol−1 for potassium nitrate and −65.5 kJ mol−1 for silver chloride.
O
One of the most sensitive ways of determining a value of H(solution) is to measure the temperature T at which a salt dissolves completely as a function of its solubility
s. A plot of ln s (as y) against 1 ÷ T (as x) is usually linear. We obtain a value
O
of H(solution) by multiplying the gradient of the graph by −R, where R is the gas constant (as described in Chapter 5, p. 210).

How does a camper’s ‘emergency heat stick’ work?
Enthalpies of Complexation
A camper is in great danger of exposure if alone on the moor or in ‘Exposure’ is a condithe desert when night falls and the weather becomes very cold. If tion of being exposed a camper has no additional heating, and knows that exposure is not to the elements, leadfar off, then he can employ an ‘emergency heat stick’. The stick ing to hypothermia, is long and thin. One of its ends contains a vial of water and, at and can lead to death. the other, a salt such as anhydrous copper sulphate, CuSO4 . Both compartments are housed within a thin-walled glass tube, itself encased in plastic.
Bending the stick breaks the glass, allowing the water to come into contact with the copper sulphate and effect the following hydration reaction:

CuSO4(s) + 5H2 O(l) − → CuSO4 · 5H2 O(s)

(3.36)

The reaction in Equation (3.36) is highly exothermic and releases 134 kJ mol−1 of energy. The camper is kept warm by this heat. The reaction in Equation (3.36) involves complexation. In this example, we could also call it ‘hydration’ or ‘adding water of crystallization’. We will call the energy released the ‘energy of complexation’
H(complexation) .
Heat is liberated when adding water to anhydrous copper sulphate because a new crystal lattice forms in response to strong, new bonds forming between the water and
Cu2+ and SO2− ions. As corroborative evidence of a change in the crystal structure,
4
note how ‘anhydrous’ copper sulphate is off-white but the pentahydrate is blue.

4

Reaction spontaneity and the direction of thermodynamic change
Introduction
We start by introducing the concept of entropy S to explain why some reactions occur spontaneously, without needing additional energy, yet others do not. The sign of S for a thermodynamic universe must be positive for spontaneity. We explore the temperature dependence of S.
In the following sections, we introduce the concept of a thermodynamic universe
(i.e. a system plus its surroundings). For a reaction to occur spontaneously in a system, we require the change in Gibbs function G to be negative. We then explore the thermodynamic behaviour of G as a function of pressure, temperature and reaction composition. Finally, we investigate the relationship between G and the respective equilibrium constant K, and outline the temperature interdependence of G and K.

4.1

The direction of physicochemical change: entropy
Why does the colour spread when placing a drop of dye in a saucer of clean water?
Reaction spontaneity and the direction of change
However gently a drop of dye solution is added to a saucer of clean, pure water, the colour of the dye soon spreads into uncoloured regions of the water. This mixing occurs inevitably without warming or any kind of external agitation – the painter with watercolour would find his art impossible without this effect. Such mixing continues

130

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Mixing occurs spontaneously, but we never see the reverse process, with dye suddenly concentrating into a coloured blob surrounded by clear, uncoloured water.
A reaction is ‘spontaneous’ if it occurs without any additional energy input.
It used to be thought reactions were spontaneous if H was negative. This simplistic idea is incorrect.

until the composition of the solution in the saucer is homogeneous, with the mixing complete. We never see the reverse process, with dye suddenly concentrating into a coloured blob surrounded by clear, uncoloured water.
In previous chapters we looked at the way heat travels from hot to cold, as described by the so called ‘minus-oneth’ law of thermodynamics, and the way net movements of heat cease at thermal equilibrium (as described by the zeroth law). Although this transfer of heat energy was quantified within the context of the first law, we have not so far been able to describe why such chemical systems occur. Thermodynamic changes only ever proceed spontaneously in one direction, but not the other. Why the difference?
In everyday life, we say the diffusion of a dye ‘just happens’ but, as scientists, we say the process is spontaneous. In years past, it was thought that all spontaneous reactions were exothermic, with non-spontaneous reactions being endothermic. There are now many exceptions to this overly simplistic rule; thus, we can confidently say that the sign of H does not dictate whether the reaction is spontaneous or not, so we need a more sophisticated way of looking at the problem of spontaneity.

When we spill a bowl of sugar, why do the grains go everywhere and cause such a mess?
Changes in the extent of disorder
Surely everyone has dropped a bowl of sugar, flour or salt, and caused a mess! The powder from the container spreads everywhere, and seems to cover the maximum area possible. Spatial distribution of the sugar granules ensures a
The granules of spilt range of energies; so, for example, some particles reside on higher sugar have a range of surfaces than others, thereby creating a range of potential energies. energies. And some granules travel faster than others, ensuring a spread of kinetic energies.
The mess caused by dropping sugar reflects the way nature always seeks to maximize disorder. Both examples so far, of dye diffusing in water and sugar causing a mess, demonstrate the achievement of greater disorder. But if we are specific, we should note how it is the energetic disorder that is maximized spontaneously.
It is easy to create disorder; it is difficult to create order. It requires effort to clean up the sugar when re-establishing order, showing in effect how reversing a spontaneous process requires an input of energy. This is why the converse situation – dropping a mess of sugar grains and creating a neat package of sugar – does not happen spontaneously in nature.

THE DIRECTION OF PHYSICOCHEMICAL CHANGE: ENTROPY

131

Why, when one end of the bath is hot and the other cold, do the temperatures equalize?
Entropy and the second law of thermodynamics
Quite often, when running a bath, the water is initially quite cold. After the hot water from the tank has had time to travel through the pipes, the water from the tap is hot. As a result, one end of the bath is hotter than the other. But a short time later, the temperature of the water is the same throughout the bath, with the hot end cooler and the cold end warmer. Temperature equilibration occurs even without stirring. Why?
We saw in Chapter 1 how the simplest way to gauge how much energy a molecule possesses is to look at its temperature. We deduce through a reasoning process such as this that molecules of water at the cold end of the bath have less energy than molecules at the hot end. Next, by combining the minus-oneth and zeroth laws of thermodynamics, we say that energy (in the form of heat) is transferred from molecules of water at the hot end of the bath to molecules at the cold end. Energy transfers until equilibrium is reached. All energy changes are adiabatic if the bath is lagged (to prevent energy loss), in accordance with the first law The word ‘entropy’ comes from the Greek of thermodynamics.
As no chemical reactions occur, we note how these thermo- en tropa, meaning dynamic changes are purely physical. But since no bonds form or ‘in change’ or ‘during transformation’. break, what is the impetus – the cause – of the transfer of energy?
We have already seen the way processes occur with an attendant increase in disorder. We now introduce the concept of entropy. The extent of energetic disorder is given the name entropy (and has the The ‘second law of thersymbol S). A bigger value of S corresponds to a greater extent of modynamics’ says a energetic disorder. process occurs spontaWe now introduce the second law of thermodynamics: a physic- neously only when the ochemical process only occurs spontaneously if accompanied by concomitant energetic an increase in the entropy S. By corollary, a non-spontaneous disorder increases. We process – one that we can force to occur by externally adding can usually approxienergy – would proceed concurrently with a decrease in the ener- mate, and talk in terms of ‘disorder’ alone. getic disorder.
We can often think of entropy merely in terms of spatial disorder, like the example of the sugar grains above; but the entropy of a substance is properly the extent of energetic disorder. Molecules of The energy is transhot and cold water in a bath exchange energy in order to maximize ferred via random, inelastic collisions betthe randomness of their energies.
Figure 4.1 depicts a graph of the number of water molecules ween the molecules of water. Such molecular having the energy E (as y) against the energy of the water molemovement is somecules E (as x). Trace (a) in Figure 4.1 shows the distribution of times called Brownian energies in a bath where half the molecules have one energy while motion; see p. 139. the other half has a different energy, which explains why the graph contains two peaks. A distribution of energies soon forms as energy

132

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE
Colder end of bath

n(E )

Hotter end of bath

Energy of water molecules E

n(E )

(a)

Energy of water molecules E
(b)

Figure 4.1 Graph of the number of water molecules of energy E against energy. (a) Soon after running the bath, so one end is hotter and the other cooler; and (b) after thermal equilibration. The
(average) energy at the peak relates to the macroscopic temperature

is transferred from one set of water molecules to other. Trace (b) in Figure 4.1 shows the distribution of energies after equilibration. In other words, the energetic disorder
S increases. The reading on a thermometer placed in the bath will represent an average energy.
The spread of energies in Figure 4.1 is a direct indication of entropy, with a wider spread indicating a greater entropy. Such energetic disorder is the consequence of having a range of energies. The spread widens spontaneously; an example of a nonspontaneous process would be the reverse process, with the molecules in a bath at, say, 50 ◦ C suddenly reverting to one having a temperature of 30 ◦ C at one end and a temperature of 70 ◦ C at the other.
The German scientist Rudolf Clausius (1822–1888) was the
In the thermodynamic first to understand the underlying physicochemical principles dicsense, an ‘engine’ is a tating reaction spontaneity. His early work aimed to understand device or machine for the sky’s blue colour, the red colours seen at sunrise and sunconverting energy into set, and the polarization of light. Like so many of the ‘greats’ work. Clausius himself of early thermodynamics, he was a mathematician. He was interwanted to devise an effiested in engines, and was determined to improve the efficiency of cient machine to consteam-powered devices, such as pumping engines used to remove vert heat energy (from water from mines, and locomotives on the railways. Clausius was a fuel) into mechanithe first to introduce entropy as a new variable into physics and cal work. chemistry. 133

THE DIRECTION OF PHYSICOCHEMICAL CHANGE: ENTROPY

Why does a room containing oranges acquire their aroma?
Spontaneity and the sign of

S

When a bowl containing fresh oranges is placed on the dining room table, the room acquires their fragrance within a few hours. The organic substance we smell after its release from the oranges is the organic terpene (+)-limonene (I), each molecule of which is small and relatively non-polar. I readily evaporates at room temperature to form a vapour.

We sometimes say these compounds volatilize. H
CH3
CH2
(I)

The process we detect when we note the intensifying smell of the oranges, is:

limonene(l) − → limonene(g)

(4.1)

so the concentration of volatile limonene in the gas phase increases with time. But why does it evaporate in this way?
Liquids can flow (and hence transfer energy by inelastic collisions), so they will have a distribution of energies. Molecules in Gaseous materials the liquid state possess a certain extent of energetic disorder and, have greater entropy therefore, have a certain extent of entropy S. By contrast, molecules than their respective in the gas phase have a greater freedom to move than do liquids, liquids. because there is a greater scope for physical movement: restrictions arising from hydrogen bonds or other physicochemical interactions are absent, and the large distances between each molecule allow for wider variations in speed, and hence in energy. Gas molecules, therefore, have greater entropy than do the liquids from which they derive. We deduce the simple result S(g) > S(l) .
We could obtain this result more rigorously. We have met the symbol ‘ ’ several times already, and recall its definition ‘final state minus initial state’, so the change in entropy S for any process is given by the simple equation
S(process) = S(final

state)

− S(initial

state)

(4.2)

If the final disorder of a spontaneous process is greater than A spontaneous process the initial disorder, then we appreciate from Equation (4.2) how a is accompanied by a spontaneous process is accompanied by S of positive sign. This positive value of S. will remain our working definition of spontaneity.
Ultimately, the sign of S explains why the smell of the oranges increases with time.

134

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Worked Example 4.1 Show mathematically how the entropy of a gas is higher than the entropy of its respective liquid.
If S(final state) is S(g) and S(initial state) is S(l) , then S = S(g) − S(l) . Because the volatilizing of the compound is spontaneous, the sign of S must be positive.
The only way to make S positive is when S(g) > S(l) .

Why do damp clothes become dry when hung outside?
Reaction spontaneity by inspection
Everyone knows damp clothes become dry when hung outside on the washing line.
Any residual water is lost by evaporation from the cloth. In fact, moisture evaporates even if the damp clothes hang limp in the absence of a breeze. The water spontaneously leaves the fabric to effect the physicochemical process H2 O(l) →
H2 O(g) .
The loss of water occurs during drying in order to increase the overall amount of entropy, because molecules of gaseous water
We obtain the energy have a greater energy than do molecules of liquid, merely as a for evaporating the water by lowering the result of being gas and liquid respectively. In summary, we could internal energy of the have employed our working definition of entropy (above), which garment fibres, so the leads to a prediction of the clothes becoming dry, given time, clothes feel cool to the as a result of the requirement to increase the entropy. Inspection touch when dry. alone allows us a shrewd guess at whether a process will occur spontaneously or not.
Worked Example 4.2 By inspection alone, decide whether the sublimation of iodine
(Equation (4.3)) will occur spontaneously or not:
I2(s) − − I2(g)
−→

(4.3)

Molecules in the gas phase have more entropy than molecules in the liquid phase; and molecules in the liquid phase have more entropy than molecules in the solid state. As an excellent generalization, the relative order of the entropies is given by
This argument says nothing about the rate of sublimation. In fact, we do not see sublimation occurring significantly at room temperature because it is so slow.

S(g)

S(l) > S(s)

(4.4)

The product of sublimation is a gas, and the precursor is a solid.
Clearly, the product has greater entropy than the starting material, so S increases during sublimation. The process is spontaneous because S is positive.

In a bottle of iodine, the space above the solid I2 always shows a slight purple hue, indicating the presence of iodine vapour.

THE DIRECTION OF PHYSICOCHEMICAL CHANGE: ENTROPY

135

SAQ 4.1 By inspection alone, decide whether the condensation of water,
H2 O(g) → H2 O(l) is spontaneous or not.

Worked Example 4.3 Now consider the chemical process
SOCl2(l) + H2 O(g) − − 2HCl(g) + SO2(g)
−→
The reaction occurs spontaneously in the laboratory without recourse to heating or catalysis. The sight of ‘smoke’ above a beaker of SOCl2(l) is ample proof of reaction spontaneity.
We see that the reaction in Equation (4.5) consumes 1 mol of gas
(i.e. water vapour) and 1 mol of liquid, and generates 3 mol of gas.
There is a small change in the number of moles: principally, the amount of gas increases. As was seen above, the entropy of a gas is greater than its respective liquid, so we see a net increase in the entropy of the reaction, making S positive.

(4.5)

This increase in the entropy S of a gas explains why an open beaker of thionyl chloride SOCl2 in the laboratory appears to be
‘smoking’.

Worked Example 4.4 By inspection alone, decide whether the formation of ammonia by the Haber process (Equation (4.6)) is spontaneous or not.
N2(g) + 3H2(g) − − 2NH3(g)
−→

(4.6)

All the species in Equation (4.6) are gases, so we cannot use the simple method of looking to see the respective phases of reactants and products (cf. Equation (4.4)).
But we notice the consumption of 4 mol of reactant to form 2 mol of product. As a crude generalization, then, we start by saying, ‘4 mol of energetic disorder are consumed during the process and 2 mol of energetic disorder are formed’. Next, with Equation (4.1) before us, we suggest the overall, crude entropy change S is roughly −2 mol of disorder per mole of reaction, so the amount of disorder decreases. We suspect the process will not be spontaneous, because S is negative.
In fact, we require heating to produce ammonia by the Haber process, so the reaction is definitely not spontaneous.
SAQ 4.2 By inspection alone, decide whether the oxidation of sulphur dioxide is thermodynamically spontaneous or not. The stoichiometry of the reaction is 1/2O2(g)
+ SO2(g) → SO3(g) .

The word ‘stoichiometry’ comes from the
Greek stoicheion, meaning ‘a part’.

Worked Example 4.5 By inspection alone, decide whether the reaction Cl2(g) + F2(g) →
2FCl(g) should occur spontaneously.
Occasionally we need to be far subtler when we look at reaction spontaneity. The reaction here involves two molecules of diatomic gas reacting to form two molecules of a different diatomic gas. Also, there is no phase change during reaction, nor any change in the numbers of molecules, so any change in the overall entropy is likely to be slight.

136

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Before we address this reaction, we need to emphasize how all these are equilibrium reactions: at completion, the reaction vessel contains product as well as unconsumed reactants. In consequence, there is a mixture at the completion of the reaction.
This change in S arises from the mixing of the elements between the two reacting species: before reaction, all atoms of chlorine were bonded only to other chlorine atoms in elemental Cl2 . By contrast, after the reaction has commenced, a choice arises with some chlorine atoms bonded to other chlorine atoms (unreacted Cl2 ) and others attached to fluorine in the product, FCl.
In fact, the experimental value of S is very small and positive.

Aside
Entropy as ‘the arrow of time’
The idea that entropy is continually increasing led many philosophers to call entropy
‘the arrow of time’. The argument goes something like this. From the Clausius equality
(see p. 142), entropy is the ratio of a body’s energy to its temperature. Entropy is generally understood to signify an inherent tendency towards disorganization.
It has been claimed that the second law means that the universe as a whole must tend inexorably towards a state of maximum entropy. By an analogy with a closed system, the entire universe must eventually end up in a state of equilibrium, with the same temperature everywhere. The stars will run out of fuel. All life will cease. The universe will slowly peter out in a featureless expanse of nothingness. It will suffer a
‘heat death’.
The idea was hugely influential. For example, it inspired the poet T. S. Eliot to write his poem The Hollow Men with perhaps his most famous lines
This the way the world ends not with a bang but a whimper.

He wrote this in 1925. Eliot’s poem, in turn, inspired others. In 1927, the astronomer
Sir Arthur Eddington said that if entropy was always increasing, then we can know the direction in which time moves by looking at the direction in which it increases. The phrase ‘entropy is the arrow of time’ gripped the popular imagination, although it is rather meaningless.
In 1928, the English scientist and idealist Sir James Jean revived the old ‘heat death’ argument, augmented with elements from Einstein’s relativity theory: since matter and energy are equivalents, he claimed, the universe must finally end up in the complete conversion of matter into energy:
The second law of thermodynamics compels materials in the universe to move ever in the same direction along the same road which ends only in death and annihilation. THE DIRECTION OF PHYSICOCHEMICAL CHANGE: ENTROPY

137

Why does crystallization of a solute occur?
Thermodynamic systems and universes
Atoms or ions of solute leave solution during the process of crystallization to form a regular repeat lattice.
The extent of solute disorder is high before crystallization, because each ion or molecule resides in solution, and thereby experiences the same freedom as a molecule of liquid. Conversely, The extent of solute disorder decreases durthe extent of disorder after crystallization will inevitably be much ing crystallization. smaller, since solute is incorporated within a solid comprising a regular repeat lattice.
The value of S can only be negative because the symbol ‘ ’ means ‘final state minus initial state’, and the extent of disorder during crystallization clearly follows the order ‘solute disorder(initial) > solute disorder(final) ’. We see how the extent of disorder in the solute decreases during crystallization in consequence of forming a lattice and, therefore, do not expect crystallization to be a spontaneous process.
But crystallization does occur, causing us to ask, ‘Why does crystallization occur even though S for the process is negative?’
To answer this question, we must consider all energetic consid- A process is thermodynamically spontaneous erations occurring during the process of crystallization, possibly only if the ‘overall’ value including phenomena not directly related to the actual processes of S is positive. inside the beaker.
Before crystallization, each particle of solute is solvated. As a simple example, a chloride ion in water is attached to six water molecules, as
[Cl(H2 O)6 ]− . Being bound to a solute species limits the freedom of solvent molecules, that is, when compared with free, unbound solvent.
Crystallization releases these six waters of solvation; see Figure 4.2:
[Cl(H2 O)6 ]− (aq) − → Cl− (in


H2O
H2O

+

H2O
H2O

Na

H2O

solid lattice)

+ 6H2 O(free,

+

Na

not solvating)

(4.7)

+ 6H2O(free)

H2O

Mobile aquo ion

Ion immobilized within a 3-D repeat lattice

Mobile water molecules Figure 4.2 Schematic representation of a crystallization process. Each solvated ion, here Na+ , releases six waters of solvation while incorporating into its crystal lattice. The overall entropy of the thermodynamic universe increases by this means

138

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

The word ‘universe’ in this context is completely different from a ‘universe’ in astronomy, so the two should not be confused. A thermodynamic ‘universe’ comprises both a ‘system’ and its ‘surroundings’.

We define a thermodynamic universe as
‘that volume large enough to enclose all the changes’; the size of the surroundings depends on the example. After their release from solvating this chloride ion, each water molecule has as much energetic disorder as did the whole chloride ion complex. Therefore, we expect a sizeable increase in the entropy of the solvent during crystallization because many water molecules are released.
When we look at the spontaneity of the crystallization process, we need to consider two entropy terms: (i) the solute (which decreases during crystallization) and (ii) the concurrent increase as solvent is freed. In summary, the entropy of the solute decreases while the entropy of the solvent increases.
The crystallization process involves a system (which we are interested in) and the surroundings. In terms of the component entropies in this example, we say S(system) is the entropy of the solute crystallizing and that S(surroundings) represents the entropy change of the solvent molecules released.
We call the sum of the system and its surroundings the thermodynamic universe (see Figure 4.3). A thermodynamic universe is described as ‘that volume large enough to enclose all the thermodynamic changes’. The entropy change of the thermodynamic universe during crystallization is S(total) , which equates to
S(total) =

S(system) +

S(surroundings)

(4.8)

The value of S(system) is negative in the example of crystallization. Accordingly, the value of S(surroundings) must be so much larger than S(system) that S(total) becomes positive. The crystallization is therefore spontaneous.
Thermodynamic
universe

Surroundings

System

Figure 4.3 We call the sum of the system and its surroundings the ‘thermodynamic universe’.
Energy is exchanged between the system and its surroundings; no energy is exchanged beyond the surrounds, i.e. outside the boundaries of the thermodynamic universe. Hence, the definition ‘a universe is that volume large enough to enclose all the thermodynamic changes’

THE TEMPERATURE DEPENDENCE OF ENTROPY

This result of S(total) being positive helps explain how considering the entropy of a system’s surroundings can obviate the apparent problems caused by only considering the processes occurring within a thermodynamic system. It also explains why crystallization is energetically feasible.
The concept of a thermodynamic system is essentially macroscopic, and assumes the participation of large numbers of molecules. Indeed, the word ‘system’ derives from the Greek sustema, meaning to set up, to stay together.

4.2

139

As a new criterion for reaction spontaneity, we say S(total) must be positive. We must consider the surroundings if we are to understand how the overall extent of energetic disorder increases during a process. The temperature dependence of entropy
Why do dust particles move more quickly by Brownian motion in warm water?
Entropy is a function of temperature
Brownian motion is the random movement of small, solid particles sitting on the surface of water. They are held in position by the surface tension γ of the meniscus.
When looking at the dust under a microscope, the dust particles appear to ‘dance’ and move randomly. But when the water is warmed, the particles, be they chalk or house dust, move faster than on cold water.
The cause of the Brownian motion is movement of water molecules, several hundred of which ‘hold’ on to the underside of each ‘Brownian motion’ is a dust particle by surface tension. These water molecules move and macroscopic observajostle continually as a consequence of their own internal energy. tion of entropy.
Warming the water increases the internal energy, itself causing the molecules to move faster than if the water was cool.
The faster molecules exhibit a greater randomness in their mo- Entropy is a function of tion than do slower molecules, as witnessed by the dust particles, temperature. which we see ‘dancing’ more erratically. The Brownian motion is more extreme. The enhanced randomness is a consequence of the water molecules having higher entropy at the higher temperature. Entropy is a function of temperature.

Why does the jam of a jam tart burn more than does the pastry?
Relationship between entropy and heat capacity
When biting into a freshly baked jam tart, the jam burns the tongue but the pastry
(which is at the same temperature) causes relatively little harm. The reason why the jam is more likely to burn is its higher ‘heat capacity’ C.

140

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Sections 3.1 and 3.2 describe heat capacity and explain how it may be determined at constant pressure Cp or at constant volume
CV . Most chemists need to make calculations with Cp , which represents the amount of energy (in the form of heat) that can be stored within a substance – the measurement having been performed at constant pressure p. For example, the heat capacity of solid water
(ice) is 39 J K−1 mol−1 . The value of Cp for liquid water is higher,
−1
−1 at 75 J K mol , so we store more energy in liquid water than when it is solid; stated another way, we need to add more energy to H2 O(l) if its temperature is to increase. Cp for steam (H2 O(g) ) is 34 J K−1 mol−1 . Cp for solid sucrose (II) – a major component of any jam – is significantly higher at 425 J K−1 mol−1 .

The heat capacity
Cp(ice) = 39 J K−1 mol−1 tells us that adding 39 J of energy increases the temperature of 1 mol of water by 1 K.

H

OH
H

OH

O

OH

HH

OH

CH2OH

O

H OH
H

CH2OH
OH

H
(II)

The heat capacity of a liquid is always greater than the heat capacity of the respective solid because the liquid, having a greater amount of energetic disorder, has a greater entropy according to
S = S2 − S1 =

T2
T1

Cp dT T

(4.9)

More energy is ‘stored’ within a liquid than in its respective solid, as gauged by the relative values of Cp implied by the connection between the heat capacity and entropy S (of a pure material). This is to be expected from everyday experience: to continue with our simplistic example, when a freshly baked jam tart is removed from the oven, the jam burns the mouth and not the pastry, because the (liquid) jam holds much more energy, i.e. has a higher Cp than does the solid pastry, even though the two are at the same temperature. The jam, in cooling to the same temperature as the tongue, gives out more energy. The tongue cannot absorb all of this energy; the energy that is not absorbed causes other processes in the mouth, and hence the burn.

141

THE TEMPERATURE DEPENDENCE OF ENTROPY

Aside
This argument is oversimplified because it is expressed in terms of jam:
(1) Jam, in comprising mainly water and sugar, will contain more moles per gram than does the pastry, which contains fats and polysaccharides, such as starch in the flour. Jam can, therefore, be considered to contain more energy per gram from a molar point of view, without even considering its liquid state.
(2) The jam is more likely to stick to the skin than does the pastry (because it is sticky liquid), thereby maximizing the possibility of heat transferring to the skin; the pastry is flaky and/or dusty, and will exhibit a lower efficiency in transferring energy.

Worked Example 4.6 Calculate the entropy change S caused by heating 1 mol of sucrose from 360 K to 400 K, which is hot enough to badly burn the mouth. Take Cp =
425 J K−1 mol−1 .
Because the value of Cp has a constant value, we place it outside the integral, which allows us to rewrite Equation (4.9), saying
S = Cp ln

T2
T1

(4.10)

We insert data into Equation (4.10) to obtain
S = 425 J K−1 mol−1 × ln

400 K
360 K

so
S = 425 J K−1 mol−1 × ln(1.11) and S = 44.8 J K−1 mol−1
SAQ 4.3 We want to warm the ice in a freezer from a temperature of
−15 ◦ C to 0 ◦ C. Calculate the change in entropy caused by the warming
(assuming no melting occurs). Take Cp for ice as 39 J K−1 mol−1 . [Hint: remember to convert to K from ◦ C.]

Aside
Cp is not independent of temperature, but varies slightly. For this reason, the approach here is only valid for relatively narrow temperature ranges of, say, 30 K. When determining S over wider temperature ranges, we can perform a calculation with Equation (4.9)

142

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

provided that we know the way Cp varies with temperature, expressed as a mathematical power series in T . For example, Cp for liquid chloroform CHCl3 is
Cp /J K−1 mol−1 = 91.47 + 7.5 × 10−2 T
Alternatively, because Equation (4.9) has the form of an integral, we could plot a graph of Cp ÷ T (as y) against T (as x) and determine the area beneath the curve. We would need to follow this approach if Cp ÷ T was so complicated a function of T that we could not describe it mathematically.

Justification Box 4.1
Entropy is the ratio of a body’s energy to its temperature according to the Clausius equality (as defined in the next section). For a reversible process, the change in entropy is defined by dq dS =
(4.11)
T where q is the change in heat and T is the thermodynamic temperature. Multiplying the right-hand side of Equation (4.11) by dT /dT (which clearly equals one), yields dS =

dT dq ×
T
dT

(4.12)

If no expansion work is done, we can safely assume that q = H . Substituting H for q, and rearranging slightly yields dS =

dH dT ×

1 dT T

(4.13)

where the term in brackets is simply Cp . We write dS =

Cp dT T

(4.14)

Solution of Equation (4.14) takes two forms: (a) the case where Cp is considered not to depend on temperature (i.e. determining the value of S over a limited range of temperatures) and (b) the more realistic case where Cp is recognized as having a finite temperature dependence.

(a) Cp is independent of temperature (over small temperature ranges).
S2
S1

dS = Cp

T2
T1

1 dT T

(4.15)

So
S = S2 − S1 = Cp [ln T ]T2
T1

(4.16)

THE TEMPERATURE DEPENDENCE OF ENTROPY

143

and hence
S = Cp ln

T2
T1

(4.17)

(b) Cp is not independent of temperature (over larger temperature ranges). We employ a similar approach to that above, except that Cp is incorporated into the integral, yielding
S2
S1

dS =

T2
T1

Cp dT T

(4.18)

which, on integration, yields Equation (4.9).

Worked Example 4.7 What is the increase in entropy when warming 1 mol of chloroform (III) from 240 K to 330 K? Take the value of Cp for chloroform from the Aside box on p. 142.
H
Cl
Cl Cl
(III)

We start with Equation (4.9), retaining the position of Cp within the integral; inserting values:
S = S330

K − S240

K

=

330 K
240 K

91.47 + 7.5 × 10−2 T dT T

Rearranging:
S=

330 K
240 K

91.47
+ 7.5 × 10−2 dT
T

Performing the integration, we obtain
S = [91.47 ln T ]330
240

K
K

+ 7.5 × 10−2 [T ]330
240

K
K

Then, we insert the variables:
S = 91.47 ln to yield so 330 K
240 K

+ 7.5 × 10−2 (330 K − 240 K)

S = (29.13 + 6.75) J K−1 mol−1
S = 35.9 J K−1 mol−1

The ‘T ’ on top and bottom cancel in the second term within the integral. 144

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

SAQ 4.4 1 mol of oxygen is warmed from 300 K to 350 K. Calculate the associated rise in entropy S if Cp(O2 ) /J K−1 mol−1 = 25.8 + 1.2 × 10−2 T /K.

4.3

Introducing the Gibbs function
Why is burning hydrogen gas in air (to form liquid water) a spontaneous reaction?
Reaction spontaneity in a system

The ‘twin’ subscript of ‘l and g’ arises because the reaction in Equation (4.19) is so exothermic that most of the water product will be steam.

Equation (4.19) describes the reaction occurring when hydrogen gas is burnt in air:

O2(g) + 2H2(g) − → 2H2 O(l

and g)

(4.19)

We notice straightaway how the number of moles decreases from three to two during the reaction, so a consideration of the system alone suggests a non-spontaneous reaction. There may also be a concurrent phase change from gas to liquid during the reaction, which confirms our original diagnosis: we expect S to be negative, and so we predict a non-spontaneous reaction.
But after a moment’s reflection, we remember that one of the simplest tests for hydrogen gas generation in a test tube is to place a lighted splint nearby, and hear the
‘pop’ sound of an explosion, i.e. the reaction in Equation (4.19) occurs spontaneously.
The ‘system’ in this example comprises the volume within which chemicals combine. The ‘surroundings’ are the volume of air around the reaction vessel or flame; because of the explosive nature of reaction, we expect this volume to be huge. The surrounding air absorbs the energy liberated during the reaction; in this example, the energy is manifested as heat and sound. For example, the entropy of the air increases as it warms up. In fact, S(surroundings) is sufficiently large and positive that the value of S(total) is positive despite the value of S(system) being negative. So we can now explain why reactions such as that in Equation (4.19) are spontaneous, although at first sight we might predict otherwise.
But, as chemists, we usually want to make quantitative predictions, which are clearly impossible here unless we can precisely determine the magnitude of
S(surroundings) , i.e. quantify the influence of the surroundings on the reaction, which is usually not a trivial problem.

How does a reflux condenser work?
Quantifying the changes in a system: the Gibbs function
All preparative chemists are familiar with the familiar Liebig condenser, which we position on top of a refluxing flask to prevent the flask boiling dry. The evaporating

INTRODUCING THE GIBBS FUNCTION

145

solvent rises up the interior passage of the condenser from the flask, cools and thence condenses (Equation (4.20)) as it touches the inner surface of the condenser. Condensed liquid then trickles back into the flask beneath.

solvent(g) − → solvent(l) + energy

(4.20)

The energy is transferred to the glass inner surface of the condenser. We maintain a cool temperature inside the condenser by running a constant flow of water through the condenser’s jacketed sleeve. The solvent releases a large amount of heat energy as it converts back to liquid, which passes to the water circulating within the jacket, and is then swept away.
Addition of heat energy to the flask causes several physicochemical changes. Firstly, energy allows the chemical reaction to proceed, The ‘Gibbs function’ G but energy is also consumed in order to convert the liquid solvent is named after Josiah into gas. An ‘audit’ of this energy is difficult, because so much of Willard Gibbs (1839– the energy is lost to the escaping solvent and thence to the sur- 1903), a humble Amerrounding water. It would be totally impossible to account for all ican who contributed to most areas of physthe energy changes without also including the surroundings as well ical chemistry. He also as the system. had a delightful sense
So we see how the heater beneath the flask needs to provide of humour: ‘A mathenergy to enable the reaction to proceed (which is what we want to ematician may say happen) in addition to providing the energy to change the surround- anything he pleases, ings, causing the evaporation of the solvent, the extent of which but a physicist must be we do not usually want to quantify, even if we could. In short, at least partially sane’. we need a simple means of taking account of all the surroundings without, for example, having to assess their spatial extent. From the second law of thermodynamics, we write
G(system) =

H(system) − T S(system)

(4.21)

(see Justification Box 4.2) where the H, T and S terms have their usual definitions, as above, and G is the ‘Gibbs function’. G is important because its value depends only on the system and not on the surroundings. By convention, a positive value of H denotes an enthalpy absorbed by the system.
H here is simply the energy given out by the system, i.e. by the reaction, or taken into it during endothermic reactions. This energy transfer affects the energy of the surroundings, which respectively absorb or receive energy from the reaction. And the change in the energy of the surroundings causes changes in the entropy of the surroundings. In effect, we can devise a ‘words-only’ definition of the Gibbs function, saying it represents ‘The energy available for reaction (i.e. the net energy), after adjusting for the entropy changes of the surroundings’.

As well as calling G the Gibbs function, it is often called the ‘Gibbs energy’ or (incorrectly)
‘free energy’.

The Gibbs function is the energy available for reaction after adjusting for the entropy changes of the surroundings.

146

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Justification Box 4.2
The total change in entropy is S(total) , which must be positive for a spontaneous process. From Equation (4.8), we say
S(total) =

S(system) +

S(surroundings) > 0

We usually know a value for S(system) from tables. Almost universally, we do not know a value for S(surroundings) .
The Clausius equality says that a microscopic process is at equilibrium if dS = dq/T where q is the heat change and T is the thermodynamic temperature (in kelvin). Similarly, for a macroThis sign change occurscopic process, S = q/T . In a chemical reaction, ring here follows since the heat energy emitted is, in fact, the enthalpy change energy is absorbed by the surroundings if of reaction H(system) , and the energy gained by the energy has been emitsurroundings of the reaction vessel will therefore be ted by the reaction,
− H(system) . Accordingly, the value of S(surroundings) and vice versa. is − H ÷ T .
Rewriting Equation (4.8) by substituting for
S(surroundings) gives
S(total) =

S(system) −

H
T

(4.22a)

The right-hand side must be positive if the process is spontaneous, so
S(system) − or H(system)
>0
T

H(system)

T

0>

(4.22b)

S(system)

Multiplying throughout by T gives
0>

H(system) − T S(system)

(4.23)

So the compound variable H(system) − T S(system) must be negative if a process is spontaneous. This compound variable occurs so often in chemistry that we will give it a symbol of its own: G, which we call the Gibbs function. Accordingly, a spontaneous process in a system is characterized by saying,
0>

G(system)

In words, the Gibbs energy must be negative if a change occurs spontaneously.

(4.24)

INTRODUCING THE GIBBS FUNCTION

The sign of

147

G

Equation (4.24) in Justification Box 4.2 shows clearly that a process only occurs spontaneously within a system if the change in
Gibbs function is negative, even if the sign of S(system) is slightly negative or if H(system) is slightly positive. Analysing the reaction in terms of our new variable G represents a great advance: previously, we could predict spontaneity if we knew that S(total) was positive – which we now realize is not necessarily a useful criterion, since we rarely know a value for S(surroundings) . It is clear from Equation (4.21) that all three variables, G, H and S, each relate to the system alone, so we can calculate the value of G by looking up values of S and H from tables, and without needing to consider the surroundings in a quantitative way.
A process occurring in a system is spontaneous if G is negative, and it is not spontaneous if G is positive, regardless of the sign of S(system) . The size of G (which is negative) is maximized for those processes and reactions for which S is positive and which are exothermic, with a negative value of H .
Worked Example 4.8 Methanol (IV) can be prepared in the gas phase by reacting carbon monoxide with hydrogen, according to Equation (4.25). Is the reaction feasible at 298 K if H O =
−90.7 kJ mol−1 and S O = −219 J K−1 mol−1 ?
−→
CO(g) + 2H2(g) − − CH3 OH(g)

The Gibbs energy must be negative if a change occurs spontaneously.

We see the analytical power of G when we realize how its value does not depend on the thermodynamic properties of the surroundings, but only on the system.

A process occurring in a system is spontaneous if G is negative, and is not spontaneous when
G is positive.

(4.25)

H
H
H
(IV)

OH

We shall use Equation (4.21), G O = H O − T S O . Inserting values (and remembering to convert from kJ to J):
G O = (−90 700 J mol−1 ) − (298 K × −219 J K−1 mol−1 )
G O = (−90 700 + 65 262) J mol−1
G O = −25.4 kJ mol−1
This value of G O is negative, so the reaction will indeed be spontaneous in this example. This is an example of where

The Gibbs function is a function of state, so values of
GO
obtained with the van’t
Hoff isotherm (see
p. 162) and routes such as Hess’s law cycles are identical.

148

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

the negative value of H overcomes the unfavourable positive − S term. In fact, although the reaction is thermodynamically feasible, the rate of reaction (see
Chapter 8) is so small that we need to heat the reaction vessel strongly to about 550 K to generate significant quantities of product to make the reaction viable.
Some zoologists believe this reaction inspired the myth of fire-breathing dragons: some large tropical lizards have glands beside their mouths to produce both H2 S and
SO2 . Under advantageous conditions, the resultant sulphur reacts so vigorously with the air that flames form as a self-defence mechanism.

SAQ 4.5 Consider the reaction
2H2 S(g) + SO2(g) →
2H2 O(g) + 3S(s) , where all species are gaseous except the sulphur. Calculate GrO for this reaction at 298 K with the thermodynamic data below:
H2 S
−1

Sf /J K
O

−1

mol

SO2

H2 O

S

−22.2

Hf /kJ mol−1
O

−296.6

−285.8

0

205.6

247.9

70.1

31.9

[Hint: it is generally easier first to determine values of Hr and Sr by constructing separate Hess’s-law-type cycles.]
SAQ 4.6 The thermodynamic quantities transfer complex formation for the reaction

of

methyl viologen + hydroquinone − − charge-transfer complex
−→

charge(4.26)

are Hr = −22.6 kJ mol−1 and Sr = −62.1 J K−1 mol−1 at 298 K. Such values have been described as ‘typical of weak charge-transfer complex interactions’. Calculate the value of Gr .

4.4

The effect of pressure on thermodynamic variables How much energy is needed?
The Gibbs–Duhem equation
‘How much energy is needed?’ is a pointless question. It is too imprecise to be useful to anyone. The amount of energy needed will depend on how much material we wish to investigate. It also depends on whether we wish to perform a chemical reaction or a physical change, such as compression. We cannot answer the question until we redefine it.
The total amount we need to pay when purchasing goods at a shop depends both on the identity of the items we buy and how many of each. When buying sweets and

THE EFFECT OF PRESSURE ON THERMODYNAMIC VARIABLES

149

apples, the total price will depend on the price of each item, and the amounts of each that we purchase. We could write it as d(money) = (price of item 1 × number of item 1)
+ (price of item 2 × number of item 2)
While more mathematical in form, we could have rewritten Equation (4.27)
∂(money)
∂(money)
× N (1) +
× N (2) + · · ·
∂(1)
∂(2)
(4.28)
where N is merely the number of item (1) or item (2), and each bracket represents the price of each item: it is the amount of money per item. An equation like Equation (4.28) is called a total differential.
In a similar way, we say that the value of the Gibbs function changes in response to changes in pressure and temperature. We write this as
G = f (p, T )
(4.29)
d(money) =

and say G is a function of pressure and temperature.
So, what is the change in G for a single, pure substance as the temperature and pressure are altered? A mathematician would start answering this question by writing out the total differential of G: dG =

∂G
∂p

T

dp +

∂G
∂T

dT p (4.30)

(4.27)
We must use the symbol ∂ (‘curly d’) in a differential when several terms are changing.
The term in the first bracket is the rate of change of one variable when all other variables are constant.

The value of G for a single, pure material is a function of both its temperature and pressure.

The small subscripted p on the first bracket tells us the differential must be obtained at constant pressure. The subscripted T indicates constant temperature.

which should remind us of Equation (4.28). The first term on the right of Equation (4.30) is the change in G per unit change in pressure, and the subsequent dp term accounts for the actual change in pressure.
The second bracket on the right-hand side is the change in G per unit change in temperature, and the final dT term accounts for the actual change in temperature.
Equation (4.30) certainly looks horrible, but in fact it’s simply a statement of the obvious – and is directly analogous to the prices of apples and sweets we started by talking about, cf. Equation (4.27).
We derived Equation (4.30) from first principles, using pure mathematics. An alternative approach is to prepare a similar equation algebraically. The result of the algebraic derivation is the Gibbs–Duhem equation: dG = V dp − S dT

(4.31)

150

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Justification Box 4.3
We start with Equation (4.21) for a single phase, and write G = H − TS . Its differential is dG = dH − T dS − S dT

(4.32)

From Chapter 2, we recall that H = U + pV , the differential of which is dH = dU + p dV + V dp
Equation (4.35) combines the first and second laws of thermodynamics: it derives from Equation (3.5) and says, in effect, dU = dq + dw. The p dV term relates to expansion work and the T dS term relates to the adiabatic transfer of heat energy.

The Gibbs–Duhem equation is also commonly (mis-)spelt
‘Gibbs–Duheme ’.

(4.33)

Substituting for dH in Equation (4.32) with the expression for dH in Equation (4.33), we obtain dG = (dU + p dV + V dp) − T dS − S dT

(4.34)

For a closed system (i.e. one in which no expansion work is possible) dU = T dS − p dV

(4.35)

Substituting for the dU term in Equation (4.34) with the expression for dU in Equation (4.35) yields dG = (T dS − p dV ) + p dV + V dp − T dS − S dT
(4.36)
The T dS and p dV terms will cancel, leaving the
Gibbs–Duhem equation, Equation (4.31).

We now come to the exciting part. By comparing the total differential of Equation (4.30) with the Gibbs–Duhem equation in Equation (4.31) we can see a pattern emerge: dG = dG = V

∂G
∂p

dp + dp ∂G dT ∂T
−S

dT

So, by direct analogy, comparing one equation with the other, we can say
V =

∂G
∂p

(4.37)

THE EFFECT OF PRESSURE ON THERMODYNAMIC VARIABLES

and
−S =

∂G
∂T

(4.38)

151

These two equations are known as the
Maxwell relations.

Equations (4.37) and (4.38) are known as the Maxwell relations.
The second Maxwell relation (Equation (4.38)) may remind us of the form of the Clausius equality (see p. 142). Although the first Maxwell relation
(Equation (4.37)) is not intuitively obvious, it will be of enormous help later when we look at the changes in G as a function of pressure.

Why does a vacuum ‘suck’?
The value of G as a function of pressure
Consider two flasks of gas connected by a small tube. Imagine also that a tap separates them, as seen by the schematic illustration in Figure 4.4. One flask contains hydrogen gas at high pressure p, for example at 2 atm. The other has such a low pressure of hydrogen that it will be called a vacuum.
As soon as the tap is opened, molecules of hydrogen move spontaneously from the high-pressure flask to the vacuum flask. The The old dictum, ‘nature movement of gas is usually so rapid that it makes a ‘slurp’ sound, abhors a vacuum’ is not which is why we often say the vacuum ‘sucks’. just an old wives tale, it
Redistributing the hydrogen gas between the two flasks is essen- is also a manifestation tially the same phenomenon as a dye diffusing, as we discussed of the second law of at the start of this chapter: the redistribution is thermodynamically thermodynamics. favourable because it increases the entropy, so S is positive.
We see how the spontaneous movement of gas always occurs from high pressure to low pressure, and also explains why a balloon Gases move sponwill deflate or pop on its own, but work is needed to blow up taneously from high the balloon or inflate a bicycle tyre (i.e. inflating a tyre is not pressure to low. spontaneous). Before

After

Figure 4.4 Two flasks are connected by a tap. One contains gas at high pressure. As soon as the tap separating the two flasks is opened, molecules of gas move spontaneously from the flask under higher pressure to the flask at lower pressure. (The intensity of the shading represents the pressure of the gas)

152

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Why do we sneeze?
The change in Gibbs function during gas movement: gas molecules move from high pressure to low
When we sneeze, the gases contained in the lungs are ejected through the throat so violently that they can move at extraordinary speeds of well over a hundred miles an hour. One of the obvious reasons for a sneeze is to expel germs, dust, etc. in the nose – which is why a sneeze can be so messy.
But a sneeze is much more sophisticated than mere germ removal: the pressure of the expelled gas is quite high because of its speed. Having left the mouth, a partial vacuum is left near the back of the throat as a result of the Bernoulli effect described below. Having a partial vacuum at the back of the throat is thermodynamically unstable, since the pressure in the nose is sure to be higher. As in the example above, gas from a region of high pressure (the nose) will be sucked into the region of low pressure (the back of the mouth) to equalize them. The nose is unblocked during this process of pressure equilibration, so one of the major reasons why we sneeze is to unblock the nose.
It is relatively easy to unblock a nose by blowing, but a sneeze is a superb means for unblocking the nose from the opposite direction.

Aside
The Bernoulli effect
Hold two corners of a piece of file paper along its narrow side. It will droop under its own weight because of the Earth’s gravitational pull acting on it. But the paper will rise and stand out almost horizontally when we blow gently over its upper surface, as if by magic (see Figure 4.5). The paper droops before blowing. Blowing induces an additional force on the paper to counteract the force of gravity.
The air pressure above the upper side of the paper decreases because the air moves over its surface faster than the air stream running past the paper’s underside. The
Partial vaccum here causes paper to lift

Blow over face of paper from here

(a)

Paper held here

(b)

Figure 4.5 The Bernoulli effect occurs when the flow of fluid over one face of a body is greater than over another, leading to pressure inequalities. Try it: (a) hold a pace of paper in both hands, and feel it sag under its own weight. (b) Blow over the paper’s upper surface, and see it lift

THE EFFECT OF PRESSURE ON THERMODYNAMIC VARIABLES

153

disparity in air speed leads to a difference in pressure. In effect, a partial vacuum forms above the paper, which ‘sucks’ the paper upwards. This is known as the Bernoulli effect.
A similar effect enables an aeroplane to fly: the curve on a plane’s wing is carefully designed such that the pressure above the wing is less than that below. The air flows over the upper face of the wing with an increased speed, leading to a decrease in pressure.
Because the upward thrust on the underside of the wing is great (because of the induced vacuum), it counterbalances the downward force due to gravity, allowing the plane to stay airborne.

How does a laboratory water pump work?
Gibbs function of pressure change
The water pump is another example of the Bernoulli effect, and is an everyday piece of equipment in most laboratories, for example being used during B¨ chner filtration. It comprises a piece of rubber u tubing to connect the flask to be evacuated to a pump. Inside the pump, a rapid flow of water past one end of a small aperture inside the head decreases the pressure of the adjacent gas, so the pressure inside the pump soon decreases.
Gas passes from the flask to the pump where the pressure is lower. The change in Gibbs function associated with these pressure changes is given by
G = RT ln

p(final) p(initial) (4.39)

where the G term represents the change in G per mole of gas. We will say here that gas enters the pump at pressure p(final) from a flask initially at pressure p(initial) . Accordingly, since p(final) < p(initial) and the term in brackets is clearly less than one, the logarithm term is negative. G is thus negative, showing that gas movement from a higher pressure p(initial) to a lower pressure p(final) is spontaneous.
It should be clear from Equation (4.39) that gas movement in the opposite direction, from low pressure (p(final) ) to high (p(initial) ) would cause G to be positive, thereby explaining why the process of gas going from low pressure to high never occurs naturally.
Stated another way, compression can only occur if energy is put into the system; so, compression involves work, which explains why pumping up a car tyre is difficult, yet the tyre will deflate of its own accord if punctured.

These highly oversimplified explanations ignore the effects of turbulent flow, and the formation of vortices.

The minimum pressure achievable with a water pump equals the vapour pressure of water, and has a value of about 28 mmHg.
G is negative for the physical process of gas moving from higher to lower pressure.
Since Equation (4.39) relies on a ratio of pressures, we say that a gas moves from ‘higher’ to ‘lower’, rather than
‘high’ to ‘low’.
Compression involves work. 154

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Equation (4.39) involves a ratio of pressures, so, although mmHg (millimetres of mercury) is not an SI unit of pressure, we are permitted to use it here.

Worked Example 4.9 The pressure inside a water pump is the same as the vapour pressure of water (28 mmHg). The pressure of gas inside a flask is the same as atmospheric pressure (760 mmHg). What is the change in Gibbs function per mole of gas that moves? Take T =
298 K.
Inserting values into Equation (4.39) yields

The pressure of vapour above a boiling liquid is the same as the atmospheric pressure.

G = 8.414 J K−1 mol−1 × 298 K × ln

28 mmHg
760 mmHg

G = 2477 J mol−1 × ln(3.68 × 10−2 )
G = 2477 J mol−1 × (−3.301)
G = −8.2 kJ mol−1

SAQ 4.7 A flask of methyl-ethyl ether (V) is being evaporated. Its boiling temperature is 298 K (the same as room temperature) so the vapour pressure of ether above the liquid is the same as atmospheric pressure,
i.e. at 100 kPa. The source of the vacuum is a water pump, so the pressure is the vapour pressure of water, 28 mmHg.

CH2
CH3

CH3
O

(V)
(1)

Convert the vacuum pressure p(vacuum) into an SI pressure, remembering that 1 atm = 101 325 kPa = 760 mmHg.

(2)

What is the molar change in Gibbs function that occurs when ether vapour is removed, i.e. when ether vapour goes from the flask at p O into the water pump at p(vacuum) ?

Justification Box 4.4
We have already obtained the first Maxwell relation (Equation (4.37)) by comparing the
Gibbs–Duhem equation with the total differential:
∂G
=V
∂p
We obtain the molar volume Vm as V ÷ n.

The ideal-gas equation says pV = nRT , or, using a molar volume for the gas (Equation (1.13)): pVm = RT

THE EFFECT OF PRESSURE ON THERMODYNAMIC VARIABLES

155

Substituting for Vm from Equation (4.37) into Equation (1.13) gives dG dp

p

= RT

(4.40)

And separation of the variables gives
1
dG
= RT dp p

(4.41)

so dG = RT

1 dp p

(4.42)

Integration, taking G1 at p(initial) and G2 at p(final) , yields
G2 − G1 = RT ln

p(final) p(initial) (4.43)

Or, more conveniently, if G2 is the final value of G and G1 the initial value of G, then the change in Gibbs function is
G = RT ln

p2 p1 i.e. Equation (4.39).

Aside
In practice, it is often found that compressing or decompressing a gas does not follow closely to the idealgas equation, particularly at high p or low T , as exemplified by the need for equations such as the van der
Waals equation or a virial expression. The equation above is a good approximation, though.
A more thorough treatment takes one of two courses:
(1)
(2)

Utilize the concept of virial coefficients; see
p. 57.
Use fugacity instead of pressure.

Fugacity f is defined as f =p×γ

(4.44)

The fugacity f can be regarded as an ‘effective’ pressure. The
‘fugacity coefficient’ γ represents the deviation from ideality. The value of γ tends to one as p tends to zero.
The word ‘fugacity’ comes from the Latin fugere, which means
‘elusive’ or ‘difficult to capture’. The modern word ‘fugitive’ comes from the same source.

156

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

where p is conventional pressure and γ is the fugacity coefficient representing the deviation from ideality. Values of γ can be measured or calculated.
We employ both concepts to compensate for gas non-ideality.

4.5

Thermodynamics and the extent of reaction
Why is a ‘weak’ acid weak?
Incomplete reactions and extent of reaction
As chemists, we should perhaps re-cast the question ‘Why is a ‘weak’ acid weak?’ by asking ‘How does the change in Gibbs function relate to the proportion of reactants that convert during a reaction to form products?’
An acid is defined as a proton donor within the Lowry–Brønsted theory (see
Chapter 6). Molecules of acid ionize in aqueous solution to form an anion and a proton, both of which are solvated. An acid such as ethanoic acid (VI) is said to be ‘weak’ if the extent to which it dissociates is incomplete; we call it ‘strong’ if ionization is complete (see Section 6.2).
O

H
H
H

O

H

(VI)

Ionization is, in fact, a chemical reaction because bonds break and form. Consider the following general ionization reaction:

HA + H2 O − → H3 O+ + A−

(4.45)

We give the extent of the reaction in Equation (4.45) the Greek symbol ξ . It should be clear that ξ has a value of zero before the reaction commences. By convention, we say that ξ = 1 mol if the reaction goes to completion. The value of ξ can take any value between these two extremes, its value increasing as the reaction proceeds. A reaction going to completion only stops when no reactant remains, which we define as ξ having a value of 1 mol, although such a situation is comparatively rare except in inorganic redox reactions. In fact, to an excellent approximation, all preparative organic reactions fail to reach completion, so 0 < ξ < 1.
The value of ξ only stops changing when the reaction stops, although the rate at which ξ changes belongs properly to the topic of kinetics (see Chapter 8). We say

We give the Greek symbol ξ (‘xi’) to the extent of reaction. ξ is commonly mispronounced as ‘exeye’.

THERMODYNAMICS AND THE EXTENT OF REACTION

Gibbs function G

it reaches its position of equilibrium, for which the value of ξ has its equilibrium value ξ(eq) . We propose perhaps the simplest of the many possible definitions of equilibrium: ‘after an initial period of reaction, no further net changes in reaction composition occur’.
So when we say that a carboxylic acid is weak, we mean that ξ(eq) is small. Note how, by saying that ξ(eq) is small at equilibrium, we effectively imply that the extent of ionization is small because
[H3 O+ ] and [A− ] are both small.
But we need to be careful when talking about the magnitudes of ξ . Consider the case of sodium ethanoate dissolved in dilute mineral acid: the reaction occurring is, in fact, the reverse of that in Equation (4.45), with a proton and carboxylate anion associating to form undissociated acid. In this case, ξ = 1 mol before the reaction occurs, and its value decreases as the reaction proceeds.
In other words, we need to define our reaction before we can speak knowledgeably about it. We can now rewrite our question, asking
‘Why is ξ
1 for a weak acid?’
The standard Gibbs function change for reaction is G O , and represents the energy available for reaction if 1 mol of reactants react until reaching equilibrium. Figure 4.6 relates G and ξ , and clearly shows how the amount of energy available for reaction
G decreases during reaction (i.e. in going from left to right as ξ increases). Stated another way, the gradient of the curve is always negative before the position of equilibrium, so any increases in ξ cause the value of G to become more negative.

∆G

157

We assume such an equilibrium is fully reversible in the sense of being dynamic – the rate at which products form is equal and opposite to the rate at which reactants regenerate via a back reaction.

Reminder: the energy released during reaction originates from the making and breaking of bonds, and the rearrangement of solvent. The full amount of energy given out is H O , but the net energy available is less that H O , being
HO − T SO.

O

x(eq)

0

Extent of reaction x

1

Figure 4.6 The value of the Gibbs function G decreases as the extent of reaction ξ until, at ξ(eq) , there is no longer any energy available for reaction, and G = 0. ξ = 0 represents no reaction and ξ = 1 mol represents complete reaction

158

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

The amount of energy liberated per incremental increase in reaction is quite large at the start of reaction, but decreases until, at equilibrium, a tiny increase in the extent of reaction would not change G(total) . The graph has reached a minimum, so the gradient at the bottom of the trough is zero.
The minimum in the graph of G against ξ is the reaction’s position of equilibrium – we call it ξ(eq) . The maximum amount of energy has already been expended at equilibrium, so G is zero.
Any further reaction beyond ξ(eq) would not only fail to liberate any further energy, but also would in fact consume energy (we would start to go ‘uphill’ on the righthand side of the figure). Any further increment of reaction would be characterized by G > 0, implying a non-spontaneous process, which is why the reaction stops at ξ(eq) .

Why does the pH of the weak acid remain constant?
The law of mass action and equilibrium constants
The amount of ethanoic acid existing as ionized ethanoate anion and solvated proton is always small (see p. 253). For that reason, the pH of a solution of weak acid is always higher than a solution of the same concentration of a strong acid. A na¨ve ı view suggests that, given time, all the undissociated acid will manage to dissociate, with the dual effect of making the acid strong, and hence lowering the pH.
We return to the graph in Figure 4.6 of Gibbs function (as y) against extent of reaction ξ (as x). At the position of the minimum, the amounts of free acid and ionized products remain constant because there is no longer any energy available for reaction, as explained in the example above.
The fundamental law of chemical equilibrium is the law of mass action, formulated in
1864 by Cato Maximilian Guldberg and Peter Waage. It has since been redefined several times. Consider the equilibrium between the four chemical species A, B, C and D: aA + bB = cC + dD

(4.46)

where the respective stoichiometric numbers are −a, −b, c and d. The law of mass action states that, at equilibrium, the mathematical ratio of the concentrations of the two reactants [A]a × [B]b and the product of the two product concentrations
[C]c × [D]d , is equal. We could, therefore, define one of two possible fractions:
[A]a [B]b
[C]c [D]d

or

[C]c [D]d
[A]a [B]b

(4.47)

This ratio of concentrations is called an equilibrium constant, and is symbolized as K.
The two ratios above are clearly related, with one being the reciprocal of the other.
Ultimately, the choice of which of these two we prefer is arbitrary, and usually relates to the way we write Equation (4.46). In consequence, the way we write this ratio is dictated by the sub-discipline of chemistry we practice. For example, in acid–base

THERMODYNAMICS AND THE EXTENT OF REACTION

159

chemistry (see Chapter 6) we write a dissociation constant, but in complexation equilibria we write a formation constant.
In fact, an equilibrium constant is only ever useful when we have carefully defined the chemical process to which it refers.

The reaction quotient
It is well known that few reactions (other than inorganic redox reactions) ever reach completion. The value of ξ(eq) is always less than one.
The quotient of products to reactants during a reaction is
Q=

[products]ν
[reactants]ν

(4.48)

The mathematical symbol means the ‘pi product’, meaning the terms are multiplied together, so (2, 3, 4) is 2 × 3 × 4 = 24.

which is sometimes called the reaction quotient. The values of ν are the respective stoichiometric numbers. The mathematical value of Q increases continually during the course of reaction because of the way it relates to concentrations during reaction. Initially, the reaction commences with a value of Q = 0, because there is no product (so the numerator is zero).

Aside
A statement such as ‘K = 0.4 mol dm−3 ’ is wrong, although we find examples in a great number of references and textbooks. We ought, rather, to say K = 0.4 when the equilibrium constant is formulated (i) in terms of concentrations, and (ii) where each concentration is expressed in the reference units of mol dm−3 . Equilibrium constants such as Kc or Kp are mere numbers.

Why does the voltage of a battery decrease to zero?
Relationships between K and

GO

The voltage of a new torch battery (AA type) is about 1.5 V. After the battery has powered the torch for some time, its voltage drops, which we see in practice as the light beam becoming dimmer. If further power is withdrawn indefinitely then the voltage from the battery eventually drops to zero, at which point we say the battery is ‘dead’ and throw it away.
A battery is a device for converting chemical energy into electrical energy (see p. 344), so the discharging occurs as a consequence of chemical reactions inside the battery. The reaction is complete

The battery produces
‘power’ W (energy per unit time) by passing a current through a resistor. The resister in a torch is the bulb filament. 160

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

when the battery is ‘dead’, i.e. the reaction has reached its equilibrium extent of reaction ξ(eq) .
The battery voltage is proportional to the change in the Gibbs function associated with the battery reaction, call it G(battery) .
Therefore, we deduce that G(battery) must decrease to zero because the battery voltage drops to zero. Figure 4.7 shows a graph of battery voltage (as y) against time of battery discharge (as x); the time of discharge is directly analogous to extent of reaction ξ . Figure 4.7 is remarkably similar to the graph of G against ξ in Figure 4.6.
The relationship between the energy available for reaction Gr and the extent of reaction (expressed in terms of the reaction quotient Q) is given by

The relationship between G and voltage is discussed in
Section 7.1.

Gr =

GrO + RT ln Q

(4.49)

where G is the energy available for reaction during chemical changes, and GrO is the standard change of Gibbs function G O , representing the change in Gibbs function from ξ = 0 to ξ = ξ(eq) .
Equation (4.49) describes the shape of the graph in Figure 4.6.
Before we look at Equation (4.46) in any quantitative sense, we
Gr = 0 is one of the note that if RT ln Q is smaller than GrO , then Gr is positive. best definitions of equilibrium.
The value of Gr only reaches zero when GrO is exactly the same as RT ln Q. In other words, there is no energy available for reaction when Gr = 0: we say the system has ‘reached equilibrium’. In fact, Gr = 0 is one of the best definitions of equilibrium.
In summary, the voltage of the battery drops to zero because the value of Gr is zero, which happened at ξ = ξ(eq) .

Battery emf vs ELi +,Li

0.8
0.6
0.4
0.2
0
10

20

30

40

50

60

Extent of discharge z

Figure 4.7 Graph of battery emf (as y) against extent of discharge (as x). Note the remarkable similarity between this figure and the left-hand side of Figure 4.6, which is not coincidental because emf ∝ G, and extent of discharge is proportional to ξ . The trace represents the ninth discharge of a rechargeable lithium–graphite battery, constructed with a solid-state electrolyte of polyethylene glycol containing LiClO4 . The shakiness of the trace reflects the difficulty in obtaining a reversible measurement. Reprinted from S. Lemont and D. Billaud, Journal of Power Sources 1995; 54: 338.
Copyright  1995, with permission from Elsevier

THERMODYNAMICS AND THE EXTENT OF REACTION

161

Justification Box 4.5
Consider again the simple reaction of Equation (4.46): aA + bB = cC + dD
We ascertain the Gibbs energy change for this reaction. We start by saying
G=

νG(products) −

νG(reactants)

where ν is the respective stoichiometric number; so
G = cGC + dGD − aGA − bGB

(4.50)

From an equation like Equation (4.43), G = G O + RT ln(p/p O ), so each G term in
Equation (4.50) may be converted to a standard Gibbs function by inserting a term like
Equation (4.43): pC G = cGC + cRT ln
O

−aRT ln

pA p O

p

pD

+ dGD + dRT ln
O

O

pO

− aGA

O

pB

O
− bGB − bRT ln

(4.51)

pO

G O by saying

We can combine the G O terms as

O
O
O
O
G O = cGC + dGD − aGA − bGB

(4.52)

So Equation (4.51) simplifies to become:
G=

G O + cRT ln
− bRT ln

pC p O

+ dRT ln

pD p O

− aRT ln

pB

pA pO (4.53)

pO

Then, using the laws of logarithms, we can simplify further:
G=

G O + RT ln

(pC /p O )c (pD /p O )d
(pA /p O )a (pB /p O )b

The bracketed term is the reaction quotient, expressed in terms of pressures, allowing us to rewrite the equation in a less intimidating form of Equation (4.49):
Gr =

GrO + RT ln Q

A similar proof may be used to derive an expression relating to G O and Kc .

(4.54)

We changed the positioned of each stoichiometric number via the laws of logarithms, saying b × ln a = ln ab .

162

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Aside
A further complication arises from the G term in Equation (4.49). The diagram above is clearer than the derivation: in reality, the differential quantity ∂G/∂ξ only corresponds to the change in Gibbs function G under certain, well defined, and precisely controlled experimental conditions.
This partial differential is called the reaction affinity in older texts and in newer texts is called the reaction free energy.

Why does the concentration of product stop changing? The van’t Hoff isotherm
Because we only ever write K (rather than
Q) at equilibrium, it is tautologous but very common to see K written as K(eq) or Ke .
The descriptor ‘isotherm’ derives from the Greek iso meaning
‘same’ and thermos meaning ‘temperature’.

Jacobus van’t Hoff was a Dutch scientist
(1852–1911). Notice the peculiar arrangement of the apostrophe, and small and capital letters in his surname. It would be beneficial if we could increase the yield of a chemical reaction by just leaving it to react longer. Unfortunately, the concentrations of reactant and product remain constant at the end of a reaction. In other words, the reaction quotient has reached a constant value.
At equilibrium, when the reaction stops, we give the reaction quotient the special name of equilibrium constant, and re-symbolize it with the letter K. The values of K and Q are exactly the same at equilibrium when the reaction stops. The value of Q is always smaller than K before equilibrium is reached, because some product has yet to form. In other words, before equilibrium, the top line of Equation (4.48) is artificially small and the bottom is artificially big.
Q and K only have the same value when the reaction has reached equilibrium, i.e. when Gr = 0. At this extent of reaction, the relationship between ξ and G O is given by the van’t Hoff isotherm:
G O = −RT ln K

(4.55)

where R and T have their usual thermodynamics meanings. The equation shows the relationship between G O and K, indicating that these two parameters are interconvertible when the temperature is held constant.

SAQ 4.8 Show that the van’t Hoff isotherm is dimensionally selfconsistent.

Worked Example 4.10 Consider the dissociation of ethanoic (acetic) acid in water to form a solvated proton and a solvated ethanoate anion, CH3 COOH + H2 O → CH3 COO−

THERMODYNAMICS AND THE EXTENT OF REACTION

+ H3 O+ . This reaction has an equilibrium constant K of about 2 ×
10−5 at room temperature (298 K) when formulated in the usual units of concentration (mol dm−3 ). What is the associated change in Gibbs function of this reaction?

Inserting values into the van’t Hoff isotherm (Equation (4.55)):

163

The correct use of the van’t Hoff isotherm necessitates using the thermodynamic temperature (expressed in kelvin).

G O = −8.314 J K−1 mol−1 × 298 K × ln(2 × 10−5 )
G O = −2478 J mol−1 × −10.8

A process occurring with a negative value of G is said to be exogenic. A process occurring with a positive value of G is said to be endogenic.

G O = +26 811 J mol−1 so G O = +26.8 kJ mol−1
Note how

G O is positive here. We say it is endogenic.

Justification Box 4.6
We start with Equation (4.49):
G=
At equilibrium, the value of

G O + RT ln Q

G is zero. Also, the value of Q is called K:
0=

G O + RT ln K

(4.56)

Subtracting the ‘−RT ln K’ term from both sides yields the van’t Hoff isotherm
(Equation (4.55)):
G O = −RT ln K
This derivation proves that equilibrium constants do exist. The value of on T , so the value of K should be independent of the total pressure.

We sometimes want to know the value of K from a value of we employ a rearranged form of the isotherm:
K = exp

− GO
RT

G O depends

G O , in which case

(4.57)

so a small change in the Gibbs function means a small value of K. Therefore, a weak acid is weak simply because G O is small.

164

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Care: we must always convert from kJ to J before calculating with
Equation (4.57).

Care: if we calculate a value of K that is extremely close to one, almost certainly we forgot to convert from kJ to J, making the fraction in the bracket a thousand times too small.

Worked Example 4.11 Consider the reaction between ethanol and ethanoic acid to form a sweet-smelling ester and water:
CH2 CH2 OH + CH3 COOH − − CH2 CH2 CO2 CH3 + H2 O (4.58)
−→
What is the equilibrium constant K at room temperature (298 K) if the associated change in Gibbs function is exogenic at −3.4 kJ mol−1 ?
Inserting values into eq. (4.57):
K = exp

−(−3400 J mol−1 )
8.314 J K−1 mol−1 × 298 K

K = exp(+1.372)
K = 3.95

A value of K greater than one corresponds to a negative value of
G O , so the esterification reaction is spontaneous and does occur to some extent without adding addition energy, e.g. by heating.
A few values of G O are summarized as a function of K in
Table 4.1 and values of K as a function of G O are listed in
Table 4.2. Clearly, K becomes larger as G O becomes more negative. Conversely, G O is positive if K is less than one.

Justification Box 4.7
We start with Equation (4.55):
G O = −RT ln K
Both sides are divided by −RT , yielding
− GO
RT

= ln K

(4.59)

Then we take the exponential of both sides to generate Equation (4.57).
Table 4.1 The relationship between G O and equilibrium constant K: values of G O as a function of K
We see from Table 4.1 that every decade increase in K causes
G O to become more negative by 5.7 kJ mol−1 per tenfold increase in K.

K
1
10
102
103
104
10−1
10−2
10−3

G O /kJ mol−1
0
−5.7
−11.4
−17.1
−22.8
+5.7
+11.4
+17.1

THERMODYNAMICS AND THE EXTENT OF REACTION

165

Table 4.2 The relationship between G O and equilibrium constant K: values of K as a function of G O
G O /kJ mol−1

K

0
−1
−10
−102
−103
+1
+10
+102

1
1.50
56.6
3.38 × 1017

0.667
0.0177
2.96 × 10−18

SAQ 4.9 What is the value of K corresponding to

O
G298

K

= −12 kJ mol−1 ?

Why do chicken eggs have thinner shells in the summer?
The effect of altering the concentration on ξ
Egg shells are made of calcium carbonate, CaCO3 . The chicken ingeniously makes shells for its eggs by a process involving carbon dioxide dissolved in its blood, yielding carbonate ions which combine chemically with calcium ions. An equilibrium is soon established between these ions and solid chalk, according to
Ca2+ (aq) + CO2− 3(aq) = CaCO3(s,

shell)

(4.60)

Unfortunately, chickens have no sweat glands, so they cannot perspire. To dissipate any excess body heat during the warm summer months, they must pant just like a dog. Panting increases the amount of carbon dioxide exhaled, itself decreasing the concentration of CO2 in a chicken’s blood. The smaller concentration [CO2− ] during
3(aq)
the warm summer causes the reaction in Equation (4.60) to shift further toward the left-hand side than in the cooler winter, i.e. the amount of chalk formed decreases.
The end result is a thinner eggshell.
Chicken farmers solve the problem of thin shells by carbonating the chickens’ drinking water in the summer. We may never know what inspired the first farmer to follow this route, but any physical chemist could have solved this problem by first writing the equilibrium constant K for Equation (4.60):
K(shell

formation)

=

[CaCO3(s) ]
[Ca

2+

2−
(aq) ][CO3(aq) ]

(4.61)

The value of K(shell formation) will not change provided the temperature is fixed. Therefore, we see that if the concentration of carbonate ions (see the bottom line) falls then

166

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

the amount of chalk on the top line must also fall. These changes must occur in tandem if K is to remain constant. In other words, decreasing the amount of CO2 in a chicken’s blood means less chalk is available for shell production. Conversely, the same reasoning suggests that increasing the concentration of carbonate – by adding carbonated water to the chicken’s drink – will increases the bottom line of Equation (4.61), and the chalk term on the top increases to maintain a constant value of K.

Le Chatelier’s principle
Arguments of this type illustrate Le Chatelier’s principle, which was formulated in
1888. It says:
Le Chatelier’s principle is named after
Henri Louis le Chatelier
(1850–1937). He also spelt his first name the
English way, as ‘Henry’.

Any system in stable chemical equilibrium, subjected to the influence of an external cause which tends to change either its temperature or its condensation (pressure, concentration, number of molecules in unit volume), either as a whole or in some of its parts, can only undergo such internal modifications as would, if produced alone, bring about a change of temperature or of condensation of opposite sign to that resulting from the external cause.

The principle represents a kind of ‘chemical inertia’, seeking to minimize the changes of the system. It has been summarized as, ‘if a constraint is applied to a system in equilibrium, then the change that occurs is such that it tends to annul the constraint’.
It is most readily seen in practice when:
(1)

(2)

4.6

The pressure in a closed system is increased (at fixed temperature) and shifts the equilibrium in the direction that decreases the system’s volume,
i.e. to decrease the change in pressure.
The temperature in a closed system is altered (at fixed pressure), and the equilibrium shifts in such a direction that the system absorbs heat from its surroundings to minimize the change in energy.

The effect of temperature on thermodynamic variables
Why does egg white denature when cooked but remain liquid at room temperature?
Effects of temperature on

G O : the Gibbs–Helmholtz equation

Boiling an egg causes the transparent and gelatinous albumen (‘egg white’) to modify chemically, causing it to become a white, opaque solid. Like all chemical reactions,

THE EFFECT OF TEMPERATURE ON THERMODYNAMIC VARIABLES

denaturing involves the rearrangement of bonds – in this case, of hydrogen bonds. For convenience, in the discussion below we say that bonds change from a spatial arrangement termed ‘1’ to a different spatial arrangement ‘2’.
From everyday experience, we know that an egg will not denature at room temperature, however long it is left. We are not saying here that the egg denatures at an almost infinitesimal rate, so the lack of reaction at room temperature is not a kinetic phenomenon; rather, we see that denaturation is energetically non-spontaneous at one temperature (25 ◦ C), and only becomes spontaneous as the temperature is raised above a certain threshold temperature, which we will call T(critical) (about 70 ◦ C for an egg).
The sign of the Gibbs function determines reaction spontaneity, so a reaction will occur if G O is negative and will not occur if
G O is positive. When the reaction is ‘poised’ at T(critical) between spontaneity and non-spontaneity, the value of G O = 0.
The changes to G O with temperature may be quantified with the Gibbs–Helmholtz equation:
O
O
G1
G2

=
T2
T1

HO

1
1

T2 T1

(4.62)

O where G2 is the change in Gibbs function at the temperature T2
O
and G1 is the change in Gibbs function at temperature T1 . Note how values of T must be expressed in terms of thermodynamic temperatures. H is the standard enthalpy of the chemical process or reaction, as determined experimentally by calorimetry or calculated via a Hess’s-law-type cycle.
The value of Hr for denaturing egg white is likely to be quite small, since it merely involves changes in hydrogen bonds.
For the purposes of this calculation, we say Hr has a value of
35 kJ mol−1 .
Additionally, we can propose an equilibrium constant of reaction, although we must call it a pseudo constant K(pseudo) because we cannot in reality determine its value. We need a value of K(pseudo) in order to describe the way hydrogen bonds change position during denaturing. We say that K(pseudo, 1) relates to hydrogen bonds in the pre-reaction position ‘1’ (i.e. prior to denaturing) and K(pseudo, 2) relates to the number of hydrogen bonds reoriented in the postreaction position ‘2’ (i.e. after denaturing). We will say here that
K(pseudo) is ‘1/10’ before the denaturing reaction, i.e. before boiling the egg at 298 K. From the van’t Hoff isotherm, K(pseudo) equates to a Gibbs function change of +5.7 kJ mol−1 .

167

This argument here has been oversimplified because the reaction is thermodynamically irreversible – after all, you cannot ‘unboil an egg’! Some reactions are spontaneous at one temperature but not at others.

The temperature dependence of the Gibbs function change is described quantitatively by the Gibbs–
Helmholtz equation.

The word ‘pseudo’ derives from the Greek stem pseudes meaning
‘falsehood’, which is often taken to mean having an appearance that belies the actual nature of a thing.

Denaturing albumen is an ‘irreversible’ process, yet the derivations below assume thermodynamic reversibility. In fact, complete reversibility is rarely essential; try to avoid making calculations if a significant extent of irreversibility is apparent.

168

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Worked Example 4.12 The white of an egg denatures while immersed in water boiling at its normal boiling temperature of 373 K. What is the value of G at this higher temperature? Take H = 35 kJ mol−1 .
It does not matter which temperature is chosen as T1 and which as T2 so long as T1 relates to G1 and T2 relates to G2 .

The value of G O at 298 K is +5.7 kJ mol−1 , the positive sign explaining the lack of a spontaneous reaction. Inserting values into the Gibbs–Helmholtz equation, Equation (4.62), yields
G373 K
5700 J mol−1

= +35 000 J mol−1
373 K
298 K

1
1

373 K 298 K

Note that T is a thermodynamic temperature, and is cited in kelvin. All energies have been converted from kJ mol−1 to J mol−1 .
O
G373 K
= (19.13 J K−1 mol−1 ) + (35 000 J mol−1 ) × (−6.75 × 10−4 K−1 )
373 K
O
G373 K
= (19.13 J K−1 mol−1 ) − (23.62 J K−1 mol−1 )
373 K
O
G373 K
= −4.4 J K−1 mol−1
373 K

so
O
G373

K

= −4.4 J K−1 mol−1 × 373 K

and
O
G373

K

= −1.67 kJ mol−1

O
G373 K has a negative value, implying that the reaction at this new, elevated temperature is now spontaneous. In summary, the Gibbs–Helmholtz equation quantifies a qualitative observation: the reaction to denature egg white is not spontaneous at room temperature, but it is spontaneous at elevated temperatures, e.g. when the egg is boiled in water.

SAQ 4.10 Consider the reaction in Equation (4.63), which occurs wholly in the gas phase:
−→
(4.63)
NH3 + 5 O2 − − NO + 3 H2 O
4
2
The value of GrO for this reaction is −239.9 kJ mol−1 at 298 K. If the enthalpy change of reaction Hr = −406.9 kJ mol−1 , then
(1)

Calculate the associated entropy change for the reaction in
Equation (4.63), and comment on its sign.

(2)

What is the value of the Gibbs function for this reaction when the temperature is increased by a further 34 K?

THE EFFECT OF TEMPERATURE ON THERMODYNAMIC VARIABLES

169

Justification Box 4.8
We will use the quantity ‘G ÷ T ’ for the purposes of this derivation. Its differential is obtained by use of the product rule. In general terms, for a compound function ab, i.e. a function of the type y = f (a, b): db da dy =a×
+b×
dx dx dx

(4.64)

so here d(G ÷ T )
1
dG
1
= ×
+G× − 2 dT T dT T
Note that the term

(4.65)

The function G ÷ T occurs so often in thermodynamics that we call it the Planck function. All standard signs O have been omitted for clarity dG is −S, so the equation becomes dT d(G ÷ T )
S
G
=− − 2 dT T
T

(4.66)

Recalling the now-familiar relationship G = H − TS , we may substitute for the −S term by saying
G−H
T
G−H
d(G ÷ T )
=
dT
T

(4.67)

−S =

×

1
G
− 2
T
T

(4.68)

The term in brackets on the right-hand side is then split up; so
G
d(G ÷ T )
H
G
= 2− 2− 2 dT T
T
T

(4.69)

On the right-hand side, the first and third terms cancel, yielding d(G ÷ T )
H
=− 2 dT T

(4.70)

Writing the equation in this way tells us that if we know the enthalpy of the system, we also know the temperature dependence of G ÷ T . Separating the variables and defining G1 as the Gibbs function change at
T1 and similarly as the value of G2 at T2 , yields
G2 /T2
G1 /T1

d(G/T ) = H

T2
T1



1 dT T2

(4.71)

This derivation assumes that both H and S are temperature invariant – a safe assumption if the variation between T1 and
T2 is small (say, 40 K or less).

170

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

so

G
T

G2 /T2
G1 /T1

=H

1
T

T2
T1

(4.72)

So, for a single chemical:
G1
G2

=H
T2
T1

1
1

T2
T1

(4.73)

And, for a chemical reaction we have Equation (4.62):
G2
G1

=
T2
T1

H

1
1

T2
T1

We call this final equation the Gibbs–Helmholtz equation.

At what temperature will the egg start to denature?
Reactions ‘poised’ at the critical temperature
Care: the nomenclature
T(critical) is employed in many other areas of physical chemistry
(e.g. see pp. 50 and
189).

The reaction is ‘poised’ at the critical temperature with G = 0.

If G O goes from positive to negative as the temperature alters, then clearly the value of G O will transiently be zero at one unique temperature. At this ‘point of reaction spontaneity’, the value of
G O = 0. We often call this the ‘critical temperature’ T(critical) .
The value of T(critical) , i.e. the temperature when the reaction first becomes thermodynamically feasible, can be determined approximately from
H
(4.74)
T(critical) =
S
Worked Example 4.13 At what temperature is the denaturation of egg albumen ‘poised’?

We will employ the thermodynamic data from Worked Example 4.12. Inserting values into Equation (4.74):
This method yields only an approximate value of T(critical) because S and H are themselves functions of temperature.

T(critical) =

35 000 J mol−1
H

=
= 356 K or 83 C
S
98.3 J K−1 mol−1

We deduce that an egg will start denaturing above about T =
83 ◦ C, confirming what every cook knows, that an egg cooks in boiling water but not in water that is merely ‘hot’.

THE EFFECT OF TEMPERATURE ON THERMODYNAMIC VARIABLES

171

SAQ 4.11 Water and carbon do not react at room temperature. Above what temperature is it feasible to prepare synthesis gas (a mixture of CO and H2 )? The reaction is
−→
C(s) + 2H2 O(g) − − CO(g) + H2(g)
Take

H O = 132 kJ mol−1 and

(4.75)

S O = 134 J K−1 mol−1 .

Justification Box 4.9
We start with Equation (4.21):
GO = 0 =

H −T S

When just ‘poised’, the value of G O is equal to zero. Accordingly, 0 =
Rearranging slightly, we obtain

H = T S.

H =T S

(4.76)

which, after dividing both sides by ‘ S’, yields Equation (4.74).

Why does recrystallization work?
The effect of temperature on K: the van’t Hoff isochore
To purify a freshly prepared sample, the preparative chemist will crystallize then recrystallize the compound until convinced it is pure. To recrystallize, we first dissolve the compound in hot solvent. The solubility s of the compound depends on the temperature T . The value of s is high at high temperature, but it decreases at lower temperatures until the solubility limit is first reached and then surpassed, and solute precipitates from solution (hopefully) to yield crystals.
The solubility s relates to a special equilibrium constant we call
We say the value of the ‘solubility product’ Ks , defined by
Ks = [solute](solution)

(4.77)

The [solute] term may, in fact, comprise several component parts if the solute is ionic, or precipitation involves agglomeration. This equilibrium constant is not written as a fraction because the ‘effective concentration’ of the undissolved solute [solute](s) can be taken to be unity.
Like all equilibrium constants, the magnitude of the equilibrium constant Ks depends quite strongly on temperature, according to

[solute](s) = 1 because its activity is unity; see
Section 7.3.

The word ‘isochore’ implies constant pressure, since iso is Greek for ‘same’ and the root chore means pressure.

172

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

the van’t Hoff isochore:
Ks(2)
ln
Ks(1)

=

O
− H(cryst)

R

1
1

T2 T1

(4.78)

O where R is the gas constant, H(cryst) is the change in enthalpy associated with crystallization, and the two temperatures are expressed in kelvin, i.e. thermodynamic temperatures. Worked Example 4.14 The solubility s of potassium nitrate is 140 g per 100 g of water at 70.9 ◦ C, which decreases to 63.6 g per 100 g of water at 39.9 ◦ C. Calculate the enthalpy
O
of crystallization, H(cryst) .
Strategy. For convenience, we will call the higher temperature T2 and the lower temperature T1 . (1) The van’t Hoff isochore, Equation (4.78), is written in terms of a ratio, so we do not need the absolute values. In other words, in this example, we can employ the solubilities s without further manipulation. We can dispense with the units of s for the same reason. (2) We convert the two temperatures to kelvin, for the van’t Hoff isochore requires thermodynamic temperatures, so T2 = 343.9 K and T1 = 312.0 K. (3) We insert values into the van’t Hoff isochore (Equation (4.78)):
140
ln
63.6

Note how the two minus signs on the right will cancel.

=−

O
− H(cryst)

−1

8.314 J K

−1

mol

− H(cryst)

×

1
1

343.9 K 312.0 K

O

ln(2.20) =

8.314 J K−1 mol−1 ln 2.20 = 0.7889

× (−2.973 × 10−4 K−1 )

We then divide both sides by ‘2.973 × 10−4 K−1 ’, so:
O
H(cryst)
0.7889
=
2.973 × 10−4 K−1
8.314 J K−1 mol−1

Only when the difference between T2 and
T1 is less than ca.
40 K can we assume the reaction enthalpy
H O is independent of temperature. We otherwise correct for the temperature dependence of H O with the Kirchhoff equation
(Equation (3.19)).

The term on the left equals 2.654 × 103 K. Multiplying both sides by
R then yields:
O
H(cryst) = 2.654 × 103 K × 8.314 J K−1 mol−1

so
O
H(cryst) = 22.1 kJ mol−1

SAQ 4.12 The simple aldehyde ethanal (VII) reacts with the di-alcohol ethylene glycol (VIII) to form a cyclic acetal (IX):

THE EFFECT OF TEMPERATURE ON THERMODYNAMIC VARIABLES

OH

CH3

CH2

O+

H

OH

CH3

O

CH2

CH2

H

O

CH2

(VIII)

(VII)

173

(IX)

Calculate the enthalpy change H O for the reaction if the equilibrium constant for the reaction halves when the temperature is raised from 300 K to 340 K.

Justification Box 4.10
We start with the Gibbs–Helmholtz equation (Equation (4.62)):
G2

T2
O

G1
=
T1
O

HO

1
1

T2
T1

Each value of G O can be converted to an equilibrium constant via the van’t Hoff
O
isotherm G O = −RT ln K, Equation (4.55). We say, G1 = −RT1 ln K1 at T1 ; and
O
G2 = −RT2 ln K2 at T2 .
Substituting for each value of G O yields
−RT1 ln K1
−RT2 ln K2

=
T2
T1

HO

1
1

T2
T1

(4.79)

which, after cancelling the T1 and T2 terms on the right-hand side, simplifies to
(−R ln K2 ) − (−R ln K1 ) =

HO

1
1

T2
T1

(4.80)

Next, we divide throughout by ‘−R’ to yield ln K2 − ln K1 = −

HO
R

1
1

T2
T1

(4.81)

and, by use of the laws of logarithms, ln a − ln b = ln(a ÷ b), so the left-hand side of the equation may be grouped, to generate the van’t Hoff isochore. Note: it is common
(but incorrect) to see H O written without its plimsoll sign ‘ O ’.

The isochore, Equation (4.81), was derived from the integrated form of the Gibbs–Helmholtz equation. It is readily shown that the van’t Hoff isochore can be rewritten in a slightly different form, as:

We often talk about
‘the isochore’ when we mean the ‘van’t Hoff isochore’. 174

REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

HO 1
× + constant
R
T

ln K =− y m

x

(4.82)

c

which is known as either the linear or graphical form of the equation. By analogy with the equation for a straight line (y = mx + c), a plot of ln K (as y) against 1 ÷ T
(as x) should be linear, with a gradient of − H O /R.
Worked Example 4.15 The isomerization of 1-butene (X) to form trans-2-butene (XI).
The equilibrium constants of reaction are given below. Determine the enthalpy of reaction
H O using a suitable graphical method.
(4.83)
(X)

T /K
K

686
1.72

702
1.63

(XI)

733
1.49

779
1.36

826
1.20

Strategy. To obtain H O : (1) we plot a graph of ln K (as y) against values of 1 ÷ T
(as x) according to Equation (4.82); (2) we determine the gradient; and (3) multiply the gradient by −R.
(1) The graph is depicted in Figure 4.8. (2) The best gradient is 1415 K. (3) H O =
‘gradient × −R’, so
H O = 1415 K × (−8.314 J K−1 mol−1 )
H O = −11.8 kJ mol−1
SAQ 4.13 The following data refer to the chemical reaction between ethanoic acid and glucose. Obtain H O from the data using a suitable graphical method. (Hint: remember to convert all temperatures to kelvin.)
T / ◦C
25
36
45
55

K
21.2 × 104
15.1 × 104
8.56 × 104
5.46 × 104

THE EFFECT OF TEMPERATURE ON THERMODYNAMIC VARIABLES

175

0.6

0.5

ln K

0.4

0.3

0.2

0.1
0.0012

0.0013

0.0014

0.0015

1 ÷ (T / K)

Figure 4.8 The equilibrium constant K for the isomerization of 1-butene depends on the temperature: van’t Hoff isochore plot of ln K (as y) against 1 ÷ T (as x) from which a value of H O can be calculated as ‘gradient × −R’

Aside
Writing table headers
Each term in the table for SAQ 4.13 has been multiplied by 104 , which is repetitious, takes up extra space, and makes the table look messy and cumbersome.
Most of the time, we try to avoid writing tables in this way, by incorporating the common factor of ‘×104 ’ into the header. We accomplish this by making use of the quantity calculus concept (see p. 13). Consider the second value of K. The table says that, at 36 ◦ C, K = 15.1 × 104 . If we divide both sides of this little equation by 104 , we obtain, K ÷ 104 = 15.1. This equation is completely correct, but is more usually written as 10−4 K = 15.1.
According, we might rewrite the first two lines of the table as:
T / ◦C
25

10−4 K
21.2

The style of this latter version is wholly correct and probably more popular than the style we started with. Don’t be fooled: a common mistake is to look at the table heading and say, ‘we need to multiply K by 10−4 ’. It has been already!

5

Phase equilibria
Introduction
Phase equilibrium describes the way phases (such as solid, liquid and/or gas) co-exist at some temperatures and pressure, but interchange at others.
First, the criteria for phase equilibria are discussed in terms of single-component systems. Then, when the ground rules are in place, multi-component systems are discussed in terms of partition, distillation and mixing.
The chapter also outlines the criteria for equilibrium in terms of the Gibbs function and chemical potential, together with the criteria for spontaneity.

5.1

Energetic introduction to phase equilibria
Why does an ice cube melt in the mouth?
Introduction to phase equilibria
The temperature of the mouth is about 37 ◦ C, so an overly simple explanation of why ice melts in the mouth is to say that the mouth is warmer than the transition temperature T(melt) . And, being warmer, the mouth supplies energy to the immobilized water molecules, thereby allowing them to break free from those bonds that hold them rigid. In this process, solid H2 O turns to liquid H2 O – the ice melts.
Incidentally, this argument also explains why the mouth feels cold after the ice has melted, since the energy necessary to melt the ice comes entirely from the mouth. In consequence, the mouth has less energy after the melting than before; this statement is wholly in accord with the zeroth law of thermodynamics, since heat energy travels from the hot mouth to the cold ice. Furthermore, if the mouth is considered as an adiabatic chamber (see p. 89), then the only way for the energy to be found for melting is for the temperature of the mouth to fall.

178

PHASE EQUILIBRIA

Further thermodynamic background: terminology
In the thermodynamic sense, a phase is defined as part of a chemical system in which all the material has the same composition and state. Appropriately, the word comes from the Greek phasis, meaning ‘appearance’. Ice, water and steam are the three simple phases of H2 O. Indeed, for almost all matter, the three
A phase is a composimple phases are solid, liquid and gas, although we must note that nent within a system, there may be many different solid phases possible since H2 O(s) existing in a precisely can adopt several different crystallographic forms. As a related defined physical state, example, the two stable phases of solid sulphur are its monoclinic
e.g. gas, liquid, or a and orthorhombic crystal forms. solid that has a single
Ice is a solid form of water, and is its only stable form below crystallographic form.
0 ◦ C. The liquid form of H2 O is the only stable form in the temperature range 0 < T < 100 ◦ C. Above 100 ◦ C, the normal, stable phase is gaseous water, ‘steam’. Water’s normal melting temperature T(melt) is 0 ◦ C
(273.15 K). The word ‘normal’ in this context implies ‘at standard pressure p O ’. The pressure p O has a value of 105 Pa. This temperature T(melt) is often called the melting point because water and ice coexist indefinitely at this temperature and pressure, but at no other temperature can they coexist. We say they reside together at equilibrium.
O
To melt the ice, an amount of energy equal to H(melt) must be added to overcome those forces that promote the water adopting
Concerning transia solid-state structure. Such forces will include hydrogen bonds. tions between the
Re-cooling the melted water to re-solidify it back to ice involves two phases ‘1’ and the same amount of energy, but this time energy is liberated, so
‘2’, Hess’s Law states
O
O that H(1→2) = −1 ×
H(melt) = − H(freeze) . The freezing process is often called fusion.
H(2→1) .
(Strictly, we ought to define the energy by saying that no pressure–volume work is performed during the melting and freezing processes, and that the melting and freezing processes occur without any changes in temperature.) Table 5.1 gives a few everyday examples of phase changes, together with some useful vocabulary.
Two or more phases can coexist indefinitely provided that we maintain certain conditions of temperature T and pressure p. The normal boiling temperature of water is 100 ◦ C, because this is the only temperature (at p = p O ) at which both liquid and
Table 5.1 Summary of terms used to describe phase changes
Phase transition

Name of transition

Everyday examples

Solid → gas
Solid → liquid
Liquid → gas
Liquid → solid
Gas → liquid
Gas → solid

Sublimation
Melting
Boiling or vaporization
Freezing, solidification or fusion
Condensation or liquification
Condensation

‘Smoke’ formed from dry ice
Melting of snow or ice
Steam formed by a kettle
Ice cubes formed in a fridge; hail
Formation of dew or rain
Formation of frost

ENERGETIC INTRODUCTION TO PHASE EQUILIBRIA

179

gaseous H2 O coexist at equilibrium. Note that this equilibrium is dynamic, because as liquid is converted to gas an equal amount of gas is also converted back to liquid.
However, the values of pressure and temperature at equilibrium depend on each other; so, if we change the pressure, then the tem- A phase diagram is a perature of equilibrium shifts accordingly (as discussed further in graph showing values
Section 5.2). If we plotted all the experimental values of pressure of applied pressure and and temperature at which equilibrium exists, to see the way they temperature at which affect the equilibrium changes, then we obtain a graph called a equilibrium exists. phase diagram, which looks something like the schematic graph in
Figure 5.1.
We call each solid line in this graph a phase boundary. If the values of p and T lie on a phase boundary, then equilibrium between A phase boundary is a two phases is guaranteed. There are three common phase bound- line on a phase diagram aries: liquid–solid, liquid–gas and solid–gas. The line separating representing values of the regions labelled ‘solid’ and ‘liquid’, for example, represents applied pressure and values of pressure and temperature at which these two phases coex- temperature at which equilibrium exists. ist – a line sometimes called the ‘melting-point phase boundary’.
The point where the three lines join is called the triple point, because three phases coexist at this single value of p and T . The triple point for water occurs at T = 273.16 K (i.e. at 0.01 ◦ C) and
The triple point on a p = 610 Pa (0.006p O ). We will discuss the critical point later. phase diagram repOnly a single phase is stable if the applied pressure and tem- resents the value of perature do not lie on a phase boundary, i.e. in one of the areas pressure and temperbetween the phase boundaries. For example, common sense tells is ature at which three that on a warm and sunny summer’s day, and at normal pressure, phases coexist at equithe only stable phase of H2 O is liquid water. These conditions of librium. p and T are indicated on the figure as point ‘D’.

Applied pressure

(b)

Liquid
D

Critical point Solid

Liquid

Solid
(c)

0.006 p O
(a)

Triple point

Freezer at
−5 °C

Mouth at
37 °C

Gas

273.16 K
Temperature T

Figure 5.1 Schematic phase diagram showing pressures and temperatures at which two phases are at equilibrium. Phase boundary (a) represents the equilibrium between steam and ice; boundary
(b) represents equilibrium between water and ice; and boundary (c) represents equilibrium between water and steam. The point D represents p and T on a warm, sunny day. Inset: warming an ice cube from −5 ◦ C to the mouth at 37 ◦ C at constant pressure causes the stable phase to convert from solid to liquid. The phase change occurs at 0 ◦ C at p O

180

PHASE EQUILIBRIA

We can predict whether an ice cube will melt just by looking carefully at the phase diagram. As an example, suppose we take
When labelling a phase an ice cube from a freezer at −5 ◦ C and put it straightaway in diagram, recall how the our mouth at a temperature of 37 ◦ C (see the inset to Figure 5.1). only stable phase at
The temperature of the ice cube is initially cooler than that of the high pressure and low mouth. The ice cube, therefore, will warm up as a consequence temperature is a solid; of the zeroth law of thermodynamics (see p. 8) until it reaches the a gas is most stable temperature of the mouth. Only then will it attain equilibrium. But, at low pressure and high temperature. The as the temperature of the ice cube rises, it crosses the phase boundphase within the crook ary, as represented by the bold horizontal arrow, and undergoes a of the ‘Y’ is therefore phase transition from solid to liquid. a liquid.
We know from Hess’s law (see p. 98) that it is often useful to consider (mentally) a physical or chemical change by dissecting it into its component parts. Accordingly, we will consider the melting of the ice cube as comprising two processes: warming from −5 ◦ C to 37 ◦ C, and subsequent melting at 37 ◦ C. During warming, the water crosses the phase boundary, implying that it changes from being a stable solid (when below 0 ◦ C) to being an unstable solid
(above 0 ◦ C). Having reached the temperature of the mouth at 37 ◦ C, the solid ice converts to its stable phase (water) in order to regain stability, i.e. the ice cube melts in the mouth. (It would be more realistic to consider three processes: warming to 0 ◦ C, melting at constant temperature, then
The Greek root meta warming from 0 to 37 ◦ C.) means ‘adjacent to’ or
Although the situation with melting in two stages appears a little
‘near to’. Something artificial, we ought to remind ourselves that the phase diagram is metastable is almost made up of thermodynamic data alone. In other words, it is possible stable . . . but not quite. to see liquid water at 105 ◦ C, but it would be a metastable phase,
i.e. it would not last long!

Aside
The arguments in this example are somewhat simplified.
Remember that the phase diagram’s y-axis is the applied pressure. At room temperature and pressure, liquid water evaporates as a consequence of entropy (e.g. see p. 134).
For this reason, both liquid and vapour are apparent even at s.t.p. The pressure of the vapour is known as the saturated vapour pressure (s.v.p.), and can be quite high.
The s.v.p. is not an applied pressure, so its magnitude is generally quite low. The
s.v.p. of water will certainly be lower than atmospheric pressure. The s.v.p. increases with temperature until, at the boiling temperature, it equals the atmospheric pressure.
One definition of boiling says that the s.v.p. equals the applied pressure.
The arguments in this section ignore the saturated vapour pressure.

ENERGETIC INTRODUCTION TO PHASE EQUILIBRIA

181

Why does water placed in a freezer become ice?
Spontaneity of phase changes

Molar Gibbs function Gm

It will be useful to concentrate on the diagram in Figure 5.2 when considering why a ‘phase change’ occurs spontaneously. We recall from Chapter 4 that one of the simplest tests of whether a thermodynamic event can occur is to ascertain whether the value of G is negative (in which case the change is indeed spontaneous) or positive (when the change is not spontaneous).
The graph in Figure 5.2 shows the molar Gibbs function Gm as a function of temperature. (Gm decreases with temperature because of increasing entropy.) The value of Gm for ice follows the line on the left-hand side of the graph; the line in the centre of the graph gives values of Gm for liquid water; and the line on the right represents Gm for gaseous water, i.e. steam. We now consider the process of an ice cube being warmed from below T(melt) to above it. The molar Gibbs functions of water and ice become comparable when the temperature reaches T(melt) . At T(melt) itself, the two values of Gm are the same – which is one definition of equilibrium.
The two values diverge once more above T(melt) .
Below T(melt) , the two values of Gm are different, implying that the two forms of water are energetically different. It should be clear that if one energy is lower than the other, then the lower energy form is the stablest; in this case, the liquid water has a higher value of Gm and is less stable than Remember how the solid ice (see the heavy vertical arrow, inset to Figure 5.2). Liquid symbol means ‘final water, therefore, is energetically unfavourable, and for that reason it state minus initial is unstable. To attain stability, the liquid water must release energy state’, so Gm = and, in the process, undergo a phase change from liquid to solid, Gm (final state) −
Gm (initial state) .
i.e. it freezes.

Solid

Gm (l)
Liquid

∆Gm
Gas

Tmelt

Gm (s)

Tmelt

Tboil

Temperature T

Figure 5.2 Graph of molar Gibbs function Gm as a function of temperature. Inset: at temperatures below T(melt) the phase transition from liquid to solid involves a negative change in Gibbs function, so it is spontaneous

182

PHASE EQUILIBRIA

It should be clear from the graph in Figure 5.2 that Gm is negative (as required for a spontaneous change) only if the final state is solid ice and the initial state is liquid water. This sign of
Gm is all that is needed to explain why liquid water freezes at temperatures below T(melt) .
Conversely, if an ice cube is warmed beyond T(melt) to the temperature of the mouth at 37 ◦ C, now it is the solid water that has excess energy; to stabilize it relative to liquid water at 37 ◦ C requires a different phase change to occur, this time from ice to liquid water. This argument again relies on the relative magnitudes of the molar Gibbs function, so Gm is only negative at this higher temperature if the final state is liquid and the initial state is solid.

These arguments represent a simple example of phase equilibria. This branch of thermodynamics tells us about the direction of change, but says nothing about the rate at which such changes occur. Why was Napoleon’s Russian campaign such a disaster?
Solid-state phase transitions
A large number of French soldiers froze to death in the winter of 1812 within a matter of weeks of their emperor Napoleon Bonaparte leading them into Russia. The loss of manpower was one of the principal reasons why Napoleon withdrew from the outskirts of Moscow, and hence lost his Russian campaign.
But why was so ruthless a general and so obsessively careful a tactician as Napoleon foolhardy enough to lead an unprepared army into the frozen wastes of Russia? In fact, he thought he was prepared, and his troops were originally well clothed with thick winter coats. The only problem was that, so the story goes, he chose at the last moment to replace the brass of the soldiers’ buttons with tin, to save money.
Metallic tin has many allotropic forms: rhombic white tin (also called β-tin) is stable at temperatures above 13 ◦ C, whereas the stable form at lower temperatures is cubic grey tin (also called α-tin). A transition such as tin(white) → tin(grey) is called a solid-state phase transition.
Figure 5.3 shows the phase diagram of tin, and clearly shows the transition from tin(white) to tin(grey) . Unfortunately, the tin allotropes have very different densities ρ, so ρ(tin, grey) = 5.8 g cm−3 but ρ(tin, white) =
The transition from
7.3 g cm−3 . The difference in ρ during the transition from white to white tin to grey was grey tin causes such an unbearable mechanical stress that the metal first noted in Europe often cracks and turns to dust – a phenomenon sometimes called during the Middle Ages,
e.g. as the pipes of
‘tin disease’ or ‘tin pest’. cathedral organs disThe air temperature when Napoleon entered Russia was apparintegrated, but the ently as low as −35 ◦ C, so the soldiers’ tin buttons converted from process was thought white to grey tin and, concurrently, disintegrated into powder. So, to be the work of if this story is true, then Napoleon’s troops froze to death because the devil. they lacked effective coat fastenings. Other common metals, such

ENERGETIC INTRODUCTION TO PHASE EQUILIBRIA

p

183

Solid
6 (grey) tin O

log(p /Pa)

4
Liquid tin

Solid
(white)
tin

2
0
−2
−4

Tin vapour

−6
0

200

400

600

800

1000

1200

Temperature/°C

Figure 5.3 Phase diagram of tin computed from thermodynamic data, showing the transition from grey tin from white tin at temperatures below 13 ◦ C. Note the logarithmic y-axis. At p O , T(white→grey) = 13 ◦ C, and T(melt) = 231.9 ◦ C. (Figure constructed from data published in Tin and its Alloys and Compounds, B. T. K. Barry and C. J. Thwaits, Ellis Horwood, Chichester, 1983)

as copper or zinc, and alloys such as brass, do not undergo phase changes of this sort, implying that the troops could have survived but for Napoleon’s last-minute change of button material.

The kinetics of phase changes
Like all spontaneous changes, the rate at which the two forms of tin interconvert is a function of temperature. Napoleon’s troops would have survived if they had entered
Russia in the summer or autumn, when the air temperature is similar to the phasetransition temperature. The rate of conversion would have been slower in the autumn, even if the air temperature had been slightly less than T(transition) – after all, the tin coating of a can of beans does not disintegrate while sitting in a cool cupboard! The conversion is only rapid enough to noticeably destroy the integrity of the buttons when the air temperature is much lower than T(transition) , i.e. when the difference between
T(air) and T(transition) is large.
Phase changes involving liquids and gases are generally fast, owing to the high mobility of the molecules. Conversely, while phase changes such as tin(white) → tin(grey) can and do occur in the solid state, the reaction is usually very much slower because it must occur wholly in the solid state, often causing any thermodynamic instabilities to remain ‘locked in’; as an example, it is clear from the phase diagram of carbon in Figure 5.4 that graphite is the stable form of carbon (cf. p. 109), yet the phase change carbon(diamond) → carbon(graphite) is so slow that a significant extent of conversion requires millions of years.
We consider chemical kinetics further in Chapter 8.

184

PHASE EQUILIBRIA

Diamond

Applied pressure p /p O

106
Liquid
104

102

1

Graphite

2000

Vapour

4000

6000

Temperature/K

Figure 5.4 The phase diagram of carbon showing the two solid-state extremes of diamond and graphite. Graphite is the thermodynamically stable form of carbon at room temperature and pressure, but the rate of the transition C(diamond) → C(graphite) is virtually infinitesimal

5.2

Pressure and temperature changes with a single-component system: qualitative discussion
How is the ‘Smoke’ in horror films made?
Effect of temperature on a phase change: sublimation

Horror films commonly show scenes depicting smoke or fog billowing about the screen during the ‘spooky’ bits. Similarly, smoke is also popular during pop concerts, perhaps to distract the fans from something occurring on or off stage. In both cases, it is the adding of dry ice to water that produces
Dry ice is solid carbon dioxide. the ‘smoke’.
Dry ice is carbon dioxide (CO2 ) in its solid phase. We call it
‘dry’ because it is wholly liquid-free at p O : such solid CO2 looks similar to normal ice
(solid water), but it ‘melts’ without leaving a puddle. We say it sublimes, i.e. undergoes a phase change involving direct conversion from solid to gas, without liquid forming as an intermediate phase. CO2(l) can only be formed at extreme pressures.
Solid CO2 is slightly denser than water, so it sinks when placed in a bucket of water. The water is likely to have a temperature of 20 ◦ C or so at room temperature, while typically the dry ice has a maximum temperature of ca −78 ◦ C (195 K). The stable phase at the temperature of the water is therefore gaseous CO2 . We should understand that the CO2(s) is thermodynamically unstable, causing the phase transition
CO2(s) → CO2(g) on immersion in the water.

PRESSURE AND TEMPERATURE CHANGES WITH A SINGLE-COMPONENT SYSTEM

185

Applied pressure p /p O

73

Solid

5.1

Liquid

Gas

1
−78.2

56.6
Temperature T /°C

31

Figure 5.5 Phase diagram of a system that sublimes at room temperature: phase diagram of carbon dioxide. (Note that the y-axis here is logarithmic)

Incidentally, the water in the bucket is essential for generating the effect of theatrical
‘smoke’ because the large volumes of CO2(g) entrap minute particles of water (which forms a colloid; see Chapter 10.2). This colloidal water is visible because it creates the same atmospheric condition known as fog, which is opaque.
Look at the phase diagram of CO2 in Figure 5.5, which is clearly similar in general form to the schematic phase diagram in Figure 5.1. A closer inspection shows that some features are different. Firstly, notice that the phase boundary between solid and liquid now has a positive gradient; in fact, water is almost unique in having a negative gradient for this line (see Section 5.1). Secondly, the conditions of room temperature
(T = 298 K and p = p O ) relate to conditions of the solid–gas phase boundary rather than the liquid–gas phase boundary.
By drawing a horizontal line across the figure at p = p O , we see how the line cuts the solid–gas phase boundary at −78.2 ◦ C. Below this temperature, the stable form of CO2 is solid dry ice, and CO2(g) is the stable form above it. Liquid CO2 is never the stable form at p O ; in fact, Figure 5.5 shows that CO2(l) will not form at pressures below 5.1 × p O . In other words, liquid CO2 is never seen naturally on Earth; which explains why dry ice sublimes rather than melts under s.t.p. conditions.

How does freeze-drying work?
Effect of pressure change on a phase change
Packets of instant coffee proudly proclaim that the product has been ‘freeze-dried’.
In practice, beans of coffee are ground, boiled in water and filtered to remove the depleted grounds. This process yields conventional ‘fresh’ coffee, as characterized by its usual colour and attractive smell. Finally, water is removed from the coffee solution to prepare granules of ‘instant’ coffee.
In principle, we could remove the water from the coffee by just boiling it off, to leave a solid residue as a form of ‘instant coffee’. In fact, some early varieties of instant coffee were made in just this way, but the flavour was generally unpleasant as

186

PHASE EQUILIBRIA

a result of charring during prolonged heating. Clearly, a better method of removing the water was required.
We now look at the phase diagram of water in Figure 5.6, which will help us follow the modern method of removing the water from coffee to yield anhydrous granules.
A low temperature is desirable to avoid charring the coffee. Water vapour can be removed from the coffee solution at any temperature, because liquids are always surrounded by their respective vapour. The pressure of the vapour is the saturated vapour pressure, s.v.p. The water is removed faster when the applied pressure decreases.
Again, a higher temperature increases the rate at which the vapour is removed. The fastest possible rate occurs when the solution boils at a temperature we call T(boil) .
Figure 5.6 shows the way in which the boiling temperature alters, with boiling becoming easier as the applied pressure decreases or
Freeze-drying is a the temperature increases, and suggests that the coffee solution will layman’s description, boil at a lower temperature when warmed in a partial vacuum. At and acknowledges that
1
a pressure of about 100 × p O , water is removed from the coffee external conditions by warming it to temperatures of about 30 ◦ C, when it boils. We may alter the conditions of a phase see that the coffee is dried and yet is never subjected to a high change, i.e. the drying temperature for long periods of time. process (removal of
It is clear that decreasing the external pressure makes boiling water) occurs at a easier. It is quite possible to remove the water from coffee at or temperature lower than near its freezing temperature – which explains the original name
100 ◦ C. of freeze-drying.
In many laboratories, a nomograph (see Figure 5.7) is pinned to the wall behind a rotary evaporator. A nomograph allows for a simple estimate of the boiling temperature as a function of pressure. Typically, pressure is expressed in the old-fashioned units of atmospheres (atm) or millimetres of mercury (mmHg).
1 atm = 760 mmHg. (The curvature of the nomograph is a consequence of the mathematical nature of the way pressure and temperature are related; see Section 5.2).

Applied pressure p

Atmospheric pressure Liquid

Solid

Gas

Pressure inside the freeze-dryer Temperature T

Figure 5.6 Freeze-drying works by decreasing the pressure, and causing a phase change; at higher pressure, the stable form of water is liquid, but the stable form at lower pressures is vapour.
Consequently, water (as vapour) leaves a sample when placed in a vacuum or low-pressure chamber: we say the sample is ‘freeze-dried’

PRESSURE AND TEMPERATURE CHANGES WITH A SINGLE-COMPONENT SYSTEM
.0

187

1

.02
.03
.04
.05
.06
°C
400

.08

°F

700

.1
°C
700

°F

.2
.3
.4

1200
.6

600

1100

.10

600
300

.8

1000
2

500

3

900

4

500
800
10

400
700
400

200

20
30
40

600

300

60
80
100

500

300

400

200

300

6
8

200
300

500
100

100

200

700

200

100

0

Observed boiling point

Boiling point corrected to
760 mmHg

Pressure p / mmHg

(a)

(b)

(c)

Figure 5.7 A typical nomograph for estimating the temperature at which a pure liquid boils when the pressure is decreased

This is how a boiling temperature at reduced pressure is estimated with a nomograph: place a straight ruler against the applied pressure as indicated on the curved right-hand scale (c). The ruler must also pass through the ‘normal’ boiling temperature on the middle scale (b). The reduced-pressure boiling temperature is then read off the left-hand scale (a). As an example, if the normal boiling temperature is 200 ◦ C, then the reduced boiling temperature may be halved to 100 ◦ C if the applied pressure is approximately 20 mmHg.
SAQ 5.1 A liquid has a normal boiling temperature of 140 ◦ C. Use the nomograph to estimate the applied pressure needed to decrease the boiling temperature to 90 ◦ C.

188

PHASE EQUILIBRIA

How does a rotary evaporator work?
Thermodynamics of phase changes
Rotary evaporators are a common feature in most undergraduate laboratories. Their primary purpose is to remove solvent following a reflux, perhaps before crystallization of a reaction product.
To operate the evaporator, we place the reaction solution in a round-bottomed flask
1
while the pressure inside the evaporator is decreased to about 30 × p O . The flask is then rotated. The solvent evaporates more easily at this low pressure than at p O .
The solvent removed under vacuum is trapped by a condenser and collected for easy re-use, or disposal in an environmentally sensitive way.
But molecules need energy if they are to leave the solution during boiling. The energy comes from the solution. The temperature of the solution would decrease rapidly if no external supply of energy was available, as a reflection of its depleted energy content (see p. 33). In fact, the solution would freeze during evaporation, so the rotating bulb is typically immersed in a bath of warm water.
An atmosphere of vapour always resides above a liquid, whether the liquid is pure, part of a mixture, or has solute dissolved within
Strictly, the term s.v.p. it. We saw on p. 180 how the pressure of this gaseous phase is applies to pure liquids. By using the term called its saturation vapour pressure, s.v.p. The s.v.p. increases with
s.v.p., we are implying increased temperature until, at the boiling point T(boil) , it equals the that all other comexternal pressure above the liquid. Evaporation occurs at temperaponents are wholly tures below T(boil) , and only above this temperature will the s.v.p. involatile, and the exceed p O . The applied pressure in a rotary evaporator is less than
s.v.p. relates only to p O , so the s.v.p. of the solvent can exceed the applied pressure the solvent.
(and allow the liquid to boil) at pressures lower than p O .
We see this phenomenon in a different way when we look back at the phase diagram in Figure 5.6. The stable phase is liquid before
Normal in the conapplying a vacuum. After turning on the water pump, to decrease text of phase equilibria means ‘performed at a the applied pressure, the s.v.p. exceeds p(applied) , and the solvent pressure of 1 bar, p O ’. boils at a lower pressure. The bold arrow again indicates how a phase change occurs during a depression of the external pressure.
We see how decreasing the pressure causes boiling of the solvent at a lower temperature than at its normal boiling temperature, i.e.
The rotary evaporator if the external pressure were p O . Such a vacuum distillation is is a simple example of desirable for a preparative organic chemist, because a lower boiling a vacuum distillation. temperature decreases the extent to which the compounds degrade.
Coffee, for example, itself does not evaporate even at low pressure, since it is a solid. Solids are generally much less volatile than liquids, owing to the stronger interactions between the particles. In consequence, the vapour pressure of a solid is several orders of magnitude smaller than that above a liquid.

PRESSURE AND TEMPERATURE CHANGES WITH A SINGLE-COMPONENT SYSTEM

189

How is coffee decaffeinated?
Critical and supercritical fluids
We continue our theme of ‘coffee’. Most coffees contain a large amount of the heterocyclic stimulant caffeine (I). Some people prefer to decrease the amounts of caffeine they ingest for health reasons, or they simply do not like to consume it at all, and they ask for decaffeinated coffee instead.
O

CH3

H 3C

N

O

N
CH3
(I)

The modern method of removing I from coffee resembles the operation of a coffee percolator, in which the water-soluble chemicals giving flavour, colour and aroma are leached from the ground-up coffee during constant irrigation with a stream of boiling water.
Figure 5.8 shows such a system: we call it a Soxhlet apparatus. Solvent is passed continually through a porous cup holding the ground coffee. The solvent removes the caffeine and trickles through the holes at the bottom of the cup, i.e. as a solution of caffeine. The solvent is then recycled: solvent at the bottom of the flask evaporates to form a gas, which condenses at the top of the column. This pure, clean solvent then irrigates the coffee a second time, and a third time, etc., until all the caffeine has been removed.
Water is a good choice of solvent in a standard kitchen percolator because it removes all the water-soluble components from the coffee – hence the flavour. Clearly, however, a different solvent is required if only the caffeine is to be removed. Such a solvent must be cheap, have a low boiling point to prevent charring of the coffee and, most importantly, should leave no toxic residues. The presence of any residue would be unsatisfactory to a customer, since it would almost certainly leave a taste; and there are also health and safety implications when residues persist.
The preferred solvent is supercritical CO2 . The reasons for this choice are many and various. Firstly, the CO2 is not hot (CO2 first CO2 is supercritical becomes critical at 31 ◦ C and 73 atm pressure; see Figure 5.5), so at temperatures and no charring of the coffee occurs during decaffeination. Furthermore, pressures above the at such a low temperature, all the components within the coffee that critical point. impart the flavour and aroma remain within the solid coffee – try soaking coffee beans in cold water and see how the water tastes afterwards! Caffeine is removed while retaining a full flavour.

190

PHASE EQUILIBRIA

Water out
Condenser

Cool, pure fluid Water in

Hot, pure fluid Coffee beans to be decaffeinated Fluid containing caffeine

Reservoir, where extracted caffeine accumulates
Heat

Figure 5.8 Coffee is decaffeinated by constantly irrigating the ground beans with supercritical carbon dioxide: schematic representation of a Soxhlet apparatus for removing caffeine from coffee

Secondly, solid CO2 is relatively cheap. Finally, after caffeine removal, any occluded
CO2 will vaporize from the coffee without the need to heat it or employ expensive vacuum technology. Again, we retain the volatile essential oils of the coffee. Even if some CO2 were to persist within the coffee granules, it is chemically inert, has no taste and would be released rapidly as soon as boiling water was added to the solid, decaffeinated coffee.

What is a critical or supercritical fluid?
We look once more at the phase diagram of CO2 in Figure 5.5. The simplest way of obtaining the data needed to construct such a figure would be to take a sample of
CO2 and determine those temperatures and pressures at which the liquid, solid and gaseous phases coexist at equilibrium. (An appropriate apparatus involves a robust container having an observation window to allow
We first looked at critius to observe the meniscus.) We then plot these values of p (as cal fluids on p. 50.
‘y’) against T (as ‘x’).

PRESSURE AND TEMPERATURE CHANGES WITH A SINGLE-COMPONENT SYSTEM

191

Let us consider more closely what happens as the conditions become more extreme inside the observation can. As heating proceeds, so the amount of CO2(l) converting to form gas increases. Accordingly, the amount of CO2 within the gaseous phase increases, which will cause the density ρ of the vapour to increase. Conversely, if we consider the liquid, at no time does its density alter appreciably, even though its volume decreases as a result of liquid It is impossible to disforming vapour. tinguish between the
From a consideration of the relative densities, we expect the liquid and gaseous liquid phase to reside at the bottom of the container, with the phases of CO2 at temless-dense gaseous phase ‘floating’ above it. The ‘critical’ point is peratures and presreached when the density of the gas has increased until it becomes sures at and above the the same as that of the liquid. In consequence, there is now no critical point. longer a lighter and a heavier phase, because ρ(liquid) = ρ(vapour) .
Accordingly, we no longer see a meniscus separating liquid at the The intensive propbottom of the container and vapour above it: it is impossible to see erties of the liquid a clear distinction between the liquid and gas components. We say and gas (density, heat that the CO2 is critical. capacity, etc.) become
Further heating or additional increases in pressure generate equal at the critisupercritical CO2 . The pressure and temperature at which the fluid cal point, which is first becomes critical are respectively termed T(critical) and p(critical) . the highest temperature and pressure at
Table 5.2 contains a few examples of T(critical) and p(critical) .
The inability to distinguish liquid from gaseous CO2 explains which both the liquid and gaseous phases why we describe critical and supercritical systems as fluids – they of a given compound are neither liquid nor gas. can coexist.
It is impossible to distinguish between the liquid and gaseous phases of CO2 at and above the critical point, which explains why a phase diagram has no phase boundary at temperatures and IUPAC defines superpressures above T(critical) . The formation of a critical fluid has an critical chromatograunusual corollary: at temperatures above T(critical) , we cannot phy as a separation cause the liquid and gaseous phases to separate by decreasing or technique in which increasing the pressure alone. The critical temperature, therefore, the mobile phase is represents the maximum values of p and T at which liquification kept above (or relTable 5.2 Critical constants T(critical) and p(critical) for some common elements and bi-element compounds
Substance
H2
He
O2
Cl2
CO2
SO2
H2 O
NH3

T(critical) /K
33.2
5.3
154.3
417
304.16
430
647.1
405.5

p(critical) /p O
12.97
2.29
50.4
77.1
73.9
78.7
220.6
113.0

atively close to) its critical temperature and pressure.

192

PHASE EQUILIBRIA

of the gas is possible. We say that there cannot be any CO2(l) at temperatures above
T(critical) .
Furthermore, supercritical CO2 does not behave as merely a mixture of liquid and gaseous CO2 , but often exhibits an exceptional ability to solvate molecules in a specific way. The removal of caffeine from coffee relies on the chromatographic separation of caffeine and the other organic substances in a coffee bean; supercritical fluid chromatography is a growing and exciting branch of chemistry.

5.3

Quantitative effects of pressure and temperature change for a single-component system
Why is ice so slippery?
Effect of p and T on the position of a solid–liquid equilibrium

We say something is ‘as slippery as an ice rink’ if it is has a tiny coefficient of friction, and we cannot get a grip underfoot. This is odd because the coefficient of friction µ for ice is quite high – try dragging a fingernail along the surface of some ice fresh from the ice box. It requires quite a lot of effort (and hence work) for a body to move over the surface of ice.
At first sight, these facts appear to represent a contradiction in terms. In fact, the reason why it is so easy to slip on ice is that ice usually has a thin layer of liquid water covering its surface: it is this water–ice combination that is treacherous and slippery.
But why does any water form on the ice if the weather is sufficiently cold for water to have frozen to form ice? Consider the ice directly beneath the blade on a skater’s ice-shoe in Figure 5.9: the edge of the blade is so sharp that an enormous pressure is exerted on the ice, as indicated by the grey tints. We now look at the phase diagram for water in Figure 5.10.
The sign of dp/dT for
Ice melts at 0 ◦ C if the pressure is p O (as represented by T1 and the liquid–solid line p1 respectively on the figure). If the pressure exerted on the ice on a phase diagram increases to p2 , then the freezing temperature decreases to T2 . (The is almost always posifreezing temperature decreases in response to the negative slope tive. Water is the only of the liquid–solid phase boundary (see the inset to Figure 5.10), common exception. which is most unusual; virtually all other substances show a positive slope of dp/dT .)
If the temperature T2 is lower than the freezing temperature of the water – and it usually is – then some of the ice converts to form liquid water; squeezing decreases the freezing temperature of the water. The water-on-ice beneath the skater’s blade is slippery enough to allow effortless skating.
The coefficient of friction µ (also called
‘friction factor’) is the quotient of the frictional force and the normal force. In other words, when we apply a force, is there a resistance to movement or not?

QUANTITATIVE EFFECTS OF PRESSURE AND TEMPERATURE CHANGE

193

Skater's blade
High
pressure

Water formed by pressure
Ice

Liquid
Solid

Gas

Pressure

Applied pressure p

Figure 5.9 Skaters apply an enormous pressure beneath the blades of their skates. This pressure causes solid ice to melt and form liquid water

p2 p1 T2
T1
Temperature

Temperature

Figure 5.10 Phase diagram of water. Inset: applying a high pressure from p1 (here p O ) to p2 causes the melting temperature of the ice to decrease from temperature T1 (here 0 ◦ C) to T2

What is ‘black ice’?
The Clapeyron equation
We give the name ‘black ice’ to the phenomenon of invisible ice on a road. In practice, anything applying a pressure to solid ice will cause a similar depression of the freezing temperature to that of the skater, so a car or heavy vehicle travelling over ice will also cause a momentary melting of the ice beneath its wheels. This water-on-ice causes the car to skid – often uncontrollably – and leads to many deaths every year. Such ice is particularly dangerous: whereas an ice skater wants the ice to be slippery, a driver does not.

194

PHASE EQUILIBRIA

We move from the qualitative argument that T(melt) decreases as p increases, and next look for a quantitative measure of the changes in melt temperature with pressure.
We will employ the Clapeyron equation: dp HO
=
dT
T V
In fact, it does not matter whether H relates to the direction of change of solid → liquid or of liquid → solid, provided that V relates to the same direction of change.

The molar change in volume Vm has units of m3 mol−1 . Values typically lie in the range 10−5 –10−6 m3 mol−1 .

The minus sign of Vm reflects the way water expands on freezing. This expansion explains why a car radiator cracks in cold weather (if it contains no ‘de-icer’): the water freezes and, in expanding, exerts a huge a pressure on the metal.

Care: following Hess’s law, we say:
O
O
H (melt) = − H(fusion) .

(5.1)

where T is the normal melting temperature, dT is the change in the melting temperature caused by changing the applied pressure by an amount of dp (in SI units of pascals), where 1 Pa is the pressure exerted by a force of 1 N over an area of 1 m2 . H is the enthalpy change associated with the melting of water and V is the change in volume on melting. Strictly, both H and V are molar quantities, and are often written as Hm and Vm , although the ‘m’ is frequently omitted.
The molar change in volume Vm has SI units of m3 mol−1 .
We should note how these volumes are molar volumes, so they refer to 1 mol of material, explaining why V is always very small. The value of Vm is usually about 10−6 to 10−5 m3 mol−1 in magnitude, equating to 1 to 10 cm3 mol−1 respectively.
We recall from Chapter 1 how the symbol means ‘final state minus initial state’, so a positive value of Vm during melting
(which is Vm (liquid) − Vm (solid) ) tells us that the liquid has a slightly larger volume than the solid from which it came. Vm (melt) is positive in the overwhelming majority of cases, but for water
Vm (melt) = −1.6 × 10−6 m3 mol−1 . This minus sign is extremely unusual: it means that ice is less dense than water. This explains why an iceberg floats in water, yet most solids sink when immersed in their respective liquid phases.
O
The enthalpy H(melt) is the energy required to melt 1 mol of material at constant pressure. We need to be careful when obtaining data from tables, because many books cite the enthalpy of fusion, which is the energy released during the opposite process of solidification. We do not need to worry, though, because we know
O
O from Hess’s law that H(melt) = − H(fusion) . The molar enthalpy of melting water is +6.0 kJ mol−1 .
Worked Example 5.1 Consider a car weighing 1000 kg (about
2200 lbs) parked on a sheet of ice at 273.15 K. Take the area under wheels in contact with the ice as 100 cm2 i.e. 10−2 m2 . What is the
O
new melting temperature of the ice – call it T(final) ? Take H(melt) =
−1
−1
−6
3
6.0 kJ mol and water Vm (melt) = −1.6 × 10 m mol .

QUANTITATIVE EFFECTS OF PRESSURE AND TEMPERATURE CHANGE

195

Strategy. (1) Calculate the pressure exerted and hence the pressure change. (2) Insert values into the Clapeyron equation (Equation (5.1)).
(1) The pressure exerted by the car is given by the equation
‘force ÷ area’. The force is simply the car weight expressed in newtons (N): force = 10 000 N, so we calculate the pressure exerted by the wheels as 10 000 N ÷ 10−2 m2 , which is 106 Pa. We see how a car exerts the astonishing pressure beneath its wheels of 106 Pa (about 10 bar).
(2) Before inserting values into the Clapeyron equation, we rearrange it slightly, first by multiplying both sides by
T V , then dividing both sides by H O , to give dT =

dp T V
HO

106 Pa × 273.15 K × (−1.6 × 10−6 ) m3 mol−1
6.0 × 103 J mol−1 dT = −0.07 K dT =

At sea level, a mass of 1 kg has a weight
(i.e. exerts a force) of approximately 10 N.

Notice how the freezing temperature of water decreases when a pressure is applied.
This decrease is directly attributable to the minus sign of V.

Next, we recall that the symbol ‘ ’ means ‘final state − initial state’.
Accordingly, we say
T = T(final) − T(initial) where the temperatures relate to the melting of ice. The normal melting temperature of ice T(initial) is 273.15 K. The final temperature T(final) of the ice with the car resting on it is obtained by rearranging, and saying
T + T(initial) = −0.07 K + 273.15 K
T(final) = 273.08 K
The new melting temperature of the ice T(final) is 273.08 K. Note how we performed this calculation with the car parked and immobile on the ice. When driving rather than parked, the pressure exerted beneath its wheels is actually considerably greater. Since
Equation (5.1) suggests that dp ∝ dT , the change in freezing temperature dT will be proportionately larger (perhaps as much as −3 K), so there will be a layer of water on the surface of the ice even if the ambient temperature is −3 ◦ C. Drive with care!
Care: the word ‘normal’ here is code: it means
‘at p = p O ’.

SAQ 5.2 Paraffin wax has a normal melting temperature
T(melt) of 320 K. The temperature of equilibrium is raised

196

PHASE EQUILIBRIA

by 1.2 K if the pressure is increased fivefold. Calculate it melts. Take H(melt) = 8.064 kJ mol−1 .

Vm for the wax as

Justification Box 5.1
Consider two phases (call them 1 and 2) that reside together in thermodynamic equilibrium. We can apply the Gibbs–Duhem equation (Equation 4.31) for each of the two phases, 1 and 2.
For phase 1 :

dG(1) = (Vm(1) dp) − (Sm(1) dT )

(5.2)

For phase 2 :

dG(2) = (Vm(2) dp) − (Sm(2) dT )

(5.3)

where the subscripts ‘m’ imply molar quantities, i.e. per mole of substance in each phase.
Now, because equilibrium exists between the two phases 1 and 2, the dG term in each equation must be the same. If they were different, then the change from phase 1 to phase 2 (G(2) − G(1) ) would not be zero at all points; but at equilibrium, the value of
G will be zero, which occurs when G(2) = G(1) . In fact, along the line of the phase boundary we say dG(1) = dG(2) .
In consequence, we may equate the two equations, saying:
(Vm(1) × dp) − (Sm(1) × dT ) = (Vm(2) × dp) − (Sm(2) × dT )
Factorizing will group together the two S and V terms to yield
(Sm(2) − Sm(1) ) dT = (Vm(2) − Vm(1) ) dp which, after a little rearranging, becomes dp = dT Finally, since equilibrium: GO =

Sm(1→2)
Vm(1→2)

H O − T S O (Equation (4.21)), and since
O
Hm
=
T

(5.4)
G O = 0 at

Sm(1→2)

Inserting this relationship into Equation (5.4) yields the Clapeyron equation in its familiar form.
O
Hm(1→2) dp = dT T V m(1→2)
In fact, Equation (5.4) is also called the Clapeyron equation. This equation holds for phase changes between any two phases and, at heart, quantitatively defines the phase

QUANTITATIVE EFFECTS OF PRESSURE AND TEMPERATURE CHANGE

197

boundaries of a phase diagram. For example: for the liquid → solid line

O
Hm (l→s) dp = dT T V m (l→s)

for the vapour → solid line

Hm (v→s) dp = dT T s Vm (v→s) v for the vapour → liquid line

Hm (v→s) dp = dT T lv Vm (v→s)

O

O

Approximations to the Clapeyron equation
We need to exercise a little caution with our terminology: we performed the calculation in Worked Example 5.1 with Equation (5.1) as written, but we should have written p rather than dp because 106 Pa is a very large change in pressure.
Similarly, the resultant change in temperature should have been The ‘d’ in Equation written as T rather than dT , although 0.07 K is not large. To (5.1) means an infiniaccommodate these larger changes in p and T , we ought to be tesimal change, whereas the ‘ ’ symbol rewrite Equation (5.1) in the related form:
O
Hm p ≈
T
T Vm

here means a large, macroscopic, change.

We are permitted to assume that dp is directly proportional to dT because H and
V are regarded as constants, although even a casual inspection of a phase diagram shows how curved the solid–gas and liquid–gas phase boundaries are. Such curvature clearly indicates that the Clapeyron equation fails to work except over extremely limited ranges of p and T . Why?
The Clapeyron equaO
We assumed in Justification Box 5.1 that H(melt) is indepen- tion fails to work for dent of temperature and pressure, which is not quite true, although phase changes involvthe dependence is usually sufficiently slight that we can legiti- ing gases, except mately ignore it. For accurate work, we need to recall the Kirchhoff over extremely limited ranges of p and T . equation (Equation (3.19)) to correct for changes in H .
Also, we saw on p. 23 how Boyle’s Law relates the volume of a gas to changes in the applied pressure. Similar expressions apply for liquids and solids (although such phases are usually much less It is preferable to compressible than gases). Furthermore, we assumed in the deriva- analyse the equilibtion of Equation (5.1) that Vm does not depend on the pressure ria of gases in terms changes, which implies that the volumes of liquid and solid phases of the related Claueach change by an identical amount during compression. This sius–Clapeyron equaapproximation is only good when (1) the pressure change is not tion; see Equation (5.5). extreme, and (2) we are considering equilibria for the solid–liquid

198

PHASE EQUILIBRIA

phase boundary, which describes melting and solidification. For these reasons, the
Clapeyron equation is most effective when dp is relatively small, i.e. 2–10 atm at most.
The worst deviations from the Clapeyron equation occur when one of the phases is a gas. This occurs because the volume of a gas depends strongly on temperature, whereas the volume of a liquid or solid does not. Accordingly, the value of Vm is not independent of temperature when the equilibrium involves a gas.

Why does deflating the tyres on a car improve its road-holding on ice?
The Clapeyron equation, continued
We saw from the Clapeyron equation, Equation (5.1), how the decrease in freezing temperature dT is proportional to the applied pressure dp, so one of the easiest ways of avoiding the lethal conversion of solid ice forming liquid water is to apply a smaller pressure – which will decrease dT in direct proportion.
The pressure change dp is caused by the additional weight of, for example, a car, lorry or ice skater, travelling over the surface
The pressure beneath of the ice. We recall our definition of ‘pressure’ as ‘force ÷ area’. the blades of an iceThere is rarely a straightforward way of decreasing the weight of skater’s shoe is enora person or car exerting the force, so the best way to decrease mous – maybe as much the pressure is to apply the same force but over a larger area. as 100 atm when the
An elementary example will suffice: cutting with a sharp knife skater twists and turns is easier than with a blunt one, because the active area along at speed. the knife-edge is greater when the knife is blunt, thus causing p to decrease.
In a similar way, if we deflate slightly the tyres on a car, we see the tyre bulge a little, causing it to ‘sag’, with more of the tyre in contact with the road surface.
So, although the weight of the car does not alter, deflating the tyre increases the area over which its weight (i.e. force) is exerted, with the result that we proportionately decrease the pressure.
In summary, we see that letting out some air from a car tyre decreases the value of dp, with the result that the change in melting temperature dT of the ice, as calculated with the Clapeyron equation (Equation (5.1)), also decreases, thereby making driving on ice much safer.
SAQ 5.3 A man is determined not to slip on the ice, so instead of wearing skates of area 10 cm2 he now wears snow shoes, with the underside of each sole having an extremely large area to spread his 100 kg mass
(equating to 1000 N). If the area of each snow shoe is 0.5 m2 , what is the depression of the freezing temperature of the ice caused by his walking over it?
Use the thermodynamic data for water given in Worked Example 5.1.

QUANTITATIVE EFFECTS OF PRESSURE AND TEMPERATURE CHANGE

199

Aside
If water behaved in a similar fashion to most other materials and possessed a positive value of Vm , then water would spontaneously freeze when pressure was applied, rather than solid ice melting under pressure. Furthermore, a positive value of Vm would instantly remove the problems discussed above, caused by vehicles travelling over ‘black’ ice, because the ice would remain solid under pressure; and remember that the slipperiness occurs because liquid water forms on top of solid ice.
Unfortunately, a different problem would present itself if Vm was positive! If Vm was positive, then Equation (5.1) shows that applying a pressure to liquid water would convert it to ice, even at temperatures slightly higher than 0 ◦ C, which provides a different source of black ice.

Why does a pressure cooker work?
The Clausius–Clapeyron equation
A pressure cooker is a sealed saucepan in which food cooks faster than it does in a simple saucepan – where ‘simple’, in this context means a saucepan that is open to the air. A pressure cooker is heated on top of a cooker or hob in the conventional way but, as the water inside it boils, the formation of steam rapidly causes the internal pressure to increase within its sealed cavity; see Figure 5.11. The internal pressure inside a good-quality pressure cooker can be as high 6 atm.
The phase diagram in Figure 5.12 highlights the pressure–temp- Remember that all erature behaviour of the boiling (gas–liquid) equilibrium. The equilibria are dynamic. normal boiling temperature T(boil) of water is 100 ◦ C, but T(boil) increases at higher pressures and decreases if the pressure decreases. As a simple example, a glass of water would boil instantly at the cold temperature of 3 K in the hard vacuum of deep space. The inset to Figure 5.12

Condensation

Vaporization

Heat

Figure 5.11 A pressure cooker enables food to cook fast because its internal pressure is high, which elevates the temperature at which food cooks

200

PHASE EQUILIBRIA

pO
Solid

Pressure

Applied pressure p

The normal boiling point of water

Liquid

p2 pO 100 °C

Gas

T2

Temperature
100 °C

Temperature T

Figure 5.12 Phase diagram to show how a pressure cooker works. Inset: applying a high pressure from p O to p2 causes the boiling temperature of the water to increase from temperature 100 ◦ C to T2

shows why the water inside the pressure cooker boils at a higher temperature as a consequence of the pan’s large internal pressure.
Having qualitatively discussed the way a pressure cooker facilitates rapid cooking, we now turn to a quantitative discussion.
The Clausius–ClapeyThe Clapeyron equation, Equation (5.1), would lead us to suppose ron equation quantifies the way a boiling temthat dp ∝ dT , but the liquid–gas phase boundary in Figure 5.12 perature changes as a is clearly curved, implying deviations from the equation. Therefunction of the applied fore, we require a new version of the Clapeyron equation, adapted pressure. At the boilto cope with the large volume change of a gas. To this end, we ing points of T1 and introduce the Clausius–Clapeyron equation:

T2 , the external pressures p1 and p2 are the same as the respective vapour pressures.

It does not matter which of the values we choose as ‘1’ and
‘2’ provided that T1 relates to p1 and T2 relates to p2 . It is permissible to swap T1 for T2 and p1 for p2 simultaneously, which amounts to multiplying both sides of the equation by ‘−1’.

p2 at T2 ln p1 at T1

=−

O
H(boil)

R

1
1

T2 T1

(5.5)

O where R is the familiar gas constant, and H(boil) is the enthalpy
O
of vaporization. H(boil) is always positive because energy must be put in to a liquid if it is to boil. T2 here is the boiling temperature when the applied pressure is p2 , whereas changing the pressure to p1 will cause the liquid to boil at a different temperature, T1 .
We need to understand that the Clausius–Clapeyron equation is really just a special case of the Clapeyron equation, and relates to phase changes in which one of the phases is a gas.

Worked Example 5.2 What is the boiling temperature of pure water inside a pressure cooker? Let T1 be the normal boiling temperature
T(boil) of water (i.e. 100 ◦ C, 373 K, at p O ) and let p2 of 6 × p O be the pressure inside the pan. The enthalpy of boiling water is 50.0 kJ mol−1 .

QUANTITATIVE EFFECTS OF PRESSURE AND TEMPERATURE CHANGE

In this example, it is simpler to insert values into Equation (5.5) and to rearrange later. Inserting values gives

ln

6 × pO pO =

−50 000 J mol−1
×
8.314 J K−1 mol−1

1
1

T2
373 K

We can omit the units of the two pressures on the left-hand side because Equation (5.5) is written as a ratio, so the units cancel: we require only a relative change in pressure. ln 6.0 = −6104 K ×

201

Notice how the ratio within the bracket on the left-hand side of the
Clausius–Clapeyron
equation permits us to dispense with absolute pressures.

1
1

T2
373 K

where ln 6.0 has a value of −1.79. Next, we rearrange slightly by dividing both sides by
6104 K, to yield:
1
−1.79
1
=

6104 K
T2
373 K so 1
1
= −2.98 × 10−4 K−1 +
T2
373 K and 1
= 2.38 × 10−3 K−1
T2
We obtain the temperature at which water boils by taking the reciprocal of both side.
T2 , is 420 K, or 147 ◦ C at a pressure of 6 × p O , which is much higher than the normal boiling temperature of 100 ◦ C.
SAQ 5.4 A mountaineer climbs Mount Everest and wishes to make a strong cup of tea. He boils his kettle, but the final drink tastes lousy because the water boiled at too low a temperature, itself because the pressure at the top of the mountain is only 0.4 × p O . Again taking the enthalpy of boiling the water to be 50 kJ mol−1 and the normal boiling temperature of water to be 373 K, calculate the temperature of the water as it boils at the top of the mountain.

The form of the Clausius–Clapeyron equation in Equation (5.5) is called the integrated form. If pressures are known for more than two temperatures, an alternative form may be employed: ln p = −

O
H(boil)

R

×

1
+ constant
T

(5.6)

202

PHASE EQUILIBRIA

so a graph of the form ‘y = mx + c’ is obtained by plotting ln p (as ‘y’) against
O
1/T (as ‘x’). The gradient of this Clapeyron graph is ‘− H(boil) ÷ R’, so we obtain
O
H(boil) as ‘gradient × −1 × R’.
The intercept of a Clapeyron graph is not useful; its value may
We employ the intebest be thought of as the pressure exerted by water boiling at infinite grated form of the temperature. This alternative of the Clausius–Clapeyron equation
Clausius–Clapeyron
is sometimes referred to as the linear (or graphical) form. equation when we know two temperatures and pressures, and the graphical form for three or more.

Worked Example 5.3 The Clausius–Clapeyron equation need not apply merely to boiling (liquid–gas) equilibria, it also describes sublimation equilibria (gas–solid).
Consider the following thermodynamic data, which concern the sublimation of iodine:

T(sublimation) /K p(I2 ) /Pa

270
50

280
133

290
334

300
787

310
1755

320
3722

330
7542

340
14 659

Figure 5.13 shows a plot of ln p(I2 ) (as ‘y’) against 1/T(sublimation) (as
O
‘x’). The enthalpy H(sublimation) is obtained via the gradient of the graph 62 kJ mol−1 (note the positive sign).

We obtain the value of H as ‘gradient×
−1 × R’.

−1
−2

ln(p /Pa)

−3
−4
−5
−6
−7
−8
0.0029

0.0031

0.0033

0.0035

0.0037

K /T

Figure 5.13 The linear form of the Clausius–Clapeyron equation: a graph of ln p (as ‘y’) against
O
1/T (as ‘x’) should be linear with a slope of − H(vap) ÷ R

QUANTITATIVE EFFECTS OF PRESSURE AND TEMPERATURE CHANGE

203

Aside
Why does food cook faster at higher pressures?
The process of cooking involves a complicated series of chemical reactions, each of which proceeds with a rate constant of k. When boiling an egg, for example, the ratelimiting process is denaturation of the proteins from which albumen is made. Such denaturation has an activation energy Ea of about 40 kJ mol−1 .
The rate constant of reaction varies with temperature, with k increasing as the temperature increases. k
We consider the Arrheis a function of T according to the well-known Arrhenius equation in appronius equation: priate detail in Chapk at T2 ln k at T1

Ea
=−
R

1
1

T2
T1

ter 8.

(5.7)

We saw in Worked Example 5.2 how the temperature of the boiling water increases from 100 ◦ C to 147 ◦ C in a pressure cooker. A simple calculation with the Arrhenius equation (Equation (5.7)) shows that the rate constant of cooking increases by a little over fourfold at the higher temperature inside a pressure cooker.
Boiling an egg takes about 4 min at 100 ◦ C, so boiling an egg in a pressure cooker takes about 1 min.

Justification Box 5.2
The Clapeyron equation, Equation (5.1), yields a quantitative description of a phase boundary on a phase diagram. Equation (5.1) works quite well for the liquid–solid phase boundary, but if the equilibrium is boiling or sublimation – both of which involve a gaseous phase – then the Clapeyron equation is a poor predictor.
For simplicity, we will suppose the phase change is the boiling of a liquid: liquid → gas. We must make three assumptions if we are to derive a variant that can accommodate the large changes in the volume of a gas:
Assumption 1: we assume the enthalpy of the phase change is independent of temperature and pressure. This assumption is good over limited ranges of both p and T , although note how the Kirchhoff equation (Equation (3.19)) quantifies changes in H .
Assumption 2: we assume the gas is perfect, i.e. it obeys the ideal-gas equation,
Equation (1.13), so pV = nRT or pV m = RT where Vm is the molar volume of the gas.

204

PHASE EQUILIBRIA

Assumption 3: Vm is the molar change in volume during the phase change. The value of Vm = Vm(g) − Vm(l) , where Vm(l) is typically 20 cm3 mol−1 and Vm(g) is
22.4 dm3 mol−1 (at s.t.p.), i.e. 22 400 cm3 mol−1 . In response to the vast discrepancy between Vm(g) and Vm(l) , we assume that Vm ≈ Vm(g) , i.e. that Vm(l) is negligible by comparison. This third approximation is generally good, and will only break down at very low temperatures.
First, we rewrite the Clapeyron equation in response to approximation 2:
O
Hm dp = dT T Vm(g)

Next, since we assume the gas is ideal, we can substitute for the Vm term via the ideal-gas equation, and say Vm = RT ÷ p: dp = dT O
Hm
p
×
T
RT

Next, we multiply together the two T terms, rearrange and separate the variables, to give: O
Hm
1
1
dp =
× 2 dT p R
T
Integrating with the limits p2 at T2 and p1 at T1 gives
We place the ‘ Hm ÷ R’ term outside the righthand integral because its value is constant.
O

p2 p1 1 dp = p O
Hm
R

T2
T1

1 dT T2

Subsequent integration yields
[ln p]p2 = − p1 O
Hm
1
×
R
T

T2
T1

Next, we insert limits: ln p2 − ln p1 = −

O
Hm
R

1
1

T2
T1

And, finally, we group together the two logarithmic terms to yield the Clausius–Clapeyron equation:
HO 1
1
p2
=−
− ln p1
R
T2
T1

PHASE EQUILIBRIA INVOLVING TWO-COMPONENT SYSTEMS: PARTITION

5.4

205

Phase equilibria involving two-component systems: partition
Why does a fizzy drink lose its fizz and go flat?
Equilibrium constants of partition
Drinks such as lemonade, orangeade or coke contain dissolved CO2 gas. As soon as the drink enters the warm interior of the mouth, CO2 comes out of solution, imparting a sensation we say is ‘fizzy’.
The CO2 is pumped into the drink at the relatively high pressure of about 3 bar.
After sealing the bottle, equilibrium soon forms between the gaseous CO2 in the space above the drink and the CO2 dissolved in the liquid drink (Figure 5.14). We say the
CO2 is partitioned between the gas and liquid phases.
The proportions of CO2 in the space above the liquid and in the liquid are fixed according to an equilibrium constant, which we call the partition constant:
K(partition) =

amount of CO2 in phase 1 amount of CO2 in phase 2

We need to note how the identities of phases 1 and 2 must be defined before K can be cited. We need to be aware that K(partition) is only ever useful if the identities of phases 1 and 2 are defined.
On opening the drink bottle we hear a hissing sound, which occurs because the pressure of the escaping CO2 gas above the liquid is greater than the atmospheric pressure. We saw in Chapter 4 that the molar change in Gibbs function for movement of a gas is given by p(final) G = RT ln p(initial) (5.8)

This equilibrium constant is often incorrectly called a ‘partition function’ – which is in fact a term from statistical mechanics.

(5.9)

Carbon dioxide in the gas phase

Carbon dioxide dissolved in solution

Figure 5.14 In a bottle of fizzy drink, carbon dioxide is partitioned between the gas and the solution phases

206

PHASE EQUILIBRIA

The value of G is only ever negative, as required by a thermodynamically spontaneous process, if the initial pressure p(initial) is greater than the final pressure p(final) ,
i.e. the fraction is less than one. In other words, Equation (5.9) shows why G is negative only if the pressure of the CO2 in the space above the liquid has a pressure that is greater than p O .
We disrupted the equilibrium in the bottle when we allowed out much of the CO2 gas that formerly resided within the space above the liquid; conversely, the CO2 dissolved in the liquid remains in solution.
After drinking a mouthful of the drink, we screw on the bottle top to stop any more CO2 being lost, and come back to the bottle later when a thirst returns. The
CO2 re-equilibrates rapidly, with some of the CO2 in the liquid phase passing to the gaseous phase. Movement of CO2 occurs in order to maintain the constant value of
K(partition) : we call it ‘re-partitioning’.
Although the value of K(partition) does not alter, the amount of
CO2 in each of the phases has decreased because some of the CO2
A fizzy drink goes ‘flat’ was lost on opening the bottle. The liquid, therefore, contains less after opening it sevCO2 than before, which is why it is perceived to be less fizzy. And eral times because the water is depleted of after opening the bottle several times, and losing gaseous CO2 each
CO2 . time, the overall amount of CO2 in the liquid is so depleted that the drink no longer sparkles, which is when we say it has ‘gone flat’.
Worked Example 5.4 A bottle of fizzy pop contains CO2 . What are the relative amounts of CO2 in the water and air if K(partition) = 4?
Firstly, we need to note that stating a value of K(partition) is useless unless we know how the equilibrium constant K(partition) was written, i.e. which of the phases ‘1’ and ‘2’ in
Equation (5.8) is the air and which is the liquid?
In fact, most of the CO2 resides in the liquid, so Equation (5.8) would be written as
K(partition) =

A bottle of fizzy drink going flat is a fairly trivial example of partition, but the principle is vital to processes such as reactions in two-phase media or the operation of a high-performance liquid chromatography column. concentration of CO2 in the drink concentration of CO2 in the air above the liquid

This partition constant has a value of 4, which means that four times as much CO2 resides in the drink as in the liquid of the space above the drink. Stated another way, four-fifths of the CO2 is in the gas phase and one-fifth is in solution (in the drink).
SAQ 5.5 An aqueous solution of sucrose is prepared. It is shaken with an equal volume of pure chloroform. The two solutions do not mix. The sucrose partitions between the two solutions, and is more soluble in the water. The value of K(partition) for this water–chloroform system is 5.3.
What percentage of the sucrose resides in the chloroform?

PHASE EQUILIBRIA INVOLVING TWO-COMPONENT SYSTEMS: PARTITION

207

How does a separating funnel work?
Partition as a function of solvent
The operation of a separating funnel depends on partition. A solvent contains some solute. A different solvent, which is immiscible with the first, contains no compound.
Because the two solvents are immiscible – which means they do not mix – the separating funnel will show two distinct layers (see Immiscible solutions
Figure 5.15). After shaking the funnel vigorously, and allowing its do not mix. The words contents to settle, some of the solute will have partitioned between ‘miscible’ and its conthe two solvents, with some sample passing from the solution into verse ‘immiscible’ derive from the Latin the previously pure solvent 1.
We usually repeat this procedure two or three times during the word miscere, meaning
‘to mix’. practice of solvent extraction, and separate the two layers after each vigorous shake (we call this procedure ‘running off’ the heavier, lower layer of liquid). Several extractions are needed because K(partition) is usually quite small, which implies that only a fraction of the solute is removed from the solution during each partition cycle.

Solvent 1

Meniscus (across which partition occurs)

Solvent 2

Figure 5.15 A separating funnel is a good example of partition: solute is partitioned between two immiscible liquids

208

PHASE EQUILIBRIA

Aside
The reason we need to shake the two solutions together when partitioning is because the solute only passes from one solvent to the other across the interface between them,
i.e. across the meniscus.
The meniscus is quite small if the funnel is kept still, and partitioning is slow.
Conversely, shaking the funnel generates a large number of small globules of solvent, which greatly increases the ‘active’ surface area of the meniscus. Therefore, we shake the funnel to increase the rate of partitioning.

Why is an ice cube only misty at its centre?
The temperature dependence of partition
Most ice cubes look misty at their centre, but are otherwise quite clear. The ice from which the ice cubes are made is usually obtained from the tap, so it contains dissolved impurities such as chlorine (to ensure its sterility) and gases from the atmosphere. The mist at the centre of the ice cube comprises millions of minute air bubbles containing these gases, principally nitrogen and oxygen.
Gaseous oxygen readily partitions with oxygen dissolved in soluA fish would not be tion, in much the same way as the partitioning of CO2 in the able to ‘breathe’ in fizzy-drink example above. The exact amount of oxygen in soluwater if it contained no tion depends on the value of K(partition) , which itself depends on the oxygen gas. temperature. Tap water is always saturated with oxygen, the amount depending on the temperature. The maximum concentration of oxygen in
Like all other equilibwater – about 0.02 mol dm−3 – occurs at a temperature of 3 ◦ C. rium constants, the
The amount of oxygen dissolved in water will decrease below value of K(partition) this temperature, since K(partition) decreases. Accordingly, much disdepends strongly on temperature. solved oxygen is expelled from solution as the water freezes, merely to keep track of the constant decreasing value of K(partition) .
The tap water in the ice tray of our fridge undergoes some interesting phase changes during freezing. Even cold water straight from a tap is warmer than the air within a freezer. Water is a poor thermal conductor and does not freeze evenly, i.e. all at once; rather, it freezes progressively. The first part of the water to freeze is that adjacent to the freezer atmosphere; this outer layer of ice gradually becomes thicker with time, causing the amount of liquid water at the cube’s core to decrease during freezing.
But ice cannot contain much dissolved oxygen, so air is expelled from solution each time an increment of water freezes. This oxygen enters any liquid water nearby, which clearly resides near the centre of the cube. We see how the oxygen from the water concentrates progressively near the cube centre during freezing.
Eventually, all the oxygen formerly in the water resides in a small volume of water near the cube centre. Finally, as the freezing process nears its completion and even

PHASE EQUILIBRIA INVOLVING TWO-COMPONENT SYSTEMS: PARTITION

209

this last portion solidifies, the amount of oxygen in solution exceeds K(partition) and leaves solution as gaseous oxygen. It is this expelled oxygen we see as tiny bubbles of gas.

Aside
Zone refining is a technique for decreasing the level of impurities in some metals, alloys, semiconductors, and other materials; this is particularly so for doped semiconductors, in which the amount of an impurity must be known and carefully controlled. The technique relies on the impurities being more soluble in a molten sample (like oxygen in water, as noted above) than in the solid state.
To exploit this observation, a cylindrical bar of material is passed slowly through an induction heater and a narrow molten ‘zone’ is moved along its length. This causes the impurities to segregate at one end of the bar and super-pure material at the other.
In general, the impurities move in the same direction as the molten zone moves if the impurities lower the melting point of the material (see p. 212).

How does recrystallization work?
Partition and the solubility product
We say the solution is saturated if solute is partitioned between a liquid-phase solution and undissolved, solid material (Figure 5.16). In other words, the solution contains as much solute as is feasible, thermodynamically, while the remainder remains as solid. The best way to tell whether a solution is saturated, therefore, is to look for undissolved solid. If K(partition) is small then we say that not much of the solute resides in solution, so most of the salt remains as solid – we say the salt is not very soluble.
Conversely, most, if not all, of the salt enters solution if K(partition) is large.
Like all equilibrium constants, the value of K(partition) depends on temperature, sometimes strongly so. It also depends on the solvent polarity. For example, K(partition)

Solution saturated with solute

Solid crystals of solute

Figure 5.16 In a saturated solution, the solute is partitioned between the solid state and solute in solution

210

PHASE EQUILIBRIA

of sodium chloride (NaCl) in water is large, so a saturated solution has a concentration of about 4 mol dm−3 ; a saturated solution of NaCl in ethanol contains less than
0.01 mol dm−3 of solute.
An alternative way of expressing the partition constant of a sparingly soluble salt is to define its ‘solubility product’ Ksp (also called
Strictly, we should the ‘solubility constant’ Ks ). Ks is defined as the product of the ion speak in terms of ionic activities rather than activities of an ionic solute in its saturated solution, each raised to concentrations; see its stoichiometric number νi . Ks is expressed with due reference to
p. 312 ff. the dissociation equilibria involved and the ions present.
We saw above how the extent of partition is temperature dependent; in that example, excess air was expelled from solution during freezing, since the solubility of air was exceeded in a cold freezer box, and the gas left solution in order for the value of K(partition) to be maintained.
Like all equilibrium constants, K(partition) is a function of temperature, thereby allowing the preparative chemist to recrystallize a
The improved purity freshly made compound. In practice, we dissolve the compound in of precipitated solute a solvent that is sufficiently hot so that K(partition) is large, as shown implies that K(partition) for the impurities is by the high solubility. Conversely, K(partition) decreases so much on different from that for cooling that much of the solute undergoes a phase change from the major solute. the solution phase to solid in order to maintain the new, lower value of K(partition) . The preparative chemist delights in the way that the precipitated solid retrieved is generally purer than that initially added to the hot solvent.
The energy necessary to dissolve 1 mol of solute is called the ‘enthalpy of solution’
O
H(solution) (cf. p. 125). A value of H can be estimated by analysing the solubility s of a solute (which is clearly a function of K(partition) ) with temperature T .
The value of K(partition) changes with temperature; the temperature dependence of an equilibrium constant is given by the van’t Hoff isochore: ln H(solution) is sometimes called ‘heat of solution’, particularly in older books. The word
‘heat’ here can mislead, and tempts us to ignore the possibility of pressure–volume work.
O

K(partition)2
K(partition)1

=−

O
H(solution)

R

1
1

T2 T1

(5.10)

O so an approximate value of H(solution) may be obtained from the gradient of a graph of an isochore plot of ln s (as ‘y’) against
1/T (as ‘x’). Since s increases with increased T , we predict that
O
H(solution) will be positive.

O
Worked Example 5.5 Calculate the enthalpy of solution H(solution) from the following solubilities s of potassium nitrate as a function of temperature T . Values of s were obtained from solubility experiments.

T /K s/g per 100 g of water

354
347.6 342
334
329
322
319
317
140.0 117.0 100.0 79.8 68.7 54.6 49.4 46.1

PHASE EQUILIBRIA INVOLVING TWO-COMPONENT SYSTEMS: PARTITION

211

ln(s /g per 100 g of water)

5

4.5

4

3.5
0.0028

0.0029

0.003

0.0031

0.0032

K /T

Figure 5.17 The solubility s of a partially soluble salt is related to the equilibrium constant
K(partition) and obeys the van’t Hoff isochore, so a plot of ln s (as ‘y’) against 1/T (as ‘x’) should
O
be linear, with a slope of ‘− H(solution) ÷ R’. Note how the temperature is expressed in kelvin; a graph drawn with temperatures expressed in Celsius would have produced a curved plot. The label
K/T on the x-axis comes from 1/T ÷ 1/K

The solubility s is a function of K(partition) ; so, from Equation (5.10), a plot of ln s
(as ‘y’) against 1/T (as ‘x’) yields the straight-line graph in Figure 5.17. A value of
O
H(solution) = 34 kJ mol−1 is obtained by multiplying the gradient ‘−1 × R’.

Why are some eggshells brown and some white?
Partition between two solid solutes
The major component within an eggshell is calcium carbonate (chalk). Binders and pigments make up the remainder of the eggshell mass, accounting for about 2–5%.
Before a hen lays its egg, the shell forms inside its body via a complicated series of precipitation reactions, the precursor for each being water soluble. Sometimes the hen’s diet includes highly coloured compounds, such as corn husk. The chemicals forming the colour co-precipitate with the calcium carbonate of the shell during shell formation, which we see as the egg shell’s colour. Any substantial change in the hen’s diet causes a different combination of chemicals to precipitate during shell formation, explaining why we see differently coloured eggs.
In summary, this simple example illustrates the partition of solutes during precipitation: the colour of an egg shell results from the partitioning of chemicals, some coloured, between the growing shell and the gut of the hen during shell growth. 212

5.5

PHASE EQUILIBRIA

Phase equilibria and colligative properties
Why does a mixed-melting-point determination work?
Effects of impurity on phase equilibria in a two-component system

The best ‘fail-safe’ way of telling whether a freshly prepared compound is identical to a sample prepared previously is to perform a mixed-melting-point experiment.
In practice, we take two samples: the first comprises material whose origin and purity we know is good. The second is fresh
A ‘mixed melting point’ from the laboratory bench: it may be pure and identical to the is the only absolutely fail-safe way of deterfirst sample, pure but a different compound, or impure, i.e. a mixmining the purity of ture. We take the melting point of each separately, and call them a sample. respectively T(melt, pure) and T(melt, unknown) . We know for sure that the samples are different if these two melting temperatures differ.
Ambiguity remains, though. What if the melting temperatures are the same but, by some strange coincidence, the new sample is different from the pure sample but has the same melting temperature? We therefore determine the melting temperature of a mixture. We mix some of the material known to be pure into the sample of unknown compound. If the two melting points are still the same then the two materials are indeed identical. But any decrease in T(melt, impure) means they are not the same. The value of T(melt, mixture) will always be lower than T(melt, pure) if the two samples are different, as evidenced by the decrease in T(melt) . We call it a depression of melting point (or depression of freezing point).

Introduction to colligative properties: chemical potential
The depression of a melting point is one of the simplest manifestations of a colligative property. Other everyday examples include pressure, osmotic pressure, vapour pressure and elevation of boiling point.
For simplicity, we will start by thinking of one compound as the
‘Colligative properties’ depend on the num‘host’ with the other is a ‘contaminant’. We find experimentally ber, rather than the that the magnitude of the depression T depends only on the nature, of the chemamount of contaminant added to the host and not on the identity of ical particles (atoms the compounds involved – this is a general finding when working or molecules) under with colligative properties. A simple example will demonstrate how study. this finding can occur: consider a gas at room temperature. The ideal-gas equation (Equation (1.13)) says pV = nRT , and holds reasonably well under s.t.p. conditions. The equation makes it clear that the pressure p depends only on n, V and T , where V and T are thermodynamic variables, and n relates to the number of the particles but does not depend on the chemical nature of the compounds from which the gas is made. Therefore, we see how pressure is a colligative property within the above definition.

PHASE EQUILIBRIA AND COLLIGATIVE PROPERTIES

213

Earlier, on p. 181, we looked at the phase changes of a single-component system
(our examples included the melting of an ice cube) in terms of changes in the molar
Gibbs function Gm . In a similar manner, we now look at changes in the Gibbs function for each component within the mixture; and because several components participate, we need to consider more variables, to describe both the host and the contaminant. We are now in a position to understand why the melting point
For a pure substance, of a mixture is lower than that of the pure host. Previously, when the chemical potential we considered the melting of a simple single-component system, µ is merely another we framed our thinking in terms of the molar Gibbs function Gm . name for the molar
In a similar way, we now look at the molar Gibbs function of each Gibbs function. component i within a mixture. Component i could be a contaminant. But because i is only one part of a system, we call the value of Gm for material i the partial molar Gibbs function. The partial molar Gibbs function is also called the chemical potential, and is symbolized with the Greek letter mu, µ.
We define the ‘mole fraction’ xi as the number of moles of component i expressed as a proportion of the total number of moles present: xi =

number of moles of component i total number of moles

The value of µi – the molar Gibbs function of the contaminant – decreases as xi decreases. In fact, the chemical potential µi of the contaminant is a function of its mole fraction within the host, according to Equation (5.11): µi = µiO + RT ln xi

(5.12)

where xi is the mole fraction of the species i, and µiO is its standard chemical potential. Equation (5.12) should remind us of
Equation (4.49), which relates G and G O .
Notice that the mole fraction x has a maximum value of unity.
The value of x decreases as the proportion of contaminant increases. Since the logarithm of a number less than one is always negative, we see how the RT ln xi term on the right-hand side of
Equation (5.12) is zero for a pure material (implying µi = µiO ). At xi < 1, causing the term RT ln xi to be negative. In other words, will always decrease from a maximum value of µiO as the amount increases. Figure 5.18 depicts graphically the relationship in Equation (5.12), and shows the partial molar Gibbs function of the host material as a function of temperature. We first consider the heavy bold lines, which relate to a pure host material, i.e. before contamination. The figure clearly shows two bold lines, one each for the material when solid and another at higher temperatures for the

(5.11)
The mole fraction x of the host DEcreases as the amount of contaminant INcreases. The sum of all the mole fractions must always equal one; and the mole fraction of a pure material is also one.
Strictly, Equation (5.12) relates to an ideal mixture at constant p and T .

all other times, the value of µ of contaminant
Remember: in this type of graph, the lines for solid and liquid intersect at the melting temperature. 214

PHASE EQUILIBRIA

Partial molar Gibbs function
(chemical potential mi)

Solid

Liquid

T(melt) mixture

T(melt) pure solvent (pure)

Temperature T

Figure 5.18 Adding a chemical to a host (mixing) causes its chemical potential µ to decrease, thereby explaining why a melting-point temperature is a good test of purity. The heavy solid lines represent the chemical potential of the pure material and the thin lines are those of the host containing impurities

respective liquid. In fact, when we remember that the chemical potential for a pure material is the same as the molar Gibbs function, we see how this graph (the bold line for the pure host material) is identical to Figure 5.2. And we recall from the start of this chapter how the lines representing Gm for solid and Gm for liquid intersect at the melting temperature, because liquid and solid are in equilibrium at T(melt) , i.e.
Gm(liquid) = Gm(solid) at T(melt) .
We look once more at Figure 5.18, but this time we concentrate on the thinner lines. These lines are seen to be parallel to the bold lines, but have been displaced down the page. These thin lines represent the values of Gm of the host within the mixture (i.e. the once pure material following contamination). The line for the solid mixture has been displaced to a lesser extent than the line for the liquid, simply because the Gibbs function for liquid phases is more
As the mole fraction of sensitive to contamination. contaminant increases
(as xi gets larger), so
The vertical difference between the upper bold line (representing we are forced to draw µ O ) and the lower thin line (which is µ) arises from Eq. (5.12): the line progressively it is a direct consequence of mixing. In fact, the mathematical lower down the figure. composition of Eq. (5.12) dictates that we draw the line for an impure material (when xi < 1) lower on the page than the line for the pure material.
It is now time to draw all the threads together, and look at the temperature at which the thin lines intersect. It is clear from
A mixed-melting-point
Figure 5.18 that the intersection temperature for the mixture occurs experiment is an ideal at a cooler temperature than that for the pure material, showing test of a material’s why the melting point temperature for a mixture is depressed relpurity since T(melt) ative to a pure compound. The depression of freezing point is a never drops unless the direct consequence of chemical potentials as defined in Equation compound is impure.
(5.12).

PHASE EQUILIBRIA AND COLLIGATIVE PROPERTIES

215

We now see why the melting-point temperature decreases following contamination, when its mole fraction deviates from unity. Conversely, the mole fraction does not change at all if the two components within the mixed-melting-point experiment are the same, in which T(melt) remains the same.

Justification Box 5.3
When we formulated the total differential of G (Equation (4.30)) in Chapter 4, we only considered the case of a pure substance, saying dG =

∂G
∂p

dp +

∂G
∂T

dT

We assumed then the only variables were temperature and pressure. We must now rewrite Equation (4.30), but we add another variable, the amount of substance ni in a mixture: dG =

∂Gi
∂p

dp +

∂Gi
∂T

dT +

∂Gi
∂ni

dni

(5.13)

We append an additional subscript to this expression for dG to emphasize that we refer to the material i within a mixture. As written, Equation (5.13) could refer to either the host or the contaminant – so long as we define which is i.
The term ∂Gi /∂ni occurs so often in second law of thermodynamics that it has its own name: the ‘chemical potential’ µ, which is defined more formally as µi =

∂Gi
∂ni

p,T ,nj

(5.14)

where the subscripts to the bracket indicate that the variables p, T , and the amounts of all other components nj in the mixture, each remain constant. The chemical potential is therefore seen to be the slope on a graph of Gibbs function G (as ‘y’) against the amount of substance ni (as ‘x’); see Figure 5.19. In general, the chemical potential varies with composition, according to Equation (5.12).
The chemical potential µ can be thought of as the constant of proportionality between a change in the amount of a species and the resultant change in the Gibbs function of a system.
The way we wrote ∂G in Equation (5.13) suggests the chemical potential µ is the
Gibbs function of 1 mol of species i mixed into an infinite amount of host material.
For example, if we dissolve 1 mol of sugar in a roomful of tea then the increase in
Gibbs function is µ(sugar) . An alternative way to think of the chemical potential µ is to consider dissolving an infinitesimal amount of chemical i in 1 mol of host.

216

PHASE EQUILIBRIA

Gibbs function, G

Slope = mi

Composition ni

Figure 5.19 The chemical potential µi (the partial molar Gibbs function) of a species in a mixture is obtained as the slope of a graph of Gibbs function G as a function of composition

We need to employ ‘mental acrobatics’ of this type merely to ensure that our definition of µ is watertight – the overall composition of the mixture cannot be allowed to change significantly. How did the Victorians make ice cream?
Cryoscopy and the depression of freezing point
The people of London and Paris in Victorian times (the second half of the nineteenth century) were always keen to experience the latest fad or novelty, just like many rich and prosperous people today. And one of their favourite ‘new inventions’ was ice cream and sorbets made of frozen fruit.
The ice cream was made this way: the fruit and/or cream to be frozen is packed into a small tub and suspended in an ice bath. Rock salt is then added to the ice, which depresses its freezing temperature (in effect causing the ice to melt). Energy is needed to melt the ice. H(melt) = 6.0 kJ mol−1 for pure water. This energy comes from the fruit and cream in the tub. As energy from the cream and fruit passes through the tub wall to the ice, it freezes. Again, we see how a body’s temperature is a good gauge of its internal energy (see p. 34).
The first satisfactory theory to explain how this cooling process works was that of
Fran¸ ois-Marie Raoult, in 1878. Though forgotten now, Raoult already knew ‘Blagc den’s law’: a dissolved substance lowers the freezing point of a
Dissolving a solute in solvent in direct proportion to the concentration of the solute. In a solvent causes a practice, this law was interpreted by saying that an ice–brine mixdepression of freezing ture (made with five cups of ice to one of rock salt) had a freezing point, in the same way point at about −2.7 ◦ C. Adding too much salt caused the temperaas mixing solids. ture to fall too far and too fast, causing the outside of the ice

PHASE EQUILIBRIA AND COLLIGATIVE PROPERTIES

cream to freeze prematurely while the core remained liquid. Adding too little salt meant that the ice did not melt, or remained at a temperature close to 0 ◦ C, so the cream and fruit juices remained liquid.
This depression of the freezing point occurs in just the same way as the lower melting point of an impure sample, as discussed previously. This determination of the depression of the freezing point is termed crysoscopy.

217

The word ‘cryoscopy’ comes from the Greek kryos, which literally means ‘frost’.

Why boil vegetables in salted water?
Ebullioscopy and the elevation of boiling point
We often boil vegetables in salted water (the concentration of table salt is usually in the range 0.01–0.05 mol dm−3 ). The salt makes the food taste nicer, although we should wash off any excess salt water if we wish to maintain a healthy blood pressure.
But salted water boils at a higher temperature than does pure water, so the food cooks more quickly. (We saw on p. 203 how The word ‘ebullioscopy’ a hotter temperature promotes faster cooking.) The salt causes an comes from the Latin elevation of boiling point, which is another colligative property. We (e)bulirre, meaning
‘bubbles’ or ‘bubbly’. call the determination of such an elevation ebullioscopy.
Look at Figure 5.20, the left-hand side of which should remind us In a related way, we of Figure 5.18; it has two intersection points. At the low-tempera- say that someone is ture end of the graph, we see again why the French ice-cream ‘ebullient’ if they have a makers added salt to the ice, to depress its freezing point. But, ‘bubbly’ personality. when we look at the right-hand side of the figure, we see a second intersection, this time between the lines for liquid and gas: the temperature at which the lines intersect gives us the boiling point T(boil) .

Chemical potential m

Solid
Liquid

Gas

Temperature T
Freezing point of water + solute

Freezing point of pure water

Boiling point of pure water

Boiling point of water + solute

Figure 5.20 Salt in water causes the water to boil at a higher temperature and freeze at a lower temperature; adding a solute to a solvent decreases the chemical potential µ of the solvent. The bold lines represent pure water and the thinner lines represent water-containing solute

218

PHASE EQUILIBRIA

The figure shows how adding salt to the water has caused both the lines for liquid and for gas to drop down the page, thus causing the intersection temperature to change.
Therefore, a second consequence of adding salt to water, in addition to changing its chemical potential, is to change the temperature at which boiling occurs. Note that the boiling temperature is raised, relative to that of pure water.

Why does the ice on a path melt when sprinkled with salt?
Quantitative cryoscopy
The ice on a path or road is slippery and dangerous, as we saw when considering black ice and ice skaters. One of the simplest ways to make a road or path safer is to sprinkle salt on it, which causes the ice to melt. In practice, rock salt is preferred to table salt, because it is cheap (it does not need to be purified) and because its coarse grains lend additional grip underfoot, even before the salt has dissolved fully.
The depression of freezing temperature occurs because ions from the salt enter the lattice of the solid ice. The contaminated ice melts at a lower temperature than does pure ice, and so the freezing point decreases. Even at temperatures below the normal melting temperatures of pure ice, salted water remains a liquid – which explains why the path or road is safer.
We must appreciate, however, that no chemical reaction occurs between the salt and the water; more or less, any ionic salt, when
The ‘molaLity’ m is the put on ice, will therefore cause it to melt. The chemical identity of number of moles of solute dissolved per the salt is irrelevant – it need not be sodium chloride at all. What unit mass of solvent; matters is the amount of the salt added to the ice, which relates
‘molaRity’ (note the eventually to the mole fraction of salt. So, what is the magnitude different spelling) is of the freezing-point depression? the number of moles
Let the depression of the freezing point be T , the magnitude of solute dissolved per of which depends entirely on the amount of solute in the solvent. unit volume.
Re-interpreting Blagden’s law gives
We prefer ‘molaLity’ m to ‘molaRity’
(i.e. concentration c) because the volume of a liquid or solution changes with temperature, whereas that of a mass does not.
Accordingly, molality is temperature independent whereas concentration is not.

T ∝ molality

(5.15)

The amount is measured in terms of the molality of the solute.
Molality (note the spelling) is defined as the amount of solute dissolved per unit mass of solvent: molality, m =

moles of solute mass of solvent

(5.16)

where the number of moles of solute is equal to ‘mass of solute ÷ molar mass of solute’. The proportionality constant in Equation
(5.15) is the cryoscopic constant K(cryoscopic) . Table 5.3 contains a few typical values of K(cryoscopic) , from which it can be seen that

PHASE EQUILIBRIA AND COLLIGATIVE PROPERTIES

219

Table 5.3 Sample values of boiling and freezing points, and cryoscopic and ebullioscopic constants
Substance
Acetic acid
Acetone
Benzene
Camphor
Carbon disulphide
Carbon tetrachloride
Chloroform
Cyclohexane
Ethanol
Ethyl acetate
Ethyl ether
Methanol
Methyl acetate n-Hexane n-Octane
Naphthalene
Nitrobenzene
Phenol
Toluene
Water

Boiling point/ ◦ C
118.5
56.1
80.2
208.0
46.3
76.5
61.2
80.74
78.3
77.1
34.5
64.7
57
68.7
125.7
217.9
210.8
181.8
110.6
100

Freezing point/ ◦ C
16.60
−94.7
5.455
179.5
−111.5
−22.99
−65.5
6.55
−114.6
−83.6
−116.2
−97.7
−98.1
−95.3
−56.8
80.3
5.7
40.9
−95.0
0

K(ebullioscopic)
/K kg mol−1

K(cryoscopic)
/K kg mol−1

3.08
1.71
2.61
5.95
2.40
5.03
3.63
2.79
1.07
2.77
2.02
0.83
2.15
2.75
4.02
6.94
5.24
3.56
3.33
0.512

3.59

5.065
40
3.83
29.8
4.70
20.0
1.99

1.79




5.80
8.1
7.27

1.858

camphor as a solvent causes the largest depression. Note that K has the units of
K kg mol−1 , whereas mass and molar mass are both expressed with the units in units of grammes, so any combination of Equations (5.15) and (5.16) requires a correction term of 1000 g kg−1 . Accordingly, Equation (5.15) becomes
T = K(cryoscopic) × 1000 ×

mass of solute molar mass of solute

×

1 mass of solvent

(5.17)

where the term in parentheses is n, the number of moles of solute.
Worked Example 5.6 10 g of pure sodium chloride is dissolved in 1000 g of water.
By how much is the freezing temperature depressed from its normal melting temperature of T = 273.15 K? Take K(cryoscopic) from Table 5.3 as 1.86 K kg mol−1 .
Inserting values into Equation (5.17) yields
T = 1.86 K kg mol−1 × 1000 g kg−1 × so T = 0.32 K

10 g
1
×
−1
1000 g
58.5 g mol

220

PHASE EQUILIBRIA

This value of T represents the depression of the freezing temperature, so it is negative

showing that the water will freeze at the lower temperature of
(273.16 − 0.32) K.

SAQ 5.6 Pure water has a normal freezing point of 273.15 K. What will be the new normal freezing point of water if 11 g of KCl is dissolved in 0.9 dm3 of water?
The cryoscopic constant of water is 1.86 K kg−1 mol−1 ; assume the density of water is 1 g cm−3 , i.e. molality and molarity are the same.

An almost identical equation relates the elevation of boiling point to the molality:
T(elevation) = K(ebullioscopic) × 1000 × molality of the salt

(5.18)

where K(ebullioscopic) relates to the elevation of boiling temperature. Table 5.3 contains a few sample values of K(ebullioscopic) . It can be seen from the relative values of K(ebullioscopic) and K(cryoscopic) in Table 5.3 that dissolving a solute in a solvent has a more pronounced effect on the freezing temperature than on the boiling temperature. Aside
The ice on a car windscreen will also melt when squirted with de-icer. Similarly, we add anti-freeze to the water circulating in a car radiator to prevent it freezing; the radiator would probably crack on freezing without it; see the note on p. 194.
Windscreen de-icer and engine anti-freeze both depress the freezing point of water via the same principle as rock salt depressing the temperature at which ice freezes on a road. The active ingredient in these cryoscopic products is ethylene glycol (II), which is more environmentally friendly than rock salt. It has two physicochemical advantages over rock salt: (1) being liquid, it can more readily enter between the microscopic crystals of solid ice, thereby speeding up the process of cryoscopic melting; (2) rock salt is impure, whereas II is pure, so we need less II to effect the same depression of freezing point.

CH2
OH

(II)

CH2
OH

Ethylene glycol is also less destructive to the paintwork of a car than rock salt is, but it is toxic to humans.

PHASE EQUILIBRIA INVOLVING VAPOUR PRESSURE

5.6

221

Phase equilibria involving vapour pressure
Why does petrol sometimes have a strong smell and sometimes not?
Dalton’s law
The acrid smell of petrol on a station forecourt is sometimes overpoweringly strong, yet at other times it is so weak as to be almost absent. The smell is usually stronger on a still day with no wind, and inspection shows that someone has spilled some petrol on the ground nearby. At the other extreme, the smell is weaker when there is a breeze, which either blows away the spilt liquid or merely dilutes the petrol in the air.
The subjective experience of how strong a smell is relates to the amount of petrol in the air; and the amount is directly proportional to the pressure of gaseous petrol.
We call this pressure of petrol the ‘partial pressure’ p(petrol) .
And if several gases exist together, which is the case for petrol in air, then the total pressure equals the sum of the partial pressures according to Dalton’s law: p(total) =

pi

(5.19)

In the case of a petrol smell near a station forecourt, the smell is strong when the partial pressure of the petrol vapour is large, and it is slight when p(petrol) is small.
These differences in p(petrol) need not mean any difference in the overall pressure p(total) , merely that the composition of the gaseous mixture we breathe is variable.
SAQ 5.7 What is the total pressure of 10 g of nitrogen gas and 15 g of methane at 298 K, and what is the partial pressure of nitrogen in the mixture? [Hint: you must first calculate the number of moles involved.]

Justification Box 5.4
The total number of moles equals the sum of its constituents, so n(total) = nA + nB + . . .
The ideal-gas equation (Equation (1.13)) says pV = nRT ; thus p(petrol) V = n(petrol) RT , so n(petrol) = p(petrol) V ÷ RT .
Accordingly, in a mixture of gases such as petrol, oxygen and nitrogen: p(petrol) V p(oxygen) V p(nitrogen) V p(total) V
=
+
+
RT
RT
RT
RT

222

PHASE EQUILIBRIA

We can cancel the gas constant R, the volume and temperature, which are all constant, to yield p(total) = p(petrol) + p(oxygen) + p(nitrogen) which is Dalton’s law, Equation (5.19).

How do anaesthetics work?
Gases dissolving in liquids: Henry’s law
‘Anaesthesia’ is the science of making someone unconscious.
The word comes from the Greek aesthesis, meaning sensation
(from which we get the modern English word ‘aesthetic’, i.e. to please the sensations).
The initial ‘ana’ makes the word negative, i.e. without sensation.

A really deep, chemically induced sleep is termed ‘narcosis’, from the Greek narke, meaning ‘numbness’.
Similarly, we similarly call a class-A drug a
‘narcotic’.
Henry’s law is named after William Henry
(1775–1836), and says that the amount of gas dissolved in a liquid or solid is in direct proportion to the partial pressure of the gas.

An anaesthetist administers chemicals such as halothane (III) to a patient before and during an operation to promote unconsciousness.
Medical procedures such as operations would be impossible for the surgeon if the patient were awake and could move; and they would also be traumatic for a patient who was aware of what the surgery entailed.
BrCl

F
F

F

H
(III)

Although the topic of anaesthesia is hugely complicated, it is clear that the physiological effect of the compounds depends on their entrapment in the blood. Once dissolved, the compounds pass to the brain where they promote their narcotic effects. It is now clear that the best anaesthetics dissolve in the lipids from which cell membranes are generally made. The anaesthetic probably alters the properties of the cell membranes, altering the rates at which neurotransmitters enter and leave the cell.
A really deep ‘sleep’ requires a large amount of anaesthetic and a shallower sleep requires less material. A trained anaesthetist knows just how much anaesthetic to administer to induce the correct depth of sleep, and achieves this by varying the relative pressures of the gases breathed by the patient.
In effect, the anaesthetist relies on Henry’s law, which states that the equilibrium amount of gas that dissolves in a liquid is proportional to the mole fraction of the gas above the liquid. Henry published his studies in 1803, and showed how the amount of gas dissolved in a liquid is directly proportional to the pressure (or

PHASE EQUILIBRIA INVOLVING VAPOUR PRESSURE

223

Table 5.4 Henry’s law constants kH for gases in water at 25 ◦ C
Gas

kH /mol dm−3 bar−1

CO2
O2
CH4
N2

3.38 × 10−2
1.28 × 10−3
1.34 × 10−3
6.48 × 10−4

partial pressure) of the gas above it. Stated in another form, Henry’s law says:
[i(soln) ] = kH pi

(5.20)

where pi is the partial pressure of the gas i, and [i(soln) ] is the concentration of the material i in solution. The constant of proportionality kH is the respective value of Henry’s constant for the gas, which relates to the solubility of the gas in the medium of choice.
Table 5.4 lists a few Henry’s law constants, which relate to the solubility of gases in water.
Worked Example 5.7 What is the concentration of molecular oxygen in water at 25 ◦ C? The atmosphere above the water has a pressure of 105 Pa and contains 21 per cent of oxygen.
Strategy. (1) We calculate the partial pressure of oxygen p(O2 ) . (2) We calculate the concentration [O2(aq) ] using Henry’s law, Equation
(5.20), [O2(aq) ] = p(O2 ) × kH(O2 ) .

One of the simplest ways of removing gaseous oxygen from water is to bubble nitrogen gas through it (a process called
‘sparging’).

Strictly, Henry’s law only holds for dilute systems, typically in the mole-fraction range
0–2 per cent. The law tends to break down as the mole fraction x increases.

(1) From the partial of oxygen p(O2 ) = x(O2 ) × the total pressure p(total) , where x is the mole fraction: p(O2 ) = 0.21 × 105 Pa p(O2 ) = 2.1 × 104 Pa or 0.21 bar
(2) To obtain the concentration of oxygen, we insert values into Henry’s law,
Equation (5.20):
[O2(aq) ] = 0.21 × p O × 1.28 × 10−3 mol dm−3 bar−1
[O2(aq) ] = 2.69 × 10−4 mol dm−3

We need to be aware that kH is an equilibrium constant, so its value depends strongly on temperature. For example, at 35 ◦ C, water only accommodates 7.03 mg of oxygen per litre, which explains why fish in warm water sometimes die from oxygen starvation. This relatively high concentration of oxygen helps explain why fish can survive in water. 224

PHASE EQUILIBRIA

How do carbon monoxide sensors work?
Henry’s law and solid-state systems
Small, portable sensors are now available to monitor the air we breathe for such toxins as carbon monoxide, CO. As soon as the air contains more than a critical concentration of CO, the sensor alerts the householder, who then opens a window or identifies the source of the gas.
At the ‘heart’ of the sensor is a slab of doped transition-metal oxide. Its mode of operation is to detect the concentration of CO within the oxide slab, which is in direct proportion to the concentration of CO gas in the air surrounding it, according to Henry’s law.
A small voltage is applied across the metal oxide. When it contains no CO, the electrical conductivity of the oxide is quite poor, so the current through the sensor is minute (we argue this corollary from Ohm’s law). But increasing the concentration of
CO in the air causes a proportionate increase in the amount of CO incorporating into the solid oxide, which has a profound influence on electrical conductivity through the slab, causing the current through the slab to increase dramatically. A microchip within the sensor continually monitors the current. As soon as the current increases above its minimum permissible level, the alarm sounds.
In general, Henry’s
So, in summary, CO gas partitions between the air and carefully law only applies over formulated solid oxides. Henry’s law dictates the amount of CO in relatively small ranges of gas pressure. the oxide.

Why does green petrol smell different from leaded petrol?
Effects of amount of material on vapour pressure
A car engine requires petrol as its source of fuel. Such petrol has a low boiling temperature of about 60 ◦ C. Being so volatile, the liquid petrol is always surrounded with petrol vapour. We say it has a high vapour pressure (also called ‘saturated vapour pressure’), which explains why we smell it so readily.
Once started, the engine carburettor squirts a mixture of air and volatile petrol into a hot engine cylinder, where the mixture is ignited with a spark. The resultant explosion (we call it ‘firing’) provides the ultimate source of kinetic energy to propel the car.
A car engine typically requires four cylinders, which fire in a carefully synchronized manner. Unfortunately, these explosions sometimes occur prematurely, before

Petrol is only useful in a car engine because it is volatile.

PHASE EQUILIBRIA INVOLVING VAPOUR PRESSURE

the spark has been applied, so the explosions cease to be synchronized. It is clearly undesirable for a cylinder to fire out of sequence, since the kinetic energy is supplied in a jerky, irreproducible manner. The engine sounds dreadful, hence the word ‘knock’.
Modern petrol contains small amounts of additives to inhibit this knocking. ‘Leaded’ petrol, for example, contains the organometallic compound lead tetraethyl, PbEt4 . Although PbEt4 is excellent at stopping knocking, the lead by-products are toxic. In fact, most
EU countries now ban PbEt4 .
So-called ‘green’ petrol is a preferred alternative to leaded petrol: it contains about 3 per cent of the aromatic hydrocarbon benzene
(C6 H6 , IV) as an additive, the benzene acting as a lead-free alternative to PbEt4 as an ‘anti-knocking’ compound.

225

We experience knocking (which we colloquially call ‘pinking’) when explosions within a car engine are not synchronized. Lead tetraethyl is the most widely made organometallic compound in the world.
It is toxic, and killed over 40 chemical workers during its early development. (IV)

The PbEt4 in petrol does not smell much because it is not volatile. By contrast, benzene is much more volatile – almost as volatile as petrol. The vapour above ‘green’ petrol, therefore, contains quite a high proportion of benzene (as detected by its cloying, sweet smell) as well as gaseous petrol. That is why green petrol has a sweeter smell than petrol on its own.

Why do some brands of ‘green’ petrol smell different from others?
Raoult’s law
The ‘petrol’ we buy comprises a mixture of naturally occurring hydrocarbons, a principal component of which is octane; but the mixture also contains a small amount of benzene. Some brands of petrol contain more benzene than others, both because of variations in the conditions with which the crude oil is distilled into fractions, and also variations in the reservoir from which the crude oil is obtained. The proportion varies quite widely: the average is presently about 3 per cent.
Petrol containing a lot of benzene smells more strongly of benzene than petrol containing less of it. In fact, the intensity of the smell is in direct proportion to the amount of benzene in the petrol: at equilibrium, the pressure of vapour above a liquid mixture

In the countries of
North America, petrol is often called ‘gas’, which is short for gasoline’.

Raoult’s law is merely a special form of Henry’s law. 226

PHASE EQUILIBRIA

depends on the liquid’s composition, according to Raoult’s law:
O
p(benzene) = p(benzene) x(benzene)

Raoult’s law states that (at constant temperature) the partial pressure of component i in the vapour residing at equilibrium above a liquid is proportional to the mole fraction xi of component in the liquid.

(5.21)

where x(benzene) is the mole fraction of the benzene in the liquid.
If we assume that liquid benzene and petrol have the same densities (which is entirely reasonable), then petrol containing 3 per cent of benzene represents a mole fraction x(benzene) = 0.03; the mole fraction of the petrol in the liquid mixture is therefore 0.97
(or 97 per cent). The vapour above the petrol mixture will also be a mixture, containing some of each hydrocarbon in the petrol.
We call the pressure due to the benzene component its partial pressure p(benzene) . The constant of proportionality in Equation (5.21) is
O
p(benzene) , which represents the pressure of gaseous benzene above pure (i.e. unmixed) liquid benzene.

Calculations with Raoult’s law
If we know the mole fraction of a liquid i (via Equation (5.11)) and the vapour pressures of the pure liquids piO , then we can ascertain the total vapour pressure of the gaseous mixture hovering at equilibrium above the liquid.
The intensity of the benzene smell is proportional to the amount of benzene in the vapour, p(benzene) . According to Equation (5.21), p(benzene) is a simple function of how much benzene resides within the liquid petrol mixture. Figure 5.21 shows a graph of the partial pressures of benzene and octane above a mixture of the two liquids. (For convenience, we assume here that the mixture comprises only these two components.) The extreme mole fractions, 0 and 1, at either end of the graph relate to pure petrol
(x = 0) and pure benzene (x = 1) respectively. The mole fractions between these values represent mixtures of the two. The solid, bold line represents the total mole fraction while the dashed lines represent the vapour pressures of the two constituent vapours. It is clear that the sum of the two dashed lines equals the bold line, and represents another way of saying Dalton’s law: the total vapour pressure above a mixture of liquids is the sum of the individual vapour pressures.

If a two-component system of A and B forms an ideal mixture, then we can calculate xA if we know xB because xA + xB = 1, so xB = (1 − xA ).

Benzene is more volatile than bromobenzene because its vapour pressure is higher.

Worked Example 5.8 The two liquids benzene and bromobenzene are mixed intimately at 298 K. At equilibrium, the pressures of the gases above beakers of the pure liquids are 100.1 kPa and 60.4 kPa respectively. What is the vapour pressure above the mixture if 3 mol of benzene are mixed with 4 mol of bromobenzene?

PHASE EQUILIBRIA INVOLVING VAPOUR PRESSURE

227

70 000

Partial pressure of benzene/Pa

Partial pressure of octane/Pa

80 000
Total vapour pressure

60 000
50 000
Vapour pressure of benzene

40 000
30 000
20 000
10 000

Vapour pressure of octane

0
0

0.2

0.4
0.6
0.8
Mole fraction of benzene

1

Figure 5.21 Petrol (‘gasoline’) is a mixture of liquid hydrocarbons. The partial pressure of benzene is nearly twice that of octane, making it much more volatile. The bold line represents the total pressure of vapour above a basin of petrol, and comprises the sum of two partial pressures: benzene (open circles) and octane (filled circles). Each partial pressure is proportional to the mole fraction of the respective liquid in the petrol mixture

From Dalton’s law, the total vapour pressure is simply the sum of the individual vapour pressures: p(total) = p(benzene) + p(bromobenzene) so, from Raoult’s law, these partial pressures may be obtained by substituting each p term with piO × xi :

Care: do not confuse p O (the standard presO sure of 105 Pa) with pi , the vapour pressure of pure i.

O
O
p(total) = (p(benzene) × x(benzene) ) + (p(bromobenzene) × x(bromobenzene))
(5.22)
We know from the question that there are 7 mol of liquid. We obtain the respective mole fractions x from Equation (5.11): the mole fraction of benzene is 3 and the mole fraction
7
of bromobenzene is 4 .
7
Substituting values of xi and piO into Equation (5.22) yields the total pressure p(total) as

p(total) = 100.1 kPa ×

3
7

+ 60.4 kPa ×

4
7

p(total) = (42.9 kPa) + (34.5 kPa) so p(total) = 77.4 kPa

228

PHASE EQUILIBRIA

An ideal mixture comprises a pair (or more) of liquids that obey
Raoult’s law.

Because these two liquids, when mixed, obey Raoult’s law, we say they form an ideal mixture. In fact, relatively few pairs of liquids form ideal mixtures: a few examples include benzene and bromobenzene, benzene and toluene, bromobenzene and chlorobenzene, n-pentane and i-pentane. Note how each set represents a pair of liquids showing a significant extent of similarity.

SAQ 5.8 Benzene and toluene form an ideal mixture, i.e. they obey
Raoult’s law. At 20 ◦ C, the pressure p O of benzene and toluene are 0.747 × p O and 0.223 × p O respectively. What is the pressure above a mixture of these two liquids that contains 12 mol% of benzene?

Worked Example 5.9 (Continuing from Worked Example 5.8.) What are the mole fractions of benzene and bromobenzene in the vapour?
From the definition of mole fraction x in Equation (5.11) above, we say x(benzene, vapour)

=

moles of benzene in the vapour total number of moles in the vapour phase

The numbers of moles ni are directly proportional to the partial pressures pi if we assume that each vapour behaves as an ideal gas (we assume here that T , R and V are constant).
Accordingly, we can say x(benzene) =

pressure of benzene total pressure

Substituting numbers from Worked Example 5.8:
Note how the units cancel to yield a dimensionless mole fraction.

x(benzene)

We need four mole fractions to define this two-component system – two for the liquid phases and two for the vapour phases.

42.9 kPa
77.4 kPa
= 0.554

x(benzene) =

The mole fraction of benzene in the vapour is 0.554, so it contains 55.4 per cent benzene. The remainder of the vapour comprises the second component bromobenzene, so the vapour contains (100 − 55.4)% =
44.6% of bromobenzene.
Note how the liquid comprises 43 per cent benzene and 57 per cent bromobenzene, but the vapour contains proportionately more of the volatile benzene. We should expect the vapour to be richer in the more volatile component.

SAQ 5.9 Continuing with the system in SAQ 5.8, what is the mole fraction of toluene in the vapour above the mixture?

In fact, most liquid mixtures do not obey Raoult’s law particularly well, owing to molecular interactions.

PHASE EQUILIBRIA INVOLVING VAPOUR PRESSURE

229

Why does a cup of hot coffee yield more steam than above a cup of boiling water at the same temperature?
The effects of poor mixing (immiscibility)
Prepare two cups: put boiling water into one and boiling coffee in the other. The temperature of each is the same because the water comes from the same kettle, yet the amount of steam coming from the coffee is seen to be greater. (We obtain a better view of the steam by placing both cups on a sunny window sill, and looking at the shadows cast on the opposing wall as the light passes through the vapour as it rises from the cups.)
When performing this little experiment, we will probably notice how the steam above the coffee has an extremely strong smell of The rate of steam procoffee, although the smell dissipates rapidly as the rate of steam duction decreases with time as the water cools production decreases.
This experiment is a simple example of steam distillation. down because energy
Adding steam promotes the volatilization of otherwise non-volatile is lost from the cup as components, simplifying their extraction. For simplicity, we will water molecules enter say that the smell derives from a single sweet-smelling chemical the gas phase.
‘coffee’. Coffee and water are not wholly miscible, with some of the essential oils from the coffee existing as tiny globules – we call the mixture a colloid (see Chapter 10). We have generated a two-phase system. Both phases, the water and the coffee, are saturated with each other. In fact, these globules would cause strong coffee to appear slightly misty, but for its strong colour blocking all light. We never see phase separation in the coffee cup, with a layer of oil floating above a layer of water, because the coffee’s concentration is never high enough.
We say a pure liquid boils when its vapour pressure equals the external, atmospheric pressure (see p. 188). Similarly, when boiling The boiling of such a a mixture, boiling occurs when the sum of the partial pressures mixture requires the
(p(water) + p(coffee) ) equals p O . It is for this reason that the steam sum of the pressures, above the coffee cup smells strongly of coffee, because the vapour not just the pressure contains the essential oils (e.g. esters) that impart the smell. But the of one component, to
O
water generates steam at a pressure of p O when the water added equal p . to the cup is boiling, so the partial pressure of the coffee p(coffee) is additional. For this reason, we produce more steam than above the cup containing only water.

How are essential oils for aromatherapy extracted from plants?
Steam distillation
The ‘essential oils’ of a plant or crop usually comprise a mixture of esters. At its simplest, the oils are extracted from a plant by distillation, as employed in a standard

230

PHASE EQUILIBRIA

undergraduate laboratory. Since plants contain such a small amount of this precious oil, a ton of plant may be needed to produce a single fluid ounce. Some flowers, such as jasmine or tuberose, contain very small amounts of essential oil, and the petals are very temperature sensitive, so heating them would destroy the blossoms before releasing the essential oils.
To add to the cost further, many of these compounds are rather sensitive to temperature and would decompose before vaporizing. For example, oil of cloves (from
Eugenia caryophyllata) is rich in the phenol eugenol (V), which has a boiling point of 250 ◦ C). We cannot extract the oils via a conventional distillation apparatus.

H2C

H
C

CH2

OCH3
OH
(V)

Heat-sensitive or water-immiscible compounds are purified by steam distillation at temperatures considerably lower than their usual boiling temperatures. Solvent extraction of essential oils tends to generate material that is contaminated with solvent (and cannot be sold); and mechanical pressing of a plant usually generates too poor a yield to be economically viable.

The most common method of extracting essential oils is steam distillation. The plant is first crushed mechanically, to ensure a high surface area, and placed in a closed still. High-pressure steam is forced through the still, with the plant pulp becoming hot as the steam yields its heat of vaporization (see p. 79). The steam forces the microscopic pockets holding the essential oils to open and to release their contents. Tiny droplets of essential oil evaporate and mix in the gas-phase mixture with the steam. The mixture is then swept through the still before condensing in a similar manner to a conventional distillation.
Such ‘steam heating’ is even, and avoids the risk of overheating and decomposition that can occur in hot spots when external heating is used. The steam condenses back into water and the droplets coagulate to form liquid oil. Esters and essential oils do not mix with water, so phase separation occurs on cooling, and we see a layer of oil forming above a layer of condensed water. The oil is decanted or skimmed off the surface of the water, dried, and packaged.
The only practical problem encountered when collecting organic compounds by steam distillation is that liquids of low volatility will usually distil slowly, since the proportion of compound in the vapour is proportional to the vapour pressure, according to n(oil) p(oil)
=
p(water) n(water) (5.23)

PHASE EQUILIBRIA INVOLVING VAPOUR PRESSURE

231

In practice, we force water vapour (steam) at high pressure through the clove pulp to obtain a significant partial pressure of eugenol (V).

Justification Box 5.5
When considering the theory behind steam distillation, we start with the ideal-gas equation (Equation (1.13)), pV = nRT . We will consider two components: oil and water. For the oil, we say p(oil) V = n(oil) RT , and for the water p(water) V = n(water) RT .
Dividing the two equations by R and V (which are both constant) yields p(oil) = n(oil) × T p(water) = n(water) × T

for the oil for the water

We then divide each pressure by the respective number of moles ni , to obtain p(oil) ÷ n(oil) = T p(water) ÷ n(water) = T

for the oil for the water

The temperature of the two materials will be T , which is the same for each as they are in thermal equilibrium. We therefore equate the two expressions, saying p(oil) ÷ n(oil) = p(water) ÷ n(water)
Dividing both sides by p(water) and multiplying both sides by n(oil) yields Equation (5.23): n(oil) p(oil)
=
p(water) n(water) so we see how the percentage of each constituent in the vapour depends only on its vapour pressure at the distillation temperature.
To extract a relatively involatile oil such as eugenol (V) without charring requires a high pressure of steam, although the steam will not be hotter than 100 ◦ C, so we generate a mixture of vapours at a temperature lower than that of the less volatile component.

6

Acids and Bases
Introduction
Equilibria involving acids and bases are discussed from within the Lowry–Brønsted theory, which defines an acid as a proton donor and a base as a proton acceptor (or
‘abstracter’). The additional concept of pH is then introduced. ‘Strong’ and ‘weak’ acids are discussed in terms of the acidity constant Ka , and then conjugate acids and bases are identified.
Acid–base buffers comprise both a weak acid or base and its respective salt. Calculations with buffers employing the Henderson–Hasselbach equation are introduced and evaluated, thereby allowing the calculation of the pH of a buffer. Next, titrations and pH indicators are discussed, and their modes of action placed into context.

6.1

Properties of Lowry–Brønsted acids and bases
Why does vinegar taste sour?
The Lowry–Brønsted theory of acids
We instantly experience a sour, bitter taste when consuming anything containing vinegar. The component within the vinegar causing the sensation is ethanoic acid,
CH3 COOH (I) (also called ‘acetic acid’ in industry). Vinegar contains between 10 and 15% by volume of ethanoic acid, the remain- The Latin word for der being water (85–90%) and small amounts of other components ‘sour’ is acidus. such as caramel, which are added to impart extra flavour.
H
H

O

C
O

H
(I)

H

234

ACIDS AND BASES

The German chemist
Liebig, in 1838, was the first to suggest mobile, replaceable, hydrogen atoms being responsible for acidic properties. Arrhenius extended the idea in
1887, when he said the hydrogen existed as a proton. The O–H bond in an acid is sometimes said to be ‘labile’, since it is so easily broken. The word derives from the
Latin labi, to lapse (i.e. to change).

The ‘Lowry–Brønsted theory’ says an acid is a proton donor.

The ethanoic acid molecule is essentially covalent, explaining why it is liquid when pure at room temperature. Nevertheless, the molecule is charged, with the O–H bond characterized by a high percentage of ionic character. Because water is so polar a solvent, it strongly solvates any solute dissolved within it. In aqueous solutions, water molecules strongly solvate the oxygen- and protoncontaining ends of the O–H bond, causing the bond to break in a significant proportion of the ethanoic acid molecules, according to the following simplistic reaction:

CH3 COOH(aq) − → CH3 COO− (aq) + H+ (aq)

(6.1)

We say the acid dissociates. The bare proton is very small, and has a large charge density, causing it to attract the negative end of the water dipole. The proton produced by Equation (6.1) is, therefore, hydrated in aqueous solutions, and is more accurately represented by saying H+ (aq) .
We see how solvated protons impart the subjective impression of a sour, bitter flavour to the ethanoic acid in vinegar. In fact, not only the sour flavour, but also the majority of the properties we typically associate with an acid (see Table 6.1) can be attributed to an acidic material forming one or more solvated protons H+ (aq) in solution. This classification of an acid is called the Lowry–Brønsted theory after the two scientists who (independently) proposed this definition of an acid in 1923. More succinctly, their theory says an acid is
Table 6.1

Typical properties of Lowry–Brønsted acids

Acid property
Acids dissolve a metal to form a salt plus hydrogen Acids dissolve a metal oxide to form a salt and water
Acids react with metal carbonates to form a salt and carbon dioxide

Acids are corrosive
Acids react with a base to form a salt and water (‘neutralization’)

Example from everyday life
Metallic sodium reacts with water, and
‘fizzes’ as hydrogen gas evolves
The ability of vinegar to clean tarnished silver by dissolving away the coloured coating of Ag2 O
The fizzing sensation in the mouth when eating sherbet (saliva is acidic, with a pH of 6.5); sherbet generally contains an organic acid, such as malic or ascorbic acids Teeth decay after eating sugar, and one of the first metabolites from sugar is lactic acid
Rubbing a dock leaf (which contains an organic base) on the site of a nettle sting
(which contains acid) will neutralize the acid and relieve the pain

PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES

235

a substance capable of donating a proton. Therefore, we describe ethanoic acid as a
‘Lowry–Brønsted acid’.

Why is it dangerous to allow water near an electrical appliance, if water is an insulator?
The solvated proton
We all need to know that electricity and water are an extremely dangerous combination, and explains why we are taught never to sit in the bath at the same time as shaving with a plug-in razor or drying our hair. Electrocution is almost inevitable, and is often fatal.
The quantity we call ‘electricity’ is a manifestation of charge Q passing through a suitable conductor. The electrical conductivity κ
Care: the symbol of of water (be it dishwater, rainwater, bathwater, etc.) must be rela- conductivity is not K tively high because electricity can readily conduct through water. but the Greek letter
Nevertheless, the value of κ for water is so low that we class water kappa, κ. as an insulator. Surely there is a contradiction here?
‘Super-pure water’ has been distilled several times, and is indeed an insulator: its conductivity κ is low at 6.2 × 10−8 S cm−1 at 298 K, and lies midway between classic insulators such as Teflon, with a conductivity of about 10−15 S cm−1 , and semiconductors such as doped silicon, for which κ = 10−2 S cm−1 . The conductivity of metallic copper is as high as 106 S cm−1 .
The value of κ cited above was for super-pure water, i.e. the product of multiple distillations, but ‘normal’ water from a tap Pure water is a mixture will inevitably contain solutes (hence the ‘furring’ inside a kettle of three components:
H2 O, and its two disor pipe). Inorganic solutes are generally ionic salts; most organic sociation products, the solutes are not ionic. The conductivity of super-pure water is low solvated proton (H O+ )
3
because the molecules of water are almost exclusively covalent, and the hydroxide ion with the extent of ionicity being very slight. But as soon as a salt (OH− ). dissolves, the extent of ionicity in the water increases dramatically, causing more extensive water dissociation.
Ignoring for the moment the solute in solution, the water dis- It is safer in many sociation involves the splitting of water itself in a process called instances to assume the solvated proton autoprotolysis. The reaction is usually represented as

2H2 O − → H3 O

+

(aq)



+ OH

(aq)

(6.2)

where the H3 O+ (aq) species is often called a solvated proton. It also has the names hydroxonium ion and hydronium ion. The complex ion H3 O+ (aq) is a more accurate representation of the proton responsible for acidic behaviour than the simplistic ‘H+ (aq) ’ we wrote in
Equation (6.1). Note how the left-hand side of Equation (6.2) is covalent and the right-hand side is ionic.

has the formula unit
[H(H2 O)4 ]+ , with four water molecules arranged tetrahedrally around a central proton, the proton being stabilized by a lone pair from each oxygen atom. 236

ACIDS AND BASES

Dissolving a solute generally shifts the reaction in Equation (6.2) from left to right, thereby increasing the concentration of ionic species in solution. This increased number of ions causes the conductivity κ of water to increase, thereby making it a fatally efficient conductor of charge.

Why is bottled water ‘neutral’?
Autoprotolysis
The word ‘criterion’ is used of a principle or thing we choose to use as a standard when judging a situation.
The plural or criterion is ‘criteria’, not
‘criterions’.

‘Autoprotolysis’ comes from proto–indicating the proton, and lysis, which is a Greek root meaning ‘to cleave or split’. The prefix auto means ‘by self’ or ‘without external assistance’. The labels of many cosmetic products, as well as those on most bottles of drinking water, emphasize how the product is ‘neutral’, implying how it is neither acidic nor alkaline. This stipulation is deemed to show how healthy the water is. But how do they know?
And, furthermore, what is their criterion for testing?
A better way of defining ‘neutral’ is to say equal numbers of protons and hydroxide ions reside in solution (both types of ion being solvated). How does this situation arise? Autoprotolysis, as mentioned above, represents the self-production of protons, which is achieved by the splitting of water according to Equation (6.2).
It is clear from Equation (6.2) how the consequence of such autoprotolytic splitting is a solution with equal numbers of protons and hydroxide ions.
When water contains no dissolved solutes, the concentrations of the solvated protons and the hydroxide ions are equal. Accordingly, from our definition of ‘neutral’ above, we see why pure water should always be neutral, since [H3 O+ (aq) ] = [OH− (aq) ].
As with all physicochemical processes, the extent of Equation
(6.2) may be quantified by an equilibrium constant K. We call it the autoprotolysis constant, as defined by
K=

[H3 O+ (aq) ][OH− (aq) ]
[H2 O]2

(6.3)

The water term in the denominator of Equation (6.3) is always large when compared with the other two concentrations on the top, so we say it remains constant.
This assumption explains why it is rare to see the autoprotolysis constant written as
Equation (6.3). Rather, we usually rewrite it as
Kw = [H3 O+ (aq) ][OH− (aq) ]

(6.4)

where K in Equation (6.3) = Kw × [H2 O]2 in Equation (6.4). We will only employ
Equation (6.4) from now on. We call Kw the autoprotolysis constant or ionic product of water.

PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES

237

Table 6.2 Values of the autoprotolysis constant Kw as a function of temperature Temperature T / ◦ C
Kw × 1014

0
0.12

18
0.61

25
1.04

34
2.05

50
5.66

Source: Physical Chemistry, W. J. Moore (4th Edn), Longmans, London, 1962,
p. 365.

The value of Kw is 1.04 × 10−14 at 298 K when expressed in concentration units of mol dm−3 . Like all equilibrium constants, its value depends on the temperature. Table 6.2 lists a few values of
Kw as a function of temperature. Note how Kw increases slightly as the temperature increases.
It should now be clear from Equation (6.4) how water splits
(dissociates) to form equal number of protons and hydroxide ions, hence its neutrality, allowing us to calculate the numbers of each from the value of Kw .

Note how Kw has units.
Kw is often re-expressed as pKw , where the ‘p’ is a mathematical operator (see
p. 246). pKw has a value of 14 at 298 K.

Worked Example 6.1 What is the concentration of the solvated proton in super-pure water? Since [H3 O+ (aq) ] = [OH− (aq) ], we could rewrite Equation (6.4) as
Kw = [H3 O+ (aq) ]2
Taking the square root of both sides of this expression, we obtain
[H3 O+ (aq) ]/mol dm−3 = so Kw = [10−14 ]1/2

[H3 O+ (aq) ] = 10−7 mol dm−3

The concentration of solvated protons in super-pure water is clearly very small.

What is ‘acid rain’?
Hydrolysis
Acid rain is one of the worst manifestations of the damage we, as humans, inflict on our planet. Chemicals combine with elemental oxygen during the burning of fossil fuels, trees and rubbish to generate large amounts of ‘acidic oxides’ such as nitrogen monoxide (NO), carbon dioxide (CO2 ) and sulphur dioxide (SO2 ).
Natural coal and oil contain many compounds of nitrogen. One of the worst products of their combustion is the acidic oxide of nitrogen, NO. At once, we are startled by this terminology, because the Lowry–Brønsted definition of an acid involves the release of a proton, yet nitrogen monoxide NO has no proton to give.

238

ACIDS AND BASES

As long ago as the
18th century, French chemists appreciated how burning elemental carbon, nitrogen or sulphur generated compounds which, when dissolved in water, yielded an acidic solution. The nitric acid in acid rain forms by a more complicated mechanism: 4NO(g) + 2H2 O(l)
+O2(g) − − 4HNO3(aq)
−→

‘Hydrolysis’ means to split water, the word coming from the two
Greek roots hydro meaning water, and lysis meaning ‘to cleave or split’.

To understand the acidity of pollutants such as NO and CO2 , we need to appreciate how the gas does not so much dissolve in water as react with it, according to

CO2(g) + 2H2 O(l) − → HCO− 3(aq) + H3 O+ (aq)

Carbonic acid, H2 CO3(aq) , never exists as a pure compound; it only exists as a species in aqueous solution, where it dissociates in just the same way as ethanoic acid in Equation (6.1) to form a solvated proton and the HCO− ion. Note how we form a sol3(aq) vated proton H3 O+ (aq) by splitting a molecule of water, rather than merely donating a proton. Carbonic acid is, nevertheless, a
Lowry–Brønsted acid.
The carbonic acid produced in Equation (6.5) is a proton donor, so the solution contains more solvated protons than hydroxide ions, resulting in rain that is (overall) an acid. To make the risk of pollution worse, ‘acid rain’ in fact contains a mixture of several water-borne acids, principally nitric acid, HNO3 (from nitrous oxide in water), and sulphurous acid, H2 SO3 (an aqueous solution of sulphur dioxide).
In summary, we see how the concentrations of H3 O+ and OH− are the same if water contains no dissolved solutes, but dissolving a solute such as NO increases the concentration of H3 O+ ; in a similar way, the concentration of OH− will increase if the water contains any species capable of consuming protons.
It is time to introduce a few new words. We say carbonic acid forms by hydrolysis, i.e. by splitting a molecule of water. We describe the extent of hydrolysis in Equation (6.5) by the following equilibrium constant:
K=

Care: the values of K from these equations are only meaningful for concentrations at equilibrium. (6.5)

[HCO3 − ][H3 O+ ]
[CO2 ][H2 O]2

(6.6)

We sometimes call Equation (6.6) the hydrolysis constant of carbon dioxide. In fact, the water term in the ‘denominator’ (the bottom line) is so large compared with all the other terms that it remains essentially constant. Therefore, we write Equation (6.6) in a different form:
[HCO3 − ][H3 O+ ]
(6.7)
K =
[CO2 ]
Note how the two K terms, K in Equation (6.6) and K in Equation
(6.7), will have different values.

PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES

239

Why does cutting an onion make us cry?
Other aqueous acids in the environment
The reason why our eyes weep copiously when peeling an onion is because the onion contains minute pockets of sulphur trioxide, SO3(g) . Cutting the onion releases this gas. A mammalian eye is covered with a thin film of water-based liquid (‘tears’) to minimize friction with the eyelid. The tears occur in response to SO3 dissolving in this layer of water to form sulphuric acid:

SO3(g) + H2 O(l) − → H2 SO4(aq)

(6.8)

The sulphuric acid produced dissociates in the water to form SO4 2− and two protons.
The eyes sting as a direct consequence of contact with this acid.

Why does splashing the hands with sodium hydroxide solution make them feel ‘soapy’?
Proton abstraction
Sodium hydroxide in solution dissociates to yield solvated cations and anions, Na+ and the hydroxide ion OH− respectively:
NaOH(s) − → Na+ (aq) + OH− (aq)


(6.9)

The solvated hydroxide ion in Equation (6.9) is formed in addition to the hydroxide ions produced during water autoprotolysis, so there are more hydroxide ions in solution than solvated protons, yielding excess hydroxide in solution. We say the solution is alkaline. As an alternative name, we say hydroxide is a base
(see p. 241).
Oils in the skin react readily with the hydroxide ions via the same chemical process occurring when spray-on oven cleaner ‘eats’ into the grime in an oven, reacting principally by the OH− (aq) ion consuming protons. Let us start, for example, with a molecule possessing a proton capable of being donated; call it HA, where
‘A’ is merely an anion of some sort. This proton must be labile.
The hydroxide ion removes this labile proton to generate water, according to

HA + OH− (aq) − → A− (aq) + H2 O

(6.10)

This proton-removing ability characterizes the reactions of hydroxide ions in aqueous solutions, and of bases in general. We

Care: It is a common mistake to call the
OH− ion a ‘hydroxyl’.
It is not: a hydroxyl is correctly a covalently bound –OH group, for example in an alcohol.
Fullers’ earth is a type of clay named after a fuller, whose job was to clean cloth, e.g. stripping wool of its grease.
Fullers’ earth removes oils and grease from cloth because of its alkalinity, just like an oven cleaner solution.

The Lowry–Brønsted theory says a base is a proton remover.

240

ACIDS AND BASES

A basic chemical consumes protons.

go further by defining hydroxide as a base because it reacts with
(i.e. consumes) labile protons. And any chemical capable of removing protons is said to be basic.

Aside
Saponification
The word ‘saponify’ comes from the Latin sapo, meaning soap.

Hydroxide ions react to split (‘hydrolyse’) natural esters in the skin to form glycerol (II) and palmitic or stearic acid – a reaction called saponification. Palmitic and stearic acids subsequently react with the base to form the respective long-chain carboxylate anions – which is soap.

H2C
OH

H
C
OH

CH2
OH

(II)

These cleansing properties of bases were appreciated in antiquity. For example, in a portion of the Bible probably written in about 1200 BC, a character called Job declares his desire to be clean, saying, ‘If I washed myself with soap and snow, and my hands with washing soda . . .’ (snow was thought to be especially pure and soda (Na2 CO3 · 10H2 O) is alkaline and has long been used as a soap). This quote may be found in full in the
Bible, see Job 9:30.
The Jewish Prophet Jeremiah writing in about 700 BC says much the same thing: look at Jeremiah 2:21–23 in the Hebrew Bible.

Why is aqueous ammonia alkaline?
Lowry–Brønsted bases
All aqueous solutions naturally contain hydroxide ions in consequence of the autoprotolytic reaction in Equation (6.2). As we have seen, there will be equal numbers of solvated protons and solvated hydroxide ions unless we add an acid or base to it. A solution containing more solvated protons than hydroxide ions is said to be an
‘acid’ within the Lowry–Brønsted theory, and a solution comprising more hydroxide ions than solvated protons is said to be a base.

PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES

But a word of caution: species other than metal hydroxides can act as bases. Ammonia is such an example, since it can abstract protons in aqueous solution according to

NH3(aq) + H2 O − → NH+ 4(aq) + OH− (aq)

(6.11)

241

We say a proton is abstracted when removed selectively.
Similarly, we call a selective summary or pr´cis of a piece of e prose ‘an abstract’.

To abstract a proton is to remove only the proton. The substantial extent of dissociation in Equation (6.11) helps explain why
‘aqueous ammonia’ is more properly called ‘ammonium hydroxide’, NH4 OH. We generate the solvated hydroxide ion OH− (aq) by abstracting a proton from water. The OH− (aq) ion in Equation (6.11) is chemically and physically identical to the solvated hydroxide ion generated by dissolving NaOH or KOH in water.

Why is there no vinegar in crisps of salt and vinegar flavour?
Conjugate acids and bases
Potato crisps come in many flavours, perhaps the most popular being ‘salt and vinegar’. Curiously, a quick glance at the packet’s list of ingredients reveals how the crisps contain unhealthy amounts of salt, but no vinegar (ethanoic acid) at all. In fact, the manufacturer dusts the crisps with powdered sodium ethanoate (NaCO2 CH3 ), because ‘real’ vinegar would soon make the crisps limp and soggy. Inside the mouth, acid from the saliva reacts with the ethanoate anion to form ethanoic acid:

CH3 CO− + H3 O+ (aq) − → CH3 CO2 H(aq) + H2 O
2(aq)

(6.12)

This reaction proceeds inside the mouth, rapidly reaching its position of equilibrium, and allowing the ethanoic acid to impart its distinctive vinegary flavour.
The solvated proton on the left of Equation (6.12) acts as an acid, since it donates a proton at the same time as the ethanoate The word ‘conjugate’ ion behaves as a base, because it accepts a proton. To complicate comes from the Latin the situation, the reaction is one half of a dynamic equilibrium, conjugare, meaning
i.e. it proceeds in both the forward and backward directions. In the ‘to yoke together’ backward direction, we notice how this time the ethanoic acid acts (the prefix con means
‘together’ and jugare as an acid and the water acts as a base.
The reaction in Equation (6.12) illustrates the coexistence of two is ‘to yoke’). Simiacids and two bases. We say the ethanoate ion and ethanoic acid larly, the English word
‘conjugal’ relates to represent a conjugate pair, and the solvated proton and the water marriage and concerns form a second conjugate pair. Within the ethanoic–ethanoate pair, the joining of husband the ethanoic acid is the conjugate acid and the ethanoate anion and wife. is the conjugate base. Similarly, H3 O+ is a conjugate acid to the

242

ACIDS AND BASES

conjugate base of H2 O. Other examples of conjugate acid–base pairs include nitric acid and nitrate ion, and ammonium ion and ammonia (the acid being cited first in each case).
We must treat with caution one further aspect of the Brønsted theory: multiple proton-donation reactions. Consider the example of the bicarbonate ion HCO3 − in water. When titrating bicarbonate with a base such as hydroxide, the ion behaves as an acid to form the carbonate anion and water:
A substance like bicarbonate, which can react as either an acid or as a base, is said to be amphoteric. The word comes from the Greek amphoteros, meaning
‘both’.


HCO3 − + OH− − → CO3 2− + H2 O

(6.13)

But, conversely, when titrating ions with an acid, the bicarbonate behaves as a base, losing its proton to form carbonic acid:

HCO3 − + H3 O+ − → H2 CO3 + H2 O

(6.14)

We see how the same ion acts as an acid or as a base, depending on the other reagents in solution. We say the bicarbonate ion is amphoteric, since it reacts either as an acid or as a base.

SAQ 6.1 Consider the following pairs, and for each decide which is the conjugate acid and which the base: (a) carbonate and bicarbonate;
(b) H2 EDTA2− and H3 EDTA− ; (c) HNO2 and NO2 − .

Aside
Related models of acids and bases
The concept of acid and base can be generalized in several ways. In liquid ammonia, for example, the ammonium and amide ions (NH4 + and NH2 − respectively) coexist. The roles of these ions are directly comparable with H3 O+ and OH− in water. In ammonia, the species NH4 Cl and NaNH2 can be considered to be the respective acid and base conjugates, just as HCl and NaOH are an acid–base pair in water. This solvent-based classification of acids and bases derived from Franklin, in 1905. His ideas are worth careful thought, although we no longer use his terminology.
Brønsted’s definition of acids and bases (see p. 234 and 240) emphasizes the complementary nature of acids and bases, but it is broader than Franklin’s model because it does not require a solvent, and can even be applied to gas-phase reactions, e.g.
HCl(g) + NH3(g) → NH4 Cl(s) .

How did soldiers avoid chlorine gas poisoning at the Second Battle of Ypres?
Neutralization reactions with acids and bases
The bloody Second Battle of Ypres was fought in France on 22 April 1915, and was the first time in modern warfare when poison gases were employed. At a crucial

PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES

243

stage in the battle, the German forces filled the air above the enemy trenches with chlorine gas.
Elemental chlorine Cl2 dissolves slightly in water, and hydrolyses some of the water to yield hypochlorous acid, HOCl, according to
Hypochlorous acid,
Cl2(aq) + H2 O(l) − → HCl(aq) + HOCl(aq)


(6.15)

The reaction in Equation (6.15) occurs readily in the lungs and eyes (the sensitive tissues of which are lined with water) to cause irreparable damage. Troops exposed to chlorine apparently experienced a particularly slow and nasty death.
The German troops did not advance, because they were not sure if the gas masks issued to their own troops could withstand the chlorine. They were also deterred by the incursion of a Canadian regiment. But one of the young Canadian soldiers knew a little chemistry: sniffing the gas, he guessed its identity correctly, and ordered the soldiers to cover their faces with handkerchiefs (or bandages) soaked in their own urine. The idea spread quickly, and the Canadians, together with two Yorkshire territorial battalions, were able to push back the German troops.
One of the major constituents of urine is the di-amine, urea
(III). Each amine group in urea should remind us of ammonia in
Equation (6.11). Solutions of urea in water are basic because the two amine moieties each abstract a proton from water, to generate an ammonium salt and a hydroxide ion:

HOCl, is one of the active components in household bleach.

After this battle, both sides showed reluctance to employ poisonous gases again, being afraid it would drift back and poison their own troops.
Cl2 gas also caused extensive corrosion of rifles and artillery breech blocks, making them unusable.

Reminder: to a chemist, the word basic does not mean ‘elementary’ or ‘fundamental’, but
‘proton abstracting’.

O
H2N

NH2
(III)

O

O
+ 2H 2 O

H2 N

NH2

H3 N

2OH −

NH3

(6.16)

The two OH− ions formed during Equation (6.16) explain why aqueous solutions of urea are alkaline.
As we saw above, chlorine forms hypochlorous acid, HOCl. The hydroxide ions generated from urea react with the hypochlorous acid in a typical acid–base reaction,

244

ACIDS AND BASES

to form a salt and water:

HOCl(aq) + OH− (aq) − → ClO− (aq) + H2 O

(6.17)

where ClO− is the hypochlorite ion.
Equation (6.17) is an example of a neutralization reaction, a topic we discuss in more depth in Section 6.3.

How is sherbet made?
Effervescence and reactions of acids
Sherbet and sweets yielding a fizzy sensation in the mouth generally contain two components, an acid and a simple carbonate
Malic acid (IV) occurs or bicarbonate. A typical reaction of an acid with a carbonate is naturally, and is the effervescence: the generation of gaseous carbon dioxide. In a wellcause of the sharp taste in over-ripe apples. known brand of British ‘fizzy lolly’, the base is sodium bicarbonate and the acid is malic acid (IV). Ascorbic acid (vitamin C) is another common acid included within sherbet.
HO
HOOC

H
CH2COOH
(IV)

The acid and the bicarbonate dissolve in saliva as soon as the ‘fizzy lolly’ is placed in the mouth. If we abbreviate the malic acid to HM (M being the maliate anion), the ‘fizzing’ reaction in the mouth is described by
HM(aq) + HCO− − → M− (aq) + CO2(g) ↑ +H2 O(l)
3(aq) −

(6.18)

where in this case the subscript ‘aq’ means aqueous saliva. The subjective sensation of ‘fizz’ derives from the evolution of gaseous carbon dioxide.
This reaction between an acid and a carbonate is one of the oldest
Soda is an old fashchemical reactions known to man. For example, it says in a portion ioned name for sodium of the Hebrew Bible written about 1000 BC, ‘. . . like vinegar poured carbonate, Na2 CO3 · on soda, is one who sings songs to a heavy heart’, i.e. inappropriate
10H2 O. and lighthearted singing can lead to a dramatic response! The quote may be found in full in the book of Proverbs 25:20.

Ossified means ‘to make rigid and hard.’
The word comes from the Latin ossis, meaning bone.

Why do steps made of limestone sometimes feel slippery?
The typical reactions of an alkali
Limestone is an ossified form of calcium carbonate, CaCO3 . Limestone surfaces soon become slippery and can be quite dangerous if

PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES

245

Table 6.3 Typical properties of Lowry–Brønsted bases
Base property

Example from everyday life

Bases react with an acid to form a salt and water (‘neutralization’)

Bases react with esters to form an alcohol and carboxylic acid (‘saponification’)
Bases can be corrosive

Rubbing a dock leaf (which contains an organic base) on the site of a nettle sting
(which contains acid) will neutralize the acid and relieve the pain
Aqueous solutions of base feel ‘soapy’ to the touch Oven cleaner comprising caustic soda
(NaOH) can cause severe burns to the skin

pools of stagnant rainwater collect. Although calcium carbonate is essentially insoluble in water, minute amounts do dissolve to form a dilute solution, which is alkaline.
The alkali in these water pools reacts with organic matter such as algae and moss growing on the stone. The most common of these reactions is saponification (see
p. 240), which causes naturally occurring esters to split, to form the respective carboxylic acid and an alcohol. Once formed, this carboxylic acid reacts with more alkaline rainwater to form a metal carboxylate, according to

2RCOOH(aq) + CaCO3(aq) − → Ca2+ (RCOO− )2(aq)
+ CO2(aq) + H2 O

(6.19)

The carbon dioxide generated by Equation (6.19) generally remains in solution as carbonic acid, although the rainwater can look a little cloudy because minute bubbles form.
Like most other metal carboxylates, the calcium carboxylate
(Ca(RCOO)2 ) formed during Equation (6.19) readily forms a
‘soap’, the name arising since its aqueous solutions feel slippery and soapy to the touch. The other commonly encountered metal carboxylates are the major components of household soap, which is typically a mixture of potassium stearate and potassium palmitate
(the salts of stearic and palmitic acids).
In summary, steps of limestone become slippery because the stagnant water on their surface is alkaline, thereby generating a solution of an organic soap. Other typical properties of bases and alkalis are listed in Table 6.3.

Metal carboxylates are called soaps because they saponify oils in the skin and decrease the surface tension γ of water, which makes the surfaces more slippery.

The serial TV programs known as ‘soap operas’ earned their name in the USA at a time when much of a programmaker’s funding came from adverts for household soap.

Why is the acid in a car battery more corrosive than vinegar? pH Car batteries generally contain sulphuric acid at a concentration of about 10 mol dm−3 .
It is extremely corrosive, and can generate horrific chemical burns. By contrast, the

246

ACIDS AND BASES

Care: the ‘H’ in pH derives from the symbol for hydrogen, and is always given a big letter. The ‘p’ is a mathematical operator, and is always small.

concentration of the solvated protons in vinegar lies in the range
10−4 –10−5 mol dm−3 .
Between these two acids, there is up to a million-fold difference in the number of solvated protons per litre. We cannot cope with the unwieldy magnitude of this difference and tend to talk instead in terms of the logarithm of the concentration. To this end, we introduce a new concept: the pH. This is defined mathematically as ‘minus the logarithm (to the base ten) of the hydrogen ion concentration’: pH = − log10 [H+ /mol dm−3 ]

An acid’s pH is defined as minus the logarithm (to the base ten) of the hydrogen ion concentration. (6.20)

The concentrations of bench acids in an undergraduate laboratory are generally less than 1 mol dm−3 , so by corollary the minus sign to Equation (6.20) suggests we generally work with positive values of pH. Only if the solution has a concentration greater than
1 mol dm−3 will the pH be negative. Contrary to popular belief,
The ‘p’ in Equation a negative pH is not impossible. (Try inserting a concentration of
(6.20) is the mathe2.0 mol dm−3 into Equation (6.20) and see what happens!) matical operator
‘− log10 ’ of something.
Notice how we generally infer the solvated proton, H3 O+ , each pH means we have time we write a concentration as [H+ ], which helps explain why applied the operator ‘p’ the concept of pH is rarely useful when considering acids disto [H+ ]. The p is short solved in non-aqueous solvents. When comparing the battery acid for potenz, German for with the bench acid, we say that the battery acid has a lower pH power. than does the bench acid, because the number of solvated protons is greater and, therefore, it is more acidic. Figure 6.1 shows the relationship between the concentration of the solvated protons and
The lower the pH, the pH. We now appreciate why the pH increases as the concentration more concentrated the decreases. acid.
Apart from the convenience of the logarithmically compressed scale, the concept of pH remains popular because one of the most popular methods of measuring the acidity of an aqueous solution is the glass electrode
(see p. 336), the measurement of which is directly proportional to pH, rather than to
[H3 O+ ].
We need to introduce a word of caution. Most modern calculators cite an answer with as many as ten significant figures, but we do not know the concentration to more than two or three significant figures. In a related way, we note how the pH of blood is routinely measured to within 0.001 of a pH unit, but most chemical applications
[H3O+]/mol dm−3 10 1 pH −1 0

10−1

10−2

1

2

10−3 . . . 10−7 . . .
3

...

7

...

10−10

10−11

10

11

Figure 6.1 The relationship between concentrations of strong acids and the solution pH

PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES

247

do not require us to cite the pH to more than 0.05 of a pH unit. In fact, we can rarely cite pH to a greater precision than 0.01 for most biological applications.
Worked Example 6.2 What is the pH of bench nitric acid having a concentration of 0.25 mol dm−3 ?
Inserting values into Equation (6.20): pH = − log[0.25 mol dm−3 /mol dm−3 ] = 0.6
The acid has a pH of 0.6.

The concentration of the solvated protons in
Equation (6.18) needs to be expressed in the familiar (but non-SI) units of mol dm−3 ; the
SI unit of concentration is mol m−3 .

SAQ 6.2 What is the pH of hydrochloric acid having a concentration of 0.2 mol dm−3 ?
SAQ 6.3 To highlight the point made above concerning numbers of significant figures, determine the (negligible) difference in [H3 O+ ] between two acid solutions, one having a pH of 6.31 and the other 6.32.

Sometimes we know the pH and wish to know the concentration of the solvated protons. Hence, we need to rewrite Equation (6.20), making [H+ ] the subject, to obtain
[H+ /mol dm−3 ] = 10−pH

(6.21)

We divide the concentration by its units to yield a dimensionless number. We are assuming the concentration of the proton H+ is the same as the concentration of
H3 O+ (aq) .

Worked Example 6.3 What is the concentration of nitric acid having a pH of 3.5?
Inserting values into Equation (6.21):
[H+ ] = 10−3.5 so [H+ ] = 3.16 × 10−4 mol dm−3

SAQ 6.4 What is the concentration of nitric acid of pH = 2.2?

Occasionally, we can merely look at a pH and say straightaway what is its concentration. If the pH is a whole number – call it x – the concentration will take the form 1 × 10−x mol dm−3 . As examples, if the pH is 6, the concentration is 10−6 mol dm−3 ; if the concentration is 10−3 mol dm−3 then the pH is 3, and so on.

Sometimes, we refer to a whole number as an integer. Worked Example 6.4 Without using a calculator, what is the concentration of HNO3 solution if its pH is 4?

248

ACIDS AND BASES

If the pH is x, then the concentration will be 10−x mol dm−3 , so the acid concentration is 10−4 mol dm−3 .
SAQ 6.5 Without using a calculator, what is the pH of hydrochloric acid of concentration 10−5 mol dm−3 ?

Aside
Units
We encounter problems when it becomes necessary to take the logarithm of a concentration (which has units), since it contravenes one of the laws of mathematics. To overcome this problem, we implicitly employ a ‘dodge’ by rewriting the equation as pH = − log10

[H3 O+ ] cO where the term c O is the standard state, which generally has the value of 1 mol dm−3 .
The c O term is introduced merely to allow the units within the bracketed term above to cancel. Throughout this chapter, concentrations will be employed relative to this standard, thereby obviating the problem inherent with concentrations having units.

Justification Box 6.1
Sørenson introduced this definition of pH in
1909.

The definition of pH is given in Equation (6.20) as pH = − log10 [H+ ]

First we multiply both sides by ‘−1’:
−pH = log10 [H+ ] and then take the antilog, to expose the concentration term. Correctly, the function
‘antilog’ means a mathematical operator, which performs the opposite job to the original function. The opposite function to log is a type of exponential. As the log is written in base 10, so the exponent must also be in base 10.
If y = log10 x, then x = 10y . This way we obtain Equation (6.21).

Sometimes we need to know the pH of basic solutions.

PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES

249

Worked Example 6.5 What is the pH of a solution of sodium hydroxide of concentration 0.02 mol dm−3 ? Assume the temperature is 298 K.
At first sight, this problem appears to be identical to those in previous Worked Examples, but we soon appreciate how it is complicated because we need first to calculate the concentration of the free protons before we can convert to a pH. However, if we know the concentration of the alkali, we can calculate the pH thus: pKw = pH + pOH

(6.22)

pKw = − log10 Kw pH = − log10 [H3 O+ ]
(aq)
pOH = − log10 [OH− (aq) ]

where pKw and pH have their usual definitions, and we define pOH as pOH = − log10 [OH− ]

(6.23)

Inserting the concentration [NaOH] = 0.02 mol dm−3 into Equation
(6.22) yields the value of pOH = 2. The value of pKw at 298 K is 14
(see Table 6.2). Therefore: pH = pKw − pOH

We must look up the value of Kw from Table
6.2 if the temperature differs from 298 K, and then calculate a different value using
Equation (6.26).

pH = 14 − 2 pH = 12
SAQ 6.6 What is the pH of a solution of potassium hydroxide of concentration 6 × 10−3 mol dm−3 . Again, assume the temperature is 298 K.

Justification Box 6.2
Water dissociates to form ions according to Equation (6.2). The ionic product of the concentrations is the autoprotolysis constant Kw , according to Equation (6.4).
Taking logarithms of Equation (6.4) yields log10 Kw = log10 [H3 O+ ] + log10 [OH− ]

(6.24)

Next, we multiply each term by ‘−1’ to yield
− log10 Kw = − log10 [H3 O+ ] − log10 [OH− ]

(6.25)

The term ‘− log10 [H3 O+ ]’ is the solution pH and the term ‘− log10 Kw ’ is defined according to pKw = − log10 Kw

(6.26)

We give the name pOH to the third term ‘− log10 [OH− ]’. We thereby obtain Equation (6.22).

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ACIDS AND BASES

Why do equimolar solutions of sulphuric acid and nitric acid have different pHs?
Mono-, di- and tri-basic acids
Equimolar means ‘of equal molarity’, so equimolar solutions have the same concentration.

Nitric acid, HNO3 , readily dissolves in water, where it dissociates according to

HNO3(aq) + H2 O − → 1H3 O+ (aq) + NO−
3(aq)

(6.27)

The stoichiometry illustrates how each formula unit generates a single solvated proton. By contrast, sulphuric acid, H2 SO4 , dissociates in solution according to

H2 SO4(aq) + 2H2 O − → 2H3 O+ (aq) + SO2−
4(aq)

(6.28)

so each formula unit of sulphuric acid generates two solvated protons. In other words, each mole of nitric acid generates only 1 mol of solvated protons but each mole of sulphuric acid generates 2 mol of solvated protons. We say nitric acid is a mono-protic acid and sulphuric acid is a di-protic acid. Tri-protic acids are rare. Fully protonated ethylene diamine tetra-acetic acid H4 EDTA (V) is a tetra-protic acid.
O

OH
O
N

CH2 CH2

N

OH
O

OH
O

OH
(V)

Equation (6.27) demonstrated how the concentration of the solvated protons equates to the concentration of a mono-protic acid from which it derived; but, from Equation
(6.28), the concentration of the solvated protons will be twice the concentration if the parent acid is di-protic. These different stoichiometries affect the pH, as demonstrated now by Worked Examples 6.6 and 6.7.
Worked Example 6.6 Nitric acid of concentration 0.01 mol dm−3 is dissolved in water.
What is its pH?
Since one solvated proton is formed per molecule of acid, the concentration [H+ (aq) ] is also 0.01 mol dm−3 .
The pH of this acidic solution is obtained by inserting values into Equation (6.20): pH = − log10 [0.01] pH = 2

PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES

251

Worked Example 6.7 What is the pH of sulphuric acid having the same concentration in water as the nitric acid in Worked Example 6.6?
This time, two solvated protons are formed per molecule of acid, so the concentration of
[H+ (aq) ] will be 0.02 mol dm−3 .
The pH of this acidic solution is obtained by inserting values into Equation (6.20): pH = − log10 [2 × 0.01] pH = 1.68

A pH electrode immersed in turn into these two solutions would register a different pH despite the concentrations of the parent acids being the same.
We need to be careful with these calculations, because the extent of dissociation may also differ; see p. 255ff.

pH electrodes and pH meters are discussed in
Chapter 7.

What is the pH of a ‘neutral’ solution? pH and neutrality
A medicine or skin lotion is often described as ‘pH neutral’ as though it was obviously a good thing. A solution is defined as neutral if it contains neither an excess of solvated protons nor an excess of hydroxide ions. Equation (6.4) tells us the autoprotolysis constant Kw of super-pure water (water containing no additional solute) is 10−14 (mol dm−3 )2 . Furthermore, we saw in
All neutral solutions
Worked Example 6.1 how the concentration of the solvated protons have a pH of 7 at was 10−7 mol dm−3 at 298 K.
298 K.
By considering both the definition of pH in Equation (6.20) and the concentration of the solvated protons from Worked
Example 6.1, we see how a sample of super-pure water – which is neutral – has a pH of 7 at 298 K. We now go further and say all The maximum pH of neutral solutions have a pH of 7. By corollary, we need to appre- an acid will be just less ciate how an acidic solution always has a pH less than 7. If the pH than 7, at 298 K. is exactly 7, then the solution is neutral.
The pH of a Lowry–Brønsted acid decreases as its concentration The pH of a Lowry– increases. Bench nitric acid of concentration 1 mol dm−3 has a Brønsted acid DEcrepH = 0. An acid of higher concentration will, therefore, have a ases as its concentranegative pH (the occasions when we need to employ such solutions tion INcreases. are, thankfully, rare).

What do we mean when we say blood plasma has a ‘pH of 7.4’?
The pH of alkaline solutions
Table 6.4 lists the pH of many natural substances, and suggests human blood plasma, for example, should have a pH in the range 7.3–7.5. The pH of many natural

252

ACIDS AND BASES

Table 6.4 The pHs of naturally occurring substances, listed in order of decreasing acidity
Human body
Gastric juices
Faeces
Duodenal contents
Urine
Saliva
Milk
Bile
Spinal fluid
Blood plasma

Common foodstuffs
1.0–3.0
4.6–8.4
4.8–8.2
4.8–8.4
6.5–7.5
6.6–7.6
6.8–7.0
7.3–7.5
7.3–7.5

Limes
Rhubarb
Apricots
Tomatoes
Spinach
Salmon
Maple syrup
Tap water
Egg white (fresh)

1.8–2.0
3.1–3.2
3.6–4.0
4.0–4.4
5.1–5.7
6.1–6.3
6.5–7.0
6.5–8.0
7.6–8.0

Source: Handbook of Chemistry and Physics (66th Edition), R. C. Weast (ed.),
CRC Press, Boca Raton, Florida, 1985, page D-146.

substances is higher than 7, so we cannot call them either ‘acidic’ or ‘neutral’. The pHs range from 1.8 for limes (which explains why they taste so sour) to 7.8 or so for fresh egg white (albumen).
Blood ‘plasma’ is that part of the blood remaining after removal of the haemoglobin cells that impart a characteristic ‘blood-red’
The word ‘product’ colour. According to Table 6.4, most people’s plasma has a pH is used here in its in the range 7.3–7.5. So, what is the concentration of solvated mathematical sense of
‘multiplied by’. protons in such plasma? We met the autoprotolysis constant Kw in Equation (6.4). Although we discussed it in terms of super-pure water, curiously the relationship still applies to any aqueous system. The product of the concentrations of solvated protons and hydroxide ions is always 10−14 at 298 K.
If Equation (6.4) applies although the water contains dissolved solute, then we can calculate the concentration of solvated protons and the concentration of hydroxide ions, and hence ascertain what a pH of more than ‘7’ actually means.
Worked Example 6.8 What is the concentration [OH− (aq) ] in blood plasma of pH 7.4?
Answer strategy. (1) We first calculate the concentration of solvated protons from the pH, via Equation (6.20). (2) Second, we compare the concentration [H3 O+ (aq) ] with that of a neutral solution, via Equation (6.4).
(1) The concentration [H3 O+ (aq) ] is obtained from Equation (6.20). Inserting values:
[H3 O+ (aq) ] = 10−7.4 so [H3 O+ (aq) ] = 4 × 10−8 mol dm−3

(2) We see how the concentration obtained in part (1) is in fact less than the 1 ×
10−7 mol dm−3 we saw for pure water, as calculated in Worked Example 6.1.

‘STRONG’ AND ‘WEAK’ ACIDS AND BASES

253

Equation (6.4) says Kw = [H3 O+ (aq) ] × [OH− (aq) ], where the product of the two concentration has a value of 10−14 (mol dm−3 )2 at 298 K. Knowing the values of Kw and
[H3 O+ (aq) ], we can calculate the concentration of hydroxide ions in the blood plasma.
Rearranging to make [OH− (aq) ] the subject, we obtain
[OH− (aq) ] =
Inserting values,
[OH− (aq) ] = so Kw
[H3 O+ (aq) ]

10−14 (mol dm−3 )2
4 × 10−8 mol dm−3

[OH− (aq) ] = 2.5 × 10−7 mol dm−3

We see how the pH of blood plasma is higher than 7, so the concentration of hydroxide ions exceeds their concentration in superpure water. We derive the generalization: aqueous solutions in which the concentration of hydroxide is greater than the concentration of solvated protons have a pH higher than 7. The pH is lower than 7 if the concentration of hydroxide is less than the concentration [H3 O+ (aq) ].

Aqueous solutions in which the concentration of hydroxide exceeds the concentration of solvated protons show a pH higher than 7.

SAQ 6.7 A solution of ammonia in water has a pH of 9. Without using a calculator, what is the concentration of solvated protons and hence what is the concentration of hydroxide ions?
SAQ 6.8 What is the pH of sodium hydroxide solution of concentration
10−2 mol dm−3 ?

6.2

‘Strong’ and ‘weak’ acids and bases
Why is a nettle sting more painful than a burn from ethanoic acid?
Introducing ‘strong’ and ‘weak’ acids
Brushing against a common nettle Urtica dioica can cause a painful sting. The active component in a nettle sting is methanoic acid
(VI), also called ‘formic acid’. The sting of a nettle also contains natural additives to ensure that the methanoic acid stays on the skin, thereby maximizing the damage to its sensitive underlying tissue known as the epidermis.

The sting of a red ant also contains methanoic acid.

254

ACIDS AND BASES

The word ‘epidermis’ derives from two Greek words, derma meaning skin, and epi meaning
‘at’, ‘at the base of’, or
‘in additional to’. The same root epi occurs in ‘epidural’, a form of pain relief in which an injection is made at the base of the dura, located in the spine.

We say an acid is strong if the extent of its ionization is high, and weak if the extent of its ionization is small.

Care: do not confuse the words strong and weak acids with everyday usage, where we usually say something is ‘strong’ if its concentration is large, and
‘weak’ if its concentration is small.

To a chemist, the words ‘strong’ and
‘weak’ relate only to the extent of ionic dissociation.

O
H
O
(VI)

H

The chemical structures of I and VI reveal the strong similarities between ethanoic and methanoic acids, yet the smaller molecule is considerably nastier to the skin. Why? Methanoic acid dissociates in water to form the solvated methanoate anion HCOO− (aq) and a solvated proton in a directly analogous fashion to ethanoic acid dissolving in water; Equation (6.1). In methanoic acid of concentration 0.01 mol dm−3 , about 0.14 per cent of the molecules have dissociated to yield a solvated proton. By contrast, in ethanoic acid of the same concentration, only 0.04 per cent of the molecules have dissociated. We say the methanoic acid is a stronger acid than ethanoic since it yields more protons per mole. Conversely, ethanoic acid is weaker.
We might rephrase this statement, and say an acid is strong if its extent of ionization is high, and weak if the extent of ionization is small. Within this latter definition, both I and VI are weak acids.
In summary, the word ‘acid’ is better applied to methanoic acid than to ethanoic acid, since it is more acidic, and so methanoic acid in a nettle sting is more able to damage the skin than the ethanoic acid in vinegar.
But we need to be careful. In everyday usage, we say often something is ‘strong’ when we mean its concentration is large; similarly, we say something is ‘weak’ if its concentration is small. As a good example, when a strong cup of tea has a dark brown colour (because the compounds imparting a colour are concentrated) we say the tea is ‘strong’. To a chemist, the words ‘strong’ and ‘weak’ relate only to the extent of ionic dissociation.

Why is ‘carbolic acid’ not in fact an acid?
Acidity constants

‘Carbolic acid’ is the old-fashioned name for hydroxybenzene
(VII), otherwise known as phenol. It was first used as an antiseptic to prevent the infection of post-operative wounds. The British surgeon Joseph
(later ‘Lord’) Lister (1827–1912) discovered these antiseptic qualities in 1867 while working as Professor of Medicine at the Glasgow Royal Infirmary. He squirted a

‘STRONG’ AND ‘WEAK’ ACIDS AND BASES

dilute aqueous solution of VII directly onto a post-operative wound and found that the phenol killed all the bacteria, thereby yielding the first reliable antiseptic in an era when medical science was in its infancy.
OH

255

Phenol is a rare example of a stable enol
(pronounced ‘ene-ol’), with a hydroxyl bonded to a C=C bond. Most enols tautomerize to form a ketone.

(VII)

The antibacterial properties of VII are no longer utilized in modern hospitals because more potent antiseptics have now been formulated. But its memory persists in the continued use of ‘carbolic soap’, which contains small amounts of phenol.
Phenol in water is relatively reactive, thereby explaining its potency against bacteria. But phenol dissolved in water contains relatively few solvated protons, so it is not particularly acidic. But its old name is carbolic acid!
Phenol (VII) can dissociate according to
PhOH(aq) + H2 O(l) − → PhO− (aq) + H3 O+ (aq)


(6.29)

where Ph is a phenyl ring and PhO− (aq) is the phenolate anion. An equilibrium constant may be written to describe this reaction:
K=

[PhO− (aq) ][H3 O+ (aq) ]
[PhOH(aq) ][H2 O]

(6.30)

In fact, the water term in the denominator remains essentially constant, since it is always huge compared with all the other terms.
Accordingly, we usually write a slightly altered version of K, crossmultiplying both sides of the equation with the concentration of water to yield
[PhO− (aq) ][H3 O+ (aq) ]
Ka =
(6.31)
[PhOH(aq) ]
The resultant (modified) equilibrium constant is called the acidity constant of phenol, and has the new symbol Ka , which has a value is 10−10 for phenol. Ka is also called the acid constant, the acid dissociation constant or just the dissociation constant. The value of Ka for phenol is clearly tiny, and quantifies just how small the extent is to which it dissociates to form a solvated proton.

The word ‘antiseptic’ comes from the Latin prefix anti meaning
‘before’ or ‘against’, and ‘septic’ comes from the Latin septis, meaning a bacterial infection. An ‘antiseptic’, therefore, prevents the processes or substances causing an infection. Historically, carbolic acid was so called because solid phenol causes nasty chemical burns to the skin. The root carbo comes from the French for ‘coal’.

Care: Do not confuse ‘Ph’ (a common abbreviation for a phenyl ring) with ‘pH’
(which is a mathematical operator meaning
− log10 [H+ (aq) ]).
Ka for phenol is 10−10 when expressing the concentrations with the units of mol dm−3 .

256

ACIDS AND BASES

Gibbs function G

The value of Ka for ethanoic acid is a hundred thousand times larger at 1.8 × 10−5 , and Ka for methanoic acid is ten times larger still, at 1.8 × 10−4 ; so methanoic acid generates more solvated protons per mole of acid than either phenol or ethanoic acid.
We discover the relative differences in Ka when walking in the country, for a nettle can give a nasty sting (i.e. a chemical burn) but vinegar does not burn the skin. We say methanoic acid is a stronger acid than ethanoic acid because its value of Ka is larger. A mole of phenol yields few protons, so we say it is a weak acid, because its value of Ka is tiny.
These descriptions of ‘strong’ and ‘weak’ acid are no longer
A strong acid has a subjective, but depend on the magnitude of Ka : a strong acid has large value of Ka , and a large value of Ka and a weak acid has a low value of Ka . Stated a weak acid has a low another way, the position of the acid-dissociation equilibrium lies value of Ka . close to the reactants for a weak acid but close to the products for a strong acid, as shown schematically in Figure 6.2.
A crude generalization
Carboxylic acids such as ethanoic acid are generally weak besuggests that inorganic cause their values of Ka are small (although see p. 261). By conacids are strong and trast, so called mineral acids such as sulphuric or nitric are classed organic acids are weak. as strong because their respective values of Ka are large. Although there is little consensus, a simplistic rule suggests we class an acid as weak if its value of Ka drops below about 10−3 . The acid is
The values of Ka genstrong if Ka > 10−3 . erally increase with
Table 6.5 contains a selection of Ka values. Acids characterized increasing temperaby large values of Ka are stronger than those with smaller values of ture, causing the acid to
Ka . Each Ka value in Table 6.5 was obtained at 298 K. Being an be stronger at high T . equilibrium constant, we anticipate temperature-dependent values of Ka , with Ka generally increasing slightly as T increases.

WEAK

STRONG

Reactants, i.e. HA + H2O

Products, i.e. H3O+ + A−

0

1

Extent of reaction x

Figure 6.2 Graph of Gibbs function G (as ‘y’) against the extent of reaction ξ (as ‘x’). The minimum of the graph corresponds to the position of equilibrium: the position of equilibrium for a weak acid, such as ethanoic acid, lies near the un-ionized reactants; the position of equilibrium for a strong acid, like sulphuric acid, lies near the ionized products

‘STRONG’ AND ‘WEAK’ ACIDS AND BASES

257

Table 6.5 Acidity (‘dissociation’) constants Ka for inorganic Lowry–Brønsted acids in water at 298 K. Values of Ka are dimensionless: all values presuppose equilibrium constants such as
Equation (6.35), and were calculated with concentrations expressed in mol dm−3
Ka(1)

Acid

Ka(2)

Ka(3)

1.2 × 10−2
1.02 × 10−7
5.61 × 10−11
6.23 × 10−8

2.2 × 10−11

−8

4.0 × 10
1.0 × 107
4.6 × 10−4
1.0 × 102
1.4 × 10−2
4.3 × 10−7
7.53 × 10−3

Hypochlorous, HOCl
Hydrochloric, HCl
Nitrous, HNO2
Sulphuric, H2 SO4
Sulphurous, H2 SO3
Carbonic, H2 CO3
Phosphoric, H3 PO4

Table 6.6 As for Table 6.5, but for inorganic acids and showing the effects of various structural changes
Acid
Effect of extent of halogenation
CH3 COOH
ClCH2 COOH
Cl2 CHCOOH
Cl3 COOH
Effect of halide
FCH2 COOH
ClCH2 COOH
BrCH2 COOH
ICH2 COOH
Effect of chain length
HCOOH
CH3 COOH
CH3 CH2 COOH
CH3 CH2 CH2 COOH
Effect of substituent in benzoic acids
C6 H5 COOH p-NO2 –C6 H4 COOH p-CH3 O–C6 H4 COOH p-NH2 –C6 H4 COOH

105 Ka
1.75
136
5530
23 200
260
136
125
67
17.7
1.75
1.35
1.51
6.3
36.0
3.3
1.4

In summary, carbolic acid (phenol, VII) is an extremely weak acid because its value of Ka is 10−10 , quantifying how small is the concentration of solvated protons in its solutions.

Basicity constants
Having categorized acids into ‘strong’ and ‘weak’ via the concept of acidity constants Ka , we now look at the strengths of various

The cause of phenol’s corrosive properties does not relate to its ability to form solvated protons (as indicated by the value of Ka ) but its ability to penetrate the skin and disrupt the chemical processes occurring within the epidermis, to painful effect. 258

ACIDS AND BASES

bases. It is possible to write an equilibrium constant K to describe the hydrolysis of bases such as ammonia (see Equation (6.12)). We write the appropriate equilibrium constant in just the same way as we wrote an expression for K to describe the acidic behaviour of phenol:
[NH4 + ][OH− ]
K=
(6.32)
We sometimes call the
[NH3 ][H2 O] equilibrium constant in Equation (6.33) a basicity constant, and symbolize it as Kb .

As with the expression in Equation (6.6), this equilibrium constant can be simplified by incorporating the water term into K, thereby yielding a new constant which we will call Kb , the basicity constant:
[NH4 + ][OH− ]
(6.33)
Kb =
[NH3 ]

where Kb in Equation (6.33) is quite different from the K in Equation (6.32). The value of Kb for ammonia is 1.74 × 10−5 , which is quite small, causing us to say ammonia is a weak base. The value of Kb for sodium hydroxide is much larger at
0.6, so we say NaOH is a strong base.
But, curiously, this new equilibrium constant Kb is redundant because we could have calculated its value from known values of Ka according to
Ka × Kb = Kw

(6.34)

where Kw is the autoprotolysis constant of water from p. 236. Older textbooks sometimes cite values of Kb , but we really do not need to employ two separate K constants.
SAQ 6.9 What is the value of Ka for the ammonium ion, NH4 + ? Take Kb from the paragraphs immediately above, and Kw = 10−14 .

Justification Box 6.3
Consider a weak acid, HA, dissociating: HA → H3 O+ + A− . Its acidity constant Ka is given by
Ka =

[A− ][H3 O+ ]
[HA]

(6.35)

and then consider a weak base (the conjugate of the weak acid) forming a hydroxide ion in solution, H2 O + A− → OH− + HA. Its basicity constant is given by
Kb =

[HA][OH− ]
[A− ]

(6.36)

‘STRONG’ AND ‘WEAK’ ACIDS AND BASES

259

Multiplying the expressions for Ka and Kb yields
Ka × Kb =

[A− ][H3 O+ ] [HA][OH− ]
×
[HA]
[A− ]

The HA and A− terms clearly cancel to yield [H3 O+ ][OH− ], which is Kw .

Why does carbonic acid behave as a mono-protic acid?
Variations in the value of Ka
Carbonic acid, H2 CO3 , is naturally occurring, and forms when carbon dioxide from the air dissolves in water. From its formula, we expect it to be a di-protic acid, but it is generally classed as mono-protic. Why?
In water at 298 K, the ionization reaction follows the equation

H2 CO3(aq) + H2 O − → HCO− + H3 O+ (aq)
3(aq)

(6.37)

The value of Ka for the reaction in Equation (6.37) is 4.3 × 10−7 , so carbonic acid is certainly a very weak acid. The hydrogen carbonate anion HCO3 − could dissociate further, according to

HCO− 3(aq) + H2 O − → CO2− + H3 O+ (aq)
3(aq)

(6.38)

but its value of Ka is low at 5.6 × 10−11 , so we conclude that the HCO− ion is too
3
weak an acid to shed its proton under normal conditions. Thus, carbonic acid has two protons: the loss of the first one is relatively easy, but the proportion of molecules losing both protons is truly minute. Only one of the protons is labile.
This situation is relatively common. If we look, for example, at the values of
Ka in Table 6.5, we see that phosphoric acid is a strong acid insofar as the loss of the first proton occurs with Ka = 7.5 × 10−3 , but the loss of the second proton, to form HPO2− , is difficult, as characterized by Ka = 6.2 × 10−8 . In other words, the
4
dihydrogen phosphate anion H2 PO− is a very weak acid. And the hydrogen phosphate
4
di-anion HPO2− has a low a value of Ka = 2.2 × 10−11 , causing us to say that the
4
PO3− anion does not normally exist. Even the loss of the second proton of sulphuric
4
acid is characterized by a modest value of Ka = 10−2 .
This formal definition of Ka can be extended to multi-protic Multi-protic acids have acids. We consider the dissociation to occur in a step-wise manner, a different value of Ka the acid losing one proton at a time. Consider, for example, the for each proton donation step, with the two-proton donation reactions of sulphuric acid:

(1) H2 SO4 − → H+ + HSO−
4
(2) HSO4 − − → H+ + SO2−

4

values of Ka decreasing with each proton donation step.

260

ACIDS AND BASES

The equilibrium constant for the dissociation of H2 SO4 :
The subscript (1) tells us we are considering the first proton to be lost. Ka(1) =

[HSO− ][H+ ]
4
[H2 SO4 ]

Ka(2) =

[SO2− ][H+ ]
4
[HSO4 − ]

Similarly,
Ka(1) is always bigger than Ka(2) .

In fact, we can also extend this treatment to bases, looking at the step-base addition of protons.
SAQ 6.10 The tetra-protic acid H4 EDTA (V) has four possible proton equilibrium constants. Write an expression for each, for Ka(1) to Ka(4) .

Why is an organic acid such as trichloroethanoic acid so strong?
Effect of structure on the Ka of a weak acid
The value of Ka for trichloroacetic acid CCl3 COOH (VIII) is very large at 0.23.
Indeed, it is stronger as an acid than the HSO− ion – quite remarkable for an organic
4
acid!

Cl
Cl
Cl

O

(VIII)

OH

Let us return to the example of ethanoic acid (I). The principal structural difference between I and VIII is the way we replace each of the three methyl protons in ethanoic acid with chlorine atoms.
The three methyl protons in I are slightly electropositive, implying that the central carbon of the –CH3 group bears a slight negative charge. This excess charge is not large, but it is sufficient to disrupt the position of the acid-dissociation equilibrium, as follows. Although the undissociated acid has no formal charge, the ethanoate anion has a full negative charge, which is located principally on the carboxyl end of the anion. It might be easier to think of this negative charge residing on just one of the oxygen atoms within the anion, but in fact both oxygen atoms and
Delocalization is a the central carbon each bear some of the charge. We say the charge means of stabilizing is delocalized, according to structure IX, which is a more accurate an ion. representation of the carboxylate anion than merely –COO− . The

TITRATION ANALYSES

261

right-hand structure of IX is effectively a mixture of the two structures to its left.
Note how the name resonance implies charge delocalization.
O
R

O
R

O

O
R

O

O

A double-headed arrow
‘↔’ indicates resonance.

(IX)

To reiterate, the hydrogen atoms in the methyl group are slightly electropositive, with each seeking to relocate their own small amounts of charge onto the central carboxyl carbon. In consequence, the ethanoic anion (cf. structure
IX with R = CH3 ) has a central carbon bearing a larger negative Ions are more likely to charge – both from the ionization reaction but also from the hydro- form if they are stable, gen atoms of the –CH3 group. In consequence, the central carbon and less likely to form of the ethanoate anion is slightly destabilized; and any chemical if unstable. species is less likely to form if it is unstable.
Next we look at the structure of trichloroethanoic acid (VIII). In contrast to the hydrogen atoms of ethanoic acid, the three chlorine atoms are powerfully electron withdrawing. The chlorine atoms cause extensive delocalization of the negative charge on the Cl3 COO− anion, with most of the negative charge absorbed by the three chlorine atoms and less on the oxygen atoms of the carboxyl. Such a relocation of charge stabilizes the anion; and any chemical species is more likely to form if it is stable.
Statistically, we find fewer ethanoate anions than trichloroethanoate anions in the respective solutions of the two acids. And if there are fewer ethanoate anions in solution per mole of ethanoic acid, then there will be fewer solvated protons. In other words, the extent to which ethanoic acid dissociates is less than the corresponding extent for trichloroethanoic acid. I is a weak acid and VIII is strong; dipping a simple pH electrode into a solution of each of the two acids rapidly demonstrates this truth.
This sort of delocalization stabilizes the ion; in fact, the Cl3 COO− anion is more stable than the parent molecule, Cl3 COOH. For this reason, the solvated anion resides in solution in preference to the acid. Ka is therefore large, making trichloroacetic acid one of the strongest of the common organic acids.
Trifluoroethanoic acid (probably better known as trifluoroacetic acid, TFA) is stronger still, with a value of Ka = 1.70.

6.3

Titration analyses
Why does a dock leaf bring relief after a nettle sting?
Introducing titrations
We first met nettle stings on p. 253, where methanoic (‘formic’) acid was identified as the active toxin causing the pain. Like its

The common dock leaf,
Rumex obtusifolia, and the yellow dock leaf,
Rumex crispus, are in fact equally common.

262

ACIDS AND BASES

The naturally occurring substance histamine causes blood capillaries to dilate and smooth muscle to contract.
Most cells release it in response to wounding, allergies, and most inflammatory conditions. Antihistamines block the production of this substance, thereby combating a painful swelling. structurally similar sister, ethanoic acid (I), methanoic acid dissociates in water to yield a solvated proton, H3 O+ (aq) .
Rubbing the site of the sting with a crushed dock leaf is a simple yet rapid way of decreasing the extent of the pain. In common with many other weeds, the sap of a dock leaf contains a mixture of natural amines (e.g. urea (III) above), as well as natural antihistamines to help decrease any inflammation. The amines are solvated and, because the sap is water based, are alkaline. Being alkaline, these amines react with methanoic acid to yield a neutral salt, according to
O

O
H

H
O
H




N

H
R

O−

+

H3 N

R

(6.39)

H

where R is the remainder of the amine molecule. We see how the process of pain removal involves a neutralization process.

Notice how the lone pair on nitrogen of the amine attracts a proton from the carboxylic acid. How do indigestion tablets work?
Calculations concerning neutralization

Excess acid in the stomach is one of the major causes of indigestion, arising from a difficulty in digesting food. The usual cause of such indigestion is the stomach simply containing too much hydrochloric acid, or the stomach acid having too high a concentration (its pH should be about 3). These failures cause acid to remain even when all the food has been digested fully. The excess acid is not passive, but tends to digest the lining of the stomach to cause an ulcer, or reacts by alternative reaction routes, generally resulting in ‘wind’, the gases of which principally comprises methane.
Most indigestion tablets are made of aluminium or magnesium hydroxides. The hydroxide in the tablet removes the excess stomSome indigestion ach acid via a simple acid–base neutralization reaction: tablets contain chalk

(CaCO3 ) but the large volume of CO2 produced (cf. Equation
(6.19)) can itself cause dyspepsia. −
3HCl(stomach) + Al(OH)3(tablet) − → 3H2 O + AlCl3(aq)

(6.40)

The cause of the indigestion is removed because the acid is consumed. Solid (unreacted) aluminium hydroxide is relatively insoluble in the gut, and does not dissolve to generate an alkaline solution.
Rather, the outer layer of the tablet dissolves slowly, with just sufficient entering solution to neutralize the acid. Tablet dissolution stops when the neutralization reaction is complete.

TITRATION ANALYSES

A similar process occurs when we spread a thick paste of zinc and castor oil on a baby’s bottom each time we change its nappy. The
‘zinc’ is in fact zinc oxide, ZnO, which, being amphoteric, reacts with the uric acid in the baby’s urine, thereby neutralizing it.
Worked Example 6.9 But how much stomach acid is neutralized by a single indigestion tablet? The tablet contains 0.01 mol of MOH, where ‘M’ is a monovalent metal and M+ its cation.
We first consider the reaction in the stomach, saying it proceeds with
1 mol of hydrochloric acid reacting with 1 mol of alkali:
−→
MOH(s) + H3 O+ (aq) − − M+ (aq) + 2H2 O

263

Some campaigners believe the AlCl3 produced by Equation
(6.40) hastens the onset of Alzheimer’s disease. Certainly, the brains of people with this nasty condition contain too much aluminium.

(6.41)

M+ is merely a cation. We say Equation (6.41) is a 1:1 reaction, occurring with a 1:1 stoichiometry. Such a stoichiometry simplifies the calculation; the 3:1 stoichiometry in
Equation (6.40) will be considered later.
From the stoichiometry of Equation (6.41), we say the neutralization is complete after equal amounts of acid and alkali react. In other words, we neutralize an amount n of hydrochloric acid with exactly the same amount of metal hydroxide, i.e. with
1 × 10−2 mol.
The tablet can neutralize 0.01 mol of stomach acid.

This simple calculation illustrates the fundamental truth underlying neutralization reactions: complete reaction requires equal amounts of acid and alkali. In fact, the primary purpose of a titration is to measure an unknown amount of a substance in a sample, as determined via a chemical reaction with a known amount of a suitable reagent. We perform the titration to ascertain when an equivalent amount of the reagent has been added to the sample. When the amount of acid and alkali are just equal, we have the equivalence point, from which we can determine the unknown amount.
In a typical titration experiment, we start with a known volume of sample, call it V(sample) . If we know its concentration c(sample) , we also know the amount of it, as V(sample) × c(sample) . During the course of the titration, the unknown reagent is added to the solution, usually drop wise, until the equivalence point is reached (e.g. determining the endpoint by adding an indicator; see p. 273ff).
At equivalence, the amounts of known and unknown reagents are the same, so n(sample) = n(unknown) . Knowing the amount of sample and the volume of solution of the unknown, we can calculate the concentration of the unknown.

The experimental technique of measuring out the amount of acid and alkali needed for neutralization is termed a titration. The amounts of acid and alkali are equal at the equivalence point.
The linguistic similarity between these two words is no coincidence!

We need equal numbers of moles of acid and alkali to effect neutralization. 264

ACIDS AND BASES

Worked Example 6.10 The methanoic acid from a nettle sting is extracted into 50 cm3 of water and neutralized in the laboratory by titrating with sodium hydroxide solution.
The concentration of NaOH is 0.010 mol dm−3 . The volume of NaOH solution needed to neutralize the acid is 34.2 cm3 . What is the concentration c of the acid?
Unlike the Worked Example 6.9, we do not know the number of moles n of either reactant, we only know the volumes of each. But we do know one of the concentrations.
Answer strategy. (1) First, we calculate the amount n of hydroxide required to neutralize the acid. (2) We equate this amount n with the amount of acid neutralized by the alkali.
(3) Knowing the amount of acid, we finally calculate its concentration.
(1) To determine the amount of alkali, we first remember the definition of concentration c as concentration, c =

We could have achieved this conversion with quantity calculus: knowing there are
1000 cm3 per dm3 (so
10−3 dm3 cm−3 ). In SI units, we write the volume as 34.2 cm3 ×
10−3 dm3 cm−3 . The units of cm3 and cm−3 cancel to yield V =
0.0342 dm3 .
Notice now the units of dm3 and dm−3 cancel out here.

(6.42)

and rearrange Equation (6.41) to make amount n the subject, i.e. n=c×V The volume V of alkali is 34.2 cm3 . As there are 1000 cm3 in a litre, this volume equates to (34.2 ÷ 1000) dm3 = 0.0342 dm3 . Accordingly n = 0.0342 dm3 × 0.010 mol dm−3 = 3.42 × 10−4 mol
(2) The reaction between the acid and alkali is a simple 1:1 reaction, so 3.42 × 10−4 mol of alkali reacts with exactly 3.42 × 10−4 mol of acid.
(3) The concentration of the acid is given by Equation (6.42) again.
Inserting values: concentration c =

Notice how we converted the volume of acid solution (50 cm3 ) to 0.05 dm3 .

amount, n volume, V

3.42 × 10−4 mol
0.05 dm3

c = 6.8 × 10−3 mol dm−3
After extraction, the concentration of the methanoic acid is 0.068 mol dm−3 .

An altogether simpler and quicker way of calculating the concentration of an acid during a titration is to employ the equation c(acid) × V(acid) = c(alkali) × V(alkali)

(6.43)

where the V terms are volumes of solution and the c terms are concentrations.

TITRATION ANALYSES

265

Worked Example 6.11 A titration is performed with 25 cm3 of NaOH neutralizing
29.4 cm3 of nitric acid. The concentration of NaOH is 0.02 mol dm−3 . Calculate the concentration of the acid.
We rearrange Equation (6.43), to make c(acid) the subject: c(acid) =

c(alkali) V(alkali)
V(acid)

(6.44)

We then insert values into Equation (6.44): c(acid) =

0.02 mol dm3 × 25 cm3
29.4 cm3

c(acid) = 0.017 mol dm−3

We obtain here a ratio of volumes (V(alkali) ÷
V(acid) ), enabling us to cancel the units of the two volumes. Units are irrelevant if both volumes have the same units. Justification Box 6.4
A definition of the point of ‘neutralization’ in words says, ‘at the neutralization point, the number of moles of acid equals the number of moles of hydroxide’. We re-express the definition as
(6.45)
n(acid) = n(alkali)
Next, from Equation (6.42), we recall how the concentration of a solution c when multiplied by its respective volume V equals the number of moles of solute: n = c × V .
Clearly, n(acid) = c(acid) × V(acid) , and n(alkali) = c(alkali) × V(alkali) .
Accordingly, substituting for n(acid) and n(alkali) into Equation (6.45) yields Equation (6.43).

SAQ 6.11 What volume of NaOH (of concentration 0.07 mol dm−3 ) is required to neutralize 12 cm3 of nitric acid of concentration 0.05 mol dm−3 ?

Aside
Equation (6.43) is a simplified version of a more general equation: c(acid) = s

c(alkali) V(alkali)
V(acid)

(6.46)

where s is the so-called stoichiometric ratio.
For the calculation of a mono-protic acid with a mono-basic base, the stoichiometry is simply 1:1 because 1 mol of acid reacts with 1 mol of base. We say the stoichiometric ratio s = 1. The value of s will be two if sulphuric acid reacts with NaOH since 2 mol of base are required to react fully with 1 mol of acid. For the reaction of NaOH with citric acid, s = 3; and s = 4 if the acid is H4 EDTA.

266

ACIDS AND BASES

The value of s when Ca(OH)2 reacts with HNO3 will be 1 , and the value when citric
2
acid reacts with Ca(OH)2 will be 3 .
2

SAQ 6.12 What volume of Ca(OH)2 (of concentration 0.20 mol dm−3 ) is required to neutralize 50 cm3 of nitric acid of concentration 0.10 mol dm−3 ?

An alternative way of determining the endpoint of a titration is to monitor the pH during a titration, and plot a graph of pH
(as ‘y’) against volume V of alkali added (as ‘x’). Typically, the concentration of the acid is unknown, but we know accurately the concentration of alkali. Figure 6.3 shows such as graph – we call it a pH curve – in schematic form. The shape is sigmoidal, with the pH changing very rapidly at the end point.
In practice, we obtain the end point by extrapolating the two linear regions of the pH curve (the extrapolants should be parallel).
A third parallel line is drawn, positioned exactly midway between the two extrapolants. The volume at which this third line crosses the pH curve indicates the end point. Knowing the volume V(end point) , we can calculate the concentration of the acid via a calculation similar to Worked
Example 6.11.
Incidentally, the end point also represents the volume at which the pH changes most dramatically, i.e. the steepest portion of the graph. For this reason, we occasionally plot a different graph of gradient (as ‘y’) against volume V (as ‘x’); see Figure 6.4.
We obtain the gradient as ‘ pH ÷ V ’. The end point in Figure 6.4 relates to the graph maximum.

pH

Sigmoidal literally means ‘shaped like a Greek sigma ς’. The name derives from the
Greek word sigmoides, meaning ‘sigma-like’.
(There are two Greek letters called sigma, used differently in word construction. The other has the shape σ .)

Volume at the end point

Volume of alkali added V

Figure 6.3 A schematic pH curve for the titration of a strong acid with a strong base. At the equivalence point, the amount of alkali added is the same as the amount of acid in solution initially, allowing for an accurate calculation of the acid’s concentration. Note how the end point is determined by extrapolating the linear regions, and drawing a third parallel line between them 267

∆pH
∆V

pH BUFFERS

Volume at the end point
Volume of alkali added V

Figure 6.4 A schematic of the first derivative of the pH curve in Figure 6.3. The end-point volume is determined as the volume at the peak. A first derivative plot such as this can yield a more accurate end point than drawing parallel lines on Figure 6.3

6.4

pH buffers
Why does the pH of blood not alter after eating pickle?
Introduction to buffers
A ‘pickle’ is a food preserved in vinegar (ethanoic acid). Pickles generally have a sharp, acidic flavour in consequence of the acid preservative. Many systems – especially living cells – require their pH to be maintained over a very restricted range in order to prevent catastrophic damage to the cell. Enzymes and proteins denature, for example, if the pH deviates by more than a fraction. Traces of the food we eat are readily detected in the blood quite soon after eating, so why does the concentration in the blood remain constant, rather than dropping substantially with the additional acid in our diet?
Before we attempt an answer, look again at Figure 6.3, which clearly shows an almost invariant pH after adding a small volume of alkali. Similarly, at the righthand side of the graph the pH does not vary much. We see an insensitivity of the solution pH to adding acid or alkali; only around the end point does the pH alter appreciably. The parts of the titration graph having an invariant pH are termed the buffer regions, and we call the attendant pH stabilization a buffer action.
In a similar way, blood does not change its pH because it contains suitable concentrations of carbonic acid and bicarbonate ion, which act as a buffer, as below.

Why are some lakes more acidic than others? Buffer action
Acid rain is the major cause of acidity in open-air lakes and ponds
(see p. 237). Various natural oxides such as CO2 dissolve in water

Pollutant gases include
SO2 , SO3 , NO and NO2 .
It is now common to write SOx and NOx to indicate this variable valency within the mixture. 268

ACIDS AND BASES

pH

to generate an acid, so the typical pH of normal rainwater is about 5.6; but rainwater becomes more acidic if pollutants, particularly SOx and NOx , in addition to natural
CO2 , dissolve in the water. As an example, the average pH of rain in the eastern United
States of America (which produces about one-quarter of the world’s pollution) lies in the range 3.9–4.5. Over a continental landmass, the partial pressure of SO2 can be as high as 5 × 10−9 × p O , representing a truly massive amount of pollution.
After rainfall, the pH of the water in some lakes does not change, whereas others rapidly become too acidic to sustain aquatic life.
Remember: ‘weak’ in
Why? The difference arises from the buffering action of the water. this sense indicates the extent to which a weak
Some lakes resist gross change in pH because they contain other acid dissociates, and chemicals that are able to take up or release protons into solution does not relate to its following the addition of acid (in the rain). These chemicals in concentration. the lake help stabilize the water pH, to form a buffer. Look at
Figure 6.5, which shows a pH curve for a weak acid titrated with an alkali. Figure 6.5 is clearly similar to Figure 6.3 after the end-point volume, but it has a much shallower curve at lower volumes. In fact, we occasionally have difficulty ascertaining a clear end point because the curvature is so pronounced.
A buffer comprises (1) a weak acid and a salt of that acid, (2) a weak base and a salt of that base, or (3) it may contain an acid salt. We define an acid–base buffer as
‘a solution whose pH does not change after adding (small amounts of) a strong acid or base’. Sodium ascorbate is a favourite buffer
A buffer is a solution in the food industry. of constant pH, which
We can think of water entering the lake in terms of a titration. resists changes in pH
A solution of alkali enters a fixed volume of acid: the alkaline following the addition of small amounts of solution is water entering from the lake’s tributary rivers, and the acid or alkali. acid is the lake, which contains the weak acid H2 CO3 (carbonic acid) deriving from atmospheric carbon dioxide. The alkali in the tributary rivers is calcium hydroxide Ca(OH)2 , which enters the

Volume at the end point pKa Half volume at the end point

Volume of alkali added V

Figure 6.5 A typical pH curve for the titration of carbonic acid (a weak acid) with a strong base.
The concentration of H2 CO3 and HCO− are the same after adding half the neutralization volume
3
of alkali. At this point, pH = pKa

pH BUFFERS

water as the river passes over the limestone floor of river basins.
Calcium hydroxide is a fairly strong base.
Figure 6.5 shows a buffering action since the pH does not change particularly while adding alkali to the solution. In fact, as soon as the alkali mixes with the acid in the lake, its hydroxide ions are neutralized by reaction with solvated protons in the lake, thereby resisting changes in the pH. Figure 6.5 shows how little the lake pH changes; we term the relatively invariant range of constant pH the buffer region of the lake water. The mid pH of the buffer region corresponds quite closely to the pKa of the weak acid (here H2 CO3 ), where the pKa is a mathematical function of Ka , as defined by pKa = − log10 Ka

(6.47)

269

Limestone or chalk dissolve in water to a limited extent. The CaCO3 decomposes naturally to form Ca(OH)2 , thereby generating alkaline water.
The natural buffers in the lake ‘mop up’ any additional alkali entering the lake from the tributary rivers, thereby restricting any changes to the pH.

As a good generalization, the buffer region extends over the range of pKa ± 1.
Only when all the acid in the lake has been consumed will the pH A buffer is only really rise significantly. In fact, the end point of such a titration is gauged effective at restrictwhen the pH rises above pH 7, i.e. the pH of acid–base neutrality. ing changes if the pH
The pH of the lake water fluctuates when not replenished by the remains in the range alkaline river water in the tributary rivers. In fact, the pH of the pKa ± 1. lake water drops significantly each time it rains (i.e. when more
H2 CO3 enters the lake). If the amounts of acid and alkali in the water remain relatively low, then the slight fluctuations in water pH will not be great enough to kill life forms in the lake.
The system above describes the addition of alkali to a lake containing a weak acid. The reverse process also occurs, with acid being added to a base, e.g. when the tributary rivers deliver acid rain to a lake and the lake basin is made of limestone or chalk. In such a case, the lake pH drops as the acid rain from the rivers depletes the amounts of natural Ca(OH)2 dissolved in the lake.
As a further permutation, adding a strong acid to a weak base also yields a buffer solution, this time with a buffer region centred on the pKa of the base. The pH at the end point will be lower than 7.

Buffers
Each species within a buffer solution participates in an equilibrium reaction, as characterized by an equilibrium constant K. Adding an acid (or base) to a buffer solution causes the equilibrium to shift, thereby preventing the number of protons from changing, itself preventing changes in the pH. The A buffer is a solution change in the reaction’s position of equilibrium is another mani- of a weak acid mixed with its conjugate base, festation of Le Chatelier’s principle (see p. 166). which restricts changes
One of the most common buffers in the laboratory is the soto the pH. called ‘phosphate buffer’, which has a pH of 7.0. It comprises salts

270

ACIDS AND BASES

of hydrogen phosphate and dihydrogen phosphate, in the following equilibrium:
H2PO42−



HPO4−

+ H+

(6.48) conjugate acid

conjugate base

If this example were to proceed in the reverse direction, then the hydrogen phosphate
(on the right) would be the base, since it receives a proton, and the dihydrogen phosphate (on the left) would be the conjugate the acid.
The equilibrium constant of the reaction in Equation (6.48) is given by
Ka =

[H+ ][HPO2− ]
4
[H2 PO− ]
4

(6.49)

Notice how the equilibrium constant K in Equation (6.49) is also an acidity constant, hence the subscripted ‘a’. The value of K remains constant provided the temperature is not altered.
Now imagine adding some acid to the solution – either by mistake or deliberately.
Clearly, the concentration of H+ will increase. To prevent the value of Ka changing, some of the hydrogen phosphate ions combine with the additional protons to form dihydrogen phosphate (i.e. Equation (6.48) in reverse). The position of the equilibrium adjusts quickly and efficiently to ‘mop up’ the extra protons in the buffer solution. In summary, the pH is prevented from changing because protons are consumed.

How do we make a ‘constant-pH solution’?
The Henderson–Hasselbach equation
We often need to prepare a solution having a constant pH. Such solutions are vital in the cosmetics industry, as well as when making foodstuffs and in the more traditional experiments performed by the biologist and physical chemist.
To make such a solution, we could calculate exactly how many moles of acid to add to water, but this method is generally difficult, since even small errors in weighing the acid can cause wide fluctuations in the pH. Furthermore, we cannot easily weigh out one of acid oxides such as NO. Anyway, the pH of a weak acid does not clearly follow the acid’s concentration (see p. 254).
The Henderson–Hasselbach equation, Equation (6.50), relates the pH of a buffer solution to the amounts of conjugate acid and
In some texts, Equaconjugate base it contains: tion (6.50) is called the
Henderson–
HasselbaLch equation.

pH = pKa + log10

[A− ]
[HA]

(6.50)

We follow the usual pattern here by making a buffer with a weak acid HA and a solution of its conjugate base, such as the sodium salt of the respective anion, A− .

pH BUFFERS

271

We can prepare a buffer of almost any pH provided we know the pKa of the acid; and such values are easily calculated from the Ka values in Table 6.5 and in most books of physical chemistry and Equation (6.50). We first choose a weak acid whose pKa is relatively close to the buffer pH we want. We then need to measure out accurately the volume of acid and base solutions, as dictated by Equation (6.50).
Worked Example 6.12 We need to prepare a buffer of pH 9.8 by mixing solutions of ammonia and ammonium chloride solution. What volumes of each are required? Take the Ka of the ammonium ion as 6 × 10−10 . Assume the two solutions have the same concentration before mixing.
Strategy: (1) We calculate the pKa of the acid. (2) We identify which component is the acid and which the base. (3) And we calculate the proportions of each according to
Equation (6.50).
(1) From Equation (6.50), we define the pKa as − log10 Ka . Inserting values, we obtain a pKa of 9.22.
(2) The action of the buffer represents the balanced reaction, NH4 Cl → NH3 + HCl, so NH4 Cl is the acid and NH3 is the base.
(3) To calculate the ratio of acid to base, we insert values into Equation (6.50):
9.8 = 9.2 + log10
0.6 = log10

[NH3 ]
[NH+ ]
4

[NH3 ]
[NH+ ]
4

Taking antilogs of both sides to remove the logarithm, we obtain
100.6 =

[NH3 ]
[NH+ ]
4

[NH3 ]
=4
[NH+ ]
4
So, we calculate the buffer requires four volumes of ammonia solution to one of ammonium (as the chloride salt, here).

We are permitted to calculate a ratio like this if the concentrations of acid and conjugate base are the same. SAQ 6.13 What is the pH of ammonia–ammonium buffer if three volumes of NH4 Cl are added to two volumes of NH3 ?

We want a buffer solution because its pH stays constant after adding small amounts of acid or base. Consider the example of adding hydrochloric acid to a buffer, as described in the following Worked Example.
Worked Example 6.13 Consider the so-called ‘acetate buffer’, made with equal volumes of sodium ethanoate and ethanoic acid solutions. The concentration of each solution is 0.1 mol dm−3 . A small volume (10 cm3 ) of strong acid (HCl of concentration
1 mol dm−3 ) is added to a litre of this buffer. The pH before adding HCl is 4.70. What is its new pH?

272

ACIDS AND BASES

Strategy: we first calculate the number of moles of hydrochloric acid added. Second, we calculate the new concentrations of ethanoic acid and ethanoate. And third, we employ the Henderson–Hasselbach equation once more.
(1) 10 cm3 represents one-hundredth of a litre. From
Equation (6.42), the number of moles is 0.01 mol.
(2) Before adding the hydrochloric acid, the concentrations of ethanoate and ethanoic acid are constant at 0.1 mol dm−3 . The hydrochloric acid added reacts with the conjugate base in the buffer
(the ethanoate anion) to form ethanoic acid. Accordingly, the concentration [CH3 COO− ] decreases and the concentration [CH3 COOH] increases. (We assume the reaction is quantitative.) Therefore, the concentration of ethanoate is (0.1 − 0.01) mol dm−3 =
0.09 mol dm−3 . The concentration of ethanoic acid is (0.1 + 0.01) mol dm−3 =
0.11 mol dm−3 .
(3) Inserting values into Equation (6.50):

The acetate buffer is an extremely popular choice in the food industry. The buffer might be described on a food packet as an acidity regulator.

pH = 4.70 + log10

0.09
0.11

pH = 4.70 + log (0.818) pH = 4.70 + (−0.09) so pH = 4.61

So, we see how the pH shifts by less than one tenth of a pH unit after adding quite a lot of acid. Adding this same amount of HCl to distilled water would change the pH from 7 to 2, a shift of five pH units.
SAQ 6.14 Consider the ammonia–ammonium buffer in Worked Example 6.12. Starting with 1 dm3 of buffer solution containing 0.05 mol dm−3 each of NH3 and NH4 Cl, calculate the pH after adding 8 cm3 of NaOH solution of concentration 0.1 mol dm−3 .

Justification Box 6.5
We start by writing the equilibrium constant for a weak acid HA dissociating in water,
HA + H2 O → H3 O+ + A− , where each ion is solvated. The dissociation constant for the acid Ka is given by Equation (6.35):
[H3 O+ ][A− ]
[HA]
where, as usual, we ignore the water term. Taking logarithms of Equation (6.35) yields
Ka =

log10 Ka = log10

[H3 O+ ][A− ]
[HA]

(6.51)

ACID–BASE INDICATORS

273

We can split the fraction term in Equation (6.51) by employing the laws of logarithms, to yield log10 Ka = log10 [H3 O+ ] + log10

[A− ]
[HA]

(6.52)

The term ‘log10 Ka ’ should remind us of pKa (Equation 6.52), and the term log10 [H3 O+ ] will remind us of pH in Equation (6.20), so we rewrite Equation (6.52) as
[A− ]
(6.53)
[HA] which, after a little rearranging, yields the Henderson–Hasselbach equation,
Equation (6.50).
−pKa = −pH + log10

6.5

Acid–base indicators
What is ‘the litmus test’? pH indicators

The name ‘litmus’ comes from the Old
Norse litmosi, which derives from litr and mosi, meaning dye and moss respectively.

Litmus is a naturally occurring substance obtained from lichen. It imparts an intense colour to aqueous solutions. In this sense, the indicator is a dye whose colour is sensitive to the solution pH.
If the solution is rich in solvated protons (causing the pH to be Much of the litmus less than 7) then litmus has an intense red colour. Conversely, a in a laboratory is solution rich in hydroxide ions (with a pH greater than 7) causes pre-impregnated on the litmus to have a blue colour. dry paper.
To the practical chemist, the utility of litmus arises from the way its colour changes as a function of pH. Placing a single drop of litmus solution into a beaker of solution allows us an instant test of Litmus is an indicator. the acidity (or lack of it). It indicates whether the pH is less than To avoid ambiguity,
7 (the litmus is red, so the solution is acidic), or the pH is greater we shall call it an than 7 (the litmus is blue, so the solution is alkaline). Accordingly, ‘acid–base indicator’ or a ‘pH indicator’. we call litmus a pH indicator.
In practical terms, we generally employ litmus during a titration.
The flask will contain a known volume of acid of unknown con- Litmus often looks centration, and we add alkali from a burette. We know we have purple–grey at the reached neutralization when the Litmus changes from red (acid neutralization point. in excess) and just starts changing to blue. We know the pH of This colour tells us we the solution is exactly 7 when neutralization is complete, and then have a mixture of both note the volume of the alkali, and perform a calculation similar to the red and blue forms of litmus.
Worked Example 6.11.
The great English scientist Robert Boyle (1627–1691) was the first to document the use of natural vegetable dyes as acid–base indicators.

274

ACIDS AND BASES

Why do some hydrangea bushes look red and others blue? The chemical basis of acid–base indicators
Hydrangeas (genus Hydrangea) are beautiful bushy plants having multiple flower heads. In soils comprising much compost the flowers have a blue colour, but in soils with much lime or bone meal the heads are pink or even crimson–pink in colour. Very occasionally, the flowers are mauve. ‘Lime’ is the old-fashioned name for calcium oxide, and is alkaline; bone meal contains a lot of phosphate, which is also likely to raise the soil pH. The colour of the hydrangea is therefore an indication of the acid content of the soil: the flower of a hydrangea is blue in acidic soil because the
The word ‘chromoplant sap is slightly acidic; red hydrangeas exist in alkaline soil phore’ comes from two because the sap transports alkali from the soil to the petals. The
Greek words, ‘khrorare mauve hydrangea indicates a soil of neutral pH. We see how mos’ meaning colour the chromophore in the flower is an acid–base indicator. and ‘phoro’, which means ‘to give’ or ‘to
The chromophore in hydrangeas is delphinidin (X), which is a impart’. A chromophore member of the anthrocyanidin class of compounds. Compound X is therefore a species reminds us of phenol (VII), indicating that delphinidin is also a imparting colour. weak acid. In fact, all pH indicators are weak acids or weak bases, and the ability to change colour is a visible manifestation of the indicator’s ability to undergo reversible changes in structure. In the
All pH indicators are laboratory, only a tiny amount of the pH indicator is added to the weak acids or weak titration solution, so it is really just a probe of the solution pH. It bases. does not participate in the acid–base reaction, except insofar as its own structure changes with the solution pH.
As an example, whereas the anthracene-based core of molecular X is relatively inert, the side-chain ‘X’ is remarkably sensitive to the pH of its surroundings (principally, to the pH of the solution in which it dissolves).
The name ‘hydrangea’ derives from classical Greek mythology, in which the
‘hydra’ was a beast with many heads.

+

OH

O

X

OH
OH
(X)

Figure 6.6 shows the structure of the side substituent as a function of pH.
The hydroxyl group placed para to the anthracene core is protonated in acidic solutions (i.e. when the hydrangea sap is slightly acidic). The proton is abstracted in alkaline sap, causing molecular rearrangement to form the quinone moiety.

ACID–BASE INDICATORS
OH

OH
OH

+

HO

275

O

O
HO

O

O−

OH
OH

OH

OH

OH
Red

Blue
OH
O

HO

O
OH
OH
OH
Mauve

Figure 6.6 Anthrocyanidins impart colour to many natural substances, such as strawberries and cherries. The choice of side chains can cause a huge change in the anthrocyanidin’s colour. If the side chain is pH sensitive then the anthrocyanidin acts as an acid–base indicator: structures of an anthrocyanidin at three pHs (red in high acidity and low pH, blue in low acidity and high pH and mauve in inter-midiate pHs)

It is also astonishing how the rich blue of a cornflower (Centaurea cyanus) and the majestic red flame of the corn poppy (Papaver rheas) each derive from the same chromophore – again based on an anthrocyanidin. The pH of cornflower and poppy sap does not vary with soil composition, which explains why we see neither red cornflowers nor blue corn poppies.

Aside
It is fascinating to appreciate the economy with which nature produces colours (elementary colour theory is outlined in Chapter 9). The trihydroxyphenyl group of the anthrocyanidin (X) imparts a colour to both hydrangeas and delphiniums. The dihydroxyphenyl group (XI) is remarkably similar, and imparts a red or blue colour to roses, cherries and blackberries. The singly hydroxylated phenyl ring in XII is the chromophore giving a red colour to raspberries, strawberries and geraniums, but it is not pH sensitive.
OH
R

OH
(XI)

R

OH
(XII)

276

ACIDS AND BASES

Why does phenolphthalein indicator not turn red until pH 8.2?
Which acid–base indicator to use?
Litmus was probably the most popular choice of acid–base indicator, but it is not a good choice for colour-blind chemists. The use of phenolphthalein as an acid–base indicator comes a close second. Phenolphthalein (XIII) is another weak organic acid.
It is not particularly water soluble, so we generally dissolve it in aqueous ethanol.
The ethanol explains the pleasant, sweet smell of phenolphthalein solutions.
OH

OH
O
O
(XIII)

Phenolphthalein is colourless and clear in acidic solutions, but imparts an intense puce pink colour in alkaline solutions of higher pH, with λ(max) = 552 nm. The coloured form of phenolphthalein contains a quinone moiety; in fact, any chromophore based on a quinone has a red colour. But if a solution is prepared at pH 7 (e.g. as determined with a pH meter), we find the phenolphthalein indicator is still colourless, and the pink colour only appears when the pH reaches 8.2. Therefore, we have a problem: the indicator has not detected neutrality, since it changes colour at too
Table 6.7 Some common pH indicators, their useful pH ranges and the changes in colour occurring as the pH increases. An increasing pH accompanies a decreasing concentration of the solvated proton
Indicator

pH range

Colour change

Methyl violet
Crystal violet
Litmus
Methyl orange
Ethyl red
Alizarin red S
3-Nitrophenol
Phenolphthalein

0.0–1.6
0.0–1.8
6.5–7.5
3.2–4.4
4.0–5.8
4.6–8.0
6.8–8.6
8.2–10.0

Yellow → blue
Yellow → blue
Red → blue
Red → yellow
Colourless → red
Yellow → red
Colourless → yellow
Colourless → pink

ACID–BASE INDICATORS

277

high a value of pH. However, the graph in Figure 6.3 shows how the pH of the titration solution changes dramatically near the end point: in fact, only a tiny incremental addition of alkali solution is needed to substantially increase the solution pH by several pH units. In other words, a fraction of a drop of alkali solution is the only difference between pH 7 (at the true volume at neutralization) and pH 8 when the phenolphthalein changes from colourless to puce pink.
Table 6.7 lists the pH changes for a series of common pH indicators. The colour changes occur over a wide range of pHs, the exact value depending on the indicator chosen. Methyl violet changes from yellow to blue as the pH increases between 0 and
1.6. At the opposite extreme, phenolphthalein responds to pH changes in the range
8.2 to 10.

7

Electrochemistry
Introduction
This chapter commences by describing cells and redox chemistry. Faraday’s laws of electrolysis describe the way that charge and current passage necessarily consume and produce redox materials. The properties of each component within a cell are described in terms of potential, current and composition.
Next, the nature of half-cells is explained, together with the necessary thermodynamic backgrounds of the theory of activity and the Nernst equation.
In the final sections, we introduce several key electrochemical applications, such as the pH electrode (a type of concentration cell), nerve cells (which rely on junction potentials) and batteries.

7.1

Introduction to cells: terminology and background
Why does putting aluminium foil in the mouth cause pain? Introduction to electrochemistry
Most people have at some time experienced a severe pain in their teeth after accidentally eating a piece of sweet wrapper. Those teeth that hurt are usually nowhere near the scrap of wrapper. The only people who escape this nasty sensation are those without metal fillings in their teeth.
The type of sweet wrapper referred to here is generally made of aluminium metal, even if we call it ‘silver paper’. Such aluminium dissolves readily in acidic, conductive electrolytes; and the pH of saliva is about 6.5–7.2.
The dissolution of aluminium is an oxidative process, so it generates several electrons. The resultant aluminium ions stay in solution Two redox states of the same material are next to the metal from which they came. We generate a redox coucalled a redox couple. ple, which we define as ‘two redox states of the same material’.

280

ELECTROCHEMISTRY

The word ‘amalgam’ probably comes from the Greek malagma meaning ‘to make soft’, because a metal becomes pliable when dissolved in mercury.
Another English word from the same root is
‘malleable’.
A cell comprises two or more half-cells in contact with a common electrolyte. The cell is the cause of the pain.

Oxidation reactions occur at the anode.

While it feels as though all the mouth fills with this pain, in fact the pain only manifests itself through those teeth filled with metal, the metal being silver dissolved in mercury to form a solid – we call it a silver amalgam. Corrosion of the filling’s surface causes it to bear a layer of oxidized silver, so the tooth filling also represents a redox couple, with silver and silver oxide coexisting.
An electrochemical cell is defined as ‘two or more half-cells in contact with a common electrolyte’. We see from this definition how a cell forms within the mouth, with aluminium as the more positive pole (the anode) and the fillings acting as the more negative pole (the cathode). Saliva completes this cell as an electrolyte. All the electrochemical processes occurring are contained within the boundaries of the cell.
Oxidation proceeds at the anode of the cell according to

Al(s) − → Al3+ (aq) + 3e−

(7.1)

and occurs concurrently with a reduction reaction at the cathode:
Ag2 O(s) + 2e− − → 2Ag0 (s) + O2−


Reduction reactions occur at the cathode.

(7.2)

The origins of the words ‘anode’ and ‘cathode’ tell us much.
‘Anode’ comes from the Greek words ana, meaning ‘up’, and hodos means ‘way’ or ‘route’, so the anode is the electrode to which electrons travel from oxidation, travelling to higher energies (i.e. energetically ‘uphill’). The word
‘cathode’ comes from the Greek hodos (as above), and cat meaning ‘descent’. The
English word ‘cascade’ comes from this same source, so a cathode is the electrode to which the electrons travel (energetically downhill) during reduction.
The oxidation and reduction reactions must occur concurrently because the electrons released by the dissolution of the aluminium are required for the reduction of the silver oxide layer on the surface of the filling. For this reason, we need to balance the two electrode reactions in Equations (7.1) and (7.2) to ensure the same number of electrons appear in each. The pain felt at the tooth’s nerve is a response to this flow of electrons. The paths of electron flow are depicted schematically in Figure 7.1.
Each electron has a ‘charge’ Q. When we quantify the number of electrons produced or consumed, we measure the overall charge flowing. Alternatively, we might measure the rate at which the electrons flow (how many flow per unit time, t): this rate is termed the current I. Equation (7.3) shows the relationship between current I and charge Q: dQ (7.3)
I=
dt
So ultimately the pain we feel in our teeth comes from a flow of current.

INTRODUCTION TO CELLS: TERMINOLOGY AND BACKGROUND
Electrons

Aluminium metal

Silver metal

− electrons
(oxidation)

Tooth interior and nerve

+ electrons
(reduction)
Silver ions

Aluminium ions

Aluminium
(‘silver foil’)

281

Electrolyte
(saliva)

Silver
(tooth filling)

Figure 7.1 Schematic illustration of the electron cycles that ultimately cause a sensation of pain in the teeth in people who have metallic fillings and who have inadvertently eaten a piece of aluminium (‘silver’) foil, e.g. while eating sweets

Why does an electric cattle prod cause pain?
The magnitude of a current
An electric cattle prod looks a little like a walking stick with an attached battery. A potential from the tip of the stick is applied to a cow’s flank, and the induced current hurts the animal. The cow moves where prompted to avoid a reapplication of the pain, thereby simplifying the job of cowherding.
Ohm’s law says that applying a potential V across an electrical resistor R induces a proportional current I . We can state this relationship mathematically as
V = IR

(7.4)

It is reported that the great Victorian scientist Michael Faraday discovered a variation of Ohm’s law some 20 years before Ohm The word ‘empirical’ himself published the law that today bears his name. Faraday, with implies a result derived from experiment rather his typical phlegm and ingenuity, grasped a resistor between his than theory. Similarly, two hands, immersed them both in a bowl of tepid water, and a chemist calculates a applied a voltage between them. It hurt. He found the empirical compound’s ‘empirical relationship ‘pain ∝ I × R’. formula’ while fully
The pain Faraday induced was in direct proportion to the magni- aware its value is based tude of the current passed. He discovered the principle underlying on experiment rather the action of an electric cattle prod. It is sobering to realize how than theory.
Faraday’s result is today employed through most of the world as the basis of torture. Despite the explicit banning of torture by the
UN Charter on Human Rights (of 1948), it is common knowledge that giving an electric shock is one of the most effective known means of causing pain.
In summary, the cattle prod causes pain because of the current formed in response to applying a voltage.

282

ELECTROCHEMISTRY

What is the simplest way to clean a tarnished silver spoon? Electrochemistry: the chemistry of electron transfer
Oxidation is loss of electrons. Reduction is gain of electrons.

Cutlery or ornaments made of silver tarnish and become black; this is a shame, because clean, shiny silver is very attractive. The
‘tarnish’ comprises a thin layer of silver that has oxidized following contact with the air to form black silver(I) oxide:

4Ag(s) + O2(g) − → 2Ag2 O(s)

In fact, the tarnish on silver usually comprises both silver(I) oxide and a little silver(I) sulphide. (7.5)

Such silver can be rather difficult to clean without abrasives
(which wear away the metal). The following is a simple electrochemical means of cleaning the silver: immerse the tarnished silver in a saucer of electrolyte, such as salt solution or vinegar, and wrap it in a piece of aluminium foil. Within a few minutes the silver is cleaner and bright, whereas the aluminium has lost some of its shininess.
The shine from the aluminium is lost as atoms on the surface of the foil are oxidized to form Al3+ ions (Equation (7.1)), which diffuse into solution. Because the aluminium touches the silver, the electrons generated by Equation (7.1) enter the silver and cause electro-reduction of the surface layer of Ag2 O (Equation (7.2)).
In summary, we construct a simple electrochemical cell in which the silver to be cleaned is the cathode (Equation (7.2)) and aluminium foil as the anode supplies the electrons via Equation (7.1).
The salt or vinegar acts as an electrolyte, and is needed since the product Al3+ requires counter ions to ensure electro-neutrality (so aluminium ethanoate forms).
The oxide ions combine with protons from the vinegar to form water. Figure 7.2 illustrates these processes occurring in schematic form.
Aluminium metal

Silver metal

− electrons
(oxidation)

+ electrons
(reduction)

Aluminium ions

Aluminium foil

Silver ions

Electrolyte
(vinegar)

Silver spoon

Figure 7.2 Illustration of the electron cycles that allow for the trouble-free cleaning of silver: we immerse the tarnished silver in an electrolyte, such as vinegar, and touch the silver with aluminium foil INTRODUCTION TO CELLS: TERMINOLOGY AND BACKGROUND

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And finally a word of caution: aluminium ethanoate is toxic, so wash the silver spoon thoroughly after removing its tarnish.

How does ‘electrolysis’ stop hair growth?
Electrochemical reactions and electrolysis
To many people, particularly the image conscious, electrolysis means removing hairs from the arms and legs – a practice sometimes called ‘electrology’. The purpose of such electrolysis is to remove the hair follicles temporarily, thereby avoiding the need to shave. We ‘treat’ each individual hair by inserting a tiny surgical
‘probe’ (in reality, an electrode) into the hair follicle. Applying a voltage to the hair root for a fraction of a second kills the root. The electrolysed hair is then removed, and will not regrow for some time. This procedure is performed repeatedly until the desired area is cleared. But how does it work?
A voltage is applied to the electrode: we say it is polarized. A current flows in response to the voltage, and electrons are consumed by electrochemical reactions around the electrodes. Electrolysis occurs. Each electron that flows through the electrode must be involved in a redox reaction, either oxidation or reduction. The electrons entering or leaving the electrode move as a result of reactions occurring in the immediate vicinity of the electrode. Conversely, electrons can travel in the opposite direction (leaving the electrode) to facilitate reduction reactions.
In summary, this form of electrolysis is effective because the charge passing through the electrode generates chemicals inside the hair follicle. The resultant trauma kills the hair root. A leg or arm treated in this way remains hairless until a new, healthy root regrows later in the previously damaged follicle.

Electrochemical means the chemistry of the electron. It is wrong to claim a permanent method of hair removal: after electrolysis, the hair will grow back, albeit thinner and finer.
The word ‘electrolysis’ derives from the
Greek words lysis, meaning ‘splitting’ or
‘cleavage’, and the root electro, meaning
‘charge’ or ‘electricity’. Strictly, then, electrolysis involves electrochemical bond cleavage. Why power a car with a heavy-duty battery yet use a small battery in a watch?
Faraday’s laws of electrolysis
A battery is a device for converting chemical energy into electrical energy. The amount of energy required by the user varies according to the application in mind. For example, a watch battery only powers the tiny display on the face of a digital watch. For this purpose,

Batteries are described in more detail in
Section 7.7.

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Table 7.1 Faraday’s laws of electrolysis
Faraday’s first law
The number of moles of a species formed at an electrode during electrolysis is proportional to the electrochemical charge passed: Q = I × t
Faraday’s second law
A given charge liberates different species in the ratio of their relative formula masses, divided by the number of electrons in the electrode reaction

it only needs to deliver a tiny current of a micro-amp or so. Conversely, a car battery
(usually a ‘lead–acid cell’, as described on p. 347) is bulky and heavy because it must deliver a massive amount of electrical energy, particularly when starting the car.
Other batteries generate currents of intermediate magnitude, such as those needed in torches, mobile phones and portable cassette and CD players.
The amount of charge generated or consumed by a battery is in direct proportion to the number of electrons involved, according to Faraday’s laws, which are given in
Table 7.1. Both electrons and ions possess charge. When a current is drawn through a cell, the charged electrons move through the conductive electrodes (as defined on
p. 300) concurrently with charged ions moving through the electrolyte. The ions are anions (which bear a negative charge) and cations (which are positive).
Underlying both of Faraday’s laws lies the fundamental truth that each electron possesses the same charge.
Worked Example 7.1 What is the charge on 1 mol of electrons?
The coulomb, C, is the
SI unit of charge.

The charge e on a single electron is 1.6 × 10−19 C and there are
6.022 × 1023 electrons per mole (the Avogadro number L), so the charge on a mole of electrons is given by the simple expression charge on one electron = L × e

(7.6)

Inserting numbers into Equation (7.6), we obtain
The charge on 1 mol of electrons is termed ‘a faraday’ F.

charge on 1 mole of electrons
= 1.6 × 10−19 C × 6.026 × 1023 mol−1
We see that 1 mol has a charge of 96 487 C mol−1 . This quantity of charge is known as the ‘Faraday’ F.

SAQ 7.1 An electrolysis needle (i.e. an electrode) delivers 1 nmol of electrons to a hair root. How many faraday’s of charge are consumed, and how many coulombs does it represent?

INTRODUCTION TO CELLS: TERMINOLOGY AND BACKGROUND

Worked Example 7.2 How much silver is generated by reductively passing 1F of charge through a silver-based watch battery?
We will answer this question in terms Faraday’s first law, which was first formulated in 1834 (see Table 7.1).
Each electron is required to effect the reduction reaction Ag+ (aq) +

e → Ag(s) , so one electron generates one atom of Ag, and 1F of charge (i.e. 1 mol of electrons) generates 1 mol of Ag atoms. The metal forms at the expense of 1 mol of Ag+ ions. Similarly, 1 mol of electrons, if passed oxidatively, would generate 1 mol of Ag+ ion from Ag metal.
We see a direct proportionality between the charge passed and the amount of material formed during electrolysis, as predicted by Faraday’s first law.

285

Silver has the symbol
‘Ag’ because its Latin name is argentium, itself derived from the
Greek for money, argurion. The Spanish colonized parts of South
America in the 16th century. They named it Argentina (dog-Latin for ‘silver land’) when they discovered its vast reserves of silver.

SAQ 7.2 10−10 F of charge are inserted into a hair pore through a fine needle electrode. Each electron generates one molecule of a chemical to poison the hair root. How much of the chemical is formed?

Worked Example 7.3 A Daniell cell (see p. 345) is constructed, and 1F of charge is passed through a solution containing copper(II) ions. What mass of copper is formed?
Assume that the charge is only consumed during the reduction reaction, and is performed with 100 percent efficiency.
We will answer this question by introducing Faraday’s second law (see Table 7.1).
The reduction of one copper(II) ion requires two electrons according to the reaction
Cu2+ (aq) + 2e− − − Cu(s)
−→

(7.7)

We need 2F of charge to generate 1 mol of Cu(s) , and 1F of charge will form 0.5 mol of copper.
1 mol of copper has a mass of 64 g, so 1F generates (64 ÷ 2) g of copper. The mass of copper generated by passing 1F is 32 g.
SAQ 7.3 How much aluminium metal is formed by passing 2F of charge through a solution of Al3+ ions? [Hint: assume the reaction at the electrode is Al3+ (aq) + 3e− → Al(s) .]

How is coloured (‘anodized’) aluminium produced?
Currents generate chemicals: dynamic electrochemistry
Saucepans and other household implements made of aluminium often have a brightly coloured, shiny coating. This outer layer comprises aluminium oxide incorporating a small amount of dye.

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Al2 O3 is also called alumina. The layer is deposited with the saucepan immersed in a vat of dye solution (usually acidified to pH 1 or 2), and made the positive terminal of a cell. As the electrolysis proceeds, so the aluminium on the surface of the saucepan is oxidized:

2Al(s) + 3H2 O − → Al2 O3(s) + 6H+ (aq)

(7.8)

The aluminium is white and shiny before applying the potential. A critical potential exists below which no electro-oxidation will commence. At more extreme potentials, the surface atoms of the aluminium oxidize to form Al3+ ions, which combine with oxide ions from the water to form Al2 O3 . This electro-precipitation of solid aluminium oxide is so rapid that molecules of dye get trapped within it, and hence its coloured aspect.
The dye resides inside the layer of alumina. Its colour persists
We say the dye occlubecause it is protected from harmful UV light, as well as mechandes within a matrix of ical abrasion and chemical attack. solid aluminium oxide.
But the chemical reaction forming this coloured layer of oxide represents only one part of the cell. A cell contains a minimum of two electrodes, so a cell comprises two reactions – we call them
A cell must comprise a minimum of two elechalf-reactions: one describes the chemical changes at the positive trodes. electrode (the anode) and the other describes the changes that occur at the negative cathode.
The same number of electrons conducts through (i.e. are conducted by) each of the two electrodes. If we think in terms of charge flowing per unit time, we would say the same ‘current’ I flows through each electrode. The electrons travel in opposite directions, insofar as they leave or enter an electrode, which explains why the current through the anode is oxidative and the current through the cathode is reductive.
We say
(7.9)
I(anode) = −I(cathode) where the minus sign reminds us that the electrons either move in or out of the electrode. Because these two currents are equal (and opposite), the same amount of reaction will occur at either electrode. We see how an electrode reaction must also occur at the cathode as well as the desired oxidative formation of alumina at the anode. (The exact nature of the reaction at the anode will depend on factors such as the choice of electrode material.)

How do we prevent the corrosion of an oil rig?
Introduction to electrochemical equilibrium
Oil rigs are often built to survive in some of the most inhospitable climates in the world. For example, the oil rigs in the North Sea between the UK and Scandinavia

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287

frequently withstand force-10 gales. Having built the rig, we appreciate how important is the need of maximizing its lifetime. And one of the major limits to its life span is corrosion.
Oil rigs are made of steel. The sea in which they stand contains vast quantities of dissolved salts such as sodium chloride, which is particularly ‘aggressive’ to ferrous metals. The corrosion reaction generally involves oxidative dissolution of the iron, to yield ferric salts, which dissolve in the sea:

Fe(s) − → Fe3+ (aq) + 3e−

(7.10)

If left unchecked, dissolution would cause thinning and hence weakening of the legs on which the rig stands.
One of the most ingenious ways in which corrosion is inhibited is to strap a power pack to each leg (just above the level of the In reality, several of sea) and apply a continuous reductive current. An electrode couple the iron compounds would form when a small portion of the iron oxidizes. The couple are solid, such as rust. would itself set up a small voltage, itself promoting further disso- This clever method of lution. The reductive current coming from the power pack reduces averting corrosion can any ferric ions back to iron metal, which significantly decreases also arrest the corrosion of rails and the the rate at which the rig leg corrodes. undersides of boats.
Clearly, we want the net current at the iron to be zero (hence no overall reaction). The rate of corrosion would be enhanced if the power pack supplied an oxidative current, and wasteful side reac- The simplest definition tions involving the seawater itself would occur if the power pack of equilibrium in an produced a large reductive current. The net current through the electrochemistry cell is iron can be positive, negative or zero, depending on the potential that no concentrations applied to the rig’s leg. The conserver of the rig wants equilibrium, change. implying no change.
All the discussions of electrochemistry so far in this chapter concern current – the flow of charged electrons. We call this branch of electrochemistry dynamic, implying that compositions change in response to the flow of electrons. Much of the time, however, we wish to perform electrochemical experiments at equilibrium.
One of our best definitions of ‘equilibrium electrochemistry’ says the net current is zero; and from Faraday’s laws (Table 7.1), a zero Electrochemical meacurrent means that no material is consumed and no products are surements at equilibrium are made at zero formed at the electrode.
But this equilibrium at the oil rig is dynamic: the phrase current.
‘dynamic equilibrium’ implies that currents do pass, but the current of the forward reaction is equal and opposite to the current of the back reaction, according to
I(forward, eq) = −I(backward, eq)
(7.11)
and the overall (net) current is the sum of these two:
I(net) = I(forward) + I(backward)

(7.12)

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ELECTROCHEMISTRY

Equation (7.11) is important, since it emphasizes how currents flow even at equilibrium.
But the value of I(net) is only ever zero at equilibrium because I(forward) = −I(backward) , which can only happen at one particular energy, neither too reductive nor too oxidative.
The voltage around the legs of the oil rig needs to be chosen carefully.

What is a battery?
The emf of cells
A battery is an electrochemical cell, and is defined as ‘a device comprising two or more redox couples’ (where each couple comprises two redox states of the same material). An oxidation reaction occurs at the negative pole of the battery in tandem with a reduction reaction at the positive pole. Both reactions proceed with the passage of current. The two redox couples are separated physically
The word ‘cell’ comes by an electrolyte. from the Latin for ‘small
The battery requires two redox couples because it is a cell. Each room’, which explains couple could be thought of as representing half of a complete cell. why a prisoner is kept
This sort of reasoning explains why the two redox couples are in a ‘cell’. called half-cells. We could, therefore, redefine a cell as a device comprising two half-cells separated with an electrolyte.
In practice, the voltage of a battery is measured when its two ends are connected to the two terminals of a voltmeter, one contact secured to the positive terminal of the battery and the other at the negative. But a voltmeter is a device to measure differences in potential, so we start to see how the ‘voltage’ cited on a battery label is simply the difference in potential between the two poles of the battery.
While the voltage of the cell represents the potential difference
The cell’s emf is a pribetween the two ‘terminals’ of the battery, in reality it relates to mary physicochemical the separation in energy between the two half-cells. We call this property, and is measeparation the emf, where the initials derive from the archaic phrase sured with a voltmeter electromotive force. An emf is defined as always being positive. or potentiometer.
We have already seen from Faraday’s laws how a zero current implies that no redox chemistry occurs. Accordingly, we stipulate
An emf is always that the meter must draw absolutely no current if we want to meadefined as being posisure the battery’s emf at equilibrium. Henceforth, we will assume tive. that all values of emf were determined at zero current.
A battery is a device for converting chemical energy into electrical energy. Aside
The term ‘emf’ follows from the archaic term ‘electromotive force’. Physicists prefer to call the emf a ‘potential difference’ or symbolize it as a ‘p.d.’.
Confusingly, potential also has the symbols of U, V and E, depending on the context.

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289

Why do hydrogen fuel cells sometimes ‘dry up’?
Cells and half-cells
Hydrogen fuel cells promise to fuel prototype cars in the near future. We define such a fuel cell as a machine for utilizing the energies of hydrogen and oxygen gases, hitherto separated, to yield a usable electric current without combustion or explosion.
Unlike the simple batteries above, the oxygen and hydrogen gases fuelling these cells are transported from large, high-pressure tanks outside the cell. The gases then feed through separate pipes onto the opposing sides of a semi-permeable membrane (see
Figure 7.3), the two sides of which are coated with a thin layer of platinum metal, and represent the anode and cathode of the fuel cell. This membrane helps explain why such cells are often called PEM fuel cells, where the acronym stands for ‘proton exchange membrane’.
When it reaches the polymer membrane, hydrogen gas is oxidized at the negative side of the cell (drawn on the left of Figure The energy necessary
7.3), forming protons according to to cleave the H–H bond
H2(g) − → 2H+ + 2e− (Pt)


(7.13)

The subscripted ‘Pt’ helps emphasize how the two electrons conduct away from the membrane through the thin layer of platinum

is provided by the energy liberated when forming the two H–Pt bonds after molecular dissociation. Load


+

Hydrogen gas

Oxygen gas

Permeable polymer membrane
Water

Thin electrodes
(layers of platinum)

Figure 7.3 A hydrogen–oxygen fuel cell. The water formed at the cathode on the right-hand side of the cell condenses and collects at the bottom of the cell, and drains through a channel at the bottom right-hand side

290

ELECTROCHEMISTRY

metal surrounding the electrolyte, and enter into the external circuit where they perform work. The platinum also catalyses the dissociation of diatomic H2 gas to form reactive Hž atoms.
Once formed, the protons diffuse through the platinum layer and
The membrane, made enter deep into the layer of semi-permeable membrane. They travel from a perfluorinated from the left-hand side of the membrane to its right extremity in polymer bearing sulresponse to a gradient in concentration. (Movement caused by a phonic acid groups, is concentration gradient will remind us of dye diffusing through a known in the trade as saucer of water, as described on p. 129.)
NafionTM .
At the positive side of the cell (drawn here on the right), oxygen gas is reduced to oxide ions, according to
Note: the electron count in Equations
(7.13) and (7.14) should balance in reality −
O2(g) + 4e− (Pt) − → 2O2− (aq)

(7.14)

The electrons necessary to effect the reduction of gaseous oxygen come from the external circuit, and enter the oxygen half-cell through the layer of platinum coating the cathode, thereby explaining why the electrons in Equation (7.14) are subscripted with ‘Pt’.
The O2− ions combine chemically with protons that have traversed the Nafion membrane and form water, which collects at the foot of the cell.
Because the overall cell reaction is exothermic, the value of the cell emf decreases with increasing temperature, so the temperature is generally kept relatively low at about 200 ◦ C. The cell emf is 1.23 V at this temperature.
One of the main advantages of this hydrogen fuel cell is the rapid rate at which hydrogen is oxidized at the platinum surface. Most
The rate of oxygen of the cell’s operational difficulties relate to the oxygen side of the reduction can be accelerated by finely dividdevice. Firstly, the reduction of gaseous oxygen in Equation (7.14) ing the platinum catais relatively slow, so the rate at which the cell operates is somewhat lyst, thereby increasing limited. But more serious is the way the cell requires a continual its effective area. flow of gas, as below. The hydrogen half-cell comprises an electrode couple because two redox states of the same material coexist there (H2 and H+ ). In a similar way, the oxygen half-cell also comprises an electrode couple, but this time of O2 and O2− .
If the flow of oxygen falters, e.g. when the surface of the cathode is covered with water, then no gaseous O2 can reach the platinum outer layer. In response, firstly no electrons are consumed to yield oxide ions, and secondly the right-hand side of the cell ‘floods’ with the excess protons that have traversed the polymer membrane and not yet reacted with O2− . Furthermore, without the reduction of oxygen, there is no redox couple at the cathode. The fuel cell ceases to operate, and can produce no more electrical energy.
This simple example helps explain why a cell requires no fewer than two halfcells. A half-cell on its own cannot exchange electrons, and cannot truly be termed a cell.

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291

Why bother to draw cells?
Cells and the ‘cell schematic’
Having defined a cell, we now want to know the best way of representing it. Undoubtedly, the simplest is to draw it diagrammatically – no doubt each picture of a cell being a work of art in miniature.
But drawing is laborious, so we generally employ a more sensible alternative: we write a cell schematic, which is a convenient abbreviation of a cell. It can be ‘read’ as though it was a cross-section, showing each interface and phase. It is, therefore, simply a shorthand way of saying which components are incorporated in the cell as cathode, anode, electrolyte, etc., and where they reside.
Most people find that a correct understanding of how to write a cell schematic also helps them understand the way a cell works. Accordingly, Table 7.2 contains a series of simple rules for constructing the schematic.
Worked Example 7.4 We construct a cell with the copper(II) | copper and zinc(II) | zinc redox couples, the copper couple being more positive than the zinc couple. What is the cell schematic?
Answer strategy: we will work sequentially through the rules in Table 7.2.
First, we note how the copper couple is the most positive, so we
For convenience, we write it on the right. The zinc is, therefore, the more negative and often omit the subscript we write it on the left. We commence the schematic by writing, descriptors and the ‘ ’
Zn(s) . . . Cu(s) ⊕. and ‘⊕’ signs.

Table 7.2

Rules for constructing a cell schematic

1. We always write the redox couple associated with the positive electrode on the right-hand side
2. We write the redox couple associated with the negative electrode on the left-hand side
3. We write a salt bridge as a double vertical line: ||
4. If one redox form is conductive and can function as an electrode, then we write it on one extremity of the schematic.
5. We represent the phase boundary separating this electrode and the solution containing the other redox species by a single vertical line:a |
6. If both redox states of a couple reside in the same solution (e.g. Pb2+ and Pb4+ ), then they share the same phase. Such a couple is written conventionally with the two redox states separated by only a comma: Pb4+ , Pb2+
7. Following from 6: we see that no electrode is in solution to measure the energy at equilibration of the two redox species. Therefore, we place an inert electrode in solution; almost universally, platinum is the choice a We write a single line | or, better, a dotted vertical line, if the salt bridge is replaced by a simple porous membrane. 292

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There is a phase boundary between the Zn and Zn2+ because the
Zn is solid but the Zn2+ is dissolved within a liquid electrolyte. A similar boundary exists in the copper half-cell.

An alternative way of looking at the schematic is to consider it as
‘the path taken by a charged particle during a walk from one electrode to the other’.

To be a redox couple, the zinc ions will be in contact with the zinc electrode, which we write as Zn2+ (aq) |Zn(s) , the vertical line emphasizing that there is a phase boundary between them. We can write the other couple as Cu2+ (aq) |Cu(s) , with similar reasoning. Note that if the two electrodes are written at the extreme ends of the cell schematic, then the redox ionic states must be located somewhere between them. The schematic now looks like Zn(s) |Zn2+ (aq) . . .
Cu2+ (aq) |Cu(s) ⊕.
Finally, we note that the two half-cells must ‘communicate’ somehow – they must be connected. It is common practice to assume that a salt bridge has been incorporated, unless stated otherwise, so we join the notations for the two half-cell with a double vertical line, as
Zn(s) |Zn2+ (aq) ||Cu2+ (aq) |Cu(s) ⊕.
SAQ 7.4 Write the cell schematic for a cell comprising the Fe3+ ,Fe (positive) and Co2+ ,Co (negative) couples.

Worked Example 7.5 Write a cell schematic for a cell comprising the couples Br2 , Br− and H+ , H2 . The bromine | bromide couple is the more positive. Assume that all solutions are aqueous.

This is a more complicated cell, because we have to consider the involvement of more phases than in the previous example.
Right-hand side: the bromine couple is the more positive couple, so we write it on the right of the schematic. Neither Br2 nor Br− is metallic, so we need an inert electrode. By convention, we employ platinum if no other choice is stipulated. The electrode at the far right of the schematic is therefore Pt, as . . . Pt(s) ⊕.
Br2 and Br− are both soluble in water – indeed, they are mutually soluble, forming a single-phase solution. Being in the same phase, we
We write the oxidized cannot write a phase boundary (as either ‘Br2 |Br− ’ or as ‘Br− |Br2 ’), form first if both redox so we write it as ‘Br2 , Br− (aq) ’. Note how we write the oxidized form states reside in the first and separate the two redox states with a comma. The right-hand same phase, sepaside of the schematic is therefore ‘Br2 , Br− (aq) |Pt(s) ⊕’. rating them with a comma. Left-hand side: neither gaseous hydrogen nor aqueous protonic solutions will conduct electrons, so again an inert electrode is required on the extremity of the schematic. We again choose platinum. The left-hand side of the cell is: Pt(s) .
Hydrogen gas is in immediate contact with the platinum inert electrode. (We bubble it through an acidic solution.) Gas and solution are different phases, so we write the hydrogen couple as H2(g) |H+ (aq) , and the left-hand side of the schematic becomes ‘ Pt(s) |
H2(g) |H+ (aq) ’.

INTRODUCTION TO CELLS: TERMINOLOGY AND BACKGROUND

Finally, we join the two half-cells via a salt bridge, as
+

Pt(s) |H2(g) |H



(aq) ||Br2 , Br (aq) |Pt(s)

⊕.

From now on, we will omit both the symbols and ⊕, and merely assume that the right-hand side is the positive pole and the left-hand side the negative.

293

As an extra check, we note how the salt bridge dips into the proton solution, so the term for H+ needs to be written adjacent to the symbol ‘||’.

Why do digital watches lose time in the winter?
The temperature dependence of emf
A digital watch keeps time by applying a tiny potential (voltage) across a crystal of quartz, causing it to vibrate at a precise frequency of ν cycles per second. The watch keeps time by counting off 1 s each time the quartz vibrates ν times, explaining why the majority of the components within the watch comprise a counting mechanism.
Unfortunately, the number of vibrations of the crystal per second is dictated by the potential applied to the quartz, so a larger voltage makes the frequency ν increase, and a smaller voltage causes ν to slow. For this reason, the potential of the watch battery must be constant.
In Section 4.6, we saw how the value of G is never independent of temperature, except in those rare cases when S(cell) = 0.
Accordingly, the value of G(cell) for a battery depends on whether someone is wearing the watch while playing outside in the cold snow or is sunbathing in the blistering heat of a tropical summer.
And the emf of the watch battery is itself a function of the change in Gibbs function, G(cell) according to
G(cell) = −nF × emf

Frequency ν has the SI unit of hertz (Hz). 1 Hz represents 1 cycle or vibration per second, so a frequency of ν
Hz means ν cycles per second. The voltage from the battery induces a minute mechanical strain in the crystal, causing it to vibrate – a property known as the piezoelectric effect.

(7.15)

where F is the Faraday constant and n is the number of moles of electrons transferred per mole of reaction. The value of G(cell) is negative if the reaction proceeds reversibly (see Section 4.3), so the emf is defined as positive to ensure that G(cell) is always negative.
In other words, the value of G(cell) relates to the spontaneous cell reaction.
So we understand that as the emf changes with temperature, so the quartz crystal vibrates at a different frequency – all because

Care: The output voltage of a battery is only an emf when measured at zero current,
i.e. when not operating the watch.

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G(cell) is a function of temperature. Ultimately, then, a digital watch loses time in the winter as a simple result of the cold.
Worked Example 7.6 The emf of a typical ‘lithium’ watch battery (which is a cell) is
3.0 V. What is G(cell) ?
Look at the units and note how 1 J = 1 C ×
1 V.

The number of electrons transferred n in a lithium battery is one, since the redox couple is Li+ , Li. Inserting values into Equation
(7.15) yields:
G(cell) = −1 × 96 485 C mol−1 × 3.0 V
G(cell) = −289 500 J mol−1 = −290 kJ mol−1

Notice how the molar energy released by a simply battery is enormous.
SAQ 7.5 A manganese dioxide battery has an emf of 1.5 V and n = 2.
Calculate G(cell) .

In the absence of any pressure–volume work, the value of G(cell) is equal to the work needed to transfer charge from the negative end of the cell to the positive. In practice, G(cell) equates to the amount of charge passed, i.e. the number of charged particles multiplied by the magnitude of that charge.

Why is a battery’s potential not constant?
Non-equilibrium measurements of emf
We define the emf as having a positive value and, strictly, it is always determined at equilibrium. The two currents (for anode and cathode) are generally different. Neither of them is related to potential in a linear way.

A healthy battery for powering a Walkman or radio has a voltage of about 1.5 V. In the terminology of batteries, this value is called its open-circuit potential, but an electrochemist talking in terms of cells will call it the emf. This voltage is read on a voltmeter when we remove the battery from the device before measurement. But the voltage would be different if we had measured it while the battery was, for example, powering a torch.
We perform work whenever we connect the two poles of the battery across a load. The ‘load’ in this respect might be a torch, calculator, car, phone or watch – anything which causes a current to pass. And this flow of current causes the voltage across the battery or cell to decrease; see Figure 7.4. We call this voltage the
‘voltage under load’.
A similar graph to Figure 7.4 could have been drawn but with the x axis being the resistance between the two electrodes: if the

Potential/V

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295

Current drawn / A

Figure 7.4 Schematic diagram showing how a cell’s potential decreases with current. We call the cell potential the emf only when the current is zero

resistance between the two electrodes is zero, which is clearly the case if they should touch, then the cell potential is zero – we say the cell has ‘shorted’.
Ohm’s law, Equation (7.16), describes the difference between the emf and a voltage under load:
V = IR
(7.16)
where I is the current flowing, R is the resistance of the load and V is the decrease in the voltage of the cell. When a current is drawn, the potential of the cell decreases by the amount V in Equation (7.16). We will call this new (smaller) voltage E(load) , and its magnitude is given by
E(load) = emf − IR
In summary, we say the voltage of a cell is the same as a cell’s emf if determined at zero current. From Faraday’s laws of electrolysis, this criterion implies that none of the compositions within the cell can change. In other words, a cell emf is an equilibrium quantity.
For this reason, it is not wise to speak of terms such as ‘anode’ of ‘cathode’ for a cell at equilibrium, because these terms relate to electrodes that give or receive charge during current flow; and our definition of equilibrium implies that no current does flows.
We therefore adopt the convention: the terms ‘anode’ or ‘cathode’ will no longer be employed in our treatment of equilibrium electrochemistry. (7.17)
The emf can only ever be determined at zero current. Why a battery’s emf decreases permanently after a current has flowed is explained on
p. 328.

What is a ‘standard cell’?
The thermodynamics of cells
A standard cell produces a precise voltage and, before the advent of reliable voltmeters, was needed to calibrate medical and laboratory equipment. It is generally agreed that the first standard cell was the Clark cell (see p. 299), but the most popular was the Weston saturated cadmium cell, patented in 1893.

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ELECTROCHEMISTRY

Edward Weston (1850–1936) was a giant in the history of electrical measuring instruments. In the field of measurement, he developed three important components: the standard cell, the manganin resistor and the electrical indicating instrument.
The main advantage of Weston’s cell was its insensitivity to temperature, and the emf of almost 1 V: to be precise, 1.0183 V at 20 ◦ C. It is usually constructed in an Hshaped glass vessel. One arm contains a cadmium amalgam electrode beneath a paste of hydrated cadmium sulphate (3CdSO4 · 5H2 O) cadmium sulphate and mercury(I) sulphate. The other arm contains elemental mercury. Its schematic is Cd(Hg)|CdSO4(aq) ,
Hg2 SO4 |Hg.
The Weston saturated cadmium cell became the international
The Clark cell was standard for emf in 1911. Weston waived his patent rights shortly patented by Latimer afterward to ensure that anyone was allowed to manufacture it.
Clark in the 1880s, and
Weston’s cell was much less temperature sensitive than the previwas the first standard ous standard, the Clark cell. We recall how the value of G changes cell. with temperature according to Equation (4.38). In a similar way, the value of G(cell) for a cell relates to the entropy change S(cell) such that the change of emf with temperature follows

Remember: the small subscripted ‘p’ indicates that the quantity is measured at constant pressure. It does not mean ‘multiplied by p’.

The temperature voltage coefficient has several names: ‘temperature coefficient’,
‘voltage coefficient’ or
‘temperature coefficient of voltage’.
Table 7.3 contains a few values of
(d (emf )/dT ).

That this value of d(emf )/dT is negative tells us that the emf DEcreases when the temperature
INcreases.

S(cell) = nF

d(emf ) dT p

(7.18)

the value of (d(emf )/dT )p is virtually zero for the Weston cell.
If we assume the differential (d (emf )/dT ) is a constant, then
Equation (7.18) has the form of a straight line, y = mx, and a graph of emf (as ‘y’) against T (as ‘x’) should be linear. Figure 7.5 shows such a graph for the Clark cell, Hg|HgSO4 , ZnSO4 (sat’d)|
Zn. Its gradient represents the extent to which the cell emf varies with temperature, and is called the temperature voltage coefficient.
The gradient may be either positive or negative depending on the cell, and typically has a magnitude in the range 10−5 to 10−4
V K−1 . We want a smaller value of (d (emf )/dT ) if the emf is to be insensitive to temperature.
Having determined the temperature dependence of emf as the gradient of a graph of emf against temperature, we obtain the value of S(cell) as ‘gradient × n × F ’.
Worked Example 7.7 The temperature voltage coefficient for a simple alkaline torch battery is −6.0 × 10−4 V K−1 . What is the entropy change associated with battery discharge? The number of electrons transferred in the cell reaction n = 2.
Inserting values into Equation (7.18):
S(cell) = 2 × 96 485 C mol−1 × −6.0 × 10−4 V K−1
S(cell) = −116 J K−1 mol−1

INTRODUCTION TO CELLS: TERMINOLOGY AND BACKGROUND

297

1.45
Gradient = ‘temperature voltage coefficient’

emf/ V

1.44

1.43

1.42

1.41
280

290

300

310

320

330

Temperature T / K

Figure 7.5 Graph of cell emf against temperature for the Clark cell Hg|HgSO4 , ZnSO4 (sat’d)|Zn.
We call the gradient of this graph the ‘temperature voltage coefficient’
Table 7.3 Temperature voltage coefficients for various cells and half cells
Cella

(d(emf )/dT )p /V K−1

Standard hydrogen electrode
Clark standard cell
Saturated calomel electrode
Silver–silver chloride
Silver–silver bromide
Weston standard cell

0 (by definition)b
6.0 × 10−4
+7 × 10−4
−8.655 × 10−5
−4.99 × 10−4
−5 × 10−5

a
Individual electrodes are cited with the SHE as the second electrode of the cell. b The potential of the SHE is defined as zero at all temperatures.

SAQ 7.6 The emf of a lithium watch battery is exactly 3.000 V at 298 K, but the value decreases to 2.985 V at 270 K. Calculate the temperature voltage coefficient and hence the change in entropy S(cell) during cell discharge. (Take n = 1.)

Occasionally, the temperature voltage coefficient is not expressed as a simple number, but as a power series in T (we generally call it a virial series, or expansion). For example,
Equation (7.19) cites such a series for the cell Pt(s) |H2(g) |HBr(aq) |AgBr(s) |Ag(s) : emf /V = 0.071 31 − 4.99 × 10−4 (T /K − 298) − 3.45 × 10−6 (T /K − 298)2 (7.19)
We insert values of temperature T into the expression to obtain a value for emf. Values of S(cell) are obtained by performing two calculations, inserting first one temperature

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ELECTROCHEMISTRY

T1 to obtain the emf at T1 , and then a second T2 to obtain another value of emf. We calculate a value of (d(emf )/dT ) using emf at T2 − emf at T1 d(emf )
=
dT
T2 − T1
The value of

(7.20)

S(cell) is then determined in the usual way via Equation (7.18).

SAQ 7.7 Insert values of T = 310 K into Equation (7.19) to calculate the potential of the cell Pt(s) |H2(g) |HBr(aq) |AgBr(s) |Ag(s) .
SAQ 7.8 Repeat the calculation in SAQ 7.7, this time with T = 360 K, and hence determine S(cell) .

Justification Box 7.1
The relationship between changes in Gibbs function and temperature (at constant pressure p) is defined using Equation (4.38):
− S=

∂ G
∂T

p

We know from Equation (7.15) that the change in G(cell) with temperature is ‘−nF × emf ’. The entropy change of the cell S(cell) is then obtained by substituting for G(cell) in Equation (7.18):
− S(cell) =

∂(−nF × emf )
∂T

(7.18)

p

Firstly, the two minus signs cancel; and, secondly, n and F are both constants. Taking them out of the differential yields Equation (7.18) in the form above.

These values of G,
H and S relate to a complete cell, because thermodynamic data cannot be measured experimentally for halfcells alone.

To obtain the change in enthalpy during the cell reaction, we recall from the second law of thermodynamics how H = G +
T S (Equation (4.21)). In this context, each term relates to the cell. We substitute for G(cell) and S(cell) via Equations (7.15) and (7.18) respectively, to yield
H(cell) = −nF × emf + T nF

d(emf ) dT p

(7.21)

so, knowing the emf as a function of temperature, we can readily obtain a value of
H(cell) .

INTRODUCTION TO CELLS: TERMINOLOGY AND BACKGROUND

299

Worked Example 7.8 The Clark cell Zn|Zn2+ , Hg2 SO4 |Hg is often employed as a standard cell since its emf is known exactly as a function of temperature. The cell emf is
1.423 V at 298 K and its temperature coefficient of voltage is −1.2 × 10−4 V K−1 . What are G(cell) , S(cell) and thence H(cell) at 298 K?
Before we commence, we note that the spontaneous cell reaction is
−→
Zn + Hg2 SO4 + 7H2 O − − ZnSO4 · 7H2 O + 2Hg0 so the cell reaction is a two-electron process.
Next, we recall from Equation (7.15) that G(cell) = −nF × emf . Inserting values for the cell at 298 K gives
G(cell) = −2 × 96 485 C mol−1 × 1.423 V
G(cell) = −275 kJ mol−1
Then, from Equation (7.18), we recall that
S(cell) = nF

d(emf ) dT p

Inserting values:
S(cell) = 2 × 96 485 C mol−1 × (−1.2 × 10−4 V K−1 )
S(cell) = −23.2 J K−1 mol−1
Finally, from Equation (4.21), we say that insert values:

H(cell) =

G(cell) + T S(cell) . We again

H(cell) = (−275 kJ mol−1 ) + (298 K × −23.2 J K−1 mol−1 ) so H(cell) = −282 kJ mol−1

SAQ 7.9 A different cell has an emf of 1.100 V at 298 K.
The temperature voltage coefficient is +0.35 mV K−1 . Calculate G(cell) , S(cell) and hence H(cell) for the cell at
298 K. Take n = 2.

Remember: 5 mV (5 milli-volts) is the same as 5 × 10−3 V.

Having performed a few calculations, we note how values of G(cell) and for
H(cell) tend to be rather large. Selections of G(cell) values are given in Table 7.4.

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Table 7.4 Table of values of emf and n

G(cell) as a function of

G(cell) /kJ mol−1
(when n = 1)

G(cell) /kJ mol−1
(when n = 2)

emf /V
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0

106
116
125
135
145
154
164
174
183
193

212
232
251
270
289
309
328
347
367
386

Aside
While electrochemical methods are experimentally easy, the practical difficulties of obtaining accurate thermodynamic data are so severe that the experimentally determined values of G(cell) , H(cell) and S(cell) can only be regarded as ‘approximate’ unless we perform a daunting series of precautions. The two most common errors are: (1) allowing current passage to occur, causing the value of the cell emf to be too small; and (2) not performing the measurement reversibly.
The most common fault under (2) is changing the temperature of the cell too fast, so the temperature inside the cell is not the same as the temperature of, for example, the water bath in which a thermometer is placed.
None of these thermodynamic equations is reliable if we fail to operate the cell reversibly, since the emf is no longer an exact thermodynamic quantity. Why aren’t electrodes made from wood?
Electrodes: redox, passive and amalgam
The impetus for electronic motion is the chance to lose energy
(see p. 60), so electrons move from high energy to low.

Electrochemical cells comprise a minimum of two half-cells, the energetic separation between them being proportional to the cell emf. Since this energy is usually expressed as a voltage, we see that the energy needs to be measured electrically as a voltage.
We have seen already how a cell’s composition changes if a charge flows through it – we argued this phenomenon in terms of Faraday’s laws. We cause electrochemical reactions to occur whenever a cell converts chemical energy into electrical energy.

INTRODUCTION TO CELLS: TERMINOLOGY AND BACKGROUND

301

For this reason, we say a battery or cell discharges during operation, with each electron from the cell flowing from high energy to low.
All electrochemical cells (including batteries) have two poles: one relates to the half-cell that is positively charged and the other Currents conduct through an electrode relates to the negatively charged pole. Negatively charged electrons by means of electrons. are produced at the anode as one of the products of the electrochemical reaction occurring at there. But if the electrons are to move then we need something through which they can conduct to and from the terminals of the cell: we need an electrode.
The phenomenon we call electricity comprises a flow of charged electrons. Wood is a poor conductor of electricity because electrons are inhibited from moving freely through it: we say the wood has a high electrical ‘resistance’ R. By contrast, most metals are good conductors of charge. We see how A redox electrode acts an electrode needs to be electrically conductive if the electrons are as a reagent as well as to move. an electron conductor,
Most electrodes are metallic. Sometimes the metal of an electrode as the metal of an can also be one component part of a redox couple. Good examples electrode can also be include metallic iron, copper, zinc, lead or tin. A tin electrode forms one component part of a couple when in contact with tin(IV) ions, etc. Such electrodes are a redox couple. called redox electrodes (or non-passive). In effect, a redox electrode has two roles: first, it acts as a reagent; and, secondly, it measures the energy of the redox couple of which it forms one part when Metallic mercury is a poor choice of inert connected to a voltmeter.
Some metals, such as aluminium or magnesium, cannot function electrode at positive as redox electrodes because of a coating of passivating oxide. Oth- potentials because it oxidizes to form Hg(II) ers, such as calcium or lithium, are simply too reactive, and would ions. dissolve if immersed in solution.
But it is also extremely common for both redox states of a redox couple to be non-conductive. Simple examples might include disWe require an inert solving bromine in an aqueous solution of bromide ions, or the electrode when both oxidation of hydrogen gas to form protons, at the heart of a hydro- parts of a redox couple gen fuel cell; see Equation (7.13). In such cases, the energy of reside in solution, or the couple must be determined through a different sort of elec- do not conduct: the trode, which we call an inert electrode. Typical examples of inert electrode measures the electrodes include platinum, gold, glassy carbon or (at negative energy of the couple. potentials) mercury. The metal of an inert electrode itself does not react in any chemical sense: such electrodes function merely as a probe of the electrode potential for measurements at zero current,
We denote an amaland as a source, or sink, of electrons for electrolysis processes if gam by writing the ‘Hg’ current is to flow. in brackets after the
The final class of electrodes we encounter are amalgam elec- symbol of the dissolved trodes, formed by ‘dissolving’ a metal in elemental (liquid) mer- element; cobalt amalcury, generally to yield a solid. We denote an amalgam with brack- gam is symbolized as ets, so the amalgam of sodium in mercury is written as Na(Hg). The Co(Hg). properties of such amalgams can be surprisingly different from their

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ELECTROCHEMISTRY

constituent parts, so Na(Hg) is solid and, when prepared with certain concentrations of sodium, does not even react with water.

Why is electricity more dangerous in wet weather?
Electrolytes for cells, and introducing ions
Most electrical apparatus is safe when operated in a dry environment, but everyone should know that water and electricity represent a lethal combination. Only a minimal amount of charge conducts through air, so cutting dry grass with an electric mower is safe. Cutting the same grass during a heavy downpour risks electrocution, because water is a good conductor of electricity.
But water does not conduct electrons, so the charges must move through water by a wholly different mechanism than through a
Currents conduct metallic electrode. In fact, the charge carriers through solutions – through an electrolyte by means of ions. aqueous or otherwise – are solvated ions. The ‘mobility’ µ of an ion in water is sufficiently high that charge conducts rapidly from a wet electrical appliance toward the person holding it: it behaves
Ultimately, the word as an electrolyte.
‘ion’ derives from the
All cells comprise half-cells, electrodes and a conductive elecGreek eimi ‘to go’, trolyte; the latter component separates the electrodes and conducts implying the arrival of ions. It is usually, although not always, a liquid and normally has someone or something. an ionic substance dissolved within it, the solid dissociating in
We get the English solution to form ions. Aqueous electrolytes are a favourite choice word ‘aim’ from this root. because the high ‘dielectric constant’ of water imparts a high
‘ionic conductivity’ κ to the solution.
Sometimes electrochemists are forced to construct electrochemIonic conductivity is ical cells without water, e.g. if the analyte is water sensitive or often given the Greek merely insoluble. In these cases, we construct the cell with an symbol kappa (κ) organic solvent, the usual choice being the liquids acetonitrile, whereas electrical conpropylene carbonate (I), N,N-dimethylformamide (DMF) or diductivity is given the methylsulphoxide (DMSO), each of which is quite polar because different Greek symbol of its high dielectric constant . sigma (σ ).

O
O

O

(I)

CH3

In some experiments, we need to enhance the ionic conductivity of a solution, so we add an additional ionic compound to it. Rather confusingly, we call both the compound and the resultant solution ‘an electrolyte’.

INTRODUCING HALF-CELLS AND ELECTRODE POTENTIALS

303

The preferred electrolytes if the solvent is water are KCl and NaNO3 . If the solvent is a non-aqueous organic liquid, then we prefer salts of tetra-alkyl ammonium, such as tetra-n-butylammonium tetrafluoroborate, n Bu4 N+ BF4− .

7.2

Introducing half-cells and electrode potentials Why are the voltages of watch and car batteries different? Relationships between emf and electrode potentials
Being a cell, a battery contains two half-cells separated by an electrolyte. The electrodes are needed to connect the half-cells to an external circuit. Each electrode may act as part of a redox couple, but neither has to be.
The market for batteries is huge, with new types and applications being developed all the time. For example, a watch battery is a type An ‘electrode potenof ‘silver oxide’ cell: silver in contact with silver oxide forms one tial’ E is the energy half-cell while the other is zinc metal and dications. Conversely, (expressed as a volta car battery is constructed with the two couples lead(IV)|lead age) when a redox and lead(IV)|lead(II). The electrolyte is sulphuric acid, hence this couple is at equilibrium. battery’s popular name of ‘lead–acid’ cell (see further discussion The value of E cannot be measured directly on p. 347). and must be calculated
The first difference between these two batteries is the voltage from an experimental they produce: a watch battery produces about 3 V and a lead–acid emf. cell about 2 V. The obvious cause of the difference in emf are the different half-cells. The ‘electrode potential’ E is the energy, expressed as a voltage, when a redox couple is at equilibrium.
Two redox states of the
As a cell comprises two half-cells, we can now define the emf same material form a according to redox couple. emf = E(positive

half-cell)

− E(negative

half-cell)

(7.22)

This definition is absolutely crucial. It does not matter if the values of E for both half-cells are negative or both are positive:
E(positive) is defined as being the more positive of the two half-cells, and E(negative) is the more negative.
We now consider the emf in more detail, and start by saying that it represents the separation in potential between the two halfcell potentials; See Equation (7.22). In order for G(cell) to remain positive for all thermodynamically spontaneous cell discharges, the emf is defined as always being positive.

It is impossible to determine the potential of a single electrode: only its potential relative to another electrode can be measured.

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Another convention dictates that we write the more positive electrode on the righthand side of a cell, so we often see Equation (7.22) written in a slightly different form: (7.23) emf = E(RHS) − E(LHS)
Being a potential, the electrode potential has the symbol E. We must exercise care in the way we cite it. E is the energy of a redox couple, since it relates to two redox species, both an oxidized and reduced form, ‘O’ and ‘R’ respectively. We supplement the symbol E with appropriate subscripts, as EO,R .
Worked Example 7.9 Consider the electrode potentials for metallic lead within a lead– acid battery. The lead has three common redox states, Pb4+ , Pb2+ and Pb0 , so there are three possible equilibria to consider:
Pb4+ (aq) + 2e− = Pb2+ (aq)

for which E(equilibrium) = EPb4+ ,Pb2+

Pb4+ (aq) + 4e− = Pb0 (s)

for which E(equilibrium) = EPb4+ ,Pb0

2+

Pb

(aq)



+ 2e = Pb

0

(s)

for which E(equilibrium) = EPb2+ ,Pb0

We now see why it is so misleading to say merely EPb or, worse, ‘the electrode potential of lead’.

We cite the oxidized form first, as EO,R .

We usually omit the superscripted ‘zero’ on uncharged redox states. When choosing between two ionic valences, the name of the higher (more oxidized) state ends with -ic and the lower (less oxidized) form ends with -ous.

We conventionally cite the oxidized form first within each symbol, which is why the general form is EO,R , so EPb4+ ,Pb2+ is correct, but EPb2+ ,Pb4+ is not. Some people experience difficulty in deciding which redox state is oxidized and which is the reduced. A simple way to differentiate between them is to write the balanced redox reaction as a reduction. For example, consider the oxidation reaction in Equation (7.1). On rewriting this as a reduction, i.e.
Al3+ (aq) + 3e− = Al(s) , the oxidized redox form will automatically precede the reduced form as we read the equation from left to right, i.e. are written in the correct order. For example, EO,R for the couple in Equation (7.1) is EAl3+ ,Al .
We usually cite an uncharged participant without a superscript.
Considering the reaction Pb2+ + 2e− = Pb, the expression EPb2+ ,Pb is correct but the ‘0’ in EPb2+ ,Pb0 is superfluous.
SAQ 7.10 Consider the cobaltous ion | cobalt redox couple. Write an expression for its electrode potential.

With more complicated redox reactions, such as 2H+ (aq) + 2e− =
H2(g) , we would not normally write the stoichiometric number, so we prefer EH+ ,H2 to E2H+ ,H2 ; the additional ‘2’ before H+ is superfluous here.

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INTRODUCING HALF-CELLS AND ELECTRODE POTENTIALS

SAQ 7.11 Write down an expression for the electrode potential of the bromine | bromide couple. [Hint: it might help to write the balanced redox reaction first.]

How do ‘electrochromic’ car mirrors work?
Introducing an orbital approach to dynamic electrochemistry
It’s quite common when driving at night to be dazzled by the
Electrochromic mirrors lights of the vehicle behind as they reflect from the driver’s neware now a common view or door mirror. We can prevent the dazzle by forming a feature in expensive layer of coloured material over the reflecting surface within an cars. electrochromic mirror. Such mirrors are sometimes called ‘smart mirrors’ or electronic ‘anti-dazzle mirrors’.
These mirrors are electrochromic if they contain a substance that changes colour according to its redox state. For example, methylene blue, MB+ (II), is a chromophore because it has an intense blue colour. II is a popular choice of electrochromic material for such mirrors: it is blue when fully oxidized, but it becomes colourless when reduced according to

MB0 − → MB+ + e− colourless blue

(7.24)

N

(H3C)2N

Cl−

S

N(CH3)2

(II)

We can now explain how an electrochromic car mirror operates. The mirror is constructed with II in its colourless form, so the mirror functions in a normal way.
The driver ‘activates’ the mirror when the ‘anti-dazzle’ state of the mirror is required, and the coloured form of methylene blue (MB+ ) is generated oxidatively according to Equation (7.24). Coloured MB+ blocks out the dazzling reflection at the mirror by absorbing about 70 per cent of the light. After our vehicle has been overtaken and we require the mirror to function normally again, we reduce MB+ back to colourless
MB0 via the reverse of Equation (7.24), and return the mirror to its colourless state. These two situations are depicted in Figure 7.6.
An electron is donated
We discuss ‘colour’ in Chapter 9, so we restrict ourselves here to to an orbital during saying the colour of a substance depends on the way its electrons reduction. The electron interact with light; crucially, absorption of a photon causes an elec- removed during oxidatron to promote between the two frontier orbitals. The separation tion is taken from an in energy between these two orbitals is E, the magnitude of which orbital. relates to the wavelength of the light absorbed λ according to the
Planck–Einstein equation, E = hc/λ, where h is the Planck constant and c is the

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ELECTROCHEMISTRY
Conventional
mirror

Electrochromic mirror Incident beam

Incident beam

Emerging beam

Emerging beam

Reflective surface (a)

Reflective surface Electrochromic layers (b)

Figure 7.6 Mirrors: (a) an ordinary car driver’s mirror reflects the lights of a following car, which can dazzle the driver; (b) in an electrochromic mirror, a layer of optically absorbing chemical is electro-generated in front of the reflector layer, thereby decreasing the scope for dazzle. The width of the arrows indicates the relative light intensity

speed of light in a vacuum. The value of E for MB0 corresponds to an absorption in the UV region, so MB0 appears colourless. Oxidation of MB0 to MB+ causes a previously occupied orbital to become empty, itself changing the energy separation
E between the two frontier orbitals. And if E changes, then the Planck–Einstein equation tells us the wavelength λ of the light absorbed must also change. E for
MB+ corresponds to λ of about 600 nm, so the ion is blue.
The reasoning above helps explain why MB0 and MB+ have different colours. To summarize, we say that the colours in an electrochromic mirror change following oxidation or reduction because different orbitals are occupied before and after the electrode reaction.

Why does a potential form at an electrode?
Formation of charged electrodes
For convenience, we will discuss here the formation of charges with the example of copper metal immersed in a solution of copper sulphate (comprising Cu2+ ions). We consider first the situation when the positive pole of a cell is, say, bromine in contact with bromide ions, causing the copper to be the negative electrode.
Let’s look at the little strip cartoon in Figure 7.7, which shows the surface of a copper electrode. For clarity, we have drawn only one of the trillion or so atoms on its surface. When the cell of which it is a part is permitted to discharge spontaneously, the copper electrode acquires a negative charge in consequence of an oxidative electron-transfer reaction (the reverse of Equation (7.7)). During the oxidation, the surface-bound atom loses the two electrons needed to bond the atom to the electrode surface, becomes a cation and diffuses into the bulk of the solution.

INTRODUCING HALF-CELLS AND ELECTRODE POTENTIALS

307

Positive ion
(cation)

Surface atom



++



Electrode

Figure 7.7

Excess surface charge

Schematic drawing to illustrate how an electrode acquires its negative charge

Positive ion
(cation)

Surface atom

+
+

Electrode

Figure 7.8

Excess surface charge

Schematic drawing to illustrate how an electrode acquires its positive charge

The two electrons previously ‘locked’ into the bond remain on the electrode surface, imparting a negative potential.
We now consider a slightly different cell in which the copper half-cell is the positive pole. Perhaps the negative electrode is zinc metal in contact with Zn2+ ions. If the cell discharges spontaneously, then the electron-transfer reaction is the reduction reaction in Equation (7.7) as depicted in the strip cartoon in Figure 7.8. A bond forms between the surface of the copper electrode and a Cu2+ cation in the solution The electrons needed to reduce the cation come from the electrode, imparting a net positive charge to its surface.
Finally, we should note that the extent of oxidation or reduction needed to cause a surface charge of this type need not be large; and the acquisition of charge, whether positive or negative, is fast and requires no more than a millisecond after immersing the electrodes in their respective half-cells.

308

7.3

ELECTROCHEMISTRY

Activity
Why does the smell of brandy decrease after dissolving table salt in it?
Real and ‘perceived’ concentrations

At the risk of spoiling a good glass of brandy, try adding a little table salt to it and notice how the intensity of the smell is not so strong after the salt dissolves.
We recall from Chapter 5 how the intensity of a smell we detect with our nose is proportional to the vapour pressure of the substance
We mention the volatile causing it. The vapour pressure of ethanol is p(ethanol) , its magnialcohol here because it is responsible for the tude being proportional to the mole fraction of ethanol in the brandy; smell. brandy typically contains about 40 per cent (by volume) of alcohol.
Although adding table salt does not decrease the proportion of the alcohol in the brandy, it does decrease the apparent amount. And because the perceived proportion is lowered, so the vapour pressure drops, and we discern the intensity of the smell has decreased. We are entering the world of ‘perceived’ concentrations.
Although the actual concentrations of the volatile components in solution remain unchanged after adding the salt, the system perThe ‘activity’ a is the ceives a decrease in the concentration of the volatile components. thermodynamically perceived concentration.
This phenomenon – that the perceived concentration differs from the real concentration – is quite common in the thermodynamics of solution-phase electrochemistry. We say that the concentration persists, but the
‘activity’ a has decreased by adding the salt.
As a working definition, the activity may be said to be ‘the perceived concentration’ and is therefore somewhat of a ‘fudge factor’. More formally, the activity a is defined by c (7.25) a = Oγ c where c is the real concentration. The concluding term γ , termed the activity coefficient, is best visualized as the ratio of a solute’s ‘perceived’ and ‘real’ concentrations.
The activity a and the activity coefficient γ are both dimensionless quantities, which explains why we must include the additional
We only add the term
‘c O ’ term, thereby ensuring that a also has no units. We say the c O in order to render the activity dimensionvalue of c O is 1 mol dm−3 when c is expressed in the usual units less. of mol dm−3 , and 1 mol m−3 if c is expressed in the SI units of mol m−3 , and so on.

Why does the smell of gravy become less intense after adding salt to it?
The effect of composition on activity
Gravy is a complicated mixture of organic chemicals derived from soluble meat extracts. Its sheer complexity forces us to simplify our arguments, so we will

ACTIVITY

309

approximate and say it contains just one component in a water-based solution. Any incursions into reality, achieved by extending our thoughts to encompass a multicomponent system, will not change the nature of these arguments at all.
Adding table salt to gravy causes its lovely smell to become less intense. This is a general result in cooking: adding a solute (particularly if the solute is ionic) decreases the smell, in just the same way as adding table salt decreased the smell of brandy in the example directly above.
The ability to smell a solute relies on it having a vapour pressure above the solution. Analysing the vapour above a gravy dish The pressure above shows that it contains molecules of both solvent (water) and solute a solution relates to
(gravy), hence its damp aroma. The vapour pressures above the the composition of gravy dish do not alter, provided that we keep the temperature con- solution, according stant and maintain the equilibrium between solution and vapour. to Henry’s law; see
The proportion of the solute in the vapour is always small because Section 5.6. most of it remains in solution, within the heavier liquid phase.
As a good approximation, the vapour pressure of each solute in the vapour above the dish is dictated by the respective mole fractions in the gravy beneath. As an example, adding water to the gravy solution dilutes it and, therefore, decreases the gravy smell, because the mole fraction of the
An electrochemist assesgravy has decreased. ses the number of ions
Putting ionic NaCl in the gravy increases the number of ions in and their relative influsolution, each of which can then interact with the water and the solute, ence by means of the which decreases the ‘perceived concentration’ of solute. In fact, we ‘ionic strength’ I (as can now go further and say the thermodynamic activity a represents defined below). the concentration of a solute in the presence of interactions.

Why add alcohol to eau de Cologne?
Changing the perceived concentration
Fragrant eau de Cologne is a dilute perfume introduced in Cologne (Germany) in 1709 by Jean Marie Farina. It was probably a modification of a popular formula made before 1700 by Paul
Feminis, an Italian in Cologne, and was based on bergamot and other citrus oils. The water of Cologne was believed to have the power to ward off bubonic plague.
Eau de Cologne perfume is made from about 80–85 per cent water and 12–15 per cent ethanol. Volatile esters make up the remainder, and provide both the smell and colour.
The vapour pressure of alcohol is higher than that of water, so adding alcohol to an aqueous perfume increases the pressure of the gases above the liquid. In this way, the activity a of the organic components imparting the smell will increase and thereby increase the perceived concentration of the esters. And increasing a(ester)

The word ‘perfume’ comes from the Latin per fumem, meaning
‘through smoke’.
These esters are stable in the dark, but degrade in strong sunlight, which explains why so many perfumes are sold in bottles of darkened or frosted glass. 310

ELECTROCHEMISTRY

has the effect of making the eau de Cologne more pungent. Stated another way, the product requires less ester because the alcohol increases its perceived concentration.
Incidentally, the manufacturer also saves money this way.

Thermodynamic activity a
Every day, electrochemists perform measurements that require a knowledge of the activity a. Measurements can be made in terms of straightforward concentrations if solutions are very dilute, but ‘very dilute’ in this context implies c ≈ 10−4 mol dm−3 , or less. Since most solutions are far more concentrated than millimoles per litre, from now on we will write all equations in terms of activities a instead of concentration c.
The values of activity a and concentration c are the same for very dilute solutions, so the ratio of a and c is one because the
The concept of activity real and perceived concentrations are the same. If a = c, then was introduced in the early 20th century by
Equation (7.25) shows how the activity coefficient γ has a value one of the giants of of unity at low concentration.
American chemistry, G.
By contrast, the perceived concentration is usually less than the
N. Lewis. real concentration whenever the solution is more concentrated, so γ < 1. To illustrate this point, Figure 7.9 shows the relationship between the activity coefficient γ (as ‘y’) and concentration (as ‘x’) for a few simple solutes in water. The graph shows clearly how the value of γ can drop quite dramatically as the concentration increases.

1

0.8

HCl [1 : 1]
KCl [1 : 1]
KOH [1 : 1]
NaBr [1 : 1]

g

0.6

0.4

0.2
H2SO4 [1 : 2]
CuSO4 [2 : 2]
In2(SO4)3 [2 : 3]

0
0

0.2

0.4

0.6

0.8

Concentration / mol dm

1

1.2

−3

Figure 7.9 The dependence of the mean ionic activity coefficient γ± on concentration for a few simple solutes

ACTIVITY

311

The activity of a solid The activity of a pure solid in its standard state is unity, so the activity of pure copper or of zinc metal elec- Unit and unity here both mean ‘one’, so trodes is one. We write this as a(Cu) or a(Zn) = 1. unit activity means
The activity of an impure solid is more complicated. Such an a = 1.
‘impure’ system might be represented by a solid metal with a dirty surface, or it might represent a mixture of two metals, either as an alloy or an amalgam with a metal ‘dissolved’ in mercury.
For example, consider the bi-metallic alloy known as bronze, which contains tin (30 mol%) and copper (70 mol%). There are An ‘alloy’ is a mixture two activities in this alloy system, one each for tin and copper. The of metals, and is not a activity of each metal is obtained as its respective mole fraction x, compound. so x(Sn) = a(Sn) = 0.3, and a(Cu) = 0.7.
Worked Example 7.10 A tooth filling is made of a silver amalgam that comprises
37 mol% silver. What is the activity of the mercury, a(Hg) ?
The activity of the mercury a(Hg) is the same as its mole fraction, x(Hg) . By definition x(Hg) + x(Ag) = 1 so x(Hg) = 1 − x(Ag) = 0.63

The sum of the mole fractions x must always add up to one because
‘the sum of the constituents adds up to the whole’.

The activity is therefore x(Hg) = a(Hg) = 0.63
The activity of a gas The activity of a pure gas is its pressure
Reminder: the value of
(in multiples of p O ), so a(H2 ) = p(H2 ) ÷ p O . The activity of pure p O is 105 Pa.
O
hydrogen gas a(H2 ) at p is therefore unity.
In fact, for safety reasons it is not particularly common to employ pure gases during electrochemical procedures, so mixtures are pre- In a mixture of gases, ferred. As an example, the hydrogen gas at the heart of the standard we call the inert gas a hydrogen electrode (SHE) is generally mixed with elemental nitro- base or bath gas. gen, with no more than 10 per cent of H2 by pressure. We call the other gas a base or bath gas. Conversely, we might also say that hydrogen dilutes the nitrogen, and so is a diluent.
In such cases, we can again approximate the activity to the mole fraction x.
Worked Example 7.11 Hydrogen gas is mixed with a nitrogen ‘bath gas’. The overall pressure is p O . If the mole fraction of the hydrogen is expressed as 10 per cent, what is its activity?
By definition, x(X) = partial pressure, p(X) , so a(H2 ) = p(H2 ) ÷ p(total) = 0.1

312

ELECTROCHEMISTRY

The activity of a solution It is unwise to speak in broad terms of ‘the activity of a solution’ because so many different situations may be considered. For example, consider the following two examples.
(1) The activity of a mixture of liquids. It is rarely a good idea to suggest that the activity of a liquid in a mixture is equal to its mole
Amalgams are liquid fraction x because of complications borne of intermolecular interwhen very dilute, but actions (e.g. see Chapter 2 and Section 5.6 concerning Raoult’s are solid if the mole fraction of mercury law). Thankfully, it is generally rare that an electrochemist wants drops below about 70 to study liquid mixtures of this sort (except amalgams diluted to a per cent. maximum mole fraction of about 1 per cent metal in Hg), so we will not consider such a situation any further.
(2) The activity of a solute in a liquid solvent. The activity a and concentration c may be considered to be wholly identical if the concentration is tiny (to a maximum of about 10−3 mol dm−3 ), provided the solution contains no other solutes. Such a concentration is so tiny, however, as to imply slightly polluted distilled water, and is not particularly useful.
For all other situations, we employ the Debye–H¨ ckel laws (as below) to calculate u the activity coefficient γ . And, knowing the value of γ , we then say that a = (c ÷ c O ) × γ (Equation (7.25)), remembering to remove the concentration units because a is dimensionless.

Why does the cell emf alter after adding LiCl?
Ionic ‘screening’
Consider the Daniell cell Zn|Zn2+ ||Cu2+ |Cu. The cell emf is about 1.1 V when prepared with clean, pure electrodes and both solutions at unit activity. The emf decreases to about 1.05 V after adding lithium chloride to the copper half-cell. Adding more
LiCl, but this time to the zinc solution, increases the emf slightly, to about 1.08 V.
No redox chemistry occurs, so no copper ions are reduced to copper metal nor is zinc metal oxidized to form Zn2+ . No comIn fact, a similar result plexes form in solution, so the changes in emf may be attributed is obtained when adding most ionic electrolytes. entirely to changing the composition of the solutions.
Lithium and chloride ions are not wholly passive, but interact with the ions originally in solution. Let us look at the copper ions, each of which can associate electrostatically with chloride ions, causing it to resemble a dandelion ‘clock’ with the central copper
We need a slightly ion looking as though it radiates chloride ions. All the ions are different form of γ solvated with water. These interactions are coulombic in nature, when working with so negatively charged chloride anions interact attractively with the electrolyte solutions: positive charges of the copper cations. Copper and lithium cations we call it the mean repel. Conversely, the additional Li+ ions attract the negatively ionic activity coefficient charged sulphates from the original solution; again, Cl− and SO2− γ± , as below.
4
anions repel.

ACTIVITY

313

The ionic atmosphere moves continually, so we consider its com- ‘Associated ions’ in position statistically. Crystallization of solutions would occur if the this context means ionic charges were static, but association and subsequent dissocia- an association species tion occur all the time in a dynamic process, so even the ions in a held together (albeit dilute solution form a three-dimensional structure similar to that in transiently) via electroa solid’s repeat lattice. Thermal vibrations free the ions by shaking static interactions. apart the momentary interactions.
The ions surrounding each copper cation are termed the ionic atmosphere. In the neighbourhood of any positively charged ion (such as a copper cation), there are likely to be more negative charges than positive (and vice versa). We say the cations are surrounded with a shell of anions, and each anion is surrounded by a shell of cations.
The ionic atmosphere can, therefore, be thought to look much like an onion, or a
Russian doll, with successive layers of alternate charges, with the result that charges effectively ‘cancel’ each other out when viewed from afar.
Having associated with other ions, we say the copper ion is screened from anything else having a charge (including the electrode), so the full extent of its charge cannot be
‘experienced’. In consequence, the magnitude of the electrostatic interactions between widely separated ions will decrease.
The electrode potential measured at an electrode relates to the
‘Coulomb potential energy’ V ‘seen’ by the electrode due to the The ‘Coulomb potential ions in solution. V relates to two charges z1 and z2 (one being the energy’ V is equal to the work that must be electrode here) separated by a distance r, according to z+ z−
V =
4π o r r

(7.26)

done to bring a charge z+ from infinity to a distance of r from the charge z− .

where o is the permittivity of free space and r is the relative permittivity of the solvent. In water at 25 ◦ C, r has a value of 78.54.
The magnitude of V relates to interactions between the electrode and nearby ions nestling within the interface separating the electrode and the ionic solution. Since the ‘effective’ (visible) charge on the ions decreases, so the electrode perceives there to be fewer of them. In other words, it perceives the concentration to have dipped below the actual concentration. This perceived decrease in the number of charges then causes the voltmeter to read a different, smaller value of ECu2+ ,Cu .
The zinc ions in the other half of the Daniell cell can similarly interact with ions added to solution, causing the zinc electrode to ‘see’ fewer Zn2+ species, and the voltmeter again reads a different, smaller value of EZn2+ ,Zn . Since the emf represents the separation between the electrode potentials of the two half-cells, any changes in the emf illustrate the changes in the constituent electrode potentials.

Background to the Debye–H¨ckel theory u The interactions between the ions originally in solution and any added LiCl are best treated within the context of the Debye–H¨ ckel theory, which derives from a u knowledge of electrostatic considerations.

314

ELECTROCHEMISTRY

Firstly, we assume the ions have an energy distribution as defined by the Boltzmann distribution law (see p. 35). Secondly, we say that electrostatic forces affect the behaviour and the mean positions of all ions in solution. It should be intuitively clear that ions having a larger charge are more likely to associate strongly than ions having a smaller charge. This explains why copper ions are more likely to associate than are sodium ions. The magnitude of the force exerted by an ion with a charge z1 on another charge z2 separated by an inter-ion distance of r in a medium of relative permittivity r is the ‘electrostatic interaction’ φ, as defined by φ= z+ z−
4π o r r 2

(7.27)

Note how this equation states that the force is inversely proportional to the square of the distance between the two charges r, so the value of φ decreases rapidly as r increases. Since cations and anions have opposite charges, φ is negative.
The force between two anions will yield a positive value of φ. We
Positive values of φ see how a positive value of φ implies an inter-ionic repulsion and imply repulsion, and a a negative value implies an inter-ionic attraction. negative value attracThe Debye–H¨ ckel theory suggests that the probability of findu tion. ing ions of the opposite charge within the ionic atmosphere increases with increasing attractive force.

Why does adding NaCl to a cell alter the emf, but adding tonic water doesn’t?
The effects of ion association and concentration on γ
Sodium chloride – table salt – is a ‘strong’ ionic electrolyte because it dissociates fully when dissolved in water (see the discussion of weak and strong acids in
Section 6.2). The only electrolytes in tonic water are sugar (which is not ionic) and sodium carbonate, which is a weak electrolyte, so very few ions are formed by adding the tonic water to a cell.
The ratio of perceived to real concentrations is called the activity coefficient γ
(because, from Equation (7.25), γ = a ÷ c). Furthermore, from the definition of activity in Equation (7.20), γ will have a value in the range zero to one. The diagram in Figure 7.9 shows the relationship between γ and concentration c for a few ionic electrolytes. Adding NaCl to solution causes γ to decrease greatly because the number of ions in solution increases. Adding tonic water does not decrease the activity coefficient much because the concentration of the ions remains largely unchanged. The change in γ varies more with ionic electrolytes because the interactions are far stronger. And if the value of γ does not change, then the real and perceived concentrations will remain essentially the same.

ACTIVITY

The extent of ionic screening depends on the extent of association. The only time that association is absent, and we can treat ions as though free and visible (‘unscreened’), is at infinite dilution.

Why does MgCl2 cause a greater decrease in perceived concentration than KCl?

315

Infinite dilution (extrapolation to zero concentration) means so small a concentration that the possibility of two ions meeting, and thence associating, is tiny to non-existent.

The mean ionic activity coefficient γ±
The value of γ depends

The extent of ionic association depends on the ions we add to on the solute employed. the solution. And the extent of association will effect the extent of screening, itself dictating how extreme the difference is between perceived and real concentration. For these reasons, the value of γ (= a ÷ c) depends on the choice of solute as well as its concentration, so we ought to cite the solute whenever we cite an activity coefficient.
The value of γ is even more difficult to predict because solutes contain both anions and cations. In fact, it is impossible to dif- We cannot know either ferentiate between the effects of each, so we measure a weighted γ+ or γ− ; we can only average. Consider a simple electrolyte such as KCl, which has know the value of their one anion per cation. (We call it a ‘1:1 electrolyte’.) In KCl, the geometric mean γ± . activity coefficient of the anions is called γ(Cl− ) and the activity coefficient of the cations is γ(K+ ) . We cannot know either γ+ or γ− ; we can only know the value of γ± . Accordingly, we modify We call KCl a 1:1 elecEquation (7.25) slightly by writing trolyte, since the ratio a= c γ± cO

(7.28)

of anions to cations is
1:1.

where the only change is the incorporation of the mean ionic activity coefficient γ± .
The mean ionic activity coefficient is obtained as a geometric mean via γ± =

γ(K+ ) × γ(Cl− )

(7.29)

By analogy, the expression for the mean ionic activity coefficient γ± for a 2:1 electrolyte such as K2 SO4 is given by γ± =

3

2 γ+ × γ−

(7.30)

where the cube root results from the stoichiometry, since K2 SO4 contains three ions

(we could have written the root term alternatively as 3 γ+ × γ+ × γ− , with one γ term per ion). Again, a 1:3 electrolyte such as FeCl3 dissolves to form four ions, so an expression for its mean ionic activity coefficient γ± will include a fourth root, etc.

316

ELECTROCHEMISTRY

SAQ 7.12 Write an expression similar to Equation (7.29) for the 2:3 electrolyte Fe2 (SO4 )3 .

Why is calcium better than table salt at stopping soap lathering? Ionic strength I and the Debye–H¨ckel laws u People whose houses are built on chalky ground find that their kettles and boilers become lined with a hard ‘scale’. We say that the water in the area is ‘hard’, meaning that minute amounts of chalk are dissolved in it. The hard layer of ‘scale’ is chalk that precipitated onto the inside surface of the kettle or boiler during heating.
It is difficult to get a good soapy froth when washing the hands in ‘hard water’ because the ions from chalk in the water associate
We look at the actions with the long-chain fatty acids in soap, preventing it from ionizing of soaps in Chapter 10. properly. Conversely, if the water contains table salt – for example, when washing the dishes after cooking salted meat – there is less of a problem with forming a good froth. Although the concentrations of sodium and calcium ions may be similar, the larger charges on the calcium and carbonate ions impart a disproportionate effect, and strongly inhibit the formation of frothy soap bubbles.
Having discussed ionic screening and its effects on the value of γ± , we now consider the ionic charge z. When assessing the
In ‘dynamic’ electrochemistry (when influence of z, we first define the extent to which a solute procurrents flow) we need motes association, and thus screening. The preferred parameter is to be careful not to the ‘ionic strength’ I, as defined by mistake ionic strength and current, since both have the symbol I.

1
I=
2

i=i
2
ci zi

(7.31)

i=1

where zi is the charge on the ion i in units of electronic charge, and ci is its concentration. We will consider three simple examples to demonstrate how ionic strengths
I are calculated.
Worked Example 7.12 Calculate the ionic strength of a simple 1:1 electrolyte, such as
NaCl, that has a concentration of c = 0.01 mol dm−3 .
Inserting values into Equation (7.31) we obtain
I=

1
2

{ [Na+] × (+1)2 + [Cl−] × (−1)2 } terms for the sodium ions

terms for the chloride ions

ACTIVITY

We next insert concentration terms, noting that one sodium ion and one chloride are formed per formula unit of sodium chloride (which is why we call it a 1:1 electrolyte). Accordingly, the concentrations of the two ions, [Na+ ] and [Cl− ], are the same as [NaCl], so
I=

1
{([NaCl] × 1) + ([NaCl] × 1)}
2

so we obtain the result for a 1:1 electrolyte that I(NaCl) = [NaCl].
Note that I has the same units as concentration: inserting the NaCl concentration [NaCl] = 0.01 mol dm−3 , we obtain I =
0.01 mol dm−3 .

317

NaCl is called a ‘1:1 electrolyte’ because the formula unit contains one anion and one cation. We obtain the result
I = c only for 1:1 (univalent) electrolytes.

Worked Example 7.13 Calculate the ionic strength of the 2:2 electrolyte FeSO4 at a concentration c = 0.01 mol dm−3 .
Inserting charges in Equation (7.31):
I=

1
{[Fe2+ ] × (+2)2 + [SO2− ] × (−2)2 }
4
2

We next insert concentrations, again noting that one ferrous ion and one sulphate ion are formed per formula unit:
I=

1
{([FeSO4 ] × 4) + ([FeSO4 ] × 4)}
2

so we obtain the result I = 4 × c for this, a 2:2 electrolyte.
Inserting the concentration c of [FeSO4 ] = 0.01 mol dm−3 , we obtain I = 0.04 mol dm−3 , which explains why hard water containing FeSO4 has a greater influence than table salt of the same concentration.
Worked Example 7.14 Calculate the ionic strength of the 1:2 electrolyte CuCl2 , again of concentration 0.01 mol dm−3 .
Inserting charges into Equation (7.31):
I=

1
{[Cu2+ ] × (+2)2 + [Cl− ] × (−1)2 }
2

We next insert concentrations. In this case, there are two chloride ions formed per formula unit of salt, so [Cl− ] = 2 × [CuCl2 ], but only one copper, so [Cu2+ ] = [CuCl2 ].
I=

1
{([CuCl2 ] × 4) + (2[CuCl2 ] × 1)}
2

Note how the calculation requires the charge per anion, rather than the total anionic charge.

so we obtain the result I = 3 × c for this, a 1:2 electrolyte. And, I = 0.03 mol dm−3 because [CuCl2 ] = 0.01 mol dm−3 .

318

ELECTROCHEMISTRY

SAQ 7.13 Calculate the relationship between concentration and ionic strength for the 1:3 electrolyte CoCl3 .
Ionic strength I is an integral multiple of concentration c, where integer means whole number. A calculation of I not yielding a whole number is wrong.

Table 7.5 summarizes all the relationships between concentration and ionic strength I for salts of the type Mx+ Xy− , listed as a function of electrolyte concentration. Notice that the figures in the table are all integers. A calculation of I not yielding a whole number is wrong.
Ions with large charges generally yield weak electrolytes, so the numbers of ions in solution are often smaller than predicted. For this reason, values of I calculated for salts represented by the bottom right-hand corner of Table 7.5 might be too high.

Why does the solubility of AgCl change after adding
MgSO4 ?
Calculating values of γ±
Silver chloride is fairly insoluble (see p. 332), with a solubility product Ksp of 1.74 × 10−10 mol2 dm−6 . Its concentration in pure distilled water will, therefore, be 1.3 × 10−5 mol dm−3 , but adding magnesium sulphate to the solution increases it solubility appreciably; see Figure 7.10.
This increase in solubility is not an example of the common ion effect, because there are no ions in common. Also impossible is the idea that the equilibrium constant has changed, because it is a constant.
Strictly, we should formulate all equilibrium constants in terms of activities rather than concentrations, so Equation (7.32) describes Ksp for dissolving partially soluble
AgCl in water:

We obtain the concentration [Ag+ ] = 1.3 ×
10−5 as the square root of 1.74 × 10−10 mol2 dm−6 .

Ksp = a(Ag+ ) a(Cl− ) = [Ag+ ][Cl− ] × γ(Ag+ ) γ(Cl− )
Table 7.5 Summary of the relationship between ionic strength I and concentration c. As an example, sodium sulfate (a 1:2 electrolyte) has an ionic strength that is three times larger than c
X−
M+
M2+
M3+
M4+

X2−

X3−

X4−

1
3
6
10

3
4
15
12

6
15
9
42

10
12
42
16

(7.32)

ACTIVITY

319

s (AgCl)/ 10−5 mol dm−3

1.7

1.6

1.5

1.4
Solubility of AgCl in pure water

1.3
0

0.2

0.4

0.6

0.8

1

1.2

[MgSO4] / 0.01 mol dm−3

Figure 7.10 The solubility s of AgCl (as ‘y’) in aqueous solutions of MgSO4 against its concentration, [MgSO4 ] (as ‘x’). T = 298.15 K

The exact structure of the equilibrium constant on the right-hand side of Equation
(7.32) follows from the definition of activity a in Equation (7.25). The product of the
2
two γ terms is γ± .
The values of the activity coefficients decrease with increasing ionic strength I (as below). The only way for Ksp to remain constant at the same time as the activity coefficient γ± decreasing is for the concentrations c to increase. And this is exactly what happens: the concentration of AgCl has increased by about 50 per cent when the concentration of MgSO4 is 1.2 mol dm−3 .
Changes in solubility product are one means of experimentally determining a value of activity coefficient, because we can independently determine the concentrations
(e.g. via a titration) and the values of all γ± will be ‘one’ at zero ionic strength.
An ionic strength of
Alternatively, we can calculate a value of γ± with the Debye–
10−3 mol dm−3 could
H¨ ckel laws. There are two such laws: the limiting and the simpli- imply a concentrau fied laws. Calculations with the limiting law are only valid at very tion as low as 10−4
−3
low ionic strengths (i.e. 0 < I ≤ 10−3 mol dm−3 ), which is very mol dm , because
I > c. dilute. The limiting law is given by

(7.33) log10 γ± = −A|z+ z− | I where A is the so-called Debye–H¨ ckel ‘A’ constant (or factor), u which has a value of 0.509 mol−1/2 dm3/2 at 25 ◦ C. z+ and z− are the charges per cation and per anion respectively. The vertical modulus lines ‘|’ signify that the charges on the ions have magnitude, but we need to ignore their signs (in practice, we call them both positive). The quantities between the two vertical modulus lines ‘|’ have magnitude alone, so we ignore the signs on the charges z+ and z−

320

ELECTROCHEMISTRY


From Equation (7.28), we expect a plot of log10 γ± (as ‘y’) against I (as ‘x’) to be linear. It generally is linear, although it deviates appreciably at higher ionic strengths. Worked Example 7.15 What is the activity coefficient of copper in a solution of copper sulphate of concentration 10−4 mol dm−3 ?
Note how we ignore the sign of the negative charge here.

Copper sulphate is a 2:2 electrolyte so, from Table 7.5, the ionic strength I is four times its concentration. We say I = 4 × 10−4 mol dm−3 .
Inserting values into Equation (7.33): log10 γ± = −0.509 | + 2 × −2|(4 × 10−4 )1/2

When calculating with
Equation (7.33), be sure to use ‘log’ (in base 10) rather than
‘ln’ (log in base e).

log10 γ± = −2.04 × (2 × 10−2 ) log10 γ± = −4.07 × 10−2
Taking the anti-log: γ± = 10−0.0407 γ± = 0.911

At extremely low ionic strengths, the simplified law becomes the limiting law. This follows since the denomi√ nator ‘I + b I’ tends to one as ionic strength
I tends to zero, causing the numerator to become one.

We calculate that the perceived concentration is 91 percent of the real concentration. For solutions that are more concentrated (i.e. for ionic strengths u in the range 10−3 < I < 10−1 ), we need the Debye–H¨ ckel simplified law:

A|z+ z− | I
(7.34)
log10 γ ± = −

1+b I where all other terms have the same meaning as above, and b is a constant having an approximate value of one. We include b because its units are mol−1/2 dm3/2 . It is usual practice to say b =
1 mol−1/2 dm3/2 , thereby making the denominator dimensionless.

SAQ 7.14 Prove that the simplified law becomes the limiting law at very low I.

Worked Example 7.16 What is the activity coefficient of a solution of CuSO4 of concentration 10−2 mol dm−3 ?
Again, we start by saying that I = 4 × c, so I = 4 × 10−2 mol dm−3 . Inserting values into Equation (7.34):

0.509 |2 × −2| 4 × 10−2 log10 γ± = −

1 + 4 × 10−2

321

HALF-CELLS AND THE NERNST EQUATION

Table 7.6 Typical activity coefficients γ± for ionic electrolytes as a function of concentration c in water
Electrolyte/
mol dm−3
HCl (1:1)
KOH (1:1)
CaCl2 (1:2)
CuSO4 (2:2)
In2 (SO4 )3 (2:3)

γ±
−3

c = 10 /mol dm

−3

−2

−3

c = 10 /mol dm

0.996


0.74


c = 10−1 /mol dm−3 c = 1.0/mol dm−3

0.904
0.90
0.903
0.41
0.142

0.796
0.80
0.741
0.16
0.035

0.809
0.76
0.608
0.047


0.4072
1 + 0.2 log10 γ± = −0.3393 log10 γ± = −

γ± = 10−0.3383 so γ± = 0.458

Table 7.6 cites a few sample values of γ± as a function of concentration. Note how multi-valent anions and cations cause γ± to vary more greatly than do mono-valent ions. The implications are vast: if an indium electrode were to be immersed in a solution of In2 (SO4 )3 of concentration 0.1 mol dm−3 , for example, then a value of γ± = 0.035 means that the activity (the perceived concentration) would be about 30 times smaller!
SAQ 7.15 From Worked Example 7.15, the mean ionic activity coefficient γ± is 0.911 for CuSO4 at a concentration of 10−4 mol dm−3 . Show that adding MgSO4 (of concentration 0.5 mol dm−3 ) causes γ± of the CuSO4 to drop to 0.06. [Hint: first calculate the ionic strength. [MgSO4 ] is high, so ignore the CuSO4 when calculating the ionic strength I.]

7.4

Half-cells and the Nernst equation
Why does sodium react with water yet copper doesn’t?
Standard electrode potentials and the E scale
O

Sodium reacts with in water almost explosively to effect the reaction
Na(s) + H+ (aq) − → Na+ (aq) + 1 H2(g)

2

(7.35)

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ELECTROCHEMISTRY

The protons on the left-hand side come from the water. Being spontaneous, the value of Gr for Equation (7.25) is negative. The value of Gr comprises two components:
(1)

G for the oxidation reaction Na − → Na+ + e− ; and


(2)

G for the reduction reaction H+ + 2e− − → 1 H2 .
− 2

Since these two equations represent redox reactions, we have effectively separated a cell into its constituent half-cells, each of which is a single redox couple.
By contrast, copper metal does not react with water to liberate hydrogen in a reaction like Equation (7.35); on the contrary, black copper(II) oxide reacts with hydrogen gas to form copper metal:

CuO(s) + H2(g) − → Cu(s) + 2H+ + O2−

(7.36)

The protons and oxide ions combine to form water. Again, the value of Gr for
Equation (7.36) is negative, because the reaction is spontaneous. G would be positive if we wrote Equation (7.36) in reverse. The change in sign follows because the
Gibbs function is a function of state (see p. 83).
The reaction in Equation (7.36) can be split into its two constituent half-cells:
(1)


G for the reaction Cu2+ + 2e− − → Cu; and

(2)

G for the reaction H2 − → 2H+ + 2e− .


Let us look at Equation (7.36) more closely. The value of components, according to Equation (7.36):

Gr comprises two

Gr = ( GCu2+ →Cu ) + ( GH2 →2H+ )

This change of sign follows from the change in direction of the second reaction.

(7.37)

If we wished to be wholly consistent, we could write both reactions as reduction processes. Reversing the direction of reaction
(2) means that we need to change the sign of its contribution toward the overall value of Gr , so
Gr = ( GCu2+ →Cu ) − ( G2H+ →H2 )

(7.38)

We remember from Equation (7.15) how G(cell) = −nF × emf . We will now invent a similar equation, Equation (7.39), which relates G for a half-cell and its respective electrode potential EO,R , saying:
GO,R = −nFE O,R
Substituting for
(7.39) gives

(7.39)

GO,R in Equation (7.38) with the invented expression in Equation
Gr = (−nFE Cu2+ ,Cu ) − (−nFE H+ ,H2 )

(7.40)

HALF-CELLS AND THE NERNST EQUATION

323

This expression does not relate to a true cell because the two electrode potentials are not measured with electrodes, nor can we relate Gr to the emf , because electrons do not flow from one half-cell via an external circuit to the other. Nevertheless,
Equation (7.40) is a kind of proof that the overall value of Gr relates to the constituent half-cells.
If we write a similar expression to that in Equation (7.40) for the reaction between sodium metal and water in Equation (7.35), then we would have to write the term for the hydrogen couple first rather than second, because the direction of change within the couple is reversed. In fact, any couple that caused hydrogen gas to form protons would be written with the hydrogen couple first, and any couple that formed hydrogen gas from protons (the reverse reaction) would be written with the hydrogen term second.
This observation led the pioneers of electrochemical thermodynamics to construct a series of cells, each with the H+ |H2 couple as one half-cell. The emf of each was measured. Unfortunately, there were always more couples than measurements, so they could never determine values for either ECu2+ ,Cu or EH+ ,H2
(nor, indeed, for any electrode potential), so they commented on
The ‘standard electrode their relative magnitudes, and compiled a form of ranking order.
O
potential’ EO,R is the
+ ,H should
These scientists then suggested that the value of EH 2 be defined, saying that at a temperature of 298 K, pumping the value of EO,R obtained hydrogen gas at a pressure of hydrogen of 1 atm through a solution at standard conditions. of protons at unit activity generates a value of EH+ ,H2 that is always zero. They called the half-cell ‘H2(g) (p = 1 atm)|H+ (a = 1)’ the
O
standard hydrogen electrode (SHE), and gave it the symbol EH+ ,H2 . A pressure of p = 1 atm is not the same as
The ‘ O ’ symbol indicates standard conditions. p O , but its use is a
O
Then, knowing EH+ ,H2 , it was relatively easy to determine values permissible deviation of electrode potentials for any other couple. With this methodology, within the SI scheme. they devised the ‘standard electrode potentials’ E O scale (often called the ‘E nought scale’, or the ‘hydrogen scale’).
Table 7.7 contains a few such values of E O , each of which was Negative values of determined with the same standard conditions as for the hydrogen E O indicate that the
O,R
couple, i.e. at T = 298 K, all activities being unity and p = 1 atm reduced form of the
(the pressure is not, therefore, p O ). couple will react with
O
Negative values of E O (such as ENa+ ,Na = −2.71 V) indicate protons to form hydrothat the reduced form of the couple will react with protons to form gen gas. hydrogen gas, as in Equation (7.35). The more negative the value of E O , the more potent the reducing power of the redox state, so
O
E O for the magnesium couple is −2.36 V, and EK+ ,K = −2.93. Positive values of E O
O,R
O
Zinc is a less powerful reducing agent, so EZn2+ ,Zn = −0.76 V, indicate that the oxiO and a feeble reducing agent like iron yields a value of EFe2+ ,Fe of only −0.44 V.
And positive values of E O indicate that the oxidized form of the redox couple will oxidize hydrogen gas to form protons, again

dized form of the redox couple will oxidize hydrogen gas to form protons. 324

ELECTROCHEMISTRY

Table 7.7 The electrode potential series (against the SHE). The electrode potential series is an arrangement of reduction systems in ascending order of their standard electrode potential E O
Couplea,b,c
Sm2+ + 2e− = Sm
Li+ + e− = Li
K+ + e− = K
Rb+ + e− = Rb
Cs+ + e− = Cs
Ra2+ + 2e− = Ra
Ba2+ + 2e− = Ba
Sr2+ + 2e− = Sr
Ca2+ + 2e− = Ca
Na+ + e− = Na
Ce3+ + 3e− = Ce
Mg2+ + 2e− = Mg
Be2+ + 2e− = Be
U3+ + 3e− = U
Al3+ + 3e− = Al
Ti2+ + 2e− = Ti
V2+ + 2e− = V
Mn2+ + 2e− = Mn
Cr2+ + 2e− = Cr
2H2 O + 2e− = H2 + 2OH−
Cd(OH)2 + 2e− = Cd + 2OH−
Zn2+ + 2e− = Zn
Cr3+ + 3e− = Cr
O2 + e− = O−
2
In3+ + e− = In2+
S + 2e− = S2−
In3+ + 2e− = In+
Fe2+ + 2e− = Fe
Cr3+ + e− = Cr2+
Cd2+ + 2e− = Cd
In2+ + e− = In+
Ti3+ + e− = Ti2+
PbSO4 + 2e− = Pb + SO2−
4
In3+ + 3e− = In
Co2+ + 2e− = Co
Ni2+ + 2e− = Ni
AgI + e− = Ag + I−
Sn2+ + 2e− = Sn
In+ + e− = In

E O /V
−3.12
−3.05
−2.93
−2.93
−2.92
−2.92
−2.91
−2.89
−2.87
−2.71
−2.48
−2.36
−1.85
−1.79
−1.66
−1.63
−1.19
−1.18
−0.91
−0.83
−0.81
−0.76
−0.74
−0.56
−0.49
−0.48
−0.44
−0.44
−0.41
−0.40
−0.40
−0.37
−0.36
−0.34
−0.28
−0.23
−0.15
−0.14
−0.14

Couplea,b,c
Pb2+ + 2e− = Pb
Fe3+ + 3e− = Fe
Ti4+ + e− = Ti3+
2 H + + 2 e− = H2 (by definition)
AgBr + e− = Ag + Br−
Sn4+ + 2e− = Sn2+
Cu2+ + e− = Cu+
Bi3+ + 3e− = Bi
AgCl + e− = Ag + Cl−
Hg2 Cl2 + 2e− = 2Hg + 2Cl−
Cu2+ + 2e− = Cu
O2 + 2H2 O + 4e− = 4OH−
NiOOH + H2 O + e− = Ni(OH)2 + OH−
Cu+ + e− = Cu
I3− + 2e− = 3I−
I2 + 2e− = 2I−
MnO− + 3e− = MnO2
4
Hg2 SO4 + 2e− = 2Hg + SO2−
4
Fe3+ + e− = Fe2+
AgF + e− = Ag + F
Hg2+ + 2e− = 2Hg
2
Ag+ + e− = Ag
2Hg2+ + 2e− = Hg2+
2
Pu4+ + e− = Pu3+
Br2 + 2e− = 2Br−
Pr2+ + 2e− = Pr
MnO2 + 4H+ + 2e− = Mn2+ + 2H2 O
O2 + 4H+ + 4e− = 2H2 O
Cl2 + 2e− = 2Cl−
Au3+ + 3e− = Au
Mn3+ + e− = Mn2+
MnO− + 8H+ + 5e− = Mn2+ + 4H2 O
4
Ce4+ + e− = Ce3+
Pb4+ + 2e− = Pb2+
Au+ + e− = Au
Co3+ + e− = Co2+
Ag2+ + e− = Ag+
S2 O2− + 2e− = 2SO2−
8
4
F2 + 2e− = 2F−

E O /V
−0.13
−0.04
0.00
0.000
0.07
0.15
0.16
0.20
0.2223
0.27
0.34
0.40
0.49
0.52
0.53
0.54
0.58
0.62
0.77
0.78
0.79
0.80
0.92
0.97
1.09
1.20
1.23
1.23
1.36
1.50
1.51
1.51
1.61
1.67
1.69
1.81
1.98
2.05
2.87

a
The more positive the value of E O , the more readily the half-reaction occurs in the direction left to right; the more negative the value, the more readily the reaction occurs in the direction right to left. b Elemental fluorine is the strongest oxidizing agent and Sm2+ is the weakest. Oxidizing power increases from
Sm2+ to F2 . c Samarium is the strongest reducing agent and F− is the weakest. Reducing power increases from F− to Sm.

HALF-CELLS AND THE NERNST EQUATION

325

O with the magnitude of E O indicating the oxidizing power: ECu2+ ,Cu = +0.34 V, but

O a powerful oxidizing agent such as bromine has a value of EBr2 ,Br− = +1.09 V.
In summary, sodium reacts with water and copper does not in consequence of their relative electrode potentials.

Why does a torch battery eventually ‘go flat’?
The Nernst equation
A new torch battery has a voltage of about 1.5 V, but the emf decreases with usage until it becomes too small to operate the torch for which we bought it. We say the battery has ‘gone flat’, and throw it away.
We need to realize from Faraday’s laws that chemicals within a battery are consumed every time the torch is switched on, and others are generated, causing the composition within the torch to change with use. Specifically, we alter the relative amounts of oxidized and reduced forms within each half-cell, causing the electrode potential to change.
The relationship between composition and electrode potential is given by the Nernst equation
Though it is relatively
EO,R

a(O)
RT
ln
= EO,R + nF a(R)
O

(7.41)

easy to formulate relations like the Nernst equation here for a cell, Equation (7.41) properly relates to a half-cell. O where EO,R is the standard electrode potential determined at s.t.p. and is a constant, EO,R is the electrode potential determined at non
s.t.p. conditions. R, T, n and F have their usual definitions.
The bracket on the right of Equation (7.41) describes the relative activities of oxidized and reduced forms of the redox couple within a half-cell. The battery goes flat because the ratio a(O) /a(R) alters with battery usage, so the value of EO,R changes until the emf is too low for the battery to be useful.

Worked Example 7.17 A silver electrode is immersed into a dilute solution of silver nitrate, [AgNO3 ] = 10−3 mol−3 . What is the electrode potential EAg+ ,Ag at 298 K? Take
O
EAg+ ,Ag = 0.799 V.
The Nernst equation, Equation (7.41), for the silver couple is
O
EAg+ ,Ag = EAg+ ,Ag +

a(Ag+ )
RT
ln
F
a(Ag)

For simplicity, we assume that the concentration and activity of silver nitrate are the same, i.e. a(Ag+ ) = 10−3 . We also assume that the silver is pure, so its activity is unity.

We use the approximation ‘concentration
= activity’ because the solution is very dilute.

326

ELECTROCHEMISTRY

The value of RT /F is
0.0257 V at 298 K.
Note how, as a consequence of the laws of arithmetic, we multiply the RT /F term with the logarithm term before adding the value of
O
EO,R .

Note how the difference between E and E O is normally quite small.

Inserting values into Equation (7.41):
EAg+ ,Ag = 0.799 V + 0.0257 V ln

0.001
1

so
EAg+ ,Ag = 0.799 V + (0.0257 V × −6.91) and EAg+ ,Ag = 0.799 V − 0.178 V
EAg+ ,Ag = 0.621 V

SAQ 7.16 A wire of pure copper is immersed into a solu= 0.34 V and ECu2+ ,Cu = tion of copper nitrate. If E O 2+
Cu

,Cu

0.24 V, what is the concentration of Cu2+ ? Assume that a(Cu2+ ) is the same as [Cu2+ ].

The Nernst equation cannot adequately describe the relationship between an electrode potential EO,R and the concentration c of the redox couple it represents, unless we substitute for the activity, saying from Equation (7.28), a = c × γ± .
The form of Equation (7.41) will remind us of the equation of a straight line, so a plot of EO,R as the observed variable (as ‘y’), against ln(a(O) ÷ a(R) ) (as ‘x’) should
O
be linear with a gradient of RT ÷ nF and with EO,R as the intercept on the y-axis.
O
Worked Example 7.18 Determine a value for the standard electrode potential EAg+ ,Ag with the data below. Assume that γ± = 1 throughout.

[AgNO3 ]/mol dm−3
EAg+ ,Ag /V

0.001
0.563

0.002
0.640

0.005
0.664

0.01
0.682

0.02
0.699

0.05
0.723

0.1
0.741

Figure 7.11 shows a Nernst graph drawn with the data in the table. The intercept of the graph is clearly 0.8 V.

Why does EAgCl,Ag change after immersing an SSCE in a solution of salt?
Further calculations with the Nernst equation
Care: In some books,
SSCE is taken to mean a sodium chloride saturated calomel electrode.

Take a rod of silver, and immerse it in a solution of potassium chloride. A thin layer of silver chloride forms on its surface when the rod is made positive, generating a redox couple of AgCl|Ag.
We have made a silver–silver chloride electrode (SSCE).
Now take this electrode together with a second redox couple (i.e. half-cell) of constant composition, and dip them together in a series

HALF-CELLS AND THE NERNST EQUATION intercept = E Ag +, Ag

327

0.8

EAg +, Ag / V

0.7

0.6

0.5
−10

−8

−6 ln([Ag −4
+] / mol

−2

0

dm−3)

Figure 7.11 Nernst graph of the electrode potential EAg+ ,Ag as ‘y’ against ln([Ag+ ]/mol dm−3 ) as ‘x’. A value of EAg+ ,Ag = 0.8 V is obtained as the intercept on the y-axis
O

of salt solutions, and measure the emf. The magnitude of the emf will depend on the concentration of the salt. The silver rod and its outer layer of silver chloride do not alter, so why does the emf change?
The electrode potential EAgCl,Ag relates to the following redox reaction:

AgCl(s) + e− − → Ag0 (s) + Cl− (aq)

(7.42)

This redox couple is more complicated than any we have encountered yet, so the
Nernst equation will appear to be a little more involved than those above:
O
EAgCl,Ag = EAgCl,Ag +

RT
F

ln

a(AgCl) a(Ag0 ) a(Cl− )

(7.43)

Silver chloride is the oxidized form, so we write it on top of the bracketed fraction, and silver metal is the reduced form, so we write it beneath. But we must also write a term for the chloride ion, because Cl− (aq) appears in the balanced reduction reaction in Equation (7.42).
If we immerse a silver electrode bearing layer of AgCl in a concentrated solution of table salt, then the activity a(Cl− ) will be high; if the solution of salt is dilute, then
EAgCl,Ag will change in the opposite direction, according to Equation (7.43).
SAQ 7.17 An SSCE electrode is immersed in a solution of [NaCl] =
O
0.1 mol dm−3 . What is the value of EAgCl,Ag ? Take EAgCl,Ag = 0.222 V. Take all γ± = 1. [Hint: the activities a(AgCl) and a(Ag0 ) are both unity because both are pure solids.]

328

ELECTROCHEMISTRY

Why ‘earth’ a plug?
Reference electrodes
An electrical plug has three connections (or ‘pins’): ‘live’, ‘neutral’ and ‘earth’. The earth pin is necessary for safety considerations. The potential of the earth pin is the same as that of the ground, so there is no potential difference if we stand on the ground and accidentally touch the earth pin in a plug or electrical appliance. We will not be electrocuted. Conversely, the potentials of the other pins are different from that of the earth – in fact, we sometimes cite their potentials with respect to the earth pin, effectively defining the potential of the ground as being zero.
The incorporation of an earth pin is not only desirable for safety, it also enables us to know the potential of the other pins, because we cite them with respect to the earth pin.
A reaction in an electrochemical cell comprises two half-cell reactions. Even when we want to focus on a single half-cell, we must construct a whole cell and determine its cell emf, which is defined as ‘E(positive electrode) − E(negative electrode) ’. Only when we know both the emf and the value of one of the two electrode potentials can we calculate the unknown electrode potential.
A device or instrument having a known, predetermined electrode potential is called a reference electrode. A reference electrode is
A reference electrode always necessary when working with a redox couple of unknown is a constant-potential
EO,R . A reference electrode acts in a similar manner to the earth device. We need such a reference to deterpin in a plug, allowing us to know the potential of any electrode mine an unknown with respect to it. And having defined the potential of the referelectrode potential. ence electrode – like saying the potential of earth is zero – we then know the potential of our second electrode.

Aside
At the heart of any reference electrode lies a redox couple of known composition: any passage of current through the reference electrode will change its composition (we argue this in terms of Faraday’s laws in Table 7.1). This explains why we must never allow a current to flow through a reference electrode, because a current will alter its potential.

The standard hydrogen electrode – the primary reference
We define the value of E(SHE) as zero at all temperatures. The internationally accepted primary reference is the standard hydrogen electrode (SHE). The potential of the SHE half-cell is defined as 0.000 V at all temperatures. We say the schematic for the half-cell is
Pt|H2 (a = 1)|H+ (a = 1, aq)|

HALF-CELLS AND THE NERNST EQUATION

The SHE is depicted in Figure 7.12, and shows the electrode immersed in a solution of hydrogen ions at unit activity (corresponding to 1.228 mol dm−3 HCl at 20 ◦ C). Pure hydrogen gas at a pressure of 1 atm is passed over the electrode. The electrode itself consists of platinum covered with a thin layer of ‘platinum black’, i.e. finely divided platinum, electrodeposited onto the platinum metal. This additional layer thereby catalyses the electrode reaction by promoting cleavage of the H–H bonds.
Table 7.8 lists the advantages and disadvantages of the SHE.

329

We employ hydrochloric acid of concentration 1.228 mol dm−3 at 20 ◦ C because the activity of H+ is less than its concentration,
i.e. γ± < 1.

Gaseous hydrogen
(at a pressure of 1 atm)

Aqueous acid with protons at unit activity

Platinized platinum electrode

Figure 7.12 Schematic depiction of the standard hydrogen electrode (SHE). The half-cell schematic is therefore Pt|H2 (a = 1)|H+ (a = 1)

Table 7.8 Advantages and disadvantages of using the standard hydrogen electrode (SHE)
Advantage of the SHE
The SHE is the international standard
Disadvantages of the SHE
Safety
Size
Cost
Accuracy
Precision

The SHE is intrinsically dangerous because H2 gas is involved
The SHE requires cumbersome apparatus, including a heavy cylinder of hydrogen
The SHE can be expensive because of using
H2 gas
With the SHE it is difficult to ensure that the activity of the protons is exactly unity
The SHE is prone to systematic errors, e.g. cyclic fluctuations in the H2 pressure

330

ELECTROCHEMISTRY

Measurement with the hydrogen electrode The SHE is the primary reference electrode, so other half-cell potentials are measured relative to its potential. In practice,
O
if we wish to determine the value of EMn+ ,M then we construct a cell of the type
Remember that the proton is always solvated to form a hydroxonium ion, so a(H3 O+ )
= a(H+ ) .
The words unit activity here mean that the activity of silver ions is one (so the system perceives the concentration to be
1 mol dm−3 ).
Just because a halfreaction appears in a table is no guarantee that it will actually work; such potentials are often calculated.
The SHE is sometimes erroneously called a normal hydrogen electrode (NHE).

Pt|H2 (a = 1)|H+ (a = 1)||Mn+ (a = 1)|M
Worked Example 7.19 We immerse a piece of silver metal into a solution of silver ions at unit activity and at s.t.p. The potential across the cell is 0.799 V when the SHE is the negative pole. What is the standard electrode potential E O of the Ag+ , Ag couple?
By definition emf = E(positive

electrode)

− E(negative

electrode)

Inserting values gives
O
0.799 V = EAg+ ,Ag − E(SHE)
O
0.799 V = EAg+ ,Ag

because E(SHE) = 0.
The value of EAg+ ,Ag in this example is the standard electrode potential because a(Ag+ ) = 1, and s.t.p. conditions apply. We say that
O
EAg+ ,Ag = EAg+ ,Ag = 0.799 V versus the SHE. Most of the values of

E O in Table 7.7 were obtained in a similar way, although some were calculated. We should be aware from Table 7.8 that the SHE is an ideal device, and the electrode potential will not be exactly 0 V with non-standard usage. Secondary reference electrodes
The SHE is experimentally inconvenient, so potentials are often measured and quoted with respect to reference electrodes other than the SHE. By far the most common reference is the saturated calomel electrode (SCE). We will usually make our choice of reference on the basis of experimental convenience.

Note that the ‘S’ of
‘SCE’ here does NOT mean ‘standard’, but
‘saturated’.

The SCE
By far the most common secondary reference electrode is the SCE:
Hg|Hg2 Cl2 |KCl(sat’d)|

HALF-CELLS AND THE NERNST EQUATION

331

Lead to voltmeter

Glass tube
Platinum contact
Saturated KCl solution
Paste of Hg and Hg2Cl2

Crystals of KCl
Porous sinter

Figure 7.13 Schematic representation of the saturated calomel electrode (SCE)

The potential of the SCE is 0.242 V at 298 K relative to the SHE.
At the ‘heart’ of the SCE is a paste of liquid mercury and mercurous chloride (Hg2 Cl2 ), which has the old-fashioned name
‘calomel’. Figure 7.13 depicts a simple representation of the SCE.
The half-cell reaction in the SCE is
Hg2 Cl2 + 2e− − → 2Cl− + 2Hg


(7.44)

So E(SCE) = EHg2 Cl2 ,Hg . From this redox reaction, the Nernst equation for the SCE is
O
EHg2 Cl2 ,Hg = EHg2 Cl2 ,Hg +

a(Hg2 Cl2 )
RT
ln
2
2
2F
a(Hg) a(Cl− )

Calomel is the oldfashioned name for mercurous chloride,
Hg2 Cl2 . Calomel was a vital commodity in the
Middle Ages because it yields elemental mercury (‘quick silver’) when roasted; the mercury was required by alchemists.

(7.45)
Oxidative currents

the square terms for mercury and the chloride ion are needed in reverse the reaction response to the stoichiometric numbers in Equation (7.44). Both in Equation (7.44). mercury and calomel are pure substances, so their activities are Cl− and Hg are conunity. If the activity of the chloride ion is maintained at a constant sumed and Hg2 Cl2 level, then E(SCE) will have a constant value, which explains why forms. The denominator of Equation (7.45) the couple forms the basis of a reference electrode.
Changing a(Cl− ) must alter EHg2 Cl2 ,Hg , since these two variables decreases, and E(SCE) are interconnected. In practice, we maintain the activity of the chlo- increases. ride ions by placing surplus KCl crystals at the foot of the tube.
The KCl solution is saturated – hence the ‘S’ in SCE. For this reason, we should avoid any SCE not showing a crust of crystals at its bottom, because its potential will be unknown. Also, currents must never be allowed to pass through an SCE, because charge will cause a redox change in E(SCE) .
Table 7.9 lists the advantages and disadvantages of the SCE reference. Despite these flaws, the SCE is the favourite secondary reference in most laboratories.

332

ELECTROCHEMISTRY

Table 7.9 The advantages and disadvantages of using the saturated calomel electrode (SCE)
Advantages of the SCE
Cost
Size
Safety

SCEs are easy to make, and hence they are cheap
SCEs can be made quite small (say, 2 cm long and 0.5 cm in diameter)
Unlike the SHE, the SCE is non-flammable

Disadvantages of the SCE
Contamination
Chloride ions can leach out through the SCE sinter
Temperature effects
The value of dE/dT is quite large at 0.7 mV K−1
Solvent
The SCE should not be used with non-aqueous solutions The silver–silver chloride electrode
We make the best films of AgCl by anodizing a silver wire in aqueous
KCl, not in HCl. The reasons for the differences in morphology are not clear.

Photolytic breakdown is a big problem when we light the laboratory with fluorescent strips.

The silver–silver chloride electrode (SSCE) is another secondary reference electrode. A schematic of its half-cell is:
Ag|AgCl|KCl(aq, sat’d)
O
The value of EAgCl,Ag = 0.222 V.
The AgCl layer has a pale beige colour immediately it is made, but soon afterwards it assumes a pale mauve and then a dark purple aspect. The colour changes reflect chemical changes within the film, caused as a result of photolytic breakdown:

AgCl + hν + electron donor − → Ag0 + Cl−


(7.46)

The purple colour is caused by colloidal silver, formed in a similar manner to the image on a black-and-white photograph after exposure to light. For this reason, an SSCE should be remade fairly frequently. (We introduce colloids in Chapter 10.)
Table 7.10 lists the advantages and disadvantages of SSCEs.

Table 7.10 The advantages and disadvantages of using the silver–silver chloride electrode (SSCE)
Advantages of the SSCE
Cost
Stability
Size
Disadvantages of the SSCE
Photochemical stability
Contamination

The SSCE is easy and extremely cheap to make
The SSCE has the smallest temperature voltage coefficient of any common reference electrode
An SSCE can be as large or small as desired; it can even be microscopic if the silver is thin enough
The layer of AgCl must be remade often
The solid AgCl loses Cl− ions during photolytic breakdown

CONCENTRATION CELLS

7.5

333

Concentration cells
Why does steel rust fast while iron is more passive?
Concentration cells
Steel is an impure form of iron, the most common contaminants being carbon (from the coke that fuels the smelting process) and sulphur from the iron oxide ore.
Pure iron is relatively reactive, so, given time and suitable conditions of water and oxygen, it forms a layer of red hydrated iron oxide (‘rust’):

4Fe(s) + 3O2 + nH2 O − → 2Fe2 O3 · (H2 O)n(s)

Many iron ores also contain iron sulphide, which is commonly called fool’s gold.

(7.47)

By contrast, steel is considerably more reactive, and rusts faster and to a greater extent. The mole fraction x of Fe in pure iron is unity, so the activity of the metallic iron is also unity. The mole fraction x of iron in steel We define a concentration cell as a cell in will be less than unity because it is impure. The carbon is evenly which the two half-cells distributed throughout the steel, so its mole fraction x(C) is con- are identical except for stant, itself ensuring that the activity is also constant. Conversely, their relative concenthe sulphur in steel is not evenly distributed, but resides in small trations.
(microscopic) ‘pockets’. In consequence, the mole fraction of the iron host x(Fe) fluctuates, with x being higher where the steel is more pure, and lower in those pockets having a high sulphur content. To summarize, there are differences in the activity of the iron, so a concentration cell forms.
The emf of a concentration cell (in this case, on the surface of the steel where the rusting reaction actually occurs) is given by
The electrolyte on the
RT
emf = nF a2 ln a1

(7.48)

Notice how this emf has no standard electrode potential E O terms
(unlike the Nernst equation from which it derives; see Justification
Box 7.2).
A voltage forms between regions of higher irons activity a1 and regions of lower iron activity a2 (i.e. between regions of high purity and low iron purity); see Figure 7.14. We can write a schematic for a microscopic portion of the iron surface as:
Fe(a2 ), S,C|O2 , H2 O|Fe(a1 ), S,C

surface of the iron comprises water containing dissolved oxygen (e.g. rain water).

The commas in this schematic indicate that the carbon and sulphur impurities reside within the same phase as the iron. 334

ELECTROCHEMISTRY

V
Humid air
Steel

Sulphur cluster: area of steel having a low mole fraction of iron

steel having a relatively high mole fraction of iron

Figure 7.14 Concentration cells: a voltage forms between regions of higher iron activity a1 and regions of lower iron activity a2 (i.e. between regions of high purity and low iron purity). The reaction at the positive ‘anode’ is 4Fe + 3O2 → 2Fe2 O3 , and the reaction at the negative ‘cathode’ is S + 2e− → S2−

There is no salt bridge or any other means of stopping current flow in the microscopic ‘circuit’ on the iron surface, so electrochemical reduction occurs at the righthand side of the cell, and oxidation occurs at the left: at the LHS, the oxidation reaction is formation of rust (Equation (7.47)); at the RHS, the reduction reaction is usually formation of sulphide, via S +
2e− → S2− .
We can draw several important conclusions from the example of rusting steel. Firstly, if the impurities of carbon and sulphur are evenly distributed throughout the steel then, whatever their concentrations, the extent of rusting will be less than if the impurities cluster, because the emf of a concentration cell is zero when the ratio of activities is unity. Secondly, it is worth emphasizing that while oxide formation would have occurred on the surface of the iron whether it was pure or not, the steel containing impurities rusts faster as a consequence of the emf, and also more extensively than pure iron alone. Thermodynamics of concentration cells
If the two half-cells were shorted then reduction would occur at the right-hand halfcell, Cu2+ (aq) + 2e− →
Cu(s) , and oxidation would proceed at the left-hand side, Cu(s) →
Cu2+ (aq) + 2e− .

A concentration cell contains the same electroactive material in both half-cells, but in different concentration (strictly, with different activities). The emf forms in response to differences in chemical potential µ between the two half-cells. Note that such a concentration cell does not usually involve different electrode reactions
(other than, of course, that shorting causes one half-cell to undergo reduction while the other undergoes oxidation).
Worked Example 7.20 Consider the simple cell Cu|Cu2+ (a =
0.002)||Cu2+ (a = 0.02)|Cu. What is its emf ?

CONCENTRATION CELLS

335

Inserting values into Equation (7.48): emf =

0.0257 V
2

0.02
0.002

ln

emf = 0.0129 V ln(10) emf = 0.0129 V × 2.303 emf = 29.7 mV so the emf for this two-electron concentration cell is about 30 mV. For an analogous one-electron concentration cell, the emf would be 59 mV.
SAQ 7.18 The concentration cell Zn|[Zn2+ ](c = 0.0112 mol dm−3 )|[Zn2+ ]
(c = 0.2 mol dm−3 )|Zn is made. Calculate its emf, assuming all activity coefficients are unity.

Justification Box 7.2
Let the redox couple in the two half-cells be O + ne− = R. An expression for the emf of the cell is emf = E(RHS) E(LHS)
The Nernst equation for the O,R couple on the RHS of the cell is:
O
EO,R = EO,R +

RT nF a(O)RHS a(R)RHS ln

and the Nernst equation for the same O,R couple on the LHS of the cell is:
O
EO,R = EO,R +

a(O)LHS
RT
ln nF a(R)LHS

Substituting for the two electrode potentials yields an emf of the cell of emf = EO,R +
O

a(O)RHS
RT
ln nF a(R)RHS

− EO,R −
O

a(O)LHS
RT
ln nF a(R)LHS

It will be seen straightaway that the two E O terms cancel to leave emf =

a(O)RHS
RT
ln nF a(R)RHS



a(O)LHS
RT
ln nF a(R)LHS

A good example of a concentration cell would be the iron system in the worked example above, in which a(R) = a(Cu) = 1; accordingly, for simplicity here, we will assume that the reduced form of the couple is a pure solid.

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ELECTROCHEMISTRY

The emf of the concentration cell, therefore, becomes emf =

RT
RT
ln[a(O)RHS ] − ln[a(O)LHS ] nF nF

which, through the laws of logarithms, simplifies readily to yield Equation (7.48).
If we assume that the activity coefficients in the left- and right-hand half-cells are the same (which would certainly be a very reasonable assumption if a swamping electrolyte was also in solution), then the activity coefficients would cancel to yield emf =

[O]RHS
RT
ln nF [O]LHS

How do pH electrodes work?
The pH–glass electrode
A pH electrode is sometimes also called a ‘membrane’ electrode. Figure 7.15 shows how its structure consists of a glass tube culminating with a bulb of glass. This bulb is filled with a solution of chloride ions, buffered to about pH 7. A slim silver wire runs down the tube centre and is immersed in the chloride solution. It bears a thin layer of silver chloride, so the solution in the bulb is saturated with AgCl.
The bulb is usually fabricated with common soda glass, i.e. glass containing a high concentration of sodium ions. Finally, a small reference electrode, such as an SCE, is positioned beside the bulb. For this reason, the pH electrode ought properly to be called a pH combination electrode, because it is combined with a reference electrode.
If the pH electrode does not have an SCE, it is termed a glass electrode (GE). The operation of a glass electrode is identical to that of a combination pH electrode, except that an external reference electrode is
Empirical means found required. from experiment, rather
To determine a pH with a pH electrode, the bulb is fully immerthan from theory. sed in a solution of unknown acidity. The electrode has fast response because a potential develops rapidly across the layer of glass

Thin-walled glass bulb
Buffer solution
(containing chloride ions)

Silver wire
Deposit of silver chloride

Figure 7.15

Schematic representation of a pH electrode (also called a ‘glass electrode’)

CONCENTRATION CELLS

337

between the inner chloride solution and the outer, unknown acid. Empirically, we find the best response when the glass is extremely thin: the optimum seems to be 50 µm or so (50 µm = 0.05 mm = 50 × 10−6 m). Unfortunately, such thin glass is particularly fragile. The glass is not so thin that it is porous, so we do not need to worry about junction potentials Ej (see Section 7.6). The non-porous nature of the glass does imply, however, that the cell resistance is extremely large, so the circuitry of a pH meter has to operate with minute currents.
The magnitude of the potential developing across the glass depends on the difference between the concentration of acid inside A pH meter is essenthe bulb (which we know) and the concentration of the acid outside tially a precalibrated the bulb (the analyte, whose pH is to be determined). In fact, the voltmeter. emf generated across the glass depends in a linear fashion on the pH of the analyte solution provided that the internal pH does not alter, which is why we buffer it. This pH dependence shows why a pH meter is really just a pre-calibrated voltmeter, which converts the measured emf into a pH. It uses the following formula: emf = K +

2.303RT pH F

(7.49)

SAQ 7.19 An emf of 0.2532 V was obtained by immersing a glass electrode in a solution of pH 4 at 25 ◦ C. Taking E(SCE) = 0.242 V, calculate the
‘electrode constant’ K.
SAQ 7.20 Following from SAQ 7.19, the same electrode was then immersed in a solution of anilinium hydrochloride of pH = 2.3. What will be the new emf ?

In practice, we do not know the electrode constant of a pH electrode.

Electrode ‘slope’
We can readily calculate from Equation (7.49) that the emf of a pH electrode should change by 59 mV per pH unit. It is common to see this stated as ‘the electrode has a slope of 59 mV per decade’. A moment’s pause shows how this is a simple statement of the obvious: a graph of emf (as ‘y’) against [H+ ] (as ‘x’) will have a gradient of 59 mV (hence ‘slope’). The words ‘per decade’ point to the way that each pH unit represents a concentration change of 10 times, so a pH of 3 means that [H+ ] = 10−3 mol dm−3 , a pH of 4 means [H+ ] = 10−4 mol dm−3 and a pH of
5 means [H+ ] = 10−5 mol dm−3 , and so on. If the glass electrode does have a slope of 59 mV, its response is said to be Nernstian, i.e. it obeys the Nernst equation. The discussion of pH in Chapter 6 makes this same point in terms of Figure 6.1.
Table 7.11 lists the principle advantages and disadvantages encountered with the pH electrode.
SAQ 7.21 Effectively, it says above: ‘From this equation, it can be readily calculated that the emf changes by 59 mV per pH unit’. Starting with the
Nernst equation (Equation (7.41)), show this statement to be true.

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ELECTROCHEMISTRY

Table 7.11 Advantages and disadvantages of the pH electrode
Advantages
1.
2.
3.
4.
5.
6.

If recently calibrated, the GE and pH electrodes give an accurate response
The response is rapid (possibly millisecond)
The electrodes are relatively cheap
Junction potentials are absent or minimal, depending on the choice of reference electrode
The electrode draws a minimal current
The glass is chemically robust, so the GE can be used in oxidizing or reducing conditions; and the internal acid solution cannot contaminate the analyte
7. The pH electrode has a very high selectivity – perhaps as high as 105 : 1 at room temperature, so only one foreign ion is detected per 100 000 protons (although see disadvantage 6 below). The selectivity does decrease a lot above ca 35 ◦ C
Disadvantages
Both the glass and pH electrodes alike have many disadvantages
1. To some extent, the constant K is a function of the area of glass in contact with the acid analyte. For this reason, no two glass electrodes will have the same value of K
2. Also, for the same reason, K contains contributions from the strains and stresses experienced at the glass.
3. (Following from 2): the electrode should be recalibrated often
4. In fact, the value of K may itself be slightly pH dependent, since the strains and stresses themselves depend on the amount of charge incorporated into the surfaces of the glass
5. The glass is very fragile and, if possible, should not be rested against the hard walls or floor of a beaker or container
6. Finally, the measured emf contains a response from ions other than the proton. Of these other ions, the only one that is commonly present is sodium. This error is magnified at very high pH (>11) when very few protons are in solution, and is known as the
‘alkaline error’

Justification Box 7.3
At heart, the pH electrode operates as a simple concentration cell. Consider the schematic
H+ (a2 )||H+ (a1 ), then the Nernst equation can be written as Equation (7.48): emf =

RT a2 ln
F
a1

which, if written in terms of logarithms in base 10, becomes emf =

2.303RT
F

log10

a2 a1 (7.50)

Subsequent splitting of the logarithm terms gives emf =

2.303RT
F

log10 a2 −

2.303RT
F

log10 a1

(7.51)

TRANSPORT PHENOMENA

If we say that a1 is the analyte of known concentration (i.e. on the inside of the bulb), then the last term in the equation is a constant. If we call the term associated with a2 ‘K’, then we obtain emf = K +

2.303RT log10 a2
F

If a2 relates to the acidic solution of unknown concentration then we can substitute for ‘log10 a2 ’, by saying that pH = − log10 [H+ ], so: emf = K +

2.303RT
× −pH
F

(7.52)

339

Care: we have assumed here that the activities and concentrations of the solvated protons are the same.

This derivation is based on the Nernst equation written in terms of ionic activities, but pH is usually discussed in terms of concentration.

which is the same as Equation (7.49)

7.6

Transport phenomena
How do nerve cells work?
Ionic transport across membranes
The brain relays information around the body by means of nerves, allowing us to register pain, to think, or to instruct the legs to walk and hands to grip. Although the way nerves operate is far from straightforward, it is nevertheless clear that the nerve pathways conduct charge around the body, with the charged particles (electrons and ions) acting as the brain’s principal messengers between the brain and body.
The brain does not send a continuous current through the nerve, but short ‘spurts’.
We call them impulses, which transfer between nerve fibres within the synapses of cells (see Figure 7.16). The cell floats within an ionic solution called plasma. The membrane separating the synapse from the solution with which the nerve fibre is in contact surrounding the cell is the axon, and is essential to the nerve’s operation.
The charge on the inside of a cell is negative with respect to the surrounding solution. A potential difference of about −70 mV No potential difference forms across the axon (cell membrane) when the cell is ‘at rest’, i.e. forms along the membefore passing an impulse – we sometimes call it a rest potential, brane surface, only which is caused ultimately by differences in concentration either across it. side of the axon (membrane).
Movement of charge across the membrane causes the potential to change. A huge difference in concentration is seen in composition between the inside of the axon and the remainder of the nerve structure. For example, consider the compositional

340

ELECTROCHEMISTRY
Cell membrane (‘axon’) to separate the cell interior and exterior

Cell exterior

Cell interior

Ions leaving cell interior

Ions entering cell interior from exterior

Figure 7.16 Schematic diagram showing a portion of a cell, the membrane (‘axon’) and the way ions diffuse across the axon
Table 7.12

Concentrations of ions inside and outside nerve components
[Na+ ]/mol dm−3

Inside the axon
Outside the axon

[K+ ]/mol dm−3

[Cl− ]/mol dm−3

0.05
0.46

0.40
0.01

0.04–0.1
0.054

Source: J. Koryta, Ions, Electrodes and Membranes, Wiley, Chichester, 1991, p. 172.

differences in Table 7.12. The data in Table 7.12 refer to the nerves of a squid (a member of the cephalopod family) data for other species show a similar trend, with massive differences in ionic concentrations between the inside and outside of the axon. These differences, together with the exact extent to which the axon membrane is selectively permeable to ions, determines the magnitude of the potential at the cell surface.
The membrane encapsulating the axon is semi-permeable, thereby allowing the transfer of ionic material into and out from the axon. Since the cell encapsulates fluid and also floats in a fluid, we say the membrane represents a ‘liquid junction’. A potential
Some texts give the name ‘diffusion potenforms across the membrane in response to this movement of ions tial’ to Ej . across the membrane, which we call a ‘junction potential’ Ej . If left unchecked, ionic movement across the membrane would occur until mixing was intimate and the two solutions were identical.
For a nerve to transmit a ‘message’ along a nerve fibre, ions traverse the axons and transiently changing the sign of the potential across the membrane, as represented schematically in Figure 7.17. We call this new voltage an action potential, to differentiate it from the rest potential. To effect this change in potential, potassium cations
In this context, permeable indicates that ions or molecules can pass through the membrane. The mode of movement is probably diffusion or migration.

TRANSPORT PHENOMENA

341

Potential/V

+

Time t / ms
Action
potential

Rest potential



Onset of impulse

Figure 7.17 The potential across the axon–cell membrane changes in response to a stimulus, causing the potential to increase from its rest potential to its action potential

move from inside the axon concurrently with sodium ions moving in from outside. With a smaller difference in composition either side of the membrane, the junction potential decreases.
A nerve consists of an immense chain of these axons. Impulses
‘conduct’ along their length as each in turn registers an action potential, with the net result that messages transmit to and from the brain.

To achieve this otherwise difficult process, chemical ‘triggers’ promote the transfer of ions. Liquid junction potentials
A liquid junction potential Ej forms when the two half-cells of a cell contain different electrolyte solutions. The magnitude of Ej depends on the concentrations (strictly, the activities) of the constituent ions in the cell, the charges of each moving ion, and on the relative rates of ionic movement across the membrane. We record a constant value of Ej because equilibrium forms within a few milliseconds of the two half-cells adjoining across the membrane.
Liquid junction potentials are rarely large, so a value of Ej as large as 0.1 V should be regarded as exceptional. Nevertheless, In most texts, the liquid junction potentials of 30 mV are common and a major cause of junction potential is experimental error, in part because they are difficult to quantify, given the symbol Ej . In some books it is written but also because they can be quite irreproducible.
We have already encountered expressions that describe the emf as E(lj) or even E(ljp) . of a cell in terms of the potentials of its constituent half-cells, e.g.
Equation (7.23). When a junction potential is also involved – and it usually is – the emf increases according to emf = E(positive

half−cell)

− E(negative

half−cell)

+ Ej

(7.53)

which explains why we occasionally describe Ej as ‘an additional source of potential’.

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ELECTROCHEMISTRY

While it is easy to measure a value of emf, we do not know the magnitude of Ej .
SAQ 7.21 illustrates why we need to minimize Ej .
SAQ 7.22 The emf of the cell SHE |Ag+ |Ag, is 0.621 V. Use the Nernst equation to show that a(Ag+ ) = 10−3 if Ej = 0 V, but only 4.6 × 10−4 if
O
Ej = 20 mV. EAg+ ,Ag = 0.799 V. [Hint: to compensate for Ej in the second calculation, say that only 0.601 V of the emf derives from the Ag+ |Ag half-cell, i.e. EAg+ ,Ag = 0.601 V.]

What is a ‘salt bridge’?
Minimizing junction potentials
In normal electrochemical usage, the best defence against a junction potential Ej is a salt bridge. In practice, the salt bridge is typically a thin strip of filter paper soaked in electrolyte, or a U-tube containing an electrolyte. The electrolyte is usually KCl or KNO3 in relatively high concentration; the U-tube contains the salt, perhaps dissolved in a gelling agent such as agar or gelatine.
We connect the two half-cells by dipping either end of the salt bridge in a half-cell solution. A typical cell might be written in schematic form as:

It’s called a bridge because it connects the two half-cells, and salt because we saturate it with a strong ionic electrolyte. Zn(s) |Zn2+ (aq) |S|Cu2+ (aq) |Cu(s)
We write the salt bridge as ‘|S|’, where the S is the electrolyte within the salt bridge.
But how does the salt bridge minimize Ej ? We recognize first how the electrolyte in the bridge is viscous and gel-like, so ionic motion through the bridge is slow.
Secondly, the ionic diffusional processes of interest involve only the two ends of the salt bridge. Thirdly, and more importantly, the concentration of the salt in the bridge should greatly exceed the concentrations of electrolyte within either half-cell (exceed, if possible, by a factor of between 10–100 times).
The experimental use of a salt bridge is depicted in Figure 7.18. The extent of diffusion from the bridge, as represented by the large arrows in the diagram, is seen to be much greater than diffusion into the bridge, as represented by the smaller of the two arrows. A liquid junction forms at both ends of the bridge, each generating its own value of Ej . If the electrolyte in the bridge is concentrated, then the diffusion of ions moving from the bridge will dominate both of these two Ej . Furthermore, these Ej will be almost equal and opposite in magnitude, causing them to cancel each other out.
Table 7.13 shows how the concentration of the salt in the bridge has a large effect on Ej : it is seen that we achieve a lower value of Ej when the bridge is constructed with larger concentrations of salt. A junction potential Ej of as little as 1–2 mV can be achieved with a salt bridge if the electrolyte is concentrated.

BATTERIES

343

Detail
Salt bridge

c1

c2

Electrolyte from solution

Electrolyte from bridge

Figure 7.18 The two half-cells in a cell are joined with a salt bridge. Inset: more ions leave the bridge ends than enter it; the relative sizes of the arrows indicate the relative extents of diffusion
Table 7.13 Values of junction potential in aqueous cells as a function of the concentration of inert KCl within a salt bridge [KCl]/c O

Ej /mV

0.1
1.0
2.5
4.2 (sat’d)

27
8.4
3.4 c2 > c1

Figure 8.6 is a similar graph to Figure 8.5, showing the concentration profiles of sulphuric acid as a function of its initial concentration. The gradients of each trace are different, with the more concentrated solutions generating the steepest traces.
In fact, we have merely rediscovered the concept of rate, because the gradient of the concentration profile is reaction rate, being the rate at which a compound or chemical reacts as a function of time. We say the rate of reaction is the gradient, after
Equation (8.22).
Graphs such as those in Figures 8.5 and 8.6 are an ideal means
We obtain the rate of determining the rates of reaction. To obtain the rate, we plot of reaction as the the concentration of a reactant or product as a function of time, gradient of a concenand measure the slope. (Strictly, since the slopes are negative for tration profile. reactants, so the rate is ‘slope × −1’.)

Aside
Because product forms at the expense of reactant, the magnitudes of the rates of forming product and consuming reactant are the same. Although the magnitudes of the rates are the same, their signs are not: the rate of forming product is positive because the concentration increases with time.
For example, consider the reaction, aA + bB = cC + dD: d[reactant] d[Product]
=−
dt dt For the individual chemical components, we say, rate =

1 d[A]
1 d[B]
×
rate = −1 × × a dt b dt
1 d[C]
1
d[D] products rate = × rate = × c dt d dt
In each case, the minus sign indicates a decreasing concentration with time. reactants rate = −1 ×

367

QUALITATIVE DISCUSSION OF CONCENTRATION CHANGES

The rate of loss of reactant is negative because the concentration decreases with time The gradient at the start of the reaction is called the initial rate. Analysing the initial rates method is an extremely powerful way of determining the order of a reaction.
Worked Example 8.6 The following kinetic data were obtained for the reaction between nitric oxide and hydrogen at 700 ◦ C. Determine: (1) the order of the reaction of the reaction at this temperature; (2) the rate constant of reaction.
Experiment A
Initial concentration of
NO/mol dm−3
Initial concentration of
H2 /mol dm−3
Initial rate/mol dm−3 s−1

Experiment B

Experiment C

0.025

0.025

0.0125

0.01

0.005

0.01

2.4 × 10−6

1.2 × 10−6

0.6 × 10−6

The rates listed in the table were obtained as the gradients of graphs like those in
Figure 8.6.
Answer Strategy
(1) To determine the order of reaction. It is always good research strategy to change only one variable at a time. That way, the measured response (if any) can be attributed unambiguously to the change in that variable. And the variable of choice in this example will be one or other of the two initial concentrations.
The rate equation for the reaction will have the following form: rate = k[NO]x [H2 ]y

(8.23)

We determine values for the exponents x and y varying [NO] and [H2 ]. First, consider experiments A and B. In going from experiment B to experiment A, the concentration
[NO] remains constant but we double [H2 ], as a consequence of which the rate doubles. There is, therefore, a linear relationship between [H2 ] and rate, so the value of y is ‘1’.
Second, consider experiments A and C. In going from C to A, we double the concentration [NO] and the rate increases by a factor of four. Accordingly, the value of x cannot be ‘1’ because the increase in rate is not linear. In fact, as 22 = 4, we see how the exponent x has a value of ‘2’.
The reaction is first order in H2 , second order in NO and, therefore, third order overall.
Inserting values into Equation (8.23), we say, rate = k[NO]2 [H2 ].
(2) To determine the rate constant of reaction. We know the rate and concentrations for several sets of experimental conditions, so we rearrange Equation (8.23) to make k the subject and insert the concentrations. k= rate
[NO]2 [H2 ]

368

CHEMICAL KINETICS

The question cites data from three separate experiments, each of which will give the same answer. We will insert data from ‘experiment A’: k= so

2.4 × 10−6 mol dm−3 s−1
[0.25 mol dm−3 ]2 [0.01 mol dm−3 ] k = 3.84 × 103 dm6 mol−2 s−1

SAQ 8.9 Iodide reacts with thiosulphate to form elemental iodine. If the reaction solution contains a tiny amount of starch solution, then this I2 is seen by eye as a blue complex. The data below were obtained at 298 K.
Determine the order of reaction, and hence its rate constant k.
Experiment A
Initial concentration of
I− /mol dm−3
Initial concentration of
S2 O8 2− /mol dm−3
Initial rate/mol dm−3 s−1

8.3

Experiment B

Experiment C

0.1

0.1

0.2

0.05

0.025

0.05

1.5 × 10−4

7.5 × 10−5

3.0 × 10−4

Quantitative concentration changes: integrated rate equations
Why do some photographs develop so slowly?
Extent of reaction and integrated rate equations
A common problem for amateur photographers who develop their own photographs is gauging the speed necessary for development. When the solution of thiosulphate is first prepared, the photographs develop very fast, but this speed decreases quite rapidly as the solution ‘ages’, i.e. the concentration of S2 O8 2− decreases because the thiosulphate is consumed. So most inexperienced photographers have, at some time, ruined a film by developing with an ‘old’ solution: they wait what seems like forever without realizing that the concentration of thiosulphate is simply too low, meaning that development will never occur. They ruin the partially processed film by illuminating it after removal from the developing bath. These amateurs need to answer the question, ‘How much thiosulphate remains in solution as a function of time?’ We need a mathematical equation to relate concentrations and time.

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

369

The concentrations of each reactant and product will vary during the course of a chemical reaction. The so-called integrated rate equation relates the amounts of reactant remaining in solution during a reaction with the time elapsing since the reaction started. The integrated rate equation has a different form according to the order of reaction.
Let us start by considering a first-order reaction. Because the reactant concentration depends on time t, we write such concentrations with a subscript, as [A]t . The initial reactant concentration (i.e. at time t = 0) is then written as [A]0 . The constant of proportionality in these equations will be the now-familiar rate constant k1 (where the subscripted ‘1’ indicates the order).
The relationship between the two concentrations [A]0 , [A]t and t is given by ln [A]0
[A]t

= k1 t

(8.24)

Equation (8.24) is the integrated first-order rate equation. Being a logarithm, the left-hand side of Equation (8.24) is dimensionless, so the right-hand side must also be dimensionless. Accordingly, the rate constant k will have the units of s−1 when the time is expressed in terms of the SI unit of time, the second.

The first-order rate constant k will have the units of s−1 .

Worked Example 8.7 Methyl ethanoate is hydrolysed when dissolved in excess hydrochloric acid at 298 K. The ester’s concentration was 0.01 mol dm−3 at the start of the reaction, but 8.09 × 10−2 after 21 min. What is the value of the first-order rate constant k1 ?
Answer Strategy. (1) we convert the time into SI units of seconds; (2) we insert values into Equation (8.24).

(1) Convert the time to SI units: 21 min = 21 min × 60 s min−1 = 1260 s
(2) Next, inserting values into Equation (8.24)
0.01 mol dm−3 ln 0.008 09 mol dm−3

= 1260 s × k1

so ln(1.236) = k1 × 1260 s
Taking the logarithm yields
0.211 = k1 × 1260 s

Notice how the units of concentration will cancel here.

370

CHEMICAL KINETICS

and rearranging to make k1 the subject gives k1 =

0.211
= 1.68 × 10−4 s−1
1260 s

This value of k1 is relatively small, indicating that the reaction is rather slow.
SAQ 8.10 (Continuing with the same chemical example): what is the concentration of the methyl ethanoate after a time of 30 min? Keep the same value of k1 – it’s a constant.

We note, when looking at the form of Equation (8.24), how the bracket on the left-hand side contains a ratio of concentrations.
This ratio implies that we do not need to know the actual concentrations of the reagent [A]0 when the reaction starts and [A]t after a time t has elapsed since the reaction commenced; all we need to know is the fractional decrease in concentration. Incidentally, this aspect of the equation also explains why we could perform the calculation in terms of a percentage (i.e. a form of ratio) rather than a ‘proper’ concentration.
In fact, because the integrated first-order rate equation (Equation (8.24)) is written in terms of a ratio of concentrations, we do not need actual concentrations in moles per litre, but can employ any physicochemical parameter that is proportional to concentration. Obvious parameters include conductance, optical absorbance, the angle through which a beam of plane-polarized light is rotated (polarimetry), titre from a titration and even mass, e.g. if a gas is evolved.

We can calculate a rate constant k without knowing an absolute value for [A]0 by following the fractional changes in the time-dependent concentration [A]t .

Worked Example 8.8 Consider the simple reaction ‘A → product’. After 3 min, 20 per cent of A has been consumed when the reaction occurs at 298 K. What is the rate constant of reaction k1 ?
If we know the amount of reactant consumed, then we will need to calculate how much remains. Remember that we should always cite a rate constant at the temperature of measurement, because k itself depends on temperature.

We start by inserting values into the integrated Equation (8.24), noting that if 20 per cent has been consumed then 80 per cent remains, so: ln 100% of [A]0
80% of [A]0

= k1 × (3 × 60 s)

The logarithmic term on the left-hand side is ln(1.25) = 0.223, so k1 =

0.223
180 s

where the term ‘180 s’ comes from the seconds within the 3 min of observation time. We see that k1 = 1.24 × 10−3 s−1 at 298 K.
SAQ 8.11 If the molecule A reacts by a first-order mechanism such that 15% is consumed after 1276 s, what is the rate constant k?

371

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

Justification Box 8.1
Integrated rate equations for a first-order reaction
The rate law of a first-order reaction has the form
‘rate = k1 [A]’. And, by ‘rate’ we mean the rate of change of the concentration of reactant A, so rate = −

d[A]
= k1 [A] dt The minus sign in
Equation (8.25) is essential to show that the concentration of A decreases with time.

(8.25)

We will start at t = 0 with a concentration [A]0 , with the concentration decreasing with time t as [A]t . The inclusion of a minus sign is crucial, and shows that species A is a reactant and thus the amount of it decreases with time.
Separating the variables (i.e. rearranging the equation) and indicating the limits, we obtain


[A]t
[A]0

1 d[A] = k1
[A]t

t

dt
0

Note how we can place the rate constant outside the integral, because it does not change with time. Integration then yields
− [ln[A]][A]t0 = k1 [t]t0
[A]

And, after inserting the limits, we obtain
− (ln[A]t − ln[A]0 ) = k1 t
Using the laws of logarithms, the equation may be tidied further to yield Equation (8.24): ln [A]0
[A]t

= k1 t

which is the integrated first-order rate equation.
Note that if a multiple-step reaction is occurring, then this equation relates only to the case where the slowest (i.e. rate-limiting) step is kinetically first order. We will return to this idea when we consider pseudo reactions in Section 8.4.

Graphical forms of the rate equations
Similar to the integrated first-order rate equation is the linear first-order rate equation: ln[A]t = −k1 t + x m y ln[A]0 c (8.26)

372

CHEMICAL KINETICS

which we recognize as having the form of the equation for a straight line: plotting ln[A]t (as ‘y’) against time (as ‘x’) will be linear for reactions that are first order.
The rate constant k1 is obtained from such a first-order rate graph as (−1 × gradient) if the time axis is given with units of seconds.
Accordingly, the units of the first-order rate constant are s−1 .

We obtain the firstorder rate constant k by drawing a graph with the integrated first-order rate equation, and multiplying its slope by −1.

Worked Example 8.9 Consider the following reaction: hydrogen peroxide decomposes in the presence of excess cerous ion CeIII (which reacts to form ceric ion CeIV ) according to a first-order rate law. The following data were obtained at 298 K:
Time t/s
[H2 O2 ]t /mol dm−3

4
4.84

6
3.76

8
3.20

10
2.60

Time t/s
[H2 O2 ]t /mol dm−3

This reaction appears to be first order because the cerium ions are in excess, so the concentration does not really change. We look at pseudo-order reactions on p. 387.

2
6.23
12
2.16

14
1.85

16
1.49

18
1.27

20
1.01

Figure 8.7 shows the way the concentration of hydrogen peroxide decreases with time. The trace is clearly curved, and Figure 8.8 shows a graph constructed with the linear form of the first-order integrated rate equation, Equation (8.26). This latter graph is clearly linear. The rate constant is obtained from the figure as (−1 × ‘gradient’), so k = 0.11 s−1 at 298 K.
10

[H2O2]t / mol dm−3

8
6
4
2
0
0

Figure 8.7

5

10
Time t / ms

15

20

Plot of [H2 O2 ]t against time. Notice the pronounced plot curvature

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

373

ln([H2O2]t / mol dm−3)

2

1.5

1

0.5

0
0

5

10
Time t / s

15

20

Figure 8.8 Graph constructed by drawing ln[H2 O2 ]t (as ‘y’) against time t (as ‘x’), i.e. with the axes of a linear first-order rate law. Notice the linearity of the trace

Justification Box 8.2
The linear form of the integrated first-order rate equation
If we start with the now-familiar integrated rate equation of Equation (8.24): ln [A]0
[A]t

= k1 t

Using the laws of logarithms, we can split the left-hand side: ln[A]0 − ln[A]t = k1 t
Next, we multiply by −1:

ln[A]t − ln[A]0 = −k1 t

Then, by adding the [A]0 term to the right-hand side we obtain Equation (8.26): ln[A]t = −k1 t + ln[A]0 which is the linear form of the equation, as desired. We note that the intercept in
Equation (8.26) is ln[A]0 .
We could have obtained Equation (8.26) alternatively by integrating without limits during the derivation in Justification Box 8.1.

374

CHEMICAL KINETICS

The SI unit of time is the second (see p. 15), but it is sometimes more convenient to cite k in terms of non-SI units, such as min−1 or even year−1 .

Worked Example 8.10 What is the relationship between the values of rate constant expressed in units of s−1 , and expressed in units of min−1 and year−1 ?

By Analogy. Let the rate constant be k. Conversion into SI is easy: k in min−1 = 60 × (k in s−1 ) because there are 60 s per minute, so 60 times as much reaction can occur during a minute.
Via Dimensional Analysis. Again, let the rate constant be k. In this example, imagine the value of k is 3.2 year−1 .
The number of seconds in a year is (60 × 60 × 24 × 365.25) = 3.16 × 107 s. We can write this result as 3.16 × 107 s yr−1 , which (by taking reciprocals) means that
2.37 × 10−8 yr s−1 .
To obtain k with units of s−1 , we say

To convert k from SI units to the time unit of choice, just multiply the value of k by the fraction of the time interval occurring during a single second.

k = (3.2 yr−1) × (2.37 × 10−8 yr s−1)

original value of k

conversion factor

so k = 1.01 × 10−7 s−1 .
SAQ 8.12 Show that the rate constants 1.244 × 104 yr−1 and 3.94 ×
10−4 s−1 are the same.

Integrated rate equations: second-order reactions
For a second-order reaction, the form of the integrated rate equation is different:
Notice that the units of the second-order rate constant k2 are dm3 mol−1 s−1 which are, in effect, (concentration)−1 s−1 .

We do not need to know the temperature in order to answer this question; but we do need to know that T remained constant, i.e. that the reaction was thermostatted. 1
1

= k2 t
[A]t
[A]0

(8.27)

where the subscripted ‘2’ on k reminds us that it represents a second-order rate constant. The other subscripts and terms retain their previous meanings.
SAQ 8.13 Show that the units of the second-order rate constant are dm3 mol−1 s−1 . [Hint: you will need to perform a simple dimensional analysis of Equation 8.27.]

Worked Example 8.11 We encountered the dimerization of methyl viologen radical cation MV+ž in Equation (8.6) and Worked
Example 8.4. Calculate the value of the second-order rate constant k2 if the initial concentration of MV+ž was 0.001 mol dm−3 and the concentration dropped to 4 × 10−4 mol dm−3 after 0.02 s. (The temperature was 298 K.)

375

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

Inserting values into Equation (8.27):

so

10−4

1
1

= k2 × 0.02 s
−3
−3 mol dm−3 mol dm
10

2500 (mol dm−3 )−1 − 1000 (mol dm−3 )−1 = k2 × 0.02 s

and

1500 (mol dm−3 )−1 = k2 × 0.02 s

Rearranging, we say k2 = so 1500 (mol dm−3 )−1
0.2 s

k2 = 7.5 × 104 dm3 mol−1 s−1 at 298 K

which is relatively fast.
SAQ 8.14 Remaining with the same system from Worked Example 8.11, ž having calculated k2 (i.e. having ‘calibrated’ the reaction), how much MV+ remains after 40 ms (0.04 s)?
SAQ 8.15 Consider a second-order reaction which consumes 15 per cent of the initial material after 12 min and 23 s. If [A]0 was 1 × 10−3 mol dm−3 , calculate k2 . [Hint: first calculate how much material remains.]

An alternative form of the integrated rate equation is the so-called linear form
1
[A]t y = k2 t + m x

1
[A]0
c

(8.28)

which we recognize as relating to the equation of a straight line, so plotting a graph of l/[A]t (as ‘y’) against time (as ‘x’) will be linear for a reaction that is second order. The rate constant k2 is obtained directly as the gradient of the graph.

We obtain the secondorder rate constant k2 as the slope of a graph drawn according to the integrated secondorder rate equation.

Worked Example 8.12 Consider the data below, which relate to the second-order racemization of a glucose in aqueous hydrochloric acid at 17 ◦ C. The concentrations of glucose and hydrochloric acid are the same, ‘[A]’.
Time t/s
[A]/mol dm−3

0
0.400

600
0.350

1200
0.311

1800
0.279

2400
0.254

376

CHEMICAL KINETICS

A graph of the concentration [A] (as ‘y’) against time (as ‘x’) is clearly not linear; see Figure 8.9(a). Conversely, a different, linear, graph is obtained by plotting 1/[A]t (as ‘y’) against time (as ‘x’); see
Figure 8.9(b). This follows the integrated second-order rate equation.
The gradient of Figure 8.9(b) is the second-order rate constant k2 , and has a value of 6.00 × 10−4 dm3 mol−1 s−1 .

Care: The gradient is only truly k if the time axis is given with the SI units of time
(the second).

0.41
0.39

[A] / mol dm−3

0.37
0.35
0.33
0.31
0.29
0.27
0.25
0

500

1000

1500

2000

2500

2000

2500

Time t / s
(a)
4.0

1/ [A]t

3.5

3.0

2.5
0

500

1000

1500

Time t / s
(b)

Figure 8.9 Kinetics of a second-order reaction: the racemization of glucose in aqueous mineral acid at 17 ◦ C: (a) graph of concentration (as ‘y’) against time (as ‘x’); (b) graph drawn according to the linear form of the integrated second-order rate equation, obtained by plotting 1/[A]t (as ‘y’) against time (as ‘x’). The gradient of trace (b) equals the second-order rate constant k2 , and has a value of 6.00 × 10−4 dm3 mol−1 s−1

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

SAQ 8.16 Consider the following data concerning the reaction between triethylamine and methyl iodide at 20 ◦ C in an inert solvent of CCl4 . The initial concentrations of
[CH3 I]0 and [N(CH3 )3 ]0 are the same. Draw a suitable graph to demonstrate that the reaction is second order, and hence determine the value of the second-order rate constant k2 .
Time t/s
[CH3 I]0 = [N(CH3 )3 ]0 /mol dm−3

0
2.112

2400
0.813

5400
0.149

377

Remember that when plotting kinetic graphs the concentration at t = 0 is also a valid data point.

9000
0.122

18 000
0.084

Warning: if a chemical process comprises several reaction steps, only the progress of the slowest step can be followed kinetically. These graphical methods of determining k are only useful for obtaining the rate-determining step (RDS) of such reactions. Although the reaction may appear kinetically simple, it is wisest to assume otherwise. Justification Box 8.3
Integrated rate equations for a second-order reaction
We will only consider the derivation of the simplest case of a second-order reaction, where the concentrations of the two reacting species are the same. Being second order, the rate law has the form rate = k2 [A]2 . The subscript
The minus sign in
‘2’ on k indicates a second-order process. Again, by
Equation (8.29) is ess‘rate’ we mean the rate of change of the concentration ential to show that of reactant A. d[A] = k2 [A]2 rate = − dt the concentration of A decreases with time.

(8.29)

As before, we shall start at t = 0 with a concentration [A]0 . The value of [A]t decreases with time t, hence a minus sign is inserted.
Rearranging the equation, and indicating the limits yields


[A]t
[A]0

1 d[A] = k2
[A]2
t

t

dt
0

Integrating gives


−1
[A]t

[A]t
[A]0

= k2 [t]t0

378

CHEMICAL KINETICS

The two minus signs on the left will cancel. Inserting limits, and rearranging slightly gives Equation (8.27):
1
1

= k2 t
[A]t
[A]0
This equation is known as the integrated second-order rate equation.

Second-order reactions of unequal concentration
We will start with the reactants A and B having the concentrations [A]0 and [B]0 respectively. If the rate constant of reaction is k2 , and if the concentrations at time t are [A]t and [B]t respectively, then it is readily shown that
1
[A]0 × [B]t
× ln
[B]0 − [A]t
[B]0 × [A]t

= k2 t

(8.30)

We will need to look further at this equation when thinking about kinetic situations in which one of the reactants is in great excess (the so-called ‘pseudo order’ reactions described in Section 8.4).
SAQ 8.17 A 1:1 reaction occurs between A and B, and is second order.
The initial concentrations of A and B are [A]0 = 0.1 mol dm−3 and [B]0 =
0.2 mol dm−3 . What is k2 if [A]t = 0.05 mol dm−3 after 0.5 h? Remember to work out a value for [B]t as well.

Why do we often refer to a ‘half-life’ when speaking about radioactivity?
Half-lives
A radioactive substance is one in which the atomic nuclei are unstable and spontaneously decay to form other elements. Because the nuclei decay, the amount of the radioactive material decreases with time. Such decreases follow the straightforward kinetic rate laws we discussed above.
But many people talk emotionally of radioactivity ‘because radioactive materials are so poisonous’, and one of the clinching
We shall look at why arguments given to explain why radioactivity is undesirable is that radioactive materials are toxic in Section 8.3, radioactive materials have long ‘half-lives’. What is a half-life? below. And why is this facet of their behaviour important? And, for that matter, is it true that radioactive materials are poisonous?

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

379

Table 8.2 Half-lives of radioactive isotopes (listed in order of increasing atomic number)
Isotope

Half-life

12

B
C
40
K
60
Co
129
I
238
U
239
Pu

Source of radioactive isotope

0.02 s
5570 years
1.3 × 109 years
10.5 min
1.6 × 107 years
4.5 × 108 years
2.4 × 104 years

14

Unnatural (manmade)
Natural
Natural: 0.011% of all natural potassium
Unnatural: made for medicinal uses
Unnatural: fallout from nuclear weapons
Natural: 99.27% of all uranium
Unnatural: by-product of nuclear energy

The half-life of radioactive decay or of a chemical reaction is A ‘half-life’ t1/2 is the the length of time required for exactly half the material under time required for the study to be consumed, e.g. by chemical reaction or radioactive amount of material decay. We often give the half-life the symbol t1/2 , and call it to halve.
‘tee half’.
The only difference between a chemical and a radioactive half-life is that the former reflects the rate of a chemical reaction and the latter reflects the rate of radioactive (i.e. nuclear) decay. Some values of radioactive half-lives are given in the Table 8.2 to demonstrate the huge range of values t1/2 can take. The difference between chemical and radioactive toxicity is mentioned in the Aside box on p. 382.
A chemical half-life is the time required for half the material to have been consumed chemically, and a radioactive half-life is the time required for half of a radioactive substance to disappear by nuclear disintegration. 60
Co is a favourite
Since most chemicals react while dissolved in a constant volume radionuclide within the of solvent, the half-life of a chemical reaction equates to the time medical profession, required for the concentration to halve. because its half life
Worked Example 8.13 The half-life of 60 Co is 10.5 min. If we start with 100 g of 60 Co, how much remains after 42 min?

is conveniently short.

Answer Strategy

1. We determine how many of the half-lives have occurred during the time interval.
2. We then successively halve the amount of

60

Co, once per half-life.

(1) The number of half-lives is obtained by dividing 10.5 min into 42 min; so four half-lives elapse during 42 min.
(2) If four half-lives have elapsed, then the original amount of 60 Co has halved, then halved again, then halved once more and then halved a fourth time:

100 g

1st half-life 50 g

2nd half-life 25 g

3rd half-life 12.5 g

4th half-life 6.25 g

380

CHEMICAL KINETICS

1
We see that one-sixteenth of the 60 Co remains after four half-lives, because ( 1 )4 = 16 .
2
In fact, a general way of looking at the amount remaining after a few half-lives is to say that
(8.31)
fraction remaining = ( 1 )n
2

where n is the number of half-lives.
SAQ 8.18 The half-life of radioactive 14 C is 5570 years. If we start with
10 g of 14 C, show that the amount of 14 C remaining after 11 140 years is 2.5 g.

Often, though, we don’t know the half-life. One of the easier ways to determine a value of t1/2 is to draw a graph of amount of substance (as ‘y’) – or, if in solution, of concentration – against time (as ‘x’).
Worked Example 8.14 The table below shows the amount of a biological metabolite
T-IDA as a function of time.
1.

What is the half-life of T-IDA?

2.

Show that the data follow the integrated first-order rate equation.

Time t/min
[T-IDA]/µmol
dm−3

0
100

10
50

20
25

30
12.5

40
6.25

50
3.13

60
1.56

70
0.781

Figure 8.10 shows a plot of the amount of material (as ‘y’) as a function of time t (as
‘x’), which is exponential. This shape should not surprise us, because Equation (8.31) is also exponential in form.
To obtain the half-life, we first choose a concentration – any concentration will do, but we will choose [T-IDA] = 50 µmol dm−3 . We then draw a horizontal arrow from this concentration on the y-axis, note the time where this arrow strikes the curve, and then read off the time on the x-axis, and call it t1 . Next, we repeat the process, drawing an arrow from half this original concentration, in this case from [T-IDA] =
As long as a reaction
25 µmol dm−3 . We note this new time, and call it t2 . The half-life is is first order, the durasimply the difference in time between t1 and t2 . It should be clear that tion of a half-life will the half-life in this example is 10 min. be the same length
But then we notice that the time needed to decrease from 60 to of time regardless of
30 µmol dm−3 , or from 2 to 1 µmol dm−3 will also be 10 min each. the initial amount of
In fact, we deduce the important conclusion that the half-life of a material present. first-order reaction is independent of the initial concentration of material. QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

381

100
90

[A] / 10−6 mol dm−3

80
70
60
50
40
30
20
10
0
0

10

20

30
40
Time t / min

50

60

70

Figure 8.10 Kinetic trace concerning the change in concentration as a function of time: graph of [T-IDA] (as ‘y’) against time (as ‘x’) to show the way half-life is independent of the initial concentration 5

ln([A] / mol dm−3)

4
3
2
1
0
−1

0

10

20

30
40
time t / min

50

60

70

Figure 8.11 Graph plotted with data from Figure 8.10, plotted with the axes of the linear form of the integrated first-order rate equation, with ln[A] as ‘y’ against time t as ‘x’

To show that this reaction is kinetically first order, we take the logarithm of the concentration, and plot ln[A]t (as ‘y’) against time t (as ‘x’); see Figure 8.11. That the graph in Figure 8.11 is linear with this set of axes demonstrates its first-order character.
The half-life of second- or third-order reactions is not independent of the initial concentration in this way (see p. 387).

382

CHEMICAL KINETICS

Aside
The difference between chemical and radiochemical toxicity
It is good that we should be concerned about the environmental impact of what we, as chemists, do to our planet. But many environmental campaigners too easily confuse radioactive toxicity and chemical toxicity. For example, the radon gas emanating from naturally occurring granite rocks is chemically inert, because it is a rare gas, but it is toxic to humans because of its radioactivity. Conversely, sodium cyanide contains no radioactive constituents yet is chemically toxic.
The conceptual problems start when considering materials such as plutonium, which is a by-product of the nuclear electricity industry. Plutonium is one of the most chemically toxic materials known to humanity, and it is also radioactive. The half-life of 238 Pu is so long at 4.5 × 108 years (see Table 8.2) that we say with some certainty that effectively none of it will disappear from the environment by radioactive decay; and if none of it decays, then it cannot have emitted ionizing α and β particles, etc. and, therefore, cannot really be said to be a radioactive hazard. Unfortunately, the long half-life also means that the 238 Pu remains more-or-less for ever to pollute the environment with its lethal chemistry.
But if we accept that plutonium is chemically toxic,
If we accept that pluthen we must also recognize that the extent of its toxictonium is chemically ity will depend on how the plutonium is bonded chemitoxic, then we also cally, i.e. in what redox and chemical form it is present. need to recognize that
As an example, note how soldiers were poisoned with the extent of its toxicity chlorine gas during the First World War (when it was will depend on how the called Mustard Gas), but chloride in table salt is vital plutonium is bonded for life. Some plutonium compounds are more toxic chemically (see p. 59). than others.
At the other extreme are materials with very short radioactive half-lives, such as 12 B (which has a relatively short t1/2 of 0.02 h). 12 B is less likely to cause chemical poisoning than 238 Pu simply because its residence time is so short that it will transmute to become a different element and, therefore, have little time to interact in a chemical sense with anything in the environment (such as us). On the other hand, its short half-life means that the speed of its radioactive decay will generate many subatomic particles (α, β and γ particles) responsible for radioactive poisoning per unit time, causing a larger dose of radioactive poisoning.
This Aside is not intended to suggest that the threat of radioactivity is to be ignored or marginalized; but we should always aim to be well informed when confronting an environmental problem.

How was the Turin Shroud ‘carbon dated’?
Quantitative studies with half-lives
The Turin Shroud is a long linen sheet housed in Italy’s Turin Cathedral. Many people believe that the surface of the cloth bears the image of Jesus Christ (see

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

383

Figure 8.12 Many people believe the Turin Shroud bears an image of Jesus Christ, imprinted soon after his crucifixion. Radiocarbon dating suggests that the flax of the shroud dates from 1345 AD

Figure 8.12), imprinted soon after he died by crucifixion at the hands of the Roman authorities in about AD 33. There has been constant debate about the Turin Shroud and its authenticity since it first came to public notice in AD 1345: some devout people want it to be genuine, perhaps so that they can know what Jesus actually looked like, while others (many of whom are equally devout) believe it to be a fake dating from the Middle Ages.
An accurate knowledge of the Turin Shroud’s age would allow us to differentiate between these two simplistic extremes of AD 33 and AD 1345, effectively distinguishing between a certain fake and a possible relic of enormous value.
The age of the cloth was ascertained in 1988 when the Vatican (which has jurisdiction over Turin Cathedral) allowed a small piece of the cloth to be analysed by radiocarbon dating. By this means, the shroud was found to date from AD 1320 ± 65.
Even after taking account of the uncertainty of ±65 years, the age of the shroud is consistent with the idea of a medieval forgery. It cannot be genuine.
But the discussion about the shroud continues, so many people now assert that the results of the test itself are part of a ‘cover up’, or that the moment of Jesus’s resurrection occurred with a burst of high-energy sub-atomic particles, which upset the delicate ratio of carbon isotopes.

384

CHEMICAL KINETICS

In 1946, Frank Libby of the Institute of Nuclear
Sciences in Chicago initiated the dating of carbon-based artifacts by analysing the extent of radioactive decay.

Radiocarbon dating

The physicochemical basis behind the technique of radiocarbon dating is the isotopic abundances of carbon’s three isotopes: 12 C is the ‘normal’ form and constitutes 98.9 per cent of all naturally occurring carbon. 13 C is the other naturally occurring isotope, with an abundance of about 1 per cent. 14 C does not occur naturally, but tiny amounts of it are formed when high-energy particles from space collide with gases in the upper atmosphere, thus causing radiochemical modification.
All living matter is organic and, therefore, contains carbon; and since all living material must breathe CO2 , all carbon-based life forms ingest 14 C. Additionally, living matter contains 14 C deriving from the food chain.
But 14 C is radioactive, meaning that atoms of 14 C occasionally self-destruct to form a beta particle, β − , and an atom of 14 N:
The beta particle emitted during radioactive decay is an energetic electron.

14

C − → 14 N + β −


(8.32)

The half-life of the process in Equation (8.32) is 5570 years.
Following death, flora and fauna alike cease to breathe and eat, so the only 14 C in a dead body will be the 14 C it died with. And because the amounts of 14 C decrease owing to radioactive decay, the amount of the 14 C in a dead plant or person decreases whereas the amounts of the 12 C and 13 C isotopes
Radiocarbon dating is also called ‘radiometric do not. We see why the proportion of 14 C decreases steadily as a dating’ or ‘radiochemifunction of time following the instant of death. cal dating’.
By corollary, if we could measure accurately the ratio of 12 C to
14
C in a once-living sample, we could then determine roughly how long since it was last breathing. This explains why we can ‘date’
Oil and oil-based proda sample by analysing the residual amounts of 14 C. ucts contain no 14 C,
In the carbon-dating experiment, a sample is burnt in pure oxybecause the creatures from which the gen and converted into water and carbon dioxide. Both gases are oil was formed died fed into a specially designed mass spectrometer, and the relative so many millions of abundances of 12 CO2 and 14 CO2 determined. The proportion of
14
years ago. Accordingly,
CO2 formed from burning an older sample will be smaller.
Equation (8.32) has
Knowing this ratio, it is a simple matter to back calculate to proceeded to its ascertain the length of time since the sample was last alive. For completion. example, we know that a time of one half-life t1/2 has elapsed if a sample contains exactly half the expected amount of 14 C, so the
Great care is needed sample died 5570 years ago. during the preparation
Great care is needed during the preparation of the sample, since of the sample before dirt, adsorbed CO2 and other impurities can all contain additional dating to eliminate the sources of carbon. The dirt may come from the sample, or it could possibility of contamihave been adsorbed during sample collection or even contamination nation with additional during the dating procedure. The more recent the contamination, sources of carbon. the higher the proportion of carbon that is radioactive 14 C that has

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

385

not yet decayed, causing the artifact to appear younger. It has been suggested, for example, that the Turin Shroud was covered with much ‘modern’ pollen and dust at the time of its radiocarbon dating, so the date of AD 1320 refers to the age of the modern pollen rather than that of the underlying cloth itself.

¨
How old is Otzi the iceman?
Calculations with half-lives
Approximately 5000 years ago, a man set out to climb the Tyrolean Alps on the
Austrian–Italian border. At death, he was between 40 and 50 years old and suffered from several medical ailments. Some scientists believe he was caught in a heavy snowfall, fell asleep, and froze to death. Others suppose he was murdered during his journey. Either way, his body was covered with snow almost immediately and, due to the freezing weather, rapidly became a mummy – ‘The Iceman’. In 1991, his
¨
body was re-exposed and discovered by climbers in the Otzal Alps, explaining why
¨
the ‘Iceman’, as he was called, was given the nickname ‘Otzi’ (or, more commonly, as just Otzi).
Scientific techniques,
His body (see Figure 8.13) was retrieved and taken to the Dep- such as radiocarbon artment of Forensic Medicine at the University of Innsbruck. Their dating, applied to analytical tests – principally radiocarbon dating – suggest that archaeology is some¨
Otzi died between 3360 and 3100 BC. Additional radiocarbon dat- times termed archaeoing of wooden artifacts found near his body show how the site of metry. his death was used as a mountain pass for millennia before and after his lifetime.
Equation (8.33) sugBut how, having defined the half-life t1/2 as the time necessary gests the half-life is for half of a substance to decay or disappear, can we quantita- independent of the
¨
tively determine the time elapsing since Otzi died? In Justification amount of material
Box 8.4, we show how the half-life and the rate constant of decay initially present, so radioactive decay folk are related according to t1/2 ln(2)
=
k

(8.33)

lows the mathematics of first-order kinetics.

¨
Figure 8.13 The body of ‘Otzi the Iceman’ was preserved in the freezing depths of a glacier.
¨
Radiocarbon dating suggests that Ozti froze to death in about BC 3360–3100

386

CHEMICAL KINETICS

¨
Worked Example 8.15 A small portion of Otzi’s clothing was removed and burnt carefully in pure oxygen. The amount of 14 C was found to be 50.93 per cent of the amount expected if the naturally occurring fabric precursors had been freshly picked. How long is it since the crop of flax was picked, i.e. what is its age?
We start by inserting the known half-life t1/2 into Equation (8.33) to obtain a ‘rate constant of radioactive decay’. ln(2) k=
5570 yr
By calculating k, we

are in effect calibrating the experiment. We only need to do this calibration once.

Using our calculators, we need to type
‘ln(0.5093)’ as the numerator rather than a percentage. The minus sign comes from the laws of logarithms.

By this means, we calculate the rate constant as k = 1.244 ×
10−4 yr−1 . (Alternatively, we could have calculated k in terms of the SI unit of time (the second), in which case k has the value
3.94 × 10−12 s−1 .)
To calculate t, the length of time since the radioactive decay commenced (i.e. since the fabric precursor died), we again insert values into the integrated form of the first-order rate equation,
Equation (8.33). We then insert our previously calculated value of k: t= ln(50.93%)
= 5420 yr
−1.244 × 10−4 yr−1

So the interval of time t since the flax was picked and thence woven into cloth is 5420 yr, so the cloth dates from about 3420 BC.
SAQ 8.19 To return to the example of the Turin Shroud. Suppose a sloppy technique caused the precision of the 14 C measurement to decrease from 92.23 per cent to 90 ± 2 per cent. Calculate the range of ages for the shroud. [Hint: perform two calculations, one for either of the extreme values of percentage.]

Justification Box 8.4
Justification Box 8.1 shows how the integrated first-order rate equation is given by
Equation (8.24): ln [A]0
[A]t

= kt

After a length of time equal to one half-life t1/2 , the concentration of A will be [A]t , which, from the definition of half-life, has a value of 1 [A]0 .
2
Inserting these two concentrations into Equation (8.24) gives ln [A]0
1
[A]0
2

= kt1/2

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

387

The two [A]0 terms cancel, causing the bracket on the left to simplify to just ‘2’.
Accordingly, Equation (8.24) becomes ln(2) = kt1/2
So, after rearranging to make t1/2 the subject, the half-life is given by Equation (8.33): t1/2 =

ln(2) k SAQ 8.20 Show from Equation (8.27) that the half-life of a second-order equation is given by the expression: t1/2 =

1
[A]0 k2

The half-life of a second-order reaction is not independent of the initial concentrations used.

Why does the metabolism of a hormone not cause a large chemical change in the body?
‘Pseudo-order’ reactions
A hormone is a chemical that transfers information and instructions between cells in animals and plants. They are often described as the body’s ‘chemical messengers’, but they also regulate growth and development, control the function of various tissues, support reproductive functions, and regulate metabolism (i.e. the process used to break down food to create energy).
Hormones are generally quite small molecules, and are chemically uncomplicated.
Examples include adrenaline (II) and β-phenylethylamine (III).
HO
HO

CH(OH)CH2NHCH3
(II)

CH2CH2NH2
(III)

Most hormones are produced naturally in the body (e.g. adrenaline (II) is formed in the adrenal glands). From there, the hormone enters the bloodstream and is consumed chemically (a physiologist would say ‘metabolized’) at the relevant sites in the body – in fact, adrenaline accumulates and is then broken down chemically in the muscles and lungs. Adrenaline is generated in equal amounts in men and women,

388

CHEMICAL KINETICS

and promotes a stronger, faster heartbeat in times of crisis or panic. The body is thus made ready for aggression (‘fight’) or necessary feats of endurance to escape
(‘flight’). Adrenaline is also administered artificially in medical emergencies, e.g. immediately following an anaphylactic shock, in order to give the heart a ‘kick start’ after a heart attack. β-Phenylethylamine (III) is a different type of hormone, and is metabolized in women’s bodies to a far higher extent than in
We say a reaction is a man’s. The mechanism of metabolism is still a mystery, but it pseudo first-order if appears to cause feelings of excitement, alert feelings and in terms it is second- or thirdof mood, perhaps a bit of a ‘high’. Unfortunately, the bodies of order, but behaves most men do not metabolize this hormone, so they do not feel the mathematically as same ‘high’. though it were first
Hormones are potent and are produced in tiny concentrations order. (generally with a concentration of nanomoles per litre). By contrast, the chemicals in the body with which the hormone reacts have a sizeable concentration. For example, reaction with oxyA pseudo-order reacgen in the blood is one of the first processes to occur during the tion proceeds with all metabolism of adrenaline. The approximate range of [O2 ](blood) is but one of the reactants
0.02–0.05 mol dm−3 , so the change in oxygen concentration is in excess. This ensures virtually imperceptible even if all the adrenaline in the blood is that the only conconsumed. As a good corollary, then, the only concentration to centration to change change is that of the hormone, because it is consumed. appreciably is that of
Although such a reaction is clearly second order, it behaves like a the minority reactant. first-order reaction because only one of the concentrations changes.
We say it is a pseudo first-order reaction.

Why do we not see radicals forming in the skin while sunbathing? Pseudo-order rate constants
One of the most common causes of skin cancer is excessive sunbathing. Radicals are generated in the skin during irradiation with high-intensity UV-light, e.g. while lying on a beach. These radicals react with other compounds in the skin, which may ultimately cause skin cancer. But we never see these radicals by eye, because their concentration is so minuscule. And the concentration is so small because the radicals react so fast. (Photo-ionization is discussed in Chapter 9.)
The accumulated conReaction intermediates are common in mechanistic studies of centration of such a organic reactions. They are called ‘intermediates’ because they fast-reacting interreact as soon as they form. Such intermediates are sometimes mediate will always described as ‘reactive’ because they react so fast they disappear be extremely low: more or less ‘immediately’. Indeed, these intermediates are so reacperhaps as low as
−3
tive, they may react with solvent or even dissolved air in solution,
−10
10 mol dm .
i.e. with other chemicals in high concentration.

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

389

Since the reaction of intermediates is so fast, the concentration Intermediates react of the radical intermediate changes dramatically, yet the concentra- fast because their tions of the natural compounds in the skin with which the interme- activation energy is diate reacts (via second-order processes) do not change perceptibly. small – see
How in practice, then, do we determine kinetic parameters for Section 8.5. pseudo-order reactions such as these?
We will call the intermediate ‘A’ and the other reagent, which is in excess, will be called ‘B’. In the example here, B will be the skin, but is more generally the solvent in which the reaction is performed, or an additional chemical in excess.
Because the concentration of B, [B], does not alter, this reaction will obey a firstorder kinetic rate law, because only one of the concentrations changes with time.
Because the reaction is first order (albeit in a pseudo sense), a plot of ln[A]t (as ‘y’) against time (as ‘x’) will be linear for all but the longest times
(e.g. see Figure 8.14). In an analogous manner to a straightforward Pseudo-order rate confirst-order reaction, the gradient of such a plot has a value of ‘rate stants are generally constant × −1’ (see Worked Example 8.9). We generally call the indicated with a prime, rate constant k , where the prime symbol indicates that the rate e.g. k . constant is not a true rate constant, but is pseudo.
The proper rate law for reaction for the second-order reaction between A and B is: rate = k2 [A][B]

(8.34)

as for any normal second-order reaction. By contrast, the perceived rate law we measure is first order, as given by rate = k [A]

(8.35)

Writing a pseudo rate constant k without an order implies that it is pseudo first order.

ln[A]

where k is the perceived rate constant.

Time t

Figure 8.14 The reaction of A and B, with B greatly in excess is a second-order reaction, but it follows a kinetic rate law for a first-order reaction. We say it is pseudo first-order reaction.
The deviation from linearity at longer times occurs because the concentration of B (which we assume is constant) does actually change during reaction, so the reaction no longer behaves as a first-order reaction

390

CHEMICAL KINETICS

Comparing Equations (8.34) and (8.35), we obtain
The concentration [B]t barely changes, so we can write [B]0 in
Equations (8.34) and
(8.36).

k = k2 [B]

(8.36)

This little relationship shows that a pseudo-order rate constant k is not a genuine rate constant, because its value changes in proportion to the concentration of the reactant in excess (in this case, with [B]).

Worked Example 8.16 The reaction of the ester ethyl methanoate and sodium hydroxide in water is performed with NaOH in great excess ([NaOH]0 = 0.23 mol dm−3 ). The reaction has a half-life that is independent of the initial concentration of ester present.
13.2 per cent of the ester remains after 14 min and 12 s. What is the second-order rate constant of reaction k2 ?
Strategy. (1) We ascertain the order of reaction, (2) we determine the pseudo rate constant k , (3) from k , we determine the value of the second-order rate constant k2 .
(1) Is the reaction a pseudo first-order process? The question says that the half-life t1/2 of reaction is independent of initial concentration of ester, so the reaction must behave as though it was a first-order reaction in terms of [ester]. In other words, NaOH is in excess and its concentration does not vary with time.
(2) What is the value of the pseudo first-order rate constant k ? We calculate the pseudo first-order rate constant k by assuming that the reaction obeys first-order kinetics.
Accordingly, we write from Equation (8.24): ln so ln 100%
13.2%

[ester]0
[ester]t

=kt

= k × [(14 × 60) + 12] s

and ln(7.57) = k × 852 s
Because ln(7.57) = 2.02 we say k =

Note how manipulating the units in the fraction yields the correct units for k2 .

2.02
= 2.37 × 10−3 s−1
852 s

(3) What is the value of the second-order rate constant k2 ? The value of rate constant k2 can be determined from Equation (8.36), as k ÷ [NaOH], so k2 =

2.37 × 10−3 s−1
= 1.03 × 10−3 dm3 mol−1 s−1
0.23 mol dm−3

QUANTITATIVE CONCENTRATION CHANGES: INTEGRATED RATE EQUATIONS

391

SAQ 8.21 Potassium hexacyanoferrate(III) in excess oxidizes an alcohol at a temperature of 298 K. The concentration of K3 [Fe(CN)6 ] is
0.05 mol dm−3 . The concentration of the alcohol drops to 45 per cent of its t = 0 value after 20 min. Calculate first the pseudo first-order rate constant k , and thence the second-order rate constant k2 .

Justification Box 8.5
When we first looked at the derivation of integrated rate equations, we looked briefly at the case where two species A and B were reacting but [A]0 does not equal [B]0 . The integrated rate equation for such a case is Equation (8.30):
1
[A]0 × [B]t
× ln
[B]0 − [A]t
[B]0 × [A]t

= k2 t

Though we do not need to remember this fearsome-looking equation, we notice a few things about it. First, we assume that [B]0
[A]0 , causing the first term on the left-hand side to behave as 1/[B]0 .
Also, the major change in the logarithm bracket is the change in [A], since the difference between [B]0
We argue this stateand [B]t will be negligible in comparison, causing a ment by saying that if cancellation of the [B] terms on top and bottom.
[B] is, say, 100 times
Accordingly, Equation (8.30) simplifies to
[A]0
1
× ln
[B]0
[A]t

= k2 t

Since ln([A]0 /[A]t ) = k t for a pseudo first-order reaction (by analogy with Equation (8.24)), we say that
1
× (k t) = k2 t
[B]0

larger than [A], then a complete consumption of [A] (i.e. a 100 per cent change in its concentration) will be associated with only a
1 per cent change in
[B] – which is tiny.

and cancelling the two t terms, and rearranging yields k = k2 [B]0 so we re-obtain Equation (8.36).

Alternatively, the value of the true second-order rate constant may be obtained by treating Equation (8.36) as the equation of a straight line, with the form y = mx: k = k2 [B]0 m x y 392

CHEMICAL KINETICS

Accordingly, we perform the kinetic experiment with a series of concentrations [B]0 , the reactant in excess, and then plot a graph of k (as ‘y’) against [B]0 (as ‘x’). The gradient will have a value of k2 .
A graphical method such as this is usually superior to a simplistic calculation of k2 = k ÷ [NaOH] (e.g. in the Worked Example 8.16), because scatter and/or chemical back or side reactions will not be detected by a single calculation. Also, the involvement of a back reaction (see next section) would be seen most straightforwardly as a non-zero intercept in a plot of k (as ‘y’) against [reagent in excess]
(as ‘x’).
Worked Example 8.17 The following kinetic data were obtained for the second-order reaction between osmium tetroxide and an alkene, to yield a 1,2-diol. Values of k are pseudo-order rate constants because the OsO4 was always in a tiny minority. Determine the second-order rate constant k2 from the data in the following table:
[alkene]0 / mol dm−3 k /10−4 s−1

0.01

0.02

0.03

3.2

6.4

0.04

9.6

12.8

0.05
16.0

0.06

0.07

19.2

22.4

0.08
25.6

0.003

Pseudo-order rate constant k ′ /s−1

0.0025

0.002

0.0015

0.001

0.0005

0
0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

[alkene]0/mol dm−3

Figure 8.15 The rate constant of a pseudo-order reaction varies with the concentration of the reactant in excess: graph of k (as ‘y’) against [alkene]0 (as ‘x’). The data refer to the formation of a 1,2-diol by the dihydrolysis of an alkene with osmium tetroxide. The gradient of the graph yields k2 , with a value of 3.2 × 10−2 dm3 mol−1 s−1

KINETIC TREATMENT OF COMPLICATED REACTIONS

The data clearly show that k is not a true rate constant, because its value varies as the concentration of the alkene increases. That k increases linearly with increased [alkene] suggests a straightforward pseudo first-order reaction. Figure 8.15 shows a graph of k (as ‘y’) against [alkene]0 (as ‘x’). The graph is linear, and has a gradient of
3.2 × 10−2 dm3 mol−1 s−1 , which is also the value of k2 .

8.4

393

Graphs such as that in Figure 8.15 generally pass through the origin.

Kinetic treatment of complicated reactions Why is arsenic poisonous?
‘Concurrent’ or ‘competing’ reactions
Arsenic is one of the oldest and best known of poisons. It is so well known, in fact, that when the wonderful Frank Capra comedy Arsenic and Old Lace was released, everyone knew that it was Care: it is unwise to call going to be a murder mystery in which someone would be poi- a complicated reaction soned. In fact, it has even been rumoured that Napoleon died from such as these a ‘comarsenic poisoning, the arsenic coming from the green dye on his plex reaction’, since wallpaper. We deduce that even a small amount of arsenic will the word ‘complex’ implies an associacause death, or at least an unpleasant and lingering illness. tion compound.
Arsenic exists in several different redox states. The characteristic energy at which one redox state converts to the other depends on its electrode potential E (see Chapter 7). The nervous system in a human body is
‘instructed’ by the brain much like a microprocessor, and regulated by electron ‘relay cycles’ as the circuitry, which consume or eject electrons. Unfortunately, the electrons acquired or released by arsenic in the blood interfere with these naturally occurring electron relay cycles, so, following arsenic poisoning, the numbers of electrons in these relay cycles is wrong – drastically so, if a large amount of arsenic is ingested.
Improper numbers of electrons in the relay cycles cause them not to work properly, causing a breakdown of those bodily functions, which require exact amounts of charge to flow. If the nervous system fails, then the lungs are not ‘instructed’ how to work, the heart is not told to beat, etc., at Arsenic and nitrogen which point death is not too far away. compete for the
But arsenic is more subtle a poison than simply a reducing electrons participating or oxidizing agent. Arsenic is a metalloid from Group in the natural electronV(B) of the periodic table, immediately below the elements relay cycles in the nitrogen and phosphorus, both of which are vital for health. body. The number of
Unfortunately, arsenic is chemically similar to both nitrogen and electrons transferred phosphorus, and is readily incorporated into body tissues following by nitrogen depends on ingestion. Arsenic effectively tricks the body into supposing the number of arsenic that straightforward incorporation of nitrogen or phosphorus has atoms competing for them. occurred. 394

CHEMICAL KINETICS
Product 1
N

Electrons
As
Product 2

‘Sequester’ means to confiscate, seize or take control of something to prevent its further use. The word comes from the Latin sequester, meaning a ‘trustee’ or ‘agent’ whose job was to seize property.

Concurrent (or competing) reactions are so called because two reactions occur at the same time. We occasionally call them simultaneous reactions, although this terminology can be confusing. Figure 8.16 Arsenic and nitrogen compete for electrons both for and from the natural relay cycles in the body. The overall rate at which electrons are transferred by the nitrogen will alter when arsenic competes for them. The arsenic is poisonous, since these two pathways yield different products

The first time the body realizes that arsenic has been incorporated is when the redox activity (as above) proceeds at potentials when nitrogen or phosphorus are inert. By the time we detect the arsenic poisoning (i.e. we feel unwell), it is generally too late, since atoms of arsenic are covalently bound within body tissue and cannot just be flushed out or treated with an antidote.
The arsenic sequesters electrons that might otherwise be involved in other relay cycles, which is a concurrent kinetic process; see
Figure 8.16.
So arsenic is toxic because it has ‘fooled’ the body into thinking it is something else.

Why is the extent of Walden inversion smaller when a secondary alkyl halide reacts than with a primary halide?
Reaction profiles for complicated reactions

Alkyl halides react by a substitution reaction with hydroxide ions to yield an alcohol. A primary halide, such as 1-bromopentane, reacts by a simple bimolecular SN 2 mechanism, where the ‘S’
A beam of planestands for substitution, the ‘N’ for nucleophilic (because the polarized light is hydroxide ion is a nucleophile) and the ‘2’ reminds us that caused to rotate by the reaction is bimolecular. Being a single-step reaction, the an angle θ as it passes substitution reaction is necessarily the rate-determining step. The through a solution of reaction is accompanied by stereochemical inversion about the a chiral compound. central tetrahedral carbon atom to which the halide is attached – we
The magnitude of call it Walden inversion; see Figure 8.17(a). Each molecule of θ depends on the concentration of the primary halide inverts during the SN 2 reaction. For this reason, we chiral compound. could monitor the rate of the SN 2 reaction by following changes in the angle of rotation θ of plane polarized light. This angle θ changes as a function of the extent of reaction ξ , so we know the reaction is complete when θ remains constant at a new value that we call θ(final) .

KINETIC TREATMENT OF COMPLICATED REACTIONS
H

H

k

Br

OH−

395

+ Br−

HO

H
R

R
H
(a)

R1

k(1)
Br

R1
OH−

−OH

R3
R2

R3

R2

k(2)
R1

R1

HO

OH
R2
R3

R2

R3

(b)

Figure 8.17 Reaction of an alkyl halide with hydroxide ion. (a) A primary halide reacts by an SN 2 mechanism, causing Walden inversion about the central, chiral carbon. (b) A tertiary halide reacts by an SN 1 mechanism (the rate-determining step of which is unimolecular dissociation, minimizing the extent of Walden inversion and maximizing the extent of racemization). Secondary alcohols often react with both SN 1 and SN 2 mechanistic pathways proceeding concurrently

By contrast, tertiary halides, such as 2-bromo-2,2-dimethylpropane, cannot participate in an SN 2 mechanism because it would be Polarimetry is the impossible to fit two methyl groups, one bromine and a hydroxide technique of following around a single carbon. The steric congestion would be too great. the rotation of planeSo the tertiary halide reacts by a different mechanism, which we polarized light. call SN 1. It’s still a nucleophilic substitution reaction (hence the
‘S’ and the ‘N’) but this time it is a unimolecular reaction, hence Genuinely first-order the ‘1’. The rate-determining step during reaction is the slow uni- reactions are unusual. molecular dissociation of the alkyl halide to form a bromide ion It is likely that the alkyl halide collides with and a carbocation that is planar around the reacting carbon.
Addition of hydroxide occurs as a rapid follow-up reaction. Even another body (such if the alkyl halide was chiral before the carbocation formed, racem- as solvent) with suffiization occurs about the central carbon atom because the hydrox- cient energy to cause bond cleavage. ide can bond to the planar central carbon from either side (see
Figure 8.17(b)). Statistically, equal numbers of each racemate are formed, so the angle through which the plane polarized light rotated during reaction will, therefore, decrease toward 0◦ , when reaction is complete.
In summary, primary halides react almost wholly by a bimolecular process and tertiary halides react by a unimolecular process. Secondary halides are structurally between these two extreme structural examples, since reaction occurs by both SN 2 and
SN 1 routes. These two mechanisms proceed in competition, and occur concurrently.
When following the (dual-route) reaction of a secondary halide with hydroxide ion, we find that the angle θ through which plane polarized light is rotated will decrease, as for primary and tertiary halides, but will not reach zero at completion. In fact, the final angle will have a value between 0◦ and θfinal because of the mixtures of products, itself a function of the mixture of SN 1 and SN 2 reaction pathways.

396

CHEMICAL KINETICS

We looked briefly at reaction profiles in Section 8.2. Before we look at the reaction profile for the concurrent reactions of hydrolysing a secondary alkyl halide, we will look briefly at the simpler reaction of a primary alkyl halide, which proceeds via a single reaction path. And for additional simplicity, we also assume that the reaction goes to completion. We will look not only at the rate of change of the reactants’ concentration but also at the rate at which product forms.
Consider the graph in Figure 8.18, which we construct with the data in Worked Example 8.12. We have seen the upper, lighter line
We include the minus before: it represents the rate of a second-order decay of molecule A sign in Equation
(8.37) to show how with time as it reacts with the stoichiometry A + B → product. The the product concentrabold line in Figure 8.18 represents the concentration of the product. tion INcreases while
It is a 1:1 reaction, so each molecule of A consumed by the reaction the reactant concentrawill generate one molecule of product, with the consequence that tion DEcreases. the two traces are mirror images. Stated another way, the rate of forming product is the same as the rate of consuming A: rate = –

d[product] d[A] = dt dt

(8.37)

where the minus sign is introduced because one concentration increases while the other decreases.
Now, to return to the hydrolysis of the secondary alkyl halides, we will call the reactions (1) and (2), where the ‘1’ relates to the SN 1 reaction and the ‘2’ relates to the SN 2 reactions. (And we write the numbers with brackets to avoid any confusion,
i.e. to prevent us from thinking that the ‘1’ and ‘2’ indicate first- and second-order reactions respectively.) We next say that the rate constants of the two concurrent reactions are k(1) and k(2) respectively. As the two reactions proceed with the same 1:1
0.45
Concentration / mol dm −3

0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0

500

1000
1500
Time t / s

2000

2500

Figure 8.18 Concentration profile for a simple reaction of a primary alkyl halide + OH− → alcohol. The bolder, lower line represents the concentration of product as a function of time, and the fainter, upper line represents the concentration of reactant

KINETIC TREATMENT OF COMPLICATED REACTIONS

397

Concentration of product / mol dm −3

0.035
Reaction (2)

0.03
0.025
0.02

Reaction (1)

0.015
0.01
0.005
0
0

500

1000
1500
Time t / s

2000

2500

Figure 8.19 Concentration profiles for a concurrent reaction, e.g. of a secondary alkyl halide + OH− → alcohol: reaction (2) is twice as fast as reaction (1) in this example

stoichiometry, the ratio of the products will relate to the respective rate constants in a very simple way, according to moles of product formed via reaction (1) k(1) = k(2) moles of product formed via reaction (2)

(8.38)

so we see that the amounts of product depend crucially on the relative magnitudes of the two rate constants. (We shall return to this theme when we look at the way the human body generates a high temperature to cure a fever in Section 8.5).
We now look at the concentration profiles for reactions (1) and
(2). For simplicity, we shall say that reaction (2) is twice as fast as reaction (1), which is likely – unimolecular reactions are often quite slow. We should note how, at the end of the reaction at the far right-hand side of the profile in Figure 8.19, the total sum of product will be the same as the initial concentration of reactant.
Also note how, at all stages during the course of the reaction, the ratio of products will be 2:1, since that was the ratio of the two rate constants.

Care: do not confuse k1 and k(1) . k1 is the rate constant of a firstorder process, and k(1) is the rate constant of the first process in a multi-step reaction.

We can inter-convert between ‘number of moles’ and ‘concentration’ here because the reactions are performed within a constant volume of solvent.

Why does ‘standing’ a bottle of wine cause it to smell and taste better?
Consecutive reactions
It is often said that a good wine, particularly if red, should ‘stand’ before serving. By
‘standing’, we mean that the bottle should be uncorked some time before consumption,

398

CHEMICAL KINETICS

to allow air into the wine. After a period of about an hour or so, the wine should taste and smell better.
Wines contain a complicated mixture of natural products, many of which are alcohols. Ethanol is the most abundant alcohol, at a concentration of 3–11 per cent by volume. The amounts of the other alcohols generally total no more than about 0.1–1 per cent.
The majority of the smells and flavours found in nature comprise
The enzymes in wine esters, which are often covalent liquids with low boiling points and are killed if the percentage of alcohol exceeds high vapour pressures. For that reason, even a very small amount about 13 per cent. Ferof an ester can be readily detected on the palate – after all, think mentation alone cannot how much ester is generated within a single rose and yet how make a stronger wine, overwhelmingly strong its lovely smell can be! so spirits are prepared
Esters are the product of reaction between an alcohol and a carby distilling a wine. boxylic acid. Although the reaction can be slow – particularly at
Adding brandy to wine lower temperatures – the equilibrium constant is sufficiently high makes a fortified wine, for the eventual yield of ester to be significant. We see how even such as sherry. a small amount of carboxylic acid and alcohol can generate a sufficient amount of ester to be detected by smell or taste.
This helps explain why a wine should be left to stand: some of the natural alcohols in the wine are oxidized by oxygen dissolved from the air to form carboxylic acids. These acids then react with the natural alcohols to generate a wide range of esters, which connoisseurs of wines will recognize as a superior taste and ‘bouquet’.
And the reason why the wine must ‘stand’ (rather than the reaction occurring ‘immediately’ the oxygen enters the bottle on
The ‘mash’ that feropening) is that the reaction to form the ester is not a straightments to form wine sometimes includes forward one-step reaction: the first step (Equation (8.39)) is quite grape skins; which slow and occurs in low yield:
A wine is also said to age and breathe, which means the same thing.

are rich in enzymes.
It is likely, therefore, that the oxidation in Equation (8.39) is mediated (i.e. catalysed) enzymatically.

first reaction

−−− alcohol(aq) + O2(aq) − − − → carboxylic acid(aq) + H2 O(l)
(8.39)
where the rate constant of Equation (8.39) is that of a secondorder reaction. Once formed, the acid reacts more rapidly to form the respective ester: second reaction

−−−− carboxylic acid(aq) + alcohol(aq) − − − − → ester(aq)

(8.40)

where the rate constant for this second step is larger than the respective rate constant for the first step, implying that the second step is faster.
In summary, we see that esterification is a two-step process. The first step – production of a carboxylic acid – is relatively slow because its rate is proportional to the concentration of dissolved oxygen; and the [O2 ](soln) is low. Only after the wine bottle has been open for some time (it has had sufficient time to

KINETIC TREATMENT OF COMPLICATED REACTIONS

‘breathe’) will [O2 ](soln) be higher, meaning that, after ‘standing’, the additional flavour and aroma are formed in higher yield.
The case of esterification is an example of a whole class of reactions, in which the product of an initial reaction will itself undergo a further reaction. We say that there is a sequence of reaction steps, so the reaction as a whole is sequential or consecutive.
Although this example comprises two reactions in a sequence, many reactions involve a vast series of steps. Some radioactive decay routes, for example, have as many as a dozen species involved in a sequence before terminating with an eventual product. For simplicity, we will denote the reaction sequence by k(1) 399

esters that lend

To remember this latter terminology, we note that for a consecutive reaction to occur, the second step proceeds as a consequence of the first.

k(2)



A−→B−→C

(8.41)

All the other reactants will be ignored here to make the analysis more straightforward, even if steps (1) and (2) are, in fact, bimolecular. We again write the reaction number within brackets to avoid confusion: we do not want to mistake the subscripted number for the order of reaction. We call the rate constant of the first reaction k(1) and the rate constant of the second will be k(2) .
The material A is the initial reactant or precursor. It is usually stable and only reacts under the necessary conditions, e.g. when The maximum rate of mixed and/or refluxed with other reactants. We know its concentra- reaction occurs when tion because we made or bought it before the reaction commenced. the concentration of
Similarly, the material C (the product) is usually easy to handle, the intermediate B so we can weigh or analyse it when the reaction is complete, and is highest. examine or use it. By contrast, the material B is not stable – if it was, then it would not react further to form C as a second reaction step. Accordingly, it is rare that we can isolate B. We call B an intermediate, because it forms during the consumption of A but before the formation of C.
The concentration profile of a simple 1:1 reaction is always easy to draw because product is formed at the expense of the reactant, so the rate at which reactant is consumed is the same as the rate of product formation. No such simple relation holds for a consecutive reaction, because two distinct rate constants are involved. Two extreme cases need to be considered when dealing with a consecutive reaction: when k(1) > k(2) and when k(1) < k(2) . We shall treat each in turn.

Why fit a catalytic converter to a car exhaust?
Consecutive reactions in which the second reaction is slow
First, we consider the case where the first reaction is very fast compared with the second, so k(1) > k(2) . This situation is the less common of the two extremes. When

400

CHEMICAL KINETICS

all the A has been consumed (to form B), the second reaction is only just starting to convert the B into the final product C. A graph of concentration against time will show a rapid decrease in [A] but a slow increase in the concentration of the eventual product,
C. More specifically, the concentration of intermediate B will initially increase and only at later times will its concentration decrease once more as the slower reaction
(the one with rate constant k(2) ) has time to occur to any significant extent. The concentration profile of B, therefore, has a maximum;
Most modern cars are see Figure 8.20. fitted with a ‘catalytic
A good example of a consecutive process in which the second converter’, one purreaction is much slower than the first is the reaction occurring pose of which is to in a car exhaust. The engine forms carbon monoxide (CO) as its speed up this second
(slower) process in the initial product, and only at later times will CO(g) oxidize to form reaction sequence.
CO2(g) . In fact, the second reaction (CO(g) + 1 O2(g) → CO2(g) ) is
2
so slow that the concentration of CO(g) is often high enough to cause serious damage to health.

Why do some people not burn when sunbathing?
Consecutive reactions in which the second reaction is fast

Concentration

We now look at the second situation, i.e. when k(1) < k(2) . The first reaction produces the intermediate B very slowly. B is conIf k(1) < k(2) , an intersumed immediately – as soon as each molecule of intermediate is mediate is generally formed – by the second reaction, which forms the final product C. termed a reactive intermediate to emphasize
Accordingly, we call it a reactive intermediate, to emphasize how how soon it reacts. rapidly it is consumed.
The rate of the first reaction in the sequence is slow because k(1) is small, so the rate of decrease of [A] is not steep. Conversely, since k(2) is fast, we would at first expect the rate d[C]/dt to be quite high. A moment’s thought shows

A

C

B

Time

Figure 8.20 Schematic graph of concentration against time (‘a concentration profile’) for a consecutive reaction for which k(1) > k(2) . Note the maximum in the concentration of the intermediate,
B. This graph was computed with k(2) being five times slower than k(1)

KINETIC TREATMENT OF COMPLICATED REACTIONS

401

that this is not necessarily the case: rate of forming the final product, C =

d[C]
= k(2) [B] dt (8.42)

We see from Equation (8.42) that the rate of forming C is quite slow because [B] is tiny no matter how fast the rate constant k(2) .
We now consider the concentration profile of the reactive intermediate B. Because the second reaction is so much faster than the first, the concentration of B is never more than minimal. Its concentration profile is virtually horizontal, although it will show a very small maximum (it must, because [B] = 0 at the start of the reaction at t = 0; and all the intermediate B has been consumed at the end of reaction, when
[B] is again zero). At all times between these two extremes, the concentration of B at time t, [B]t , is not zero. The concentration profile is shown in Figure 8.21.
Sunbathing to obtain a tan, or simply to soak up the heat, is an inadvertent means of studying photochemical reactions in the skin. It is also a good example of a consecutive reaction for which k(1) < k(2) .
Small amounts of organic radicals are formed continually in the skin during photolysis (in a process with rate constant k(1) ). The radicals are consumed ‘immediately’ by natural substances in the An antioxidant is a chemical that prevents skin, termed antioxidants (in a different process with rate constant oxidation reactions. k(2) ). Vitamin C (L-(+)-ascorbic acid, IV) is one of the best naturally occurring antioxidants. Red wine and tea also contain efficient antioxidants. H
O

O

OH
CH2OH

OH

OH

Concentration

(IV)

C

A

B
Time

Figure 8.21 Schematic graph of concentration against time (‘a concentration profile’) for a consecutive reaction for which k(2) > k(1) . This graph was computed with k(2) being five times faster than k(1)

402

CHEMICAL KINETICS

The vast majority of organic radical reactions involve the radical as a reactive intermediate, since their values of k(2) are so high, although we need to note that the second reaction need not be particularly fast: it only has to be fast in relation to the first reaction. As a good generalization, the intermediate may be treated as a reactive intermediate if k(2) /k(1) > 10−3 .

Integrated rate equations for consecutive reactions
Our consecutive reaction here has the general form A → B → C. Deriving an integrated rate equation for a consecutive reaction is performed in much the same way as for a simple one-step reaction (see Section 8.2), although its complexity will prevent us from attempting a full derivation for ourselves here.
For the precursor A, the rate of change of [A] is given by rate = −

d[A]
= k(1) [A] dt (8.43)

where the minus sign reflects the way that [A] is consumed, meaning its concentration decreases with time.
The rate of change of C has been given already as Equation
(8.42). Equations (8.42) and (8.43) show why the derivation of
Writing an integrated integrated rate equations can be difficult for consecutive reactions: rate expression for a complicated reacwhile we can readily write an expression for the rate of forming tion is difficult because
C, the rate expression requires a knowledge of [B], which first we don’t readily know increases, then decreases. The problem is that [B] is itself a function the time-dependent of time. concentration of the
The rate of change of [B] has two components. Firstly, we form intermediate, [B]t .
B from A with a rate of ‘+k(1) × [A]’. The second part of the rate equation concerns the subsequent removal of B, which occurs with a rate of ‘−k(2) × [B]’. The minus sign here reminds us that [B] decreases in consequence of this latter process.
Combining the two rate terms in Equations (8.42) and (8.43) yields The rate expression of a complicated reaction comprises one term for each reaction step.
In this case, species
B is involved in two reactions, so the rate equation comprises two terms.

d[B]
= k(1)[A] − k(2)[B] dt (8.44) rate of the process forming the B

rate of the process removing the B

Equation (8.44) helps explain why the concentration profile in
Figure 8.20 contains a maximum. Before the peak, and at short times, the second term on the right-hand side of Equation (8.44) is tiny because [B] is small. Therefore, the net rate d[B]/dt is positive, meaning that the concentration of
B increases. Later – after the peak maximum – much of the [A] has been consumed,

KINETIC TREATMENT OF COMPLICATED REACTIONS

403

meaning more [B] resides in solution. At this stage in the reaction, the first term on the right-hand side of Equation (8.44) is relatively small, because [A] is substantially depleted, yet the second term is now quite large in response to a higher value of [B].
The second term in Equation (8.44) is consequently larger than the first, causing the overall rate to be negative, which means that [B] decreases with time.
Table 8.3 lists the concentrations [A]t , [B]t and [C]t , which are exact, and will give the correct concentrations of A, B and C at any stages of a reaction. We will not derive them here, nor will we need to learn them. But it is a good idea to recognize the interdependence between [A]t , [B]t and [C]t .

How do Reactolite sunglasses work?
The kinetics of reversible reactions
Reactolite sunglasses are photochromic: they are colourless in the dark, but become dark grey–black when strongly illuminated, e.g. on a bright summer morning. The reaction is fully reversible (in the thermodynamic sense), so when energy is removed from the system, e.g. by allowing the lenses to cool in the dark, the photochemical reaction reverses, causing the lenses to become uncoloured and fully transparent again.
The photochromic lenses contain a thin layer of a silvercontaining glass, the silver being in its +1 oxidation state. Absorption of a photon supplies the energy for an electron-transfer reaction in the glass, the product of which is atomic silver:

Ag+ + e− + hν − → Ag0

(8.45)

where the electron comes from some other component within the glass. It is the silver that we perceive as colour. In effect, the colour
Table 8.3 Mathematical equations to describe the concentrationsa of the three species A, B and C involved in the consecutive reaction, A → B → C
[A]t = [A]0 exp(−k(1) t)
[B]t =

k(1)
[A]0 {exp(k(1) t) − exp(k(2) t)} k(2) − k(1)

[C]t = [A]0 1 −

The word ‘photochromic’ comes from the Greek words photos, meaning
‘light’ and khromos meaning ‘colour’.
A photochromic substance acquires colour when illuminated.

k(2) exp(−k(1) t) + k(1) t k(2) − k(1)

a
Rate constants are denoted as k, where the subscripts indicate either the first or the second reaction in the sequence.
The subscript ‘0’ indicates the concentration at the commencement of the reaction. Concentrations at other times are denoted with a subscript t.

Excited states are defined in Chapter 9.

404

CHEMICAL KINETICS

indicates a very long-lived excited state, the unusually long time is achieved because colouration occurs in the solid state, so the reaction medium is extremely viscous.
Atomic silver is constantly being formed during illumination, but, at the same time, the reverse reaction also occurs during ‘cooling’ (also termed ‘relaxation’ – see
Chapter 9) in a process called charge recombination. Such recombination is only seen when removing the bright light that caused the initial coloration reaction, so the reaction proceeds in the opposite direction.
Reversible reactions are also termed equilibrium or opposing reactions. The reaction might be so slow that the reaction never actually reaches the RHS during a realistic time scale, but the direction of the reaction is still the same.
For a thermodynamically reversible reaction, the rate constants of the forward and reverse reactions are kn and k−n respectively.

Rate laws for reversible reactions
All the reactions considered so far in this chapter have been irreversible reactions, i.e. they only go in one direction, from fully reactants on the left-hand side to fully products on the right-hand side. (They might stop before the reaction is complete.)
We now consider the case of a reversible reaction: k −
A −→ B k−1 ←−


(8.46)

where k1 is the rate constant for the forward reaction and k−1 is the rate constant for the reverse reaction. The minus sign is inserted to tell us that the reaction concerned is that of the reverse reaction.
When writing a rate expression for such a reaction, we note that two arrows involve the species B, so, straightaway, we know that the rate expression for species B has two terms. This follows since the rate of change of the concentration of B involves two separate processes: one reaction forms the B (causing [B] to increase with time) while the other reaction is consuming the B (causing [B] to decrease). The rate is given by rate of change of [B] =

d[B]
= k1 [A] − k−1 [B] dt (8.47)

where the minus and (implicit) plus signs indicate that the concentrations decrease or increase with time respectively. Note that a new situation has arisen whereby the expression to describe the rate of change of [B] itself involves [B] – this is a general feature of rate expressions for simple reversible reactions. Similarly: rate of change of [A] =
The concentrations stop changing when the reaction reaches equilibrium. d[A]
= −k1 [A] + k−1 [B] dt (8.48)

Note that the rate of change of [A] is equal but opposite to the rate of change of [B], which is one way of saying that A is consumed at the expense of B; and B is formed at the expense of A.

KINETIC TREATMENT OF COMPLICATED REACTIONS

405

When the reaction has reached equilibrium, the rate of change of both species must be zero, since the concentrations do not alter any more – that is what we mean by true ‘equilibrium’. From Equation (8.48):
0 = k1 [A](eq) − k−1 [B](eq) so k1 [A](eq) = k−1 [B](eq) which, after a little algebraic rearranging, gives a rather surprising result:
[B](eq)
k1
=
(8.49) k−1 [A](eq)
We recognize the right-hand side of the equation as the equilibrium constant K. We give the term microscopic reversibility to the idea that the ratio of rate constants equals the equilibrium constant K.

The principle of
‘microscopic reversibility’ demonstrates how the ratio of rate constants (forward to back) for a reversible reaction equals the reaction’s equilibrium constant K.

Worked Example 8.18 Consider the reaction between pyridine and heptyl bromide, to make 1-heptylpyridinium bromide. It is an equilibrium reaction with an equilibrium constant K = 40. What is the rate constant of back reaction k−1 if the value of the forward rate constant k1 = 2.4 × 103 dm3 mol−1 s−1 ?
We start with Equation (8.49):
K=

k1 k−1 and rearrange it to make k−1 the subject, to yield k−1 =
We then insert values:

k1
K

2.4 × 103 dm3 mol−1 s−1
40

so k−1 = 60 dm3 mol−1 s−1 .
SAQ 8.22 A simple first-order reaction has a forward rate constant of
120 s−1 while the rate constant for the back reaction is 0.1 s−1 . Calculate the equilibrium constant K of this reversible reaction by invoking the principle of microscopic reversibility.
In kinetics, the equi-

Integrated rate equations for reversible reactions
In kinetics, we often term the concentrations at equilibrium ‘the infinity concentration’. The Reactolite glasses do not become

librium concentration
[A](eq) is often termed the infinity concentration, and cited with an infinity sign as [A]∞ .

406

CHEMICAL KINETICS

We should never throw away a reaction solution without measuring a value of [A](eq) .

progressively darker with time because the concentration of atomic silver reaches its infinity value [Ag](eq) . product. We will consider the case of a first-order reaction, A
Following integration of an expression similar to Equation (8.48), we arrive at ln [A]0 − [A](eq)
[A]t − [A](eq)

= (k1 + k−1 )t

(8.50)

which is very similar to the equation we had earlier for a simple reaction (i.e. one proceeding in a single direction), Equation (8.24). There are two simple differences.
Firstly, within the logarithmic bracket on the left-hand side, each term on top and bottom has the infinity reading subtracted from it. Secondly, the time t is not multiplied by a single rate constant term, but by the sum of both rate constants, forward and back.
This equation can be rearranged slightly, by separating the logarithm: ln([A]t − [A](eq) ) −(k1 + k−1 ) t y m x + ln([A]0 − [A](eq) ) c (8.51)

Note that the final term on the right-hand side is a constant. Accordingly, a plot of ln([A]t − [A](eq) ) (as ‘y’) against time (as ‘x’) will yield a straight line of gradient
−(k1 + k−1 ).
Worked Example 8.19 The data below relate to the first-order isomerization of 2hexene at 340 K, a reaction for which the equilibrium constant is known from other studies to be 10.0. What are the rate constants k1 and k−1 ?
Time t/min
([A]t − [A](eq) )/mol dm−3

0
0.114

20
0.103

47
0.091

80
107
140
0.076
0.066
0.055

Strategy. (1) we need to plot a graph of ln([A]t − [A](eq) ) (as ‘y’) against time t (as ‘x’);
(2) determine its gradient; (3) then, knowing the equilibrium constant K, we will be able to determine the two rate constants algebraically.
(1)

Figure 8.22 shows a graph for a reversible first-order reaction with the axes for an integrated rate equation ln([A]t − [A](eq) ) (as ‘y’) against time (as ‘x’). The gradient is −5.26 × 10−3 min−1 .

The microscopicreversibility relationship K = k1 ÷ k−1 cannot be applied unless we know we have a simple reversible reaction.

(2) The gradient is −(k1 + k−1 ), so (k1 + k−1 ) = +5.26 × 10−3 min−1 .
(3) Next, we perform a little algebra, and start by saying that
K = k1 ÷ k−1 , i.e. we invoke the principle of microscopic reversibility. Multiplying the bracket (k1 + k−1 ) by k−1 ÷ k−1 ,
i.e. by 1, yields gradient = k−1 (K + 1)
(8.52)

KINETIC TREATMENT OF COMPLICATED REACTIONS

407

−2

ln([A]t – [A]eq)

−2.2
−2.4
−2.6
−2.8
−3
0

50

100

150

Time t / min

Figure 8.22 Kinetic graph for a reversible first-order reaction with the axes for an integrated rate equation ln([A]t − [A](eq) ) (as ‘y’) against time (as ‘x’). The gradient is −5.26 × 10−3 min−1

Substituting values into Equation (8.52) gives
5.26 × 10−3 min−1 = k−1 (10.0 + 1) so k−1 =

5.26 × 10−3 min−1
= 4.78 × 10−4 min−1
11

Next, since we know both K and k−1 , we can calculate k1 . We know from the gradient of the graph that
(k1 + k−1 ) = 5.26 × 10−3 min−1 which, after a little rearranging, gives k1 = 5.26 × 10−3 min−1 − k−1 and, after inserting values k1 = (5.26 × 10−3 − 4.78 × 10−4 ) min−1 = 4.78 × 10−3 min−1 .
In summary, k1 = 4.78 × 10−3 min−1 and k−1 = 4.78 × 10−4 min−1 .

Notice how the ratio
‘k1 ÷ k−1 ’ yields the value of K.

408

CHEMICAL KINETICS

It is wrong (but common) to see a reversible reaction written with a double-headed arrow, as ‘A ↔ B’. Such an arrow implies resonance, e.g. between the two extreme valence-bond structures of
Kekul´ benzene. e 8.5

SAQ 8.23 Consider a reversible first-order reaction. Its integrated rate equation is given by Equation (8.50). People with poor mathematical skills often say (erroneously!) that taking away the infinity reading from both top and bottom is a waste of time because the two infinity concentration terms will cancel. Show that the infinity terms cannot be cancelled in this way; take [A](eq) = 0.4 mol dm−3 , [A]0 = 1 mol dm−3 and [A]t = 0.7 mol dm−3 .

Thermodynamic considerations: activation energy, absolute reaction rates and catalysis
Why prepare a cup of tea with boiling water?
The temperature dependence of reaction rate

The instructions printed on the side of a tea packet say, ‘To make a perfect cup of tea, add boiling water to the tea bag and leave for a minute’. The stipulation for
‘one minute’ suggests the criterion for brewing tea in water at 100 ◦ C is a kinetic requirement. In fact, it reflects the rate at which flavour is extracted from the tea bag and enters the water.
Pure water boils at 100 ◦ C (273.15 K). If the tea is prepared with cooler water, then the time taken to achieve a good cup of tea is
Remember that the longer (and by ‘good’, here, we mean a solution of tea having a temperature of boiling water T(boil) is sufficiently high concentration). If the water is merely tepid, then a itself a function of duration as long as 10 min might be required to make a satisfactory the external prescup of tea; and if the water is cold, then it is possible that the sure, according to the tea will never brew, and will always remain dilute. In summary,
Clausius–Clapeyron
the rate of flavour extraction depends on the temperature because equation (see the rate constant of flavour extraction increases with increasing
Section 5.3). temperature. Why store food in a fridge?
The temperature dependence of rate constants
Food is stored in a fridge to prevent (or slow down) the rate at which it perishes.
Foods such as milk or butter will remain fresh for longer if stored in a fridge, but they decompose or otherwise ‘go off’ more quickly if stored in a warmer environment.
The natural processes that cause food to go bad occur because of enzymes and microbes, which react with the natural constituents of the food, and multiply. When

THERMODYNAMIC CONSIDERATIONS

these biological materials have reached a certain concentration, the tastes bad, and is also likely to be toxic.
The growth of each microbe and enzyme occurs with its own unique rate. A fridge acts by cooling the food in order to slow these rates to a more manageable level. At constant temperature, the rate of each reaction equals the respective rate constant k multiplied by the concentrations of all reacting species. For example, the rate of the reaction causing milk to ‘go off’ occurs between lactic acid and an enzyme. The rate of the process is written formally as

food smells and

The rate constant k2 is truly a constant at a fixed temperature, but can vary significantly.

(8.53)

rate = k2 [lactose][enzyme] where, as usual, the subscripted ‘2’ indicates that the reaction is second order. Neither [lactose] nor [enzyme] will vary with temperature, so any variations in rate caused by cooling must, therefore, arise from changes in k2 as the temperature alters.
The rate constant k2 is truly a constant at a fixed temperature, but can vary significantly: increasing as the temperature increases and decreasing as the temperature decreases. This result explains why rate of reaction depends so strongly on temperature.

409

The rate constant increases as the temperature increases and decreases as the temperature decreases.

Why do the chemical reactions involved in cooking require heating?
Activation energy Ea and the Arrhenius theory
Cooking is an applied form of organic chemistry, since the molecules in the food occur naturally. We heat the food because the ‘Naturally occurring’ reactions occurring in, say, a pie dish require energy; and an oven was the old-fashioned is simply an excellent means of supplying large amounts of energy definition of ‘organic chemistry’, and perover extended periods of time.
The natural ingredients in food are all organic chemicals, and sisted until nearly the it is rare for organic reactions to proceed without an additional end of the 19th cenmeans of energy, which explains why we usually need to reflux a tury. reaction mixture.
It is easy to see why an endothermic reaction requires energy to react – the energy to replace the bonds, etc. must be supplied from the surroundings. But why does an exothermic reaction require additional energy? Why do we need to add any energy, since it surely seeks to lose energy?
At the heart of this form of kinetic theory is the activated complex. In this context, the word ‘activated’ simply means a species brimming with energy, and which will react as soon as possible in order to decrease that energy content.
A reaction can be thought of as a multi-step process: first the reactants approach and then they collide. Only after touching do they react. One of the more useful definitions

410

CHEMICAL KINETICS

of reaction is ‘a rearrangement of bonds’. We are saying that, as a good generalization, the atomic nuclei remain stationary during the reaction while the electrons move. This idea is important, since it is the electrons that act as the ‘glue’ between the nuclei.
Such movement occurs in such a way that the bonds between the atoms are different in the product than in the reactant.
This simple yet profound notion, that atomic nuclei are stationary during the reaction and that only electrons have time to move, is
The Franck–Condon called the Franck–Condon principle. We shall see its important principle states that atomic nuclei are staconsequences later, in Chapter 9. tionary during a reacWe now move on slightly, conceptually. Consider a single pair tion, with only elecof reactant molecules combining to form a product. As electrons trons moving – rearrange as the reaction commences, we pass smoothly from a see p. 451. structure that is purely reactant to one of pure product. The transition from one to the other is seamless; see Figure 8.23.
There will soon come a point where some bonds are almost broken and others almost formed. We have neither reactant nor product: it is a hybrid, being a mixture of both reactant and product. It is extremely unstable, and hence of extremely high energy (i.e. with respect to initial reactants or the eventual products). We call it the transition-state complex, and often give it the initials TS. To a first approximation, the character of the complex is predominantly reactant before the TS is formed, and predominantly product afterwards.
The transition-state complex TS is only ever formed in minute concentrations and for a mere fraction of a second, e.g. 10−12 s, so we do not expect to ‘see’ it except by the most sophisticated of spectroscopic techniques, such as laser flash photolysis.
Forming the TS is like pushing a marble over a large termite hill: most of the marbles cannot ascend the slope and, however high they rise up the hill’s slope, they do not ascend as far as the summit. Those rare marbles that do reach the summit appear to stay immobile for a mere moment in time, and are then propelled by their own momentum (and their own instability, in terms of potential energy) down over the termite hill and onto the other side. A chemical reaction is energetically similar: the reaction commences when molecules of reactant collide with ‘sufficient energy’.
If sufficient energy is available, then the two or more reactants join to form the transition-state complex TS, i.e. the electrons rearrange with the net results that, in effect, atoms or groups of atoms move their positions (the bonds change).
A

B

A

B

A
+

+
C
D
Reactants

B

C

D

Transition-state complex

C

D

Products

Figure 8.23 During a reaction, the participating species approach, collide and then interact. A seamless transition exists between pure reactants and pure products. The rearrangement of electrons requires large amounts of energy, which is lost as product forms. The highest energy on the activation energy graph corresponds to the formation of the transition-state complex. The relative magnitudes of the bond orders are indicated by the heaviness of the lines

THERMODYNAMIC CONSIDERATIONS

411

TS

Energy

Ea

DH(reaction)

Products

Reactants

Reaction coordinate

Figure 8.24 Reaction profile of energy (as ‘y’) against reaction coordinate (as ‘x’). The activation energy Ea is obtained as the vertical difference between the reactants and the peak of the graph, at ‘TS’

Figure 8.24 shows a graph of energy as a function of reaction progress. The transition complex is formed at the energy maximum. The figure will remind us of
Figures 3.1 and 3.2, except with the peak on top. It is similar. The enthalpy of reaction is obtained as the vertical difference between ‘REACTANTS’ and ‘PRODUCTS’ on the graph.
The ‘sufficient energy’ we mentioned as needed to form the transition-state complex is termed the activation energy, which is The activation energy given the symbol Ea (with the E denoting ‘energy’ and the sub- Ea is always posiscripted ‘a’ for ‘activation’). The word ‘activation’ ought to suggest tive, so the formation additional energy is required; in fact, the activation energy is always of a transition-state positive because the TS is always higher in energy than the reac- complex is always endothermic. tants. Stated another way, its formation is always endothermic.

Why does a reaction speed up at higher temperature?
The Arrhenius equation
In Chapters 1 and 2, we met the idea that the simplest way to increase the energy of a chemical, material or body is to raise its temperature. So heating a reaction mixture gives more energy to its molecules. Although only a tiny proportion of these molecules will ever have sufficient energy to collide successfully and form an activated complex TS, even a small increase in the amounts of energy possessed by a molecule will increase the proportion that

The simplest way to increase the energy of a chemical, material or body is to raise its temperature – see p. 34.

412

CHEMICAL KINETICS

have sufficient energy to form the TS. Therefore, heating a reaction mixture, e.g. by reflux, increases the number of successful collisions between reactant species, increasing the amount of product formed per unit time.
By increasing the temperature T , we have not changed the magnitude of the activation energy, nor have we changed the value of H of reaction. The increased rate is a kinetic result: we have enhanced the number of successful reaction collisions per unit time.
The simplest relationship between temperature T and rate constant k is given by the Arrhenius equation (Equation (8.54)), which relates the rate constant of reaction k with the thermodynamic temperature T at which the reaction is performed:

Heating a reaction mixture increases the number of successful collisions between the reactant species, increasing the amount of product formed per unit time.

k = A exp −

The Arrhenius equation is written in terms of thermodynamic temperature.

Ea
RT

where R is the gas constant and Ea is the activation energy (above), which is a constant for any particular reaction. T is the thermodynamic temperature (in kelvin), and A is called usually called the
Arrhenius ‘pre-exponential’ factor. The value of Ea depends on the reaction being studied.
The logarithmic form of Equation (8.54) ln k = ln A −

The activation energy is obtained as ‘−1 × R × gradient’, where ‘gradient’ refers to the slope of the Arrhenius plot.

(8.54)

Ea
RT

reminds us of the equation of a straight line, ‘y = c + mx’, so a plot of ln(rate constant) (as ‘y’) against 1/T (as ‘x’) will yield a straight-line graph of gradient ‘−Ea ÷ R’. In practice, we repeatedly perform the experiment to determine a value of its rate constant k, each determined at a single value of T .
We should not attempt to memorize the Arrhenius equation until we can ‘read’ it, and have satisfied ourselves that it is reasonable. Firstly, we note that R, Ea and the ‘constant’ term will not vary. We are, therefore, looking at the effect of
T on k. Next, we see that the first term on the right-hand side of the logarithmic form of Equation (8.54) decreases as the temperature increases and so the logarithmic term on the left must also decrease. However, since there is a minus sign on the right-hand side of the equation, we are saying that as T increases, so the right-hand side becomes less negative (more positive). In other words, as the temperature increases, so the logarithm of the rate constant also increases, and hence k gets larger.
Worked Example 8.20 Consider the following data that relate to the rate of removing a naturally occurring protein with bleach on a kitchen surface. What is the activation

THERMODYNAMIC CONSIDERATIONS

413

2

1.8

1.6

In k

1.4

1.2

1

0.8

0.6
0.003

0.0031

0.0032
1/T ÷ 1/K

0.0033

0.0034

Figure 8.25 An Arrhenius plot of ln(rate constant) (as ‘y’) against 1/T (as ‘x’). The data relate to the rate of removing a naturally occurring protein with bleach on a kitchen surface

energy of reaction?
Temperature T / ◦ C
Rate constant k/s−1

20
2.20

30
2.89

40
3.72

50
4.72

Answer Strategy: before we can plot anything, we first convert the temperatures from Celsius to thermodynamic temperatures in kelvin.
Then we plot ln(k/s−1 ) (as ‘y’) against 1/T (as ‘x’).
Such a plot is seen in Figure 8.25. Its gradient is equal to −Ea /R.
This graph is seen to be linear, with a gradient of −2400