...CHAPTER 14 CHEMICAL KINETICS PRACTICE EXAMPLES 1A (E) The rate of consumption for a reactant is expressed as the negative of the change in molarity divided by the time interval. The rate of reaction is expressed as the rate of consumption of a reactant or production of a product divided by its stoichiometric coefficient. A 0.3187 M 0.3629 M 1min rate of consumption of A = = = 8.93 105 M s 1 t 8.25 min 60 s rate of reaction = rate of consumption of A2 = 8.93 105 M s 1 4.46 105 M s 1 2 1B (E) We use the rate of reaction of A to determine the rate of formation of B, noting from the balanced equation that 3 moles of B form (+3 moles B) when 2 moles of A react (–2 moles A). (Recall that “M” means “moles per liter.”) 0.5522 M A 0.5684 M A 3moles B rate of B formation= 1.62 104 M s 1 60s 2 moles A 2.50 min 1min 2A (M) (a) The 2400-s tangent line intersects the 1200-s vertical line at 0.75 M and reaches 0 M at 3500 s. The slope of that tangent line is thus 0 M 0.75 M slope = = 3.3 104 M s 1 = instantaneous rate of reaction 3500 s 1200 s The instantaneous rate of reaction = 3.3 104 M s 1 . (b) At 2400 s, H 2 O 2 = 0.39 M. At 2450 s, H 2 O 2 = 0.39 M + rate t At 2450 s, H 2 O 2 = 0.39 M + 3.3 10 4 mol H 2 O 2 L1s 1 50s = 0.39 M 0.017 M = 0.37 M 2B (M) With only the data of Table 14.2 we can use only the reaction rate during the first...
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...Determination of the Reaction Rate of Mg and HCl Alex Skartvedt Alex Skartvedt, Melanie Ho, Christina Olson, Thor Swerdykiak November 11, 2015 Fall 2015 Semester Abstract: The experimental rate law of the reaction: 2HCl(aq)+Mg(s)→MgCl2(aq)+H2(g) is being determined because kinetics play a large role in the reactions happening in everyday life. The rate law, k value will be obtained in the experiment. To obtain the information a pressure probe was connected to an Erlenmeyer flask in order to record the rate of the reaction of magnesium ribbon and hydrochloric acid. The order of HCl and Mg were then determined using the information found in from the reaction. The order in HCl and Mg were found to be less than the expected...
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...CAcT HomePage Rate and Order of Reactions Skills to develop Derive the integrated rate laws from differential rate laws. Describe the variation of concentration vs. time for 1st order reactions. Describe the variation of concentration vs. time for 2nd order reactions. Figure out order of reaction from concentration vs. time plots. Rate and Order of Reactions The rate of a chemical reaction is the amount of substance reacted or produced per unit time. The rate law is an expression indicating how the rate depends on the concentrations of the reactants and catalysts. The power of the concentration in the rate law expression is called the order with respect to the reactant or catalyst. This page deals specifically with first- and second-order reaction kinetics, but you should know that other orders such as zeroth-order and 3rd order may also be involved in chemical kinetics. Reaction Rates and Stoichiometry In acidic solutions, hydrogen peroxide and iodide ion react according to the equation: H2O2 + 2H+ + 3I- = 2 H2O + I3-. In this reaction, the reaction Rate can be expressed as decreasing Rate of H2O2, - d[H2O2]/dt decreasing Rate of H+, - d[H+]/dt decreasing Rate of I-, - d[I-]/dt increasing Rate of H2O, + d[H2O] /dt increasing Rate of I3-, d[I3-]/dt However, from the stoichiometry, you can easily see the following relationship: d[H2O2] 1 d[H+] 1 d[I-] 1 d[H2O] d[I3-] - ------- = - - ------ = - - ------- = - ------ = -------- ...
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...A Kinetics Experiment The Rate of a Chemical Reaction: A Clock Reaction Andrea Deoudes February 2, 2010 Introduction: The rates of chemical reactions and the ability to control those rates are crucial aspects of life. Chemical kinetics is the study of the rates at which chemical reactions occur, the factors that affect the speed of reactions, and the mechanisms by which reactions proceed. The reaction rate depends on the reactants, the concentrations of the reactants, the temperature at which the reaction takes place, and any catalysts or inhibitors that affect the reaction. If a chemical reaction has a fast rate, a large portion of the molecules react to form products in a given time period. If a chemical reaction has a slow rate, a small portion of molecules react to form products in a given time period. This experiment studied the kinetics of a reaction between an iodide ion (I-1) and a peroxydisulfate ion (S2O8-2) in the first reaction: 2I-1 + S2O8-2 I2 + 2SO4-2. This is a relatively slow reaction. The reaction rate is dependent on the concentrations of the reactants, following the rate law: Rate = k[I-1]m[S2O8-2]n. In order to study the kinetics of this reaction, or any reaction, there must be an experimental way to measure the concentration of at least one of the reactants or products as a function of time. This was done in this experiment using a second reaction, 2S2O3-2 + I2 S4O6-2 + 2I-1, which occurred simultaneously with the reaction under investigation. Adding...
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...Ch. 15: Kinetics 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9 Week 15 Reaction Rates Rate Laws Forms of Rate Laws Integrated Rate Law Rate Laws: A Summary Reaction Mechanisms Steady-State Approximation A Model for Chemical Kinetics Catalysis CHEM 1310 - Sections L and M 1 Reaction Rates: Example Recap: Avg. Reaction rate = +∆[Product]/∆t Avg. Reaction rate = -∆[Reactant]/∆t Butane, C4 H 10, burns in oxygen to give CO2 and H2 O vapor. If the butane concentration is decreasing at a rate of 0.20 M/s, what is the rate at which the oxygen is decreasing? What is the rate at which the products are increasing? 2C4H10(g) + 13O2(g) Week 15 8CO2(g) + 10H2O(g) 2 CHEM 1310 - Sections L and M Reaction Rates: Example Butane, C4 H 10, burns in oxygen to give CO2 and H2 O vapor. If the butane concentration is decreasing at a rate of 0.20 M/s, what is the rate at which the oxygen is decreasing? What is the rate at which the products are increasing? 2C4H10(g) + 13O2(g) 0.20 M/s For oxygen: 0.20 mol C4 H 10 Ls Week 15 8CO2(g) + 10H2O(g) ? ? ? x 13 mol O2 2 mol C4 H 10 CHEM 1310 - Sections L and M = 1.3 M O2 /s 3 1 Reaction Rates: Example Butane, C4 H 10, burns in oxygen to give CO2 and H2 O vapor. If the butane concentration is decreasing at a rate of 0.20 M/s, what is the rate at which the oxygen is decreasing? What is the rate at which the products are increasing? 2C4H10(g) + 13O2(g) 0.20 M/s For CO2: 0.20 mol...
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...frequently occurs many times, and in some cases many thousands of times per second. • The miracle of life: a myriad chemical reactions in the cell occur simultaneously with great accuracy and at astonishing speed. • Without the proper enzymes to process the food you eat, it might take you 50 years to digest your breakfast. • Catalysis is probably the most important function of proteins. • Catalysts that serve this function in organism are called enzymes. • With the exception of some recently discovered RNAs that have catalytic activity, all enzymes are proteins. • Enzymes are the most efficient specific catalysts known (increase rate reaction up to 10²º). • Nonenzymatic catalysts (up to 10²-104). • Enzymes are highly specific,even to the point of distinguishing stereoisomers of a given compound. 1)Enzymes are Effective Biological Catalysts 2)Difference between Kinetic and Thermodynamic Aspects of Reactions • The rate of the reaction and its thermodynamic favorability are two different topics although so related. • The standard free energy change (ΔG°) is the difference between the energies of the reactants (initial state) and the energies of the products (final state). • Enzymes like all catalysts speed up reaction but cannot alter the equilibrium constant of ΔG°. • Reaction rate depends on activation energy ΔG°‡ (energy input required to initiate the reaction). • Uncatalyzed reaction require more energy to get started,so its rate is slower than that...
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...clear burnout temperature. The first peak at the temperature of 100℃ is observed due to the evaporation of moisture and volatilization of small organic molecules deposited onto rice husk surfaces. The second peak with a maximum weight loss rate of 12.56%/min at temperature of 296℃results from continuous evaporation of moisture and loss of volatiles contained in rice husk. The third peak occurs between 420℃ and 590℃, with a maximum weight loss rate at 505℃. The results indicate that char gasification, loss of volatiles, and fixed carbon and combustion of hydrocarbons are included. It is possible because rice husk is made of various hydrocarbons species with a wide range in boiling point. TG and DTG curves show that the combustion process of sewage sludge is very different from rice husk. The shapes of peaks for DTG profiles of sewage sludge are not sharp. It is due to weight loss of sewage sludge shows a slow process and higher ash content during combustion. The first peak near the temperature of 100℃ is observed due to the evaporation of moisture. The second peak with a maximum weight loss rate of 2.24%/min at...
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...David Lipscomb College I II The Methclnolysir of Acetal A chemical kinetics a n d g a r chromatography experiment Nashville, Tennessee The methanolysis of acetal and gas chromatography complement perfectly to provide a rare and interesting experiment for the physical chemistry student. The reaction of acetal with methanol is a consecutive reaction in which diethyl acetal (aretal) replaces its two ethoxy groups with methoxy groups forming methy1et)hylacetal first and dimethyl acetal in the second step. Investigating the reaction kinetics for this type of reaction mas almost impossible before the advent of gas chromatography (I). There are many advantages to using the methanolysis of acetal in this study: Sufficient data for a complete analysis can be collected within three hours. The reaction requires reagents which are readily available in high purity. The reactants, intermediate, and products are liquids at temperatures easily accessible for kinetic runs. The primary reactant, intermediate, and product are sepe rated rleanly on a typical gas chromatographic column. The peak heights are linear with concentration in the range studied (2). Analysis of the data affords an application of firshorder kin~ticsand the steady state hypothesis. The experiment presented herc is an adaptation of the detailed study of the reaction by Juvet and Chiu (2), who have shown it to be acid catalyzed and first order in diethyl acetal and methylethyl acetal. By decreasing the elution time...
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...Introduction: In this experiment, the rate law and rate constant for a reaction of oxalic acid with permanganate were determined. Every chemical reaction ranges from hours to femtoseconds to occur. The rate of a chemical reaction can be determined by assessing the change in either the reactant or product in a given time or point (Connors, 1990). It is important to take account of the stoichiometric ratios of each component, regardless as to which compound is chosen to determine the rate. The study of the rates of chemical reactions is called Chemical Kinetics (Soustelle, 2011). The following is an example of a generic chemical reaction: aA + bB → cC + dD The rate of the reaction can be expressed as a function of each reactant and each product: Rate = -1a∆A∆t= -1b∆B∆t= +1c∆C∆t= +1d∆D∆t For each reaction, the concentration is decreasing as the reaction occurs; therefore representing the rate as a positive value and a negative sign in the rate definition is used. The concentration of the products is conversely increasing as the reaction proceeds; therefore changing in concentration is positive (Connors, 1990). The reaction rate can be measured by accurately measuring the change in concentration of one of the reaction species over time. This experiment utilized a visual change in color of one of the reactants used; however, since visual indications can be subjective and it is important to perform the experiment multiple times. Also, since the color change would indicate that...
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...need be understood for the reason that it is needed to understand later concepts. There are six players on the court and each one has a designated position. For example there are; left-side hitters, middle hitters, right-side hitters, and the same pattern follows for the defensive players. However, because there is a set rotation, a player will not always begin in their set position and they have to move to that place. This is called displacement. Displacement is defined as the change in position of an object ad can be either positive or negative depending on the starting point and the direction. The equation to represent displacement is d=d2-d1.D is the total movement between positions. The first floor position would be d1 and the second would be d2. Beyond the fact that players have frequent displacement, the ball itself has displacement. In fact this happens constantly as it moves from one player to another and from one side of the net to the other. Now that the ideas of displacement are understood velocity and acceleration can be found. Average velocity and average acceleration are complimentary of each other and are both important to the concept of volleyball. Velocity is defined as displacement in a given time. Once can find velocity by dividing distance by time. So the equation for velocity is:...
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...Michaelis-Menten Kinetics: The Enzyme-Catalyzed Hydrolysis of Sucrose1 Background Of the numerous chemical reactions whose kinetics have been studied, some of the most interesting and complex are those which take place in living organisms. These reactions, when studied under non-biological conditions (lack of organisms and their various biomolecules), are often exceptionally slow: often hours or days, and sometimes even years are required before equilibrium is reached. Yet life requires seconds or even fractions of seconds to produce the same results, and it does this through the use of biological catalysts called enzymes. Enzymes are proteins with special clefts or pockets in their structures in which the reactions occur: the tremendous advantage that enzymes have is that these pockets, by nature of their shape and chemical environment, force the reactants to form transition states much lower in energy than the transitions states the reactants would form under other circumstances, say in solution. Thus, life can subsist at a reasonable pace and it does not take us a week to walk across a room or digest an apple. Of the many kinetic models for enzymes currently used, the first (and simplest) developed is the Michaelis-Menten Model, developed in 1913 by Leonor Michaelis and Maud Menten, which is still quite useful today. One of its early triumphs was the successful explanation of the kinetics of sucrase, or β-fructofuranidase, an enzyme found in yeast which cleaves...
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...After it was first discovered in 1886 by Hans Heinrich Landolt, the iodine clock reaction has become one of the most used classroom experiments for demonstrating kinetics (SuperchargedScience, 2018). A clock reaction is a reaction where one of the chemical species (clock chemical) starts off with a low concentration and then has a rapid increase which can then lead to a dramatic colour change (UnitverityNottingham, 1999). The iodine clock works by using iodine as the clock chemical that then reacts with clear starch to produce a deep blue colour. For the Landolt chemical clock in two clear liquids are mixed with one containing KIO3 and the other NaHSO4 they then follow the following chemical reactions. Equation 1 is known as the rate determining...
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...Unit: Reaction Rates This Unit Activity will help you meet these educational goals: • Science Inquiry—You will explore the ozone layer by doing online research and collecting information. You will then communicate your results in written form. • STEM—You will use your knowledge of kinetics to analyze scientific investigations. • 21st Century Skills—You will employ online tools for research and analysis. Introduction You have probably heard about the ozone layer and/or the hole in the ozone layer. Ozone is an allotrope of oxygen (O3) which is a very important part of our lives. In this activity, you will learn more about ozone and how it is formed. You will also investigate what causes the ozone layer to be depleted over time in the upper atmosphere. ________________________________________________________________________________ Directions and Analysis Task 1: The Ozone Layer Read about the ozone layer and answer the following questions. You can use the key word “ozone” or “ozone layer” in a search engine to learn more about it. Key search terms: ozone, ozone layer, benefits of ozone 1. Describe a few of the properties of ozone and compare these to the properties of diatomic oxygen. Type your response here: 2. Describe the process by which ozone can be formed in large cities. Type your response here: 3. Even though ozone may be dangerous to our health here on Earth, how can it be beneficial in the upper atmosphere...
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...activation energy required for a chemical reaction to occur and therefore increases the rate of the reaction. Activation Energy is the minimum energy barrier needed to be overcome before a reaction can occur by providing an alternative reaction pathway. The beneficial aspect of enzymes is that they are extremely efficient and may be used repeatedly. One enzyme may be used to catalyze thousands of reactions every second. The two factors that affect the efficiency of how enzymes...
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...BIOL 3380 Name:_____________________________________ Circle Session: T-PM W-AM W-PM R-AM R-PM F-AM F-PM Experiment 9 – Pre-lab Homework Enzyme Kinetics of LDH This pre-lab homework assignment is due at the beginning of your lab session. You are provided with the following portion of a protocol: • Determine concentration of enzyme stock solution, if unknown, by taking an A280 nm reading of a 1:100 dilution (in water). Use a total volume of 1 ml in the cuvette. • Dilute some of the enzyme stock with buffer A to make a 4 mg/ml solution. • Serially dilute the 4 mg/ml solution with buffer A to make working solutions of 400 µg/ml and 40 µg/ml. • Prepare 30 µl of each working solution for every sample The PI of the lab gives you a tube of enzyme and tells you the following before disappearing into the office to write more grant proposals: ➢ There is 50 µl of enzyme stock solution. The enzyme is expensive to purify, so follow the protocol exactly, using as little of the stock solution as possible. ➢ The concentration of the stock solution is currently not known, but a 1 mg/ml concentration of the pure enzyme has an A280 nm of 2.0. ➢ You’ll be performing the assay on 12 samples. ➢ Make enough of each working solution so that you have at least 400 ul to work with when you do the assay (to cover any waste and/or inefficiencies in pippetting). Using the spectrophotometer to read the absorbance at 280 nm, you get...
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