...Terms Hemoglobin Fetal Hemoglobin Dynamic Equilibrium – when rates of forward and reverse reactions are equal, rendering static concentration(s) of reactants and products Kc – Equilibrium constant with respect to concentration in molarity (M) Kp – Equilibrium constant with respect to partial pressures in atmospheres Equilibrium constant – ratio of concentration of products raised to their stoichiometric coefficients divided by reactants raised to their stoichiometric coefficients (reaction quotient) Heterogeneous equilibrium Reaction quotient Le Châtelier’s Principle Concepts Law of Mass Action when aA+bB↔cC+dD Kc=CcDdAaBb when A↔2B, K1=B2A when 2B↔3C, K2=C3B2 A↔2B2B↔3CA↔3C Koverall=C3A nA+2nB↔3nC therefore K'=C3nAnB2n=Kn Heterogeneous reactions – concentrations of pure solids (s) are constant; both pure solids and pure liquids (l) are excluded from equilibrium expression Calculations Equilibrium constant K=CcDdAaBb Equilibrium equation A+2B↔3C, therefore… Kforward=C3AB2 Kreverse=AB2C3 Kreverse=1Kforward Kc=CcDdAaBb PA=ART eq 14.1 Kp=Kc(RT)∆n eq. 14.2 Key Tables/Figures: Table 14.1 Initial and Equilibrium Concentrations for the Reaction H2g+I2g↔2HIg at 445°C Initial | Equilibrium | Equilibrium Constant | H2 H2 | | I2 I2 | | HI HI | | 0.50 | 0.50 | 0.0 | 0.0 | 0.0 | 0.50 | 0.50 | 0.50 | 0.50 | 1.0 | 0.50 | 0.0 | 0.50 | 1.0 | 0.0 | | H2 H2 | | I2 I2 | | HI HI | | 0.11 | 0.11 | 0.78 | 0.055 | 0.055 | 0.39 | 0.165 |...
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...Chemical Equilibrium in Solution 2/5/13 Jim Jacobs (Partner: Jessica Janiga) Abstract The purpose of this experiment was to determine the equilibrium established between iodine species [I3-, I2, and I-] dissolved in an aqueous KI solution. The values obtained for equilibrium were as follows: Kc (run#1) = 620; Kc (run#2) = 599; Kc (run#3) = 545. In effort to distinguish between the species of iodine the distribution coefficient ratio between two immiscible solvents (dichloromethane and water) was determined at multiple concentrations. I. Introduction The purpose of this experiment is to determine the equilibrium constant of iodine species at various concentrations. The equilibrium equation of the iodine species is as follows: I2+I_=I3- (1) The equilibrium constant is calculated as: KC=(I3-)I2I- (2) Where the equilibrium constant KC is calculated from the concentrations of the species and does not take into consideration the activities of the species. Knowing Kc will indicate if the reaction favors products (if Kc>1) or if it favors products (Kc<1). Sodium thiosulfate is used to determine the concentration of Iodine in solution via titration in the following reaction: 2S2O32-+ I3- → S4O62-+ 3I- Before titration excess KI is added to each solution to ensure that I2 is converted completely to I3-. This is done to ensure that molecular iodine is not loss to evaporation while in the aqueous phase (I3- is non-volatile). However...
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...185 EQUILIBRIUM 185 UNIT 7 EQUILIBRIUM After studying this unit you will be able to • identify dynamic nature of equilibrium involved in physical and chemical processes; • state the law of equilibrium; • explain characteristics of equilibria involved in physical and chemical processes; • write expressions for equilibrium constants; • establish a relationship between Kp and K c ; • explain various factors that affect the equilibrium state of a reaction; • classify substances as acids or bases according to Arrhenius, Bronsted-Lowry and Lewis concepts; • classify acids and bases as weak or strong in terms of their ionization constants; • explain the dependence of degree of ionization on concentration of the electrolyte and that of the common ion; • describe pH scale for representing hydrogen ion concentration; • explain ionisation of water and its duel role as acid and base; • describe ionic product (Kw ) and pKw for water; • appreciate use of buffer solutions; • calculate solubility product constant. Chemical equilibria are important in numerous biological and environmental processes. For example, equilibria involving O2 molecules and the protein hemoglobin play a crucial role in the transport and delivery of O2 from our lungs to our muscles. Similar equilibria involving CO molecules and hemoglobin account for the toxicity of CO. When a liquid evaporates in a closed container, molecules with relatively higher kinetic energy escape the liquid surface into the vapour...
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...Determining the equilibrium constant from pH values of changing concentrations of ethanoic acid Design Research question: How will altering the concentrations of ethanoic acid affect the pH value, and, in-turn, the equilibrium constant? Background information: When weak acids react, the reaction typically does not go to completion. Rather, the system goes to an intermediate state in which the rates of the forward and reverse reactions are equal. Such a system is said to be in chemical equilibrium. When equilibrium is reached, the reactants and the products have concentrations which do not change with time. When in equilibrium at a particular temperature, a reaction mixture obeys the Law of Chemical Equilibrium, which imposes a condition on the concentrations of reactants and products. This condition is expressed in the equilibrium constant Kc for the reaction. In this experiment, I will study the equilibrium properties of the reaction between ethanoic acid, otherwise known as acetic acid (CH3COOH) and water (H2O): CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) When solutions containing ethanoic acid and water are mixed, they react to some extent, forming CH3COO- and H3O+. As a result of the reaction, the equilibrium amounts of CH3COOH and H2O will be less; for every mole of CH3COO- formed, one mole of CH3COOH and one mole of H2O will react. The equilibrium constant expression Kc for Reaction 1 is: Kc = [CH3COOH] / [H3O+][CH3COO-] The value of...
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...Effect of a factor on the activity of an enzyme Research Question: How does pH effect the activity of the enzyme amylase and the hydrolysis of starch? Hypothesis: Amylase is an enzyme that acts on starch. pH is capable of altering the structure of the active site centre in the enzyme leading to denaturation. At each pH, the enzyme activity would be relatively different. Ideally the optimum pH is 6.0 when the enzyme works best and the fastest, however if the pH is higher or lower the hydrolysis of starch will be slower. Changing the pH above the optimum will sometimes lead to denaturation of the enzyme and the change in shape of the active site. (http://www.nuffieldfoundation.org/practical-biology/investigating-effect-ph-amylase-activity) Variables: Independent: The pH of the buffer solution. Dependent: The time taken for the solution to turn an opaque (lighter) blue from dark blue. Controlled: The volume of iodine solution; the volume of starch solution; the volume of buffer solutions; the volume of amylase solution; the temperature at 40°c; the time it is left in the water bath; and the concentration of the solutions. I will be using a water bath to keep the temperature constant, and to manipulate my independent variable, I will be using pH 4, 5, 7, 9, 10 by adding different pH buffer solutions with the amylase. * 50ml Amylase solution * 100ml Starch solution * 100ml Buffer solutions with pH 4, 5, 7, 9, 10 * 20ml Iodine solution Apparatus: ...
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...Abstract: Introduction: Materials: 12 x 5 cm3 plastic tubes 2 x 5 cm3 graduated pipette 1 x 20 cm3 of stock NADH (80 g cm-3) 1 x 20 cm3 of 50 mM phosphate buffer pH 7.2 2 x semi micro cuvettes Distilled water wash bottle Access to a Whirly mixer Access to Novaspec digital spectrophotometer 1 x 5 cm3 of a solution of NADH of unknown concentration. Method: 1. Using the protocol depicted in Table 1 set up 6 standard concentrations of NADH in buffer ranging from 0 to 80 µg cm-3. Table 1: Protocol for NADH standard curve. Tube Number. | 1 | 2 | 3 | 4 | 5 | 6 | Volume of Stock NADH (cm3 ) | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | Volume of Buffer (cm3 ) | 2.5 | 2.0 | 1.5 | 1.0 | 0.5 | 0 | 2. When you have prepared 1 to 6, mix the contents of each tube with a Whirlymixer. 3. Set up and "blank" the Novaspec with buffer ready to read absorbance at 340 nm (see Introduction, page 3). If the spectrophotometer is not set on 340nm, ask a demonstrator to help. 4. Read the absorbances of each tube at 340 nm using buffer as the blank. 5. Record the A340 readings for each series of tubes in Table 6. Calculate the concentration of NADH in each tube by using the following : 7. Enter the concentrations you have calculated into Table 2 8. Plot the mean A340 against the concentration of NADH on the graph paper to produce a calibration curve. NB you should get a straight line that passes through the origin and the point for tube 6 Results: ...
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...molecule of nitrogen to 3 of hydrogen. That is the proportion demanded by the equation. In some reactions you might choose to use an excess of one of the reactants. You would do this if it is particularly important to use up as much as possible of the other reactant - if, for example, it was much more expensive. That doesn't apply in this case. There is always a down-side to using anything other than the equation proportions. If you have an excess of one reactant there will be molecules passing through the reactor which can't possibly react because there isn't anything for them to react with. This wastes reactor space - particularly space on the surface of the catalyst. The temperature Equilibrium considerations You need to shift the position of the equilibrium as far as possible to the right in order to produce the maximum possible amount of...
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...CHEMICAL EQUILIBRIUM Equilibrium occurs when the rate of the forward reaction is equal to the rate of reverse reaction. Concentration of the reactants and the products remain the same. Dynamic equilibrium occurs when the processes do not stop and forward and reverse reactions still continue in both directions at the same rate. A homogeneous reaction is a reaction wherein all the reactants and products are in the same phase e.g. all are gases. A heterogeneous reaction is a reaction wherein all the reactants and products are not in the same phase. If there is no interaction between the system and its surroundings, i.e., a system from which no substances in the reaction can escape, we say it is a closed system. Calculation of the substance amount and concentration at equilibrium We can determine the concentration of the substance by dividing the number of mol of the substance at that stage of the reaction by the volume of the container in dm3, i.e.: c= n/V where c or [ ] = concentration of the substance in mol.dm-3 n = number of mol of the substance V = volume of the container in dm3 It is handy to use a table to determine these values. Example 1 A mixture of 5 mol H2(g) and 6 mol I2(g) are placed in a sealed container of 2 dm3 at a temperature of 4580C. The balanced chemical equation for the reaction is: Equilibrium is reached after a certain time. At equilibrium there is 4mol HI(g) in the container. Calculate the concentrations...
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...SCH 4U1 | Equilibrium and Le Châtelier’s Principle | | Introduction: Chemical equilibrium is the state of a reaction in which all reactants and products have reached constant concentrations in a closed system (DiGiuseppe, Haberer, Salciccioli, Sanader, & Vavitsas, 2012, p. 420). Chemical reactions will occur until the reaction reaches a point where the concentrations of the products and reactants become constant. Le Châtelier's principle states that chemical systems at equilibrium shift in the direction that opposes the change when a change occurs that disturbs the equilibrium (DiGiuseppe, Haberer, Salciccioli, Sanader, & Vavitsas, 2012, p. 439). In the lab activity performed, the effects of changing temperature and volume of a system were observed and recorded for the endothermic reaction of dinitrogen tetroxide with 57.2 kJ of energy: N2O4(g) + 57.2 kJ ↔2 NO2(g) Colourless Reddish-Brown As well, the effects of modifying the concentrations of the reactants and products of a system were observed and recorded for the reaction of iron (II) thiocyanide: Fe3+(aq)+SCN1-(aq)↔Fe(SCN)2+(aq) In each case, as our hypotheses predicted, the reactions shifted in the direction that restored equilibrium to the system. Discussion: Concentration: The effects that would be observed on a system at equilibrium all depend on whether products or reactants are added or removed. If products or reactants are added, the equilibrium shifts in the direction...
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...The equilibrium between nitrogen dioxide, NO2, and dinitrogen tetroxide, N2O4, is shown below. 2NO2(g) [pic] N2O4(g) Kc = 0.01 What happens when the volume of a mixture at equilibrium is decreased at a constant temperature? I. The value of Kc increases II. More N2O4 is formed III. The ratio of [pic] decreases A. I and II only B. I and III only C. II and III only D. I, II and III (Total 1 mark) 2. Which statement about chemical equilibria implies they are dynamic? A. The position of equilibrium constantly changes. B. The rates of forward and backward reactions change. C. The reactants and products continue to react. D. The concentrations of the reactants and products continue to change. (Total 1 mark) 3. Methanol may be produced by the exothermic reaction of carbon monoxide gas and hydrogen gas. CO(g) + 2H2(g) [pic] CH3OH(g) ∆HO = –103 kJ (a) State the equilibrium constant expression, Kc, for the production of methanol. ...................................................................................................................................... ...................................................................................................................................... (1) (b) State and explain the effect of changing the following conditions on the amount of methanol present at equilibrium: ...
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...LAB Report 14 The Determination of an Equilibrium Constant Zengping Li Introduction: Our objective is to determine the equilibrium constant for the reaction Fe3++SCN- = FeSCN2+ and too see if the equilibrium is same or not under different conditions. Equilibrium constant can tell us which direction the reaction shift to. If eq less than 1, the reaction shift to left. If eq equal to zero, the reaction is equal. If the eq greater than 1, the reaction shift to right. For beaker 1 the M of FeSCN is mole of FeSCN/ total volume = volume of SCN/1000X0.002/0.050=0.00016 Breaker number FeSCN 1 0.00016 2 0.00012 3 0.00008 4 0.00004 Procedure: First, mixing the known amount of Fe(NO3)3 in to known amount of SCN.( assume they react compeletely). Second ,fill the cuvette with the FeSCN mixture that we just got, place it in the spectrometer and start the data collection. Then the absorbance of the mixture will show on the spectrophotometer. With it we can calculate the equilibrium constant that we are looking for. Discussion : Data table Part I: Temperature: 21.9C FeSCN Absorbance 1 0.00016 0.603318 2 0.00012 0.449558 3 0.00008 0.295798 4 0.00004 0.142038 Y=mx+b we have m=3844 and b=-0.011722 and we are looking for absorbance y For beaker 1: y= 3844x0.00016+(-0.011722)=0.603318 Linear Regression equation: y=mx+b PART II Beaker Absorbance FeSCN at eq A 0.253 0.00006277 B 0.356 ...
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...following information (at 298 K): 2 NO(g) + Br2(g) 2 NOBr(g) Kc = 2.0 4 NO(g) 2 N2(g) + 2 O2(g) Kc = 2.1*108 5. Given the elementary reactions: HF(aq) H+(aq) + F-(aq) Kc = 6.8*10-4 H2C2O4(aq) 2 H+(aq) + C2O42-(aq) Kc = 3.8*10-6 Determine the value of the Kc for the reaction 2 HF(aq) + C2O42-(aq) 2 F-(aq) + H2C2O4(aq) 6. At 2000 C the equilibrium constant for the reaction: 2 NO(g) N2(g) + O2(g) is Kc = 2.4* 103. If the initial concentration of NO is 0.175 M, what are the equilibrium concentrations of NO, N2, and O2? 7. At 100 C the equilibrium constant for the reaction COCl2(g) CO(g) + Cl2(g) has the value Kc = 2.19 * 10-10. Are the following mixtures of COCl2, CO, and Cl2 at 100 C at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium. (a) [COCl2] = 2.00 * 10-3 M . [CO] = 3.3 * 10-6 M, [Cl2] = 6.62 * 10-6 (b) [COCl2] = 0.0100 M , [CO] = [Cl2] = 1.48 * 10-6 M 8. Consider the following equilibrium: 2 SO2(g) + O2(g) ↔ 2 SO3(g) for which ∆H < 0:...
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...list of Frequently Used Symbols and Notation A text such as Intermediate Financial Theory is, by nature, relatively notation intensive. We have adopted a strategy to minimize the notational burden within each individual chapter at the cost of being, at times, inconsistent in our use of symbols across chapters. We list here a set of symbols regularly used with their specific meaning. At times, however, we have found it more practical to use some of the listed symbols to represent a different concept. In other instances, clarity required making the symbolic representation more precise (e.g., by being more specific as to the time dimension of an interest rate). Roman Alphabet a Amount invested in the risky asset; in Chapter 14, fraction of wealth invested in the risky asset or portfolio AT Transpose of the matrix (or vector)A c Consumption; in Chapter 14 only, consumption is represented by C, while c represents ln C ck Consumption of agent k in state of nature θ θ CE Certainty equivalent CA Price of an American call option CE Price of a European call option d Dividend rate or amount ∆ Number of shares in the replicating portfolio (Chapter xx E The expectations operator ek Endowment of agent k in state of nature θ θ f Futures position (Chapter 16); pf Price of a futures contract (Chapter 16) F, G Cumulative distribution functions associated with densities: f, g Probability density functions K The strike or exercise price of an option K(˜) Kurtosis of the random variable x x ˜ L A lottery...
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...Questions 1 As seen from the excel sheet we conclude that the NVP for Project 1 results in negative amounts, as a result we do not choose project 1. As for project 2 amongst the two options we will choose Project two with DR at 10% as it yields a higher NVP value. Question 2 a) What is the equilibrium price and quantity of fertilizer in an unregulated, competitive market? 6 tons per day b) What is the efficient quantity of fertilizer? 4 tons per day c) Suppose government imposes a tax equal to the marginal external cost. What is the equilibrium price paid by consumers and the equilibrium quantity after implementation of the tax? Consumers will pay $1,200 per ton, and the efficient level of output at 4 tons per day will be achieved. d) At the output level in part (6), how much is the tax per one ton? $1200 - $800 = $400 e) How much tax revenue does government collect? $400 per ton x 4 tons per day = $1,600 per day f) What is the deadweight loss borne by society if the externality is left uncorrected? (6-4)x(1,600-1,000)/2 = $600 Question 3 a) According to Coase theorem it doesn’t matter who builds the fence as long as it is less than the cost of the crop damage. However, to answer the question; the rancher builds the fence, because the cost of the fence ($100) is less than the cost of compensating the farmer for crop damage. Together, the rancher and the farmer are $200 better off ($300 crop damage less $100...
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...Market Equilibration Process Jeremiah D. Wood ECO/561 April 19, 2014 Professor John Lindvall Market Equilibration Process Economic equilibrium is defined as a condition or state in which the economic forces are at a balance. In this particular discussion, one will discuss equilibration, the process of moving between two different points that is affected by a change in demand or supply. One will cover how a specific world event, Hurricane Katrina, caused home prices in Baton Rouge, Louisiana to fluctuate between two equilibrium states. Also to be covered is how the process of said movement occurred using the behaviors of both supply firms and consumers. In the late summer of 2005, Hurricane Katrina bared down on the City of New Orleans and the surrounding areas. This storm caused a surge that caused the storm levees to break that in turn, flooded the City of New Orleans and took most of the city’s housing with it. Because of the destruction, about two hundred and fifty thousand people were relocated to nearby Baton Rouge, making it the largest city in Louisiana. Let us start the discussion by stating that the average price of a single-family home in Baton Rouge before Katrina was one hundred thirty thousand dollars, shown by point A on the graph (O'Sullivan & Sheffrin, 2002). With the explosion of the population, the average price jumped to one hundred and fifty six thousand dollars within six months, point B, and the market shrunk from three thousand six...
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