CHAPTER 13 Chemical Equilibrium In a championship match, the muscles of tennis players need extra oxygen. The transfer of O2 from the lungs to the blood and then to the muscle tissues depends on chemical equilibria involving hemoglobin, an oxygencarrying protein. CONTENTS 492 13.1 The Equilibrium State 13.2 The Equilibrium Constant Kc 13.3 The Equilibrium Constant Kp 13.4 Heterogeneous Equilibria 13.5 Using the Equilibrium Constant 13.6 Factors That Alter the Composition of an Equilibrium Mixture: Le Châtelier’s Principle 13.7 Altering an Equilibrium Mixture: Changes in Concentration 13.8 Altering an Equilibrium Mixture: Changes in Pressure and Volume 13.9 Altering an Equilibrium Mixture: Changes in Temperature 13.10 The Effect of a Catalyst on Equilibrium 13.11 The Link between Chemical Equilibrium and Chemical Kinetics INQUIRY How Does Equilibrium Affect Oxygen Transport in the Bloodstream? 13.1 THE EQUILIBRIUM STATE 493 A t the beginning of Chapter 12, we raised three key questions about chemical reactions: What happens? How fast and by what mechanism does it happen? To what extent does it happen? The answer to the first question is given by the stoichiometry of the balanced chemical equation, and the answer to the second question is given by the kinetics of the reaction. In this chapter, we’ll look at the answer to the third question: How far does a reaction proceed toward completion before it reaches a state of chemical equilibrium—a state in which the concentrations of reactants and products no longer change? Chemical Equilibrium The state reached when the concentrations of reactants and products remain constant over time We’ve already touched on the concept of equilibrium in connection with our study of the evaporation of liquids (Section 10.5). When a liquid evaporates in a closed container, it soon gives rise to a constant vapor pressure because of a dynamic equilibrium in which the number of molecules leaving the liquid equals the number returning from the vapor. Chemical reactions behave similarly. They can occur in both forward and reverse directions, and when the rates of the forward and reverse reactions become equal, the concentrations of reactants and products remain constant. At that point, the chemical system is at equilibrium. 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 oxygen from our lungs to cells throughout our body. Similar equilibria involving CO molecules and hemoglobin account for the toxicity of carbon monoxide. A mixture of reactants and products in the equilibrium state is called an equilibrium mixture. In this chapter, we’ll address a number of important questions about the composition of equilibrium mixtures: What is the relationship between the concentrations of reactants and products in an equilibrium mixture? How can we determine equilibrium concentrations from initial concentrations? What factors can we exploit to alter the composition of an equilibrium mixture? This last question is particularly important when choosing conditions for the synthesis of industrial chemicals such as hydrogen, ammonia, and methanol (CH3OH). 䉱 In a closed container, liquid bromine and its vapor are in a dynamic equilibrium. 13.1 THE EQUILIBRIUM STATE In previous chapters, we’ve generally assumed that chemical reactions result in complete conversion of reactants to products. Many reactions, however, do not go to completion. Take, for example, the decomposition of the colorless gas dinitrogen tetroxide (N2O4) to the dark brown gas nitrogen dioxide (NO2). N2O4(g) ∆ 2 NO 2(g) Colorless Brown + Figure 13.1 shows the results of two experiments at 25 °C that illustrate the interconversion of N2O4 and NO2. In the first experiment (Figure 13.1a), we begin with N2O4 at an initial concentration of 0.0400 M. The formation of NO2 is indicated by the appearance of a brown color, and its concentration can be monitored by measuring the intensity of the color with a spectrophotometer. According to the balanced equation, 2.0 mol of NO2 forms for each mole of N2O4 that disappears, so the concentration of N2O4 at any time equals the initial concentration of N2O4 minus half the concentration of NO2. As time passes, the concentration of N2O4 decreases and the 䉱 Mexico City on a smoggy day and on a clear day. The brown color of the smog is due primarily to NO2 that results from automobile exhausts. 494 Chapter 13 CHEMICAL EQUILIBRIUM (b) Only NO2 is present initially. (a) Only N2O4 is present initially. 0.08 0.08 Chemical Equilibrium Chemical Equilibrium 0.06 [NO2] increases as [N2O4] decreases. 0.04 N2O4 0.02 NO2 0.00 Concentration (M) Concentration (M) 0.06 [N2O4] increases as [NO2] decreases. 0.04 N2O4 0.02 NO2 0.00 Time Time In both experiments, a state of chemical equilibrium is reached when the concentrations level off at constant values: [N2O4] = 0.0337 M; [NO2] = 0.0125 M. Figure 13.1 Change in the concentrations of N2O4 and NO2 with time in two experiments at 25 °C. concentration of NO2 increases until both concentrations level off at constant, equilibrium values: [N2O4] = 0.0337 M; [NO2] = 0.0125 M. In the second experiment, shown in Figure 13.1b, we begin with NO2 as the sole reactant at a concentration of 0.0800 M. The conversion of NO2 to N2O4 proceeds until the concentrations level off at the same values as obtained in the first experiment. Taken together, the two experiments demonstrate that the interconversion of N2O4 and NO2 is reversible and that the same equilibrium state is reached starting from either substance. N2O4(g) ∆ 2 NO2(g) Remember... The reaction rate is given by the rate law and ordinarily depends on the concentrations of at least some of the reacting species, usually increasing with increasing concentration and decreasing with decreasing concentration. (Section 12.2) Reaction occurs in both directions To indicate that the reaction can proceed in both forward and reverse directions, we write the balanced equation with two arrows, one pointing from reactants to products and the other pointing from products to reactants. (The terms “reactants” and “products” could be confusing in this context because the products of the forward reaction are the reactants in the reverse reaction. To avoid confusion, we’ll restrict the term reactants to the substances on the left side of the chemical equation and the term products to the substances on the right side of the equation.) Strictly speaking, all chemical reactions are reversible. What we sometimes call irreversible reactions are simply those that proceed nearly to completion, so that the equilibrium mixture contains almost all products and almost no reactants. For such reactions, the reverse reaction is often too slow to be detected. Why do the reactions of N2O4 and NO2 appear to stop after the concentrations reach their equilibrium values? We’ll explore that question in more detail in Section 13.11, but note for now that the concentrations reach constant values, not because the reactions stop, but because the rates of the forward and reverse reactions become equal. Take, for example, the experiment in which N2O4 is converted to an equilibrium mixture of NO2 and N2O4. Because reaction rates depend on concentrations (Section 12.2), the rate of the forward reaction (N2O4 : 2 NO2) decreases as the concentration of N2O4 decreases, while the rate of the reverse reaction 13.2 THE EQUILIBRIUM CONSTANT KC 495 (N2O4 ; 2 NO2) increases as the concentration of NO2 increases. Eventually, the decreasing rate of the forward reaction and the increasing rate of the reverse reaction become equal. At that point, there are no further changes in concentrations because N2O4 and NO2 both disappear as fast as they’re formed. Thus, chemical equilibrium is a dynamic state in which forward and reverse reactions continue at equal rates so that there is no net conversion of reactants to products (Figure 13.2). As N2O4 is consumed, the rate of the forward reaction decreases. Reaction rate Rate of forward reaction (N2O4 2 NO2) Chemical equilibrium (forward and reverse rates are equal) Rate of reverse reaction (N2O4 䉱 If the rate at which people move from the first floor to the second equals the rate at which people move from the second floor to the first, the number of people on each floor remains constant and the two populations are in dynamic equilibrium. 2 NO2) As NO2 is formed, the rate of the reverse reaction increases. Time When the two rates become equal, an equilibrium state is attained and there are no further changes in concentrations. Figure 13.2 Rates of the forward and reverse reactions for the decomposition of N2O4 to NO2. 13.2 THE EQUILIBRIUM CONSTANT Kc Table 13.1 lists concentration data for the experiments in Figure 13.1 along with data for three additional experiments. In experiments 1 and 2, the equilibrium mixtures have identical compositions because the initial concentration of N2O4 in experiment 1 is half the initial concentration of NO2 in experiment 2; that is, the total number of N and O atoms is the same in both experiments. In experiments 3–5, different initial concentrations of N2O4 and/or NO2 give different equilibrium concentrations. In all the experiments, however, the equilibrium concentrations are related. The last column of Table 13.1 shows that, at equilibrium, the expression [NO2]2/[N2O4] has a constant value of approximately 4.64 * 10-3 M. TABLE 13.1 Concentration Data at 25 °C for the Reaction N2O4(g) ∆ 2 NO2(g) Initial Concentrations (M) Equilibrium Concentrations (M) Equilibrium Constant Expression Experiment [N2O4] [NO2] [N2O4] [NO2] [NO2]2/[N2O4] 1 0.0400 0.0000 0.0337 0.0125 4.64 * 10-3 2 0.0000 0.0800 0.0337 0.0125 4.64 * 10-3 3 0.0600 0.0000 0.0522 0.0156 4.66 * 10-3 4 0.0000 0.0600 0.0246 0.0107 4.65 * 10-3 5 0.0200 0.0600 0.0429 0.0141 4.63 * 10-3 496 Chapter 13 CHEMICAL EQUILIBRIUM The expression [NO2]2/[N2O4] appears to be related to the balanced equation for the reaction N2O4(g) ∆ 2 NO 2(g) in that the concentration of the product is in the numerator, raised to the power of its coefficient in the balanced equation, and the concentration of the reactant is in the denominator. Is there an analogous expression with a constant value for every chemical reaction? If so, how is the form of that expression related to the balanced equation for the reaction? To answer those questions, let’s consider a general reversible reaction: aA + bB ∆ cC + dD where A and B are the reactants, C and D are the products, and a, b, c, and d are their respective stoichiometric coefficients in the balanced chemical equation. In 1864 on the basis of experimental studies of many reversible reactions, the Norwegian chemists Cato Maximilian Guldberg and Peter Waage proposed that the concentrations in an equilibrium mixture are related by the following equilibrium equation, where Kc is the equilibrium constant and the expression on the right side is called the equilibrium constant expression. Equilibrium equation: Kc = Equilibrium constant [C]c[D]d [A]a[B]b Products Reactants Equilibrium constant expression As usual, square brackets indicate the molar concentration of the substance within the brackets, hence the subscript c for “concentration” in Kc. The substances in the equilibrium constant expression may be gases or molecules and ions in a solution, but may not be pure solids or pure liquids for reasons that we’ll explain in Section 13.4. The equilibrium equation is also known as the law of mass action because in the early days of chemistry, concentration was called “active mass.” The equilibrium constant Kc is the number obtained by multiplying the equilibrium concentrations of all the products and dividing by the product of the equilibrium concentrations of all the reactants, with the concentration of each substance raised to the power of its coefficient in the balanced chemical equation. No matter what the individual equilibrium concentrations may be in a particular experiment, the equilibrium constant for a reaction at a particular temperature always has the same value. Thus, the equilibrium equation for the decomposition reaction N2O4(g) ∆ 2 NO 2 is Kc = Remember... The thermodynamic standard state is the set of conditions under which thermodynamic measurements are reported: 1 M concentration for each solute in solution, 1 atm pressure for each gas, and a specified temperature, usually 25 °C. (Section 8.5) [NO2]2 = 4.64 * 10-3 [N2O4] at 25 °C where the equilibrium constant expression is [NO2]2/[N2O4] and the equilibrium constant Kc has a value of 4.64 * 10-3 at 25 °C (Table 13.1). Values of Kc are generally reported without units because the concentrations in the equilibrium constant expression are considered to be concentration ratios in which the molarity of each substance is divided by its molarity (1 M) in the thermodynamic standard state (Section 8.5). Because the units cancel, the concentration ratios and the values of Kc are dimensionless. For experiment 1 in Table 13.1, for example, 0.0125 M 2 b [NO2]2 1M Kc = = = 4.64 * 10-3 [N2O4] 0.0337 M a b 1M a at 25 °C Equilibrium constants are temperature-dependent, so the temperature must be given when citing a value of Kc. For example, Kc for the decomposition of N2O4 increases from 4.64 * 10-3 at 25 °C to 1.53 at 127 °C. 13.2 THE EQUILIBRIUM CONSTANT KC The form of the equilibrium constant expression and the numerical value of the equilibrium constant depend on the form of the balanced chemical equation. Look again at the chemical equation and the equilibrium equation for a general reaction: aA + bB ∆ cC + dD Kc = [C]c[D]d [A]a[B]b If we write the chemical equation in the reverse direction, the new equilibrium constant expression is the reciprocal of the original expression and the new equilibrium constant Kc’ is the reciprocal of the original equilibrium constant Kc. (The prime distinguishes Kc’ from Kc.) cC + dD ∆ aA + bB Kc ¿ = [A]a[B]b c d [C] [D] = 1 Kc Because the equilibrium constants Kc and Kc’ have different numerical values, it’s important to specify the form of the balanced chemical equation when quoting the value of an equilibrium constant. WORKED EXAMPLE 13.1 WRITING EQUILIBRIUM EQUATIONS FOR GAS-PHASE REACTIONS Write the equilibrium equation for each of the following reactions: (a) N2(g) + 3 H 2(g) ∆ 2 NH 3(g) (b) 2 NH 3(g) ∆ N2(g) + 3 H 2(g) STRATEGY Put the concentrations of the reaction products in the numerator of the equilibrium constant expression and the concentrations of the reactants in the denominator. Then raise the concentration of each substance to the power of its coefficient in the balanced chemical equation. SOLUTION Coefficient of NH3 (a) Kc = [NH3]2 [N2][H2]3 Coefficient of H2 Coefficient of H2 (b) Kc′ = [N2][H2] [NH3]2 3 Kc′ = Coefficient of NH3 1 Kc Because the balanced equation in part (b) is the reverse of that in part (a), the equilibrium constant expression in part (b) is the reciprocal of the expression in part (a) and the equilibrium constant Kc’ is the reciprocal of Kc. WORKED EXAMPLE 13.2 CALCULATING THE EQUILIBRIUM CONSTANT Kc The following concentrations were measured for an equilibrium mixture at 500 K: [N2] = 3.0 * 10-2 M; [H 2] = 3.7 * 10-2 M; [NH 3] = 1.6 * 10-2 M. Calculate the equilibrium constant at 500 K for each of the reactions in Worked Example 13.1. STRATEGY To calculate the value of the equilibrium constant, substitute the equilibrium concentrations into the equilibrium equation. continued on next page 497 498 Chapter 13 CHEMICAL EQUILIBRIUM SOLUTION [NH 3]2 (a) Kc = 3 [N2][H 2] [N2][H 2]3 (b) K c¿ = [NH 3]2 (1.6 * 10-2) 2 = = 1.7 * 102 -2 3 -2 (3.0 * 10 )(3.7 * 10 ) = (3.0 * 10-2)(3.7 * 10-2) 3 = 5.9 * 10-3 (1.6 * 10-2) 2 Note that Kc’ is the reciprocal of Kc. That is, 5.9 * 10-3 = 1 1.7 * 102 WORKED EXAMPLE 13.3 WRITING EQUILIBRIUM EQUATIONS FOR SOLUTION REACTIONS Methyl tert-butyl ether (MTBE), a substance once used as a gasoline additive but now being phased out because of safety concerns, can be synthesized by heating methanol and tert-butyl alcohol with a catalytic amount of sulfuric acid: H+ CH 3OH(soln) + C4H 9OH(soln) ∆ C4H 9OCH 3(soln) + H 2O(soln) catalyst tert-Butyl alcohol Methanol Methyl tert-butyl ether In this equation, soln denotes a largely organic solution that also contains water. Write the equilibrium constant expression for the reaction. STRATEGY The rules for writing the equilibrium constant expression apply to reactions in liquid solutions as well as to gas-phase reactions (reactions in gaseous solutions). Put the concentrations of the products in the numerator and the concentrations of the reactants in the denominator. No exponents are needed because all the coefficients in the balanced chemical equation equal 1. SOLUTION Kc = 䉱 MTBE is a component of some gasolines but is being phased out because it has been found to be an unsafe contaminant of groundwater. [C4H9OCH3] [H2O] Products [CH3OH] [C4H9OH] Reactants 쑺 PROBLEM 13.1 The oxidation of sulfur dioxide to give sulfur trioxide is an important step in the industrial process for the synthesis of sulfuric acid. Write the equilibrium equation for each of the following reactions: (a) 2 SO 2(g) + O 2(g) ∆ 2 SO 3(g) (b) 2 SO3(g) ∆ 2 SO 2(g) + O 2(g) 쑺 PROBLEM 13.2 The following equilibrium concentrations were measured at 800 K: [SO2] = 3.0 * 10-3 M; [O2] = 3.5 * 10-3 M; [SO3] = 5.0 * 10-2 M. Calculate the equilibrium constant at 800 K for each of the reactions in Problem 13.1. 쑺 PROBLEM 13.3 Lactic acid, which builds up in muscle tissue upon strenuous exercise, is partially dissociated in aqueous solution: H CH3 C H CO2H(aq) OH Lactic acid Lactic acid H+(aq) + CH3 C CO2–(aq) OH Lactate ion (a) Write the equilibrium constant expression for Kc. (b) What is the value of Kc if the extent of dissociation in 0.100 M lactic acid is 3.65% at 25 °C? 13.3 THE EQUILIBRIUM CONSTANT KP 499 WORKED CONCEPTUAL EXAMPLE 13.4 JUDGING WHETHER A MIXTURE IS AT EQUILIBRIUM The following pictures represent mixtures of A molecules (red spheres) and B molecules (blue spheres), which interconvert according to the equation A ∆ B. If mixture (1) is at equilibrium, which of the other mixtures are also at equilibrium? Explain. (1) (2) (3) (4) STRATEGY The equilibrium constant for the reaction is given by Kc = [B]/[A], where the concentrations are equilibrium concentrations in units of mol/L. Since the equilibrium constant expression has the same number of concentration terms in the numerator and denominator, the volume cancels and Kc = (moles of B)/(moles of A). Because the number of moles is directly proportional to the number of molecules, Kc = (molecules of B)/ (molecules of A) in the equilibrium mixture, mixture (1). To determine whether the other mixtures are at equilibrium, count the number of molecules and compare the B/A ratio in mixtures (2)–(4) with the B/A ratio in the equilibrium mixture. SOLUTION For mixture (1), Kc = [B]/[A] = 2/6 = 1/3. For mixture (2), [B]/[A] = 4/4 = 1 Z Kc. For mixture (3), [B]/[A] = 3/9 = 1/3 = Kc. For mixture (4), [B]/[A] = 9/3 = 3 Z Kc. Mixture (3) is at equilibrium, but mixtures (2) and (4) are not at equilibrium because their equilibrium constant expression [B]/[A] does not equal Kc. CONCEPTUAL PROBLEM 13.4 The following pictures represent mixtures that contain A atoms (red), B atoms (blue), and AB and B2 molecules, which interconvert according to the equation A + B2 ∆ AB + B. If mixture (1) is at equilibrium, which of the other mixtures are also at equilibrium? Explain. (1) (2) (3) (4) 13.3 THE EQUILIBRIUM CONSTANT Kp Because gas pressures are easily measured, equilibrium equations for gas-phase reactions are often written using partial pressures (Section 9.5) rather than molar concentrations. For example, the equilibrium equation for the decomposition of N2O4 can be written as (PNO2)2 Kp = for the reaction N2O4(g) ∆ 2 NO 2(g) PN2O4 where PN2O4 and PNO2 are the partial pressures (in atmospheres) of reactants and products at equilibrium, and the subscript p on K reminds us that the equilibrium constant Kp is defined using partial pressures. As for Kc, values of Kp are dimensionless because Remember... The partial pressure of a gas in a mixture is the pressure the gas would exert if it were the only one present. Its partial pressure is independent of the partial pressures of the other gases in the mixture. (Section 9.5) 500 Chapter 13 CHEMICAL EQUILIBRIUM the partial pressures in the equilibrium equation are actually ratios of partial pressures in atmospheres to the standard-state partial pressure of 1 atm. Thus, the units cancel. Note that the equilibrium equations for Kp and Kc have the same form except that the expression for Kp contains partial pressures instead of molar concentrations. The constants Kp and Kc for the general gas-phase reaction a A + b B ∆ c C + d D are related because the pressure of each component in a mixture of ideal gases is directly proportional to its molar concentration. For component A, for example, PAV = nART PA = so nA RT = [A]RT V Similarly, PB = [B]RT, PC = [C]RT, and PD = [D]RT. The equilibrium equation for Kp is therefore given by Kp = (PC)c(PD)d (PA)a(PB)b = ([C]RT)c([D]RT)d ([A]RT)a([B]RT)b = [C]c[D]d [A]a[B]b * (RT) (c + d) - (a + b) Because the first term on the right side equals Kc, the values of Kp and Kc are related by the equation Kp = Kc(RT)¢n for the reaction a A + b B ∆ c C + d D Here, R is the gas constant, 0.082 06 (L # atm)/(K # mol), T is the absolute temperature, and ¢n = (c + d) - (a + b) is the number of moles of gaseous products minus the number of moles of gaseous reactants. For the decomposition of 1 mol of N2O4 to 2 mol of NO2, ¢n = 2 - 1 = 1, and Kp = Kc(RT): N2O4(g) ∆ 2 NO 2(g) Kp = Kc(RT) For the reaction of 1 mol of hydrogen with 1 mol of iodine to give 2 mol of hydrogen iodide, ¢n = 2 - (1 + 1) = 0, and Kp = Kc(RT)0 = Kc: Kp = Kc H 2(g) + I 2(g) ∆ 2 HI(g) + + In general, Kp equals Kc only if the same number of moles of gases appear on both sides of the balanced chemical equation so that ¢n = 0. WORKED EXAMPLE 13.5 DETERMINING THE EQUILIBRIUM CONSTANT Kp Methane (CH4) reacts with hydrogen sulfide to yield H2 and carbon disulfide, a solvent used in manufacturing rayon and cellophane: CH 4(g) + 2 H 2S(g) ∆ CS 2(g) + 4 H 2(g) + + What is the value of Kp at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of CH4, 0.25 atm of H2S, 0.52 atm of CS2, and 0.10 atm of H2? STRATEGY Write the equilibrium equation by setting Kp equal to the equilibrium constant expression using partial pressures. Put the partial pressures of products in the numerator and 13.3 THE EQUILIBRIUM CONSTANT KP the partial pressures of reactants in the denominator, with the pressure of each substance raised to the power of its coefficient in the balanced chemical equation. Then substitute the partial pressures into the equilibrium equation and solve for Kp. SOLUTION Coefficient of H2 4 Kp = (PCS2)(PH2) (PCH4)(PH2S)2 Coefficient of H2S Kp = (PCS2)(PH2)4 (PCH4)(PH2S)2 = (0.52)(0.10)4 (0.20)(0.25)2 = 4.2 × 10 –3 Note that the partial pressures must be in units of atmospheres (not mm Hg) because the standard-state partial pressure for gases is 1 atm. WORKED EXAMPLE 13.6 RELATING THE EQUILIBRIUM CONSTANTS Kp AND Kc Hydrogen is produced industrially by the steam–hydrocarbon re-forming process. The reaction that takes place in the first step of this process is H 2O(g) + CH 4(g) ∆ CO(g) + 3 H 2(g) + + (a) If Kc = 3.8 * 10-3 at 1000 K, what is the value of Kp at the same temperature? (b) If Kp = 6.1 * 104 at 1125 °C, what is the value of Kc at 1125 °C? STRATEGY To calculate Kp from Kc, or vice versa, use the equation Kp = Kc(RT)¢n, where R must be in units of (L # atm)/(K # mol), T is the temperature in kelvin, and ¢n is the number of moles of gaseous products minus the number of moles of gaseous reactants. SOLUTION (a) For this reaction, ¢n = (1 + 3) - (1 + 1) = 2. Therefore, Kp = Kc(RT)¢n = Kc(RT)2 = (3.8 * 10-3)[(0.082 06)(1000)]2 = 26 (b) Solving the equation Kp = Kc(RT)2 for Kc gives Kp 6.1 * 104 Kc = = = 4.6 (RT)2 [(0.082 06)(1398)]2 Note that the temperature in these equations is the absolute temperature; 1125 °C corresponds to 1125 + 273 = 1398 K. 쑺 PROBLEM 13.5 In the industrial synthesis of hydrogen, mixtures of CO and H2 are enriched in H2 by allowing the CO to react with steam. The chemical equation for this so-called water-gas shift reaction is CO(g) + H 2O(g) ∆ CO 2(g) + H 2(g) What is the value of Kp at 700 K if the partial pressures in an equilibrium mixture at 700 K are 1.31 atm of CO, 10.0 atm of H2O, 6.12 atm of CO2, and 20.3 atm of H2? 쑺 PROBLEM 13.6 Nitric oxide reacts with oxygen to give nitrogen dioxide, an important reaction in the Ostwald process for the industrial synthesis of nitric acid: 2 NO(g) + O2(g) ∆ 2 NO 2(g) If Kc = 6.9 * 105 at 227 °C, what is the value of Kp at this temperature? If Kp = 1.3 * 10-2 at 1000 K, what is the value of Kc at 1000 K? 501 502 Chapter 13 CHEMICAL EQUILIBRIUM 13.4 HETEROGENEOUS EQUILIBRIA Thus far we’ve been discussing homogeneous equilibria, in which all reactants and products are in a single phase, usually either gaseous or solution. Heterogeneous equilibria, by contrast, are those in which reactants and products are present in more than one phase. Take, for example, the thermal decomposition of solid calcium carbonate, a reaction used in manufacturing cement: CaCO 3(s) ∆ CaO(s) + CO 2(g) Limestone Lime When the reaction is carried out in a closed container, three phases are present at equilibrium: solid calcium carbonate, solid calcium oxide, and gaseous carbon dioxide. If we were to write the usual equilibrium equation for the reaction, including all the reactants and products, we would have Kc = [CaO][CO2] [CaCO3] Once again, though, the concentrations in the equilibrium equation are ratios of actual concentrations to concentrations in the standard state. Because the standard state of a pure solid is the pure solid itself, the concentration ratio for a pure solid (and also for a pure liquid) is equal to 1. Thus, [CaO] = 1 and [CaCO3] = 1, and so these concentrations can be omitted from the equilibrium equation: Kc = 䉱 The manufacture of cement begins with the thermal decomposition of limestone, CaCO3, in large kilns. [CaO][CO2] (1)[CO2] = = [CO 2] [CaCO3] (1) The analogous equilibrium equation in terms of pressure is Kp = PCO2 where PCO2 is the equilibrium pressure of CO2 in atmospheres: Kc = [CO2] Kp = PCO2 As a general rule, the concentrations of pure solids and pure liquids are not included when writing an equilibrium equation. We include only the concentrations of gases and the concentrations of solutes in solutions because only those concentrations can be varied. To establish equilibrium between solid CaCO3, solid CaO, and gaseous CO2, all three components must be present. It follows from the equations Kc = [CO2] and Kp = PCO2, however, that the concentration and pressure of CO2 at equilibrium are constant, independent of how much solid CaO and CaCO3 is present (Figure 13.3). If the temperature is changed, however, the concentration and pressure of CO2 will also change because the values of Kc and Kp depend on temperature. (a) Small amount of CaCO3; (b) Large amount of CaCO3; large amount of CaO small amount of CaO CO2 CO2 PCO PCO2 2 Figure 13.3 Thermal decomposition of calcium carbonate: CaCO3(s) ∆ CaO(s) ⴙ CO2(g). CaCO3 CaO CaCO3 CaO At the same temperature, the equilibrium pressure of CO2 is the same in (a) and (b), independent of how much solid CaCO3 and CaO is present. 13.5 USING THE EQUILIBRIUM CONSTANT WORKED EXAMPLE 13.7 WRITING EQUILIBRIUM EQUATIONS FOR HETEROGENEOUS EQUILIBRIA Write the equilibrium equation for each of the following reactions: (a) CO 2(g) + C(s) ∆ 2 CO(g) (b) Hg(l) + Hg 2+(aq) ∆ Hg 2 2+(aq) STRATEGY Write the usual equilibrium constant expressions but omit the pure solid carbon in part (a) and the pure liquid mercury in part (b) because the ratio of their concentrations to their concentrations in the standard state is equal to 1. SOLUTION (a) Kc = [CO]2 [CO2] Alternatively, because CO and CO2 are gases, the equilibrium equation can be written using partial pressures: Kp = (PCO)2 PCO2 The relationship between Kp and Kc is Kp = Kc(RT)¢n = Kc(RT), because ¢n = 2 - 1 = 1. (b) Kc = [Hg 22 + ] [Hg 2 + ] In this case, it’s not appropriate to write an expression for Kp because none of the reactants and products is a gas. 쑺 PROBLEM 13.7 For each of the following reactions, write the equilibrium constant expression for Kc. Where appropriate, also write the equilibrium constant expression for Kp and relate Kp to Kc. (a) 2 Fe(s) + 3 H 2O(g) ∆ Fe2O3(s) + 3 H 2(g) (b) 2 H 2O(l) ∆ 2 H 2(g) + O 2(g) (c) SiCl4(g) + 2 H 2(g) ∆ Si(s) + 4 HCl(g) (d) Hg 22 + (aq) + 2 Cl - (aq) ∆ Hg 2Cl2(s) 13.5 USING THE EQUILIBRIUM CONSTANT Knowing the value of the equilibrium constant for a chemical reaction lets us judge the extent of the reaction, predict the direction of the reaction, and calculate equilibrium concentrations from initial concentrations. Let’s look at each possibility. Judging the Extent of Reaction The numerical value of the equilibrium constant for a reaction indicates the extent to which reactants are converted to products; that is, it measures how far the reaction proceeds before the equilibrium state is reached. Consider, for example, the reaction of H2 with O2, which has a very large equilibrium constant (Kc = 2.4 * 1047 at 500 K): 2 H 2(g) + O 2(g) ∆ 2 H 2O(g) Kc = [H 2O]2 [H 2]2[O2] = 2.4 * 1047 at 500 K Because products appear in the numerator of the equilibrium constant expression and reactants are in the denominator, a very large value of Kc means that the equilibrium ratio of products to reactants is very large. In other words, the reaction proceeds nearly to completion. For example, if stoichiometric amounts of H2 and O2 are 503 504 Chapter 13 CHEMICAL EQUILIBRIUM allowed to react and [H 2O] = 0.10 M at equilibrium, then the concentrations of H2 and O2 that remain at equilibrium are negligibly small: [H 2] = 4.4 * 10-17 M and [O2] = 2.2 * 10-17 M. (Try substituting these concentrations into the equilibrium equation to show that they satisfy the equation.) By contrast, if a reaction has a very small value of Kc, the equilibrium ratio of products to reactants is very small and the reaction proceeds hardly at all before equilibrium is reached. For example, the reverse of the reaction of H2 with O2 gives the same equilibrium mixture as obtained from the forward reaction ([H 2] = 4.4 * 10-17 M, [O2] = 2.2 * 10-17 M, [H 2O] = 0.10 M). The reverse reaction does not occur to any appreciable extent, however, because its equilibrium constant is so small: Kc’ = 1/Kc = 1/(2.4 * 1047) = 4.2 * 10-48. 2 H 2O(g) ∆ 2 H 2(g) + O 2(g) Kc = [H 2]2[O2] [H 2O]2 = 4.2 * 10-48 at 500 K If a reaction has an intermediate value of Kc—say, a value in the range of 103 to 10 —then appreciable concentrations of both reactants and products are present in the equilibrium mixture. The reaction of hydrogen with iodine, for example, has Kc = 57.0 at 700 K: -3 H 2(g) + I 2(g) ∆ 2 HI(g) Kc = [HI]2 = 57.0 [H 2][I 2] at 700 K If the equilibrium concentrations of H2 and I2 are both 0.010 M, then the concentration of HI at equilibrium is 0.075 M: [HI]2 = Kc[H 2][I 2] [HI] = 3Kc[H 2][I 2] = 3(57.0)(0.010)(0.010) = 0.075 M Thus, the concentrations of both reactants and products—0.010 M and 0.075 M—are appreciable. (Note that pressing the 1x key on a calculator gives a positive number. Remember, though, that the square root of a positive number can be positive or negative. Of the two roots for the concentration of HI, ; 0.075 M, we choose the positive one because the concentration of a chemical substance is always a positive quantity.) The gas-phase decomposition of N2O4 to NO2 is another reaction with a value of Kc that is neither large nor small: Kc = 4.64 * 10-3 at 25 °C. Accordingly, equilibrium mixtures contain appreciable concentrations of both N2O4 and NO2, as shown previously in Table 13.1. We can make the following generalizations concerning the composition of equilibrium mixtures: • If Kc 7 103, products predominate over reactants. If Kc is very large, the reaction proceeds nearly to completion. • If Kc 6 10-3, reactants predominate over products. If Kc is very small, the reaction proceeds hardly at all. • If Kc is in the range 10-3 to 103, appreciable concentrations of both reactants and products are present. These points are illustrated in Figure 13.4. 13.5 USING THE EQUILIBRIUM CONSTANT Figure 13.4 Judging the extent of a reaction. The larger the value of the equilibrium constant Kc, the farther the reaction proceeds to the right before reaching the equilibrium state: Kc Very small 10−3 Reaction proceeds hardly at all. 1 Very large 103 Appreciable concentrations of both reactants and products are present at equilibrium. Reaction proceeds nearly to completion. 쑺 PROBLEM 13.8 The value of Kc for the dissociation reaction H 2(g) ∆ 2 H(g) is 1.2 * 10-42 at 500 K. Does the equilibrium mixture contain mainly H2 molecules or H atoms? Explain in light of the electron configuration of hydrogen. Predicting the Direction of Reaction Let’s look again at the gaseous reaction of hydrogen with iodine: H 2(g) + I 2(g) ∆ 2 HI(g) Kc = 57.0 at 700 K Suppose that we have a mixture of H 2(g), I 2(g), and HI(g) at 700 K with concentrations [H 2]t = 0.10 M, [I 2]t = 0.20 M, and [HI]t = 0.40 M. (The subscript t on the concentration symbols means that the concentrations were measured at some arbitrary time t, not necessarily at equilibrium.) If we substitute these concentrations into the equilibrium constant expression, we obtain a value called the reaction quotient Qc. Reaction quotient Qc = [HI]t 2 (0.40)2 = = 8.0 [H 2]t[I 2]t (0.10)(0.20) The reaction quotient Qc is defined in the same way as the equilibrium constant Kc except that the concentrations in Qc are not necessarily equilibrium values. For the case at hand, the numerical value of Qc (8.0) does not equal Kc (57.0), so the mixture of H 2(g), I 2(g), and HI(g) is not at equilibrium. As time passes, though, reaction will occur, changing the concentrations and thus changing the value of Qc in the direction of Kc. After a sufficiently long time, an equilibrium state will be reached and Qc = Kc. The reaction quotient Qc is useful because it lets us predict the direction of reaction by comparing the values of Qc and Kc. If Qc is less than Kc, movement toward equilibrium increases Qc by converting reactants to products (that is, net reaction proceeds from left to right). If Qc is greater than Kc, movement toward equilibrium decreases Qc by converting products to reactants (that is, net reaction proceeds from right to left). If Qc equals Kc, the reaction mixture is already at equilibrium, and no net reaction occurs. Thus, we can make the following generalizations concerning the direction of the reaction: • If Qc 6 Kc, net reaction goes from left to right (reactants to products). • If Qc 7 Kc, net reaction goes from right to left (products to reactants). • If Qc = Kc, no net reaction occurs. These points are illustrated in Figure 13.5. Hydrogen 505 506 Chapter 13 CHEMICAL EQUILIBRIUM Figure 13.5 Predicting the direction of reaction. The direction of net reaction depends on the relative values of Qc and Kc. Kc Qc Reactants Qc ent vem d o M ar tow rium ib l i equ Products Kc ent vem Mo ard tow rium ilib equ Reactants and products are at equilibrium. Qc Reactants Kc Products Movement toward equilibrium changes the value of Qc until it equals Kc, but the value of Kc remains constant. WORKED EXAMPLE 13.8 PREDICTING THE DIRECTION OF REACTION A mixture of 1.57 mol of N2, 1.92 mol of H2, and 8.13 mol of NH3 is introduced into a 20.0 L reaction vessel at 500 K. At this temperature, the equilibrium constant Kc for the reaction N2(g) + 3 H 2(g) ∆ 2 NH 3(g) is 1.7 * 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction? STRATEGY To determine whether the reaction mixture is at equilibrium, we need to calculate the value of the reaction quotient Qc and then compare it with the equilibrium constant Kc. If the mixture is not at equilibrium, the relative values of Qc and Kc tell us the direction of the net reaction. Because we are given amounts in moles, we must first convert moles to molar concentrations before substituting into the expression for Qc. SOLUTION The initial concentration of N2 is (1.57 mol)/(20.0 L) = 0.0785 M. Similarly, [H 2] = 0.0960 M and [NH 3] = 0.406 M. Substituting these concentrations into the equilibrium constant expression gives Qc = [NH 3]t 2 [N2]t[H 2]t 3 = (0.406)2 (0.0785)(0.0960) 3 = 2.37 * 103 Because Qc does not equal Kc (1.7 * 102), the reaction mixture is not at equilibrium. Because Qc is greater than Kc, net reaction will proceed from right to left, decreasing the concentration of NH3 and increasing the concentrations of N2 and H2 until Qc = Kc = 1.7 * 102. BALLPARK CHECK Approximate initial concentrations can be calculated by dividing rounded values of the number of moles of each substance by the volume; [N2] L (1.6 mol)/(20 L) L 0.08 M, [H 2] L (2 mol)/(20 L) L 0.1 M, and [NH 3] L (8 mol)/(20 L) L 0.4 M. Substituting these concentrations into the expression for Qc gives a ballpark estimate of Qc: Qc = [NH 3]t2 [N2]t[H 2]t3 L (0.4)2 (0.08)(0.1)3 L 2 * 103 You can calculate this value without a calculator because it equals (16 * 10-2)/ (8 * 10-5). The ballpark estimate of Qc, like the more exact value (2.37 * 103), exceeds Kc, so the reaction mixture is not at equilibrium. 13.5 USING THE EQUILIBRIUM CONSTANT 쑺 PROBLEM 13.9 The equilibrium constant Kc for the reaction 2 NO(g) + O2(g) ∆ 2 NO2(g) is 6.9 * 105 at 500 K. A 5.0 L reaction vessel at 500 K was filled with 0.060 mol of NO, 1.0 mol of O2, and 0.80 mol of NO2. (a) Is the reaction mixture at equilibrium? If not, in which direction does the net reaction proceed? (b) What is the direction of the net reaction if the initial amounts are 5.0 * 10-3 mol of NO, 0.20 mol of O2, and 4.0 mol of NO2? CONCEPTUAL PROBLEM 13.10 The reaction A2 + B2 ∆ 2 AB has an equilibrium constant Kc = 4. The following pictures represent reaction mixtures that contain A2 molecules (red), B2 molecules (blue), and AB molecules: (1) (2) (3) (a) Which reaction mixture is at equilibrium? (b) For those reaction mixtures that are not at equilibrium, will the net reaction go in the forward or reverse direction to reach equilibrium? Calculating Equilibrium Concentrations If the equilibrium constant and all the equilibrium concentrations but one are known, the unknown concentration can be calculated directly from the equilibrium equation. To illustrate, let’s consider the following problem: What is the concentration of NO in an equilibrium mixture of gaseous NO, O2, and NO2 at 500 K that contains 1.0 * 10-3 M O2 and 5.0 * 10-2 M NO2? At this temperature, the equilibrium constant Kc for the reaction 2 NO(g) + O2(g) ∆ 2 NO 2(g) is 6.9 * 105. In this problem, Kc and all the equilibrium concentrations except one are known and we’re asked to calculate the unknown equilibrium concentration. First, we write the equilibrium equation for the reaction and solve for the unknown concentration: Kc = [NO 2]2 [NO] = [NO]2[O2] [NO 2]2 C [O2]Kc Then we substitute the known values of Kc, [O2], and [NO2] into the expression for [NO]. Taking the positive square root because the concentration of NO is a positive quantity, we obtain: [NO] = C (1.0 * 10-3)(6.9 * 105) (5.0 * 10-2)2 = 23.6 * 10-6 = 1.9 * 10-3 M To be sure that we haven’t made any errors, it’s a good idea to check the result by substituting it into the equilibrium equation: Kc = 6.9 * 105 = [NO2]2 2 [NO] [O2] = (5.0 * 10-2)2 -3 2 -3 (1.9 * 10 ) (1.0 * 10 ) = 6.9 * 105 Another type of problem is one in which we know the initial concentrations but do not know any of the equilibrium concentrations. To solve this kind of problem, follow the series of steps summarized in Figure 13.6 and illustrated in Worked Examples 13.9 and 13.10. The same approach can be used to calculate equilibrium partial pressures from initial partial pressures and Kp, as shown in Worked Example 13.11. 507 508 Chapter 13 CHEMICAL EQUILIBRIUM Figure 13.6 Steps to follow in calculating equilibrium concentrations from initial concentrations. Step 1. Write the balanced equation for the reaction. Step 2. Under the balanced equation, make a table that lists for each substance involved in the reaction: (a) The initial concentration (b) The change in concentration on going to equilibrium (c) The equilibrium concentration In constructing the table, define x as the concentration (mol/L) of one of the substances that reacts on going to equilibrium and then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of x. Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you must solve a quadratic equation, choose the mathematical solution that makes chemical sense. Step 4. Calculate the equilibrium concentrations from the calculated value of x. Step 5. Check your results by substituting them into the equilibrium equation. WORKED EXAMPLE 13.9 CALCULATING EQUILIBRIUM CONCENTRATIONS FROM INITIAL CONCENTRATIONS The equilibrium constant Kc for the reaction of H2 with I2 is 57.0 at 700 K: H 2(g) + I 2(g) ∆ 2 HI(g) ? Kc = 57.0 at 700 K If 1.00 mol of H2 is allowed to react with 1.00 mol of I2 in a 10.0 L reaction vessel at 700 K, what are the concentrations of H2, I2, and HI at equilibrium? What is the composition of the equilibrium mixture in moles? STRATEGY Initial state Equilibrium state We need to calculate equilibrium concentrations from initial concentrations, so we use the method outlined in Figure 13.6. SOLUTION Step 1. The balanced equation is given: H 2(g) + I 2(g) ∆ 2 HI(g). Step 2. The initial concentrations are [H 2] = [I 2] = (1.00 mol)/(10.0 L) = 0.100 M. For convenience, define an unknown, x, as the concentration (mol/L) of H2 that reacts. According to the balanced equation for the reaction, x mol/L of H2 reacts with x mol/L of I2 to give 2x mol/L of HI. This reduces the initial concentrations of H2 and I2 from 0.100 mol/L to (0.100 - x) mol/L at equilibrium. Let’s summarize these results in a table under the balanced equation: H2(g) Initial concentration (M) Change (M) Equilibrium concentration (M) 0.100 ⴙ I2(g) ∆ 2 HI(g) 0.100 0 -x -x + 2x (0.100 - x) (0.100 - x) 2x 509 13.5 USING THE EQUILIBRIUM CONSTANT Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction: Kc = 57.0 = [HI]2 [H 2][I 2] = (2x) 2 = a (0.100 - x)(0.100 - x) 2 2x b 0.100 - x Because the right side of this equation is a perfect square, we can take the square root of both sides: 2x 157.0 = ; 7.55 = 0.100 - x Solving for x, we obtain two solutions. The equation with the positive square root of 57.0 gives + 7.55(0.100 - x) = 2x 0.755 = 2x + 7.55x x = 0.755 = 0.0791 M 9.55 The equation with the negative square root of 57.0 gives - 7.55(0.100 - x) = 2x -0.755 = 2x - 7.55x x = - 0.755 = 0.136 M - 5.55 Because the initial concentrations of H2 and I2 are 0.100 M, x can’t exceed 0.100 M. Therefore, discard x = 0.136 M as chemically unreasonable and choose the first solution, x = 0.0791 M. Step 4. Calculate the equilibrium concentrations from the calculated value of x: [H 2] = [I 2] = 0.100 - x = 0.100 - 0.0791 = 0.021 M [HI] = 2x = (2)(0.0791) = 0.158 M Step 5. Check the results by substituting them into the equilibrium equation: Kc = 57.0 = [HI]2 [H 2][I 2] = (0.158)2 (0.021)(0.021) = 57 The number of moles of each substance in the equilibrium mixture is obtained by multiplying each concentration by the volume of the reaction vessel: Moles of H 2 = Moles of I 2 = (0.021 mol/L)(10.0 L) = 0.21 mol Moles of HI = (0.158 mol/L)(10.0 L) = 1.58 mol WORKED EXAMPLE 13.10 CALCULATING EQUILIBRIUM CONCENTRATIONS FROM INITIAL CONCENTRATIONS Calculate the equilibrium concentrations of H2, I2, and HI at 700 K if the initial concentrations are [H 2] = 0.100 M and [I 2] = 0.200 M. The equilibrium constant Kc for the reaction H 2(g) + I 2(g) ∆ 2 HI(g) is 57.0 at 700 K. ? STRATEGY This problem is similar to Worked Example 13.9 except that the initial concentrations of H2 and I2 are unequal. Again, we follow the steps in Figure 13.6. SOLUTION Step 1. The balanced equation is H 2(g) + I 2(g) ∆ 2 HI(g). Step 2. Again, define x as the concentration of H2 that reacts. Set up a table of concentrations under the balanced equation: H2(g) Initial concentration (M) Change (M) Equilibrium concentration (M) ⴙ I2(g) ∆ 2 HI(g) 0.100 0.200 0 -x -x + 2x (0.100 - x) (0.200 - x) 2x continued on next page Initial state Equilibrium state 510 Chapter 13 CHEMICAL EQUILIBRIUM Step 3. Substitute the equilibrium concentrations into the equilibrium equation: [HI]2 Kc = 57.0 = [H 2][I 2] = (2x)2 (0.100 - x)(0.200 - x) Because the right side of this equation is not a perfect square, we must put the equation into the standard quadratic form, ax 2 + bx + c = 0, and then solve for x using the quadratic formula (Appendix A.4): - b ; 2b 2 - 4ac 2a Rearranging the equilibrium equation gives x = (57.0)(0.0200 - 0.300x + x 2) = 4x 2 53.0x 2 - 17.1x + 1.14 = 0 or Substituting the values of a, b, and c into the quadratic formula gives two solutions: x = 17.1 ; 3( - 17.1)2 - 4(53.0)(1.14) 2(53.0) = 17.1 ; 7.1 = 0.228 and 0.0943 106 Discard the solution that uses the positive square root (x = 0.228) because the H2 concentration can’t change by more than its initial value (0.100 M). Therefore, choose the solution that uses the negative square root (x = 0.0943). Step 4. Calculate the equilibrium concentrations from the calculated value of x: [H 2] = 0.100 - x = 0.100 - 0.0943 = 0.006 M [I 2] = 0.200 - x = 0.200 - 0.0943 = 0.106 M [HI] = 2x = (2)(0.0943) = 0.189 M Step 5. Check the results by substituting them into the equilibrium equation: Kc = 57.0 = [HI]2 [H 2][I 2] = (0.189)2 (0.006)(0.106) = 56.2 The calculated value of Kc (56.2), which should be rounded to one significant figure (6 * 101), agrees with the value given in the problem (57.0). WORKED EXAMPLE 13.11 CALCULATING EQUILIBRIUM PARTIAL PRESSURES FROM INITIAL PARTIAL PRESSURES One reaction that occurs in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and carbon dioxide. The equilibrium constant Kp for the reaction at 1000 K is 0.259. Kp = 0.259 at 1000 K FeO(s) + CO(g) ∆ Fe(s) + CO 2(g) What are the equilibrium partial pressures of CO and CO2 at 1000 K if the initial partial pressures are PCO = 1.000 atm and PCO2 = 0.500 atm? STRATEGY 䉱 The steel used in making automobiles is produced by the reaction of iron ore with carbon monoxide. We can calculate equilibrium partial pressures from initial partial pressures and Kp in the same way that we calculate equilibrium concentrations from initial concentrations and Kc. Follow the steps in Figure 13.6, but substitute partial pressures for concentrations. SOLUTION Step 1. The balanced equation is FeO(s) + CO(g) ∆ Fe(s) + CO2(g). Step 2. Define x as the partial pressure of CO that reacts. Set up a table of partial pressures of the gases under the balanced equation: FeO(s) Initial pressure (atm) Change (atm) Equilibrium pressure (atm) ⴙ CO(g) ∆ 1.000 Fe(s) ⴙ CO2(g) 0.500 -x +x (0.1000 - x) (0.500 + x) 13.6 FACTORS THAT ALTER THE COMPOSITION OF AN EQUILIBRIUM MIXTURE: LE CHÂTELIER’S PRINCIPLE Step 3. Substitute the equilibrium partial pressures into the equilibrium equation for Kp: Kp = 0.259 = PCO2 PCO = 0.500 + x 1.000 - x As usual for a heterogeneous equilibrium, we omit the pure solids from the equilibrium equation. Rearranging the equilibrium equation and solving for x gives 0.259 - 0.259x = 0.500 + x - 0.241 x = = - 0.191 1.259 Step 4. Calculate the equilibrium partial pressures from the calculated value of x: PCO = 1.000 - x = 1.000 - ( - 0.191) = 1.191 atm PCO2 = 0.500 + x = 0.500 + ( - 0.191) = 0.309 atm Step 5. Check the results by substituting them into the equilibrium equation: Kp = 0.259 = PCO2 PCO = 0.309 = 0.259 1.191 A negative value for x means that the reaction goes from products to reactants to reach equilibrium. This makes sense because the initial reaction quotient, Qp = 0.500/1.000 = 0.500, is greater than the equilibrium constant, Kp = 0.259. When Qp 7 Kp, the net reaction always goes from products to reactants (right to left). 쑺 PROBLEM 13.11 In Problem 13.8, we found that an equilibrium mixture of H2 molecules and H atoms at 500 K contains mainly H2 molecules because the equilibrium constant for the dissociation reaction H 2(g) ∆ 2 H(g) is very small (Kc = 1.2 * 10-42). (a) What is the molar concentration of H atoms at equilibrium if the equilibrium concentration of H2 is 0.10 M? (b) How many H atoms and H2 molecules are present in 1.0 L of 0.10 M H2 at 500 K? 쑺 PROBLEM 13.12 The H2/CO ratio in mixtures of carbon monoxide and hydrogen (called synthesis gas) is increased by the water-gas shift reaction CO(g) + H 2O(g) ∆ CO 2(g) + H 2(g), which has an equilibrium constant Kc = 4.24 at 800 K. Calculate the equilibrium concentrations of CO2, H2, CO, and H2O at 800 K if only CO and H2O are present initially at concentrations of 0.150 M. 쑺 PROBLEM 13.13 Calculate the equilibrium concentrations of N2O4 and NO2 at 25 °C in a vessel that contains an initial N2O4 concentration of 0.0500 M. The equilibrium constant Kc for the reaction N2O4(g) ∆ 2 NO 2(g) is 4.64 * 10-3 at 25 °C. 쑺 PROBLEM 13.14 Calculate the equilibrium concentrations at 25 °C for the reaction in Problem 13.13 if the initial concentrations are [N2O4] = 0.0200 M and [NO 2] = 0.0300 M. 쑺 PROBLEM 13.15 The equilibrium constant Kp for the reaction C(s) + H 2O(g) ∆ CO(g) + H 2(g) is 2.44 at 1000 K. What are the equilibrium partial pressures of H2O, CO, and H2 if the initial partial pressures are PH2O = 1.20 atm, PCO = 1.00 atm, and PH2 = 1.40 atm? 13.6 FACTORS THAT ALTER THE COMPOSITION OF AN EQUILIBRIUM MIXTURE: LE CHÂTELIER’S PRINCIPLE One of the main goals of chemical synthesis is to maximize the conversion of reactants to products while minimizing the expenditure of energy. This objective is achieved easily if the reaction goes nearly to completion at mild temperature and pressure. If the reaction gives an equilibrium mixture that is rich in reactants and poor in products, however, then the experimental conditions must be adjusted. 511 512 Chapter 13 CHEMICAL EQUILIBRIUM For example, in the Haber process for the synthesis of ammonia from N2 and H2 (Figure 13.7), the choice of experimental conditions is of real economic importance. Annual worldwide production of ammonia is about 120 million metric tons, primarily for use as fertilizer. Figure 13.7 Representation of the Haber process for the industrial production of ammonia. A mixture of gaseous N2 and H2 at 130–300 atm pressure is passed over a catalyst at 400–500 °C, and ammonia is produced by the reaction N2(g) + 3 H 2(g) ∆ 2 NH 3(g). The NH3 in the gaseous mixture of reactants and products is liquefied, and the unreacted N2 and H2 are recycled. Pump to circulate gases N 2 , H2 inlet Pump to compress gases Expanding gases cool Heat exchanger Heat exchanger Recycled unreacted N2 and H2 Catalyst (400–500 °C) Heating coil Refrigeration unit Preheated feed gases (130–300 atm) NH3 outlet to storage tank Several factors can be exploited to alter the composition of an equilibrium mixture: • The concentration of reactants or products can be changed. • The pressure and volume can be changed. • The temperature can be changed. A possible fourth factor, addition of a catalyst, increases only the rate at which equilibrium is reached. As we’ll see in Section 13.10, a catalyst does not affect the equilibrium concentrations. The qualitative effect of the listed changes on the composition of an equilibrium mixture can be predicted using a principle first described by the French chemist Henri-Louis Le Châtelier (pronounced Li Sha–tell–yea): Le Châtelier’s Principle If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress. The word “stress” in this context means a change in concentration, pressure, volume, or temperature that disturbs the original equilibrium. Reaction then occurs to change 13.7 ALTERING AN EQUILIBRIUM MIXTURE: CHANGES IN CONCENTRATION 513 the composition of the mixture until a new state of equilibrium is reached. The direction that the reaction takes (reactants to products or vice versa) is the one that reduces the stress. In the next three sections, we’ll look at the different kinds of stress that can change the composition of an equilibrium mixture. 13.7 ALTERING AN EQUILIBRIUM MIXTURE: CHANGES IN CONCENTRATION Let’s consider the equilibrium that occurs in the Haber process for the synthesis of ammonia: N2(g) + 3 H2(g) 2 NH3(g) Kc = 0.291 at 700 K + Suppose that we have an equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 at 700 K and that we disturb the equilibrium by increasing the N2 concentration to 1.50 M. Le Châtelier’s principle tells us that reaction will occur to relieve the stress of the increased concentration of N2 by converting some of the N2 to NH3. As the N2 concentration decreases, the H2 concentration must also decrease and the NH3 concentration must increase in accord with the stoichiometry of the balanced equation. These changes are illustrated in Figure 13.8. Figure 13.8 Net conversion of N2 and H2 to NH3 occurs until a new equilibrium is established. That is, the N2 and H2 concentrations decrease, while the NH3 concentration increases. Initial equilibrium Qc = Kc Concentration (M) 3.00 New equilibrium Qc < Kc Qc = Kc H2 H2 2.00 NH3 NH3 N2 1.00 N2 N2 added at this time Time In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Châtelier’s principle predicts that • The concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance. • The concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance. Changes in concentrations when N2 is added to an equilibrium mixture of N2, H2, and NH3. 514 Chapter 13 CHEMICAL EQUILIBRIUM If these rules are applied to the equilibrium N2(g) + 3 H 2(g) ∆ 2 NH 3(g), then the yield of ammonia is increased by an increase in the N2 or H2 concentration or by a decrease in the NH3 concentration (Figure 13.9). In the industrial production of ammonia, the concentration of gaseous NH3 is decreased by liquefying the ammonia (bp -33 °C) as it’s formed, and so more ammonia is produced. Any of the changes marked in blue shifts the equilibrium to the left. Any of the changes marked in red shifts the equilibrium to the right. decrease decrease decrease N2 + 3 H2 2 NH3 increase increase increase Figure 13.9 Effect of concentration changes on the equilibrium N2(g) ⴙ 3 H2(g) ∆ 2 NH3(g). An increase in the N2 or H2 concentration or a decrease in the NH3 concentration shifts the equilibrium from left to right. A decrease in the N2 or H2 concentration or an increase in the NH3 concentration shifts the equilibrium from right to left. Le Châtelier’s principle is a handy rule for predicting changes in the composition of an equilibrium mixture, but it doesn’t explain why those changes occur. To see why Le Châtelier’s principle works, let’s look again at the reaction quotient Qc. For the initial equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 at 700 K, Qc equals the equilibrium constant Kc (0.291) because the system is at equilibrium: Qc = [NH 3]t 2 = [N2]t[H 2]t 3 (1.98)2 (0.50)(3.00)3 = 0.29 = Kc When we disturb the equilibrium by increasing the N2 concentration to 1.50 M, the denominator of the equilibrium constant expression increases and Qc decreases to a value less than Kc: Qc = [NH 3]t 2 [N2]t[H 2]t 3 = (1.98)2 (1.50)(3.00)3 = 0.0968 6 Kc For the system to move to a new state of equilibrium, Qc must increase; that is, the numerator of the equilibrium constant expression must increase and the denominator must decrease. This implies the net conversion of N2 and H2 to NH3, just as predicted by Le Châtelier’s principle. When the new equilibrium is established (Figure 13.8), the concentrations are 1.31 M N2, 2.43 M H2, and 2.36 M NH3, and Qc again equals Kc: Qc = [NH 3]t2 [N2]t[H 2]t3 = (2.36)2 (1.31)(2.43)3 = 0.296 = Kc As another example of how a change in concentration affects an equilibrium, let’s consider the reaction in aqueous solution of iron(III) ions and thiocyanate (SCN -) ions to give an equilibrium mixture that contains the Fe–N bonded red complex ion FeNCS 2+ : Fe 3 + (aq) + SCN -(aq) ∆ FeNCS 2 + (aq) Pale yellow Colorless Red Shifts in the position of this equilibrium can be detected by observing how the color of the solution changes when we add various reagents (Figure 13.10). If we add aqueous FeCl3, the red color gets darker, as predicted by Le Châtelier’s principle. The 13.7 ALTERING AN EQUILIBRIUM MIXTURE: CHANGES IN CONCENTRATION (a) Original solution: Fe3+(pale yellow), SCN–(colorless), and FeNCS2+(red). (b) After adding FeCl3 to (a): [FeNCS2+] (c) After adding KSCN to (a): [FeNCS2+] increases. increases. (d) After adding H2C2O4 to (a): [FeNCS2+] decreases as [Fe(C2O4)33–] increases. 515 (e) After adding HgCl2 to (a): [FeNCS2+] decreases as [Hg(SCN)42–] increases. Figure 13.10 Color changes produced by adding various reagents to an equilibrium mixture of Fe 3+ (pale yellow), SCN - (colorless), and FeNCS2+ (red). concentration stress of added Fe 3+ is relieved by net reaction from left to right, which consumes some of the Fe 3+ and increases the concentration of FeNCS 2+ . (Note that the Cl - ions are not involved in the reaction.) Similarly, if we add aqueous KSCN, the stress of added SCN - shifts the equilibrium from left to right and again the red color gets darker. The equilibrium can be shifted in the opposite direction by adding reagents that remove Fe 3+ or SCN - ions. For example, oxalic acid (H2C2O4), a poisonous substance present in the leaves of plants such as rhubarb, reacts with Fe 3+ to form the stable, yellow complex ion Fe(C2O4)33- , thus decreasing the concentration of free Fe 3+(aq). In accord with Le Châtelier’s principle, the concentration stress of removed Fe 3+ is relieved by the dissociation of FeNCS 2+ to replenish the Fe 3+ ions. Because the concentration of FeNCS 2+ decreases, the red color disappears. 3 H 2C2O4(aq) + Fe 3 + (aq) ¡ Fe(C2O4)3 3 - (aq) + 6 H +(aq) FeNCS 2 + (aq) ¡ Fe 3 + (aq) + SCN - (aq) Addition of aqueous HgCl2 also eliminates the red color because HgCl2 reacts with SCN - ions to form the stable, colorless Hg–S bonded complex ion Hg(SCN)4 2- . Removal of free SCN -(aq) results in dissociation of the red FeNCS 2+ ions so as to replenish the SCN - ions. HgCl2(aq) + 4 SCN - (aq) ¡ Hg(SCN)4 2 - (aq) + 2 Cl - (aq) FeNCS 2 + (aq) ¡ Fe 3 + (aq) + SCN - (aq) WORKED EXAMPLE 13.12 APPLYING LE CHÂTELIER’S PRINCIPLE TO CONCENTRATION CHANGES The reaction of iron(III) oxide with carbon monoxide occurs in a blast furnace when iron ore is reduced to iron metal: Fe2O3(s) + 3 CO(g) ∆ 2 Fe(l) + 3 CO 2(g) Use Le Châtelier’s principle to predict the direction of the net reaction when an equilibrium mixture is disturbed by: (a) Adding Fe2O3 (b) Removing CO2 (c) Removing CO; also account for the change using the reaction quotient Qc. continued on next page 䉱 The extremely sour leaves of rhubarb contain toxins such as oxalic acid, but the stalks and roots are nutritious. 516 Chapter 13 CHEMICAL EQUILIBRIUM STRATEGY To predict the direction of net reaction, recall that a concentration stress is relieved by reaction in the direction that consumes an added substance or replenishes a removed substance. This rule does not apply to pure solids or pure liquids because their concentrations have a constant value equal to 1. SOLUTION (a) Because Fe2O3 is a pure solid, its concentration is equal to 1 and doesn’t change when more Fe2O3 is added. Therefore, there is no concentration stress and the original equilibrium is undisturbed. (b) Le Châtelier’s principle predicts that the concentration stress of removed CO2 will be relieved by net reaction from reactants to products to replenish the CO2. (c) Le Châtelier’s principle predicts that the concentration stress of removed CO will be relieved by net reaction from products to reactants to replenish the CO. The reaction quotient is Qc = [CO2]t 3 [CO]t 3 When the equilibrium is disturbed by reducing [CO], Qc increases, so that Qc 7 Kc. For the system to move to a new state of equilibrium, Qc must decrease—that is, [CO2] must decrease and [CO] must increase. Therefore, the net reaction goes from products to reactants, as predicted by Le Châtelier’s principle. 쑺 PROBLEM 13.16 Consider the equilibrium for the water-gas shift reaction: CO(g) + H 2O(g) ∆ CO 2(g) + H 2(g) Use Le Châtelier’s principle to predict how the concentration of H2 will change when the equilibrium is disturbed by: (a) Adding CO (b) Adding CO2 (c) Removing H2O (d) Removing CO2; also account for the change using the reaction quotient Qc. 13.8 ALTERING AN EQUILIBRIUM MIXTURE: CHANGES IN PRESSURE AND VOLUME To illustrate how an equilibrium mixture is affected by a change in pressure as a result of a change in the volume, let’s return to the Haber synthesis of ammonia. The balanced equation for the reaction has 4 mol of gas on the reactant side of the equation and 2 mol on the product side: N2(g) + 3 H 2(g) ∆ 2 NH 3(g) Kc = 0.291 at 700 K What happens to the composition of the equilibrium mixture if we increase the pressure by decreasing the volume? (Recall from Sections 9.2 and 9.3 that the pressure of an ideal gas is inversely proportional to the volume at constant temperature and constant number of moles of gas; P = nRT/V.) According to Le Châtelier’s principle, net reaction will occur in the direction that relieves the stress of the increased pressure, which means that the number of moles of gas must decrease. Therefore, we predict that the net reaction will proceed from left to right because the forward reaction converts 4 mol of gaseous reactants to 2 mol of gaseous products. In general, Le Châtelier’s principle predicts that • An increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas. • A decrease in pressure by expanding the volume will bring about net reaction in the direction that increases the number of moles of gas. 13.8 ALTERING AN EQUILIBRIUM MIXTURE: CHANGES IN PRESSURE AND VOLUME To see why Le Châtelier’s principle works for pressure (volume) changes, let’s look again at the reaction quotient for the equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 at 700 K: Qc = [NH 3]t 2 [N2]t[H 2]t 3 = (1.98)2 (0.50)(3.00)3 = 0.29 = Kc If we disturb the equilibrium by reducing the volume by a factor of 2, we not only double the total pressure, we also double the partial pressure and thus the molar concentration of each reactant and product (because molarity = n/V = P/RT). Because the balanced equation has more moles of gaseous reactants than gaseous products, the increase in the denominator of the equilibrium constant expression is greater than the increase in the numerator and the new value of Qc is less than the equilibrium constant Kc: Qc = [NH 3]t 2 [N2]t[H 2]t 3 = (3.96)2 (1.00)(6.00)3 = 0.0726 6 Kc For the system to move to a new state of equilibrium, Qc must increase, which means that the net reaction must go from reactants to products, as predicted by Le Châtelier’s principle (Figure 13.11). In practice, the yield of ammonia in the Haber process is increased by running the reaction at high pressure, typically 130–300 atm. (a) A mixture of gaseous N2, H2, and NH3 at equilibrium (Qc = Kc). (b) When the pressure is increased (c) Net reaction occurs from reactants by decreasing the volume, the mixture is no longer at equilibrium (Qc < Kc). to products, decreasing the total number of gaseous molecules until equilibrium is re-established (Qc = Kc). = N2 P increases as Net reaction V decreases to form products Figure 13.11 Qualitative effect of pressure and volume on the equilibrium N2(g) ⴙ 3 H2(g) ∆ 2 NH3(g). The composition of an equilibrium mixture is unaffected by a change in pressure if the reaction involves no change in the number of moles of gas. For example, the reaction of hydrogen with gaseous iodine has 2 mol of gas on both sides of the balanced equation: H 2(g) + I 2(g) ∆ 2 HI(g) If we double the pressure by halving the volume, the numerator and denominator of the reaction quotient change by the same factor and Qc remains unchanged: Qc = [HI]t 2 [H 2]t[I 2]t = H2 = NH3 517 518 Chapter 13 CHEMICAL EQUILIBRIUM In applying Le Châtelier’s principle to a heterogeneous equilibrium, the effect of pressure changes on solids and liquids can be ignored because the volume (and concentration) of a solid or a liquid is nearly independent of pressure. Consider, for example, the high-temperature reaction of carbon with steam, the first step in converting coal to gaseous fuels: C(s) + H 2O(g) ∆ CO(g) + H 2(g) Ignoring the carbon because it’s a solid, we predict that a decrease in volume (increase in pressure) will shift the equilibrium from products to reactants because the reverse reaction decreases the amount of gas from 2 mol to 1 mol. Throughout this section, we’ve been careful to limit the application of Le Châtelier’s principle to pressure changes that result from a change in volume. What happens, though, if we keep the volume constant but increase the total pressure by adding a gas that is not involved in the reaction—say, an inert gas such as argon? In that case, the equilibrium remains undisturbed because adding an inert gas at constant volume does not change the partial pressures or the molar concentrations of the substances involved in the reaction. Only if the added gas is a reactant or product does the reaction quotient change. WORKED EXAMPLE 13.13 APPLYING LE CHÂTELIER’S PRINCIPLE TO PRESSURE AND VOLUME CHANGES Does the number of moles of reaction products increase, decrease, or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume? (a) PCl5(g) ∆ PCl3(g) + Cl2(g) (b) CaO(s) + CO 2(g) ∆ CaCO 3(s) (c) 3 Fe(s) + 4 H 2O(g) ∆ Fe3O4(s) + 4 H 2(g) STRATEGY According to Le Châtelier’s principle, the stress of a decrease in pressure is relieved by net reaction in the direction that increases the number of moles of gas. SOLUTION (a) Because the forward reaction converts 1 mol of gas to 2 mol of gas, net reaction will go from reactants to products, thus increasing the number of moles of PCl3 and Cl2. (b) Because there is 1 mol of gas on the reactant side of the balanced equation and none on the product side, the stress of a decrease in pressure is relieved by net reaction from products to reactants. The number of moles of CaCO3 therefore decreases. (c) Because there are 4 mol of gas on both sides of the balanced equation, the composition of the equilibrium mixture is unaffected by a change in pressure. The number of moles of Fe3O4 and H2 remains the same. 쑺 PROBLEM 13.17 Does the number of moles of products increase, decrease, or remain the same when each of the following equilibria is subjected to an increase in pressure by decreasing the volume? (a) CO(g) + H 2O(g) ∆ CO 2(g) + H 2(g) (b) 2 CO(g) ∆ C(s) + CO 2(g) (c) N2O4(g) ∆ 2 NO 2(g) CONCEPTUAL PROBLEM 13.18 The following picture represents the equilibrium mixture for the gas-phase reaction A2 ∆ 2 A: Draw a picture that shows how the concentrations change when the pressure is increased by reducing the volume. 13.9 ALTERING AN EQUILIBRIUM MIXTURE: CHANGES IN TEMPERATURE 519 13.9 ALTERING AN EQUILIBRIUM MIXTURE: CHANGES IN TEMPERATURE When an equilibrium is disturbed by a change in concentration, pressure, or volume, the composition of the equilibrium mixture changes because the reaction quotient Qc no longer equals the equilibrium constant Kc. As long as the temperature remains constant, however, concentration, pressure, or volume changes don’t change the value of the equilibrium constant. By contrast, a change in temperature nearly always changes the value of the equilibrium constant. For the Haber synthesis of ammonia, which is an exothermic reaction, the equilibrium constant Kc decreases by a factor of 1011 over the temperature range 300–1000 K (Figure 13.12). N2(g) + 3 H 2(g) ∆ 2 NH 3(g) + 92.2 kJ ¢H° = - 92.2 kJ Figure 13.12 Temperature dependence of the equilibrium constant for the reaction N2(g) ⴙ 3 H2(g) ∆ 2 NH3(g). Kc is plotted on a logarithmic scale. Equilibrium constant, Kc 10 Temp (K) 8 Kc 300 2.6 × 10 8 400 3.9 × 10 4 500 1.7 × 10 2 600 4.2 700 2.9 × 10−1 1 800 3.9 × 10−2 10−2 900 8.1 × 10−3 1000 2.3 × 10−3 10 6 Kc decreases by a factor of 1011 on raising T from 300 K to 1000 K. 10 4 10 2 10−4 200 400 600 800 N2O4(g) 2 NO2(g); ΔH > 0 1000 Temperature (K) At low temperatures, the equilibrium mixture is rich in NH3 because Kc is large. At high temperatures, the equilibrium shifts in the direction of N2 and H2. In general, the temperature dependence of an equilibrium constant depends on the sign of ¢H° for the reaction. • The equilibrium constant for an exothermic reaction (negative ¢H°) decreases as the temperature increases. • The equilibrium constant for an endothermic reaction (positive ¢H°) increases as the temperature increases. You can predict the way in which Kc depends on temperature by using Le Châtelier’s principle. Take the endothermic decomposition of N2O4, for example: N2O4(g) + 55.3 kJ ∆ 2 NO2(g) Colorless ¢H° = + 55.3 kJ Brown Le Châtelier’s principle says that if heat is added to an equilibrium mixture, thus increasing its temperature, net reaction occurs in the direction that relieves the stress of the added heat. For an endothermic reaction, such as the decomposition of N2O4, heat is absorbed by reaction in the forward direction. The equilibrium therefore shifts to the product side at the higher temperature, which means that Kc increases with increasing temperature. Because N2O4 is colorless and NO2 has a brown color, the effect of temperature on the N2O4 -NO2 equilibrium is readily apparent from the color of the mixture (Figure 13.13). For an exothermic reaction, such as the Haber synthesis of NH3, heat is absorbed by net reaction in the reverse direction, so Kc decreases with increasing temperature. Kc increases as T increases The darker brown color of the sample at the highest temperature indicates that the equilibrium N2O4(g) 2 NO2(g) shifts from reactants to products with increasing temperature, as expected for an endothermic reaction. Figure 13.13 Sample tubes containing an equilibrium mixture of N2O4 and NO2 immersed in ice water (left), at room temperature (center), and immersed in hot water (right). 520 Chapter 13 CHEMICAL EQUILIBRIUM WORKED EXAMPLE 13.14 APPLYING LE CHÂTELIER’S PRINCIPLE TO TEMPERATURE CHANGES In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction 4 NH 3(g) + 5 O 2(g) ∆ 4 NO(g) + 6 H 2O(g) ¢H° = - 901.2 kJ How does the equilibrium amount of NO vary with an increase in temperature? STRATEGY Le Châtelier’s principle predicts that the stress of added heat when the temperature is increased will be relieved by net reaction in the direction that absorbs the heat. It’s helpful to include the heat in the balanced equation—on the reactant side if the reaction is endothermic, or on the product side if the reaction is exothermic. SOLUTION Because the oxidation of ammonia is exothermic, we include the heat (901.2 kJ) on the product side: 4 NH 3(g) + 5 O 2(g) ∆ 4 NO(g) + 6 H 2O(g) + 901.2 kJ The stress of added heat when the temperature is increased will be relieved by net reaction from products to reactants, which absorbs the added heat. The equilibrium will therefore shift to the reactant side (Kc will decrease) with an increase in temperature. Consequently, the equilibrium mixture will contain less NO at higher temperatures. 쑺 PROBLEM 13.19 When air is heated at very high temperatures in an automobile engine, the air pollutant nitric oxide is produced by the reaction ¢H° = + 182.6 kJ N2(g) + O 2(g) ∆ 2 NO(g) How does the equilibrium amount of NO vary with an increase in temperature? 쑺 PROBLEM 13.20 Ethyl acetate, a solvent used in many fingernail-polish removers, is made by the reaction of acetic acid with ethanol: CH 3CO2H(soln) + CH 3CH 2OH(soln) ∆ CH 3CO2CH 2CH 3(soln) + H 2O(soln) Acetic acid Ethanol ¢H° = - 2.9 kJ Ethyl acetate Does the amount of ethyl acetate in an equilibrium mixture increase or decrease when the temperature is increased? How does Kc change when the temperature is decreased? Justify your answers using Le Châtelier’s principle. WORKED CONCEPTUAL EXAMPLE 13.15 APPLYING LE CHÂTELIER’S PRINCIPLE TO TEMPERATURE CHANGES The following pictures represent the composition of the equilibrium mixture at 600 K and 650 K for the combination of two A molecules, 2 A(g) ∆ A2(g): 䉱 Fingernail polish can be removed by dissolving it in ethyl acetate. T = 600 K T = 650 K Is the reaction endothermic or exothermic? Explain using Le Châtelier’s principle. STRATEGY We can determine the direction of net reaction on raising the temperature by counting the number of A and A2 molecules at each temperature. According to Le Châtelier’s principle, if the net reaction converts reactants to products on raising the temperature, heat is on the reactant side of the chemical equation and the reaction is endothermic. Conversely, if the net reaction converts products to reactants on raising the temperature, heat is on the product side and the reaction is exothermic. 13.10 THE EFFECT OF A CATALYST ON EQUILIBRIUM 521 SOLUTION Two A and five A2 molecules are present at 600 K, and six A and three A2 molecules are present at 650 K. On raising the temperature, the net reaction converts products to reactants, and so heat is on the product side of the chemical equation: 2 A(g) ∆ A2(g) + heat The reaction is therefore exothermic, as expected for a reaction in which a chemical bond is formed. Note that Le Châtelier’s principle predicts that net reaction will occur in the direction that uses up the added heat. CONCEPTUAL PROBLEM 13.21 The following pictures represent the composition of the equilibrium mixture for the reaction A(g) + B(s) ∆ AB(g) at 400 K and 500 K: T = 400 K T = 500 K Is the reaction endothermic or exothermic? Explain using Le Châtelier’s principle. 13.10 THE EFFECT OF A CATALYST ON EQUILIBRIUM Recall from Section 12.14 that a catalyst increases the rate of a chemical reaction by making available a new, lower-energy pathway for the conversion of reactants to products. Because the forward and reverse reactions pass through the same transition state, a catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount. As a result, the rates of the forward and reverse reactions increase by the same factor (Figure 13.14). Remember... A catalyst is a substance that increases the rate of a transformation without being consumed in the process. (Section 12.14) The activation energy for the catalyzed pathway (red curve) is lower than that for the uncatalyzed pathway (blue curve) by an amount ∆Ea. Potential energy ∆Ea Ea (forward) without catalyst Ea (reverse) without catalyst Ea (forward) with catalyst Ea (reverse) with catalyst Reactants Products Reaction progress Because the forward and reverse reactions pass through the same transition state, the catalyst lowers the activation energy barrier for the forward and reverse reactions by the same amount. The catalyst therefore accelerates the forward and reverse reactions by the same factor, and the composition of the equilibrium mixture is unchanged. Figure 13.14 Potential energy profiles for a reaction whose activation energy is lowered by the presence of a catalyst. 522 Chapter 13 CHEMICAL EQUILIBRIUM If a reaction mixture is at equilibrium in the absence of a catalyst (that is, the forward and reverse rates are equal), it will still be at equilibrium after a catalyst is added because the forward and reverse rates, though faster, remain equal. If a reaction mixture is not at equilibrium, a catalyst accelerates the rate at which equilibrium is reached, but it does not affect the composition of the equilibrium mixture. Because a catalyst has no effect on the equilibrium concentrations, it does not appear in the balanced chemical equation or in the equilibrium constant expression. Even though a catalyst doesn’t change the position of an equilibrium, it can nevertheless significantly influence the choice of optimum conditions for a reaction. Look again at the Haber synthesis of ammonia. Because the reaction N2(g) + 3 H 2(g) ∆ 2 NH 3(g) is exothermic, its equilibrium constant decreases with increasing temperature, and optimum yields of NH3 are obtained at low temperatures. At those low temperatures, however, the rate at which equilibrium is reached is too slow for the reaction to be practical. We thus have what appears to be a no-win situation: Low temperatures give good yields but slow rates, whereas high temperatures give satisfactory rates but poor yields. The answer to the dilemma is to find a catalyst. In the early 1900s, the German chemist Fritz Haber discovered that a catalyst consisting of iron mixed with certain metal oxides causes the reaction to occur at a satisfactory rate at temperatures where the equilibrium concentration of NH3 is reasonably favorable. The yield of NH3 can be improved further by running the reaction at high pressures. Typical reaction conditions for the industrial synthesis of ammonia are 400–500 °C and 130–300 atm. 쑺 PROBLEM 13.22 A platinum catalyst is used in automobile catalytic converters to hasten the oxidation of carbon monoxide: Pt 2 CO(g) + O2(g) ERF 2 CO2(g) ¢H° = - 566 kJ Suppose that you have a reaction vessel containing an equilibrium mixture of CO(g), O 2(g), and CO2(g). Under the following conditions, will the amount of CO increase, decrease, or remain the same? (a) A platinum catalyst is added. (b) The temperature is increased. (c) The pressure is increased by decreasing the volume. (d) The pressure is increased by adding argon gas. (e) The pressure is increased by adding O2 gas. 13.11 THE LINK BETWEEN CHEMICAL EQUILIBRIUM AND CHEMICAL KINETICS We emphasized in Section 13.1 that the equilibrium state is a dynamic one in which reactant and product concentrations remain constant, not because the reaction stops but because the rates of the forward and reverse reactions are equal. To explore this idea further, let’s consider the general, reversible reaction A + B ∆ C + D Remember... Because an elementary reaction describes an individual molecular event, its rate law follows directly from its stoichiometry. (Sections 12.9 and 12.10) Let’s assume that the forward and reverse reactions occur in a single bimolecular step; that is, they are elementary reactions (Section 12.9). We can then write the following rate laws: Rate forward = kf[A][B] Rate reverse = kr[C][D] 13.11 THE LINK BETWEEN CHEMICAL EQUILIBRIUM AND CHEMICAL KINETICS 523 If we begin with a mixture that contains all reactants and no products, the initial rate of the reverse reaction is zero because [C] = [D] = 0. As A and B are converted to C and D by the forward reaction, the rate of the forward reaction decreases because [A] and [B] are getting smaller. At the same time, the rate of the reverse reaction increases because [C] and [D] are getting larger. Eventually, the decreasing rate of the forward reaction and the increasing rate of the reverse reaction become equal, and thereafter the concentrations remain constant; that is, the system is at chemical equilibrium (Figure 13.2, page 495). Because the forward and reverse rates are equal at equilibrium, we can write kf[A][B] = kr[C][D] at equilibrium which can be rearranged to give [C][D] kf = kr [A][B] The right side of this equation is the equilibrium constant expression for the forward reaction, which equals the equilibrium constant Kc since the reaction mixture is at equilibrium. Kc = [C][D] [A][B] Therefore, the equilibrium constant is simply the ratio of the rate constants for the forward and reverse reactions: Kc = kf kr In deriving this equation for Kc, we have assumed a single-step mechanism. For a multistep mechanism, each step has a characteristic rate constant ratio, kf/kr. When equilibrium is reached, each step in the mechanism must be at equilibrium, and Kc for the overall reaction is equal to the product of the rate constant ratios for the individual steps. The equation relating Kc to kf and kr provides a fundamental link between chemical equilibrium and chemical kinetics: The relative values of the rate constants for the forward and reverse reactions determine the composition of the equilibrium mixture. When kf is much larger than kr, Kc is very large and the reaction goes almost to completion. Such a reaction is sometimes said to be irreversible because the reverse reaction is often too slow to be detected. When kf and kr have comparable values, Kc has a value near 1, and comparable concentrations of both reactants and products are present at equilibrium. This is the usual situation for a reversible reaction. Addition of a catalyst to a reaction mixture increases both rate constants kf and kr because the reaction takes place by a different, lower-energy mechanism. Because kf and kr increase by the same factor, though, the ratio kf/kr is unaffected, and the value of the equilibrium constant Kc = kf/kr remains unchanged. Thus, addition of a catalyst does not alter the composition of an equilibrium mixture. The equation Kc = kf/kr also helps explain why equilibrium constants depend on temperature. Recall from Section 12.12 that the rate constant increases as the temperature increases, in accord with the Arrhenius equation k = Ae - Ea/RT. In general, the forward and reverse reactions have different values of the activation energy, so kf and kr increase by different amounts as the temperature increases. The ratio kf/kr = Kc is therefore temperature-dependent. For an exothermic reaction, Ea for the reverse reaction is greater than Ea for the forward reaction. Consequently, as the temperature increases, kr increases by more than kf increases, and so Kc = kf/kr for an exothermic reaction decreases as the temperature increases. Conversely, Kc for an endothermic reaction increases as the temperature increases. These results are in accord with Le Châtelier’s principle (Section 13.9). Remember... Because the fraction of collisions with sufficient energy for reaction is given by e-Ea /RT, the Arrhenius equation indicates that the rate constant decreases as Ea increases and increases as T increases. (Section 12.12) 524 Chapter 13 CHEMICAL EQUILIBRIUM WORKED EXAMPLE 13.16 EXPLORING THE LINK BETWEEN EQUILIBRIUM AND KINETICS The equilibrium constant Kc for the reaction of hydrogen with iodine is 57.0 at 700 K, and the reaction is endothermic (¢E = 9 kJ). kf H 2(g) + I 2(g) ERF 2 HI(g) kr Kc = 57.0 at 700 K (a) Is the rate constant kf for the formation of HI larger or smaller than the rate constant kr for the decomposition of HI? (b) The value of kr at 700 K is 1.16 * 10-3 M -1 s -1. What is the value of kf at the same temperature? (c) How are the values of kf, kr, and Kc affected by the addition of a catalyst? (d) How are the values of kf, kr, and Kc affected by an increase in temperature? STRATEGY Remember... The greater the activation energy, the steeper the slope of an Arrhenius plot (a graph of ln k versus 1/T) and the greater the increase in k for a given increase in T. (Section 12.13) To answer these questions, make use of the relationship Kc = kf/kr. Also, remember that a catalyst increases kf and kr by the same factor, and recall that the temperature dependence of a rate constant increases with increasing value of the activation energy (Section 12.13). SOLUTION (a) Because Kc = kf/kr = 57.0, the rate constant for the formation of HI (forward reaction) is larger than the rate constant for the decomposition of HI (reverse reaction) by a factor of 57.0. (b) Because Kc = kf/kr, kf = (Kc)(kr) = (57.0)(1.16 * 10-3 M -1 s -1) = 6.61 * 10-2 M -1 s -1 (c) A catalyst lowers the activation energy barrier for the forward and reverse reactions by the same amount, thus increasing the rate constants kf and kr by the same factor. Because the equilibrium constant Kc equals the ratio of kf to kr, the value of Kc is unaffected by the addition of a catalyst. (d) Because the reaction is endothermic, Ea for the forward reaction is greater than Ea for the reverse reaction. Consequently, as the temperature increases, kf increases by more than kr increases, and therefore Kc = kf/kr increases, consistent with Le Châtelier’s principle. 쑺 PROBLEM 13.23 Nitric oxide emitted from the engines of supersonic aircraft can contribute to the destruction of stratospheric ozone: kf NO(g) + O3(g) ERF NO2(g) + O 2(g) k r 䉱 Nitric oxide emissions from supersonic aircraft can contribute to destruction of the ozone layer. This reaction is highly exothermic (¢E = - 201 kJ), and its equilibrium constant Kc is 3.4 * 1034 at 300 K. (a) Which rate constant is larger, kf or kr? (b) The value of kf at 300 K is 8.5 * 106 M -1 s -1. What is the value of kr at the same temperature? (c) A typical temperature in the stratosphere is 230 K. Do the values of kf, kr, and Kc increase or decrease when the temperature is lowered from 300 K to 230 K? 525 INQUIRY HOW DOES EQUILIBRIUM AFFECT OXYGEN TRANSPORT IN THE BLOODSTREAM? INQUIRY HOW DOES EQUILIBRIUM AFFECT OXYGEN TRANSPORT IN THE BLOODSTREAM? Humans, like all animals, need oxygen. The oxygen comes from breathing: About 500 mL of air is drawn into the lungs of an average person with each breath. When the freshly inspired air travels through the bronchial passages and enters the approximately 150 million alveolar sacs of the lungs, it picks up moisture and mixes with air remaining from the previous breath. As it mixes, the concentrations of both water vapor and carbon dioxide increase. These gas concentrations are measured by their partial pressures, with the partial pressure of oxygen in the lungs usually around 100 mm Hg (Table 13.2). Oxygen then diffuses through the delicate walls of the lung alveoli and into arterial blood, which transports it to all body tissues. Only about 3% of the oxygen in blood is dissolved; the rest is chemically bound to hemoglobin molecules (Hb), large proteins that contain heme groups embedded in them. Each hemoglobin molecule contains four heme groups, and each heme group contains an iron atom that can bind to one O2 molecule. Thus, a single hemoglobin molecule can bind four molecules of oxygen. Arterial blood vessel PO2 (mm Hg) Dry air Alveolar air Arterial blood Venous blood H3C C Alveolar wall Red blood cell O2 H C C N C N CO2 Capillaries 1. Air enters the lungs and reaches the alveoli, where gases are picked up by capillaries. C C C C CH3 N C Fe CH3 Alveoli – CH2CH2CO2 C C C H2C CH C 159 100 95 40 H O2CCH2CH2 Capillary Venous blood vessel Partial Pressure of Oxygen in Human Lungs and Blood at Sea Level Source – 2. Oxygen in air binds to hemoglobin molecules in red blood cells. Bronchiole TABLE 13.2 C N C C C H H C C C CH CH3 CH2 Heme — an O2 molecule binds to the central iron atom. 3. Each hemoglobin molecule can bind four molecules of oxygen. The entire system of oxygen transport and delivery in the body depends on the pickup and release of O2 by hemoglobin according to the following series of equilibria: Hb + O2 ∆ Hb(O 2) Hb(O2) + O2 ∆ Hb(O 2)2 Hb(O2)2 + O2 ∆ Hb(O2)3 Hb(O2)3 + O2 ∆ Hb(O 2)4 The positions of the different equilibria depend on the partial pressures of O2 (PO2) in the various tissues. In hard-working, oxygen-starved muscles, where PO2 is low, oxygen is released from hemoglobin as the equilibria shift toward the left, according to Le Châtelier’s principle. In the lungs, where PO2 is high, oxygen is absorbed by hemoglobin as the equilibria shift toward the right. 526 Chapter 13 CHEMICAL EQUILIBRIUM The amount of oxygen carried by hemoglobin at any given value of PO2 is usually expressed as a percent saturation and can be found from the curve shown in Figure 13.15. The saturation is 97.5% in the lungs, where PO2 = 100 mm Hg, meaning that each hemoglobin molecule is carrying close to its maximum possible amount of 4 O2 molecules. When PO2 = 26 mm Hg, however, the saturation drops to 50%. 100 % Saturation 80 The saturation is 97.5% in the lungs, where PO2 = 100 mm Hg, meaning that each hemoglobin molecule is carrying close to its maximum possible amount of 4 O2 molecules. 60 40 When PO2 = 26 mm Hg, however, the saturation drops to 50%. 20 0 0 20 40 60 80 100 120 PO (mm Hg) 2 Figure 13.15 An oxygen-carrying curve for hemoglobin. The percent saturation of the oxygen-binding sites on hemoglobin depends on the partial pressure of oxygen (PO2). What about people who live at high altitudes? In Leadville, Colorado, for example, where the altitude is 10,156 ft, the partial pressure of O2 in the lungs is only about 68 mm Hg. Hemoglobin is only 90% saturated with O2 at this pressure, so less oxygen is available for delivery to the tissues. People who climb suddenly from sea level to high altitude thus experience a feeling of oxygen deprivation, or hypoxia, as their bodies are unable to supply enough oxygen to their tissues. The body soon copes with the situation, though, by producing more hemoglobin molecules, which both provide more capacity for O2 transport and also drive the Hb + O2 equilibria to the right. The time required to adapt to the lower O2 pressures is typically days to weeks, so athletes and hikers must train at high altitudes for some time. 쑺 PROBLEM 13.24 The affinity of hemoglobin (Hb) for CO is greater than its affinity for O2. Use Le Châtelier’s principle to predict how CO affects the equilibrium Hb + O2 ∆ Hb(O2). Suggest a reason for the toxicity of CO. 䉱 The bodies of mountain dwellers produce increased amounts of hemoglobin to cope with the low O2 pressures at high altitudes. 쑺 PROBLEM 13.25 How many O2 molecules are drawn into the lungs of an average person with each breath? Assume that the ambient air pressure is 1.00 atm and the temperature is 25 °C. CONCEPTUAL PROBLEMS 527 SUMMARY Chemical equilibrium is a dynamic state in which the concentrations of reactants and products remain constant because the rates of the forward and reverse reactions are equal. For the general reaction a A + b B ∆ c C + d D, concentrations in the equilibrium mixture are related by the equilibrium equation: Kc = [C]c[D]d [A]a[B]b The quotient on the right side of the equation is called the equilibrium constant expression. The equilibrium constant Kc is the number obtained when equilibrium concentrations (in mol/L) are substituted into the equilibrium constant expression. The value of Kc varies with temperature and depends on the form of the balanced chemical equation. The equilibrium constant Kp can be used for gas-phase reactions. It is defined in the same way as Kc except that the equilibrium constant expression contains partial pressures (in atmospheres) instead of molar concentrations. The constants Kp and Kc are related by the equation Kp = Kc(RT)¢n, where ¢n = (c + d) - (a + b). Homogeneous equilibria are those in which all reactants and products are in a single phase; heterogeneous equilibria are those in which reactants and products are present in more than one phase. The equilibrium equation for a heterogeneous equilibrium does not include concentrations of pure solids or pure liquids. The value of the equilibrium constant for a reaction makes it possible to judge the extent of reaction, predict the direction of reaction, and calculate equilibrium concentrations (or partial pressures) from initial concentrations (or partial pressures). The farther the reaction proceeds toward completion, the larger the value of Kc. The direction of a reaction not at equilibrium depends on the relative values of Kc and the reaction quotient Qc, which is defined in the same way as Kc except that the concentrations in the equilibrium constant expression are not necessarily equilibrium concentrations. If Qc 6 Kc, the net reaction goes from reactants to products to attain equilibrium; if Qc 7 Kc, the net reaction goes from products to reactants; if Qc = Kc, the system is at equilibrium. The composition of an equilibrium mixture can be altered by changes in concentration, pressure (volume), or temperature. The qualitative effect of these changes is predicted by Le Châtelier’s principle, which says that if a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress. Temperature changes affect equilibrium concentrations because Kc is temperature-dependent. As the temperature increases, Kc for an exothermic reaction decreases and Kc for an endothermic reaction increases. A catalyst increases the rate at which chemical equilibrium is reached, but it does not affect the equilibrium constant or the equilibrium concentrations. The equilibrium constant for a single-step reaction equals the ratio of the rate constants for the forward and reverse reactions: Kc = kf/kr. KEY WORDS chemical equilibrium 493 equilibrium constant Kc 496 equilibrium constant Kp 499 equilibrium equation 496 equilibrium mixture 493 heterogeneous equilibria 502 homogeneous equilibria 502 Le Châtelier’s principle 512 reaction quotient Qc 505 CONCEPTUAL PROBLEMS Increasing time Problems 13.1–13.25 appear within the chapter. 13.26 Consider the interconversion of A molecules (red spheres) and B molecules (blue spheres) according to the reaction A ∆ B. Each of the series of pictures at the right represents a separate experiment in which time increases from left to right: (a) Which of the experiments has resulted in an equilibrium state? (b) What is the value of the equilibrium constant Kc for the reaction A ∆ B? (1) (2) (c) Explain why you can calculate Kc without knowing the volume of the reaction vessel. (3) 528 Chapter 13 CHEMICAL EQUILIBRIUM 13.27 The following pictures represent the equilibrium state for three different reactions of the type A2 + X2 ∆ 2 AX (X = B, C, or D): A2 + B2 2 AB A 2 + C2 2 AC A2 + D2 2 AD (a) Which reaction has the largest equilibrium constant? (b) Which reaction has the smallest equilibrium constant? 13.28 The reaction A2 + B ∆ A + AB has an equilibrium constant Kc = 2. The following pictures represent reaction mixtures that contain A atoms (red), B atoms (blue), and A2 and AB molecules: (1) (2) 13.31 The following pictures represent the composition of the equilibrium mixture for the reaction A + B ∆ AB at 300 K and at 400 K: T = 300 K Is the reaction exothermic or endothermic? Explain using Le Châtelier’s principle. 13.32 The following pictures represent equilibrium mixtures at 325 K and 350 K for a reaction involving A atoms (red), B atoms (blue), and AB molecules: (3) T = 325 K (a) Which reaction mixture is at equilibrium? (b) For those mixtures that are not at equilibrium, will the reaction go in the forward or reverse direction to reach equilibrium? 13.29 The following pictures represent the initial state and the equilibrium state for the reaction of A2 molecules (red) with B atoms (blue) to give AB molecules: Initial state T = 400 K T = 350 K (a) Write a balanced equation for the reaction that occurs on raising the temperature. (b) Is the reaction exothermic or endothermic? Explain using Le Châtelier’s principle. (c) If the volume of the container is increased, will the number of A atoms increase, decrease, or remain the same? Explain. 13.33 The following picture represents an equilibrium mixture of solid BaCO3, solid BaO, and gaseous CO2 obtained as a result of the endothermic decomposition of BaCO3: Equilibrium state (a) Write a balanced chemical equation for the reaction. (b) If the volume of the equilibrium mixture is decreased, will the number of AB molecules increase, decrease, or remain the same? Explain. 13.30 Consider the reaction A + B ∆ AB. The vessel on the right contains an equilibrium mixture of A molecules (red spheres), B molecules (blue spheres), and AB molecules. If the stopcock is opened and the contents of the two vessels are allowed to mix, will the reaction go in the forward or reverse direction? Explain. (a) Draw a picture that represents the equilibrium mixture after addition of four more CO2 molecules. (b) Draw a picture that represents the equilibrium mixture at a higher temperature. 13.34 The following picture represents the composition of the equilibrium mixture for the endothermic reaction A2 ∆ 2 A at 500 K: Draw a picture that represents the equilibrium mixture after each of the following changes: SECTION PROBLEMS (a) Adding a catalyst (b) Increasing the volume (c) Decreasing the temperature 13.35 The following picture represents the equilibrium state for the reaction 2 AB ∆ A2 + B2: 529 (b) Will the number of A molecules in the equilibrium mixture increase, decrease, or remain the same after each of the following changes? Explain. (1) Increasing the temperature (2) Decreasing the volume (3) Increasing the pressure by adding an inert gas (4) Adding a catalyst 13.37 The following pictures represent the initial and equilibrium states for the exothermic reaction of solid A (red) with gaseous B2 (blue) to give gaseous AB: Which rate constant is larger, kf or kr? Explain. 13.36 The following pictures represent the initial and equilibrium states for the exothermic decomposition of gaseous A molecules (red) to give gaseous B molecules (blue): Initial state Initial state Equilibrium state (a) Write a balanced chemical equation for the reaction. Equilibrium state (a) Write a balanced chemical equation for the reaction. (b) Will the number of AB molecules in the equilibrium mixture increase, decrease, or remain the same after each of the following changes? Explain. (1) Increasing the partial pressure of B2 (2) Adding more solid A (3) Increasing the volume (4) Increasing the temperature SECTION PROBLEMS Equilibrium Constant Expressions and Equilibrium Constants (Sections 13.1–13.4) 13.38 For the reaction A2 + 2 B ∆ 2 AB, the rate of the forward reaction is 18 M/s and the rate of the reverse reaction is 12 M/s. The reaction is not at equilibrium. Will the reaction proceed in the forward or reverse direction to attain equilibrium? 13.39 For the reaction 2 A3 + B2 ∆ 2 A3B, the rate of the forward reaction is 0.35 M/s and the rate of the reverse reaction is 0.65 M/s. The reaction is not at equilibrium. Will the reaction proceed in the forward or reverse direction to attain equilibrium? 13.40 For each of the following equilibria, write the equilibrium constant expression for Kc: (a) CH 4(g) + H 2O(g) ∆ CO(g) + 3 H 2(g) (b) 3 F2(g) + Cl2(g) ∆ 2 ClF3(g) (c) H 2(g) + F2(g) ∆ 2 HF(g) 13.41 For each of the following equilibria, write the equilibrium constant expression for Kc: (a) 2 C2H 4(g) + O2(g) ∆ 2 CH 3CHO(g) (b) 2 NO(g) ∆ N2(g) + O2(g) (c) 4 NH 3(g) + 5 O2(g) ∆ 4 NO(g) + 6 H 2O(g) 13.42 For each of the equilibria in Problem 13.40, write the equilibrium constant expression for Kp and give the equation that relates Kp and Kc. 13.43 For each of the equilibria in Problem 13.41, write the equilibrium constant expression for Kp and give the equation that relates Kp and Kc. 13.44 Diethyl ether, used as an anesthetic, is synthesized by heating ethanol with concentrated sulfuric acid. Write the equilibrium constant expression for Kc. 2 C2H5OH(soln) C2H5OC2H5(soln) + H2O(soln) Ethanol Diethyl ether 13.45 Ethylene glycol, used as antifreeze in automobile radiators, is manufactured by the hydration of ethylene oxide. Write the equilibrium constant expression for Kc. O H 2C CH2(soln) + H2O(soln) Ethylene oxide HOCH2CH2OH(soln) Ethylene glycol 530 Chapter 13 CHEMICAL EQUILIBRIUM 13.46 If Kc = 7.5 * 10-9 at 1000 K for the reaction N2(g) + O2(g) ∆ 2 NO(g), what is Kc at 1000 K for the reaction 2 NO(g) ∆ N2(g) + O2(g)? 13.47 At 400 K, Kp = 50.2 for the reaction N2O4(g) ∆ 2 NO 2(g) What is Kp at 400 K for the reaction 2 NO2(g) ∆ N2O4(g)? 13.48 An equilibrium mixture of PCl5, PCl3, and Cl2 at a certain temperature contains 8.3 * 10-3 M PCl5, 1.5 * 10-2 M PCl3, and 3.2 * 10-2 M Cl2. Calculate the equilibrium constant Kc for the reaction PCl5(g) ∆ PCl3(g) + Cl2(g). 13.49 The partial pressures in an equilibrium mixture of NO, Cl2, and NOCl at 500 K are as follows: PNO = 0.240 atm; PCl2 = 0.608 atm; PNOCl = 1.35 atm. What is Kp at 500 K for the reaction 2 NO(g) + Cl2(g) ∆ 2 NOCl(g)? 13.50 A sample of HI (9.30 * 10-3 mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentration of I2 was 6.29 * 10-4 M. Calculate the value of Kc at 1000 K for the reaction H 2(g) + I 2(g) ∆ 2 HI(g). 13.51 Vinegar contains acetic acid, a weak acid that is partially dissociated in aqueous solution: CH 3CO2H(aq) ∆ H +(aq) + CH 3CO2 - (aq) 13.57 Naphthalene, a white solid used to make mothballs, has a vapor pressure of 0.10 mm Hg at 27 °C. Calculate the values of Kp and Kc at 27 °C for the equilibrium C10H 8(s) ∆ C10H 8(g). Naphthalene 13.58 For each of the following equilibria, write the equilibrium constant expression for Kc. Where appropriate, also write the equilibrium constant expression for Kp. (a) Fe2O3(s) + 3 CO(g) ∆ 2 Fe(l) + 3 CO2(g) (b) 4 Fe(s) + 3 O 2(g) ∆ 2 Fe2O3(s) (c) BaSO4(s) ∆ BaO(s) + SO 3(g) (d) BaSO4(s) ∆ Ba2+(aq) + SO4 2-(aq) 13.59 For each of the following equilibria, write the equilibrium constant expression for Kc. Where appropriate, also write the equilibrium constant expression for Kp. (a) WO 3(s) + 3 H 2(g) ∆ W(s) + 3 H 2O(g) Acetic acid (a) Write the equilibrium constant expression for Kc. (b) What is the value of Kc if the extent of dissociation in 1.0 M CH3CO2H is 0.42%? 13.52 The industrial solvent ethyl acetate is produced by the reaction of acetic acid with ethanol: CH 3CO2H(soln) + CH 3CH 2OH(soln) ∆ Acetic acid Ethanol CH 3CO2CH 2CH 3(soln) + H 2O(soln) Ethyl acetate (a) Write the equilibrium constant expression for Kc. (b) A solution prepared by mixing 1.00 mol of acetic acid and 1.00 mol of ethanol contains 0.65 mol of ethyl acetate at equilibrium. Calculate the value of Kc. Explain why you can calculate Kc without knowing the volume of the solution. 13.53 A characteristic reaction of ethyl acetate is hydrolysis, the reverse of the reaction in Problem 13.52. Write the equilibrium equation for the hydrolysis of ethyl acetate, and use the data in Problem 13.52 to calculate Kc for the hydrolysis reaction. 5 13.54 At 298 K, Kc is 2.2 * 10 for the reaction F(g) + O2(g) ∆ O2F(g). What is the value of Kp at this temperature? 13.55 At 298 K, Kp is 1.6 * 10-6 for the reaction 2 NOCl(g) ∆ 2 NO(g) + Cl2(g). What is the value of Kc at this temperature? 13.56 The vapor pressure of water at 25 °C is 0.0313 atm. Calculate the values of Kp and Kc at 25 °C for the equilibrium H 2O(l) ∆ H 2O(g). (b) Ag +(aq) + Cl -(aq) ∆ AgCl(s) (c) 2 FeCl3(s) + 3 H 2O(g) ∆ Fe2O3(s) + 6 HCl(g) (d) MgCO3(s) ∆ MgO(s) + CO 2(g) Using the Equilibrium Constant (Section 13.5) 13.60 When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? (a) H 2(g) + S(s) ∆ H 2S(g); Kc = 7.8 * 105 (b) N2(g) + 2 H 2(g) ∆ N2H 4(g); Kc = 7.4 * 10-26 13.61 Which of the following reactions yield appreciable equilibrium concentrations of both reactants and products? (a) 2 Cu(s) + O2(g) ∆ 2 CuO(s); Kc = 4 * 1045 (b) H 3PO4(aq) ∆ H +(aq) + H 2PO4 -(aq); Kc = 7.5 * 10-3 (c) 2 HBr(g) ∆ H 2(g) + Br2(g); Kc = 2 * 10-19 13.62 When wine spoils, ethanol is oxidized to acetic acid as O2 from the air reacts with the wine: CH 3CH 2OH(aq) + O2(aq) ∆ CH 3CO2H(aq) + H 2O(l) Ethanol Acetic acid The value of Kc for this reaction at 25 °C is 1.2 * 1082. Will much ethanol remain when the reaction has reached equilibrium? Explain. 13.63 The value of Kc for the reaction 3 O2(g) ∆ 2 O 3(g) is 1.7 * 10-56 at 25 °C. Do you expect pure air at 25 °C to contain much O3 (ozone) when O2 and O3 are in equilibrium? If the equilibrium concentration of O2 in air at 25 °C is 8 * 10-3 M, what is the equilibrium concentration of O3? 531 SECTION PROBLEMS 13.64 At 1400 K, Kc = 2.5 * 10-3 for the reaction CH 4(g) + 2 H 2S(g) ∆ CS 2(g) + 4 H 2(g). A 10.0 L reaction vessel at 1400 K contains 2.0 mol of CH4, 3.0 mol of CS2, 3.0 mol of H2, and 4.0 mol of H2S. Is the reaction mixture at equilibrium? If not, in which direction does the reaction proceed to reach equilibrium? 13.65 The first step in the industrial synthesis of hydrogen is the reaction of steam and methane to give synthesis gas, a mixture of carbon monoxide and hydrogen: H 2O(g) + CH 4(g) ∆ CO(g) + 3 H 2(g) Kc = 4.7 at 1400 K A mixture of reactants and products at 1400 K contains 0.035 M H2O, 0.050 M CH4, 0.15 M CO, and 0.20 M H2. In which direction does the reaction proceed to reach equilibrium? 13.73 Recalculate the equilibrium concentrations in Problem 13.72 if the initial concentrations are 2.24 M N2 and 0.56 M O2. (This N2/O2 concentration ratio is the ratio found in air.) 13.74 The interconversion of L-a-lysine and L- b -lysine, for which Kc = 7.20 at 333 K, is catalyzed by the enzyme lysine 2,3-aminomutase. HO O H H H H H C C C C C C H H H H O H NH2 H H H C C C C C C H H H H H H2N L-α-Lysine HO 13.66 Phosphine (PH3) decomposes at elevated temperatures, yielding gaseous P2 and H2: 2 PH 3(g) ∆ P2(g) + 3 H 2(g) Kp = 398 at 873 K If the initial partial pressures are PPH3 = 0.0260 atm, PP2 = 0.871 atm, PH2 = 0.517 atm, calculate Qp and determine the direction of reaction to attain equilibrium. 13.67 When a mixture of PH3, P2, and H2 comes to equilibrium at 873 K according to the reaction in Problem 13.66, PP2 = 0.412 atm and PH2 = 0.822 atm. What is PPH3? 13.68 Gaseous indium dihydride is formed from the elements at elevated temperature: ln(g) + H 2(g) ∆ lnH 2(g) 13.69 The following reaction, which has Kc = 0.145 at 298 K, takes place in carbon tetrachloride solution: 2 BrCl(soln) ∆ Br2(soln) + Cl2(soln) A measurement of the concentrations shows [BrCl] = 0.050 M, [Br2] = 0.035 M, and [Cl2] = 0.030 M. (a) Calculate Qc, and determine the direction of reaction to attain equilibrium. (a) Determine the equilibrium concentrations of BrCl, Br2, and Cl2. 13.70 An equilibrium mixture of N2, H2, and NH3 at 700 K contains 0.036 M N2 and 0.15 M H2. At this temperature, Kc for the reaction N2(g) + 3 H 2(g) ∆ 2 NH 3(g) is 0.29. What is the concentration of NH3? 13.71 An equilibrium mixture of O2, SO2, and SO3 contains equal concentrations of SO2 and SO3. Calculate the concentration of O2 if Kc = 2.7 * 102 for the reaction 2 SO2(g) + O2(g) ∆ 2 SO3(g). 13.72 The air pollutant NO is produced in automobile engines from the high-temperature reaction N2(g) + O2(g) ∆ 2 NO(g); Kc = 1.7 * 10-3 at 2300 K. If the initial concentrations of N2 and O2 at 2300 K are both 1.40 M, what are the concentrations of NO, N2, and O2 when the reaction mixture reaches equilibrium? NH2 L-β-Lysine L-a-Lysine occurs in proteins while L- b -lysine is a precursor to certain antibiotics. At 333 K, a solution of L-a-lysine at a concentration of 3.00 * 10-3 M is placed in contact with lysine 2,3-aminomutase. What are the equilibrium concentrations of L-a-lysine and L- b -lysine? 13.75 Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation Kp = 1.48 at 973 K Partial pressures measured in a reaction vessel are: Pln = 0.0600 atm, PH2 = 0.0350 atm, PlnH2 = 0.0760 atm. (a) Calculate Qp, and determine the direction of reaction to attain equilibrium. (b) Determine the equilibrium partial pressures of all the gases. NH2 Lysine 2,3-aminomutase 2 ClO(g) ∆ Cl2O2(g) for which Kc is 4.96 * 1011 at 253 K. For an equilibrium mixture in which [Cl2O2] is 6.00 * 10-6 M, what is [ClO]? 13.76 The value of Kc for the reaction of acetic acid with ethanol is 3.4 at 25 °C: CH 3CO2H(soln) + CH 3CH 2OH(soln) ∆ Acetic acid Ethanol CH 3CO2CH 2CH 3(soln) + H 2O(soln) Kc = 3.4 Ethyl acetate (a) How many moles of ethyl acetate are present in an equilibrium mixture that contains 4.0 mol of acetic acid, 6.0 mol of ethanol, and 12.0 mol of water at 25 °C? (b) Calculate the number of moles of all reactants and products in an equilibrium mixture prepared by mixing 1.00 mol of acetic acid and 10.00 mol of ethanol. 13.77 In a basic aqueous solution, chloromethane undergoes a substitution reaction in which Cl - is replaced by OH - : CH 3Cl(aq) + OH -(aq) ∆ CH 3OH(aq) + Cl -(aq) Chloromethane Methanol The equilibrium constant Kc is 1 * 1016. Calculate the equilibrium concentrations of CH3Cl, CH3OH, OH - , and Cl - in a solution prepared by mixing equal volumes of 0.1 M CH3Cl and 0.2 M NaOH. (Hint: In defining x, assume that the reaction goes 100% to completion, and then take account of a small amount of the reverse reaction.) 532 Chapter 13 CHEMICAL EQUILIBRIUM 13.78 At 700 K, Kp = 0.140 for the reaction ClF3(g) ∆ ClF(g) + F2(g). Calculate the equilibrium partial pressures of ClF3, ClF, and F2 if only ClF3 is present initially, at a partial pressure of 1.47 atm. (c) The solution is diluted with water (d) The temperature is increased 13.79 The reaction of iron(III) oxide with carbon monoxide is important in making steel. At 1000 K, Kp = 19.9 for the reaction Fe2O3(s) + 3 CO(g) ∆ 2 Fe(s) + 3 CO 2(g). What are the equilibrium partial pressures of CO and CO2 if CO is the only gas present initially, at a partial pressure of 0.978 atm? Le Châtelier’s Principle (Sections 13.6–13.10) 13.80 Consider the following equilibrium: Ag +(aq) + Cl -(aq) ∆ AgCl(s). Use Le Châtelier’s principle to predict how the amount of solid silver chloride will change when the equilibrium is disturbed by: (a) Adding NaCl (b) Adding AgNO3 (c) Adding NH3, which reacts with Ag + to form the complex ion Ag(NH 3)2+ (d) Removing Cl - ; also account for the change using the reaction quotient Qc. 13.81 Will the concentration of NO2 increase, decrease, or remain the same when the equilibrium NO2Cl(g) + NO(g) ∆ NOCl(g) + NO2(g) is disturbed by the following changes? (a) Adding NOCl (b) Adding NO (c) Removing NO (d) Adding NO2Cl; also account for the change using the reaction quotient Qc. 13.82 When each of the following equilibria is disturbed by increasing the pressure as a result of decreasing the volume, does the number of moles of reaction products increase, decrease, or remain the same? (a) 2 CO2(g) ∆ 2 CO(g) + O2(g) (b) N2(g) + O2(g) ∆ 2 NO(g) (c) Si(s) + 2 Cl2(g) ∆ SiCl4(g) 13.83 For each of the following equilibria, use Le Châtelier’s principle to predict the direction of reaction when the volume is increased. (a) C(s) + H 2O(g) ∆ CO(g) + H 2(g) (b) 2 H 2(g) + O2(g) ∆ 2 H 2O(g) (c) 2 Fe(s) + 3 H 2O(g) ∆ Fe2O3(s) + 3 H 2(g) 13.84 For the water-gas shift reaction CO(g) + H 2O(g) ∆ CO2(g) + H 2(g), ¢H° = - 41.2 kJ, does the amount of H2 in an equilibrium mixture increase or decrease when the temperature is increased? How does Kc change when the temperature is decreased? Justify your answers using Le Châtelier’s principle. 13.85 The value of ¢H° for the reaction 3 O2(g) ∆ 2 O3(g) is +285 kJ. Does the equilibrium constant for this reaction increase or decrease when the temperature increases? Justify your answer using Le Châtelier’s principle. 13.86 Consider the exothermic reaction CoCl4 2-(aq) + 6 H 2O(l) ∆ Co(H 2O)6 2+(aq) + 4 Cl -(aq), which interconverts the blue CoCl4 2- ion and the pink Co(H 2O)6 2+ ion. Will the equilibrium concentration of CoCl4 2- increase or decrease when the following changes occur? (a) HCl is added (b) Co(NO3)2 is added 13.87 Consider the endothermic reaction Fe 3+(aq) + Cl -(aq) ∆ FeCl2+(aq). Use Le Châtelier’s principle to predict how the equilibrium concentration of the complex ion FeCl2+ will change when: (a) Fe(NO3)3 is added (b) Cl - is precipitated as AgCl by addition of AgNO3 (c) The temperature is increased (d) A catalyst is added 13.88 Methanol (CH3OH) is manufactured by the reaction of carbon monoxide with hydrogen in the presence of a Cu/ZnO/Al2O3 catalyst: Cu/ZnO/Al2O3 CO(g) + 2 H 2(g) ERRRRRF CH 3OH(g) catalyst ¢H° = - 91 kJ Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes? (a) The temperature is increased (b) The volume is decreased (c) Helium is added (d) CO is added (e) The catalyst is removed 13.89 In the gas phase at 400 °C, isopropyl alcohol (rubbing alcohol) decomposes to acetone, an important industrial solvent: (CH 3)2CHOH(g) ∆ (CH 3)2CO(g) + H 2(g) Isopropyl alcohol ¢H ° = + 57.3 kJ Acetone Does the amount of acetone increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes? (a) The temperature is increased (b) The volume is increased (c) Argon is added (d) H2 is added (e) A catalyst is added 13.90 The following reaction is important in gold mining: 4 Au(s) + 8 CN - (aq) + O 2(g) + 2 H 2O(l) ∆ 4 Au(CN)2 - (aq) + 4 OH For a reaction mixture at equilibrium, in which direction would the reaction go to re-establish equilibrium after each of the following changes? (a) Adding gold (b) Increasing the hydroxide concentration CHAPTER PROBLEMS (c) Increasing the partial pressure of oxygen (d) Adding Fe 3+(aq), which reacts with cyanide to form Fe(CN)6 3-(aq) 13.91 The following reaction, catalyzed by iridium, is endothermic at 700 K: CaO(s) + CH 4(g) + 2 H 2O(g) ∆ CaCO3(s) + 4 H 2(g) For a reaction mixture at equilibrium at 700 K, how would the following changes affect the total quantity of CaCO3 in the reaction mixture once equilibrium is re-established? (a) Increasing the temperature (b) Adding calcium oxide (c) Removing methane (CH4) (d) Increasing the total volume (e) Adding iridium Chemical Equilibrium and Chemical Kinetics (Section 13.11) 13.92 Consider a general, single-step reaction of the type A + B ∆ C. Show that the equilibrium constant is equal to the ratio of the rate constants for the forward and reverse reactions, Kc = kf/kr. 13.93 Which of the following relative values of kf and kr results in an equilibrium mixture that contains large amounts of reactants and small amounts of products? (a) kf 7 kr (b) kf = kr (c) kf 6 kr 13.94 Consider the gas-phase hydration of hexafluoroacetone, (CF3)2CO: kf 533 13.95 Consider the reaction of chloromethane with OH - in aqueous solution: kf CH 3Cl(aq) + OH - (aq) ERF CH 3OH(aq) + Cl -(aq) kr At 25 °C, the rate constant for the forward reaction is 6 * 10-6 M -1 s -1, and the equilibrium constant Kc is 1 * 1016. Calculate the rate constant for the reverse reaction at 25 °C. 13.96 In automobile catalytic converters, the air pollutant nitric oxide is converted to nitrogen and oxygen. Listed in the table are forward and reverse rate constants for the reaction 2 NO(g) ∆ N2(g) + O2(g). Temperature (K) kf ( M -1 s -1) kr ( M -1 s -1) 1400 0.29 1.1 * 10-6 1500 1.3 1.4 * 10-5 Is the reaction endothermic or exothermic? Explain in terms of kinetics. 13.97 Forward and reverse rate constants for the reaction CO2(g) + N2(g) ∆ CO(g) + N2O(g) exhibit the following temperature dependence: Temperature (K) kf ( M -1 s -1) kr ( M -1 s -1) 1200 9.1 * 10-11 1.5 * 105 1300 2.7 * 10 -9 2.6 * 105 (CF3)2CO(g) + H 2O(g) ERF (CF3)2C(OH)2(g) kr At 76 °C, the forward and reverse rate constants are kf = 0.13 M -1 s -1 and kr = 6.2 * 10-4 s -1. What is the value of the equilibrium constant Kc? Is the reaction endothermic or exothermic? Explain in terms of kinetics. CHAPTER PROBLEMS 13.98 The equilibrium concentrations in a gas mixture at a particular temperature are 0.13 M H2, 0.70 M I2, and 2.1 M HI. What equilibrium concentrations are obtained at the same temperature when 0.20 mol of HI is injected into an empty 500.0 mL container? 13.99 A 5.00 L reaction vessel is filled with 1.00 mol of H2, 1.00 mol of I2, and 2.50 mol of HI. Calculate the equilibrium concentrations of H2, I2, and HI at 500 K. The equilibrium constant Kc at 500 K for the reaction H 2(g) + I 2(g) ∆ 2 HI(g) is 129. 13.100 At 1000 K, the value of Kc for the reaction C(s) + H 2O(g) ∆ CO(g) + H 2(g) is 3.0 * 10-2. Calculate the equilibrium concentrations of H2O, CO2, and H2 in a reaction mixture obtained by heating 6.00 mol of steam and an excess of solid carbon in a 5.00 L container. What is the molar composition of the equilibrium mixture? 13.101 The equilibrium constant Kp for the reaction PCl5(g) ∆ PCl3(g) + Cl2(g) is 3.81 * 102 at 600 K and 2.69 * 103 at 700 K. (a) Is the reaction endothermic or exothermic? (b) How are the equilibrium amounts of reactants and products affected by (i) an increase in volume, (ii) addition of an inert gas, and (iii) addition of a catalyst? 13.102 Consider the following gas-phase reaction: 2 A(g) + B(g) ∆ C(g) + D(g). An equilibrium mixture of reactants and products is subjected to the following changes: (a) A decrease in volume (b) An increase in temperature (c) Addition of reactants (d) Addition of a catalyst (e) Addition of an inert gas Which of these changes affect the composition of the equilibrium mixture but leave the value of the equilibrium constant Kc unchanged? Which of the changes affect the value of Kc? Which affect neither the composition of the equilibrium mixture nor Kc? 13.103 Baking soda (sodium bicarbonate) decomposes when it is heated: 2 NaHCO3(s) ∆ Na 2CO3(s) + CO 2(g) + H 2O(g) ¢H° = + 136 kJ Consider an equilibrium mixture of reactants and products in a closed container. How does the number of moles of CO2 change when the mixture is disturbed by the following: (a) Adding solid NaHCO3 (b) Adding water vapor (c) Decreasing the volume of the container (d) Increasing the temperature 534 Chapter 13 CHEMICAL EQUILIBRIUM 13.104 The reaction has 2 AsH 3(g) ∆ As2(g) + 3 H 2(g) Kp = 7.2 * 107 at 1073 K. At the same temperature, what is Kp for each of the following reactions? (a) As2(g) + 3 H 2(g) ∆ 2 AsH 3(g) (b) 4 AsH 3(g) ∆ 2 As2(g) + 6 H 2(g) (c) 9 H 2(g) + 3 As2(g) ∆ 6 AsH 3(g) 13.105 The reaction 2 PH 3(g) + As2(g) ∆ 2 AsH 3(g) + P2(g) has Kp = 2.9 * 10-5 at 873 K. At the same temperature, what is Kp for each of the following reactions? (a) 2 AsH 3(g) + P2(g) ∆ 2 PH 3(g) + As2(g) (b) 6 PH 3(g) + 3 As2(g) ∆ 3 P2(g) + 6 AsH 3(g) (c) 2 P2(g) + 4 AsH 3(g) ∆ 2 As2(g) + 4 PH 3(g) 13.106 When 1.000 mol of PCl5 is introduced into a 5.000 L container at 500 K, 78.50% of the PCl5 dissociates to give an equilibrium mixture of PCl5, PCl3, and Cl2: 13.109 Consider the reaction C(s) + CO2(g) ∆ 2 CO(g). When 1.50 mol of CO2 and an excess of solid carbon are heated in a 20.0 L container at 1100 K, the equilibrium concentration of CO is 7.00 * 10-2 M. (a) What is the equilibrium concentration of CO2? (b) What is the value of the equilibrium constant Kc at 1100 K? 13.110 The equilibrium constant Kp for the gas-phase thermal decomposition of tert-butyl chloride is 3.45 at 500 K: (CH 3)3CCl(g) ∆ (CH 3)2C “ CH 2(g) + HCl(g) PCl5(g) ∆ PCl3(g) + Cl2(g) (a) Calculate the values of Kc and Kp. (b) If the initial concentrations in a particular mixture of reactants and products are [PCl5] = 0.500 M, [PCl3] = 0.150 M, and [Cl2] = 0.600 M, in which direction does the reaction proceed to reach equilibrium? What are the concentrations when the mixture reaches equilibrium? 13.107 Heavy water, symbolized D2O (D = 2H) finds use as a neutron moderator in nuclear reactors. In a mixture with ordinary water, exchange of isotopes occurs according to the following equation: H 2O + D2O ∆ 2 HDO Kc = 3.86 at 298 K When 1.00 mol of H2O is combined with 1.00 mol of D2O, what are the equilibrium amounts of H2O, D2O, and HDO (in moles) at 298 K? Assume the density of the mixture is constant at 1.05 g/cm3. 13.108 Refining petroleum involves cracking large hydrocarbon molecules into smaller, more volatile pieces. A simple example of hydrocarbon cracking is the gas-phase thermal decomposition of butane to give ethane and ethylene: H H H C H H C H H C H H C H H C H H C H H + H C C H H H H Butane, C4H10 Ethane, C2H6 Ethylene, C2H4 (a) Write the equilibrium constant expressions for Kp and Kc. (b) The value of Kp at 500 °C is 12. What is the value of Kc? (c) A sample of butane having a pressure of 50 atm is heated at 500 °C in a closed container at constant volume. When equilibrium is reached, what percentage of the butane has been converted to ethane and ethylene? What is the total pressure at equilibrium? (d) How would the percent conversion in part (c) be affected by a decrease in volume? tert-Butyl chloride Isobutylene (a) Calculate the value of Kc at 500 K. (b) Calculate the molar concentrations of reactants and products in an equilibrium mixture obtained by heating 1.00 mol of tert-butyl chloride in a 5.00 L vessel at 500 K. (c) A mixture of isobutylene (0.400 atm partial pressure at 500 K) and HCl (0.600 atm partial pressure at 500 K) is allowed to reach equilibrium at 500 K. What are the equilibrium partial pressures of tert-butyl chloride, isobutylene, and HCl? 13.111 As shown in Figure 13.14, a catalyst lowers the activation energy for the forward and reverse reactions by the same amount, ¢Ea. (a) Apply the Arrhenius equation, k = Ae -Ea/RT, to the forward and reverse reactions, and show that a catalyst increases the rates of both reactions by the same factor. (b) Use the relation between the equilibrium constant and the forward and reverse rate constants, Kc = kf/kr, to show that a catalyst does not affect the value of the equilibrium constant. 13.112 Given the Arrhenius equation, k = Ae -Ea/RT, and the relation between the equilibrium constant and the forward and reverse rate constants, Kc = kf/kr, explain why Kc for an exothermic reaction decreases with increasing temperature. 13.113 At 1000 K, Kp = 2.1 * 106 and ¢H° = - 107.7 kJ for the reaction H 2(g) + Br2(g) ∆ 2 HBr(g). (a) A 0.974 mol quantity of Br2 is added to a 1.00 L reaction vessel that contains 1.22 mol of H2 gas at 1000 K. What are the partial pressures of H2, Br2, and HBr at equilibrium? (b) For the equilibrium in part (a), each of the following changes will increase the equilibrium partial pressure of HBr. Choose the change that will cause the greatest increase in the pressure of HBr, and explain your choice. (i) Adding 0.10 mol of H2 (ii) Adding 0.10 mol of Br2 (iii) Decreasing the temperature to 700 K. CHAPTER PROBLEMS 13.114 Consider the gas-phase decomposition of NOBr: 2 NOBr(g) ∆ 2 NO(g) + Br2(g) (a) When 0.0200 mol of NOBr is added to an empty 1.00 L flask and the decomposition reaction is allowed to reach equilibrium at 300 K, the total pressure in the flask is 0.588 atm. What is the equilibrium constant Kc for this reaction at 300 K? (b) What is the value of Kp for this reaction at 300 K? 13.115 At 100 °C, Kc = 4.72 for the reaction 2 NO 2(g) ∆ N2O4(g). An empty 10.0 L flask is filled with 4.60 g of NO2 at 100 °C. What is the total pressure in the flask at equilibrium? 13.116 Halogen lamps are ordinary tungsten filament lamps in which the lamp bulb contains a small amount of a halogen (often bromine). At the high temperatures of the lamp, the halogens dissociate and exist as single atoms. (a) In an ordinary tungsten lamp, the hot tungsten filament is constantly evaporating and the tungsten condenses on the relatively cool walls of the bulb. In a Br-containing halogen lamp, the tungsten reacts with the Br atoms to give gaseous WBr4: W(s) + 4 Br(g) ∆ WBr4(g) At the walls of the lamp, where the temperature is about 900 K, this reaction has an equilibrium constant Kp of about 100. If the equilibrium pressure of Br(g) is 0.010 atm, what is the equilibrium pressure of WBr4(g) near the walls of the bulb? (b) Near the tungsten filament, where the temperature is about 2800 K, the reaction in part (a) has a Kp value of about 5.0. Is the reaction exothermic or endothermic? (c) When the WBr4(g) diffuses back toward the filament, it decomposes, depositing tungsten back onto the filament. Show quantitatively that the pressure of WBr4 from part (a) will cause the reaction in part (a) to go in reverse direction at 2800 K. [The pressure of Br(g) is still 0.010 atm.] Thus, tungsten is continually recycled from the walls of the bulb back to the filament, allowing the bulb to last longer and burn brighter. 13.119 At 500 °C, F2 gas is stable and does not dissociate, but at 840 °C, some dissociation occurs: F2(g) ∆ 2 F(g). A flask filled with 0.600 atm of F2 at 500 °C was heated to 840 °C, and the pressure at equilibrium was measured to be 0.984 atm. What is the equilibrium constant Kp for the dissociation of F2 gas at 840 °C? 13.120 The reaction NO(g) + NO2(g) ∆ N2O3(g) takes place in the atmosphere with Kc = 13 at 298 K. A gas mixture is prepared with 2.0 mol NO and 3.0 mol NO2 and an initial total pressure of 1.65 atm. (a) What are the equilibrium partial pressures of NO, NO2, and N2O3 at 298 K? (b) What is the volume of the container? 13.121 Phosgene (COCl2) is a toxic gas that damages the lungs. At 360 °C, Kc = 8.4 * 10-4 for the decomposition of phosgene: COCl2(g) ∆ CO(g) + Cl2(g) If an empty 50.0 L container is charged with 1.00 mol of phosgene at 360 °C, what is the total pressure in the container after the system comes to equilibrium? 13.122 The equilibrium constant Kc for the reaction N2(g) + 3 H 2(g) ∆ 2 NH 3(g) is 4.20 at 600 K. When a quantity of gaseous NH3 was placed in a 1.00 L reaction vessel at 600 K and the reaction was allowed to reach equilibrium, the vessel was found to contain 0.200 mol of N2. How many moles of NH3 were placed in the vessel? 13.123 At 45 °C, Kc = 0.619 for the reaction N2O4(g) ∆ 2 NO2(g). If 46.0 g of N2O4 is introduced into an empty 2.00 L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45 °C? 13.124 When 9.25 g of ClF3 was introduced into an empty 2.00 L container at 700.0 K, 19.8% of the ClF3 decomposed to give an equilibrium mixture of ClF3, ClF, and F2. ClF3(g) ∆ ClF(g) + F2(g) (a) What is the value of the equilibrium constant Kc at 700.0 K? (b) What is the value of the equilibrium constant Kp at 700.0 K? 13.117 The decomposition of solid ammonium carbamate, (NH4)(NH2CO2), to gaseous ammonia and carbon dioxide is an endothermic reaction. (c) In a separate experiment, 39.4 g of ClF3 was introduced into an empty 2.00 L container at 700.0 K. What are the concentrations of ClF3, ClF, and F2 when the mixture reaches equilibrium? (NH 4)(NH 2CO2)(s) ∆ 2 NH 3(g) + CO2(g) (a) When solid (NH4)(NH2CO2) is introduced into an evacuated flask at 25 °C, the total pressure of gas at equilibrium is 0.116 atm. What is the value of Kp at 25 °C? (b) Given that the decomposition reaction is at equilibrium, how would the following changes affect the total quantity of NH3 in the flask once equilibrium is reestablished? (i) Adding CO2 (ii) Adding (NH4)(NH2CO2) (iii) Removing CO2 (iv) Increasing the total volume (v) Adding neon (vi) Increasing the temperature 13.118 At 25 °C, Kc = 216 for the reaction 2 NO2(g) ∆ N2O4(g). A 1.00 L flask containing a mixture of NO2 and N2O4 at 25 °C has a total pressure of 1.50 atm. What is the partial pressure of each gas? 535 13.125 The following reaction in aqueous solution is catalyzed by the enzyme aspartase and has Kc = 6.95 * 10-3 at 37 °C: H – O2C + NH3 C C H H L-Aspartate CO2– – O2C Aspartase H C H C + NH4+ CO2– Fumarate If the initial concentration of L-aspartate is 8.32 * 10-3 M, what are the equilibrium concentrations of L-aspartate, fumarate, and ammonium ion at 37 °C? 536 Chapter 13 CHEMICAL EQUILIBRIUM 13.126 The reaction of fumarate (Problem 13.125) with water to form L-malate is catalyzed by the enzyme fumarase; Kc = 3.3 at 37 °C. – O2C H C H C + H2O Fumarase – O2C CO2– Fumarate H OH C C H H CO2– When a reaction mixture with [fumarate] = 1.56 * 10-3 M and [L-malate] = 2.27 * 10-3 M comes to equilibrium in the presence of fumarase at 37 °C, what are the equilibrium concentrations of fumarate and L-malate? (Water can be omitted from the equilibrium equation because its concentration in dilute solutions is essentially the same as that in pure water.) L-Malate MULTICONCEPT PROBLEMS 13.127 The F ¬ F bond in F2 is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; Kp = 7.83 at 1500 K for the reaction F2(g) ∆ 2 F(g). (a) If the equilibrium partial pressure of F2 molecules at 1500 K is 0.200 atm, what is the equilibrium partial pressure of F atoms in atm? (b) What fraction of the F2 molecules dissociate at 1500 K? (c) Why is the F ¬ F bond in F2 weaker than the Cl ¬ Cl bond in Cl2? 13.128 When 0.500 mol of N2O4 is placed in a 4.00 L reaction vessel and heated at 400 K, 79.3% of the N2O4 decomposes to NO2. (a) Calculate Kc and Kp at 400 K for the reaction N2O4(g) ∆ 2 NO2(g). (b) Draw an electron-dot structure for NO2, and rationalize the structure of N2O4. 13.129 The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene is 1.0 * 105 at 500 K: CH2 H2C CH2 Cyclopropane CH3 CH CH2 Kc = 1.0 × 10 5 Propene (a) What is the value of Kp at 500 K? (b) What is the equilibrium partial pressure of cyclopropane at 500 K when the partial pressure of propene is 5.0 atm? (c) Can you alter the ratio of the two concentrations at equilibrium by adding cyclopropane or by decreasing the volume of the container? Explain. (d) Which has the larger rate constant, the forward reaction or the reverse reaction? (e) Why is cyclopropane so reactive? (Hint: Consider the hybrid orbitals used by the C atoms.) 13.130 Acetic acid tends to form dimers, (CH3CO2H)2, because of hydrogen bonding: O ······H O 2 CH3 C CH3 O Monomer H O C C O CH3 H ······O Dimer The equilibrium constant Kc for this reaction is 1.51 * 102 in benzene solution, but only 3.7 * 10-2 in water solution. (a) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in benzene. (b) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in water. (c) Why is Kc for the water solution so much smaller than Kc for the benzene solution? 13.131 A 125.4 g quantity of water and an equal molar amount of carbon monoxide were placed in an empty 10.0 L vessel, and the mixture was heated to 700 K. At equilibrium, the partial pressure of CO was 9.80 atm. The reaction is CO(g) + H 2O(g) ∆ CO2(g) + H 2(g) (a) What is the value of Kp at 700 K? (b) An additional 31.4 g of water was added to the reaction vessel, and a new state of equilibrium was achieved. What are the equilibrium partial pressures of each gas in the mixture? What is the concentration of H2 in molecules/cm3? 13.132 A 79.2 g chunk of dry ice (solid CO2) and 30.0 g of graphite (carbon) were placed in an empty 5.00 L container, and the mixture was heated to achieve equilibrium. The reaction is CO2(g) + C(s) ∆ 2 CO(g) (a) What is the value of Kp at 1000 K if the gas density at 1000 K is 16.3 g/L? (b) What is the value of Kp at 1100 K if the gas density at 1100 K is 16.9 g/L? (c) Is the reaction exothermic or endothermic? Explain. 13.133 The amount of carbon dioxide in a gaseous mixture of CO2 and CO can be determined by passing the gas into an aqueous solution that contains an excess of Ba(OH)2. The CO2 reacts, yielding a precipitate of BaCO3, but the CO does not react. This method was used to analyze the equilibrium MULTICONCEPT PROBLEMS composition of the gas obtained when 1.77 g of CO2 reacted with 2.0 g of graphite in a 1.000 L container at 1100 K. The analysis yielded 3.41 g of BaCO3. Use these data to calculate Kp at 1100 K for the reaction CO2(g) + C(s) ∆ 2 CO(g) 13.134 A 14.58 g quantity of N2O4 was placed in a 1.000 L reaction vessel at 400 K. The N2O4 decomposed to an equilibrium mixture of N2O4 and NO2 that had a total pressure of 9.15 atm. (a) What is the value of Kc for the reaction N2O4(g) ∆ 2 NO2(g) at 400 K? (b) How much heat (in kilojoules) was absorbed when the N2O4 decomposed to give the equilibrium mixture? (Standard heats of formation may be found in Appendix B.) 13.135 Consider the sublimation of mothballs at 27 °C in a room having dimensions 8.0 ft * 10.0 ft * 8.0 ft. Assume that the mothballs are pure solid naphthalene (density 1.16 g/cm3) and that they are spheres with a diameter of 12.0 mm. The equilibrium constant Kc for the sublimation of naphthalene is 5.40 * 10-6 at 27 °C. C10H 8(s) ∆ C10H 8(g) (a) When excess mothballs are present, how many gaseous naphthalene molecules are in the room at equilibrium? (b) How many mothballs are required to saturate the room with gaseous naphthalene? 13.136 Ozone is unstable with respect to decomposition to ordinary oxygen: 2 O3(g) ∆ 3 O 2(g) Kp = 1.3 * 1057 How many O3 molecules are present at equilibrium in 10 million cubic meters of air at 25 °C and 720 mm Hg pressure? 13.137 The equilibrium constant for the dimerization of acetic acid in benzene solution is 1.51 * 102 at 25 °C (see Problem 13.130). 2 CH 3CO2H ∆ (CH 3CO2H)2 Kc = 1.51 * 102 at 25 °C 537 (a) What are the equilibrium concentrations of monomer and dimer at 25 °C in a solution prepared by dissolving 0.0300 mol of pure acetic acid in enough benzene to make 250.0 mL of solution? (b) What is the osmotic pressure of the solution at 25 °C? 13.138 For the decomposition reaction PCl5(g) ∆ PCl3(g) + Cl2(g), Kp = 381 at 600 K and Kc = 46.9 at 700 K. (a) Is the reaction endothermic or exothermic? Explain. Does your answer agree with what you would predict based on bond energies? (b) If 1.25 g of PCl5 is introduced into an evacuated 0.500 L flask at 700 K and the decomposition reaction is allowed to reach equilibrium, what percent of the PCl5 will decompose and what will be the total pressure in the flask? (c) Write electron-dot structures for PCl5 and PCl3, and indicate whether these molecules have a dipole moment. Explain. 13.139 Propanol (PrOH) and methyl methacrylate (MMA) associate in solution by an intermolecular force, forming an adduct represented as PrOH # MMA. The equilibrium constant for the association reaction is Kc = 0.701 at 298 K. H3C H H C C H H PrOH CH2 O H + H3C C C O CH3 O MMA PrOH MMA (a) What is the predominant intermolecular force accounting for the interaction between PrOH and MMA? (b) Draw a plausible structure for the PrOH # MMA adduct. Use # # # to signify an intermolecular interaction. (c) If the initial concentrations are [PrOH] = 0.100 M and [MMA] = 0.0500 M, what are the equilibrium concentrations of PrOH, MMA, and PrOH # MMA?