Chapt14

Accounting

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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 14 Chemical Equilibrium INTRODUCTION EQUILIBRIUM 14.1 THE CONCEPT OF EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT IS A STATE IN WHICH THERE ARE NO OBSERVABLE CHANGES AS TIME GOES BY. WHEN A CHEMICAL REACTION HAS REACHED THE 14.2 WRITING EQUILIBRIUM CONSTANT EXPRESSIONS EQUILIBRIUM STATE, THE CONCENTRATIONS OF REACTANTS AND PRODUCTS REMAIN CONSTANT OVER TIME, AND THERE ARE NO VISIBLE CHANGES IN THE SYSTEM. HOWEVER, 14.3 THE RELATIONSHIP BETWEEN CHEMICAL KINETICS AND CHEMICAL EQUILIBRIUM THERE IS MUCH ACTIVITY AT THE MOLECU- 14.4 WHAT DOES THE EQUILIBRIUM CONSTANT TELL US? LAR LEVEL BECAUSE REACTANT MOLECULES CONTINUE TO FORM PRODUCT MOLECULES WHILE PRODUCT MOLECULES REACT TO YIELD REACTANT MOLECULES. WE THIS 14.5 FACTORS THAT AFFECT CHEMICAL EQUILIBRIUM DYNAMIC SITUATION IS THE SUBJECT OF THIS CHAPTER. WILL DISCUSS DIFFERENT TYPES OF EQUILIBRIUM REACTIONS, THE MEANING OF THE EQUILIBRIUM CONSTANT AND ITS RELATIONSHIP TO THE RATE CONSTANT, AND FACTORS THAT CAN DISRUPT A SYSTEM AT EQUILIBRIUM. 559 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 560 CHEMICAL EQUILIBRIUM 14.1 THE CONCEPT OF EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT Few chemical reactions proceed in only one direction. Most are reversible, at least to some extent. At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse process begins to take place and reactant molecules are formed from product molecules. Chemical equilibrium is achieved when the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant. Chemical equilibrium is a dynamic process. As such, it can be likened to the movement of skiers at a busy ski resort, where the number of skiers carried up the mountain on the chair lift is equal to the number coming down the slopes. Although there is a constant transfer of skiers, the number of people at the top and the number at the bottom of the slope do not change. Note that chemical equilibrium involves different substances as reactants and products. Equilibrium between two phases of the same substance is called physical equilibrium because the changes that occur are physical processes. The vaporization of water in a closed container at a given temperature is an example of physical equilibrium. In this instance, the number of H2O molecules leaving and the number returning to the liquid phase are equal: H2O(l ) 3:4 H2O(g) (Recall from Chapter 4 that the double arrow means that the reaction is reversible.) The study of physical equilibrium yields useful information, such as the equilibrium vapor pressure (see Section 11.8). However, chemists are particularly interested in chemical equilibrium processes, such as the reversible reaction involving nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4) (Figure 14.1). The progress of the reaction Liquid water in equilibrium with its vapor at room temperature. N2O4(g) 3:4 2NO2(g) can be monitored easily because N2O4 is a colorless gas, whereas NO2 has a darkbrown color that makes it sometimes visible in polluted air. Suppose that N2O4 is injected into an evacuated flask. Some brown color appears immediately, indicating the formation of NO2 molecules. The color intensifies as the dissociation of N2O4 continues until eventually equilibrium is reached. Beyond that point, no further change in color is evident because the concentrations of both N2O4 and NO2 remain constant. We can also bring about an equilibrium state by starting with pure NO2. As some of the NO2 molecules combine to form N2O4, the color fades. Yet another way to create an equilibrium state is to start with a mixture of NO2 and N2O4 and monitor the system until the color stops changing. These studies demonstrate that the above reaction is indeed reversible, because a pure component (N2O4 or NO2) reacts to give the other gas. The important thing to keep in mind is that at equilibrium, the conversions of N2O4 to FIGURE 14.1 A reversible reaction between N2O4 and NO2 molecules. + Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.1 THE CONCEPT OF EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT 561 N2O4 N2O4 Concentration Concentration Concentration N2O4 NO2 NO2 NO2 Time (a) FIGURE 14.2 Change in the concentrations of NO2 and N2O4 with time, in three situations. (a) Initially only NO2 is present. (b) Initially only N2O4 is present. (c) Initially a mixture of NO2 and N2O4 is present. Note that even though equilibrium is reached in all cases, the equilibrium concentrations of NO2 and N2O4 are not the same. Time (b) Time (c) NO2 and NO2 to N2O4 are still going on. We do not see a color change because the two rates are equal—the removal of NO2 molecules takes place as fast as the production of NO2 molecules, and N2O4 molecules are formed as quickly as they dissociate. Figure 14.2 summarizes these three situations. THE EQUILIBRIUM CONSTANT Table 14.1 shows some experimental data for the NO2 –N2O4 system at 25°C. The gas concentrations are expressed in molarity, which can be calculated from the number of moles of the gases present initially and at equilibrium and the volume of the flask in liters. Analysis of the equilibrium data shows that although the ratio [NO2]/[N2O4] gives scattered values, the ratio [NO2]2/[N2O4] gives a nearly constant value that averages 4.63 10 3: K [NO2]2 [N2O4] 4.63 10 3 (14.1) where K is a constant for the equilibrium reaction N2O4(g) 3:4 2NO2(g) at 25°C. Note that the exponent 2 for [NO2] in this expression is the same as the stoichiometric coefficient for NO2 in the reversible reaction. We can generalize this phenomenon with the following reversible reaction: TABLE 14.1 The NO2–N2O4 System at 25°C INITIAL CONCENTRATIONS (M) [NO2] Forward Main Menu [N2O4] [NO2] [N2O4] [NO2] [N2O4] [NO2]2 [N2O4] 0.000 0.0500 0.0300 0.0400 0.200 Back EQUILIBRIUM CONCENTRATIONS (M) 0.670 0.446 0.500 0.600 0.000 0.0547 0.0457 0.0475 0.0523 0.0204 0.643 0.448 0.491 0.594 0.0898 0.0851 0.102 0.0967 0.0880 0.227 4.65 4.66 4.60 4.60 4.63 TOC Study Guide TOC RATIO OF CONCENTRATIONS AT EQUILIBRIUM Textbook Website 10 10 10 10 10 3 3 3 3 3 MHHE Website 562 CHEMICAL EQUILIBRIUM aA bB 3:4 cC dD where a, b, c, and d are the stoichiometric coefficients for the reacting species A, B, C, and D. For the reaction at a particular temperature K The signs and signs mean “much greater than” and “much smaller than”, respectively. 14.2 [C]c[D]d [A]a[B]b (14.2) where K is the equilibrium constant. Equation (14.2) was formulated by two Norwegian chemists, Cato Guldberg† and Peter Waage,‡ in 1864. It is the mathematical expression of their law of mass action, which holds that for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, K (the equilibrium constant). Note that although the concentrations may vary, as long as a given reaction is at equilibrium and the temperature does not change, according to the law of mass action, the value of K remains constant. The validity of Equation (14.2) and the law of mass action has been established by studying many reversible reactions. The equilibrium constant, then, is defined by a quotient, the numerator of which is obtained by multiplying together the equilibrium concentrations of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation. Applying the same procedure to the equilibrium concentrations of reactants gives the denominator. The magnitude of the equilibrium constant tells us whether an equilibrium reaction favors the products or reactants. If K is much greater than 1 (that is, K 1), the equilibrium will lie to the right and favors the products. Conversely, if the equilibrium constant is much smaller than 1 (that is, K 1), the equilibrium will lie to the left and favor the reactants. In this context, any number greater than 10 is considered to be much greater than 1, and any number less than 0.1 is much less than 1. Although the use of the words “reactants” and “products” may seem confusing because any substance serving as a reactant in the forward reaction also is a product of the reverse reaction, it is in keeping with the convention of referring to substances on the left of the equilibrium arrows as “reactants” and those on the right as “products.” WRITING EQUILIBRIUM CONSTANT EXPRESSIONS The concept of equilibrium constants is extremely important in chemistry. As you will soon see, equilibrium constants are the key to solving a wide variety of stoichiometry problems involving equilibrium systems. For example, an industrial chemist who wants to maximize the yield of sulfuric acid, say, must have a clear understanding of the equilibrium constants for all the steps in the process, starting from the oxidation of sulfur and ending with the formation of the final product. A physician specializing in clinical cases of acid-base imbalance needs to know the equilibrium constants of weak acids and bases. And a knowledge of equilibrium constants of pertinent gas-phase reactions will help an atmospheric chemist better understand the process of ozone destruction in the stratosphere. † Cato Maximilian Guldberg (1836–1902). Norwegian chemist and mathematician. Guldberg’s research was mainly in thermodynamics. ‡ Peter Waage (1833–1900). Norwegian chemist. Like that of his coworker, Guldberg, Waage’s research was primarily in thermodynamics. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.2 WRITING EQUILIBRIUM CONSTANT EXPRESSIONS 563 To use equilibrium constants, we must express them in terms of the reactant and product concentrations. Our only guide is the law of mass action [Equation (14.2)], which is the general formula for finding equilibrium concentrations. However, because the concentrations of the reactants and products can be expressed in different units and because the reacting species are not always in the same phase, there may be more than one way to express the equilibrium constant for the same reaction. To begin with, we will consider reactions in which the reactants and products are in the same phase. HOMOGENEOUS EQUILIBRIA The term homogeneous equilibrium applies to reactions in which all reacting species are in the same phase. An example of homogeneous gas-phase equilibrium is the dissociation of N2O4. The equilibrium constant, as given in Equation (14.1), is Kc [NO2]2 [N2O4] Note that the subscript in Kc indicates that the concentrations of the reacting species are expressed in molarity or moles per liter. The concentrations of reactants and products in gaseous reactions can also be expressed in terms of their partial pressures. From Equation (5.6) we see that at constant temperature the pressure P of a gas is directly related to the concentration in mol/L of the gas; that is, P (n/V)RT. Thus, for the equilibrium process N2O4(g) 3:4 2NO2(g) we can write KP 2 PNO2 PN2O4 (14.3) where PNO2 and PN2O4 are the equilibrium partial pressures (in atm) of NO2 and N2O4, respectively. The subscript in KP tells us that equilibrium concentrations are expressed in terms of pressure. In general, Kc is not equal to KP, since the partial pressures of reactants and products are not equal to their concentrations expressed in moles per liter. A simple relationship between KP and Kc can be derived as follows. Let us consider the following equilibrium in the gas phase: aA(g) 3:4 bB(g) where a and b are stoichiometric coefficients. The equilibrium constant Kc is given by Kc and the expression for KP is [B]b [A]a KP Pb B Pa A where PA and PB are the partial pressures of A and B. Assuming ideal gas behavior, PAV PA Back Forward Main Menu TOC nART nART V Study Guide TOC Textbook Website MHHE Website 564 CHEMICAL EQUILIBRIUM where V is the volume of the container in liters. Also PBV nBRT PB nBRT V Substituting these relations into the expression for KP, we obtain nBRT V nART V KP b a b nB V nA V a (RT)b a Now both nA/V and nB/V have units of mol/L and can be replaced by [A] and [B], so that KP [B]b (RT) [A]a Kc(RT) n n (14.4) where n b a moles of gaseous products moles of gaseous reactants Since pressures are usually expressed in atm, the gas constant R is given by 0.0821 L atm/K mol, and we can write the relationship between KP and Kc as To use this equation, the pressures in KP must be in atm. KP Kc(0.0821T ) n (14.5) In general, KP Kc except in the special case in which n 0 as in the equilibrium mixture of molecular hydrogen, molecular bromine, and hydrogen bromide: H2(g) Br2(g) 3:4 2HBr(g) In this case Equation (14.5) can be written as KP Any number raised to the zero power is equal to 1. Kc(0.0821T )0 Kc As another example of homogeneous equilibrium, let us consider the ionization of acetic acid (CH3COOH) in water: CH3COOH(aq) H2O(l ) 3:4 CH3COO (aq) H3O (aq) The equilibrium constant is Kc ′ [CH3COO ][H3O ] [CH3COOH][H2O] (We use the prime for Kc here to distinguish it from the final form of equilibrium constant to be derived below.) In 1 L, or 1000 g, of water, there are 1000 g/(18.02 g/mol), or 55.5 moles, of water. Therefore, the “concentration” of water, or [H2O], is 55.5 mol/L, or 55.5 M. This is a large quantity compared to the concentrations of other species in solution (usually 1 M or smaller), and we can assume that it does not change appreciably during the course of a reaction. Thus we may treat [H2O] as a constant and rewrite the equilibrium constant as Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.2 WRITING EQUILIBRIUM CONSTANT EXPRESSIONS Kc 565 [CH3COO ][H3O ] [CH3COOH] where K c [H2O] ′ Kc Note that it is general practice not to include units for the equilibrium constant. In thermodynamics, K is defined as having no units because every concentration (molarity) or pressure (atm) term is actually a ratio to a standard value, which is 1 M or 1 atm. This procedure eliminates all units but does not alter the numerical parts of the concentration or pressure. Consequently, K has no units. Later we will extend this practice to acid-base equilibria and solubility equilibria (to be discussed in Chapters 15 and 16). The following examples illustrate the procedure for writing equilibrium constant expressions and calculating equilibrium constants and equilibrium concentrations. EXAMPLE 14.1 Write expressions for Kc, and KP if applicable, for the following reversible reactions at equilibrium: (a) HF(aq) H2O(l ) 3:4 H3O (aq) F (aq) (b) 2NO(g) O2(g) 3:4 2NO2(g) (c) CH3COOH(aq) C2H5OH(aq) 3:4 CH3COOC2H5(aq) H2O(l ) Answer K c: ′ (a) Since there are no gases present, KP does not apply and we have only Kc ′ [H3O ][F ] [HF][H2O] HF is a weak acid, so that the amount of water consumed in acid ionizations is negligible compared with the total amount of water present as solvent. Thus we can rewrite the equilibrium constant as Kc (b) Kc [H3O ][F ] [HF] [NO2]2 KP [NO]2[O2] 2 PNO2 P2 PO2 NO (c) The equilibrium constant K c is given by ′ Kc ′ [CH3COOC2H5][H2O] [CH3COOH][C2H5OH] Because the water produced in the reaction is negligible compared with the water solvent, the concentration of water does not change. Thus we can write the new equilibrium constant as Similar problems: 14.7, 14.8. Kc [CH3COOC2H5] [CH3COOH][C2H5OH] PRACTICE EXERCISE Write Kc and KP for the decomposition of nitrogen pentoxide: 2N2O5(g) 3:4 4NO2(g) Back Forward Main Menu TOC Study Guide TOC O2(g) Textbook Website MHHE Website 566 CHEMICAL EQUILIBRIUM EXAMPLE 14.2 The following equilibrium process has been studied at 230°C: 2NO(g) O2(g) 3:4 2NO2(g) In one experiment the concentrations of the reacting species at equilibrium are found to be [NO] 0.0542 M, [O2] 0.127 M, and [NO2] 15.5 M. Calculate the equilibrium constant (Kc) of the reaction at this temperature. Answer The equilibrium constant is given by Kc [NO2]2 [NO]2[O2] Substituting the concentrations, we find that (15.5)2 (0.0542)2(0.127) Kc Similar problem: 14.14. 6.44 105 Comment Note that Kc is given without units. Also, the large magnitude of Kc is consistent with the high product (NO2) concentration relative to the concentrations of the reactants (NO and O2). PRACTICE EXERCISE Carbonyl chloride (COCl2), also called phosgene, was used in World War I as a poisonous gas. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form carbonyl chloride CO(g) Cl2(g) 3:4 COCl2(g) at 74°C are [CO] 1.2 10 2 M, [Cl2] Calculate the equilibrium constant (Kc). 0.054 M, and [COCl2] 0.14 M. EXAMPLE 14.3 The equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine PCl5(g) 3:4 PCl3(g) Cl2(g) is found to be 1.05 at 250°C. If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250°C? Answer First, we write KP in terms of the partial pressures of the reacting species. PPCl3PCl2 PPCl5 KP Knowing the partial pressures, we write 1.05 or Similar problem: 14.17. Back Forward PCl2 Comment Main Menu (0.463)(PCl2) (0.875) (1.05)(0.875) (0.463) 1.98 atm Note that we have added atm as the unit for PCl2. TOC Study Guide TOC Textbook Website MHHE Website 14.2 WRITING EQUILIBRIUM CONSTANT EXPRESSIONS 567 PRACTICE EXERCISE The equilibrium constant KP for the reaction 2NO2(g) 3:4 2NO(g) is 158 at 1000 K. Calculate PO2 if PNO2 O2(g) 0.400 atm and PNO 0.270 atm. EXAMPLE 14.4 The equilibrium constant (Kc) for the reaction N2O4(g) 3:4 2NO2(g) 3 is 4.63 10 at 25°C. What is the value of KP at this temperature? Answer From Equation (14.5) we write KP Since T 298 K and n 1 2 KP Kc(0.0821T) n 1, we have (4.63 10 3)(0.0821 298) 0.113 Similar problems: 14.15, 14.19. Comment Note that KP, like Kc, is a dimensionless quantity. This example shows that we can get quite a different value for the equilibrium constant for the same reaction, depending on whether we express the concentrations in moles per liter or in atmospheres. PRACTICE EXERCISE For the reaction N2(g) KP is 4.3 10 4 3H2(g) 3:4 2NH3(g) at 375°C. Calculate Kc for the reaction. HETEROGENEOUS EQUILIBRIA As you might expect, a heterogeneous equilibrium results from a reversible reaction involving reactants and products that are in different phases. For example, when calcium carbonate is heated in a closed vessel, the following equilibrium is attained: CaCO3(s) 3:4 CaO(s) CO2(g) The two solids and one gas constitute three separate phases. At equilibrium, we might write the equilibrium constant as The mineral calcite is made of calcium carbonate, as are chalk and marble. Kc ′ Forward Main Menu (14.6) (Again, the prime for Kc here is to distinguish it from the final form of equilibrium constant to be derived below.) However, the “concentration” of a solid, like its density, is an intensive property and does not depend on how much of the substance is present. For example, the “molar concentration” of copper (density: 8.96 g/cm3) at 20°C is the same, whether we have 1 gram or 1 ton of the metal: [Cu] Back [CaO][CO2] [CaCO3] TOC 8.96 g 1 cm3 1 mol 63.55 g Study Guide TOC 0.141 mol/cm3 141 mol/L Textbook Website MHHE Website 568 CHEMICAL EQUILIBRIUM FIGURE 14.3 In (a) and (b) the equilibrium pressure of CO2 is the same at the same temperature, despite the presence of different amounts of CaCO3 and CaO. CaCO3 CaCO3 CaO CaO (a) (b) For this reason, the terms [CaCO3] and [CaO] are themselves constants and can be combined with the equilibrium constant. We can simplify Equation (14.6) by writing [CaCO3] Kc ′ [CaO] Kc [CO2] (14.7) where Kc, the “new” equilibrium constant, is conveniently expressed in terms of a single concentration, that of CO2. Note that the value of Kc does not depend on how much CaCO3 and CaO are present, as long as some of each is present at equilibrium (Figure 14.3). Alternatively, we can express the equilibrium as KP PCO2 (14.8) The equilibrium constant in this case is numerically equal to the pressure of CO2 gas, an easily measurable quantity. What has been said about solids also applies to liquids. Thus, if a liquid is a reactant or a product, we can treat its concentration as constant and omit it from the equilibrium constant expression. Reactions involving heterogeneous equilibria are the subject of the following examples. EXAMPLE 14.5 Write the equilibrium constant expression Kc, and KP if applicable, for each of the following heterogeneous systems: (a) (NH4)2Se(s) 3:4 2NH3(g) H2Se(g) (b) AgCl(s) 3:4 Ag (aq) Cl (aq) (c) P4(s) 6Cl2(g) 3:4 4PCl3(l) Answer (a) The equilibrium constant is given by Kc ′ [NH3]2[H2Se] [(NH4)2Se] However, since (NH4)2Se is a solid, we write the new equilibrium constant as Kc [NH3]2[H2Se] where Kc K c[(NH4)2Se]. Alternatively, we can express the equilibrium constant ′ KP in terms of the partial pressures of NH3 and H2Se: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.2 WRITING EQUILIBRIUM CONSTANT EXPRESSIONS KP P2 3PH2Se NH Kc ′ [Ag ][Cl ] [AgCl] Kc (b) 569 [Ag ][Cl ] Again, we have incorporated [AgCl] into Kc because AgCl is a solid. (c) The equilibrium constant is [PCl3]4 [P4][Cl2]6 Kc ′ Since pure solids and pure liquids do not appear in the equilibrium constant expression, we write Kc 1 [Cl2]6 Alternatively, we can express the equilibrium constant in terms of the pressure of Cl2: KP Similar problem: 14.8. 1 P6 2 Cl PRACTICE EXERCISE Write equilibrium constant expressions for Kc and KP for the formation of nickel tetracarbonyl, which is used to separate nickel from other impurities: Ni(s) 4CO(g) 3:4 Ni(CO)4(g) EXAMPLE 14.6 Consider the following heterogeneous equilibrium: CaCO3(s) 3:4 CaO(s) CO2(g) At 800°C, the pressure of CO2 is 0.236 atm. Calculate (a) KP and (b) Kc for the reaction at this temperature. Answer (a) Using Equation (14.8), we write KP PCO2 0.236 (b) From Equation (14.5), we know Kp In this case, T 800 tion and obtain 273 Kc Kc(0.0821 2.68 n 1, so we substitute these in the equa- 1073 K and n 0.236 Similar problem: 14.20. Kc(0.0821T ) 10 1073) 3 PRACTICE EXERCISE Consider the following equilibrium at 295 K: NH4HS(s) 3:4 NH3(g) H2S(g) The partial pressure of each gas is 0.265 atm. Calculate KP and Kc for the reaction. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 570 CHEMICAL EQUILIBRIUM MULTIPLE EQUILIBRIA The reactions we have considered so far are all relatively simple. A more complicated situation is one in which the product molecules in one equilibrium system are involved in a second equilibrium process: A B 3:4 C D C D 3:4 E F The products formed in the first reaction, C and D, react further to form products E and F. At equilibrium we can write two separate equilibrium constants: Kc ′ Kc ″ and [C][D] [A][B] [E][F] [C][D] The overall reaction is given by the sum of the two reactions A D Kc ′ C D 3:4 E F Kc ″ A Overall reaction: B 3:4 C B 3:4 E F Kc and the equilibrium constant Kc for the overall reaction is [E][F] [A][B] Kc We obtain the same expression if we take the product of the expressions for Kc and K c: ′ ″ Kc K ″ ′c [C][D] [A][B] [E][F] [C][D] [E][F] [A][B] Therefore, Kc Kc K c ′″ (14.9) We can now make an important statement about multiple equilibria: If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. Among the many known examples of multiple equilibria is the ionization of diprotic acids in aqueous solution. The following equilibrium constants have been determined for carbonic acid (H2CO3) at 25°C: H2CO3(aq) 3:4 H (aq) HCO3 (aq) Kc ′ HCO3 (aq) 3:4 H (aq) CO2 (aq) 3 Kc ″ [H ][HCO3 ] [H2CO3] [H ][CO2 ] 3 [HCO3 ] 4.2 4.8 10 10 7 11 The overall reaction is the sum of these two reactions H2CO3(aq) 3:4 2H (aq) CO2 (aq) 3 and the corresponding equilibrium constant is given by Kc Back Forward Main Menu TOC 2 [H ]2[CO3 ] [H2CO3] Study Guide TOC Textbook Website MHHE Website 14.2 WRITING EQUILIBRIUM CONSTANT EXPRESSIONS 571 Using Equation (14.9) we arrive at Kc Kc K c ′″ (4.2 2.0 10 7)(4.8 10 10 11 ) 17 THE FORM OF K AND THE EQUILIBRIUM EQUATION Before concluding this section, let us look at two important rules for writing equilibrium constants: • The reciprocal of x is 1/x. When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. Thus if we write the NO2–N2O4 equilibrium as N2O4(g) 3:4 2NO2(g) then [NO2]2 [N2O4] Kc 4.63 10 3 However, we can represent the equilibrium equally well as 2NO2(g) 3:4 N2O4(g) and the equilibrium constant is now given by Kc ′ [N2O4] [NO2]2 1 Kc 1 4.63 10 3 216 You can see that Kc 1/K c or KcK c 1.00. Either Kc or K c is a valid equilibrium ′ ′ ′ constant, but it is meaningless to say that the equilibrium constant for the NO2 N2O4 system is 4.63 10 3, or 216, unless we also specify how the equilibrium equation is written. • The value of K also depends on how the equilibrium equation is balanced. Consider the following ways of describing the same equilibrium: N2O4(g) 3:4 NO2(g) Kc ′ [NO2] [N2O4] N2O4(g) 3:4 2NO2(g) 1 2 Kc [NO2]2 [N2O4] 1 2 Looking at the exponents we see that K c ′ Kc. In Table 14.1 we find Kc 4.63 10 3; therefore K c 0.0680. ′ According to the law of mass action, each concentration term in the equilibrium constant expression is raised to a power equal to its stoichiometric coefficient. Thus if you double a chemical equation throughout, the corresponding equilibrium constant will be the square of the original value; if you triple the equation, the equilibrium constant will be the cube of the original value, and so on. The NO2–N2O4 example illustrates once again the need to write the chemical equation when quoting the numerical value of an equilibrium constant. The following example deals with the relationship between the equilibrium constants for differently balanced equations describing the same reaction. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 572 CHEMICAL EQUILIBRIUM EXAMPLE 14.7 The reaction for the production of ammonia can be written in a number of ways: (a) N2(g) 3H2(g) 3:4 2NH3(g) (b) 1 N2(g) 3 H2(g) 3:4 NH3(g) 2 2 (c) 1 N2(g) H2(g) 3:4 2 NH3(g) 3 3 Write the equilibrium constant expression for each formulation. (Express the concentrations of the reacting species in mol/L.) (d) How are the equilibrium constants related to one another? Answer (a) Ka (b) Kb [NH3]2 [N2][H2]3 [NH3] 1 2 [N2] [H2] 3 2 2 3 (c) Kc (d) Ka Ka K2 b Similar problem: 14.18. K2 b K3 c K3 c [NH3] [N2] [H2] 1 3 or Kb 3 2 Kc PRACTICE EXERCISE Write the equilibrium expression (Kc) for each of the following reactions and show how they are related to each other: (a) 3O2(g) 3:4 2O3(g), (b) O2(g) 3:4 2 O3(g). 3 SUMMARY OF GUIDELINES FOR WRITING EQUILIBRIUM CONSTANT EXPRESSIONS • • • • • 14.3 The concentrations of the reacting species in the condensed phase are expressed in mol/L; in the gaseous phase, the concentrations can be expressed in mol/L or in atm. Kc is related to KP by a simple equation [Equation (14.5)]. The concentrations of pure solids, pure liquids (in heterogeneous equilibria), and solvents (in homogeneous equilibria) do not appear in the equilibrium constant expressions. The equilibrium constant (Kc or KP) is a dimensionless quantity. In quoting a value for the equilibrium constant, we must specify the balanced equation and the temperature. If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. THE RELATIONSHIP BETWEEN CHEMICAL KINETICS AND CHEMICAL EQUILIBRIUM We have seen that K, defined in Equation (14.2), is constant at a given temperature regardless of variations in individual equilibrium concentrations (review Table 14.1). We can find out why this is so and at the same time gain insight into the equilibrium process by considering the kinetics of chemical reactions. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.3 To review reaction mechanisms, see Section 13.5. THE RELATIONSHIP BETWEEN CHEMICAL KINETICS AND CHEMICAL EQUILIBRIUM 573 Let us suppose that the following reversible reaction occurs via a mechanism consisting of a single elementary step in both the forward and reverse directions: kf 2B 3:4 AB2 kr A The forward rate is given by kf [A][B]2 ratef and the reverse rate is given by rater kr[AB2] where kf and kr are the rate constants for the forward and reverse directions. At equilibrium, when no net changes occur, the two rates must be equal: ratef kf [A][B]2 kf kr or rater kr[AB2] [AB2] [A][B]2 Since both kf and kr are constants at a given temperature, their ratio is also a constant, which is equal to the equilibrium constant Kc. kf kr Kc [AB2] [A][B]2 So Kc is always a constant regardless of the equilibrium concentrations of the reacting species because it is always equal to kf /kr, the quotient of two quantities that are themselves constant at a given temperature. Because rate constants are temperature-dependent [see Equation (13.8)], it follows that the equilibrium constant must also change with temperature. Now suppose the same reaction has a mechanism with more than one elementary step. Suppose it occurs via a two-step mechanism as follows: k′ f 2B 3:4 B2 kr ′ Step 1: Step 2: A k″ f B2 3:4 AB2 k″ r Overall reaction: A 2B 3:4 AB2 This is an example of multiple equilibria, discussed in Section 14.2. We write the expressions for the equilibrium constants: K′ k′ f k′ r [B2] [B]2 (14.10) K″ k″ f k″ r [AB2] [A][B2] (14.11) Multiplying Equation (14.10) by Equation (14.11), we get K′K ″ Back Forward Main Menu TOC [B2][AB2] [B]2[A][B2] Study Guide TOC [AB2] [A][B]2 Textbook Website MHHE Website 574 CHEMICAL EQUILIBRIUM Kc [AB2] [A][B]2 Since both K′ and K ″ are constants, Kc is also a constant. This result lets us generalize our treatment of the reaction aA bB 3:4 cC dD Regardless of whether this reaction occurs via a single-step or a multistep mechanism, we can write the equilibrium constant expression according to the law of mass action shown in Equation (14.2): K [C]c[D]d [A]a[B]b In summary, we see that in terms of chemical kinetics, the equilibrium constant of a reaction can be expressed as a ratio of the rate constants of the forward and reverse reactions. This analysis explains why the equilibrium constant is a constant and why its value changes with temperature. 14.4 WHAT DOES THE EQUILIBRIUM CONSTANT TELL US? We have seen that the equilibrium constant for a given reaction can be calculated from known equilibrium concentrations. Once we know the value of the equilibrium constant, we can use Equation (14.2) to calculate unknown equilibrium concentrations — remembering, of course, that the equilibrium constant has a constant value only if the temperature does not change. In general, the equilibrium constant helps us to predict the direction in which a reaction mixture will proceed to achieve equilibrium and to calculate the concentrations of reactants and products once equilibrium has been reached. These uses of the equilibrium constant will be explored in this section. PREDICTING THE DIRECTION OF A REACTION The equilibrium constant Kc for the formation of hydrogen iodide from molecular hydrogen and molecular iodine in the gas phase H2(g) I2(g) 3:4 2HI(g) is 54.3 at 430°C. Suppose that in a certain experiment we place 0.243 mole of H2, 0.146 mole of I2, and 1.98 moles of HI all in a 1.00-L container at 430°C. Will there be a net reaction to form more H2 and I2 or more HI? Inserting the starting concentrations in the equilibrium constant expression, we write [HI]2 0 [H2]0[I2]0 (1.98)2 (0.243)(0.146) 111 where the subscript 0 indicates initial concentrations (before equilibrium is reached). 2 Because the quotient [HI]0 /[H2]0[I2]0 is greater than Kc, this system is not at equilibrium. Consequently, some of the HI will react to form more H2 and I2 (decreasing the value of the quotient). Thus the net reaction proceeds from right to left to reach equilibrium. For reactions that have not reached equilibrium, such as the formation of HI considered above, we obtain the reaction quotient (Qc), instead of the equilibrium constant by substituting the initial concentrations into the equilibrium constant expression. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.4 WHAT DOES THE EQUILIBRIUM CONSTANT TELL US? 575 To determine the direction in which the net reaction will proceed to achieve equilibrium, we compare the values of Qc and Kc. The three possible cases are as follows: • Qc • Qc • Qc Kc The ratio of initial concentrations of products to reactants is too large. To reach equilibrium, products must be converted to reactants. The system proceeds from right to left (consuming products, forming reactants) to reach equilibrium. Kc The initial concentrations are equilibrium concentrations. The system is at equilibrium. Kc The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds from left to right (consuming reactants, forming products) to reach equilibrium. A comparison of Kc with Qc for a reaction is shown in the following example. EXAMPLE 14.8 At the start of a reaction, there are 0.249 mol N2, 3.21 10 2 mol H2, and 6.42 10 4 mol NH3 in a 3.50-L reaction vessel at 375°C. If the equilibrium constant (Kc) for the reaction N2(g) 3H2(g) 3:4 2NH3(g) is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed. Answer The initial concentrations of the reacting species are [N2]0 [H2]0 [NH3]0 0.249 mol 3.50 L 0.0711 M 3.21 10 2 mol 3.50 L 9.17 10 3 M 6.42 10 4 mol 3.50 L 1.83 10 4 M Next we write 2 [NH3]0 [N2]0[H2]3 0 Similar problems: 14.37, 14.38. (1.83 10 4)2 (0.0711)(9.17 10 3)3 0.611 Qc Since Qc is smaller than Kc (1.2), the system is not at equilibrium. The net result will be an increase in the concentration of NH3 and a decrease in the concentrations of N2 and H2. That is, the net reaction will proceed from left to right until equilibrium is reached. PRACTICE EXERCISE The equilibrium constant (Kc) for the formation of nitrosyl chloride, an orangeyellow compound, from nitric oxide and molecular chlorine 2NO(g) Cl2(g) 3:4 2NOCl(g) is 6.5 104 at 35°C. In a certain experiment, 2.0 10 2 mole of NO, 8.3 10 3 mole of Cl2, and 6.8 moles of NOCl are mixed in a 2.0-L flask. In which direction will the system proceed to reach equilibrium? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 576 CHEMICAL EQUILIBRIUM CALCULATING EQUILIBRIUM CONCENTRATIONS If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the equilibrium mixture from the initial concentrations. Commonly, only the initial reactant concentrations are given. Let us consider the following system involving two organic compounds, cis-stilbene and trans-stilbene, in a nonpolar hydrocarbon solvent (Figure 14.4): cis-stilbene 3:4 trans-stilbene The equilibrium constant (Kc) for this system is 24.0 at 200°C. Suppose that initially only cis-stilbene is present at a concentration of 0.850 mol/L. How do we calculate the concentrations of cis- and trans-stilbene at equilibrium? From the stoichiometry of the reaction we see that for every mole of cis-stilbene converted, 1 mole of trans-stilbene is formed. Let x be the equilibrium concentration of trans-stilbene in mol/L; therefore, the equilibrium concentration of cis-stilbene must be (0.850 x) mol/L. It is useful to summarize the changes in concentration as follows: FIGURE 14.4 The equilibrium between cis-stilbene and transstilbene. Note that both molecules have the same molecular formula (C14H12) and also the same type of bonds. However, in cisstilbene, the benzene rings are on one side of the CPC bond and the H atoms are on the other side, whereas in trans-stilbene the benzene rings and the H atoms are across from the CPC bond. These compounds have different melting points and dipole moments. Initial (M): Change (M): cis-stilbene 3:4 trans -stilbene 0.850 0 x x Equilibrium (M): (0.850 x) x A positive ( ) change represents an increase and a negative ( ) change a decrease in concentration at equilibrium. Next we set up the equilibrium constant expression Kc 24.0 x [trans-stilbene] [cis-stilbene] x 0.850 x 0.816 M Having solved for x, we calculate the equilibrium concentrations of cis-stilbene and trans-stilbene as follows: [cis-stilbene] [trans-stilbene] (0.850 0.816) M 0.034 M 0.816 M To check the results we could use the equilibrium concentrations to calculate Kc. We summarize our approach to solving equilibrium constant problems as follows: 1. 2. 3. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. Having solved for x, calculate the equilibrium concentrations of all species. The following examples illustrate the application of this three-step procedure. The mechanism for this reaction was discussed in Section 13.5. For simplicity, we will ignore the formation of an iodine intermediate in calculating the equilibrium concentrations here. Back Forward EXAMPLE 14.9 A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L stainless-steel flask at 430°C. The equilibrium constant Kc for the reaction H2(g) I2(g) 3:4 2HI(g) is 54.3 at this temperature. Calculate the concentrations of H2, I2, and HI at equilibrium. Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.4 WHAT DOES THE EQUILIBRIUM CONSTANT TELL US? 577 Answer Step 1: The stoichiometry of the reaction is 1 mol H2 reacting with 1 mol I2 to yield 2 mol HI. Let x be the depletion in concentration (mol/L) of H2 and I2 at equilibrium. It follows that the equilibrium concentration of HI must be 2x. We summarize the changes in concentrations as follows: H2 0.500 x Initial (M): Change (M): Equilibrium (M): I2 0.500 x (0.500 x) (0.500 3:4 2HI 0.000 2x x) 2x Step 2: The equilibrium constant is given by [HI]2 [H2][I2] Kc Substituting, we get 54.3 (0.500 (2x)2 x)(0.500 x) Taking the square root of both sides, we get 7.37 x 2x 0.500 x 0.393 M Step 3: At equilibrium, the concentrations are [H2] (0.500 0.393) M 0.107 M [I2] (0.500 0.393) M 0.107 M [HI] 2 0.393 M 0.786 M You can check your answers by calculating Kc using the equilibrium concentrations. Comment Similar problem: 14.46. PRACTICE EXERCISE Consider the reaction in Example 14.9. Starting with a concentration of 0.040 M for HI, calculate the concentrations of HI, H2, and I2 at equilibrium. EXAMPLE 14.10 For the same reaction and temperature as in Example 14.9, suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium. Following the procedure in Example 14.8, we first show that the initial concentrations do not correspond to the equilibrium concentrations. Then we proceed as follows. Step 1: Let x be the depletion in concentration (mol/L) of H2 and I2 at equilibrium. From the stoichiometry of the reaction it follows that the increase in concentration for HI must be 2x. Next we write Answer Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 578 CHEMICAL EQUILIBRIUM H2 0.00623 x Initial (M ): Change (M ): Equilibrium (M ): I2 0.00414 x x) (0.00623 2HI 0.0224 2x 3:4 (0.00414 x) (0.0224 2x) Step 2: The equilibrium constant is [HI]2 [H2][I2] Kc Substituting, we get (0.0224 2x)2 (0.00623 x)(0.00414 54.3 x) It is not possible to solve this equation by the square root shortcut, as the starting concentrations [H2] and [I2] are unequal. Instead, we must first carry out the multiplications 54.3(2.58 10 5 x2) 0.0104x 5.02 10 4 0.0896x 4x2 Collecting terms, we get 50.3x2 0.654x 8.98 10 2 This is a quadratic equation of the form ax a quadratic equation (see Appendix 4) is Here we have a 50.3, b 0.654 x x b2 2a b x 4 0 bx x 10 4, so that 8.98 ( 0.654)2 4(50.3)(8.98 2 50.3 or 0. The solution for 4ac 0.654, and c 0.0114 M c 10 4) 0.00156 M The first solution is physically impossible since the amounts of H2 and I2 reacted would be more than those originally present. The second solution gives the correct answer. Note that in solving quadratic equations of this type, one answer is always physically impossible, so choosing a value for x is easy. Step 3: At equilibrium, the concentrations are [H2] [I2] (0.00414 [HI] Similar problem: 14.86. (0.00623 (0.0224 0.00156) M 0.00467 M 0.00156) M 0.00258 M 2 0.00156) M 0.0255 M PRACTICE EXERCISE At 1280°C the equilibrium constant (Kc) for the reaction Br2(g) 3:4 2Br(g) 3 is 1.1 10 . If the initial concentrations are [Br2] 6.3 10 2 M and [Br] 1.2 10 2 M, calculate the concentrations of these species at equilibrium. Examples 14.9 and 14.10 show that we can calculate the concentrations of all the reacting species at equilibrium if we know the equilibrium constant and the initial con- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.5 FACTORS THAT AFFECT CHEMICAL EQUILIBRIUM 579 centrations. This information is valuable if we need to estimate the yield of a reaction. For example, if the reaction between H2 and I2 to form HI were to go to completion, the number of moles of HI formed in Example 14.9 would be 2 0.500 mol, or 1.00 mol. However, because of the equilibrium process, the actual amount of HI formed can be no more than 2 0.393 mol, or 0.786 mol, a 78.6 percent yield. 14.5 FACTORS THAT AFFECT CHEMICAL EQUILIBRIUM Chemical equilibrium represents a balance between forward and reverse reactions. In most cases, this balance is quite delicate. Changes in experimental conditions may disturb the balance and shift the equilibrium position so that more or less of the desired product is formed. When we say that an equilibrium position shifts to the right, for example, we mean that the net reaction is now from left to right. Variables that can be controlled experimentally are concentration, pressure, volume, and temperature. Here we will examine how each of these variables affects a reacting system at equilibrium. In addition, we will examine the effect of a catalyst on equilibrium. LE CHATELIER’S PRINCIPLE There is a general rule that helps us to predict the direction in which an equilibrium reaction will move when a change in concentration, pressure, volume, or temperature occurs. The rule, known as Le Chatelier’s† principle, states that if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset. The word “stress” here means a change in concentration, pressure, volume, or temperature that removes a system from the equilibrium state. We will use Le Chatelier ’s principle to assess the effects of such changes. CHANGES IN CONCENTRATION Iron(III) thiocyanate [Fe(SCN)3] dissolves readily in water to give a red solution. The red color is due to the presence of hydrated FeSCN2 ion. The equilibrium between undissociated FeSCN2 and the Fe3 and SCN ions is given by FeSCN2 (aq) 3:4 Fe3 (aq) red pale yellow SCN (aq) colorless What happens if we add some sodium thiocyanate (NaSCN) to this solution? In this case, the stress applied to the equilibrium system is an increase in the concentration of SCN (from the dissociation of NaSCN). To offset this stress, some Fe3 ions react with the added SCN ions, and the equilibrium shifts from right to left: FeSCN2 (aq) m88 Fe3 (aq) Both Na and NO3 are colorless spectator ions. SCN (aq) Consequently, the red color of the solution deepens (Figure 14.5). Similarly, if we added iron(III) nitrate [Fe(NO3)3] to the original solution, the red color would also deepen because the additional Fe3 ions [from Fe(NO3)3] would shift the equilibrium from right to left. † Henry Louis Le Chatelier (1850–1936). French chemist. Le Chatelier did work on metallurgy, cements, glasses, fuels, and explosives. He was also noted for his skills in industrial management. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 580 CHEMICAL EQUILIBRIUM (a) (b) (c) (d) FIGURE 14.5 Effect of concentration change on the position of equilibrium. (a) An aqueous Fe(SCN)3 solution. The color of the solution is due to both the red FeSCN2 and the yellow Fe3 species. (b) After the addition of some NaSCN to the solution in (a), the equilibrium shifts to the left. (c) After the addition of some Fe(NO3)3 to the solution in (a), the equilibrium shifts to the left. (d) After the addition of some H2C2O4 to the solution in (a), the equilibrium shifts to the right. The yellow color is due to the Fe(C2O4)3 ions. 3 Oxalic acid is sometimes used to remove bathtub rings that consist of rust, or Fe2O3. Now suppose we add some oxalic acid (H2C2O4) to the original solution. Oxalic acid ionizes in water to form the oxalate ion, C2O2 , which binds strongly to the Fe3 4 ions. The formation of the stable yellow ion Fe(C2O4)3 removes free Fe3 ions in so3 lution. Consequently, more FeSCN2 units dissociate and the equilibrium shifts from left to right: FeSCN2 (aq) 88n Fe3 (aq) Since Le Chatelier ’s principle simply summarizes the observed behavior of equilibrium systems, it is incorrect to say that a given equilibrium shift occurs “because of ” Le Chatelier ’s principle. SCN (aq) The red solution will turn yellow due to the formation of Fe(C2O4)3 ions. 3 This experiment demonstrates that all reactants and products are present in the reacting system at equilibrium. Second, increasing the concentrations of the products (Fe3 or SCN ) shifts the equilibrium to the left, and decreasing the concentration of the product Fe3 shifts the equilibrium to the right. These results are just as predicted by Le Chatelier ’s principle. The effect of a change in concentration on the equilibrium position is shown in the following example. EXAMPLE 14.11 At 720°C, the equilibrium constant Kc for the reaction N2(g) 3H2(g) 3:4 2NH3(g) is 2.37 10 3. In a certain experiment, the equilibrium concentrations are [N2] 0.683 M, [H2] 8.80 M, and [NH3] 1.05 M. Suppose some of the NH3 is added to the mixture so that its concentration is increased to 3.65 M. (a) Use Le Chatelier ’s principle to predict the shift in direction of the net reaction to reach a new equilibrium. (b) Confirm your prediction by calculating the reaction quotient Qc and comparing its value with Kc. Answer (a) The stress applied to the system is the addition of NH3. To offset this stress, some NH3 reacts to produce N2 and H2 until a new equilibrium is established. The net reaction therefore shifts from right to left; that is, N2(g) Back Forward Main Menu TOC 3H2(g) m88 2NH3(g) Study Guide TOC Textbook Website MHHE Website 14.5 Initial Final equilibrium Change equilibrium FACTORS THAT AFFECT CHEMICAL EQUILIBRIUM (b) At the instant when some of the NH3 is added, the system is no longer at equilibrium. The reaction quotient is given by [NH3]2 0 [N2]0[H2]3 0 Qc H2 Concentration 581 (3.65)2 (0.683)(8.80)3 2.86 10 2 Since this value is greater than 2.37 10 3, the net reaction shifts from right to left until Qc equals Kc. Figure 14.6 shows qualitatively the changes in concentrations of the reacting species. NH3 PRACTICE EXERCISE N2 At 430°C, the equilibrium constant (KP) for the reaction 2NO(g) Time FIGURE 14.6 Changes in concentration of H2, N2, and NH3 after the addition of NH3 to the equilibrium mixture. O2(g) 3:4 2NO2(g) is 1.5 105. In one experiment, the initial pressures of NO, O2, and NO2 are 2.1 10 3 atm, 1.1 10 2 atm, and 0.14 atm, respectively. Calculate QP and predict the direction that the net reaction will shift to reach equilibrium. CHANGES IN VOLUME AND PRESSURE Changes in pressure ordinarily do not affect the concentrations of reacting species in condensed phases (say, in an aqueous solution) because liquids and solids are virtually incompressible. On the other hand, concentrations of gases are greatly affected by changes in pressure. Let us look again at Equation (5.6): PV P nRT n RT V Note that P and V are related to each other inversely: The greater the pressure, the smaller the volume, and vice versa. Note, too, that the term (n/V) is the concentration of the gas in mol/L, and it varies directly with pressure. Suppose that the equilibrium system N2O4(g) 3:4 2NO2(g) is in a cylinder fitted with a movable piston. What happens if we increase the pressure on the gases by pushing down on the piston at constant temperature? Since the volume decreases, the concentration (n/V) of both NO2 and N2O4 increases. Because the concentration of NO2 is squared in the equilibrium constant expression, the increase in pressure increases the numerator more than the denominator. The system is no longer at equilibrium, so we write Qc The shift in equilibrium can also be predicted using Le Chatelier ’s principle. Back Forward Main Menu [NO2]2 0 [N2O4]0 Thus Qc Kc, and the net reaction will shift to the left until Qc Kc. Conversely, a decrease in pressure (increase in volume) would result in Qc Kc, and the net reaction would shift to the right until Qc Kc. TOC Study Guide TOC Textbook Website MHHE Website 582 CHEMICAL EQUILIBRIUM In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases (the reverse reaction, in this case), and a decrease in pressure (increase in volume) favors the net reaction that increases the total number of moles of gases (here, the forward reaction). For reactions in which there is no change in the number of moles of gases, a pressure (or volume) change has no effect on the position of equilibrium. It is possible to change the pressure of a system without changing its volume. Suppose the NO2–N2O4 system is contained in a stainless-steel vessel whose volume is constant. We can increase the total pressure in the vessel by adding an inert gas (helium, for example) to the equilibrium system. Adding helium to the equilibrium mixture at constant volume increases the total gas pressure and decreases the mole fractions of both NO2 and N2O4; but the partial pressure of each gas, given by the product of its mole fraction and total pressure (see Section 5.6), does not change. Thus the presence of an inert gas in such a case does not affect the equilibrium. The following example illustrates the effect of a pressure change on the equilibrium position. EXAMPLE 14.12 Consider the following equilibrium systems: (a) 2PbS(s) 3O2(g) 3:4 2PbO(s) 2SO2(g) (b) PCl5(g) 3:4 PCl3(g) Cl2(g) (c) H2(g) CO2(g) 3:4 H2O(g) CO(g) Predict the direction of the net reaction in each case as a result of increasing the pressure (decreasing the volume) on the system at constant temperature. (a) Consider only the gaseous molecules. In the balanced equation there are 3 moles of gaseous reactants and 2 moles of gaseous products. Therefore the net reaction will shift toward the products (to the right) when the pressure is increased. (b) The number of moles of products is 2 and that of reactants is 1; therefore the net reaction will shift to the left, toward the reactant. (c) The number of moles of products is equal to the number of moles of reactants, so a change in pressure has no effect on the equilibrium. Answer Similar problem: 14.54. Comment In each case, the prediction is consistent with Le Chatelier ’s principle. PRACTICE EXERCISE Consider the equilibrium reaction involving nitrosyl chloride, nitric oxide, and molecular chlorine 2NOCl(g) 3:4 2NO(g) Cl2(g) Predict the direction of the net reaction as a result of decreasing the pressure (increasing the volume) on the system at constant temperature. CHANGES IN TEMPERATURE A change in concentration, pressure, or volume may alter the equilibrium position, but it does not change the value of the equilibrium constant. Only a change in temperature can alter the equilibrium constant. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.5 FACTORS THAT AFFECT CHEMICAL EQUILIBRIUM 583 FIGURE 14.7 (a) Two bulbs containing a mixture of NO2 and N2O4 gases at equilibrium. (b) When one bulb is immersed in ice water (left), its color becomes lighter, indicating the formation of colorless N2O4 gas. When the other bulb is immersed in hot water, its color darkens, indicating an increase in NO2. (a) (b) The formation of NO2 from N2O4 is an endothermic process: N2O4(g) 88n 2NO2(g) H° 58.0 kJ and the reverse reaction is exothermic: 2NO2(g) 88n N2O4(g) H° 58.0 kJ At equilibrium the net heat effect is zero because there is no net reaction. What happens if the following equilibrium system N2O4(g) 3:4 2NO2(g) is heated at constant volume? Since endothermic processes absorb heat from the surroundings, heating favors the dissociation of N2O4 into NO2 molecules. Consequently, the equilibrium constant, given by Kc [NO2]2 [N2O4] increases with temperature (Figure 14.7). As another example, consider the equilibrium between the following ions: CoCl2 4 blue 6H2O 3:4 Co(H2O)2 6 pink 4Cl The formation of CoCl2 is endothermic. On heating, the equilibrium shifts to the left 4 and the solution turns blue. Cooling favors the exothermic reaction [the formation of Co(H2O)2 ] and the solution turns pink (Figure 14.8). 6 In summary, a temperature increase favors an endothermic reaction, and a temperature decrease favors an exothermic reaction. THE EFFECT OF A CATALYST We know that a catalyst enhances the rate of a reaction by lowering the reaction’s activation energy (Section 13.9). However, as Figure 13.20 shows, a catalyst lowers the activation energy of the forward reaction and the reverse reaction to the same extent. We can therefore conclude that the presence of a catalyst does not alter the equilibrium constant, nor does it shift the position of an equilibrium system. Adding a catalyst to a reaction mixture that is not at equilibrium will simply cause the mixture to Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 584 CHEMICAL EQUILIBRIUM FIGURE 14.8 (Left) Heating favors the formation of the blue CoCl 2 ion. (Right) Cooling fa4 vors the formation of the pink Co(H2O)2 ion. 6 reach equilibrium sooner. The same equilibrium mixture could be obtained without the catalyst, but we might have to wait much longer for it to happen. SUMMARY OF FACTORS THAT MAY AFFECT THE EQUILIBRIUM POSITION We have considered four ways to affect a reacting system at equilibrium. It is important to remember that, of the four, only a change in temperature changes the value of the equilibrium constant. Changes in concentration, pressure, and volume can alter the equilibrium concentrations of the reacting mixture, but they cannot change the equilibrium constant as long as the temperature does not change. A catalyst can speed up the process, but it has no effect on the equilibrium constant or on the equilibrium concentrations of the reacting species. Two processes that illustrate the effects of changed conditions on equilibrium processes are discussed in Chemistry in Action essays on pp. 585 and 586. The effects of temperature, concentration, and pressure change, as well as the addition of an inert gas, on an equilibrium system are treated in the following example. EXAMPLE 14.13 Consider the following equilibrium process between dinitrogen tetrafluoride (N2F4) and nitrogen difluoride (NF2): N2F4(g) 3:4 2NF2(g) H° 38.5 kJ Predict the changes in the equilibrium if (a) the reacting mixture is heated at constant volume; (b) NF2 gas is removed from the reacting mixture at constant temperature and volume; (c) the pressure on the reacting mixture is decreased at constant temperature; and (d) an inert gas, such as helium, is added to the reacting mixture at constant volume and temperature. (a) Since the forward reaction is endothermic, an increase in temperature favors the formation of NF2. The equilibrium constant Answer Kc [NF2]2 [N2F4] will therefore increase with increasing temperature. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 14.5 Similar problems: 14.55, 14.56. FACTORS THAT AFFECT CHEMICAL EQUILIBRIUM 585 (b) The stress here is the removal of NF2 gas. To offset it, more N2F4 will decompose to form NF2. The equilibrium constant Kc remains unchanged, however. (c) A decrease in pressure (which is accompanied by an increase in gas volume) favors the formation of more gas molecules, that is, the forward reaction. Thus, more NF2 gas will be formed. The equilibrium constant will remain unchanged. (d) Adding helium to the equilibrium mixture at constant volume will not shift the equilibrium. PRACTICE EXERCISE Consider the equilibrium between molecular oxygen and ozone 3O2(g) 3:4 2O3(g) H° 284 kJ What would be the effect of (a) increasing the pressure on the system by decreasing the volume, (b) increasing pressure by adding O2 to the system, (c) decreasing the temperature, and (d) adding a catalyst? Back The Haber Process Knowing the factors that affect chemical equilibrium has great practical value for industrial applications, such as the synthesis of ammonia. The Haber process for synthesizing ammonia from molecular hydrogen and nitrogen uses a heterogeneous catalyst to speed up the reaction (see p. 540). Let us look at the equilibrium reaction for ammonia synthesis to determine whether there are factors that could be manipulated to enhance the yield. Suppose, as a prominent industrial chemist at the turn of the twentieth century, you are asked to design an efficient procedure for synthesizing ammonia from hydrogen and nitrogen. Your main objective is to obtain a high yield of the product while keeping the production costs down. Your first step is to take a careful look at the balanced equation for the production of ammonia: N2(g) 3H2(g) 3:4 2NH3(g) H° 92.6 kJ Two ideas strike you: First, since 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3, a higher yield of NH3 can be obtained at equilibrium if the reaction is carried out under high pressures. This is indeed the case, as shown by the plot of mole percent of NH3 versus the total pressure of the reacting system. Second, the exothermic nature of the forward Forward Main Menu TOC 100 Mole percent of NH3 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry 80 60 40 20 0 1000 2000 3000 4000 Pressure (atm) Mole percent of NH3 as a function of the total pressures of the gases at 425°C. reaction tells you that the equilibrium constant for the reaction will decrease with increasing temperature. Thus, for maximum yield of NH3, the reaction should be run at the lowest possible temperature. The graph on p. 586 shows that the yield of ammonia increases with decreasing temperature. A low-temperature operation (say, 220 K or 53°C) is desirable in other respects too. The boiling point of NH3 is 33.5°C, so as it formed it would quickly condense to a liquid, which could be conveniently removed from the react- Study Guide TOC Textbook Website MHHE Website 586 CHEMICAL EQUILIBRIUM H2 + N2 NH3 80 Mole percent Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chem 100 Compressor N2 + H2 Reaction chamber (catalysts) NH3 + H2 + N2 60 Unreacted 40 H2 + N2 20 Ammonia condenser Liquid NH3 0 200 300 400 500 Storage tanks Temperature (° C) The composition (mole percent) of H2 N2 and NH3 at equilibrium (for a certain starting mixture) as a function of temperature. ing system. (Both H2 and N2 are still gases at this temperature.) Consequently, the net reaction would shift from left to right, just as desired. On paper, then, these are your conclusions. Let us compare your recommendations with the actual conditions in an industrial plant. Typically, the operating pressures are between 500 atm and 1000 atm, so you are right to advocate high pressure. Furthermore, in the industrial process the NH3 never reaches its equilibrium value but is constantly removed from the reaction mixture in a continuous process operation. This design makes sense, too, as you had anticipated. The Schematic diagram of the Haber process for ammonia synthesis. The heat generated from the reaction is used to heat the incoming gases. only discrepancy is that the operation is usually carried out at about 500°C. This high-temperature operation is costly and the yield of NH3 is low. The justification for this choice is that the rate of NH3 production increases with increasing temperature. Commercially, faster production of NH3 is preferable even if it means a lower yield and higher operating cost. For this reason, the combination of high-pressure, high-temperature conditions and the proper catalyst is the most efficient way to produce ammonia on a large scale. Chemistry in Action Chemistry in Action Chemistry Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Life at High Altitudes and Hemoglobin Production In the human body, countless chemical equilibria must be maintained to ensure physiological well-being. If environmental conditions change, the body must adapt to keep functioning. The consequences of a sudden change in altitude dramatize this fact. Flying from San Francisco, which is at sea level, to Mexico City, where the elevation is 2.3 km (1.4 mi), or scaling a 3-km mountain in two days can cause headache, nausea, extreme fatigue, and other discomforts. These conditions are all symptoms of hypoxia, a deficiency in the amount of oxygen reaching body tissues. In serious cases, the victim may slip into a coma and die if not Forward Main Menu TOC treated quickly. And yet a person living at a high altitude for weeks or months gradually recovers from altitude sickness and adjusts to the low oxygen content in the atmosphere, so that he or she can function normally. The combination of oxygen with the hemoglobin (Hb) molecule, which carries oxygen through the blood, is a complex reaction, but for our purposes it can be represented by a simplified equation: Hb(aq) O2(aq) 3:4 HbO2(aq) where HbO2 is oxyhemoglobin, the hemoglobinoxygen complex that actually transports oxygen to tis- Study Guide TOC Textbook Website MHHE Website SUMMARY OF FACTS AND CONCEPTS 587 Chemistry in Action Chemistry in Action Chemistry in Action sues. The equilibrium constant is Kc [HbO2] [Hb][O2] At an altitude of 3 km the partial pressure of oxygen is only about 0.14 atm, compared with 0.2 atm at sea level. According to Le Chatelier’s principle, a decrease in oxygen concentration will shift the equilibrium shown in the equation above from right to left. This change depletes the supply of oxyhemoglobin, causing hypoxia. Given enough time, the body copes with this problem by producing more hemoglobin molecules. The equilibrium will then gradually shift back toward the formation of oxyhemoglobin. It takes two to three weeks for the increase in hemoglobin production to meet the body’s basic needs adequately. A return to full capacity may require several years to occur. Studies show that long-time residents of highaltitude areas have high hemoglobin levels in their blood — sometimes as much as 50 percent more than individuals living at sea level! SUMMARY OF FACTS AND CONCEPTS •K [C]c[D]d [A]a[B]b • KP • Kc SUMMARY OF KEY EQUATIONS Kc(0.0821T ) KcK c ′″ (14.2) n (14.5) (14.9) Mountaineers need weeks or even months to become acclimatized before scaling summits such as Mount Everest. Law of Mass Action. General expression of equilibrium constant. Relationship between KP and Kc. The equilibrium constant for the overall reaction is given by the product of the equilibrium constants for the individual reactions. 1. Dynamic equilibria between phases are called physical equilibria. Chemical equilibrium is a reversible process in which the rates of the forward and reverse reactions are equal and the concentrations of reactants and products do not change with time. 2. For the general chemical reaction aA bB 3:4 cC dD the concentrations of reactants and products at equilibrium (in moles per liter) are related by the equilibrium constant expression [Equation (14.2)]. 3. The equilibrium constant for gases, KP, expresses the relationship of the equilibrium partial pressures (in atm). 4. A chemical equilibrium process in which all reactants and products are in the same phase is homogeneous. If the reactants and products are not all in the same phase, the equilibrium is heterogeneous. The concentrations of pure solids, pure liquids, and solvents are constant and do not appear in the equilibrium constant expression of a reaction. 5. If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 588 CHEMICAL EQUILIBRIUM 6. The value of K depends on how the chemical equation is balanced, and the equilibrium constant for the reverse of a particular reaction is the reciprocal of the equilibrium constant of that reaction. 7. The equilibrium constant is the ratio of the rate constant for the forward reaction to that for the reverse reaction. 8. The reaction quotient Q has the same form as the equilibrium constant, but it applies to a reaction that may not be at equilibrium. If Q K, the reaction will proceed from right to left to achieve equilibrium. If Q K, the reaction will proceed from left to right to achieve equilibrium. 9. Le Chatelier ’s principle states that if an external stress is applied to a system at chemical equilibrium, the system will adjust to partially offset the stress. 10. Only a change in temperature changes the value of the equilibrium constant for a particular reaction. Changes in concentration, pressure, or volume may change the equilibrium concentrations of reactants and products. The addition of a catalyst hastens the attainment of equilibrium but does not affect the equilibrium concentrations of reactants and products. KEY WORDS Chemical equilibrium, p. 560 Equilibrium constant (K ), p. 562 Heterogeneous equilibrium, p. 567 Homogeneous equilibrium, p. 563 Law of Mass Action, p. 562 Le Chatelier ’s principle, p. 579 Physical equilibrium, p. 560 Reaction quotient (Q), p. 574 QUESTIONS AND PROBLEMS THE CONCEPT OF EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT Review Questions 14.1 Define equilibrium. Give two examples of a dynamic equilibrium. 14.2 Explain the difference between physical equilibrium and chemical equilibrium. Give two examples of each. 14.3 What is the law of mass action? 14.4 Briefly describe the importance of equilibrium in the study of chemical reactions. EQUILIBRIUM CONSTANT EXPRESSIONS 14.9 14.10 14.11 Review Questions 14.5 Define homogeneous equilibrium and heterogeneous equilibrium. Give two examples of each. 14.6 What do the symbols Kc and KP represent? 14.7 Write the expressions for the equilibrium constants KP of the following thermal decomposition reactions: (a) 2NaHCO3(s) 3:4 Na2CO3(s) CO2(g) H2O(g) (b) 2CaSO4(s) 3:4 2CaO(s) 2SO2(g) O2(g) 14.8 Write equilibrium constant expressions for Kc, and for KP if applicable, for the following processes: (a) 2CO2(g) 3:4 2CO(g) O2(g) (b) 3O2(g) 3:4 2O3(g) Back Forward Main Menu TOC 14.12 (c) CO(g) Cl2(g) 3:4 COCl2(g) (d) H2O(g) C(s) 3:4 CO(g) H2(g) (e) HCOOH(aq) 3:4 H (aq) HCOO (aq) (f) 2HgO(s) 3:4 2Hg(l ) O2(g) Write the equilibrium constant expressions for Kc and KP, if applicable, for the following reactions: (a) 2NO2(g) 7H2(g) 3:4 2NH3(g) 4H2O(l) (b) 2ZnS(s) 3O2(g) 3:4 2ZnO(s) 2SO2(g) (c) C(s) CO2(g) 3:4 2CO(g) (d) C6H5COOH(aq) 3:4 C6H5COO (aq) H (aq) Write the equation relating Kc to KP, and define all the terms. What is the rule for writing the equilibrium constant for the overall reaction involving two or more reactions? Give an example of a multiple equilibria reaction. Problems 14.13 The equilibrium constant (Kc) for the reaction 2HCl(g) 3:4 H2(g) Cl2(g) 34 is 4.17 10 at 25°C. What is the equilibrium constant for the reaction H2(g) Cl2(g) 3:4 2HCl(g) at the same temperature? Study Guide TOC Textbook Website MHHE Website 589 QUESTIONS AND PROBLEMS 14.14 Consider the following equilibrium process at 700°C: 2H2(g) S2(g) 3:4 2H2S(g) Analysis shows that there are 2.50 moles of H2, 1.35 10 5 mole of S2, and 8.70 moles of H2S present in a 12.0-L flask. Calculate the equilibrium constant Kc for the reaction. 14.15 What is KP at 1273°C for the reaction 2CO(g) O2(g) 3:4 2CO2(g) if Kc is 2.24 1022 at the same temperature? 14.16 The equilibrium constant KP for the reaction 2SO3(g) 3:4 2SO2(g) O2(g) 5 is 1.8 10 at 350°C. What is Kc for this reaction? 14.17 Consider the following reaction: N2(g) O2(g) 3:4 2NO(g) I2(g) 3:4 2I(g) is 3.8 10 equilibrium at 727°C. Calculate Kc and KP for the Starting with only the solid, it is found that at 40°C the total gas pressure (NH3 and CO2) is 0.363 atm. Calculate the equilibrium constant KP. 14.23 Consider the following reaction at 1600°C: Br2(g) 3:4 2Br(g) When 1.05 moles of Br2 are put in a 0.980-L flask, 1.20 percent of the Br2 undergoes dissociation. Calculate the equilibrium constant Kc for the reaction. 14.24 Pure phosgene gas (COCl2), 3.00 10 2 mol, was placed in a 1.50-L container. It was heated to 800 K, and at equilibrium the pressure of CO was found to be 0.497 atm. Calculate the equilibrium constant KP for the reaction CO(g) Cl2(g) 3:4 COCl2(g) 2NOBr(g) 3:4 2NO(g) 2NOCl(g) 3:4 2NO(g) S (aq) Kc ′ Cl2(g) is 1.05 at 250°C. The reaction starts with a mixture of PCl5, PCl3, and Cl2 at pressures 0.177 atm, 0.223 atm, and 0.111 atm, respectively, at 250°C. When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why. 14.22 Ammonium carbamate, NH4CO2NH2, decomposes as follows: Back Forward Main Menu TOC HS (aq) 3:4 H (aq) 9.5 10 8 2 S (aq) Kc ″ 1.0 10 19 Calculate the equilibrium constant for the following reaction at the same temperature: CO2(g) is 0.105 atm at 350°C. Calculate KP and Kc for this reaction. 14.21 The equilibrium constant KP for the reaction PCl5 (g) 3:4 PCl3(g) Cl2(g) Calculate the equilibrium constant Kc for the reaction. 14.27 The following equilibrium constants have been determined for hydrosulfuric acid at 25°C: H2S(aq) 3:4 H (aq) at the same temperature. 14.20 At equilibrium, the pressure of the reacting mixture Br2(g) If nitrosyl bromide, NOBr, is 34 percent dissociated at 25°C and the total pressure is 0.25 atm, calculate KP and Kc for the dissociation at this temperature. 14.26 A 2.50-mole quantity of NOCl was initially in a 1.50-L reaction chamber at 400°C. After equilibrium was established, it was found that 28.0 percent of the NOCl had dissociated: 2I(g) 3:4 I2(g) CaCO3(s) 3:4 CaO(s) CO2(g) 14.25 Consider the equilibrium If the equilibrium partial pressures of N2, O2, and NO are 0.15 atm, 0.33 atm, and 0.050 atm, respectively, at 2200°C, what is KP? 14.18 A reaction vessel contains NH3, N2, and H2 at equilibrium at a certain temperature. The equilibrium concentrations are [NH3] 0.25 M, [N2] 0.11 M, and [H2] 1.91 M. Calculate the equilibrium constant Kc for the synthesis of ammonia if the reaction is represented as (a) N2(g) 3H2(g) 3:4 2NH3(g) (b) 1 N2(g) 3 H2(g) 3:4 NH3(g) 2 2 14.19 The equilibrium constant Kc for the reaction 5 NH4CO2NH2(s) 3:4 2NH3(g) H2S(aq) 3:4 2H (aq) S2 (aq) 14.28 The following equilibrium constants have been determined for oxalic acid at 25°C: C2H2O4(aq) 3:4 H (aq) C2HO4 (aq) 3:4 H (aq) C2HO4 (aq) Kc 6.5 ′ 10 2 C2O2 (aq) 4 K c 6.1 ″ 10 5 Calculate the equilibrium constant for the following reaction at the same temperature: Study Guide TOC C2H2O4(aq) 3:4 2H (aq) Textbook Website C2O2 (aq) 4 MHHE Website 590 CHEMICAL EQUILIBRIUM 14.29 The following equilibrium constants were determined at 1123 K: C(s) CO2(g) 3:4 2CO(g) CO(g) Cl2(g) 3:4 COCl2(g) KP ′ 1.3 1014 KP ″ 6.0 10 3 Write the equilibrium constant expression KP, and calculate the equilibrium constant at 1123 K for C(s) CO2(g) 2S(s) Kc ′ 4.2 1052 3O2(g) 3:4 2SO3(g) Kc ″ 9.8 10128 Calculate the equilibrium constant Kc for the following reaction at that temperature: O2(g) 3:4 2SO3(g) THE RELATIONSHIP BETWEEN CHEMICAL KINETICS AND CHEMICAL EQUILIBRIUM Review Questions 14.31 Based on rate constant considerations, explain why the equilibrium constant depends on temperature. 14.32 Explain why reactions with large equilibrium constants, such as the formation of rust (Fe2O3), may have very slow rates. Problems 14.33 Water is a very weak electrolyte that undergoes the following ionization (called autoionization): k1 H2O(l) 3:4 H (aq) k1 OH (aq) (a) If k1 2.4 10 5 s 1 and k 1 1.3 1011/M s, calculate the equilibrium constant K where K [H ][OH ]/[H2O]. (b) Calculate the product [H ] [OH ] and [H ] and [OH ]. 14.34 Consider the following reaction, which takes place in a single elementary step: 2A k1 B 3:4 A2B k1 If the equilibrium constant Kc is 12.6 at a certain temperature and if kr 5.1 10 2 s 1, calculate the value of kf. WHAT DOES THE EQUILIBRIUM CONSTANT TELL US? Review Questions 14.35 Define reaction quotient. How does it differ from equilibrium constant? Back Forward 14.37 The equilibrium constant KP for the reaction 2SO2( g) O2(g) 3:4 SO2(g) 2SO2(g) Problems 2Cl2(g) 3:4 2COCl2(g) 14.30 At a certain temperature the following reactions have the constants shown: S(s) 14.36 Outline the steps for calculating the concentrations of reacting species in an equilibrium reaction. Main Menu TOC O2( g) 3:4 2SO3( g) 4 is 5.60 10 at 350°C. The initial pressures of SO2 and O2 in a mixture are 0.350 atm and 0.762 atm, respectively, at 350°C. When the mixture equilibrates, is the total pressure less than or greater than the sum of the initial pressures (1.112 atm)? 14.38 For the synthesis of ammonia N2(g) 3H2(g) 3:4 2NH3(g) the equilibrium constant Kc at 375°C is 1.2. Starting with [H2]0 0.76 M, [N2]0 0.60 M, and [NH3]0 0.48 M, which gases will have increased in concentration and which will have decreased in concentration when the mixture comes to equilibrium? 14.39 For the reaction H2(g) CO2(g) 3:4 H2O(g) CO(g) at 700°C, Kc 0.534. Calculate the number of moles of H2 that are present at equilibrium if a mixture of 0.300 mole of CO and 0.300 mole of H2O is heated to 700°C in a 10.0-L container. 14.40 At 1000 K, a sample of pure NO2 gas decomposes: 2NO2(g) 3:4 2NO(g) O2(g) The equilibrium constant KP is 158. Analysis shows that the partial pressure of O2 is 0.25 atm at equilibrium. Calculate the pressure of NO and NO2 in the mixture. 14.41 The equilibrium constant Kc for the reaction H2(g) Br2(g) 3:4 2HBr(g) 6 is 2.18 10 at 730°C. Starting with 3.20 moles of HBr in a 12.0-L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium. 14.42 The dissociation of molecular iodine into iodine atoms is represented as I2(g) 3:4 2I(g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 10 5. Suppose you start with 0.0456 mole of I2 in a 2.30-L flask at 1000 K. What are the concentrations of the gases at equilibrium? 14.43 The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63 10 3 at 527°C: Study Guide TOC COCl2(g) 3:4 CO(g) Cl2(g) Textbook Website MHHE Website QUESTIONS AND PROBLEMS Calculate the equilibrium partial pressure of all the components, starting with pure phosgene at 0.760 atm. 14.44 Consider the following equilibrium process at 686°C: CO2(g) H2(g) 3:4 CO(g) H2O(g) The equilibrium concentrations of the reacting species are [CO] 0.050 M, [H2] 0.045 M, [CO2] 0.086 M, and [H2O] 0.040 M. (a) Calculate Kc for the reaction at 686°C. (b) If we add CO2 to increase its concentration to 0.50 mol/L, what will the concentrations of all the gases be when equilibrium is reestablished? 14.45 Consider the heterogeneous equilibrium process: C(s) CO2(g) 3:4 2CO(g) At 700°C, the total pressure of the system is found to be 4.50 atm. If the equilibrium constant KP is 1.52, calculate the equilibrium partial pressures of CO2 and CO. 14.46 The equilibrium constant Kc for the reaction H2(g) CO2(g) 3:4 H2O(g) CO(g) is 4.2 at 1650°C. Initially 0.80 mol H2 and 0.80 mol CO2 are injected into a 5.0-L flask. Calculate the concentration of each species at equilibrium. FACTORS THAT AFFECT EQUILIBRIUM Review Questions 14.47 Explain Le Chatelier ’s principle. How can this principle help us maximize the yields of reactions? 14.48 Use Le Chatelier ’s principle to explain why the equilibrium vapor pressure of a liquid increases with increasing temperature. 14.49 List four factors that can shift the position of an equilibrium. Only one of these factors can alter the value of the equilibrium constant. Which one is it? 14.50 Does the addition of a catalyst have any effects on the position of an equilibrium? 2NaHCO3(s) 3:4 Na2CO3(s) 14.51 Consider the following equilibrium system involving SO2, Cl2, and SO2Cl2 (sulfuryl dichloride): SO2(g) Cl2(g) 3:4 SO2Cl2(g) Predict how the equilibrium position would change if (a) Cl2 gas were added to the system; (b) SO2Cl2 were removed from the system; (c) SO2 were removed from the system. The temperature remains constant. 14.52 Heating solid sodium bicarbonate in a closed vessel establishes the following equilibrium: Back Forward Main Menu TOC CO2(g) What would happen to the equilibrium position if (a) some of the CO2 were removed from the system; (b) some solid Na2CO3 were added to the system; (c) some of the solid NaHCO3 were removed from the system? The temperature remains constant. 14.53 Consider the following equilibrium systems: (a) A 3:4 2B (b) A B 3:4 C (c) A 3:4 B H° H° H° 20.0 kJ 5.4 kJ 0.0 kJ Predict the change in the equilibrium constant Kc that would occur in each case if the temperature of the reacting system were raised. 14.54 What effect does an increase in pressure have on each of the following systems at equilibrium? The temperature is kept constant, and in each case, the reactants are in a cylinder fitted with a movable piston. (a) A(s) 3:4 2B(s) (b) 2A(l) 3:4 B(l) (c) A(s) 3:4 B(g) (d) A(g) 3:4 B(g) (e) A(g) 3:4 2B(g) 14.55 Consider the equilibrium 2I(g) 3:4 I2(g) What would be the effect on the position of equilibrium of (a) increasing the total pressure on the system by decreasing its volume; (b) adding I2 to the reaction mixture; and (c) decreasing the temperature? 14.56 Consider the following equilibrium process: PCl5(g) 3:4 PCl3(g) Cl2(g) H° 92.5 kJ Predict the direction of the shift in equilibrium when (a) the temperature is raised; (b) more chlorine gas is added to the reaction mixture; (c) some PCl3 is removed from the mixture; (d) the pressure on the gases is increased; (e) a catalyst is added to the reaction mixture. 14.57 Consider the reaction 2SO2(g) Problems H2O(g) 591 O2(g) 3:4 2SO3(g) H° 198.2 kJ Comment on the changes in the concentrations of SO2, O2, and SO3 at equilibrium if we were to (a) increase the temperature; (b) increase the pressure; (c) increase SO2; (d) add a catalyst; (e) add helium at constant volume. 14.58 In the uncatalyzed reaction N2O4(g) 3:4 2NO2(g) the pressure of the gases at equilibrium are PN2O4 0.377 atm and PNO2 1.56 atm at 100°C. What would happen to these pressures if a catalyst were added to the mixture? Study Guide TOC Textbook Website MHHE Website 592 CHEMICAL EQUILIBRIUM 14.59 Consider the gas-phase reaction 2CO(g) A(g) 3:4 2B(g) O2(g) 3:4 2CO2(g) Predict the shift in the equilibrium position when helium gas is added to the equilibrium mixture (a) at constant pressure and (b) at constant volume. 14.60 Consider the following equilibrium reaction in a closed container: CaCO3(s) 3:4 CaO(s) ADDITIONAL PROBLEMS 14.61 Consider the statement: “The equilibrium constant of a reacting mixture of solid NH4Cl and gaseous NH3 and HCl is 0.316.” List three important pieces of information that are missing from this statement. 14.62 Pure nitrosyl chloride (NOCl) gas was heated to 240°C in a 1.00-L container. At equilibrium the total pressure was 1.00 atm and the NOCl pressure was 0.64 atm. Forward Main Menu TOC O2(g) is 2 10 42 at 25°C. (a) What is Kc for the reaction at the same temperature? (b) The very small value of KP (and Kc) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change. 14.67 Consider the following reacting system: 2NO(g) Cl2(g) 3:4 2NOCl(g) What combination of temperature and pressure would maximize the yield of nitrosyl chloride (NOCl)? [Hint: H °(NOCl) 51.7 kJ/mol. You will f also need to consult Appendix 3.] 14.68 At a certain temperature and a total pressure of 1.2 atm, the partial pressures of an equilibrium mixture are PA 0.60 atm and PB 0.60 atm. (a) Calculate the KP for the reaction at this temperature. (b) If the total pressure were increased to 1.5 atm, what would be the partial pressures of A and B at equilibrium? 14.69 The decomposition of ammonium hydrogen sulfide NH4HS(s) 3:4 NH3(g) H2O(g) Would we obtain more CO2 and H2O by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel? 14.65 Consider the following reaction at equilibrium: Back 2H2O(g) 3:4 2H2(g) 2A(g) 3:4 B(g) (a) Calculate the partial pressure of NO under these conditions if the partial pressures of nitrogen and oxygen are 3.0 atm and 0.012 atm, respectively. (b) Repeat the calculation for atmospheric conditions where the partial pressures of nitrogen and oxygen are 0.78 atm and 0.21 atm and the temperature is 25°C. (The KP for the reaction is 4.0 10 31 at this temperature.) (c) Is the formation of NO endothermic or exothermic? (d) What natural phenomenon promotes the formation of NO? Why? 14.64 Baking soda (sodium bicarbonate) undergoes thermal decomposition as follows: CO2(g) [B] 0.843 0.764 0.724 14.66 The equilibrium constant KP for the reaction O2(g) 3:4 2NO(g) 2NaHCO3(s) 3:4 Na2CO3(s) [A] 0.0125 0.171 0.250 200 300 400 Cl2(g) (a) Calculate the partial pressures of NO and Cl2 in the system. (b) Calculate the equilibrium constant KP. 14.63 The equilibrium constant (KP) for the formation of the air pollutant nitric oxide (NO) in an automobile engine at 530°C is 2.9 10 11: N2(g) TEMPERATURE (°C) CO2(g) What will happen if (a) the volume is increased; (b) some CaO is added to the mixture; (c) some CaCO3 is removed; (d) some CO2 is added to the mixture; (e) a few drops of a NaOH solution are added to the mixture; (f) a few drops of a HCl solution are added to the mixture (ignore the reaction between CO2 and water); (g) temperature is increased? 2NOCl(g) 3:4 2NO(g) From the following data, calculate the equilibrium constant (both KP and Kc) at each temperature. Is the reaction endothermic or exothermic? H2S(g) is an endothermic process. A 6.1589-g sample of the solid is placed in an evacuated 4.000-L vessel at exactly 24°C. After equilibrium has been established, the total pressure inside is 0.709 atm. Some solid NH4HS remains in the vessel. (a) What is the KP for the reaction? (b) What percentage of the solid has decomposed? (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel? 14.70 Consider the reaction 2NO(g) O2( g) 3:4 2NO2( g) At 430°C, an equilibrium mixture consists of 0.020 mole of O2, 0.040 mole of NO, and 0.96 mole of NO2. Calculate KP for the reaction, given that the total pressure is 0.20 atm. 14.71 When heated, ammonium carbamate decomposes as follows: Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS NH4CO2NH2(s) 3:4 2NH3(g) CO2(g) At a certain temperature the equilibrium pressure of the system is 0.318 atm. Calculate KP for the reaction. 14.72 A mixture of 0.47 mole of H2 and 3.59 moles of HCl is heated to 2800°C. Calculate the equilibrium partial pressures of H2, Cl2, and HCl if the total pressure is 2.00 atm. For the reaction H2(g) Cl2( g) 3:4 2HCl( g) KP is 193 at 2800°C. 14.73 Consider the reaction between NO2 and N2O4 in a closed container: be obtained from the ratio of observed pressure over calculated pressure, assuming no dissociation.] 14.79 Eggshells are composed mostly of calcium carbonate (CaCO3) formed by the reaction Ca2 (aq) 2H2S(g) 3:4 2H2(g) N2(g) S2(g) 4 is 2.25 10 . If [H2S] 4.84 10 3 M and [H2] 1.50 10 3 M, calculate [S2]. 14.76 A quantity of 6.75 g of SO2Cl2 was placed in a 2.00-L flask. At 648 K, there is 0.0345 mole of SO2 present. Calculate Kc for the reaction SO2Cl2(g) 3:4 SO2(g) Cl2(g) 14.77 The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for the acid rain phenomenon. The equilibrium constant KP for the reaction 2SO2(g) O2(g) 3:4 2SO3(g) is 0.13 at 830°C. In one experiment 2.00 mol SO2 and 2.00 mol O2 were initially present in a flask. What must the total pressure at equilibrium be in order to have an 80.0 percent yield of SO3? 14.78 Consider the dissociation of iodine: I2(g) 3:4 2I(g) A 1.00-g sample of I2 is heated to 1200°C in a 500mL flask. At equilibrium the total pressure is 1.51 atm. Calculate KP for the reaction. [Hint: Use the result in 14.73(a). The degree of dissociation can Back Forward Main Menu TOC CO2 (aq) 3:4 CaCO3(s) 3 The carbonate ions are supplied by carbon dioxide produced as a result of metabolism. Explain why eggshells are thinner in the summer when the rate of panting by chickens is greater. Suggest a remedy for this situation. 14.80 The equilibrium constant KP for the following reaction is 4.31 10 4 at 375°C: N2O4( g) 3:4 2NO2( g) Initially there is 1 mole of N2O4 present. At equilibrium, mole of N2O4 has dissociated to form NO2. (a) Derive an expression for KP in terms of and P, the total pressure. (b) How does the expression in (a) help you predict the shift in equilibrium due to an increase in P? Does your prediction agree with Le Chatelier ’s principle? 14.74 One mole of N2 and three moles of H2 are placed in a flask at 375°C. Calculate the total pressure of the system at equilibrium if the mole fraction of NH3 is 0.21. The KP for the reaction is 4.31 10 4. 14.75 At 1130°C the equilibrium constant (Kc) for the reaction 593 3H2( g) 3:4 2NH3( g) In a certain experiment a student starts with 0.862 atm of N2 and 0.373 atm of H2 in a constantvolume vessel at 375°C. Calculate the partial pressures of all species when equilibrium is reached. 14.81 A quantity of 0.20 mole of carbon dioxide was heated to a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: C(s) CO2( g) 3:4 2CO( g) Under these conditions, the average molar mass of the gases was 35 g/mol. (a) Calculate the mole fractions of CO and CO2. (b) What is KP if the total pressure is 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.) 14.82 When dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose 3:4 glucose A chemist prepared a 0.244 M fructose solution at 25°C. At equilibrium, it was found that its concentration had decreased to 0.113 M. (a) Calculate the equilibrium constant for the reaction. (b) At equilibrium, what percentage of fructose was converted to glucose? 14.83 At room temperature, solid iodine is in equilibrium with its vapor through sublimation and deposition (see p. 452). Describe how you would use radioactive iodine, in either solid or vapor form, to show that there is a dynamic equilibrium between these two phases. 14.84 At 1024°C, the pressure of oxygen gas from the decomposition of copper(II) oxide (CuO) is 0.49 atm: 4CuO(s) 3:4 2Cu2O(s) O2(g) (a) What is KP for the reaction? (b) Calculate the fraction of CuO that will decompose if 0.16 mole of it is placed in a 2.0-L flask at 1024°C. (c) What Study Guide TOC Textbook Website MHHE Website 594 CHEMICAL EQUILIBRIUM would the fraction be if a 1.0 mole sample of CuO were used? (d) What is the smallest amount of CuO (in moles) that would establish the equilibrium? 14.85 A mixture containing 3.9 moles of NO and 0.88 mole of CO2 was allowed to react in a flask at a certain temperature according to the equation NO(g) CO2( g) 3:4 NO2( g) subsequently. How does this “thought” experiment convince you that no such catalyst can exist? CO( g) At equilibrium, 0.11 mole of CO2 was present. Calculate the equilibrium constant Kc of this reaction. 14.86 The equilibrium constant Kc for the reaction H2( g) A(g) 3:4 B(g) Catalyst N2(g) 2A(g) 3:4 B(g) is established inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction (2A 88n B) but does not affect the reverse process (B 88n 2A). Suppose the catalyst is suddenly exposed to the equilibrium system as shown here. Describe what would happen Main Menu 14.90 The equilibrium constant Kc for the following reaction is 1.2 at 375°C. C(g) In an experiment, A was heated at a certain temperature until its equilibrium pressure reached 0.14P, where P is the total pressure. Calculate the equilibrium constant KP of this reaction. 14.88 When a gas was heated under atmospheric conditions, its color deepened. Heating above 150°C caused the color to fade, and at 550°C the color was barely detectable. However, at 550°C, the color was partially restored by increasing the pressure of the system. Which of the following best fits the above description? Justify your choice. (a) A mixture of hydrogen and bromine, (b) pure bromine, (c) a mixture of nitrogen dioxide and dinitrogen tetroxide. (Hint: Bromine has a reddish color and nitrogen dioxide is a brown gas. The other gases are colorless.) 14.89 In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type Forward TOC String B I2( g) 3:4 2HI( g) is 54.3 at 430°C. At the start of the reaction there are 0.714 mole of H2, 0.984 mole of I2, and 0.886 mole of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium. 14.87 When heated, a gaseous compound A dissociates as follows: Back 2A 3H2(g) 3:4 2NH3(g) (a) What is the value of KP for this reaction? (b) What is the value of the equilibrium constant Kc for 2NH3(g) 3:4 N2(g) (c) What is Kc for 1 2 N2(g) 3H2(g)? 3 2 H2(g) 3:4 NH3(g)? (d) What are the values of KP for the reactions de- scribed in (b) and (c)? 14.91 A sealed glass bulb contains a mixture of NO2 and N2O4 gases. Describe what happens to the following properties of the gases when the bulb is heated from 20°C to 40°C: (a) color, (b) pressure, (c) average molar mass, (d) degree of dissociation (from N2O4 to NO2), (e) density. Assume that volume remains constant. (Hint: NO2 is a brown gas; N2O4 is colorless.) 14.92 At 20°C, the vapor pressure of water is 0.0231 atm. Calculate KP and Kc for the process H2O(l ) 3:4 H2O(g) 14.93 Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathe 88n Na. We might expect that ode is Na potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at 892°C: Study Guide TOC Na(g) KCl(l ) 3:4 NaCl(l ) Textbook Website K(g) MHHE Website QUESTIONS AND PROBLEMS In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are 892°C and 770°C, respectively.) 14.94 In the gas phase, nitrogen dioxide is actually a mixture of nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4). If the density of such a mixture is 2.3 g/L at 74°C and 1.3 atm, calculate the partial pressures of the gases and KP for the dissociation of N2O4. 14.95 About 75 percent of hydrogen for industrial use is produced by the steam-reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at 800°C to give hydrogen and carbon monoxide: CH4(g) H2O(g) 3:4 CO(g) 3H2(g) H° 206 kJ The secondary stage is carried out at about 1000°C, in the presence of air, to convert the remaining methane to hydrogen: CH4(g) 1 2 O2(g) 3:4 CO(g) 2H2(g) H° 35.7 kJ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stage? (b) The equilibrium constant Kc for the primary stage is 18 at 800°C. (i) Calculate KP for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium? 14.96 Photosynthesis can be represented by 6CO2(g) 6H2O(l ) 3:4 C6H12O6(s) 6O2(g) H° 2801 kJ Explain how the equilibrium would be affected by the following changes: (a) partial pressure of CO2 is increased, (b) O2 is removed from the mixture, (c) C6H12O6 (sucrose) is removed from the mixture, (d) more water is added, (e) a catalyst is added, (f ) temperature is decreased. 14.97 Consider the decomposition of ammonium chloride at a certain temperature: NH4Cl(s) 3:4 NH3(g) Calculate the equilibrium constant KP if the total pressure is 2.2 atm at that temperature. 14.98 At 25°C, the equilibrium partial pressures of NO2 and N2O4 are 0.15 atm and 0.20 atm, respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established. Back Forward Main Menu TOC 14.99 In 1899 the German chemist Ludwig Mond developed a process for purifying nickel by converting it to the volatile nickel tetracarbonyl [Ni(CO)4] (b.p. 42.2°C): Ni(s) 4CO(g) 3:4 Ni(CO)4(g) (a) Describe how you can separate nickel and its solid impurities. (b) How would you recover nickel? [ H ° for Ni(CO)4 is 602.9 kJ/mol.] f 14.100 Consider the equilibrium reaction described in Problem 14.21. A quantity of 2.50 g of PCl5 is placed in an evacuated 0.500 L flask and heated to 250°C. (a) Calculate the pressure of PCl5, assuming it does not dissociate. (b) Calculate the partial pressure of PCl5 at equilibrium. (c) What is the total pressure at equilibrium? (d) What is the degree of dissociation of PCl5? (The degree of dissociation is given by the fraction of PCl5 that has undergone dissociation.) 14.101 Consider the equilibrium system 3A 3:4 B. Sketch the changes in the concentrations of A and B over time for the following situations: (a) initially only A is present; (b) initially only B is present; (c) initially both A and B are present (with A in higher concentration). In each case, assume that the concentration of B is higher than that of A at equilibrium. 14.102 The vapor pressure of mercury is 0.002 mmHg at 26°C. (a) Calculate Kc and KP for the process Hg(l ) 3:4 Hg(g). (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring 6.1 m long, 5.3 m wide, and 3.1 m high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor in mg/m3. Does this concentration exceed the safety limit of 0.05 mg/m3? (Ignore the volume of furniture and other objects in the laboratory.) Answers to Practice Exercises: [NO2]4[O2] ; KP [N2O5]2 4 PNO2PO2 14.2 2.2 102 2 PN2O5 PNi(CO)4 [Ni(CO )4] 14.3 347 atm 14.4 1.2 14.5 Kc ; KP [CO]4 P4 O C 4 14.6 KP 0.0702; Kc 1.20 10 14.1 Kc 14.7 (a) Ka HCl(g) 595 [O3]2 (b) Kb [O2]3 [O3]2/3 ; Ka [O2] K3 b 14.8 From right to left. 14.9 [HI] 0.031 M, [H2] 4.3 10 3 M, [I2] 4.3 10 3 M 14.10 [Br2] 0.065 M, [Br] 8.4 10 3 M 14.11 QP 4.0 105; the net reaction will shift from right to left. 14.12 Left to right. 14.13 The equilibrium will shift from (a) left to right, (b) left to right, and (c) right to left. (d) A catalyst has no effect on the equilibrium. Study Guide TOC Textbook Website MHHE Website ...
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