Chemistry Past Paper - Chemical Equilibrium PDF

15.5 CALCULATING EQUILIBRIUM CONSTANTS determine the direction in which a reaction mixture must proceed We see that the value of an equilibrium constant can be calculated to achieve equilibrium. from equilibrium concentrations of reactants and products....

15.5 CALCULATING EQUILIBRIUM CONSTANTS determine the direction in which a reaction mixture must proceed We see that the value of an equilibrium constant can be calculated to achieve equilibrium. from equilibrium concentrations of reactants and products. 15.7 LE CHÂTELIER’S PRINCIPLE 15.6 APPLICATIONS OF EQUILIBRIUM CONSTANTS We discuss Le Châtelier’s principle, which predicts how a system We also see that equilibrium constants can be used to predict at equilibrium responds to changes in concentration, volume, equilibrium concentrations of reactants and products and to pressure, and temperature. CHEMICAL EQUILIBRIUM TO BE IN EQUILIBRIUM IS to be in a state of balance. A tug of war in which the two sides pull with equal force so that the rope does not move is an example of a static equilibrium, one in which an object is at rest. Equilibria can also be dynamic, as illustrated in the chapter-opening photograph, which shows cars traveling in both directions over a bridge that serves as the entry to a city. If the rate at which cars leave the city equals the rate at which they enter, the two opposing processes are in balance, and the net number of cars in the city is constant. We have already encountered several instances of dynamic equilibrium. For example, the vapor above a liquid in a closed container is in equilibrium with the liquid phase (Section 11.5), which means that the rate at which molecules escape from the liquid into the gas phase equals the rate at which molecules in the gas phase become part of the liquid. Similarly, in a saturated sodium chloride solution in contact with undissolved sodium chloride, the solid is in equilibrium with the ions dispersed in water. (Section 13.2) The rate at which ions leave the solid surface equals the rate at which other ions leave the liquid and become part of the solid. In this chapter we consider dynamic equilibria in chemical reactions. Chemical equilibrium occurs when opposing reactions proceed at equal rates: The rate at which the products form from the reactants equals the rate at which the reactants form from the products. As a result, concentrations cease to change, making the reaction appear to be 611 612 CHAPTER 15 Chemical Equilibrium stopped. Chemical equilibria are involved in many natural phenomena and play important roles in many industrial processes. In this and the next two chapters, we will explore chemical equilibrium in some detail. Later, in Chapter 19, we will learn how to relate chemical equilibria to thermodynamics. Here we learn how to express the equilibrium state of a reaction in quantitative terms and study the factors that determine the relative concentrations of reactants and products in equilibrium mixtures. 15.1 | THE CONCEPT OF EQUILIBRIUM Let’s examine a simple chemical reaction to see how it reaches an equilibrium state—a mixture of reactants and products whose concentrations no longer change with time. We begin with N2O4, a colorless substance that dissociates to form brown NO2. 쑼 FIGURE 15.1 shows a sample of frozen N2O4 inside a sealed tube. The solid N2O4 va- porizes as it is warmed above its boiling point (21.2 °C), and the gas turns darker as the colorless N2O4 gas dissociates into brown NO2 gas. Eventually, even though there is still N2O4 in the tube, the color stops getting darker because the system reaches equilibrium. We are left with an equilibrium mixture of N2O4 and NO2 in which the concentrations of the gases no longer change as time passes. Because the reaction is in a closed system, where no gases can escape, equilibrium will eventually be reached. GO FIGURE How can you tell if you are at equilibrium? N2O4 NO2 Frozen N2O4 sample is Warmed N2O4 dissociates Colors stop changing, equilibrium nearly colorless to brown NO2(g) reached: rate of reaction N2O4(g) 2 NO2(g)  rate of reaction 2 NO2(g) N2O4(g) 쑿 FIGURE 15.1 The equilibrium between NO2 and N2O4.  SECTION 15.1 The Concept of Equilibrium 613 The equilibrium mixture results because the reaction is reversible: N2O4 can form NO2, and NO2 can form N2O4. This situation is represented by writing the equation for the reaction with two half arrows pointing in opposite directions: (Section 4.1) N2O4(g) Δ 2 NO2(g) [15.1] Colorless Brown We can analyze this equilibrium using our knowledge of kinetics. Let’s call the de- composition of N2O4 the forward reaction and the formation of N2O4 the reverse reaction. In this case, both the forward reaction and the reverse reaction are elementary reactions. As we learned in Section 14.6, the rate laws for elementary reactions can be written from their chemical equations: Forward reaction: N2O4(g) ¡ 2 NO2(g) Rate f = kf[N2O4] [15.2] Reverse reaction: 2 NO2(g) ¡ N2O4(g) Rate r = kr[NO2]2 [15.3] At equilibrium, the rate at which NO2 forms in the forward reaction equals the rate at which N2O4 forms in the reverse reaction: kf [N2O4] = kr[NO2]2 [15.4] Forward reaction Reverse reaction Rearranging this equation gives [NO2]2 kf = = a constant [15.5] [N2O4] kr From Equation 15.5 we see that the quotient of two rate constants is another constant. We also see that, at equilibrium, the ratio of the concentration terms equals this same constant. (We consider this constant, called the equilibrium constant, in Section 15.2.) It makes no difference whether we start with N2O4 or with NO2, or even with some mixture of the two. At equilibrium, at a given temperature, the ratio equals a specific value. Thus, there is an important constraint on the proportions of N2O4 and NO2 at equilibrium. Once equilibrium is established, the concentrations of N2O4 and NO2 no longer change, as shown in 쑼 FIGURE 15.2(a). However, the fact that the composition of the equilibrium mixture remains constant with time does not mean that N2O4 and NO2 stop reacting. On the contrary, the equilibrium is dynamic—which means some N2O4 is always converting to NO2 and some NO2 is always converting to N2O4. At equilibrium, however, the two processes occur at the same rate, as shown in Figure 15.2(b). We learn several important lessons about equilibrium from this example: At equilibrium, the concentrations of reactants and products no longer change with time. For equilibrium to occur, neither reactants nor products can escape from the system. At equilibrium, a particular ratio of concentration terms equals a constant. GO FIGURE At equilibrium, are the concentrations of NO2 and N2O4 equal? NO2 kf [N2O4] Concentration Rate N2O4 Equilibrium kr[NO2]2 achieved Equilibrium 씱 FIGURE 15.2 Achieving chemical achieved (rates are equal) equilibrium in the N2O4(g) Δ 2 NO2(g) 0 0 Time Time reaction. Equilibrium occurs when the rate of the forward reaction equals the rate of the (a) (b) reverse reaction. 614 CHAPTER 15 Chemical Equilibrium GIVE IT SOME THOUGHT a. Which quantities are equal in a dynamic equilibrium? b. If the rate constant for the forward reaction in Equation 15.1 is larger than the rate constant for the reverse reaction, will the constant in Equation 15.5 be greater than 1 or smaller than 1? 15.2 | THE EQUILIBRIUM CONSTANT A reaction in which reactants convert to products and products convert to reactants in the same reaction vessel naturally leads to an equilibrium, regardless of how compli- cated the reaction is and regardless of the nature of the kinetic processes for the forward and reverse reactions. Consider the synthesis of ammonia from nitrogen and hydrogen: N2(g) + 3 H2(g) Δ 2 NH3(g) [15.6] This reaction is the basis for the Haber process, which is critical for the production of fertilizers and therefore critical to the world’s food supply. In the Haber process, N2 and H2 react at high pressure and temperature in the presence of a catalyst to form ammo- nia. In a closed system, however, the reaction does not lead to complete consumption of the N2 and H2. Rather, at some point the reaction appears to stop with all three compo- nents of the reaction mixture present at the same time. How the concentrations of H2, N2, and NH3 vary with time is shown in 쑼 FIGURE 15.3. Notice that an equilibrium mixture is obtained regardless of whether we begin with N2 and H2 or with NH3. The equilibrium condition is reached from either direction. GIVE IT SOME THOUGHT How do we know when equilibrium has been reached in a chemical reaction? An expression similar to Equation 15.5 governs the concentrations of N2, H2, and NH3 at equilibrium. If we were to systematically change the relative amounts of the three gases in the starting mixture and then analyze each equilibrium mixture, we could determine the relationship among the equilibrium concentrations. Chemists carried out studies of this kind on other chemical systems in the nine- teenth century before Haber’s work. In 1864, Cato Maximilian Guldberg (1836–1902) and Peter Waage (1833–1900) postulated their law of mass action, which expresses, for any reaction, the relationship between the concentrations of the reactants and products present at equilibrium. Suppose we have the general equilibrium equation aA + bB Δ dD + eE [15.7] where A, B, D, and E are the chemical species involved and a, b, d, and e are their coeffi- cients in the balanced chemical equation. According to the law of mass action, the equilibrium condition is described by the expression [D]d[E]e — products Kc = [15.8] [A]a[B]b — reactants We call this relationship the equilibrium-constant expression (or merely the equilibrium expression) for the reaction. The constant Kc , the equilibrium constant, is the numerical value obtained when we substitute molar equilibrium concentrations Equilibrium achieved Equilibrium achieved Concentration Concentration 씰 FIGURE 15.3 The same equilibrium H2 H2 is reached whether we start with only NH3 NH3 reactants (N2 and H2) or with only N2 N2 product (NH3). Time Time  SECTION 15.2 The Equilibrium Constant 615 CHEMISTRY PUT TO WORK The Haber Process The Haber process provides a historically interesting example of the complex impact of chemistry on our lives. At the start of World The quantity of food required to feed the ever-increasing War I, in 1914, Germany depended on nitrate deposits in Chile for human population far exceeds that provided by the nitrogen-containing compounds needed to manufacture explo- nitrogen-fixing plants. (Section 14.7) Therefore, sives. During the war, the Allied naval blockade of South America cut human agriculture requires substantial amounts of off this supply. However, by using the Haber reaction to fix nitrogen ammonia-based fertilizers for croplands. Of all from air, Germany was able to continue to produce explosives. Ex- the chemical reactions that humans have learned to control for their perts have estimated that World War I would have ended before 1918 own purposes, the synthesis of ammonia from hydrogen and atmos- had it not been for the Haber process. pheric nitrogen is one of the most important. From these unhappy beginnings as a major factor in interna- In 1912 the German chemist Fritz Haber (1868–1934) devel- tional warfare, the Haber process has become the world’s principal oped the Haber process (Equation 15.6). The process is sometimes source of fixed nitrogen. The same process that prolonged World War also called the Haber–Bosch process to honor Karl Bosch, the engineer I has enabled the manufacture of fertilizers that have increased crop who developed the industrial process on a large scale. The engineer- yields, thereby saving millions of people from starvation. About 40 ing needed to implement the Haber process requires the use of billion pounds of ammonia are manufactured annually in the United temperatures and pressures (approximately 500 °C and 200 to 600 States, mostly by the Haber process. The ammonia can be applied atm) that were difficult to achieve at that time. directly to the soil (쑼 FIGURE 15.4), or it can be converted into am- monium salts that are also used as fertilizers. Haber was a patriotic German who gave enthusiastic support to his nation’s war effort. He served as chief of Germany’s Chemical Warfare Service during World War I and developed the use of chlorine as a poison-gas weapon. Consequently, the deci- sion to award him the Nobel Prize in Chemistry in 1918 was the subject of con- siderable controversy and criticism. The ultimate irony, however, came in 1933 when Haber was expelled from Germany because he was Jewish. RELATED EXERCISES: 15.46 and 15.76 씱 FIGURE 15.4 Liquid ammonia used as fertilizer by direct injection into soil. into the equilibrium-constant expression. The subscript c on the K indicates that con- centrations expressed in molarity are used to evaluate the constant. The numerator of the equilibrium-constant expression is the product of the concentrations of all substances on the product side of the equilibrium equation, each raised to a power equal to its coefficient in the balanced equation. The denomi- nator is similarly derived from the reactant side of the equilibrium equation. Thus, for the Haber process, N2(g) + 3 H2(g) Δ 2 NH3(g), the equilibrium-constant expression is [NH3]2 Kc = [15.9] [N2][H2]3 Once we know the balanced chemical equation for a reaction that reaches equilib- rium, we can write the equilibrium-constant expression even if we do not know the reaction mechanism. The equilibrium-constant expression depends only on the stoichiom- etry of the reaction, not on its mechanism. 616 CHAPTER 15 Chemical Equilibrium The value of the equilibrium constant at any given temperature does not depend on the initial amounts of reactants and products. It also does not matter whether other sub- stances are present, as long as they do not react with a reactant or a product. The value of Kc depends only on the particular reaction and on the temperature. SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for Kc for the following reactions: (a) 2 O3(g) Δ 3 O2(g) (b) 2 NO(g) + Cl 2(g) Δ 2 NOCl(g) (c) Ag +(aq) + 2 NH3(aq) Δ Ag(NH3)2 +(aq) SOLUTION Analyze We are given three equations and are asked to write an equilibrium-constant expression for each. Plan Using the law of mass action, we write each expression as a quotient having the product concentration terms in the numerator and the reactant concentration terms in the denominator. Each concentration term is raised to the power of its coefficient in the balanced chemical equation. Solve [O2]3 [NOCl]2 [Ag(NH3)2+] (a) Kc = (b) Kc = (c) Kc = [Ag +][NH3]2 2 2 [O3] [NO] [Cl 2] PRACTICE EXERCISE Write the equilibrium-constant expression Kc for (a) H2(g) + I2(g) Δ 2 HI(g), (b) Cd2+(aq) + 4 Br -(aq) Δ CdBr42- (aq). [HI]2 [CdBr42-] Answers: (a) Kc = (b) Kc = [H2][I2] [Cd2+][Br -]4 Evaluating Kc We can illustrate how the law of mass action was discovered empirically and demon- strate that the equilibrium constant is independent of starting concentrations by examining a series of experiments involving dinitrogen tetroxide and nitrogen dioxide: [NO2]2 N2O4(g) Δ 2 NO2(g) Kc = [15.10] [N2O4] We start with several sealed tubes containing different concentrations of NO2 and N2O4. The tubes are kept at 100 °C until equilibrium is reached. We then analyze the mixtures and determine the equilibrium concentrations of NO2 and N2O4, which are shown in 쑼 TABLE 15.1. To evaluate Kc , we insert the equilibrium concentrations into the equilibrium- constant expression. For example, using Experiment 1 data, [NO2] = 0.0172 M and [N2O4] = 0.00140 M, we find [NO2]2 [0.0172]2 Kc = = = 0.211 [N2O4] 0.00140 TABLE 15.1 Initial and Equilibrium Concentrations of N2O4(g) and NO2(g) at 100 °C Initial Initial Equilibrium Equilibrium Experiment [N2O4] (M) [NO2] (M) [N2O4] (M) [NO2] (M) Kc 1 0.0 0.0200 0.00140 0.0172 0.211 2 0.0 0.0300 0.00280 0.0243 0.211 3 0.0 0.0400 0.00452 0.0310 0.213 4 0.0200 0.0 0.00452 0.0310 0.213  SECTION 15.2 The Equilibrium Constant 617 Proceeding in the same way, the values of Kc for the other samples are calculated. Note from Table 15.1 that the value for Kc is constant (within the limits of experimental error) even though the initial concentrations vary. Furthermore, Experiment 4 shows 0.0400 that equilibrium can be achieved beginning with N2O4 rather than with NO2. That is, Experiment 3 equilibrium can be approached from either direction. 씰 FIGURE 15.5 shows how [NO2] (M) Experiments 3 and 4 result in the same equilibrium mixture even though the two 0.0300 experiments start with very different NO2 concentrations. Notice that no units are given for Kc either in Table 15.1 or in the calculation we just 0.0200 did using Experiment 1 data. It is common practice to write equilibrium constants with- Experiment 4 out units for reasons that we address later in this section. Recall that we began our discussion of equilibrium in terms of rates. Equation 15.5 0.0100 shows that Kc is equal to kf >kr , the ratio of the forward rate constant to the reverse rate constant. For the N2O4>NO2 reaction, Kc = 0.212, which means that kr is 4.72 times as 0 large as kf (since 1>0.212 = 4.72). It is not possible to obtain the absolute value of either Time rate constant knowing only the value of Kc. 쑿 FIGURE 15.5 The same equilibrium mixture is produced regardless of the GIVE IT SOME THOUGHT initial NO2 concentration. The How does the value of Kc in Equation 15.10 depend on the starting concentration of NO2 either increases or concentrations of NO2 and N2O4? decreases until equilibrium is reached. Equilibrium Constants in Terms of Pressure, Kp When the reactants and products in a chemical reaction are gases, we can formulate the equilibrium-constant expression in terms of partial pressures. When partial pressures in atmospheres are used in the expression, we denote the equilibrium constant Kp (where the subscript p stands for pressure). For the general reaction in Equation 15.7, we have (PD)d(PE)e Kp = [15.11] (PA)a(PB)b where PA is the partial pressure of A in atmospheres, PB is the partial pressure of B in atmospheres, and so forth. For example, for our N2O4>NO2 reaction we have (PNO2)2 Kp = PN2O4 GIVE IT SOME THOUGHT What is the difference between the equilibrium constant Kc and the equilibrium constant Kp? For a given reaction, the numerical value of Kc is generally different from the nu- merical value of Kp. We must therefore take care to indicate, via subscript c or p, which constant we are using. It is possible, however, to calculate one from the other using the ideal-gas equation: (Section 10.4) n PV = nRT, so P = RT [15.12] V The usual units for n>V are mol>L, which equals molarity, M. For substance A in our generic reaction, we therefore see that nA PA = RT = [A]RT [15.13] V When we substitute Equation 15.13 and like expressions for the other gaseous components of the reaction into Equation 15.11, we obtain a general expression relating Kp and Kc : Kp = Kc(RT)¢n [15.14] The quantity ¢n is the change in the number of moles of gas in the balanced chemical equation. It equals the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants: ¢n = (moles of gaseous product) - (moles of gaseous reactant) [15.15] 618 CHAPTER 15 Chemical Equilibrium For example, in the N2O4(g) Δ 2 NO2(g) reaction, there are two moles of product NO2 and one mole of reactant N2O4. Therefore, ¢n = 2 - 1 = 1, and Kp = Kc(RT) for this reaction. From Equation 15.14, we see that Kp = Kc only when the same number of moles of gas appears on both sides of the balanced chemical equation, so that ¢n = 0. SAMPLE EXERCISE 15.2 Converting between Kc and Kp For the Haber process, N2(g) + 3 H2(g) Δ 2 NH3(g) Kc = 9.60 at 300 °C. Calculate Kp for this reaction at this temperature. SOLUTION Analyze We are given Kc for a reaction and asked to calculate Kp. Plan The relationship between Kc and Kp is given by Equation 15.14. To apply that equation, we must determine ¢n by comparing the number of moles of product with the number of moles of reactants (Equation 15.15). Solve With 2 mol of gaseous products (2 NH3) and 4 mol of gaseous reactants (1 N2 + 3 H2), ¢n = 2 - 4 = -2. (Remember that ¢ functions are always based on products minus reactants.) The temperature is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.08206 L-atm>mol-K. Using Kc = 9.60, we therefore have (9.60) Kp = Kc(RT)¢n = (9.60)(0.08206 * 573)-2 = = 4.34 * 10-3 (0.08206 * 573)2 PRACTICE EXERCISE For the equilibrium 2 SO3(g) Δ 2 SO2(g) + O2(g), Kc is 4.08 * 10-3 at 1000 K. Calculate the value for Kp. Answer: 0.335 Equilibrium Constants and Units You may wonder why equilibrium constants are reported without units. The equilib- rium constant is related to the kinetics of a reaction as well as to the thermodynamics. (We explore this latter connection in Chapter 19.) Equilibrium constants derived from thermodynamic measurements are defined in terms of activities rather than concentra- tions or partial pressures. The activity of any substance in an ideal mixture is the ratio of the concentration or pressure of the substance either to a reference concentration (1 M) or to a reference pressure (1 atm). For example, if the concentration of a substance in an equilibrium mixture is 0.010 M, its activity is 0.010 M>1 M = 0.010. The units of such ratios always cancel and, consequently, activities have no units. Furthermore, the numerical value of the activity equals the concentration. For pure solids and pure liquids, the situation is even simpler because the activities then merely equal 1 (again with no units). In real systems, activities are also ratios that have no units. Even though these activ- ities may not be exactly numerically equal to concentrations, we will ignore the differences. All we need to know at this point is that activities have no units. As a result, the thermodynamic equilibrium constants derived from them also have no units. It is therefore common practice to write all types of equilibrium constants without units, a practice that we adhere to in this text. In more advanced chemistry courses, you may make more rigorous distinctions between concentrations and activities. GIVE IT SOME THOUGHT If the concentration of N2O4 in an equilibrium mixture is 0.00140 M, what is its activity? (Assume the solution is ideal.)  SECTION 15.3 Understanding and Working with Equilibrium Constants 619 | 15.3 UNDERSTANDING AND WORKING WITH EQUILIBRIUM CONSTANTS Before doing calculations with equilibrium constants, it is valuable to understand what the magnitude of an equilibrium constant can tell us about the relative concentrations of reactants and products in an equilibrium mixture. It is also useful to consider how the magnitude of any equilibrium constant depends on how the chemical equation is expressed. The Magnitude of Equilibrium Constants The magnitude of the equilibrium constant for a reaction gives us important informa- tion about the composition of the equilibrium mixture. For example, consider the GO FIGURE experimental data for the reaction of carbon monoxide gas and chlorine gas at 100 °C to What would this figure look like for form phosgene (COCl2), a toxic gas used in the manufacture of certain polymers and a reaction in which K « 1? insecticides: [COCl 2] CO(g) + Cl 2(g) Δ COCl 2(g) Kc = = 4.56 * 109 [CO][Cl 2] For the equilibrium constant to be so large, the numerator of the equilibrium-constant Reactants Products expression must be approximately a billion (109) times larger than the denominator. Thus, the equilibrium concentration of COCl2 must be much greater than that of CO or K >> 1, equilibrium Cl2, and in fact this is just what we find experimentally. We say that this equilibrium lies “lies to the right” to the right (that is, toward the product side). Likewise, a very small equilibrium con- stant indicates that the equilibrium mixture contains mostly reactants. We then say that the equilibrium lies to the left. In general, If K W 1 (large K): Equilibrium lies to right, products predominate Reactants Products If K V 1 (small K): Equilibrium lies to left, reactants predominate K L product and 0.40 mol>L reactant, giving Kc = 0.60>0.40 = 1.5. (You will get the same result by merely dividing the number of spheres of each kind: 6 spheres>4 spheres = 1.5.) In (ii) we have 0.10 mol>L product and 0.90 mol>L reactant, giv- ing Kc = 0.10>0.90 = 0.11 (or 1 sphere>9 spheres = 0.11). In (iii) we have 0.80 mol>L prod- uct and 0.20 mol>L reactant, giving Kc = 0.80>0.20 = 4.0 (or 8 spheres> 2 spheres = 4.0). These calculations verify the order in (a). Comment Imagine a drawing that represents a reaction with a very small or very large value of Kc. For example, what would the drawing look like if Kc = 1 * 10-5? In that case there would need to be 100,000 reactant molecules for only 1 product molecule. But then, that would be impractical to draw. PRACTICE EXERCISE For the reaction H2(g) + I2(g) Δ 2 HI(g), Kp = 794 at 298 K and Kp = 55 at 700 K. Is the formation of HI favored more at the higher or lower temperature? Answer: at the lower temperature because Kp is larger at the lower temperature The Direction of the Chemical Equation and K We have seen that we can represent the N2O4>NO2 equilibrium as [NO2]2 N2O4(g) Δ 2 NO2(g) Kc = = 0.212 (at 100 °C) [15.16] [N2O4] We could equally well consider this equilibrium in terms of the reverse reaction: 2 NO2(g) Δ N2O4(g) The equilibrium expression is then [N2O4] 1 Kc = = = 4.72 (at 100 °C) [15.17] [NO2]2 0.212 Equation 15.17 is the reciprocal of the expression in Equation 15.16. The equilibrium-con- stant expression for a reaction written in one direction is the reciprocal of the expression for the reaction written in the reverse direction. Consequently, the numerical value of the equilibrium constant for the reaction written in one direction is the reciprocal of that for the reverse reaction. Both expressions are equally valid, but it is meaningless to say that the equilibrium constant for the equilibrium between NO2 and N2O4 is “0.212” or “4.72” unless we indicate how the equilibrium reaction is written and specify the tem- perature. Therefore, whenever you are using an equilibrium constant, you should always write the associated balanced chemical equation. SAMPLE EXERCISE 15.4 Evaluating an Equilibrium Constant When an Equation Is Reversed For the reaction that is run at 25 °C, Kc = 1 * 10-30. Use this information to write the equilibrium-constant expression and calculate the equilibrium N2(g) + O2(g) Δ 2 NO(g) constant for the reaction 2 NO(g) Δ N2(g) + O2(g) SOLUTION Analyze We are asked to write the equilibrium-constant expression Plan The equilibrium-constant expression is a quotient of products for a reaction and to determine the value of Kc given the chemical over reactants, each raised to a power equal to its coefficient in the equation and equilibrium constant for the reverse reaction. balanced equation. The value of the equilibrium constant is the recip- rocal of that for the reverse reaction.  SECTION 15.3 Understanding and Working with Equilibrium Constants 621 Solve [N2][O2] Writing products over reactants, we have Kc = [NO]2 Both the equilibrium-constant expression and the numerical value of the equilibrium [N2][O2] 1 constant are the reciprocals of those for the formation of NO from N2 and O2: Kc = = = 1 * 1030 [NO]2 1 * 10-30 Comment Regardless of the way we express the equilibrium among NO, N2, and O2, at 25 °C it lies on the side that favors N2 and O2. Thus, the equilibrium mixture will contain mostly N2 and O2 with very little NO present. PRACTICE EXERCISE For N2(g) + 3 H2(g) Δ 2 NH3(g), Kp = 4.34 * 10-3 at 300 °C. What is the value of Kp for the reverse reaction? Answer: 2.30 * 102 Relating Chemical Equation Stoichiometry and Equilibrium Constants There are many ways to write a balanced chemical equation for a given reaction. For example, if we multiply Equation 15.1, N2O4(g) Δ 2 NO2(g) by 2, we have 2 N2O4(g) Δ 4 NO2(g) This chemical equation is balanced and might be written this way in some contexts. Therefore, the equilibrium-constant expression for this equation is [NO2]4 Kc = [N2O4]2 which is the square of the equilibrium-constant expression given in Equation 15.10 for the reaction as written in Equation 15.1: [NO2]2>[N2O4]. Because the new equilibrium- constant expression equals the original expression squared, the new equilibrium constant Kc equals the original constant squared: 0.2122 = 0.0449 (at 100 °C). Once again, it is important to remember that you must relate each equilibrium constant you work with to a specific balanced chemical equation. The concentrations of the sub- stances in the equilibrium mixture will be the same no matter how you write the chemical equation, but the value of Kc you calculate depends completely on how you write the reaction. GIVE IT SOME THOUGHT How does the magnitude of Kp for the reaction 2 HI(g) Δ H2(g) + I2(g) change if the equilibrium is written 6 HI(g) Δ 3 H2(g) + 3 I2(g)? It is also possible to calculate the equilibrium constant for a reaction if we know the equilibrium constants for other reactions that add up to give us the one we want, simi- lar to Hess’s law. (Section 5.6) For example, consider the following two reactions, their equilibrium-constant expressions, and their equilibrium constants at 100 °C: [NO]2[Br2] 1. 2 NOBr(g) Δ 2 NO(g) + Br2(g) Kc = = 0.014 [NOBr]2 [BrCl]2 2. Br2(g) + Cl 2(g) Δ 2 BrCl(g) Kc = = 7.2 [Br2][Cl 2] The net sum of these two equations is 3. 2 NOBr(g) + Cl 2(g) Δ 2 NO(g) + 2 BrCl(g) 622 CHAPTER 15 Chemical Equilibrium You can prove algebraically that the equilibrium-constant expression for reaction 3 is the product of the expressions for reactions 1 and 2: [NO]2[BrCl]2 [NO]2[Br2] [BrCl]2 Kc = = * [NOBr]2[Cl 2] [NOBr]2 [Br2][Cl 2] Thus, Kc3 = (Kc1)(Kc2) = (0.014)(7.2) = 0.10 To summarize: 1. The equilibrium constant of a reaction in the reverse direction is the inverse (or reciprocal) of the equilibrium constant of the reaction in the forward direction: A + B Δ C + D K1 C + D Δ A + B K = 1>K1 2. The equilibrium constant of a reaction that has been multiplied by a number is equal to the original equilibrium constant raised to a power equal to that number. A + B Δ C + D K1 nA + nB Δ nC + nD K = K1 n 3. The equilibrium constant for a net reaction made up of two or more reactions is the product of the equilibrium constants for the individual reactions: 1. A + B Δ C + D K1 2. C + F Δ G + A K2 3. B + F Δ D + G K3 = (K1)(K2) SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions Given the reactions determine the value of Kc for the reaction + - -4 HF(aq) Δ H (aq) + F (aq) Kc = 6.8 * 10 2 HF(aq) + C2O42-(aq) Δ 2 F -(aq) + H2C2O4(aq) H2C2O4(aq) Δ 2 H + (aq) + C2O42 - (aq) Kc = 3.8 * 10-6 SOLUTION Analyze We are given two equilibrium equations and the corre- Plan We cannot simply add the first two equations to get the third. sponding equilibrium constants and are asked to determine the equi- Instead, we need to determine how to manipulate the equations to librium constant for a third equation, which is related to the first two. come up with the steps that will add to give us the desired equation. Solve If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get 2 HF(aq) Δ 2 H+(aq) + 2 F -(aq) Kc = (6.8 * 10-4)2 = 4.6 * 10-7 Reversing the second equation and again making the corresponding 1 change to its equilibrium constant 2 H+(aq) + C2O42-(aq) Δ H2C2O4(aq) Kc = = 2.6 * 105 (taking the reciprocal) gives 3.8 * 10-6 Now we have two equations that sum to give the net equation, and 2 HF(aq) Δ 2 H + (aq) + 2 F -(aq) Kc = 4.6 * 10-7 we can multiply the individual Kc 2 H+(aq) + C2O42 - (aq) Δ H2C2O4(aq) Kc = 2.5 * 105 values to get the desired equilib- rium constant. 2 HF(aq) + C2O42 - (aq) Δ 2 F - (aq) + H2C2O4(aq) Kc = (4.6 * 10-7)(2.6 * 105) = 0.12 PRACTICE EXERCISE Given that, at 700 K, Kp = 54.0 for the reaction H2(g) + I2(g) Δ 2 HI(g) and Kp = 1.04 * 10-4 for the re- action N2(g) + 3 H2(g) Δ 2 NH3(g), determine the value of Kp for the reaction 2 NH3(g) + 3 I2(g) Δ 6 HI(g) + N2(g) at 700 K. (54.0)3 Answer: = 1.51 * 109 1.04 * 10-4  SECTION 15.4 Heterogeneous Equilibria 623 15.4 | HETEROGENEOUS EQUILIBRIA GO FIGURE Many equilibria involve substances that are all in the same phase, usually gas or liquid. Imagine starting with only CaO in a Such equilibria are called homogeneous equilibria. In some cases, however, the sub- bell jar and adding CO2(g) to make stances in equilibrium are in different phases, giving rise to heterogeneous equilibria. its pressure the same as it is in As an example of the latter, consider the equilibrium that occurs when solid lead(II) these two bell jars. How does the chloride dissolves in water to form a saturated solution: equilibrium concentration of CO2(g) in your jar compare with the CO2(g) PbCl 2(s) Δ Pb 2+(aq) + 2 Cl-(aq) [15.18] equilibrium concentration in these two jars? This system consists of a solid in equilibrium with two aqueous species. If we want to write the equilibrium-constant expression for this process, we encounter a problem CaCO3(s) CaO(s)  CO2(g) we have not encountered previously: How do we express the concentration of a solid? Although we can express that concentration in moles per unit volume, it is unneces- sary to do so in writing equilibrium-constant expressions. Whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium-constant expression. Thus, the equilibrium-constant expression for the reaction of Equation 15.18 is CO2 (g) Kc = [Pb 2+][Cl-]2 [15.19] Even though PbCl2(s) does not appear in the equilibrium-constant expression, it must be present for equilibrium to occur. The fact that pure solids and pure liquids are excluded from equilibrium-constant expressions can be explained in two ways. First, the concentration of a pure solid or liq- uid has a constant value. If the mass of a solid is doubled, its volume also doubles. Thus, its concentration, which relates to the ratio of mass to volume, stays the same. Because CaCO3 CaO equilibrium-constant expressions include terms only for reactants and products whose Large amount of CaCO3, concentrations can change during a chemical reaction, the concentrations of pure solids small amount of CaO, and pure liquids are omitted. gas pressure P The omission can also be rationalized in a second way. Recall from Section 15.2 that what is substituted into a thermodynamic equilibrium expression is the activity of each substance, which is a ratio of the concentration to a reference value. For a pure sub- stance, the reference value is the concentration of the pure substance, so that the activity of any pure solid or liquid is always 1. CO2 (g) GIVE IT SOME THOUGHT Write the equilibrium-constant expression for the evaporation of water, H2O(l) Δ H2O(g), in terms of partial pressures. Decomposition of calcium carbonate is another example of a heterogeneous reaction: CaCO3(s) Δ CaO(s) + CO2(g) CaCO3 CaO Omitting the concentrations of the solids from the equilibrium-constant expression gives Small amount of CaCO3 , Kc = [CO2] and Kp = PCO2 large amount of CaO, gas pressure still P These equations tell us that at a given temperature, an equilibrium among CaCO3, CaO, 쑿 FIGURE 15.7 At a given temperature, and CO2 always leads to the same CO2 partial pressure as long as all three components the equilibrium pressure of CO2 in the bell are present. As shown in 씰 FIGURE 15.7, we have the same CO2 pressure regardless of jars is the same no matter how much of the relative amounts of CaO and CaCO3. each solid is present. SAMPLE EXERCISE 15.6 Writing Equilibrium-Constant Expressions for Heterogeneous Reactions Write the equilibrium-constant expression Kc for (a) CO2(g) + H2(g) Δ CO(g) + H2O(l) (b) SnO2(s) + 2 CO(g) Δ Sn(s) + 2 CO2(g) 624 CHAPTER 15 Chemical Equilibrium SOLUTION Analyze We are given two chemical equations, both for heterogeneous equilibria, and asked to write the corresponding equilibrium-constant expressions. Plan We use the law of mass action, remembering to omit any pure solids and pure liquids from the expressions. Solve [CO] (a) The equilibrium-constant expression is Kc = [CO2][H2] Because H2O appears in the reaction as a liquid, its concentration does not appear in the equilibrium-constant expression. [CO2]2 (b) The equilibrium-constant expression is Kc = [CO]2 Because SnO2 and Sn are pure solids, their concentrations do not appear in the equilibrium-constant expression. PRACTICE EXERCISE Write the following equilibrium-constant expressions: (a) Kc for Cr(s) + 3 Ag +(aq) Δ Cr 3+(aq) + 3 Ag(s) (b) Kp for 3 Fe(s) + 4 H2O(g) Δ Fe 3O4(s) + 4 H2(g) [Cr 3+] (PH2)4 Answers: (a) Kc = (b) Kp = [Ag +]3 (PH2O)4 SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium Each of these mixtures was placed in a closed container and allowed to stand: (a) CaCO3(s) (b) CaO(s) and CO2(g) at a pressure greater than the value of Kp (c) CaCO3(s) and CO2(g) at a pressure greater than the value of Kp (d) CaCO3(s) and CaO(s) Determine whether or not each mixture can attain the equilibrium CaCO3(s) Δ CaO(s) + CO2(g) SOLUTION Analyze We are asked which of several combinations of species can establish an equilibrium between calcium carbonate and its decomposition products, calcium oxide and carbon dioxide. Plan For equilibrium to be achieved, it must be possible for both the forward process and the reverse process to occur. For the forward process to occur, there must be some calcium carbon- ate present. For the reverse process to occur, there must be both calcium oxide and carbon dioxide. In both cases, either the necessary compounds may be present initially or they may be formed by reaction of the other species. Solve Equilibrium can be reached in all cases except (c) as long as sufficient quantities of solids are present. (a) CaCO3 simply decomposes, forming CaO(s) and CO2(g) until the equi- librium pressure of CO2 is attained. There must be enough CaCO3, however, to allow the CO2 pressure to reach equilibrium. (b) CO2 continues to combine with CaO until the partial pres- sure of the CO2 decreases to the equilibrium value. (c) There is no CaO present, so equilibrium cannot be attained because there is no way the CO2 pressure can decrease to its equilibrium value (which would require some of the CO2 to react with CaO). (d) The situation is essen- tially the same as in (a): CaCO3 decomposes until equilibrium is attained. The presence of CaO initially makes no difference. PRACTICE EXERCISE When added to Fe3O4(s) in a closed container, which one of the following substances — H2(g), H2O(g), O2(g) — allows equilibrium to be established in the reaction 3 Fe(s) + 4 H2O(g) Δ Fe 3O4(s) + 4 H2(g)? Answer: H2(g)  SECTION 15.5 Calculating Equilibrium Constants 625 When a solvent is a reactant or product in an equilibrium, its concentration is omit- ted from the equilibrium-constant expression, provided the concentrations of reactants and products are low, so that the solvent is essentially a pure substance. Applying this guideline to an equilibrium involving water as a solvent, H2O(l) + CO32-(aq) Δ OH-(aq) + HCO3 -(aq) [15.20] gives an equilibrium-constant expression that does not contain [H2O]: [OH-][HCO3 -] Kc = [15.21] [CO32-] GIVE IT SOME THOUGHT Write the equilibrium-constant expression for the reaction NH3(aq) + H2O(l) Δ NH4 +(aq) + OH -(aq) 15.5 | CALCULATING EQUILIBRIUM CONSTANTS If we can measure the equilibrium concentrations of all the reactants and products in a chemical reaction, as we did with the data in Table 15.1, calculating the value of the equilibrium constant is straightforward. We simply insert all the equilibrium concentra- tions into the equilibrium-constant expression for the reaction. SAMPLE EXERCISE 15.8 Calculating K When All Equilibrium Concentrations Are Known After a mixture of hydrogen and nitrogen gases in a reaction vessel is allowed to attain equilib- rium at 472 °C, it is found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data, calculate the equilibrium constant Kp for the reaction N2(g) + 3 H2(g) Δ 2 NH3(g) SOLUTION Analyze We are given a balanced equation and equilibrium partial pressures and are asked to calculate the value of the equilibrium constant. Plan Using the balanced equation, we write the equilibrium-constant expression. We then substitute the equilibrium partial pressures into the expression and solve for Kp. Solve (PNH3)2 (0.166)2 Kp = 3 = = 2.79 * 10-5 PN2(PH2) (2.46)(7.38)3 PRACTICE EXERCISE An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 °C: [CH3COOH] = 1.65 * 10-2 M; [H+] = 5.44 * 10-4 M; and [CH3COO - ] = 5.44 * 10-4 M. Calculate the equilibrium constant Kc for the ionization of acetic acid at 25 °C. The reaction is CH3COOH(aq) Δ H+(aq) + CH3COO -(aq) Answer: 1.79 * 10-5 Often we do not know the equilibrium concentrations of all species in an equilib- rium mixture. If we know the equilibrium concentration of at least one species, however, we can generally use the stoichiometry of the reaction to deduce the equilib- rium concentrations of the others. The following steps outline the procedure: 1. Tabulate all known initial and equilibrium concentrations of the species that appear in the equilibrium-constant expression. 2. For those species for which initial and equilibrium concentrations are known, cal- culate the change in concentration that occurs as the system reaches equilibrium. 626 CHAPTER 15 Chemical Equilibrium 3. Use the stoichiometry of the reaction (that is, the coefficients in the balanced chem- ical equation) to calculate the changes in concentration for all other species in the equilibrium-constant expression. 4. Use initial concentrations from step 1 and changes in concentration from step 3 to calculate any equilibrium concentrations not tabulated in step 1. 5. Determine the value of the equilibrium constant. SAMPLE EXERCISE 15.9 Calculating K from Initial and Equilibrium Concentrations A closed system initially containing 1.000 * 10-3 M H2 and 2.000 * 10-3 M I2 at 448 °C is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87 * 10-3 M. Calculate Kc at 448 °C for the reaction taking place, which is H2(g) + I2(g) Δ 2 HI(g) SOLUTION Analyze We are given the initial concentrations of H2 and l2 and the Plan We construct a table to find equilibrium concentrations of all equilibrium concentration of HI. We are asked to calculate the equilib- species and then use the equilibrium concentrations to calculate the rium constant Kc for H2(g) + I2(g) Δ 2 Hl(g). equilibrium constant. Solve First, we tabulate the initial and equi- H2(g) + I2(g) Δ 2 HI(g) librium concentrations of as many species as -3 -3 we can. We also provide space in our table for Initial concentration (M) 1.000 * 10 2.000 * 10 0 listing the changes in concentrations. As Change in concentration (M) shown, it is convenient to use the chemical equation as the heading for the table. Equilibrium concentration (M) 1.87 * 10-3 Second, we calculate the change in HI con- centration, which is the difference between the equilibrium and initial values: Change in [HI] = 1.87 * 10-3 M - 0 = 1.87 * 10-3 M Third, we use the coefficients in the bal- mol HI 1 mol H2 mol H2 a 1.87 * 10-3 ba b = 0.935 * 10-3 anced equation to relate the change in [HI] L 2 mol HI L to the changes in [H2] and [I2]: mol HI 1 mol I2 mol I2 a 1.87 * 10-3 ba b = 0.935 * 10-3 L 2 mol HI L Fourth, we calculate the equilibrium con- centrations of H2 and I2, using initial concentrations and changes in concentra- [H2] = 1.000 * 10-3 M - 0.935 * 10-3 M = 0.065 * 10-3 M tion. The equilibrium concentration equals the initial concentration minus that [I2] = 2.000 * 10-3 M - 0.935 * 10-3 M = 1.065 * 10-3 M consumed: Our table now looks like this (with equili- H2(g) + I2(g) Δ 2 HI(g) brium concentrations in blue for emphasis): -3 -3 Initial concentration (M) 1.000 * 10 2.000 * 10 0 Change in concentration (M) -0.935 * 10 -3 -0.935 * 10 -3 +1.87 * 10-3 Equilibrium concentration (M) 0.065 * 10-3 1.065 * 10-3 1.87 * 10-3 Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed. Finally, we use the equilibrium-constant expression to calculate the equilibrium [HI]2 (1.87 * 10-3)2 Kc = = = 51 constant: [H2][I2] (0.065 * 10-3)(1.065 * 10-3) Comment The same method can be applied to gaseous equilibrium problems to calculate Kp , in which case partial pressures are used as table entries in place of molar concentrations. Your instructor may refer to this kind of table as an ICE chart, where ICE stands for Initial – Change – Equilibrium. PRACTICE EXERCISE Sulfur trioxide decomposes at high temperature in a sealed container: 2 SO3(g) Δ 2 SO2(g) + O2(g). Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K. Answer: 0.338  SECTION 15.6 Applications of Equilibrium Constants 627 | 15.6 APPLICATIONS OF EQUILIBRIUM CONSTANTS We have seen that the magnitude of K indicates the extent to which a reaction proceeds. If K is very large, the equilibrium mixture contains mostly substances on the product side of the equation for the reaction. (That is, the reaction proceeds far to the right.) If K is very small (that is, much less than 1), the equilibrium mixture contains mainly sub- stances on the reactant side of the equation. The equilibrium constant also allows us to (1) predict the direction in which a reaction mixture achieves equilibrium and (2) cal- culate equilibrium concentrations of reactants and products. Predicting the Direction of Reaction For the formation of NH3 from N2 and H2 (Equation 15.6), Kc = 0.105 at 472 °C. Sup- pose we place 2.00 mol of H2, 1.00 mol of N2, and 2.00 mol of NH3 in a 1.00-L container at 472 °C. How will the mixture react to reach equilibrium? Will N2 and H2 react to form more NH3, or will NH3 decompose to N2 and H2? To answer this question, we substitute the starting concentrations of N2, H2, and NH3 into the equilibrium-constant expression and compare its value to the equilibrium constant: [NH3]2 (2.00)2 3 = = 0.500 whereas Kc = 0.105 [15.22] [N2][H2] (1.00)(2.00)3 To reach equilibrium, the quotient [NH3]2>[N2][H2]3 must decrease from the starting value of 0.500 to the equilibrium value of 0.105. Because the system is closed, this change can happen only if [NH3] decreases and [N2] and [H2] increase. Thus, the reac- tion proceeds toward equilibrium by forming N2 and H2 from NH3; that is, the reaction as written in Equation 15.6 proceeds from right to left. This approach can be formalized by defining a quantity called the reaction quotient. The reaction quotient, Q, is a number obtained by substituting reactant and product con- centrations or partial pressures at any point during a reaction into an equilibrium-constant expression. Therefore, for the general reaction aA + bB Δ dD + eE the reaction quotient in terms of molar concentrations is [D]d[E]e Qc = [15.23] [A]a[B]b (A related quantity Qp can be written for any reaction that involves gases by using par- tial pressures instead of concentrations.) Although we use what looks like the equilibrium-constant expression to calculate the reaction quotient, the concentrations we use may or may not be the equilibrium concentrations. For example, when we substituted the starting concentrations into the equilibrium-constant expression of Equation 15.22, we obtained Qc = 0.500 whereas Kc = 0.105. The equilibrium constant has only one value at each temperature. The reac- tion quotient, however, varies as the reaction proceeds. Of what use is Q? One practical thing we can do with Q is tell whether our reaction really is at equilibrium, which is an especially valuable option when a reaction is very slow. We can take samples of our reaction mixture as the reaction proceeds, separate the components, and measure their concentrations. Then we insert these numbers into Equation 15.23 for our reaction. To determine whether or not we are at equilibrium, or in which direction the reaction proceeds to achieve equilibrium, we compare the values of Qc and Kc or Qp and Kp. Three possible situations arise: Q = K: The reaction quotient equals the equilibrium constant only if the system is at equilibrium. 628 CHAPTER 15 Chemical Equilibrium At equilibrium Q 7 K: The concentration of products is too large and that of reactants too small. Substances on the right side of the chemical equation react to form substances on Q QK the left; the reaction proceeds from right to left to approach equilibrium. K Reaction forms Q 6 K: The concentration of products is too small and that of reactants too large. products The reaction achieves equilibrium by forming more products; it proceeds from left to right. Q QK These relationships are summarized in 씱 FIGURE 15.8. K Equilibrium SAMPLE EXERCISE 15.10 Predicting the Direction of Approach Q QK to Equilibrium K Reaction forms At 448 °C the equilibrium constant Kc for the reaction reactants H2(g) + I2(g) Δ 2 HI(g) is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 쑿 FIGURE 15.8 Predicting the direction 2.0 * 10-2 mol of HI, 1.0 * 10-2 mol of H2, and 3.0 * 10-2 mol of I2 in a 2.00-L container. of a reaction by comparing Q and K at a given temperature. SOLUTION Analyze We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium. Plan We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient, Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed. Solve The initial concentrations are [HI] = 2.0 * 10-2 mol>2.00 L = 1.0 * 10-2 M [H2] = 1.0 * 10-2 mol>2.00 L = 5.0 * 10-3 M [I2] = 3.0 * 10-2 mol>2.00 L = 1.5 * 10-2 M [HI]2 (1.0 * 10-2)2 The reaction quotient is therefore Qc = = = 1.3 [H2][I2] (5.0 * 10-3)(1.5 * 10-2) Because Qc 6 Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium. PRACTICE EXERCISE At 1000 K the value of Kp for the reaction 2 SO3(g) Δ 2 SO2(g) + O2(g) is 0.338. Calculate the value for Qp, and predict the direction in which the reaction proceeds toward equilibrium if the initial partial pressures are PSO3 = 0.16 atm; PSO2 = 0.41 atm; PO2 = 2.5 atm. Answer: Q p = 16; Q p 7 Kp, and so the reaction will proceed from right to left, forming more SO3. Calculating Equilibrium Concentrations Chemists frequently need to calculate the amounts of reactants and products present at equilibrium in a reaction for which they know the equilibrium constant. The ap- proach in solving problems of this type is similar to the one we used for evaluating equilibrium constants: We tabulate initial concentrations or partial pressures, changes in those concentrations or pressures, and final equilibrium concentrations or partial pressures. Usually we end up using the equilibrium-constant expression to derive an equation that must be solved for an unknown quantity, as demonstrated in Sample Exercise 15.11.  SECTION 15.6 Applications of Equilibrium Constants 629 SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations For the Haber process, N2(g) + 3 H2(g) Δ 2 NH3(g), Kp = 1.45 * 10-5 at 500 °C. In an equilibrium mixture of the three gases at 500 °C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture? SOLUTION Analyze We are given an equilibrium constant, Kp, and the equilib- Plan We can set Kp equal to the equilibrium-constant expression and rium partial pressures of two of the three substances in the equation substitute in the partial pressures that we know. Then we can solve for (N2 and H2), and we are asked to calculate the equilibrium partial the only unknown in the equation. pressure for the third substance (NH3). Solve We tabulate the equilibrium pressures: N2(g) + 3 H2(g) Δ 2 NH3(g) Equilibrium pressure (atm) 0.432 0.928 x Because we do not know the equilibrium pressure of NH3, we represent it with x. At equilibrium the pressures must satisfy the (PNH3)2 x2 equilibrium-constant expression: Kp = 3 = = 1.45 * 10-5 PN2(PH2) (0.432)(0.928)3 We now rearrange the equation to solve for x: x = (1.45 * 10-5)(0.432)(0.928)3 = 5.01 * 10-6 2 x = 25.01 * 10-6 = 2.24 * 10-3 atm = PNH3 Check We can always check our answer by using it to recalculate (2.24 * 10-3)2 the value of the equilibrium constant: Kp = = 1.45 * 10-5 (0.432)(0.928)3 PRACTICE EXERCISE At 500 K the reaction PCl 5(g) Δ PCl 3(g) + Cl 2(g) has Kp = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium mixture? Answer: 1.22 atm In many situations we know the value of the equilibrium constant and the initial amounts of all species. We must then solve for the equilibrium amounts. Solving this type of problem usually entails treating the change in concentration as a variable. The stoichiometry of the reaction gives us the relationship between the changes in the amounts of all the reactants and products, as illustrated in Sample Exercise 15.12. The calculations frequently involve the quadratic formula, as you will see in this exercise. SAMPLE EXERCISE 15.12 Calculating Equilibrium Concentrations from Initial Concentrations A 1.000-L flask is filled with 1.000 mol of H2(g) and 2.000 mol of I2(g) at 448 °C. The value of the equilibrium constant Kc for the reaction H2(g) + I2(g) Δ 2 HI(g) at 448 °C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter? SOLUTION Analyze We are given the volume of a container, an equilibrium Plan In this case we are not given any of the equilibrium concentra- constant, and starting amounts of reactants in the container and are tions. We must develop some relationships that relate the initial con- asked to calculate the equilibrium concentrations of all species. centrations to those at equilibrium. The procedure is similar in many regards to that outlined in Sample Exercise 15.9, where we calculated an equilibrium constant using initial concentrations. Solve First, we note the initial concentra- tions of H2 and I2: [H2] = 1.000 M and [I2] = 2.000 M Second, we construct a table in which we tabulate the initial concentrations: H2(g) + I2(g) Δ 2 HI(g) Initial concentration (M) 1.000 2.000 0 Change in concentration (M) Equilibrium concentration (M) 630 CHAPTER 15 Chemical Equilibrium Third, we use the stoichiometry of the reaction to determine the changes in concen- tration that occur as the reaction proceeds to equilibrium. The H2 and I2 concentrations will decrease as equilibrium is established and that of HI will increase. Let’s represent the change in concentration of H2 by x. The H2(g) + I2(g) Δ 2 HI(g) balanced chemical equation tells us the Initial concentration (M) 1.000 2.000 0 relationship between the changes in the con- centrations of the three gases. For each x mol Change in concentration (M) -x -x +2x of H2 that reacts, x mol of I2 are consumed and Equilibrium concentration (M) 2x mol of HI are produced: H2(g) + I2(g) Δ 2 HI(g) Fourth, we use initial concentrations and changes in concentrations, as dictated by Initial concentration (M) 1.000 2.000 0 stoichiometry, to express the equilibrium Change in concentration (M) -x -x + 2x concentrations. With all our entries, our table Equilibrium concentration (M) 1.000 - x 2.000 - x 2x now looks like this: Fifth, we substitute the equilibrium concentra- tions into the equilibrium-constant expression [HI]2 (2x)2 and solve for x: Kc = = = 50.5 [H2][I2] (1.000 - x)(2.000 - x) If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quad- ratic equation in x: 4x 2 = 50.5(x 2 - 3.000x + 2.000) 46.5x 2 - 151.5x + 101.0 = 0 Solving the quadratic equation (Appendix - (- 151.5) ; 2(- 151.5)2 - 4(46.5)(101.0) A.3) leads to two solutions for x: x = = 2.323 or 0.935 2(46.5) When we substitute x = 2.323 into the ex- pressions for the equilibrium concentrations, we find negative concentrations of H2 and I2. Because a negative concentration is not [H2] = 1.000 - x = 0.065 M chemically meaningful, we reject this solu- tion. We then use x = 0.935 to find the [I2] = 2.000 - x = 1.065 M equilibrium concentrations: [HI] = 2x = 1.87 M Check We can check our solution by put- ting these numbers into the equilibrium- constant expression to assure that we correctly [HI]2 (1.87)2 calculate the equilibrium constant: Kc = = = 51 [H2][I2] (0.065)(1.065) Comment Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions to the equation will give you a value that leads to negative concentrations and thus is not chemically mean- ingful. Reject this solution to the quadratic equation. PRACTICE EXERCISE For the equilibrium PCl 5(g) Δ PCl 3(g) + Cl 2(g), the equilibrium constant Kp is 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium pres- sures of PCl5, PCl3, and Cl2 at this temperature? Answer: PPCl5 = 0.967 atm, PPCl3 = PCl2 = 0.693 atm 15.7 | LE CHÂTELIER’S PRINCIPLE Many of the products we use in everyday life are obtained from the chemical industry. Chemists and chemical engineers in industry spend a great deal of time and effort to maximize the yield of valuable products and minimize waste. For example, when Haber developed his process for making ammonia from N2 and H2, he examined how reaction conditions might be varied to increase yield. Using the values of the equilibrium con- stant at various temperatures, he calculated the equilibrium amounts of NH3 formed under a variety of conditions. Some of Haber’s results are shown in 씰 FIGURE 15.9.  SECTION 15.7 Le Châtelier’s Principle 631 GO FIGURE At what combination of pressure and temperature should you run the reaction to maximize NH3 yield? Percent of NH3 Percent of NH3 increases with decreases with increasing 8.8 increasing % 12 pressure 18.9% temperature.9% 16 26.9% 27.0%.4% 32 20 35.2%.8% 38.9% 37.7% 42.8% 47.9% Percent NH3.8% 48 54.8% produced.9% 550 60.6% ) °C 500 200 re( tu To 씱 FIGURE 15.9 Effect of temperature tal 300 450 ra pr pe and pressure on NH3 yield in the Haber ess 400 m ur process. Each mixture was produced by Te e( 400 atm 500 starting with a 3:1 molar mixture of H2 ) and N2. Notice that the percent of NH3 present at equilibrium decreases with increasing temper- ature and increases with increasing pressure. We can understand these effects in terms of a principle first put forward by Henri- Louis Le Châtelier* (1850–1936), a French industrial chemist: If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position so as to counteract the effect of the disturbance. Le Chatelier’s Principle If a system at equilibrium is disturbed by a change in concentration, pressure, or temperature, the system will shift its equilibrium position so as to counter the effect of the disturbance. Concentration: adding or removing a reactant or product If a substance is added to a system at equilibrium, the system reacts to consume some of the substance. If a substance is removed from a system, the system reacts to produce more of substance. Initial equilibrium Substance added Equilibrium reestablished    Substances react Pressure: changing the pressure by changing the volume Pressure At constant temperature, reducing the volume of a gaseous equilibrium Initial volume System shifts mixture causes the system to shift in the direction that reduces the to direction of number of moles of gas. fewer moles Temperature: If the temperature of a system at equilibrium is increased, the system reacts as if we added a reactant to an endothermic reaction or a product to an exothermic reaction. The equilibrium shifts in the direction that consumes the “excess reactant,” namely heat. Endothermic Exothermic Increasing T Decreasing T Decreasing T Increasing T Reaction shifts right Reaction shifts left Reaction shifts right Reaction shifts left *Pronounced “le-SHOT-lee-ay.” 632 CHAPTER 15 Chemical Equilibrium In this section we use Le Châtelier’s principle to make qualitative predictions about how a system at equilibrium responds to various changes in external conditions. We consider three ways in which a chemical equilibrium can be disturbed: (1) adding or re- moving a reactant or product, (2) changing the pressure by changing the volume, and (3) changing the temperature. Change in Reactant or Product Concentration A system at dynamic equilibrium is in a state of balance. When the concentrations