• • Think About It It isn't nece ssary for every species in the reaction to be a gas only those species that appear in the equilibrium . expressIOn. which simplifies to where D.n = b - a. In general, Equation 15 .3 D.n = moles of gaseous products - moles of gaseous reactants Because pressures are usually expressed in atmospheres, the gas constant R is 0.08206 L . atmlmol . K, and we can write the relationship between Kp and Kc as Equation 15.4 Kp = K c [(0 .08206 L . atmlK . mol) X T]~ n Kp is equal to Kc only in the special case where D.n = 0, as in the following equilibrium reaction: Hig) + Br2(g) +=. ==:z' 2HBr(g) In this case, Equation 15.4 can be written as Kp = Kc [(0.08206 L . atmlK . mol) X T]0 = K c Keep in mind that Kp expressions can only be written for reactions in which every species in the equilibrium expression is a gas. (Remember that solids and pure liquids do not appear in the equilibrium expression.) Sample Problems 15.5 and 15.6 let you practice writing Kp expressions and illustrate the conversion between Kc and Kp . Write Kp expressions for (a) PCI 3 (g) + CI 2 (g) ;;: .• = ::!:' PCl s(g), (b) 0 2(g) + 2H 2 (g) :;:. = ::!:' 2H 2 0(I), and (c) Fig) + H2(g) • ' 2HF(g). Strategy Write equilibrium expressions for each equation, expressing the concentrations of the gases in partial pressures. Setup (a) All the species in this equation are gases, so they will all appear in the Kp expression. (b) Only the reactants are gases. (c) All species are gases. Solution Checkpoint 15.3 Equilibrium Expressions 15. 3.1 Select the correct equilibrium expression for the reaction 15.3.2 Select the corr ec t equilibrium expression for the reaction H +(aq) + OW (aq) • ' H 2 0( I) CaO(s) + CO 2 (g) • ' CaC0 3 (s) a) [H 2 O l eq Kc = + [H leq[OWl eq • a) K = 1 c [C0 2 l eq b) K = [H+l eq[ OWl eq c [H 2 O l eq b) [CaC0 3 l eq K= c [CaOl eq [C0 2l eq c) Kc = [H +l eq[ OH- l eq c) Kc = [C0 2 l eq d) K = 1 c + _ [H l eq [OH l eq d) K = [CaOl eq [C0 2l eq c [ CaC0 3 l eq e) Kc = [H+l eq [OH- leq[ H20l eq e) [C0 2 l eq K= c [CaOl eq ][CaC0 3l eq 604 15.3.3 • Practice Problem A Write Kp expressions for (a) 2CO(g) + Oz(g) :;:, =~. 2CO z (g), (b) CaC0 3 (s) • • CaO (s) + CO z(g), and (c) Nz(g) + 3H z (g) • • 2NH 3 (g). Practice Problem B Write the equation for the ga seous equilibrium corresponding to each of the following Kp expressions: (P NO / (a) Kp = z (P NO ) (Po) 2 2 The equilibrium constant, Kc, for the reaction (PHI i ( c) K - ' ' = - p - (P L, )(P R ,) is 4.63 X 10- 3 at 25°C. What is the value of Kp at this temperature? Strategy Use Equation 15.4 to convert f rom Kc to Kp. Be sure to convert temperature in degrees Celsius to kelvins. Setup Using Equation 15.3, T = 298 K. Solution Kp = [K c (0.08206 L . atm ) X T] K'mol = (4.63 X 10- 3 )(0.08206 X 298) = 0.113 Practice Problem A Por the reaction Kc is 2.3 X lO - z at 375°C. Calculate Kp for the reaction at this temperature. Practice Problem B Kp = 2.79 X 10- 5 for the reaction in Practice Problem A at 472°C. What is Kc for this reaction at 472 ° C? Given the following information, HP(aq) • • H +(aq) + P- (aq) 15.3.4 Kc for the reaction Think About It Note that we have essentially disregarded the units of Rand T so that the resulting equilibrium constant, K p , is unitless. Equilibrium constants commonly are treated as unitless quantities. Br 2(g) :;:. =~. 2Br (g) Kc = 6.8 X 10- 4 (at 25°C) H Z C Z 0 4 (aq) :;:.=~' 2H +(aq) + C 1 0 ~ - ( aq ) is 1.1 X 10- 3 at 1280°C. Calculate the value of Kp for this reaction at this temperature. Kc = 3.8 X 10- 6 (at 25°C) a) 1.1 X 10- 3 determine the value of Kc for the following reaction at 25°C: b) 18 C 2 0 ~ - + 2HF. • 2P- + H 1 C 2 0 4 c) 0.14 a) 2.6 X 10- 9 d) 9.1 X 10 2 b) 1.8 X 10- 11 e) 8.3 X 10- 6 c) 1.2 X 10- 1 d) 2.6 X 10 5 e) 6.8 X 10- 4 605 I' I , \ \ 606 CHAPTER 15 Chemical Equilibrium Qc is calcula t ed us i ng the initial concentrations of reactants and products. Simil a rly, Q p can be calcul ated us ing the initial pa rti al pr essu res of re actants and products. Remem ber t ha t calc u lati ng Q c is just like ca lc ul ating Kc: prod ucts ov er r eact ants, each rais ed to the appro pr iate pow er excep t that the co n ce n tra ti ons we use to calcu la te Q c are the starting conce n tra ti ons . To calculate Kc we must us e equilibrium conce ntrat io n s. The co m par is on of Q with K can re f er ei ther to Q c an d Kc or Qp an d Kp . U sing Equilibrium Expressions to Solve Problems We have already us ed equilibrium expressions to determine the value of an equilibrium constant using equilibrium concentrations. In this section, we will learn how to use equilibrium expressions to predict the direction of a reaction and to calculate equilibrium concentrations. Predicting the Direction of a Reaction If we start an experiment with only reactants, we know that the reactant concentrations will decrease and the product concentrations will increase; that is, the reaction must proceed in the forward direction in order for equilibrium to be established. Likewise, if we start an experiment with only product s, we know that the product concentrations will decrease and the reactant con- centrations will increase. In this case, the reaction mu st proceed in the reverse direction to achieve equilibrium, But often we mu st predict the direction in which a reaction will proceed when we start with a mixture of reactants and products. For this situation, we calculate the value of the reac- . . . . . . . . . . . . . . . . . . . . . . . . . tion quotient, Q e, and compare it to the value of the equilibrium constant, Ke. The equilibrium constant, Ke, for the ga seous formation of hydrogen iodide from molecular hydrogen and molecular iodine, is 54.3 at 430 °C. If we were to conduct an experiment starting with a mixture of 0.243 mole of H?, 0.146 mole of 1 2 , and 1.98 moles of III in a 1.00-L container at 430 °C, would more HI form or would III be consumed and more H2 and 12 form? Using the starting concentrations, we can . calC u la te' the reaction quotient as follows: (1.98)2 -:-::- :: :: , :-::- -:-: :::- = 111 (0.243)(0.146) where the subscript "i" indicates initial concentration. Because the reaction quotient does not equal Ke (Q e = Ill , Ke = 54.3), the reaction is not at equilibrium. In order to establish equilib- rium, the reaction will pro cee d to the left, consuming III and producing H2 and 1 2 , decreasing the value of the numerator and increasing the value in the denominator until the value of the reaction quotient equals that of the equilibrium constant. Thu s, the reaction proceeds in the reverse direc- tion (from right to left) in order to reach equilibrium. . . . . . . . . . . . . . . . . . . . . . . . There are three possibilities when we compare Q with K: Q < K The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants mu st be converted to products. The system proceeds in the forward direction (from left to right ). Q = K The initial concentrations are equilibrium concentrations. The system is already at equilibrium, and there will be no net movement in either direction. Q > K The ratio of initial concentrations of products to reactants is too large. To reach equilibrium, products mu st be converted to reactants, The system proceeds in the reverse dir ection (f rom right to left). Sample Problem 15.7 shows how the value of Q is used to determine the direction of a reac- tion that is not at equilibrium. Sample Problem 15.7 At 375°C, the equilibrium constant for the reaction is 1. 2. At the start of a reac ti on, the concentrations of N 2 , H z, and NH3 are 0.071 M, 9.2 X 10- 3 M, and 1. 83 X 10- 4 M, respec ti vel y. Determine whether this system is at equilibrium, and if not, determine in which direction it must proceed to establish equilibrium. Strategy Use the initial concentrations to calculate Qc, and then compare Qc with Kc. SECTION 15.4 Using Equilibrium Expressions to Solve Problems 607 Setup (1.83 X 1O- 4 )z - - , ,- = 0.61 (0.071)(9.2 X 10- 3 )3 Solution The calculated value of Qc is less than Kc. Therefore, the reaction is not at equilibrium and must proceed to the right to establish eqUilibrium. Practice Problem A The equilibrium constant, Kc, for the formation of nitros yl chloride from nitric oxide and chlorine, 2NO(g) + Cl z(g) +=, ~. 2NOCl(g) is 6.5 X 10 4 at 35 ° C. In which direction will the reaction proceed to reach equilibrium if the starting concentrations of NO, Clz, and NOCI are 1.1 X 10 - 3 M, 3.5 X 10- 4 M, and 1.9 M, respectively? • Practice Problem B Calculate Kp for the formation of nitrosyl chloride from nitric oxide and chlorine at 35° C, and determine whether the reaction will proceed to the right or the left to achieve equilibrium when the starting pre ss ures are P NO = 1.01 atm, P CI = 0.42 atm, and P NOCI = 1.76 atm. 2 Calculating Equilibrium Concentrations If we know the equilibrium constant for a reaction , we can calculate the concentrations in the equi- 1ibrium mixture from the initial reactant concentrations. Consider the following system involving two organic compounds, cis- and trans-stilbene: , , cis-Stilbene H / l =C H trans-Stilbene The equilibrium constant (K e) for this s ystem is 24.0 at 200° C. If we know that the starting con- centration of cis-stilbene is 0.850 M, we can use the equilibrium expre ss ion to determine the equilibrium concentrations of both specie s. The s toichiometry of the reaction tells us that for every mole of cis-stilbene converted, 1 mole of trans-stilbene is produced. We will let x be the equi- librium concentration of trans-stilbene in mollL; therefore , the equilibrium concentration of cis- stilbene mu st be (0.850 - x) mollL. It is us eful to summarize the se change s in concentration s in an equilibrium table: cis-stilbene +, ==' trans-stilbene Initial concentration (M): 0.850 o Change in concentration (M): -x +x Equilibrium concentration (M): 0.850 - x x We then use the equilibrium concentration s, defined in term s of x, in the equilibrium expre ssion: [trans-stilbene] Ke = . [cis-stIlbene] 24.0 = x 0.850 - x x = 0.816 M Think About It In proceeding to the right, a reaction consumes reactants and produces more produc ts. This increases the numerator in the reaction quotient and decreases the denominator. The result is an increase in Qc until it is equal to Kc, at which point equilibrium will be established. 608 CHAPTER 15 Chemical Equilibrium Think About It Always check your answer by inserting the calculated concentrations into the equilibrium expression: [HIl ~ q (0.378)2 [H2le q[I 2 leq (0.051)2 = 54.9 = Kc The small difference between the calculated Kc and the one given in the problem statement is due to rounding. Having solved for x, we calculate the equilibrium concentrations of cis- and trans-stilbene as follows: [cis-stilbene] = (0.850 - x) M = 0.034 M [trans-stilbene] = x M = 0.816 M A good way to check the answer to a problem such as this is to use the calculated equilibrium con- centrations in the equilibrium expression and make sure that we get the correct Kc value. K = 0.816 = 24 c 0.034 Sample Problems 15.8 and 15.9 provide additional examples of this kind of problem. Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, is 54.3 at 430°C. What will the concentrations be at equilibrium if we start with 0.240 M concentrations of both H2 and 1 2 ') Strategy Construct an equilibrium table to determine the equilibrium concentration of each species in terms of an unknown (x); solve for x, and use it to calculate the equilibrium molar concentrations. Setup Insert the starting concentrations that we know into the equilibrium table: H 2 (g) + I 2 (g) :;::. =::!:' 2HI(g) Initial concentration (M): 0.240 0.240 0 Change in concentration (M): Equilibrium concentration (M): Solution We define the change in concentration of one of the reactants as x. Because there is no product at the start of the reaction, the reactant concentration must decrease; that is, this reaction must proceed in the forward direction to reach equilibrium. According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x), and the product concentration will increase by twice that amount (2x). Combining the initial concentration and the change in concentration for each species, we get expressions (in terms of x) for the equilibrium concentrations: + • , 2HI(g) o Initial concentration (M): 0.240 ~ ~ -x + 2x Change in concentration (M): - x + ~ Equilibrium concentration (M): 0.240 - x 0.240 - x 2x • Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x. ? [HIl~q K = =-:: c=-': c [H2le q[I 2leq (2X) 2 54.3 = c:: ~ , c: :-::-cc:  (0.240 - x)(0.240 - x) "54.3 = 2x 0.240 - x x = 0.189 (0.240 - X)2 Using the calculated value of x, we can determine the equilibrium concentration of each species as follows: [H2leq = (0.240 - x) M = 0.051 M [I2leq = (0.240 - x) M = 0.051 M [HIl eq = 2x = 0.378 M SECTION 15.4 Using Equilibrium Expressions to Solve Problems 609 Practice Problem A Calculate the equilibrium concentrations of H 2 , 1 2 , and HI at 430°C if the initial concentrations are [H2l; = [I2li = 0 M, and [HIli = 0.525 M. Practice Problem B Calculate the equilibrium concentrations of H2o 120 and HI at 430°C if the initial concentrations are [H 2 ]; = [I2l; = 0.100 M, and [HIl; = 0.200 M. For the same reaction and temperature as in Sample Problem 15.8, calculate the equilibrium concentrations of all three species if the starting concentrations are as follows: [H 2 l; = 0.00623 M, [1 2 ]; = 0.00414 M, and [HIl; = 0.0424 M. Strategy Using the initial concentrations, calculate the reaction quotient, Q c, and compare it to the value of Kc (given in the problem statement of Sample Problem 15.8) to determine which direction the reaction will proceed in order to establish equilibrium. Then, construct an equilibrium table to determine the equilibrium concentrations. · Setup [HIl! [H 2 ];[I 2 l; (0.0424i (0.00623)(0.00414) = 69.7 Therefore, Qc > Kc, so the system will have to proceed to the left (reverse) to reach equilibrium. The equilibrium table is H2(g) + I2(g) :;:, ==' 2HI(g) Initial concentration (M): 0.00623 0.00414 0.0424 Change in concentration (M): Equilibrium concentration (M): Solution Because we know the reaction must proceed from right to left, we know that the concentration of HI will decrease and the concentrations of H2 and 12 will increase. Therefore, the table should be filled in as follows: Initial concentration (M): Change in concentration (M): Equilibrium concentration (M): Hig) 0.00623 +x 0.00623 + x + I 2 (g) 0.00414 +x 0.00414 + x , , 2HI(g) 0.0424 -2x 0.0424 - 2x Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x. (0.0424 - 2xi 5 4 . 3 = c::-::-::-::-' :: , ,-:: ::-::- -'-: c c- (0.00623 + x)(0.00414 + x) It isn't possible to solve this equation the way we did in Practice Problem 15.8 (by taking the square root of both sides) because the concentrations of H2 and 12 are unequal. Instead, we have to carry out the multiplications, 54.3(2.58 X 10- 5 + 1.04 X 1O - 2 x + x 2 ) = 1.80 X 10- 3 - 1.70 X 1O- 1 x + 4x 2 Collecting terms we get 50.3x 2 + 0.735x - 4.00 X 10- 4 = 0 This is a quadratic equation of the form ax 2 + bx + c = O. The solution for the quadratic equation (see Appendix 1) is x= -b ± ~ b 2 - 4ac 2a (Continued) 610 CHAPTER 15 Chemical Equilibrium Think About It Checking this result gives [HI]2 Kc = [H 2 ][12] (0.0414i : , :- :: :c : :-:: :-:: = 54. 3 (0.00676)(0.00467) If Q < K the reaction will occur as wr itten. If Q > K the reverse reaction will occur. For this reaction, 1M = 0; therefore, Kc is equal to Kp. Here we have a = 50.3, b = 0.735, and c = -4.00 X 1O~4, so -0.735 ± ~(0.735)2 4(50.3)(- 4.00 X 1O ~ 4) x = 2(50.3) x = 5.25 X 1O~4 or x = -0.0151 Only the first of these values, 5.25 X 1O~4, makes sense because concentration cannot be a negative number. Using the calculated value of x, we can determine the equilibrium concentration of each species as follows: [H 2 ]e q = (0.00623+ x) M = 0.00676 M [1 2] eq = (0.00414 + x) M = 0.00467 M [HI]eq = (0.0424 - 2x) = 0.0414 M Practice Problem At 1280°C the equilibrium constant Kc for the reaction Br 2(g) :;::. =::t, 2Br(g) is 1.1 X 1O ~3 . If the initial concentrations are [Br2] = 6.3 X 1O~2 M and [Br] = 1.2 X 1O~2 M, calculate the concentrations of these two species at equilibrium. Here is a summary of the use of initial reactant concentrations to determine equilibrium concentrations: 1. Construct an equilibrium table, and fill in the initial concentrations (including any that are zero). 2. Use initial concentrations to calculate the reaction quotient, Q, and compare Q to K to deter- mine ' the ' cfirectloi11ll ' whiCh ' th 'e ' reacdoil will proceed. 3. Define x as the amount of a particular species consumed, and use the stoichiometry of the reaction to define (in terIllS of x) the amount of other species consumed or produced. 4. For each species in the equilibrium, add the change in concentration to the initial concentra- tion to get the equilibrium concentration. 5. Use the equilibrium concentrations and the equilibrium expression to solve for x. 6. Using the calculated value of x, determine the concentrations of all species at equilibrium. 7. Check your work by plugging the calculated equilibrium concentrations into the equilibrium expression. The result should be very close to the Kc stated in the problem. The same procedure applies to Kp. Sample Problem 15.10 shows how to solve an equilibrium problem using partial pressures . . Sample Problem 15.10 A mixture of 5.75 atm of H2 and 5.75 atm OfI 2 is contained in a 1.0-L vessel at 430°C. The eqiiiili:iii'uiii con·s·t'ant' (Kp) for the reaction at this temperature is 54.3. Determine the equilibrium partial pressures of Hb Ib and HI. Strategy Construct an equilibrium table to determine the equilibrium partial pressures. Setup The equilibrium table is H 2 (g) + 12(g) , 2HI(g) • Ini tial partial pressure (atm): 5.75 5.75 0 Change in partial pressure (atm): - x - x +2x Equilibrium partial pressure (atm): 5.75 - x 5.75 - x 2x SECTION 15.5 Factors That Affect Chemical Equilibrium 611 Solution Setting the equilibrium expression equal to K p , ( 2X )2 54.3 = - ~~ , (5 .75 - xi Taking the square root of both sides of the equation gives / 54.3 = 2x 5.75 - x 7.369 = 2x 5.75 - x 7.369(5.75 - x) = 2x 42.37 - 7.369x = 2x 42.37 = 9.369x x = 4.52 The equilibrium partial pressures are PH = PI = 5.75 - 4.52 = 1.23 atm, and Pm = 9.04 atm. 2 2 • Practice Problem A Determine the equilibrium partial pressures of H 2 , 1 20 and HI if we begin the experiment with 1.75 atm each of H2 and 12 at 430 °C. ·················· · ············ · ······· · ···························· Practice Problem B. Determine the equilibrium partial pressures of H 2 , 1 2 , and HI if we be gin the experiment with 2.75 atm HI at 430 °C. Checkpoint 15.4 Using Equilibrium Expressions to Solve Problems Use the following information to answer questions 15.4.1 and 15.4.2: Kc for the reaction is 1.7 X 10- 2 at 250°C. 15.4.1 What will the equilibrium concentration of A, B, and C be at this temperature if [Ali = [Bli = 0.7 50 M . ([Cl = OJ? a) 6.1 X 10- 3 M, 6.1 X 10- 3 M, 0.092M b) 0.046 M, 0.046 M, 0.092 M c) 0.70 M, 0.70 M, 0.046 M d) 0.70 M, 0.70 M, 0.092 M e) 0.087 M, 0.087 M, 0.66 M A + B. • 2C 15.4.2 What will the equilibrium concentrations of A, B, and C be at this temperature if [C]i = 0.875 M ([Ali = [Bl = O J? a) 0.41 M, 0.41 M, 0.82 M b) 0.41 M, 0.41 M, 0.054 M c) 0.43 M, 0.43 M, 0.00 74 M d) 0.43 M, 0.43 M, 0.44 M e) 0.43 M, 0.43 M, 0.43 M Factors That Affect Chemical Equilibrium One of the interesting and useful features of chemical equilibria is that they can be manipulated in specific ways to maximize production of a desired substance. Consider, for example, the industrial production of ammonia from its constituent elements by the Haber proces s: More than 100 million tons of ammonia is produced annually by this reaction, with most of the resulting ammonia being used for fertilizers to enhance crop production. Clearly it would be in the best interest of industry to maximize the yield of NH 3. In this section, we willleam about the vari- 0us ways in which an equilibrium can be manipulated in order to accomplish this goal. Le Chiitelier's principle states that when a stress is applied to a system at equilibrium, the system will respond by shifting in the direction that minimizes the effect of the stress. In this con- text, "stress" refers to a disturbance of the system at equilibrium by an y of the following means: Think About It Plugging the calculated partial pressures into the equilibrium expression gives (9.04)2 :c = 54.0 (1.23)2 The small difference between this result a nd the equilibrium constant given in the problem statement is due to rounding. Wh en a r eaction starts with re actan ts a nd p ro du cts , be su re to c al c ul at e Q a nd co m pa re it to K to d eterm ine whi ch dir ecti on the re a cti on will proceed to re ach equilibri um . 612 CHAPTER 15 Chemical Equilibrium Re mem ber that at equ ili br iu m, the reactio n quoti ent, Q" is eq ual to the eq uilibrium constan t, K,. While ree stabli sh ing equilibrium ca u se s a decrease in the N2 conc en t ra tion, t he final co nc entr at i on will still be h ig he r th an that in the or ig inal equilib rium mixtur e. Th e system re sp ond s to t he stre ss of the a dde d re actan t by cons umi ng part of it. • The addition of a reactant or product • The removal of a reactant or product • A change in volume of the system, resulting in a change in concentration or partial pressure of the reactants and products • A change in temperature "Shifting" refers to the occunence of either the forward or reverse reaction such that the effect of the stress is partially offset as the system reestablis he s equiliblium. An equilibrium that shifts to the right is one in which more products are produced by the forward reaction. An equi'librium that shifts to the left is one in which more reactants are produced by the reverse reaction. Using Le Chatelier's principle, we can predict the direction in which an equilibrium will shift, given the specific stress that is applied. Addition or Removal of a Substance Again using the Haber process as an example, consider a system at 700 K, in which the equiliblium concentrations are as follows: . . . . . . . . , . . . . . . . . . . . . . . , Using these concentrations in the reaction quotient expression, we can calculate the value of Kc for the reaction at this temperature as follows: [ NH 3f [N 2 ] [H 2 ] 3 ( 1.52)2 =- = 0.297 = Kc (2.05)(1.56)3 If we were to apply stress to this system by adding more N 2 , increasing its concentration from 2.05 M to 3.51 M, the system would no longer be at equilibrium. To see that this is true, use the new concentration of nitrogen in the reaction quotient expression. The new calculated value of Qc (0.173) is no longer equal to the value of Kc (0.297). [ NH 3]2 [N 2 ] [H 2 ] 3 ( 1.52)2 ,. = 0.173 =I=- Kc (3.51)(1.56)3 For this system to reestablish equilibrium, the net reaction will have to shift in such a way that Qc is again equal to Ke, which is constant at a given temperature. Recall from Section 15.4 that when Q is less than K, the reaction proceeds to the light in order to achieve equilibrium. Likewise, an equilibrium that is stres sed in such a way that Q becomes less than K will shift to the right in order to reestablish equiliblium. This means that the forward reaction, the consumption of N2 and H2 to . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . produce NH 3 , will occur. The result will be a net decrease in the concentrations of N2 and H2 (thus making the denominator of the reaction quotient smaller), and a net increase in the concentration of NH 3 (thus making the numerator larger). When the concentrations of all species are such that Qc is again equal to K c, the system will have established a new equilibrium position, meaning that it will have shifted in one direction or the other, resulting in a new equilibrium concentration for each species. Figure 15.5 shows how the concentrations of N 2 , H 2 , and NH 3 change when N2 is added to the original equiliblium mixture. Conversely, if we were to remove N2 from the original equiliblium mixture, the lower con- centration in the denominator of the reaction quotient would result in Qc being greater than K e . In this case the reaction will shift to the left. That is, the reverse reaction will take place, thereby increasing the concentrations of N2 and H2 and decreasing the concentration of NH 3 until Qc is once again equal to K c . The addition or removal of NH 3 will cause a shift in the equilibrium, too. The addition of NH 3 will cause a shift to the left; the removal of NH 3 will cause a shift to the right. Figure 15.6(a) shows the additions and removals that cause this equiliblium to shift to the right. Figure 15.6(b) shows those that cause it to shift to the left. In essence, a system at equiliblium will respond to addition of a species by consuming some of that species, and it will respond to the removal of a species by producing more of that species. It is important to remember that addition or removal of a species from an equilibrium mixture does not change the value of the equiliblium constant, K. Rather, it changes temporarily the value of the reaction quotient, Q. Furthermore, in order to cause a shift in the equilibrium, the species added or removed must be one that appears in the reaction quotient expression. In the case of a heteroge- SECTION 1S.5 Factors That Affect Chemica l Equilibri um 613 ~ S c 0 .~ ~ '" l:J c " u " 0 U 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 - - - - - - - - . I . Original equili brium mixture Addition Addition ~ ~ [ NH 3f [N z l[ H z I 3 • I Time ( 1.52)z :; = 0.173 (3.51)(1.56)3 Immedi atel y after addition of N z N z NH 3 Hz I [NH Jz ( 164 )z - - - - 0 = 0.297 [Nz J[HzI3 (3.45)(1.38 )3 After eq uilibrium has been reestablished Ad di tion / Nz(g) + 3H z (g) , ' 2N H3(g) Nz(g) + 3H ig) +: , :::::; , :- 2N H 3 (g) ~ / / Remov al Removal Rem ova l (a) (b) neou s equilibrium, altering the amount of a so lid or liquid spec i es does n ot cha ng e the p os iti on of th e e quilibrium b eca use doing so does not change the va lue of Q. Sample Problem 15.11 s how s the effects of st re ss on a sys tem at equilibrium . . . Sample Problem 15.11 Hydrogen sulfide (Hz S) is a contaminant commonly found in natural gas. It is removed by reaction with oxygen to produce elemental sulfur. 2H z S( g) + O zCg) :;:, ==' 2S(s ) + 2H z O(g) For each of the fo llowing scenarios, determine whether the equilibrium will shift to the right, sh ift to the left, or neithe r: (a) addition of Oz (g), (b) removal of H 2 S( g), (c) removal of H zO(g), and (d) addition of S(s). Strategy Use Le Chiitelier's principle to pr edi ct the direction of shift for each case. Remember that the position of th e equilibrium is o nl y changed by the addition or removal of a species that appears in the reaction quotient expression. Setup Begin by writing the reaction quotient expression: [H z O] z Qc = ? [J:lzSnO z] Because sulfur is a solid, it does not appear in the expression. Changes in the concentrati on of any of the other species will cause a change in the equilibrium position. Addition of a reactant or removal of a product that appears in the expression for Qc wi ll shift the equilibrium to the right: Addition Addition \ \ 2H2S(g) + 0 2(g) 2S(s) + 2H20(g) , - \ Removal (Continued) Figure 15.5 Adding more of a reactant to a system at equilibrium causes the equilibrium position to shift toward product. Th e system responds to the addition of N z by consuming some of the added N z (and some of the other reactant, H z) to produce more NH 3 . Figure 15.6 (a) Addition of a reactant or removal of a product will cause an equilibrium to shift to the right. (b) Addition of a product or rem oval of a reactant will cause an equilibrium to shift to the l ef t. [...]... reaction between CO 2 and water), (g) temperature is increased? 15.64 + Oig) :;:.==' 2NO(g) (a) Calculate the partial pressure of NO under these conditions if the partial pressures of nitrogen and oxygen are 3.0 and 0.012 atm, respectively (b) Repeat the calculation for atmospheric conditions where the partial pressures of nitrogen and oxygen are 0.78 and 0.21 atm and the temperature is 25°e (The Kp for the... constant Kp for the reaction At a certain temperature and a total pressure of 1.2 atm, the partial pressures of an equilibrium mixture 2A(g) :;::.====:' B(g) are PA = 0.60 atm and P B = 0.60 atm (a) Calculate the K p 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? 15.74 629 The decomposition of ammonium... Hb0 2 is oxyhemoglobin, the hemoglobin-oxygen complex that actually transports oxygen to tissues The equilibrium expression for this process is [Hb0 2] Kc = -[Hb ] [- ?] - OAt an altitude of 3 km, the partial pressure of oxygen is only about 0.14 atm, compared with 0.20 atrn at sea level Accorcling to Le Chfttelier's principle, a decrease in oxygen concentration will shift the hemoglobin-oxyhemoglobin... equilibrium constants as dimensionless quantities The reason is as follows: Plior to being used in an equilibrium expression, each molar concentration is divided by a reference concentration of 1 M, and each partial pressure is divided by a reference pressure of 1 atm The reference concentration (1 M) and reference pressure (1 atm) are known as the standard states for aqueous and gaseous species, respectively... 4 and N0 2 (at 25 °C) are 0.491 M and 0.0475 M, respectively We can calculate the value of Kc as follows: 0.0475 M ( K = c 1M 0.491M 1M 2 = 4.60 X 10- 3 We can then convert the molar concentrations to partial pressures using the ideal gas equation: I1NO P NO = 2 V 2 RT = [N0 2 ](0.08206)(298) Finally, we can calculate the value of Kp as = 1.16 atm ! 1.16 atm 2 I atm Kp = :: - = 0.112 12.0 atm I atm... is the reason why there are no units associated with equilibrium constants Division of each entry in an equilibrium expression by a reference value actually enables us to use molar concentrations and partial pressures in the same equilibrium expression For example, we can write the equilibrium expression for the reaction as [Zn2+] K= PH2 1M 1 atm [H+] 2 1M Each of the entries in the expression is divided... of reactants and products • Equilibrium expressions that contain only gases can be written either as Ke expressions or as K p expressions Kp expressions have the same fonn as Kc expressions but contain partial pressures rather than molar concentrations The reaction quotient, Q, can also be expressed in terms of the pressures of products and reactants In this case it is labeled Qp Equilibrium is the condition... < 10- 3) indicates that reactants are favored at equilibrium Section 15.5 • According to Le Chiitelier's principle, a system at equilibrium will react to stress by shifting in the direction that will partially offset the effect of the stress ~ • The stresses that can be applied to a system at equilibrium include the addition or removal of a substance, changes in the volume of the reaction vessel, and... 3H2 (g) :;:.=~ 2NH3(g) (b) ~Nig) + ~H2(g) • NH 3(g) 15.28 + K~ = 6.5 X 10- 2 K~ = 6.1 X 10- 5 Calculate the equilibrium constant for the following reaction at the same temperature: If the equilibrium partial pressures of N 2, O 2, and NO are 0 15 , 0.33, and 0.050 atm, respectively, at 2200°C, what is Kp? 15.27 + HC 2 0 4 (aq) At equilibrium, the pressure of the reacting rrilxture CaC0 3(s) :;:.=~... catalyst have any effects on the position of an equilibrium? At 1000 K, a sample of pure NO z gas decomposes: 2NO z(g) +.=:!:' 2NO(g) + OzCg) The equilibrium constant Kp is 158 Analysis shows that the partial pressure of O 2 is 0.25 atm at equilibrium Calculate the pressure of NO and NO z in the mixture 15.43 The equilibrium constant Kc for the reaction Hz(g) + BrzCg) +.=:!:' 2HBr(g) is 2.18 X 106 at . is H 2 (g) + 12(g) , 2HI(g) • Ini tial partial pressure (atm): 5.75 5.75 0 Change in partial pressure (atm): - x - x +2x Equilibrium partial pressure (atm): 5.75 - x 5.75 - x 2x. temperature is 54.3. Determine the equilibrium partial pressures of Hb Ib and HI. Strategy Construct an equilibrium table to determine the equilibrium partial pressures. Setup The equilibrium. 9.369x x = 4.52 The equilibrium partial pressures are PH = PI = 5.75 - 4.52 = 1.23 atm, and Pm = 9.04 atm. 2 2 • Practice Problem A Determine the equilibrium partial pressures of H 2 ,