CHAPTER 15 CHEMICAL EQUILIBRIUM 15.7 (a) 2 2 c 2 2 [CO] [O ] [CO ] =K 2 2 2 CO O 2 CO ⋅ = P PP K P (b) 2 3 c 3 2 [O ] [O ] =K 3 2 2 O 3 O = P P K P (c) 2 c 2 [COCl ] [CO][Cl ] =K 2 2 COCl CO Cl = ⋅ P P K PP (d) 2 c 2 [CO][H ] [H O] =K 2 2 CO H HO ⋅ = P PP K P (e) c [H ][HCOO ] [HCOOH] +− =K (f) c2 [O ]=K 2 O = P K P 15.8 (a) 22 CO H O = P KPP (b) 2 2 2 O SO = P K PP 15.9 (a) 3 22 2 2 NH 3 c 27 27 22 NO H [NH ] [NO ] [H ] == P P KK PP (b) 2 2 2 2 SO 2 c 33 2 O [SO ] [O ] == P P KK P (c) 2 22 CO c 2CO [CO] [CO ] == P P KK P (d) 65 c 65 [C H COO ][H ] [C H COOH] −+ =K 15.11 c [B] [A] =K (1) With K c = 10, products are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of B to be 10 times that of A at equilibrium. Choice (a) is the best choice with 10 B molecules and 1 A molecule. CHAPTER 15: CHEMICAL EQUILIBRIUM 315 (2) With K c = 0.10, reactants are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of A to be 10 times that of B at equilibrium. Choice (d) is the best choice with 10 A molecules and 1 B molecule. You can calculate K c in each case without knowing the volume of the container because the mole ratio between A and B is the same. Volume will cancel from the K c expression. Only moles of each component are needed to calculate K c . 15.12 Note that we are comparing similar reactions at equilibrium – two reactants producing one product, all with coefficients of one in the balanced equation. (a) The reaction, A + C U AC has the largest equilibrium constant. Of the three diagrams, there is the most product present at equilibrium. (b) The reaction, A + D U AD has the smallest equilibrium constant. Of the three diagrams, there is the least amount of product present at equilibrium. 15.13 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 34 11 4.17 10 − == = × × 33 '2.401 K K 15.14 The problem states that the system is at equilibrium, so we simply substitute the equilibrium concentrations into the equilibrium constant expression to calculate K c . Step 1: Calculate the concentrations of the components in units of mol/L. The molarities can be calculated by simply dividing the number of moles by the volume of the flask. 2 2.50 mol [H ] 0.208 12.0 L ==M 5 6 2 1.35 10 mol [S ] 1.13 10 12.0 L − − × ==×M 2 8.70 mol [H S] 0.725 12.0 L ==M Step 2: Once the molarities are known, K c can be found by substituting the molarities into the equilibrium constant expression. 2 2 2 226 22 [H S] (0.725) [H ] [S ] (0.208) (1.13 10 ) − == = × 7 c 1.08 10K × If you forget to convert moles to moles/liter, will you get a different answer? Under what circumstances will the two answers be the same? CHAPTER 15: CHEMICAL EQUILIBRIUM 316 15.15 Using Equation (15.5) of the text: K P = K c (0.0821 T) Δn where, Δn = 2 − 3 = −1 and T = (1273 + 273) K = 1546 K K P = (2.24 × 10 22 )(0.0821 × 1546) −1 = 1.76 × 10 20 15.16 Strategy: The relationship between K c and K P is given by Equation (15.5) of the text. What is the change in the number of moles of gases from reactant to product? Recall that Δn = moles of gaseous products − moles of gaseous reactants What unit of temperature should we use? Solution: The relationship between K c and K P is given by Equation (15.5) of the text. K P = K c (0.0821 T) Δn Rearrange the equation relating K P and K c , solving for K c . c (0.0821 ) Δ = P n K K T Because T = 575 K and Δn = 3 − 2 = 1, we have: 4 5.0 10 (0.0821)(575 K) (0.0821 ) − Δ × == = 5 c 1.1 10 P n K T K − × 15.17 We can write the equilibrium constant expression from the balanced equation and substitute in the pressures. 22 2 2 NO NO (0.050) (0.15)(0.33) == =0.051 P PP P K Do we need to know the temperature? 15.18 The equilibrium constant expressions are: (a) 2 3 c 3 22 [NH ] [N ][H ] =K (b) 31 22 3 c 22 [NH ] [N ] [H ] =K Substituting the given equilibrium concentration gives: (a) 2 3 (0.25) (0.11)(1.91) == c 0.082K (b) 31 22 (0.25) (0.11) (1.91) == c 0.29K Is there a relationship between the K c values from parts (a) and (b)? CHAPTER 15: CHEMICAL EQUILIBRIUM 317 15.19 The equilibrium constant expression for the two forms of the equation are: 2 ' 2 cc 2 2 [I ][I] and [I ] [I] ==KK The relationship between the two equilibrium constants is '4 c 5 c 11 2.6 10 3.8 10 − == =× × K K K P can be found as shown below. K P = K c '(0.0821 T) Δn = (2.6 × 10 4 )(0.0821 × 1000) −1 = 3.2 × 10 2 15.20 Because pure solids do not enter into an equilibrium constant expression, we can calculate K P directly from the pressure that is due solely to CO 2 (g). 2 CO = 0.105=P P K Now, we can convert K P to K c using the following equation. K P = K c (0.0821 T) Δn c (0.0821 ) Δ = P n K K T (1 0) 0.105 (0.0821 623) − == × 3 c 2.05 10K − × 15.21 We substitute the given pressures into the reaction quotient expression. 32 5 PCl Cl PCl (0.223)(0.111) 0.140 (0.177) == = P PP Q P The calculated value of Q P is less than K P for this system. The system will change in a way to increase Q P until it is equal to K P . To achieve this, the pressures of PCl 3 and Cl 2 must increase, and the pressure of PCl 5 must decrease. Could you actually determine the final pressure of each gas? 15.22 Strategy: Because they are constant quantities, the concentrations of solids and liquids do not appear in the equilibrium constant expressions for heterogeneous systems. The total pressure at equilibrium that is given is due to both NH 3 and CO 2 . Note that for every 1 atm of CO 2 produced, 2 atm of NH 3 will be produced due to the stoichiometry of the balanced equation. Using this ratio, we can calculate the partial pressures of NH 3 and CO 2 at equilibrium. Solution: The equilibrium constant expression for the reaction is 2 3 2 CO NH = P KPP The total pressure in the flask (0.363 atm) is a sum of the partial pressures of NH 3 and CO 2 . 32 TNHCO 0.363 atm=+=PP P CHAPTER 15: CHEMICAL EQUILIBRIUM 318 Let the partial pressure of CO 2 = x. From the stoichiometry of the balanced equation, you should find that 32 NH CO 2.=PP Therefore, the partial pressure of NH 3 = 2x. Substituting into the equation for total pressure gives: 32 TNHCO 23=+=+=PP P xx x 3x = 0.363 atm 2 CO 0.121 atm==xP 3 NH 2 0.242 atm==Px Substitute the equilibrium pressures into the equilibrium constant expression to solve for K P . 2 3 22 CO NH (0.242) (0.121)== = 3 7.09 10PP P K − × 15.23 Of the original 1.05 moles of Br 2 , 1.20% has dissociated. The amount of Br 2 dissociated in molar concentration is: 2 1.05 mol [Br ] 0.0120 0.0129 0.980 L =× =M Setting up a table: Br 2 (g) U 2Br(g) Initial (M): 1.05 mol 1.07 0.980 L = M 0 Change (M): −0.0129 +2(0.0129) Equilibrium (M): 1.06 0.0258 22 2 [Br] (0.0258) [Br ] 1.06 − == =× 4 c 6.3 10K 15.24 If the CO pressure at equilibrium is 0.497 atm, the balanced equation requires the chlorine pressure to have the same value. The initial pressure of phosgene gas can be found from the ideal gas equation. 2 (3.00 10 mol)(0.0821 L atm/mol K)(800 K) 1.31 atm (1.50 L) − ×⋅⋅ == = nRT P V Since there is a 1:1 mole ratio between phosgene and CO, the partial pressure of CO formed (0.497 atm) equals the partial pressure of phosgene reacted. The phosgene pressure at equilibrium is: CO(g) + Cl 2 (g) U COCl 2 (g) Initial (atm): 0 0 1.31 Change (atm): +0.497 +0.497 −0.497 Equilibrium (atm): 0.497 0.497 0.81 The value of K P is then found by substitution. 2 2 COCl 2 CO Cl 0.81 (0.497) ===3.3 P PP P K CHAPTER 15: CHEMICAL EQUILIBRIUM 319 15.25 Let x be the initial pressure of NOBr. Using the balanced equation, we can write expressions for the partial pressures at equilibrium. P NOBr = (1 − 0.34)x = 0.66x P NO = 0.34x P Br 2 = 0.17x The sum of these is the total pressure. 0.66x + 0.34x + 0.17x = 1.17x = 0.25 atm x = 0.21 atm The equilibrium pressures are then P NOBr = 0.66(0.21) = 0.14 atm P NO = 0.34(0.21) = 0.071 atm P Br 2 = 0.17(0.21) = 0.036 atm We find K P by substitution. 2 2 2 NO Br 3 22 NOBr () (0.071) (0.036) 9.3 10 ( ) (0.14) − == =× P PP K P The relationship between K P and K c is given by K P = K c (0.0821 T) Δn We find K c (for this system Δn = +1) 3 1 9.3 10 ( ) (0.0821 298) − − Δ × === =× × 4 c 3.8 10 PP n KK RT RT K 15.26 In this problem, you are asked to calculate K c . Step 1: Calculate the initial concentration of NOCl. We carry an extra significant figure throughout this calculation to minimize rounding errors. 0 2.50 mol [NOCl] 1.667 1.50 L ==M Step 2: Let's represent the change in concentration of NOCl as −2x. Setting up a table: 2NOCl(g) U 2NO(g) + Cl 2 (g) Initial (M): 1.667 0 0 Change (M): −2x +2x +x Equilibrium (M): 1.667 − 2x 2x x If 28.0 percent of the NOCl has dissociated at equilibrium, the amount reacted is: (0.280)(1.667 M) = 0.4668 M CHAPTER 15: CHEMICAL EQUILIBRIUM 320 In the table above, we have represented the amount of NOCl that reacts as 2x. Therefore, 2x = 0.4668 M x = 0.2334 M The equilibrium concentrations of NOCl, NO, and Cl 2 are: [NOCl] = (1.67 − 2x)M = (1.667 − 0.4668)M = 1.200 M [NO] = 2x = 0.4668 M [Cl 2 ] = x = 0.2334 M Step 3: The equilibrium constant K c can be calculated by substituting the above concentrations into the equilibrium constant expression. 2 2 2 22 [NO] [Cl ] (0.4668) (0.2334) [NOCl] (1.200) == = c 0.0353K 15.29 Given: 3 2 2 2 SO 4 2 O SO 5.60 10==× P P K PP Initially, the total pressure is (0.350 + 0.762) atm or 1.112 atm. As the reaction progresses from left to right toward equilibrium there will be a decrease in the number of moles of molecules present. (Note that 2 moles of SO 2 react with 1 mole of O 2 to produce 2 moles of SO 3 , or, at constant pressure, three atmospheres of reactants forms two atmospheres of products.) Since pressure is directly proportional to the number of molecules present, at equilibrium the total pressure will be less than 1.112 atm. 15.30 Strategy: We are given the initial concentrations of the gases, so we can calculate the reaction quotient (Q c ). How does a comparison of Q c with K c enable us to determine if the system is at equilibrium or, if not, in which direction the net reaction will proceed to reach equilibrium? Solution: Recall that for a system to be at equilibrium, Q c = K c . Substitute the given concentrations into the equation for the reaction quotient to calculate Q c . 2 2 30 c 33 20 20 [NH ] [0.48] 0.87 [N ] [H ] [0.60][0.76] == =Q Comparing Q c to K c , we find that Q c < K c (0.87 < 1.2). The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds from left to right (consuming reactants, forming products) to reach equilibrium. Therefore, [NH 3 ] will increase and [N 2 ] and [H 2 ] will decrease at equilibrium. 15.31 The balanced equation shows that one mole of carbon monoxide will combine with one mole of water to form hydrogen and carbon dioxide. Let x be the depletion in the concentration of either CO or H 2 O at equilibrium (why can x serve to represent either quantity?). The equilibrium concentration of hydrogen must then also be equal to x. The changes are summarized as shown in the table H 2 + CO 2 U H 2 O + CO Initial (M): 0 0 0.0300 0.0300 Change (M): +x +x −x −x Equilibrium (M): x x (0.0300 − x) (0.0300 − x) CHAPTER 15: CHEMICAL EQUILIBRIUM 321 The equilibrium constant is: 2 c 22 [H O][CO] 0.534 [H ][CO ] ==K Substituting, 2 2 (0.0300 ) 0.534 − = x x Taking the square root of both sides, we obtain: (0.0300 ) 0.534 0.731 − == x x x = 0.0173 M The number of moles of H 2 formed is: 0.0173 mol/L × 10.0 L = 0.173 mol H 2 15.32 Strategy: The equilibrium constant K P is given, and we start with pure NO 2 . The partial pressure of O 2 at equilibrium is 0.25 atm. From the stoichiometry of the reaction, we can determine the partial pressure of NO at equilibrium. Knowing K P and the partial pressures of both O 2 and NO, we can solve for the partial pressure of NO 2 . Solution: Since the reaction started with only pure NO 2 , the equilibrium concentration of NO must be twice the equilibrium concentration of O 2 , due to the 2:1 mole ratio of the balanced equation. Therefore, the equilibrium partial pressure of NO is (2 × 0.25 atm) = 0.50 atm. We can find the equilibrium NO 2 pressure by rearranging the equilibrium constant expression, then substituting in the known values. 2 2 2 NO O 2 NO = P P P K P 2 2 2 NO O (0.50) (0.25) 158 == = 2 NO 0.020 atm P PP K P 15.33 Notice that the balanced equation requires that for every two moles of HBr consumed, one mole of H 2 and one mole of Br 2 must be formed. Let 2x be the depletion in the concentration of HBr at equilibrium. The equilibrium concentrations of H 2 and Br 2 must therefore each be x. (Why?) The changes are shown in the table H 2 + Br 2 U 2HBr Initial (M): 0 0 0.267 Change (M): +x +x −2x Equilibrium (M): x x (0.267 − 2x) The equilibrium constant relationship is given by: 2 c 22 [HBr] [H ][Br ] =K CHAPTER 15: CHEMICAL EQUILIBRIUM 322 Substitution of the equilibrium concentration expressions gives 2 6 c 2 (0.267 2 ) 2.18 10 − ==× x K x Taking the square root of both sides we obtain: 3 0.267 2 1.48 10 − =× x x x = 1.80 × 10 −4 The equilibrium concentrations are: [H 2 ] = [Br 2 ] = 1.80 × 10 −4 M [HBr] = 0.267 − 2(1.80 × 10 −4 ) = 0.267 M If the depletion in the concentration of HBr at equilibrium were defined as x, rather than 2x, what would be the appropriate expressions for the equilibrium concentrations of H 2 and Br 2 ? Should the final answers be different in this case? 15.34 Strategy: We are given the initial amount of I 2 (in moles) in a vessel of known volume (in liters), so we can calculate its molar concentration. Because initially no I atoms are present, the system could not be at equilibrium. Therefore, some I 2 will dissociate to form I atoms until equilibrium is established. Solution: We follow the procedure outlined in Section 15.3 of the text to calculate the equilibrium concentrations. Step 1: The initial concentration of I 2 is 0.0456 mol/2.30 L = 0.0198 M. The stoichiometry of the problem shows 1 mole of I 2 dissociating to 2 moles of I atoms. Let x be the amount (in mol/L) of I 2 dissociated. It follows that the equilibrium concentration of I atoms must be 2x. We summarize the changes in concentrations as follows: I 2 (g) U 2I(g) Initial (M): 0.0198 0.000 Change (M): −x +2x Equilibrium (M): (0.0198 − x) 2x Step 2: Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 22 5 c 2 [I] (2 ) 3.80 10 [I ] (0.0198 ) − == =× − x K x 4x 2 + (3.80 × 10 −5 )x − (7.52 × 10 −7 ) = 0 The above equation is a quadratic equation of the form ax 2 + bx + c = 0. The solution for a quadratic equation is 2 b b4ac 2a −± − =x CHAPTER 15: CHEMICAL EQUILIBRIUM 323 Here, we have a = 4, b = 3.80 × 10 −5 , and c = −7.52 × 10 −7 . Substituting into the above equation, 552 7 ( 3.80 10 ) (3.80 10 ) 4(4)( 7.52 10 ) 2(4) −− − −× ± × − −× =x 53 ( 3.80 10 ) (3.47 10 ) 8 −− −× ± × =x x = 4.29 × 10 −4 M or x = −4.39 × 10 −4 M The second solution is physically impossible because you cannot have a negative concentration. The first solution is the correct answer. Step 3: Having solved for x, calculate the equilibrium concentrations of all species. [I] = 2x = (2)(4.29 × 10 −4 M) = 8.58 × 10 −4 M [I 2 ] = (0.0198 − x) = [0.0198 − (4.29 × 10 −4 )] M = 0.0194 M Tip: We could have simplified this problem by assuming that x was small compared to 0.0198. We could then assume that 0.0198 − x ≈ 0.0198. By making this assumption, we could have avoided solving a quadratic equation. 15.35 Since equilibrium pressures are desired, we calculate K P . K P = K c (0.0821 T) Δn = (4.63 × 10 −3 )(0.0821 × 800) 1 = 0.304 COCl 2 (g) U CO(g) + Cl 2 (g) Initial (atm): 0.760 0.000 0.000 Change (atm): −x +x +x Equilibrium (atm): (0.760 − x) x x 2 0.304 (0.760 ) = − x x x 2 + 0.304x − 0.231 = 0 x = 0.352 atm At equilibrium: (0.760 0.352)atm=− = 2 COCl 0.408 atmP P CO = 0.352 atm = 2 Cl 0.352 atmP 15.36 (a) The equilibrium constant, K c , can be found by simple substitution. 2 22 [H O][CO] (0.040)(0.050) [CO ][H ] (0.086)(0.045) == = c 0.52K (b) The magnitude of the reaction quotient Q c for the system after the concentration of CO 2 becomes 0.50 mol/L, but before equilibrium is reestablished, is: CHAPTER 15: CHEMICAL EQUILIBRIUM 324 c (0.040)(0.050) 0.089 (0.50)(0.045) ==Q The value of Q c is smaller than K c ; therefore, the system will shift to the right, increasing the concentrations of CO and H 2 O and decreasing the concentrations of CO 2 and H 2 . Let x be the depletion in the concentration of CO 2 at equilibrium. The stoichiometry of the balanced equation then requires that the decrease in the concentration of H 2 must also be x, and that the concentration increases of CO and H 2 O be equal to x as well. The changes in the original concentrations are shown in the table. CO 2 + H 2 U CO + H 2 O Initial (M): 0.50 0.045 0.050 0.040 Change (M): −x −x +x +x Equilibrium (M): (0.50 − x) (0.045 − x) (0.050 + x) (0.040 + x) The equilibrium constant expression is: 2 c 22 [H O][CO] (0.040 )(0.050 ) 0.52 [CO ][H ] (0.50 )(0.045 ) ++ == = −− xx K 0.52(x 2 − 0.545x + 0.0225) = x 2 + 0.090x + 0.0020 0.48x 2 + 0.373x − (9.7 × 10 −3 ) = 0 The positive root of the equation is x = 0.025. The equilibrium concentrations are: [CO 2 ] = (0.50 − 0.025) M = 0.48 M [H 2 ] = (0.045 − 0.025) M = 0.020 M [CO] = (0.050 + 0.025) M = 0.075 M [H 2 O] = (0.040 + 0.025) M = 0.065 M 15.37 The equilibrium constant expression for the system is: 2 2 CO CO () = P P K P The total pressure can be expressed as: 2 total CO CO = +PPP If we let the partial pressure of CO be x, then the partial pressure of CO 2 is: 2 CO total (4.50 )atm=−=−PPx x Substitution gives the equation: 2 2 2 CO CO () 1.52 (4.50 ) == = − P P x K Px This can be rearranged to the quadratic: x 2 + 1.52x − 6.84 = 0 CHAPTER 15: CHEMICAL EQUILIBRIUM 325 The solutions are x = 1.96 and x = −3.48; only the positive result has physical significance (why?). The equilibrium pressures are P CO = x = 1.96 atm (4.50 1.96)=−= 2 CO 2.54 atmP 15.38 The initial concentrations are [H 2 ] = 0.80 mol/5.0 L = 0.16 M and [CO 2 ] = 0.80 mol/5.0 L = 0.16 M. H 2 (g) + CO 2 (g) U H 2 O(g) + CO(g) Initial (M): 0.16 0.16 0.00 0.00 Change (M): −x −x +x +x Equilibrium (M): 0.16 − x 0.16 − x x x 2 2 c 2 22 [H O][CO] 4.2 [H ][CO ] (0.16 ) === − x K x Taking the square root of both sides, we obtain: 2.0 0.16 = − x x x = 0.11 M The equilibrium concentrations are: [H 2 ] = [CO 2 ] = (0.16 − 0.11) M = 0.05 M [H 2 O] = [CO] = 0.11 M 15.43 (a) Addition of more Cl 2 (g) (a reactant) would shift the position of equilibrium to the right. (b) Removal of SO 2 Cl 2 (g) (a product) would shift the position of equilibrium to the right. (c) Removal of SO 2 (g) (a reactant) would shift the position of equilibrium to the left. 15.44 (a) Removal of CO 2 (g) from the system would shift the position of equilibrium to the right. (b) Addition of more solid Na 2 CO 3 would have no effect. [Na 2 CO 3 ] does not appear in the equilibrium constant expression. (c) Removal of some of the solid NaHCO 3 would have no effect. Same reason as (b). 15.45 (a) This reaction is endothermic. (Why?) According to Section 15.4 of the text, an increase in temperature favors an endothermic reaction, so the equilibrium constant should become larger. (b) This reaction is exothermic. Such reactions are favored by decreases in temperature. The magnitude of K c should decrease. (c) In this system heat is neither absorbed nor released. A change in temperature should have no effect on the magnitude of the equilibrium constant. 15.46 Strategy: A change in pressure can affect only the volume of a gas, but not that of a solid or liquid because solids and liquids are much less compressible. The stress applied is an increase in pressure. According to Le Châtelier's principle, the system will adjust to partially offset this stress. In other words, the system will CHAPTER 15: CHEMICAL EQUILIBRIUM 326 adjust to decrease the pressure. This can be achieved by shifting to the side of the equation that has fewer moles of gas. Recall that pressure is directly proportional to moles of gas: PV = nRT so P ∝ n. Solution: (a) Changes in pressure ordinarily do not affect the concentrations of reacting species in condensed phases because liquids and solids are virtually incompressible. Pressure change should have no effect on this system. (b) Same situation as (a). (c) Only the product is in the gas phase. An increase in pressure should favor the reaction that decreases the total number of moles of gas. The equilibrium should shift to the left, that is, the amount of B should decrease and that of A should increase. (d) In this equation there are equal moles of gaseous reactants and products. A shift in either direction will have no effect on the total number of moles of gas present. There will be no change when the pressure is increased. (e) A shift in the direction of the reverse reaction (left) will have the result of decreasing the total number of moles of gas present. 15.47 (a) A pressure increase will favor the reaction (forward or reverse?) that decreases the total number of moles of gas. The equilibrium should shift to the right, i.e., more I 2 will be produced at the expense of I. (b) If the concentration of I 2 is suddenly altered, the system is no longer at equilibrium. Evaluating the magnitude of the reaction quotient Q c allows us to predict the direction of the resulting equilibrium shift. The reaction quotient for this system is: 20 c 2 0 [I ] [I] =Q Increasing the concentration of I 2 will increase Q c . The equilibrium will be reestablished in such a way that Q c is again equal to the equilibrium constant. More I will form. The system shifts to the left to establish equilibrium. (c) The forward reaction is exothermic. A decrease in temperature will shift the system to the right to reestablish equilibrium. 15.48 Strategy: (a) What does the sign of ΔH° indicate about the heat change (endothermic or exothermic) for the forward reaction? (b) The stress is the addition of Cl 2 gas. How will the system adjust to partially offset the stress? (c) The stress is the removal of PCl 3 gas. How will the system adjust to partially offset the stress? (d) The stress is an increase in pressure. The system will adjust to decrease the pressure. Remember, pressure is directly proportional to moles of gas. (e) What is the function of a catalyst? How does it affect a reacting system not at equilibrium? at equilibrium? Solution: (a) The stress applied is the heat added to the system. Note that the reaction is endothermic (ΔH° > 0). Endothermic reactions absorb heat from the surroundings; therefore, we can think of heat as a reactant. heat + PCl 5 (g) U PCl 3 (g) + Cl 2 (g) The system will adjust to remove some of the added heat by undergoing a decomposition reaction (from left to right) (b) The stress is the addition of Cl 2 gas. The system will shift in the direction to remove some of the added Cl 2 . The system shifts from right to left until equilibrium is reestablished. CHAPTER 15: CHEMICAL EQUILIBRIUM 327 (c) The stress is the removal of PCl 3 gas. The system will shift to replace some of the PCl 3 that was removed. The system shifts from left to right until equilibrium is reestablished. (d) The stress applied is an increase in pressure. The system will adjust to remove the stress by decreasing the pressure. Recall that pressure is directly proportional to the number of moles of gas. In the balanced equation we see 1 mole of gas on the reactants side and 2 moles of gas on the products side. The pressure can be decreased by shifting to the side with the fewer moles of gas. The system will shift from right to left to reestablish equilibrium. (e) The function of a catalyst is to increase the rate of a reaction. If a catalyst is added to the reacting system not at equilibrium, the system will reach equilibrium faster than if left undisturbed. If a system is already at equilibrium, as in this case, the addition of a catalyst will not affect either the concentrations of reactant and product, or the equilibrium constant. 15.49 (a) Increasing the temperature favors the endothermic reaction so that the concentrations of SO 2 and O 2 will increase while that of SO 3 will decrease. (b) Increasing the pressure favors the reaction that decreases the number of moles of gas. The concentration of SO 3 will increase. (c) Increasing the concentration of SO 2 will lead to an increase in the concentration of SO 3 and a decrease in the concentration of O 2 . (d) A catalyst has no effect on the position of equilibrium. (e) Adding an inert gas at constant volume has no effect on the position of equilibrium. 15.50 There will be no change in the pressures. A catalyst has no effect on the position of the equilibrium. 15.51 (a) If helium gas is added to the system without changing the pressure or the temperature, the volume of the container must necessarily be increased. This will decrease the partial pressures of all the reactants and products. A pressure decrease will favor the reaction that increases the number of moles of gas. The position of equilibrium will shift to the left. (b) If the volume remains unchanged, the partial pressures of all the reactants and products will remain the same. The reaction quotient, Q c , will still equal the equilibrium constant, and there will be no change in the position of equilibrium. 15.52 For this system, K P = [CO 2 ]. This means that to remain at equilibrium, the pressure of carbon dioxide must stay at a fixed value as long as the temperature remains the same. (a) If the volume is increased, the pressure of CO 2 will drop (Boyle's law, pressure and volume are inversely proportional). Some CaCO 3 will break down to form more CO 2 and CaO. (Shift right) (b) Assuming that the amount of added solid CaO is not so large that the volume of the system is altered significantly, there should be no change at all. If a huge amount of CaO were added, this would have the effect of reducing the volume of the container. What would happen then? (c) Assuming that the amount of CaCO 3 removed doesn't alter the container volume significantly, there should be no change. Removing a huge amount of CaCO 3 will have the effect of increasing the container volume. The result in that case will be the same as in part (a). (d) The pressure of CO 2 will be greater and will exceed the value of K P . Some CO 2 will combine with CaO to form more CaCO 3 . (Shift left) CHAPTER 15: CHEMICAL EQUILIBRIUM 328 (e) Carbon dioxide combines with aqueous NaOH according to the equation CO 2 (g) + NaOH(aq) → NaHCO 3 (aq) This will have the effect of reducing the CO 2 pressure and causing more CaCO 3 to break down to CO 2 and CaO. (Shift right) (f) Carbon dioxide does not react with hydrochloric acid, but CaCO 3 does. CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + CO 2 (g) + H 2 O(l) The CO 2 produced by the action of the acid will combine with CaO as discussed in (d) above. (Shift left) (g) This is a decomposition reaction. Decomposition reactions are endothermic. Increasing the temperature will favor this reaction and produce more CO 2 and CaO. (Shift right) 15.53 (i) The temperature of the system is not given. (ii) It is not stated whether the equilibrium constant is K P or K c (would they be different for this reaction?). (iii) A balanced equation is not given. (iv) The phases of the reactants and products are not given. 15.54 (a) Since the total pressure is 1.00 atm, the sum of the partial pressures of NO and Cl 2 is 1.00 atm − partial pressure of NOCl = 1.00 atm − 0.64 atm = 0.36 atm The stoichiometry of the reaction requires that the partial pressure of NO be twice that of Cl 2 . Hence, the partial pressure of NO is 0.24 atm and the partial pressure of Cl 2 is 0.12 atm. (b) The equilibrium constant K P is found by substituting the partial pressures calculated in part (a) into the equilibrium constant expression. 2 2 2 NO Cl 22 NOCl (0.24) (0.12) (0.64) == =0.017 PP P P K 15.55 Since K P increases with temperature, it is an endothermic reaction. 15.56 The equilibrium expression for this system is given by: 22 CO H O = P KPP (a) In a closed vessel the decomposition will stop when the product of the partial pressures of CO 2 and H 2 O equals K P . Adding more sodium bicarbonate will have no effect. (b) In an open vessel, CO 2 (g) and H 2 O(g) will escape from the vessel, and the partial pressures of CO 2 and H 2 O will never become large enough for their product to equal K P . Therefore, equilibrium will never be established. Adding more sodium bicarbonate will result in the production of more CO 2 and H 2 O. 15.57 The relevant relationships are: 2 c [B] [A] =K and 2 B A = P P K P K P = K c (0.0821 T) Δn = K c (0.0821 T) Δn = +1 CHAPTER 15: CHEMICAL EQUILIBRIUM 329 We set up a table for the calculated values of K c and K P . T (°C) K c K P 200 2 (0.843) 56.9 (0.0125) = 56.9(0.0821 × 473) = 2.21 × 10 3 300 2 (0.764) 3.41 (0.171) = 3.41(0.0821 × 573) = 1.60 × 10 2 400 2 (0.724) 2.10 (0.250) = 2.10(0.0821 × 673) = 116 Since K c (and K P ) decrease with temperature, the reaction is exothermic. 15.58 (a) The equation that relates K P and K c is: K P = K c (0.0821 T) Δn For this reaction, Δn = 3 − 2 = 1 42 210 (0.0821 ) (0.0821 298) − × == = × 44 c 810 P K T K − × (b) Because of a very large activation energy, the reaction of hydrogen with oxygen is infinitely slow without a catalyst or an initiator. The action of a single spark on a mixture of these gases results in the explosive formation of water. 15.59 Using data from Appendix 2 we calculate the enthalpy change for the reaction. fff2 2 (NOCl) 2 (NO) (Cl ) 2(51.7 kJ/mol) 2(90.4 kJ/mol) (0) 77.4 kJ/molΔ°=Δ −Δ −Δ = − − =−HH H H ooo The enthalpy change is negative, so the reaction is exothermic. The formation of NOCl will be favored by low temperature. A pressure increase favors the reaction forming fewer moles of gas. The formation of NOCl will be favored by high pressure. 15.60 (a) Calculate the value of K P by substituting the equilibrium partial pressures into the equilibrium constant expression. B 22 A (0.60) (0.60) == =1.7 P P P K (b) The total pressure is the sum of the partial pressures for the two gaseous components, A and B. We can write: P A + P B = 1.5 atm and P B = 1.5 − P A CHAPTER 15: CHEMICAL EQUILIBRIUM 330 Substituting into the expression for K P gives: A 2 A (1.5 ) 1.7 − == P P K P 2 AA 1.7 1.5 0+ −=PP Solving the quadratic equation, we obtain: P A = 0.69 atm and by difference, P B = 0.81 atm Check that substituting these equilibrium concentrations into the equilibrium constant expression gives the equilibrium constant calculated in part (a). B 22 A 0.81 1.7 (0.69) == = P P K P 15.61 (a) The balanced equation shows that equal amounts of ammonia and hydrogen sulfide are formed in this decomposition. The partial pressures of these gases must just be half the total pressure, i.e., 0.355 atm. The value of K P is 32 2 NH H S (0.355)===0.126PP P K (b) We find the number of moles of ammonia (or hydrogen sulfide) and ammonium hydrogen sulfide. 3 NH (0.355 atm)(4.000 L) 0.0582 mol (0.0821 L atm/K mol)(297 K) == = ⋅⋅ PV n RT 4 NH HS 1mol 6.1589 g 0.1205 mol (before decomposition) 51.12 g =×=n From the balanced equation the percent decomposed is 0.0582 mol 100% 0.1205 mol ×=48.3% (c) If the temperature does not change, K P has the same value. The total pressure will still be 0.709 atm at equilibrium. In other words the amounts of ammonia and hydrogen sulfide will be twice as great, and the amount of solid ammonium hydrogen sulfide will be: [0.1205 − 2(0.0582)]mol = 0.0041 mol NH 4 HS 15.62 Total number of moles of gas is: 0.020 + 0.040 + 0.96 = 1.02 mol of gas You can calculate the partial pressure of each gaseous component from the mole fraction and the total pressure. NO NO T 0.040 0.20 0.0078 atm 1.02 ==×=PPΧ 22 OOT 0.020 0.20 0.0039 atm 1.02 ==×=PPΧ CHAPTER 15: CHEMICAL EQUILIBRIUM 331 22 NO NO T 0.96 0.20 0.19 atm 1.02 ==×=PPΧ Calculate K P by substituting the partial pressures into the equilibrium constant expression. 2 2 2 2 NO 22 NO O (0.19) (0.0078) (0.0039) == = 5 1.5 10 P PP P K × 15.63 Since the reactant is a solid, we can write: 32 2 NH CO ()= P KPP The total pressure is the sum of the ammonia and carbon dioxide pressures. 32 total NH CO =+PPP From the stoichiometry, 32 NH CO 2=PP Therefore: 22 2 total CO CO CO 2 3 0.318 atm=+==PPPP 2 CO 0.106 atm=P 3 NH 0.212 atm=P Substituting into the equilibrium expression: K P = (0.212) 2 (0.106) = 4.76 × 10 −3 15.64 Set up a table that contains the initial concentrations, the change in concentrations, and the equilibrium concentration. Assume that the vessel has a volume of 1 L. H 2 + Cl 2 U 2HCl Initial (M): 0.47 0 3.59 Change (M): +x +x −2x Equilibrium (M): (0.47 + x) x (3.59 − 2x) Substitute the equilibrium concentrations into the equilibrium constant expression, then solve for x. Since Δn = 0, K c = K P. 22 c 22 [HCl] (3.59 2 ) 193 [H ][Cl ] (0.47 ) − == = + x K xx Solving the quadratic equation, x = 0.10 Having solved for x, calculate the equilibrium concentrations of all species. [H 2 ] = 0.57 M [Cl 2 ] = 0.10 M [HCl] = 3.39 M Since we assumed that the vessel had a volume of 1 L, the above molarities also correspond to the number of moles of each component. CHAPTER 15: CHEMICAL EQUILIBRIUM 332 From the mole fraction of each component and the total pressure, we can calculate the partial pressure of each component. Total number of moles = 0.57 + 0.10 + 3.39 = 4.06 mol 0.57 2.00 4.06 =×= 2 H 0.28 atmP 0.10 2.00 4.06 =×= 2 Cl 0.049 atmP 3.39 2.00 4.06 =×= HCl 1.67 atmP 15.65 (a) From the balanced equation N 2 O 4 U 2NO 2 Initial (mol): 1 0 Change (mol): −α +2α Equilibrium (mol): (1 − α) 2α The total moles in the system = (moles N 2 O 4 + moles NO 2 ) = [(1 − α) + 2α] = 1 + α. If the total pressure in the system is P, then: 24 2 NO NO 12 and 11 −α α == + α+ PPPP 2 24 2 2 2 NO NO 2 1 1 1 ⎛⎞α ⎜⎟ +α ⎝⎠ == ⎛⎞−α ⎜⎟ +α ⎝⎠ P P P K P P 2 4 1 1 ⎛⎞ α ⎜⎟ +α ⎝⎠ == −α − 2 2 4 1 P P K P α α (b) Rearranging the K P expression: 4α 2 P = K P − α 2 K P α 2 (4P + K P ) = K P 2 4 α= + P P K PK 4 α= + P P K PK K P is a constant (at constant temperature). Thus, as P increases, α must decrease, indicating that the system shifts to the left. This is also what one would predict based on the Chatelier's principle. CHAPTER 15: CHEMICAL EQUILIBRIUM 333 15.66 This is a difficult problem. Express the equilibrium number of moles in terms of the initial moles and the change in number of moles (x). Next, calculate the mole fraction of each component. Using the mole fraction, you should come up with a relationship between partial pressure and total pressure for each component. Substitute the partial pressures into the equilibrium constant expression to solve for the total pressure, P T . The reaction is: N 2 + 3 H 2 U 2 NH 3 Initial (mol): 1 3 0 Change (mol): −x −3x 2x Equilibrium (mol): (1 − x) (3 − 3x) 2x 3 3 mol of NH Mole fraction of NH total number of moles = 3 NH 22 (1 ) (3 3 ) 2 4 2 == −+− + − x x x xx x Χ 2 0.21 42 = − x x x = 0.35 mol Substituting x into the following mole fraction equations, the mole fractions of N 2 and H 2 can be calculated. 2 N 1 1 0.35 0.20 42 42(0.35) −− == = x x Χ 2 H 33 33(0.35) 0.59 42 42(0.35) −− == = x x Χ The partial pressures of each component are equal to the mole fraction multiplied by the total pressure. 3 NH T 0.21=PP 2 NT 0.20=PP 2 HT 0.59=PP Substitute the partial pressures above (in terms of P T ) into the equilibrium constant expression, and solve for P T . 3 2 2 2 NH 3 N H = P P K PP 22 4 T 3 TT (0.21) 4.31 10 (0.59 ) (0.20 ) − ×= P PP 4 2 T 1.07 4.31 10 − ×= P P T = 5.0 × 10 1 atm 15.67 For the balanced equation: 2 22 c 2 2 [H ] [S ] [H S] =K 2 2 3 42 c 23 2 [H S] 4.84 10 (2.25 10 ) [H ] 1.50 10 − −− − ⎛⎞ × == ×=×⎜⎟ × ⎝⎠ 3 2 [S ] 2.34 10K M CHAPTER 15: CHEMICAL EQUILIBRIUM 334 15.68 We carry an additional significant figure throughout this calculation to minimize rounding errors. The initial molarity of SO 2 Cl 2 is: 22 22 22 22 1molSO Cl 6.75 g SO Cl 135.0 g SO Cl [SO Cl ] 0.02500 2.00 L × ==M The concentration of SO 2 at equilibrium is: 2 0.0345 mol [SO ] 0.01725 2.00 L ==M Since there is a 1:1 mole ratio between SO 2 and SO 2 Cl 2 , the concentration of SO 2 at equilibrium (0.01725 M) equals the concentration of SO 2 Cl 2 reacted. The concentrations of SO 2 Cl 2 and Cl 2 at equilibrium are: SO 2 Cl 2 (g) U SO 2 (g) + Cl 2 (g) Initial (M): 0.02500 0 0 Change (M): −0.01725 +0.01725 +0.01725 Equilibrium (M): 0.00775 0.01725 0.01725 Substitute the equilibrium concentrations into the equilibrium constant expression to calculate K c . 22 22 [SO ][Cl ] (0.01725)(0.01725) [SO Cl ] (0.00775) == = 2 c 3.84 10K − × 15.69 For a 100% yield, 2.00 moles of SO 3 would be formed (why?). An 80% yield means 2.00 moles × (0.80) = 1.60 moles SO 3 is formed. The amount of SO 2 remaining at equilibrium = (2.00 − 1.60) mol = 0.40 mol The amount of O 2 reacted = 1 2 × (amount of SO 2 reacted) = ( 1 2 × 1.60) mol = 0.80 mol The amount of O 2 remaining at equilibrium = (2.00 − 0.80) mol = 1.20 mol Total moles at equilibrium = moles SO 2 + moles O 2 + moles SO 3 = (0.40 + 1.20 + 1.60 ) mol = 3.20 mol 2 SO total total 0.40 0.125 3.20 ==PP P 2 Ototaltoal 1.20 0.375 3.20 ==PP P 3 SO total total 1.60 0.500 3.20 ==PP P 3 22 2 SO 2 SO O = P P K PP 2 total 2 total total (0.500 ) 0.13 (0.125 ) (0.375 ) = P PP P total = 328 atm CHAPTER 15: CHEMICAL EQUILIBRIUM 335 15.70 I 2 (g) U 2I(g) Assuming 1 mole of I 2 is present originally and α moles reacts, at equilibrium: [I 2 ] = 1 − α, [I] = 2α. The total number of moles present in the system = (1 − α) + 2α = 1 + α. From Problem 15.65(a) in the text, we know that K P is equal to: 2 2 4 1 α = −α P K P (1) If there were no dissociation, then the pressure would be: 1mol L atm 1.00 g 0.0821 (1473 K) 253.8 g mol K 0.953 atm 0.500 L ⎛⎞⋅ ⎛⎞ × ⎜⎟⎜⎟ ⋅ ⎝⎠ ⎝⎠ == = nRT P V observed pressure 1.51 atm 1 calculated pressure 0.953 atm 1 + α == α = 0.584 Substituting in equation (1) above: 22 4 (4)(0.584) 1.51 11.584) α == ×= −α − 3.13P P K 15.71 Panting decreases the concentration of CO 2 because CO 2 is exhaled during respiration. This decreases the concentration of carbonate ions, shifting the equilibrium to the left. Less CaCO 3 is produced. Two possible solutions would be either to cool the chickens' environment or to feed them carbonated water. 15.72 According to the ideal gas law, pressure is directly proportional to the concentration of a gas in mol/L if the reaction is at constant volume and temperature. Therefore, pressure may be used as a concentration unit. The reaction is: N 2 + 3H 2 U 2NH 3 Initial (atm): 0.862 0.373 0 Change (atm): −x −3x +2x Equilibrium (atm): (0.862 − x) (0.373 − 3x) 2x 3 2 2 2 NH 3 N H = P P K PP 2 4 3 (2 ) 4.31 10 (0.373 3 ) (0.862 ) − ×= −− x x x At this point, we need to make two assumptions that 3x is very small compared to 0.373 and that x is very small compared to 0.862. Hence, 0.373 − 3x ≈ 0.373 and 0.862 − x ≈ 0.862 CHAPTER 15: CHEMICAL EQUILIBRIUM 336 2 4 3 (2 ) 4.31 10 (0.373) (0.862) − ×≈ x Solving for x. x = 2.20 × 10 −3 atm The equilibrium pressures are: 3 [0.862 (2.20 10 )]atm − =−× = 2 N 0.860 atmP 3 [0.373 (3)(2.20 10 )]atm − =− × = 2 H 0.366 atmP 3 (2)(2.20 10 atm) − =× = 3 3 NH 4.40 10 atmP − × Was the assumption valid that we made above? Typically, the assumption is considered valid if x is less than 5 percent of the number that we said it was very small compared to. Is this the case? 15.73 (a) The sum of the mole fractions must equal one. 2 CO CO 1+=ΧΧ and 2 CO CO 1= −Χ Χ According to the hint, the average molar mass is the sum of the products of the mole fraction of each gas and its molar mass. (Χ CO × 28.01 g) + [(1 − Χ CO ) × 44.01 g] = 35 g Solving, X CO = 0.56 and = 2 CO 0.44Χ (b) Solving for the pressures 2 total CO CO 11 atm=P=P+P P CO = Χ CO P total = 0.56 × 11 atm = 6.2 atm 22 CO CO total (0.44)(11 atm) 4.8 atm= ==PPΧ 2 2 2 CO CO (6.2) 4.8 == =8.0 P P P K 15.74 (a) The equation is: fructose U glucose Initial (M): 0.244 0 Change (M): −0.131 +0.131 Equilibrium (M): 0.113 0.131 Calculating the equilibrium constant, [glucose] 0.131 [fructose] 0.113 === c 1.16K (b) amount of fructose converted 100% original amount of fructose =×Percent converted 0.131 100% 0.244 =×=53.7% CHAPTER 15: CHEMICAL EQUILIBRIUM 337 15.75 If you started with radioactive iodine in the solid phase, then you should find radioactive iodine in the vapor phase at equilibrium. Conversely, if you started with radioactive iodine in the vapor phase, you should find radioactive iodine in the solid phase. Both of these observations indicate a dynamic equilibrium between solid and vapor phase. 15.76 (a) There is only one gas phase component, O 2 . The equilibrium constant is simply 2 O = P K P so, K P = 0.49 (b) From the ideal gas equation, we can calculate the moles of O 2 produced by the decomposition of CuO. 2 3 O 2 (0.49 atm)(2.0 L) 9.2 10 mol O (0.0821 L atm/K mol)(1297 K) − == =× ⋅⋅ PV n RT From the balanced equation, 32 2 2 4molCuO (9.2 10 mol O ) 3.7 10 mol CuO decomposed 1molO −− ××=× amount of CuO lost original amount of CuO =Fraction of CuO decomposed 2 3.7 10 mol 0.16 mol − × ==0.23 (c) If a 1.0 mole sample were used, the pressure of oxygen would still be the same (0.49 atm) and it would be due to the same quantity of O 2 . Remember, a pure solid does not affect the equilibrium position. The moles of CuO lost would still be 3.7 × 10 −2 mol. Thus the fraction decomposed would be: 0.037 1.0 = 0.037 (d) If the number of moles of CuO were less than 3.7 × 10 −2 mol, the equilibrium could not be established because the pressure of O 2 would be less than 0.49 atm. Therefore, the smallest number of moles of CuO needed to establish equilibrium must be slightly greater than 3.7 × 10 −2 mol. 15.77 If there were 0.88 mole of CO 2 initially and at equilibrium there were 0.11 mole, then (0.88 − 0.11) mole = 0.77 mole reacted. NO + CO 2 U NO 2 + CO Initial (mol): 3.9 0.88 0 0 Change (mol): −0.77 −0.77 +0.77 +0.77 Equilibrium (mol): (3.9 − 0.77) 0.11 0.77 0.77 Solving for the equilibrium constant: (0.77)(0.77) (3.9 0.77)(0.11) == − c 1.7K In the balanced equation there are equal number of moles of products and reactants; therefore, the volume of the container will not affect the calculation of K c . We can solve for the equilibrium constant in terms of moles. CHAPTER 15: CHEMICAL EQUILIBRIUM 338 15.78 We first must find the initial concentrations of all the species in the system. 20 0.714 mol [H ] 0.298 2.40 L ==M 20 0.984 mol [I ] 0.410 2.40 L ==M 0 0.886 mol [HI] 0.369 2.40 L ==M Calculate the reaction quotient by substituting the initial concentrations into the appropriate equation. 2 2 0 c 20 20 [HI] (0.369) 1.11 [H ] [I ] (0.298)(0.410) == =Q We find that Q c is less than K c . The equilibrium will shift to the right, decreasing the concentrations of H 2 and I 2 and increasing the concentration of HI. We set up the usual table. Let x be the decrease in concentration of H 2 and I 2 . H 2 + I 2 U 2 HI Initial (M): 0.298 0.410 0.369 Change (M): −x −x +2x Equilibrium (M): (0.298 − x) (0.410 − x) (0.369 + 2x) The equilibrium constant expression is: 22 c 22 [HI] (0.369 2 ) 54.3 [H ][I ] (0.298 )(0.410 ) + == = −− x K xx This becomes the quadratic equation 50.3x 2 − 39.9x + 6.48 = 0 The smaller root is x = 0.228 M. (The larger root is physically impossible.) Having solved for x, calculate the equilibrium concentrations. [H 2 ] = (0.298 − 0.228) M = 0.070 M [I 2 ] = (0.410 − 0.228) M = 0.182 M [HI] = [0.369 + 2(0.228)] M = 0.825 M 15.79 Since we started with pure A, then any A that is lost forms equal amounts of B and C. Since the total pressure is P, the pressure of B + C = P − 0.14 P = 0.86 P. The pressure of B = C = 0.43 P. BC A (0.43 )(0.43 ) 0.14 == =1.3 PP PP P K P 15.80 The gas cannot be (a) because the color became lighter with heating. Heating (a) to 150°C would produce some HBr, which is colorless and would lighten rather than darken the gas. The gas cannot be (b) because Br 2 doesn't dissociate into Br atoms at 150°C, so the color shouldn't change. CHAPTER 15: CHEMICAL EQUILIBRIUM 339 The gas must be (c). From 25°C to 150°C, heating causes N 2 O 4 to dissociate into NO 2 , thus darkening the color (NO 2 is a brown gas). N 2 O 4 (g) → 2NO 2 (g) Above 150°C, the NO 2 breaks up into colorless NO and O 2 . 2NO 2 (g) → 2NO(g) + O 2 (g) An increase in pressure shifts the equilibrium back to the left, forming NO 2 , thus darkening the gas again. 2NO(g) + O 2 (g) → 2NO 2 (g) 15.81 Given the following: 2 3 c 3 22 [NH ] 0.65 [N ][H ] ==K (a) Temperature must have units of Kelvin. K P = K c (0.0821 T) Δn K P = (0.65)(0.0821 × 668) (2−4) = 2.2 × 10 −4 (b) Recalling that, forward reverse 1 =K K Therefore, 1 0.65 == ' c 1.5K (c) Since the equation 1 2 N 2 (g) + 3 2 H 2 (g) U NH 3 (g) is equivalent to 1 2 [N 2 (g) + 3H 2 (g) U 2NH 3 (g)] then, K c ' for the reaction: 1 2 N 2 (g) + 3 2 H 2 (g) U NH 3 (g) equals 1 2 c ()K for the reaction: N 2 (g) + 3H 2 (g) U 2NH 3 (g) Thus, 1 2 c ( ) 0.65=== ' c 0.81KK (d) For K P in part (b): K P = (1.5)(0.0821 × 668) +2 = 4.5 × 10 3 and for K P in part (c): K P = (0.81)(0.0821 × 668) −1 = 0.015 CHAPTER 15: CHEMICAL EQUILIBRIUM 340 15.82 Decomposition of N 2 O 4 is an endothermic process, N 2 O 4 U 2NO 2 . As the temperature is increased, the system shifts to the right to reestablish equilibrium. (a) Color deepens (b) increases (c) decreases (d) increases (e) unchanged (conservation of mass) 15.83 The vapor pressure of water is equivalent to saying the partial pressure of H 2 O(g). 2 HO ==0.0231P P K 1 0.0231 (0.0821 ) (0.0821 293) Δ == = × 4 c 9.60 10 P n K T K − × 15.84 Potassium is more volatile than sodium. Therefore, its removal shifts the equilibrium from left to right. 15.85 This problem involves two types of problems: Calculations using density or molar mass and Dalton’s Law of Partial Pressures. We can calculate the average molar mass of the gaseous mixture from the density. = dRT P M Let M be the average molar mass of NO 2 and N 2 O 4 . The above equation becomes: (2.9 g/L)(0.0821 L atm/K mol)(347 K) 1.3 atm ⋅⋅ == dRT P M M = 63.6 g/mol The average molar mass is equal to the sum of the molar masses of each component times the respective mole fractions. Setting this up, we can calculate the mole fraction of each component. 2 2 24 24 NO NO NO NO 63.6 g/mol=+ =ΧΧMM M 22 NO NO (46.01 g/mol) (1 )(92.01 g/mol) 63.6 g/mol+− = 2 NO 0.618=Χ We can now calculate the partial pressure of NO 2 from the mole fraction and the total pressure. 22 NO NO T =PPΧ (0.618)(1.3 atm)== 2 NO 0.80 atmP We can calculate the partial pressure of N 2 O 4 by difference. 24 2 NO T NO =−PPP (1.3 0.80)atm= −= 24 NO 0.50 atmP Finally, we can calculate K P for the dissociation of N 2 O 4 . 2 24 2 2 NO NO (0.80) 0.50 == =1.3 P P P K CHAPTER 15: CHEMICAL EQUILIBRIUM 341 15.86 (a) Since both reactions are endothermic (ΔH° is positive), according to Le Châtelier’s principle the products would be favored at high temperatures. Indeed, the steam-reforming process is carried out at very high temperatures (between 800°C and 1000°C). It is interesting to note that in a plant that uses natural gas (methane) for both hydrogen generation and heating, about one-third of the gas is burned to maintain the high temperatures. In each reaction there are more moles of products than reactants; therefore, we expect products to be favored at low pressures. In reality, the reactions are carried out at high pressures. The reason is that when the hydrogen gas produced is used captively (usually in the synthesis of ammonia), high pressure leads to higher yields of ammonia. (b) (i) The relation between K c and K P is given by Equation (15.5) of the text: K P = K c (0.0821 T) Δn Since Δn = 4 − 2 = 2, we write: K P = (18)(0.0821 × 1073) 2 = 1.4 × 10 5 (ii) Let x be the amount of CH 4 and H 2 O (in atm) reacted. We write: CH 4 + H 2 O U CO + 3H 2 Initial (atm): 15 15 0 0 Change (atm): −x −x +x +3x Equilibrium (atm): 15 − x 15 − x x 3x The equilibrium constant is given by: 2 42 3 CO H CH H O = P PP K PP 34 5 2 ()(3) 27 1.4 10 (15 )(15 ) (15 ) ×= = −− − xx x xx x Taking the square root of both sides, we obtain: 2 2 5.2 3.7 10 15 ×= − x x which can be expressed as 5.2x 2 + (3.7 × 10 2 x) − (5.6 × 10 3 ) = 0 Solving the quadratic equation, we obtain x = 13 atm (The other solution for x is negative and is physically impossible.) At equilibrium, the pressures are: (15 13)= −= 4 CH 2atmP (15 13)= −= 2 HO 2atmP CHAPTER 15: CHEMICAL EQUILIBRIUM 342 P CO = 13 atm 3(13 atm)== 2 H 39 atmP 15.87 (a) shifts to right (b) shifts to right (c) no change (d) no change (e) no change (f) shifts to left (g) shifts to right 15.88 3 NH HClP K =P P 3 NH HCl 2.2 1.1 atm 2 ===PP K P = (1.1)(1.1) = 1.2 15.89 The equilibrium is: N 2 O 4 (g) U 2NO 2 (g) 2 24 2 2 NO NO () 0.15 0.113 0.20 === P P K P Volume is doubled so pressure is halved. Let’s calculate Q P and compare it to K P . 2 0.15 2 0.0563 0.20 2 ⎛⎞ ⎜⎟ ⎝⎠ ==< ⎛⎞ ⎜⎟ ⎝⎠ PP QK Equilibrium will shift to the right. Some N 2 O 4 will react, and some NO 2 will be formed. Let’s let x = amount of N 2 O 4 reacted. N 2 O 4 (g) U 2NO 2 (g) Initial (atm): 0.10 0.075 Change (atm): −x +2x Equilibrium (atm): 0.10 − x 0.075 + 2x Substitute into the K P expression to solve for x. 2 (0.075 2 ) 0.113 0.10 + == − P x K x 4x 2 + 0.413x − 5.67 × 10 −3 = 0 x = 0.0123 At equilibrium: 0.075 2(0.0123) 0.0996=+ = ≈ 2 NO 0.100 atmP 0.10 0.0123=− = 24 NO 0.09 atmP Check: 2 (0.10) 0.111 0.09 == P K close enough to 0.113 CHAPTER 15: CHEMICAL EQUILIBRIUM 343 15.90 (a) React Ni with CO above 50°C. Pump away the Ni(CO) 4 vapor (shift equilibrium to right), leaving the solid impurities behind. (b) Consider the reverse reaction: Ni(CO) 4 (g) → Ni(s) + 4CO(g) ff4 4(CO) [Ni(CO)]Δ°=Δ −ΔHH H oo ΔH° = (4)(−110.5 kJ/mol) − (1)(−602.9 kJ/mol) = 160.9 kJ/mol The decomposition is endothermic, which is favored at high temperatures. Heat Ni(CO) 4 above 200°C to convert it back to Ni. 15.91 (a) Molar mass of PCl 5 = 208.2 g/mol () 1mol L atm 2.50 g 0.0821 523 K 208.2 g mol K 0.500 L ⎛⎞⋅ ⎛⎞ × ⎜⎟⎜⎟ ⋅ ⎝⎠ ⎝⎠ == =1.03 atm nRT V P (b) PCl 5 U PCl 3 + Cl 2 Initial (atm) 1.03 0 0 Change (atm) −x +x +x Equilibrium (atm) 1.03 − x x x 2 1.05 1.03 == − P x K x x 2 + 1.05x − 1.08 = 0 x = 0.639 At equilibrium: 1.03 0.639=− = 5 PCl 0.39 atmP (c) P T = (1.03 − x) + x + x = 1.03 + 0.639 = 1.67 atm (d) 0.639 atm 1.03 atm = 0.620 15.92 (a) K P = P Hg = 0.0020 mmHg = 2.6 × 10 −6 atm = 2.6 × 10 −6 (equil. constants are expressed without units) 6 1 2.6 10 (0.0821 ) (0.0821 299) − Δ × == = × 7 c 1.1 10 P n K T K − × (b) Volume of lab = (6.1 m)(5.3 m)(3.1 m) = 100 m 3 [Hg] = K c CHAPTER 15: CHEMICAL EQUILIBRIUM 344 3 7 3 3 1.1 10 mol 200.6 g 1 L 1 cm 100 m 1L 1mol 0.01m 1000 cm − ⎛⎞× =××××= ⎜⎟ ⎝⎠ Total mass of Hg vapor 2.2 g The concentration of mercury vapor in the room is: 3 3 2.2 g 0.022 g/m 100 m == 3 22 mg/m Yes! This concentration exceeds the safety limit of 0.05 mg/m 3 . Better clean up the spill! 15.93 (a) A catalyst speeds up the rates of the forward and reverse reactions to the same extent. (b) A catalyst would not change the energies of the reactant and product. (c) The first reaction is exothermic. Raising the temperature would favor the reverse reaction, increasing the amount of reactant and decreasing the amount of product at equilibrium. The equilibrium constant, K, would decrease. The second reaction is endothermic. Raising the temperature would favor the forward reaction, increasing the amount of product and decreasing the amount of reactant at equilibrium. The equilibrium constant, K, would increase. (d) A catalyst lowers the activation energy for the forward and reverse reactions to the same extent. Adding a catalyst to a reaction mixture will simply cause the mixture to reach equilibrium sooner. The same equilibrium mixture could be obtained without the catalyst, but we might have to wait longer for equilibrium to be reached. If the same equilibrium position is reached, with or without a catalyst, then the equilibrium constant is the same. 15.94 First, let's calculate the initial concentration of ammonia. 3 3 3 1molNH 14.6 g 17.03 g NH [NH ] 0.214 4.00 L × ==M Let's set up a table to represent the equilibrium concentrations. We represent the amount of NH 3 that reacts as 2x. 2NH 3 (g) U N 2 (g) + 3H 2 (g) Initial (M): 0.214 0 0 Change (M): −2x +x +3x Equilibrium (M): 0.214 − 2x x 3x Substitute into the equilibrium constant expression to solve for x. 3 22 c 2 3 [N ][H ] [NH ] =K 34 22 ()(3) 27 0.83 (0.214 2 ) (0.214 2 ) == −− xx x x x Taking the square root of both sides of the equation gives: 2 5.20 0.91 0.214 2 = − x x Rearranging, 5.20x 2 + 1.82x − 0.195 = 0 CHAPTER 15: CHEMICAL EQUILIBRIUM 345 Solving the quadratic equation gives the solutions: x = 0.086 M and x = −0.44 M The positive root is the correct answer. The equilibrium concentrations are: [NH 3 ] = 0.214 − 2(0.086) = 0.042 M [N 2 ] = 0.086 M [H 2 ] = 3(0.086) = 0.26 M 15.95 Since the catalyst is exposed to the reacting system, it would catalyze the 2A → B reaction. This shift would result in a decrease in the number of gas molecules, so the gas pressure decreases. The piston would be pushed down by the atmospheric pressure. When the cover is over the box, the catalyst is no longer able to favor the forward reaction. To reestablish equilibrium, the B → 2A step would dominate. This would increase the gas pressure so the piston rises and so on. Conclusion: Such a catalyst would result in a perpetual motion machine (the piston would move up and down forever) which can be used to do work without input of energy or net consumption of chemicals. Such a machine cannot exist. 15.96 Initially, at equilibrium: [NO 2 ] = 0.0475 M and [N 2 O 4 ] = 0.491 M. At the instant the volume is halved, the concentrations double. [NO 2 ] = 2(0.0475 M) = 0.0950 M and [N 2 O 4 ] = 2(0.491 M) = 0.982 M. The system is no longer at equilibrium. The system will shift to the left to offset the increase in pressure when the volume is halved. When a new equilibrium position is established, we write: N 2 O 4 U 2NO 2 0.982 M + x 0.0950 M – 2x 2 2 2 c 24 [NO ] (0.0950 2 ) [N O ] (0.982 ) − == + x K x 4x 2 – 0.3846x + 4.478 × 10 −3 = 0 Solving x = 0.0826 M (impossible) and x = 0.0136 M At the new equilibrium, [N 2 O 4 ] = 0.982 + 0.0136 = 0.996 M [NO 2 ] = 0.0950 – (2 × 0.0136) = 0.0678 M As we can see, the new equilibrium concentration of NO 2 is greater than the initial equilibrium concentration (0.0475 M). Therefore, the gases should look darker! 15.97 To determine ΔH°, we need to plot ln K P versus 1/T (y vs. x). ln K P 1/T 4.93 0.00167 1.63 0.00143 −0.83 0.00125 −2.77 0.00111 −4.34 0.00100 CHAPTER 15: CHEMICAL EQUILIBRIUM 346 y = 1.38E+04x - 1.81E+01 -5 -4 -3 -2 -1 0 1 2 3 4 5 0.0010 0.0011 0.0012 0.0013 0.0014 0.0015 0.0016 0.0017 1/T (K −1 ) ln K P The slope of the plot equals −ΔH°/R. 4 1.38 10 K 8.314 J/mol K Δ° ×=− ⋅ H ΔH° = −1.15 × 10 5 J/mol = −115 kJ/mol 15.98 (a) We start by writing the van’t Hoff equation at two different temperatures. 1 1 ln −Δ ° =+ H K C RT 2 2 ln −Δ ° =+ H K C RT 12 12 ln ln −Δ° −Δ° −= H H KK RT RT 1 221 1 ln ⎛⎞Δ° 1 =− ⎜⎟ ⎝⎠ K H K RT T Assuming an endothermic reaction, ΔH° > 0 and T 2 > T 1 . Then, 21 1⎛⎞Δ °1 − ⎜⎟ ⎝⎠ H RT T < 0, meaning that 1 2 ln K K < 0 or K 1 < K 2 . A larger K 2 indicates that there are more products at equilibrium as the temperature is raised. This agrees with LeChatelier’s principle that an increase in temperature favors the forward endothermic reaction. The opposite of the above discussion holds for an exothermic reaction. (b) Treating H 2 O(l) U H 2 O(g) ΔH vap = ? as a heterogeneous equilibrium, 2 HO = P KP. CHAPTER 15: CHEMICAL EQUILIBRIUM 347 We substitute into the equation derived in part (a) to solve for ΔH vap . 1 221 1 ln ⎛⎞Δ °1 =− ⎜⎟ ⎝⎠ K H K RT T 31.82 mmHg 1 ln 92.51 mmHg 8.314 J/mol K 323 K 303 K ⎛⎞Δ° 1 =− ⎜⎟ ⋅ ⎝⎠ H −1.067 = ΔH°(−2.458 × 10 −5 ) ΔH° = 4.34 × 10 4 J/mol = 43.4 kJ/mol 15.99 Using Equation (14.10) of the text, we can calculate k −1 . a /− = E RT kAe Then, we can calculate k 1 using the expression 1 c 1− = k K k (see Section 15.1 of the text) 3 41 10 J/mol (8.314 J/mol K)(298 K) 12 1 1 (1.0 10 s ) ⎛⎞ × −⎜⎟ ⋅ − ⎝⎠ − =×ke k −1 = 6.5 × 10 4 s −1 1 c 1− = k K k 3 1 41 9.83 10 6.5 10 s − ×= × k k 1 = 6.4 × 10 8 s −1 15.100 We start with a table. A 2 + B 2 U 2AB Initial (mol): 1 3 0 Change (mol): 2 − x 2 − x +x Equilibrium (mol): 1 2 − x 3 2 − x x After the addition of 2 moles of A, A 2 + B 2 U 2AB Initial (mol): 3 2 − x 3 2 − x x Change (mol): 2 − x 2 − x +x Equilibrium (mol): 3 − x 3 − x 2x CHAPTER 15: CHEMICAL EQUILIBRIUM 348 We write two different equilibrium constants expressions for the two tables. 2 22 [AB] [A ][B ] =K (2 ) and (3 )(3 ) 13 22 == −−⎛⎞⎛⎞ −− ⎜⎟⎜⎟ ⎝⎠⎝⎠ xx KK xx x x We equate the equilibrium constant expressions and solve for x. 22 (2 ) (3 )(3 ) 13 22 = −−⎛⎞⎛⎞ −− ⎜⎟⎜⎟ ⎝⎠⎝⎠ xx xx x x 2 2 14 1 69 (812) 4 = − + −+ xx xx −6x + 9 = −8x + 12 x = 1.5 We substitute x back into one of the equilibrium constant expressions to solve for K. 22 (2 ) (3) (3 )(3 ) (1.5)(1.5) === −− 4.0 x xx K Substitute x into the other equilibrium constant expression to see if you obtain the same value for K. Note that we used moles rather than molarity for the concentrations, because the volume, V, cancels in the equilibrium constant expressions. 15.101 (a) First, we calculate the moles of I 2 . 42 22 2 1molI mol I 0.032 g I 1.26 10 mol 253.8 g I − =× =× Let x be the number of moles of I 2 that dissolves in CCl 4 , so (1.26 × 10 −4 − x)mol remains dissolved in water. We set up expressions for the concentrations of I 2 in CCl 4 and H 2 O. 4 22 (1.26 10 ) mol mol [I ( )] and [I (CCl )] 0.200 L 0.030 L − ×− == xx aq Next, we substitute these concentrations into the equilibrium constant expression and solve for x. 24 2 [I (CCl )] [I ( )] =K aq 4 0.030 83 (1.26 10 ) 0.200 − = ×− x x 83(1.26 × 10 −4 − x) = 6.67x x = 1.166 × 10 −4 CHAPTER 15: CHEMICAL EQUILIBRIUM 349 The fraction of I 2 remaining in the aqueous phase is: 44 4 (1.26 10 ) (1.166 10 ) fraction ( ) 1.26 10 −− − ×−× == × 0.075f (b) The first extraction leaves only 7.5% I 2 in the water. The next extraction with 0.030 L of CCl 4 will leave only (0.075)(0.075) = 5.6 × 10 −3 . This is the fraction remaining after the second extraction which is only 0.56%. (c) For a single extraction using 0.060 L of CCl 4 , we let y be the number of moles of I 2 in CCl 4 . 4 0.060 83 (1.26 10 ) 0.200 − = ×− y y 83(1.26 × 10 −4 − y) = 3.33y y = 1.211 × 10 −4 The fraction of I 2 remaining in the aqueous phase is: 44 4 (1.26 10 ) (1.211 10 ) fraction ( ) 1.26 10 −− − ×−× == × 0.039f The fraction of I 2 remaining dissolved in water is 0.039 or 3.9%. The extraction with 0.060 L of CCl 4 is not as effective as two separate extractions of 0.030 L each. bjc Microsoft Word - c15prob5eIM.doc