4. CHEMICAL EQUILIBRIUM n Equilibrium Constants, Slides of Chemistry

Download 4. CHEMICAL EQUILIBRIUM n Equilibrium Constants and more Slides Chemistry in PDF only on Docsity! 4. CHEMICAL EQUILIBRIUM n Equilibrium Constants 4.1. 2A →← Y + 2Z Initial amounts: 4 0 0 mol Amounts at equilibrium: 1 1.5 3.0 mol Concentrations at equilibrium: 1 5 1.5 5 3.0 5 mol dm–3 Kc = (1.5/5) (3.0/5)2 (1/5)2 mol dm–3 = (1.5) (3)2 (5) mol dm–3 = 2.7 mol dm–3 4.2. A + B →← Y + Z Initial amounts: x 3 0 0 mol Amounts at equilibrium: x – 2 1 2 2 mol 2 2 2 1 × − ×( )x = 0.1; x – 2 = 40 x = 42 Thus, initially there must be 42 mol of A. 4.3. A + 2B →← Z Initial amounts: x 4 0 mol Amounts at equilibrium: x – 1 2 1 mol Concentrations at equilibrium: x –1 5 2 5 1 5 mol dm–3 CHEMICAL EQUILIBRIUM n 73 1/5 [(x – 1)/5] (2/5)2 = 0.25 = 25 4(x – 1) 25 = x – 1 x = 26 Thus, initially there must be 26 mol of A. 4.4. Two moles of SO3(g) produce 3 mol of product; thus Σν = +1 mol. Then, from Eq. 4.26, KP = Kc(RT)+1 = (0.0271 mol dm–3) (8.3145 J K–1 mol–1 × 1100 K) = 247.8 J dm–3 = 2.478 × 105 J m–3 = 2.478 × 105 Pa = 2.48 bar 4.5 I2 →← 2I Initial amounts: 0.0061 0 mol Equilibrium amounts: 0.0061(1 – 0.0274) 0.0061 × 2 × 0.0274 mol = 5.93 × 10–3 = 3.3428 × 10– 4 mol Kc = ( . / . ) . / . 3 3428 10 0 5 5 93 10 0 5 4 2 3 × × mol dm–3 = 3.77 × 10–5 mol dm–3 KP = 3.77 × 10–5 mol dm–3 (8.3145 J K–1 mol–1 × 900 K) = 0.282 Pa = 2.82 × 10–6 bar 4.6. Addition of N2 at constant volume and temperature necessarily requires the equilibrium to shift to the right. Kc = [NH3]2 [N2][H2]3 = n2 NH3 V2 nN2 n3 H2 If nN2 is increased at constant V, the equilibrium must shift so as to produce more ammonia. If the pressure (as well as the temperature) is held constant, however, addition of N2 requires that V is increased. If the proportional increase in V2 is greater than the increase in nN2 the equilibrium will shift to the left when N2 is added. The volume V is proportional to nN2 + nH2 + nNH3 , and V2 is proportional to (nN2 + nH2 + nNH3 )2 If nN2 is very much larger than nH2 + nNH3 , V2 will increase approximately with n2 N2 and therefore increases more strongly than nN2 . If nN2 is not much larger than nH2 + nNH3 , an increase in nN2 will have a relatively smaller effect on V2 . The increase in ammonia dissociation when N2 is added is therefore expected when N2 is in excess, but not otherwise. 76 n CHAPTER 4 4.10. H2 + I2 →← 2HI Initially: 1 3 0 mol At equilibrium: 1 – x 2 3 – x 2 x mol After addition of 2 mol H2: 3 – x 3 – x 2x mol K = x2  1 – x 2  3 – x 2 = 4x2 (3 – x)2 x = 3/2 K = 4(3/2)2 (3/2)2 = 4 4.11. 12.7 g iodine = 0.05 mol I2 When all of the solid iodine has just gone, the iodine pressure is 0.10 atm. The consumption of 0.05 mol I2 leads to the formation of 0.10 mol HI, which exerts a pressure of PHI = 0.1 mol × 0.0831 dm3 bar K–1 mol–1 × 313.15 K 10 dm3 = 0.260 bar Then, if PH2 is the partial pressure of H2 after equilibrium is established, 2 2 H (0.260 bar) 0.10 barP × = 20 PH2 = 0.0338 bar nH2 = 0.0338 bar × 10 dm3 0.0831 × 313.15 bar dm3 mol–1 = 0.0130 mol Thus, 0.0130 mol of H2 is produced in the equilibrium mixture, and 0.05 mol of H2 is required to remove the 0.05 mol of I2. Therefore, 0.065 mol of H2 must be added. 4.12. Assuming that we start with one mole of N2O4(g), N2O4(g) š 2NO2 (g) 1– α 2α We obtain (1+α) moles of gas at equilibrium. Therefore, equilibrium partial pressures are 2 4 2N O NO 1 2 ; , 1 1 α α α α −= = + + P P P P and CHEMICAL EQUILIBRIUM n 77 2 1 2 4 ; ( ) ; / . 1 α α −= = = −P c P x PK P K K RT K K P Therefore, at 0.597 bar, KP = 1.08 bar, Kc = 4.35 × 10–2 mol dm–3, Kx = 1.81, and at 6.18 bar, KP = 1.45 bar, Kc = 5.84 × 10–2 mol dm–3, Kx = 0.234. 4.13. Rewriting the reaction in terms of one mole of HCl, we get HCl(g) + 1 4O 2(g) →← 1 2Cl2(g) + 1 2H2O(g) (1 – y) PO2 y/2 y/2 From examining the equation above, it is possible to establish the following relationships: It is important to remember that the same ratios hold for partial pressures also. Now, 4.14. Assuming ideal behavior, the partial pressure of oxygen is 0.51 bar. Therefore, 4.15. Since no reactants are initially present, we write H2(g) + I2(g) →← HI(g) x x n – 2x where n is the initial amount of HI: n = 10.0 g/(127.912 g mol–1) = 7.8179 × 10–2 mol. Since 2 2 ( 2 ) 2 65.0 , or 65 ,P n x n x K xx − −= = = we can solve for x. The solution is x = 7.7695 × 10–3 mol. Therefore, the equilibrium mole fractions are xH2 = xI2 = x/n = 9.94 × 10–2; xHI = (n – 2x)/n = 0.801. .1 ; )1(2 2 22 Cl OH HCl Cl = − = x x y y x x . 1 )1(2 4/1 O 4/1 OHCl Cl 4/1 OHCl 2/1 OH 2/1 Cl 22 2 2 22 Py y PP P PP PP KP × − = × = × × = .bar 87.1 51.0 1 )76.01(2 76.0 41 25.0 / PK −=× − = 78 n CHAPTER 4 n Equilibrium Constants and Gibbs Energy Changes 4.16. a. ∆G°/J mol–1 = –8.3145 × 283.15 ln (2.19 × 10–3) ∆G° = 14 417 J mol–1 = 14.42 kJ mol–1 b. (C6H5COOH)2 →← 2C6H5COOH Initially: 0 0.1 mol dm–3 At equilibrium: x 0.1 – 2x mol dm–3 (0.1 – 2x)2 x = 2.19 × 10–3 4x2 – 0.402 19x + 0.01 = 0 x = 0.402 19 ± 0.001 76 8 = 0.402 19 ± 0.041 95 8 = 0.0555 or 0.04503 mol Only the second answer is possible, and this leads to Dimer: 0.045 mol dm–3; Monomer: 0.01 mol dm–3 4.17. KP at 3000 K = (0.4)2 × 0.2 (0.6)2 = 0.0889 atm ∆G°/J mol–1 = –8.3145 × 3000 ln 0.0889 ∆G° = 60 370 J mol–1 = 60.4 kJ mol–1 4.18. a. ln Kc = – 2930 8.3145 × 310.15 = –1.136; Kc = 0.321 b. ln Kc = + 15 500 8.3145 × 310.15 = 6.01; Kc = 408 c. Kc = 0.321 × 408 = 130.9 ∆G° = 2.93 – 15.5 = –12.6 kJ mol–1 4.19. a. ∆G° = ∆G°Products – ∆G°Reactants = 2(–16.63) – 0 – (3 × 0) = –33.26 kJ mol–1 KP = exp(–∆G°/RT) = 6.71 × 105 bar–2 b. ∆G° = –32.82 – 209.20 = –242.02 kJ mol–1 KP = 2.5 × 1042 bar–2 c. ∆G° = –32.82 – 68.15 = –100.97 kJ mol–1 KP = 4.9 × 1017 bar–1 CHEMICAL EQUILIBRIUM n 81 b. ln (KP/bar–1) = – ∆G° RT = +100 970 8.3145 × 298.15 = 40.73 KP = 4.89 × 1017 bar–1 c. Kc = KP (RT) = 4.89 × 1017 × 0.0831 × 298.15 = 1.21 × 1019 dm3 mol–1 d. ∆G° = –RT ln Kc = –8.3145 × 298.15 ln (1.21 × 1019) = –108 925 J mol–1 = –108.9 kJ mol–1 e. ∆S° = ∆H° –∆G° T = –136 940 + 108 925 298.15 = –93.96 J K–1 mol–1 f. ∆G°(100 °C) = ∆H° – T∆S° = –136 940 + (373.15 ×120.6) = –91 938 J mol–1 ln (KP/bar–1) = –∆G° RT = 91 938 8.3145 × 373.15 = 29.63 KP(100°C) = 7.38 × 1012 bar–1 4.28. a. 2H2(g) + O2(g) →← 2H2O(g) ∆fG° = 2∆fG°H2O –2∆fG°H2 – ∆fG°O2 = 2(–228.57) – 2(0) – 0 = –457.14 kJ mol–1 ∆fH° = 2∆fH°H2O – 2∆fH°H2 – ∆fH°O2 = 2(–241.82) – 2(0) –0 = –483.64 kJ mol–1 ∆G° = ∆H° – T∆S° ∆fS° = – –457.14 – (–483.64) 298.15 = –0.08888 kJ mol–1= –88.88 J K–1 mol–1 b. ∆G° = –RT ln K° P ln (K° P /bar–1) = 457 140 J mol–1 8.3145 J K–1 mol–1 298.15 K = 184.4 K° P = 1.222 × 1080 bar–1 c. ∆G(2000°C)/kJ mol–1 = ∆fH° – 2273.15 ∆fS° = –483.64 – 2273.15 (–0.08888) = –281.6 kJ mol–1 ln (KP/bar–1) = 281 600 8.3145 × 2273.15 = 14.90 KP = 2.96 × 106 bar–1 82 n CHAPTER 4 4.29. ∆rG° = 2∆fG°(O3, g) – 3∆fG °(O2, g) = 2(163.2) – 3(0) = 326.4 kJ mol–1. K = exp(–∆rG°/RT) = exp[–326 400 /(8.3145 × 400)] = 3.371 × 10–43. 4.30. a. ∆S° = ∆H° – ∆G° T = –20 100 + 31 000 310.15 = 35.1 J K–1 mol–1 b. ln Kc = 31 000 8.3145 × 310.15 = 12.02 Kc = 1.66 × 105 mol dm–3 c. ∆G°(25°C) = –20 100 – (298.15 × 35.1) = –30 570 J mol–1 ln Kc = 30 570 8.3145 × 298.15 = 12.33 Kc = 2.27 × 105 mol dm–3 4.31. a. ∆H° = –165.98 + 146.44 = –19.54 kJ mol–1 ∆S° = 306.4 – 349.0 = –42.6 J K–1 mol–1 ∆G° = –19 540 + (42.6 × 298.15) = –6839 J mol–1 = –6.84 kJ mol–1 b. ln KP = 6839 8.3145 × 298.15 = 2.759; KP = 15.78 If partial pressure of neopentane = x bar, partial pressure of n-pentane = (1 – x) bar x 1 – x = 15.78 x = 15.78 – 15.78x; x = 0.940 1 – x = 0.060 Thus P(neopentane) = 0.940 bar and P(n-pentane) = 0.060 bar. 4.32. a. Slope of plot of ln Kc against 1/T is ln 3    1 313.15 – 1 298.15 =    1.0986 1.607 × 10– 4 = –6836 From Eq. 4.73, ∆H° = 6836 × 8.3145 = 56 840 J mol–1 CHEMICAL EQUILIBRIUM n 83 = 56.8 kJ mol–1 b. –56.8 kJ mol–1 4.33. a. Slope of plot of ln Kc against 1/T is – ln 1.45 1 298.15 – 1 303.15 = – 0.372 5.532 × 10–5 = –6724.6 K ∆H° = 6724.6 × 8.3145 = 55 910 J mol–1 = 55.9 kJ mol–1 At 25 °C, ∆G° = –RT ln 1.00 × 10–14 = –8.3145 × 298.15 × (–32.236) = 79 912 J mol–1 ∆S° = ∆H° – ∆G° T = 55 910 – 79 912 298.15 = –80.5 J K–1 mol–1 b. The difference between the reciprocals of the absolute temperatures corresponding to 25 °C and 37 °C is 3.3540 × 10–3 – 3.2242 × 10–3 = 1.298 × 10– 4 The slope of the plot ln Kw against 1/T was –6724.6 K, and in going from 25 °C to 37 °C ln Kw is thus increased by 6724.6 × 1.298 × 10–4 = 0.873 At 25°C, ln Kw is –32.244 and at 37 °C it is therefore –32.244 + 0.873 = – 31.371 Kw at 37 °C is therefore 2.38 × 10–14 mol2 dm–6. 4.34. a. At 400 °C, log10(KP/bar) = 7.55 – 4844 673.15 = 0.354 KP = 2.259 bar ∆G°/kJ mol–1 = –8.3145 × 673.15 ln 2.259 ∆G° = –4549 J mol–1 = –4.55 kJ mol–1 From Figure 4.2 (b), ∆H° = 4844 × 8.3145 × 2.303 J mol–1 = 92 750 J mol–1 = 92.75 kJ mol–1 ∆S° = 92 800 + 4549 673.15 J K–1 mol–1 86 n CHAPTER 4 = – 5697 T/K – 1.51 ln (T/K) + 1.41 × 10– 4 (T/K) + 4.61 × 104 (T/K)2 + I I is obtained from the fact that at 298.15, ln K = –11.55 I = –11.55 + 19.11 + 8.60 – 0.042 – 0.519 = 15.60 ln K = 15.60 – 5697 T/K – 1.51 ln (T/K) + 1.41 × 10– 4 (T/K) + 4.61 × 104 (T/K)2 e. ln KP at 1000 K = 15.60 – 5697 1000 – 1.51 ln 1000 + 0.141 + 4.61 × 104 10002 = –0.34 KP (1000 K) = 0.71 4.39. Partial pressures at 1395 K are: CO: 0.000 140 atm; CO2: (1 – 0.000 140) atm; O2: 0.000 070 atm KP = (0.000 140)2 × 0.000 070 (0.9999)2 = 1.372 × 10–12 atm = 1.39 × 10–12 bar At 1443 K, KP = (0.000 250)2 × 0.000 125 (0.999 75)2 = 7.814 × 10–12 atm = 7.92 × 10–12 bar At 1498 K, KP = (0.000 471)2 × 0.000 2355 (0.999 529)2 = 5.227 × 10–11 atm = 5.30 × 10–11 bar Then T/K 1012KP/atm 104/(T/K) ln(1012KP/atm) 1395 1.372 7.168 0.3163 1443 7.814 6.930 2.0559 1498 52.27 6.676 3.956 From a plot of ln (KP/atm) against 1/(T/K), slope = –7.39 × 104 K and ∆H° = –R × slope = 609 kJ mol–1. ∆G°(1395 K) = –8.3145 × 1395 ln (1.372 × 10–12) = 316.8 kJ mol–1 ∆S° = ∆H° – ∆G° T = 209 J K–1 mol–1 (standard state 1 atm) CHEMICAL EQUILIBRIUM n 87 4.40. Suppose that there are present x mol of I2 and y mol of I: x + y 2 = 1.958 × 10–3 P = (x + y) mol RT V At 800° C 558.0 760.0 atm = (x + y)0.082 05 dm3 atm K–1 mol–1 249.8 × 10–3 dm3 × 1073.15 K x + y = 2.0829 × 10–3 y 2 = 0.1249 × 10–3; y = 2.498 × 10– 4 x = 1.833 × 10–3 Degree of dissociation, α = 1 – x 1.958 × 10–3 = 1 – 1.833 × 10–3 1.958 × 10–3 = 0.0638 At 1000°C, x + y = 2.3535 × 10–3 y 2 = 3.955 × 10–4; y = 7.91 × 10– 4 x = 1.5625 × 10–3; α = 0.202 At 1200°C, x + y = 2.7715 × 10–3 y 2 = 8.135 × 10–4; y = 1.627 × 10–3 x = 1.1445 × 10–3; α = 0.415 a. α = 0.0638, 0.202, 0.415 at the three temperatures b. At 800 °C, Kc = (2.498 × 10–4 mol)2 1.833 × 10–3 mol × 0.2498 dm3 = 1.363 × 10–4 mol dm–3 At 1000 °C, Kc = (7.91 × 10– 4)2 1.5625 × 10–3 × 0.2498 = 16.0 × 10–4 mol dm–3 At 1200 °C, Kc = (1.627 × 10–3)2 1.1445 × 10–3 × 0.2498 = 92.59 × 10–4 mol dm–3 88 n CHAPTER 4 c. KP = Kc (RT)Σν (Eq. 4.26) = KcRT since Σν = 1 At 800 °C, KP = 1.363 × 10–4 mol dm–3 × 100 dm3 m–3 × 8.3145 J K–1 mol–1 × 1073.15 K = 1.216 kPa = 9.12 Torr = 0.0122 bar At 1000 °C, KP = 16.97 kPa At 1200 °C, KP = 113.4 kPa d. T/K 104Kc/mol dm–3 104/(T/K) ln 104Kc mol dm–3 1073.15 1.363 9.318 0.310 1273.15 16.03 7.855 2.774 1473.15 92.59 6.788 4.528 Slope of a plot of ln(Kc/mol dm–3) against 1/(T/K) is –1.67 × 104 K–1. ∆U° = 139 kJ mol–1 ∆H° = ∆U° + RT = 139 000 + (8.3145 × 1273.15) = 149 586 J mol–1 = 150 kJ mol–1 e. At 1000°C, KP = 16.97 kPa = 0.169 bar. ∆G° = –(8.3145 × 1273.15 J mol–1) ln(0.169/bar) = 18 819 J mol–1 = 18.8 kJ mol–1 (standard state: 1 bar) ∆S° = ∆H° – ∆G° T = 103.0 J K–1 mol–1 4.41. ∆G° = –RT ln K° = –8.3145 × 300 ln 5.7 × 10–3 = 12 889 J mol–1 = 12.9 kJ mol–1 Slope of plot = ln (5.7 × 10–3/7.8 × 10– 4) (1/300) – (1/340) = 1.989 3.922 × 10– 4 = 5071 K ∆H° = –8.3145 × 5071 = –42 160 J mol–1 = –42.16 kJ mol–1 ∆S° = (∆H° – ∆G°)/T = –183 J K–1 mol–1 4.42. ∆G° = ∆H° – T∆S° = –85 200 + 300 × 170.2 = –34 140 J mol–1 CHEMICAL EQUILIBRIUM n 91 Kx = KP P–Σν, so that at 2 bar pressure, Kx = 1.36 × 10–57 bar–1 × 2 bar = 2.72 × 10–57 This equilibrium constant (Eq. 4.31) is x(O3)2/x(O2)3, and since it is so small, x(O2) is almost exactly unity. Thus x(O3)2 = 2.72 × 10–57 and x(O3) = 5.22 × 10–29 4.48. At 500 K, by Eq. 2.46, ∆H° = ∆H°(300 K) + ∆CP(500 – 300) = –9600 –(7.11 × 200) = –11 022 J mol–1 For the entropy change at 500 K, the corresponding equation is ∆S°(T2) = ∆S°(T1) + ∫T2 T1 ∆CP dT T = ∆S°(T1) + ∆CP ln (T2/T1) At 500 K, ∆S° is therefore ∆S° = 22.18 – 7.11 ln(500/300) = 18.55 J K–1 mol–1 Therefore ∆G° = –11 022 – 18.55 × 500 = –20 297 J mol–1 The equilibrium constant KP is therefore exp(20 297/8.3145 × 500) = 131.9 Since the reaction involves no change in the number of molecules, Kc and Kx also have this value (see Eqs. 4.26 and 4.32), and they are not affected by the pressure. Let the mole fraction of HI be x; then the mole fractions of H2 and I2 are both (1 – x)/2, and the expression for Kx is therefore Kx = 4x2/(1 – x)2 = 131.9 Taking square roots 11.48 = 2x/(1 – x) whence x = 0.85 Pressure has no effect on the above values. 4.49. Values of ∆G° and of K[= exp(–∆G°/RT)u] are Temperature θ /°C T/K ∆G°/kJ mol–1 K 92 n CHAPTER 4 40.0 313.15 3.98 0.22 42.0 315.15 2.20 0.43 44.0 317.15 0.42 0.85 46.0 319.15 –1.36 1.67 48.0 321.15 –3.14 3.24 50.0 323.15 –4.93 6.27 ∆G° = ∆H° – T∆S° = 0 when T = 317.6 and K = 44.4°C At this temperature there will be equal concentrations of P and D. n Binding to Protein Molecules 4.50. If the concentration of M is [M], that of sites occupied and unoccupied is n[M]. The association reactions may be formulated in terms of the sites, S, S + A Ks→← SA Ks = [SA] [S][A] where [S] is the concentration of unoccupied sites and [SA] is the concentration of occupied sites. The total concentration of sites, n[M], is n[M] = [S] + [SA] = [SA]      1 Ks[A] + 1 The average number of occupied sites per molecule is the total concentration of occupied sites divided by the total concentration of M: –ν = [SA] [M] = n 1 Ks[A] + 1 = nKs[A] 1 + Ks[A] 4.51. The total concentration of the molecule M is [M]0 = [M] + [MA] + [MA2] + . . . + [MAn] The total concentration of occupied sites is the total concentration of bound A molecules: [A]b = [MA] + 2[MA2] + . . . + n[MAn] Expressing every term in terms of [A]: [M]0 = [M]{1 + K1[A] + K1K2[A]2 + . . . + (K1K2 . . . Kn)[A]n} [A]b = [M]{K1[A] + 2K1K2[A]2 + . . . + n(K1K2 . . . Kn)[A]n} Thus CHEMICAL EQUILIBRIUM n 93 –ν = [A]b [M]0 = K1[A] + 2K1K2[A]2 + . . . + n(K1K2 . . . Kn)[A]n 1 + K1[A] + K1K2[A]2 + . . . + (K1K2 . . . Kn)[A]n 4.52. With K1 = nKs, K2 = (n – 1)Ks/2, K3 = (n – 2)Ks/3 . . . Kn = Ks/n, the preceding equation becomes –ν = nKs[A] + n(n – 1)K2 s[A]2 + . . . + nKn s[A]n 1 + nKs[A] + n(n –1)K2 s[A]2/2 + . . . + Kn s[A]n = nKs[A] {1 + (n – 1)Ks[A] + . . . + Kn s –1[A]n–1} 1 + nKs[A] + n(n – 1)K2 s[A]2/2 + . . . + Kn s[A]n The coefficients are the binomial coefficients, –ν = nKs[A](1 + Ks[A])n–1 (1 + Ks[A])n = nKs[A] 1 + Ks[A] which are the expressions obtained in Problem 49. To test the equation, plot 1/–ν against 1/[A]: 1 –ν = 1 n + 1 nKs[A] One of the intercepts is 1/n. Alternatively, plot –ν against –ν /[A]: –ν = n – –ν Ks[A] 4.53. If Kn is very much larger than K1, K2, and so on, the equation obtained in Problem 51 reduces as follows: –ν = n(K1K2 . . . Kn)[A]n 1 + (K1K2 . . . Kn)[A]n = nK[A]n 1 + K[A]n where K = K1K2 . . . Kn is the overall equilibrium constant for the binding of n molecules: nA + M K →← MAn The fraction of sites occupied, θ, is –ν /n: θ = K[A]n 1 + K[A]n or θ 1 – θ = K[A]n