Answers to Assigned Problems from Chapter 2

Answers to Assigned Problems from Chapter 2 2.2. 1 mol of ice has a volume of 18.01 g/0.9168 g cm 3 = 19.64 cm 3 1 mol of water has a volume of 18.01 g/0.9998 g cm 3 = 18.01 cm 3 V(ice water) = 1.63 cm 3 mol 1 (PV) = 0.001 63 atm dm 3 = 0.165 J mol 1 H = 6025 J = U + (PV) = U (0.165 J mol 1 ) U = 6025 J mol 1 = 6.025 kj mol 1 (Difference between H and U is only 0.165 J mol 1.) Work done on the system = 0.165 J mol 1 2.3. 1 mol of water at 100 C has a volume of 18.01 g 0.9584 g cm 3 = 18.79 cm3 18.01 g Volume of steam = 0.000 596 g cm 3 = 30 218.1 cm3 Volume increase = 30 199 cm 3 mol 1 = 30.20 dm 3 mol 1 (PV) = 30.20 atm dm 3 mol 1 = 3059.3 J mol 1 = 3.06 kj mol 1 H = 4063 J mol 1 = U + (3059.3 J mol 1 ) U = 37.6 kj mol 1 Work done by the system = w = 3.06 kj 2.4. Heat the water from 10 C to 0 C: (1) q 1 = C P dt = C P (T 2 T 1 ) = 753 J mol 1 Freeze the water at 0 C: (2) q 2 = 6025 J mol 1 Cool the ice from 0 C to 10 C: (3) q 3 = 377 J mol 1 Net q = 753 6025 377 = 5649 J mol 1 = H = 5.65 kj mol 1 2.5. Heat evolved = 1.69 6937 = 11 724 J Molar mass of CH 3 COCH 3 = 3 12.01 + 6 1.008 + 16.00 = 58.08 g mol 1 Heat evolved in the combustion of 1 mol

Chapter 2 2 = 11 724 58. 08 J = 972.7 kj 0. 700 a. U = 972.7 kj mol 1 b. CH 3 COCH 3 (l) + 4O 2 (g) 3CO 2 (g) + 3H 2 O(l) n(gases) = 1 H = 972 700 8.3145 299.8 J mol 1 = 975.2 kj mol 1 2.6. a. Heat capacity of man = 292 600 J K 1 10 460 000 J Temperature rise = 292 600 J K 1 = 35.7 C Final temperature = 37 + 35.7 = 72.7 C b. H = 43 400 J mol 1 /18.0 g mol 1 = 2411 J g 1 2.7. The reaction is Mass of water required = Zn + H 2 SO 4 ZnSO 4 + H 2 (g) 10 460 000 J 2411 J g 1 = 4340 g = 4.34 kg Thus, 1 mol of gas is liberated by each mole of Zn, i.e., by 65.37 g. One hundred grams therefore liberates (100/65.37)mol = 1.53 mol of H 2. The work done by the system is P V: w = P V = n H2 RT = 1.53 mol 8.3145 J K 1 mol 1 298.15 K = 3790 J = 3.79 kj The work in a sealed vessel ( V = 0) is zero. 2.14 Enthalpy for this reaction at 298 K is the enthalpy of formation of two moles of water, i.e., H (298 K) = 2 mol ( 241.82 kj mol 1 ) = 483.64 kj. At 800 K, using Eq. (2.52), we obtain H (800 K)/J = H (298 K) + d(800 298) + 1 2 e(8002 298 2 ) f (800 1 298 1 ), where d = 2d H2 O (2d H + d 2 O2 ), and e and f are defined similarly. d/(j K 1 ) = 2 30.54 (2 27.28 + 29.96) = 23.44 e/(j K 2 ) = 2 10.29 10 3 (2 3.26 10 3 + 4.18 10 3 ) = 9.88 10 3 f/(j K) = 2 0 (2 5.0 10 4 1.67 10 5 ) = 6.7 10 4. Therefore, H (800 K)/J = 483640 23.44(800 298) + 1 2 9.88 10 3 (800 2 298 2 ) 6.7 10 4 (800 1 298 1 ) = 4.9254 10 5 or 492.54 kj.

Chapter 2 3 2.15. Molar mass of benzene = 6 12.01 + 6 1.008 = 78.11 g mol 1 Heat evolved in the combustion of 1 mol = 26. 54 kj 7811. g mol 0633. g 1 = 3274.9 kj a. U = 3274.9 kj mol 1 b. C 6 H 6 (l) + 15 2 O 2 (g) 6CO 2 (g) + 3H 2 O(l) v(gases) = 1.5 H = 3 274 900 1.5 8.3145 298.15 J mol 1 = 3278.6 kj mol 1 2.16. H = H (C 2 H 6 ) + H (H 2 ) 2[ H (CH 4 )] H = 84.68 2( 74.81) = 64.94 kj mol 1 2.17. Molar mass of CH 3 OH = 12.01 + 4 1.008 + 16.00 = 32.04 g mol 1 Amount of methanol = 5.27 g/32.04 g mol 1 = 0.164 mol Heat evolved = 119. 50 kj 32. 04 g mol 527. g 1 = 726.5 kj mol 1 = c U a. c U = 726.5 kj mol 1 CH 3 OH(l) + 3 2 O 2 (g) CO 2 (g) + 2H 2 O(l) v(gases) = 0.5 c H = 726 500 0.5 8.3145 298.15 J mol 1 = 727.7 kj mol 1 b. CH 3 OH(l) + 3 2 O 2 (g) CO 2 (g) + 2H 2O(l) (1) H = 727.7 kj mol 1 H 2 (g) + 1 2 O 2 (g) H 2O(l) (2) f H = 285.85 kj mol 1 C(s) + O 2 (g) CO 2 (g) (3) f H = 393.51 kj mol 1 2 (2) + (3) (1) gives C(s) + 2H 2 (g) + 1 2 O 2 (g) CH 3OH(l) (4) f H = 2( 285.85) 393.51 + 727.7 = 237.5 kj mol 1

Chapter 2 4 This value is slightly different from the value listed in Appendix D. Experimental data is constantly being evaluated causing changes in listed values. This problem may have used older data. c. CH 3 OH(l) CH 3 OH(g) v H = 35.27 kj mol 1 (5) (4) + (5) gives C(s) + 2H 2 (g) + 1 2 O 2 (g) CH 3 OH(g) f H = 202.2 kj mol 1 2.18. 2C(graphite) + 3H 2 (g) C 2 H 6 (g) (1) f H = 84.68 kj mol 1 C(graphite) + O 2 (g) CO 2 (g) (2) f H = 393.51 kj mol 1 H 2 (g) + 1 2 O 2 (g) H 2O(l) (3) f H = 285.85 kj mol 1 3 (3) + 2 (2) (1) gives C 2 H 6 (g) + 7 2 O 2 (g) 2CO 2 (g) + 3H 2 O(l) c H = 1559.9 kj mol 1 2.19. First, perform a multiple regression on z = a + bx + cy using the definitions z = C P,m ; x = T and y = 1/T 2. The result is z = 1.7267 + 9.3424 10 2 x 871.4y. In other words, we find that d = 1.7267 J K 1 mol 1 ; e = 9.3424 10 2 J K 2 mol 1 ; f = 8.714 10 2 J K mol 1. Below, we present two plots of this function, one in the range 15 T 275, and another, in the range 10 T 25. It can be seen that the function becomes negative at T 16.1 K. A negative heat capacity is obviously unphysical. This is an indication that the temperature dependence of heat capacities of solids at low temperature cannot be expressed using the model used here (see Chapter 16, Section 6). 30.0 Function Plot 4 Function Plot 25.0 2 20.0 0 z 15.0 10.0 z -2 5.0-4 0.0-6 -5.0 0 50 100 150 200 250 300 x -8 10.0 15.0 20.0 25.0 x

Chapter 2 5 2.21. C 3 H 6 (g) + 9 2 O 2 (g) 3CO 2 (g) + 3H 2O(l) (1) c H = 2091.2 kj mol 1 C(graphite) + O 2 (g) CO 2 (g) (2) f H = 393.51 kj mol 1 H 2 (g) + 1 2 O 2 (g) H 2O(l) (3) f H = 285.85 kj mol 1 3 (2) + 3 (3) (1) gives 3C(graphite) + 3H 2 (g) C 3 H 6 (g) f H = 53.1 kj mol 1 2.22. Perform a multiple regression on z = a + bx + cy using the definitions z = C P,m ; x = T and y = 1/T 2. The result is z = 0.80053 + 1.303 10 3 x 21991.0y. In other words, we find that d = 0.801 J K 1 mol 1 ; e = 1.303 10 3 J K 2 mol 1 ; f = 2.199 10 4 J K mol 1. 1.50 Function Plot 1.00 CP,m 0.50 0.00 150 200 250 300 350 400 450 T 500 A plot of the fit is 2.23 H = 277.69 52.26 + 285.85 = 44.1 kj mol 1 2.24. C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O (1) c H = 1370.7 kj mol 1 CH 3 CHO + 5 2 O 2 2CO 2 + 2H 2 O (2) c H = 1167.3 kj mol 1

Chapter 2 6 CH 3 COOH + 2O 2 2CO 2 + 2H 2 O (3) c H = 876.1 kj mol 1 Reaction (a) is (1) (2); H = 1370.7 + 1167.3 = 203.4 kj mol 1 Reaction (b) is (2) (3); H = 1167.3 + 876.1 = 291.2 kj mol 1 2.25. Let us list all the reactions involved: 1. CH 2 CHCN + 15 4 O 2 (g) 3 CO 2 (g) + 3 2 H 2 O(g) + (1/2)N 2 (g); ch = 1760.90 kj mol 1 2. C(graphite) + O 2 (g) CO 2 (g); f H = 393.50 kj mol 1 3. H 2 (g) + 1 2 O 2 (g) H 2 O(g); fh = 241.83 kj mol 1 4. 2C(graphite) + H 2 (g) C 2 H 2 (g); f H = 226.73 kj mol 1 5. C(graphite) + 1 2 H 2 (g) + 1 2 N 2 (g) HCN(g); fh = 135.10 kj mol 1 The desired reaction is HCN(g) + C 2 H 2 (g) CH 2 CHCN. To generate this equation from the five given above, we need to manipulate them to get 3(Eq. 2) + (3/2)(Eq. 3) (Eq. 1) (Eq. 4) (Eq. 5). Performing the same manipulations with the enthalpies, we obtain 2.26. From Appendix D H = 3( 393.50) + (3/2)( 241.83) ( 1760.90) 226.73 135.10 = 144.2 kj mol 1. 2C + 3H 2 + 1 2 O 2 C 2 H 5OH (1) f H = 277.69 kj mol 1 2C + 2H 2 + O 2 CH 3 COOH (2) f H = 484.5 kj mol 1 H 2 + 1 2 O 2 H 2 O (3) f H = 285.85 kj mol 1 Reaction is (2) + (3) (1); H = 492.7 kj mol 1 This differs slightly from the result of Problem 2.19, namely 203.4 291.2 = 494.6 kj mol 1 2.31. From Appendix D, f H for ethanol and acetic acid are 277.69 and 484.5 kj mol 1, respectively. Therefore, the enthalpy change for the reaction C 2 H 5 OH(l) + O 2 (g) CH 3 COOH(l) is 484.5 ( 277.69) = 206.8 kj mol 1. Since ethanol is fed in at the rate of 45.00 kg h 1, and only 42 mole % of ethanol is converted, the actual heat released in the reaction per hour is (MW of ethanol = 46.069 g mol 1 ) 1 1 45. 00 kg h 0. 046069 kg mol 1 1 ( 206. 8 kj mol ) 0.42 = 84840 kj h Therefore, heat will have to be removed at the rate of 84840 kj h 1.

Chapter 2 7 2.32. a. Assume that all the ice melts. The process would absorb heat from the 1000 g water. The water produced by the melting ice (100 g of it) will be initially at 0 C. It will warm up by absorbing heat from the 1000 g water present. Suppose that the final temperature is t C; then the total heat gained by the 100 g water is 100 g Heat gained = 18.02 g mol 1 [6 025 J mol 1 + 75.3 J K 1 mol 1 (t 0) K] = 33 345 J + 417.87 J K 1 t K. Finally, all of the water is at the temperature t. Using the relationship heat lost = heat gained, we get 33 345 J + 417.87 J K 1 t K = [75.3 J K 1 mol 1 1000 g 18.02 g mol 1 (t 20) K]. Solving for t, we get t = 10.9 C. Since this is not below 0 C, it follows that all of the ice in fact melts. The final temperature is 10.9 C. b. It is obvious that now not all of the ice will melt. (If we assumed it all melted, we would find the final temperature to be below 0 C.) The final temperature of the water is now 0 C, and if we suppose that x g of the ice melts, the heat balance equation is x g 6025 J mol 1 = 75.3 J K 1 mol 1 1000 g 20 K x = 250 g 2.35. a. Since a bomb calorimeter is a constant-volume instrument, the heat evolved is U. The heat evolved per gram is thus 2186.0 J/0.1328 g = 16 461 J g 1 The molecular weight of sucrose is 342.30, and therefore c U m = 16 461 J g 1 342.3 g mol 1 = 5 634 600 J mol 1 = 5635 kj mol 1 The equation for the combustion reaction is C 12 H 22 O 11 (s) + 12O 2 (g) 12CO 2 (g) + 11H 2 O(l) The change n is therefore zero, and by Eq. 2.41, H = U; thus c H m = 5635 kj mol 1 b. From the equation, with the use of Hess s law, it can be deduced that f H m = 12 f H m (CO 2,g) + 11 f H m (H 2 O,l) c H m (sucrose) = (12 393.51) + (11 285.85) + 5635 kj mol 1 = 4722.12 3144.35 + 5635 = 2231 kj mol 1 2.37. Note that we need to find the intersection of the isotherm that passes through the initial state and the adiabat that passes through the final state. Let this point be (P 0,V 0 ) at the temperature of the isotherm T i. For adiabatic processes, (Eq. 2.90), T f / T i = (V f / V 0 ) γ 1, where 3 3 γ = R + R / R = 5/ 3. 2 2

Chapter 2 8 Therefore, γ 1 = 2/3. The final volume is V f = RT f /P f = (0.08314 dm 3 bar K 1 mol 1 253.2 K)/2.0 bar = 10.526 dm 3. Therefore, V 0 = V f (T f /T i ) 3/2 = 10.526 (253.2/298.0) 3/2 = 8.244 dm 3. 2.38 a. 1 mol in 22.7 dm 3 at 273 K exerts a pressure of 1 bar; 2 mol in 11.35 dm 3 exert a pressure of 4 bar. b. PV = 4 11.35 = 45.40 bar dm 3 1 bar dm 3 = 100 J 45.40 bar dm 3 = 4.540 kj c. C V,m = C P,m R = 29.4 8.3 = 21.1 J K 1 mol 1 2.39. a. Zero b. 2 21.1 100 = 4220 J = 4.22 kj c. 4220 J = 4.22 kj d. P = 4 373/273 = 5.47 bar = 547 kpa e. P 2 V 2 = 5.47 11.35 = 62.08 bar dm 3 = 6208 J = 6.208 kj f. H = 4220 + 6208 4540 = 5888 J = 5.89 kj 2.40. a. V = 11.35 373/273 = 15.5 dm 3 b. w = P V = 4 (15.5 11.35) = 16.6 bar dm 3 c. q = 2 29.4 100 = 5880 J = 5.88 kj d. H = 5880 J = 5.88 kj = 1660 J = 1.66 kj e. U = H P V = 5880 1660 = 4220 J = 4.22 kj (= 2C V,m T = 21.1 2 100) 2.41. a. Zero b. P 2 = 8 bar = 800 kpa c. w = 2 8.3145 273 ln 2 = 3150 J = 3.15 kj d. q = 3147 J = 3.15 kj e. Zero 2.42. Initial volume of gas = nrt 2 0 083 09 27315 =.. = 4.539 dm P 3 10 Final volume = 22.70 dm 3 V = 18.16 dm 3

Chapter 2 9 a. Work done by the gas = 2.0 18.16 = 36.32 bar dm 3 = 3632 J = energy transferred to surroundings. b. U = H = 0 c. q = 3632 J 2.43. a. Work done by gas = nrt ln V final V initial = 2 8.3145 273.15 ln 5 = 7310 J b. U = H = 0 c. q = 7310 J = 7.31 kj 2.44. C V,m = 28.80 8.3145 = 20.49 J K 1 mol 1 γ = 28.80 20.49 = 1.406 a. P 2 = P 1 V γ = 3.0 1.5 2 5.0 = 0.552 bar 1.406 T 2 = T 1 V γ 1 = 298.15 1.5 2 5.0 = 182.9 K b. U m = C V,m (T 2 T 1 ) 0.406 = 20.49(182.9 298.15) = 2361 J mol 1 Amount of H 2, n = 30. 15. = 0.182 mol 0. 083 09 29815. U = 430 J for 0.182 mol H m = 28.80(182.9 298.15) = 3319 J mol 1 H = 604 J for 0.182 mol 2.45. a. Initial volume, = nrt P P 1 1/γ P 2 V 2 = = 2.934 (10) 0.763 = 17.00 dm 3 b. T 2 = T 1 P 2 V 2 P 1 = = 0.1 0.083 09 353.15 = 2.934 dm3 35315. 17. 00 = 204.6 K 10 2. 934

Chapter 2 10 c. C P,m C V,m = 8.3145 J K 1 mol 1 C P,m C P,m = 1.31; C V,m C V,m 1 = 0.31; C P,m C V,m = 0.31 C V,m 0.31 C V,m = 8.3145 J K 1 mol 1 C V,m = 26.82 J K 1 mol 1 C P,m = 35.13 J K 1 mol 1 U m = 26.82(204.6 353.15) = 3984 J K 1 mol 1 U = 398.4 J K 1 for 0.1 mol H m = 35.13 (204.6 353.15) = 5219 J K 1 mol 1 H = 522 J K 1 for 0.1 mol 2.46. a. C P,m C V,m = R = 8.3145 J K 1 mol 1 C P,m = 29.83 + 8.2 10 3 (T/K) b. T = 0, U = H = 0. It could be made to occur adiabatically by allowing free expansion, with w = q = 0. 2.50. C V,m = 5 2 R and C P,m = 7 2 R γ = 7 5 T 2 γ 1 V 2 T 1 = (Eq. 2.90) Therefore, T 2 = T 1 V 2 γ 1 = 298.15 K 1 2 2/5 = 226.0 K U m = C V,m (T 2 T 1 ) = 5 R(226.0 298.15) K 2 = 1501 J mol 1 = 1.5 kj mol 1 H m = C P,m (T 2 T 1 ) = 7 R(226.0 298.15) K 2 = 2100 J mol 1 = 2.1 kj mol 1 2.62. For 1 mol of a van der Waals gas P + a V 2 m Therefore (V m b) = RT (Eq. 1.100)

Chapter 2 11 T = P(V m b) R a + V 2 m R (V m b) and T P V = V m b R 2.63. a. Work done on the system = V 2 PdV = RT ln V 2 = 8.3145 300 ln 10 = 9757 J = 9.76 kj mol 1 0.2 b. w = RT V 2 dv V b Put V b = x; dv = dx = RT V 2 dx x = RT ln(v b)v 2 = RT ln b V 2 b = 8.3145 300 ln 9.97 0.17 = 10 155 J mol 1 = 10.16 kj mol 1 More work is done in (b) because of the greater ratio of free volumes. 2.64. a. w = RT ln = 8.3145 100 ln 20 V 2 5 = 1153 J mol 1 = 1.15 kj mol 1 b. P = RT V a V 2 w = V 2 PdV = RT V 2 dv V + a 2 dv V V 2 = +RT ln V 2 a 1 1 V 2 = 1153 3.84 101.3 1 20 1 5 = 1211 J = 1.21 kj mol 1 In (b), additional work has to be done against the molecular repulsions. 2.67. The reversible work done on the system is w rev = V 2 PdV = nrt V 2 dv V nb = nrt ln V 2 nb nb 1.00 0.08 = 2 8.3145 300 (J) ln 10.00 0.08 = 4 989 J ln(0.92/9.92) = 4 989 2.378 = 11 860 J = 11.9 kj

Chapter 2 12 From Eq. 2.125, U is zero since a is zero. To obtain H we must calculate (PV): P 1 = 2RT/( 0.08) = 2 0.0831 300/9.92 = 5.03 bar P 2 = 2RT/(V 2 0.08) = 2 0.0831 300/0.92 = 54.2 bar P 1 = 5.03 10.0 = 50.3 bar dm 3 = 5030.0 Pa m 3 = 5030.0 J P 2 V 2 = 54.2 1.00 = 54.2 bar dm 3 = 5420.0 J (PV) = P 2 V 2 P 1 = 390 J 2.68. The reversible work is H = U + (PV) = 0 + 390 = 390 J w rev = V 2 PdV = V 2 = nrt ln V 2 nrt V n 2 a n2 a 1 1 V 2 V 2 dv = 3 8.3145 300 (J) ln (1/20) (3.00) 2 0.55 (Pa m 6 ) = 22 417 4703 = 17 714 J = 17.7 kj The change in internal energy is obtained from the relationship: du = U V T dv = n2 a V 2 dv U = V 2 n 2 a V 2 dv = n 2 a 1 V V 2 = n 2 a 1 1 V 2 = (3.00) 2 0.55 (Pa m 6 ) = 4703 J = 4.70 kj To obtain H we calculate P 1 and P 2 V 2 : = 20 dm 3 P 1 = (nrt/ ) n 2 a/ 2 1 10 3 m 3 1 20 10 3 m 3 1 10 3 m 3 1 20 10 3 m 3 = [3 0.0831 300 (bar)/20] (10 5 Pa/bar) [ 9 0.55 (Pa m 6 )]/(20 10 3 m 3 ) 2 = 374 000 Pa 12 375 Pa = 361.6 kpa V 2 = 1 dm 3

Chapter 2 13 P 2 = [3 0.0831 300 (bar)/1] (10 5 Pa/bar) [ 9 0.55 (Pa m 6 )]/(10 3 m 3 ) 2 = 7 479 000 Pa 4 950 000 Pa = 2 529 000 Pa = 2529 kpa P 1 P 2 V 2 = (20 10 3 m 3 ) 361 600 Pa = 7232 J = (10 3 m 3 ) 2 529 000 Pa = 2 529 J (PV) = 2 529 7232 = 4703 J H = 4703 4703 = 9406 J = 9.41 kj.