MSJ2e_Ch18_ISM2_June26_p803

MSJ2e_Ch18_ISM2_June26_p803 - Chapter 18: Thermodynamics:...

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Unformatted text preview: Chapter 18: Thermodynamics: Directionality of Chemical Reactions (b) G = (2 mol) G (CO 2 ) (2 mol) G (CO) (1 mol) G (O 2 ) = (2 mol) (394.359 kJ/mol) (2 mol) (137.168 kJ/mol) (1 mol) (0 kJ/mol) = 514.382 kJ The reaction can be harnessed to do useful work. (c) G = (1 mol) G (C 2 H 4 ) + (1 mol) G (H 2 ) (1 mol) G (C 2 H 6 ) = (1 mol) (68.15 kJ/mol) + (1 mol) (0 kJ/mol) (1 mol) (32.82 kJ/mol) = 100.97 kJ The reaction requires work to be done. 85. Adapt the method described in the answer to Question 6.83. Find the Gibbs free energy of the graphite reaction and use that as a thermochemical conversion factor to supply the Gibbs free energy of the endergonic reactions in Questions 83 (b) and (c) and 83 (a) and (c). C(graphite) + O 2 (g) CO 2 (g) This reaction is identical to the formation reaction for CO 2 , so G = 394.359 kJ/mol The calculation for 83(b) looks like this: 166.4 kJ endergonic reaction h 1 394.359 12.011 1 = 5.068 The rest of the results are described in the table below. Endergonic Reaction Reference Calculated G (X) (kJ) Mass of graphite needed (g) 83(b) 166.4 kJ 5.068 83(c) 884.5 kJ 26.94 84(a) 1582.3 kJ 48.192 84(c) 100.97 kJ 3.0752 86. Balance the equation, then use the method described in the answer to Question 57. (a) 2 C(s) + 2 Cl 2 (g) + TiO 2 (s) TiCl 4 (g) + 2 CO(g) (b) X = (1 mol) X(TiCl 4 ) + (2 mol) X(CO) (2 mol) X(C) (2 mol) X(Cl 2 ) (1 mol) X(TiO 2 ) (X = H f , S, or G f ) H = (1 mol) (763.2 kJ/mol) + (2 mol) (110.525 kJ/mol) (2 mol) (0 kJ/mol) (2 mol) (0 kJ/mol) (1 mol) (939.7 kJ/mol) = 44.6 kJ S = (1 mol) (354.9 J K 1 mol 1 ) + (2 mol) (197.674 J K 1 mol 1 ) (2 mol) (5.740 J K 1 mol 1 ) (2 mol) (223.066 K 1 mol 1 ) (1 mol) (49.92 J K 1 mol 1 ) = 242.7 J K 1 G = (1 mol) (726.7 kJ/mol) + (2 mol) (137.168 kJ/mol) (2 mol) (0 kJ/mol) (2 mol) (0 kJ/mol) (1 mol) (884.5 kJ/mol) = 116.5 kJ (c) G is negative, so the reaction is product-favored. 17 Chapter 18: Thermodynamics: Directionality of Chemical Reactions 87 . Balance the equations, then use the method described in the answer to Question 56. (a) 2 CuO(s) 2 Cu(s) + O 2 (g) CuO(s) + C(graphite) Cu(s) + CO(g) G = (1 mol) G (Cu) + (1 mol) G (CO) (1 mol) G (CuO) (1 mol) G (C) = (1 mol) (0 kJ/mol) + (1 mol) (137.168 kJ/mol) (1 mol) (129.7 kJ/mol) (1 mol) (0 kJ/mol) = 7.5 kJ (b) 2 Ag 2 O(s) 4 Ag(s) + O 2 (g) Ag 2 O(s) + C(graphite) 2 Ag(s) + CO(g) G = (2 mol) G (Ag) + (1 mol) G (CO) (1 mol) G (Ag 2 O) (1 mol) G...
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MSJ2e_Ch18_ISM2_June26_p803 - Chapter 18: Thermodynamics:...

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