chapter18_part2

# chapter18_part2 - 852 Chapter 18 Thermodynamics...

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Chapter 18: Thermodynamics: Directionality of Chemical Reactions 852 89 . Answer: Reaction (b) needs 5.068 g; reactions (c) needs 26.94 g. Strategy and Explanation: Adapt the method described in the solution to Question 6.85. 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 reactions that require work be done in Question 87 (b) and (c). C(graphite) + O 2 (g) CO 2 (g) This equation is identical to the equation describing the formation reaction for CO 2 , so Δ G° = –394.359 kJ/mol The calculation for 87(b) looks like this: 166.4 kJ endergonic reaction " 1 mol C 394.359 kJ provided " 12.011 g C 1 mol C = 5.068 g C The rest of the results are described in the table below. Endergonic Reaction Reference Calculated Δ G° (X) (kJ) Mass of graphite needed (g) 87(b) 166.4 kJ 5.068 87(c) 884.5 kJ 26.94 90. Answer: Reaction (a) needs 13.945 g; reactions (c) needs 0.89046 g. Strategy and Explanation: Adapt the method described in the solution to Question 6.85. Find the Gibbs free energy of the hydrogen reaction and use that as a thermochemical conversion factor to supply the Gibbs free energy of the reactions that require work be done in Question 88 (a) and (c). H 2 (g) + O 2 (g) H 2 O(g) This equation is identical to the formation reaction equation for H 2 O(g), so Δ G° = –228.572 kJ/mol. The calculation for 88(a) looks like this: 1582.3 kJ endergonic reaction " 1 mol H 2 228.572 kJ provided " 2.0158 g H 2 1 mol H 2 = 13.954 g H 2 The rest of the results are described in the table below. Endergonic Reaction Reference Calculated Δ G° (X) (kJ) Mass of H 2 needed (g) 88(a) 1582.3 kJ 13.945 88(c) 100.97 kJ 0.89046 91 . Answer: (a) 2 C(s) + 2 Cl 2 (g) + TiO 2 (s) TiCl 4 (g) + 2 CO(g) (b) Δ H° = –44.6 kJ; Δ S° = 242.7 J K –1 ; Δ G° = –116.5 kJ (c) product-favored (d) more reactant-favored Strategy and Explanation: Balance the equation, then use the method described in the solution to Questions 51 and 57. (a) 2 C(s) + 2 Cl 2 (g) + TiO 2 (s) TiCl 4 (g) + 2 CO(g) Check: 2 C, 4 Cl, 1 Ti, 2 O (b) Δ 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 = Δ H f , S, or Δ G f ) Δ 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 Δ 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 Δ 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

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Chapter 18: Thermodynamics: Directionality of Chemical Reactions 853 (c) Δ G° is negative, so the reaction is product-favored. (d) As the reaction temperature increases, heat energy is added to the system. The reaction is
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chapter18_part2 - 852 Chapter 18 Thermodynamics...

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