chapter14_part1

chapter14_part1 - Chapter 14: Chemical Equilibrium 633...

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Chapter 14: Chemical Equilibrium 633 End-of-Chapter Solutions for Chapter 14 Summary Problem 1. Answer: see equations below (other answers are possible) Strategy and Explanation: The production of H 2 needs to be accomplished with a minimum production of soot (solid carbon) and CO. A couple obvious choices of reactions that would probably not be suitable are: Reaction 1a: C 8 H 18 ( l ) 8 C(s) + 9 H 2 (g) Reaction 1b: C 8 H 18 ( l ) + 8 H 2 O(g) 8 CO(g) + 17 H 2 (g) They would be ill-advised because of the identity of the second product. A reaction that might be suitable could be an adaptation of the hydrocarbon reaction with water optimized to produce CO 2 instead of CO: Reaction 1c: C 8 H 18 ( l ) + 16 H 2 O(g) 8 CO 2 (g) + 25 H 2 (g) The reaction would need to be done at high temperature with a specially-designed metal catalyst that could help capture and assist in the oxidation of any CO formed. An 1997 article describing this reaction, called octane- steam reforming, by Birdsell, Willms and Dye, at the Los Alamos National Laboratory may be found at this website: http://willms.lanl.gov/Willms%20publications/Pure%20hydrogen%20production%20from%20octane,%20ethan ol,%20methanol%20and%20methane%20reforming%20using%20a%20palladium%20membrane%20reactor% 20(1997).pdf 2. Answer: ill-advised reactions: 249.952 kJ, 1300.296 kJ, preferred reaction: 3147.330 kJ (Note: answers depend on the reactions selected in 1.) Strategy and Explanation: This is a Hess’ law question. Use Equation 6.11, the molar enthalpies of formation, ± H f o (from Appendix J), and the stoichiometry of the balanced equations to determine the molar enthalpy change for each reaction. We will start with the calculation for the ill-advised reactions: ± 1a = (8 mol) ² ± H f o (C(s)) + (9 mol) ² ± H f o (H 2 (g)) – (1 mol) ² ± H f o (C 8 H 18 ( l )) ± 1a = (8 mol) ² (0 kJ/mol) + (9 mol) ² (0 kJ/mol) – (1 mol) ² (–249.952 kJ/mol) = 249.952 kJ ± 1b = (8 mol) ² ± H f o (CO) + (17 mol) ² ± H f o (H 2 (g)) – (1 mol) ² ± H f o (C 8 H 18 ) – (8 mol) ² ± H f o (H 2 O(g)) ± 1b = (8 mol) ² (–110.525 kJ/mol) + (25 mol) ² (0 kJ/mol) – (1 mol) ² (–249.952 kJ/mol) – (8 mol) ² (–241.818 kJ/mol) = 1300.296 kJ Then do the calculation for the preferred reaction: ± 1c = (8 mol) ² ± H f o (CO 2 ) + (25 mol) ² ± H f o (H 2 (g)) – (1 mol) ² ± H f o (C 8 H 18 ) – (16 mol) ² ± H f o (O 2 ) ± 1c = (8 mol) ² (–393.509 kJ/mol) + (25 mol) ² (0 kJ/mol) – (1 mol) ² (–249.952 kJ/mol) – (16 mol) ² (–241.818 kJ/mol) = 3147.330 kJ As discussed in Section 14.7, when a reaction produces more molecules of products than it has as reactants, we will predict that the entropy increases. Reaction 1a: 1 molecule 17 molecules entropy increases Reaction 1b: 9 molecules 25 molecules entropy increases Reaction 1c: 17 molecules 33 molecules entropy increases
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Chapter 14: Chemical Equilibrium 634 At higher temperatures, the entropy effect will become more important than the energy effect in determining the position of equilibrium. Because each of these reactions has a predicted increase in the entropy, they will each become product-favored at high enough temperature.
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chapter14_part1 - Chapter 14: Chemical Equilibrium 633...

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