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Chapter 18 Free Energy and Thermodynamics 415 (c) Given: 2H₂(g) + 2NO(g) + 2H₂O(g) Find: and effect of increasing T on Conceptual Plan: = Solution: Reactant/Product from Appendix IIB) H₂(g) 0.0 NO(g) 87.6 N₂(g) 0.0 -228.6 Be sure to pull data for the correct formula and phase. = = + + = [1(0.0kJ) + [2(0.0kJ) + 2(87.6kJ)] = [-457.2kJ] [175.2kJ] = -632.4kJ Because the number of moles of gas is decreasing, the entropy change is negative; so G will become less negative with increasing temperature. Check: The units (kJ) are correct. The free energy change is negative because water has such a low free energy of formation. (d) Given: N₂O(g) + 3H₂O(g) Find: and effect of increasing T on AG Conceptual Plan: = Solution: Reactant/Product from Appendix IIB) NH₃(g) -16.4 0.0 103.7 -228.6 Be sure to pull data for the correct formula and phase. = = + + = [1(103.7kJ) + 3(-228.6kJ)] [2(-16.4kJ) + 2(0.0kJ)] = [-582.1kJ] [-32.8kJ] = -549.3kJ Because the number of moles of gas is constant, the entropy change will be small and slightly negative; so the magnitude of G will decrease with increasing temperature. Check: The units (kJ) are correct. The free energy change is negative because water has such a low free energy of formation. The entropy change is negative once the values are reviewed = -9.6J/K). 18.95 With one exception, the formation of any oxide of nitrogen at 298 K requires more moles of gas as reactants than are formed as products. For example, 1 mole of requires 0.5 mole of O₂ and 1 mole of N₂. One mole of requires 1 mole of and 1.5 moles of O₂, and on. The exception is NO, where 1 mole of NO requires 0.5 mole of O₂ and 0.5 mole of + NO(g). This reaction has a positive because what is essentially mixing of the N and O has taken place in the product. 18.97 Given: 2X(g); Pinitial = 755 torr, = 103 torr at 298 K; Pinitial = 748 torr, X = 532 torr at 755 K Find: Conceptual Plan: At each temperature Pinitial K then K1, T₂ Use stoichiometry to calculate final and K = Solution: Use stoichiometry of 2X(g); = Pinitial X₂ that at T₁ = 298 K, = Pinitial X₂ 1/2 = 755 torr torr) = 703.5 torr and at T₂ = 755 K, = Pinitial 1/2 Copyright © 2017 Pearson Education, Inc.