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Chapter 18 Free Energy and Thermodynamics 399 Solution: Reactant/Product (kJ/mol from Appendix IIB) NH₄Cl(s) -314.4 HCl(g) -92.3 -45.9 Be sure to pull data for the correct formula and phase. = = + - = + 1(-45.9kJ)] [1(-314.4kJ)] = [-138.2kJ] [-314.4kJ] = +176.2kJ then Reactant/Product K from Appendix IIB) 94.6 HCl(g) 186.9 192.8 Be sure to pull data for the correct formula and phase. = (reactants) = + - = (1(186.9 J/K) + 1(192.8J/K)] [1(94.6J/K)] = [379.7J/K] [94.6J/K] = then T = 273.15 + 25 °C = 298 K then +285.1 K X 1000 J = +0.2851 kJ/K then = - = +176.2kJ = +91.2 kJ = +9.12 X 10⁴ J; so the reaction is nonspontaneous. It can be made spontaneous by raising the temperature. Check: The units (kJ, J/K, and kJ) are correct. The reaction requires the breaking of a bond, we expect that this will be an endothermic reaction. We expect a positive entropy change because we are increasing the number of moles of gas. Because the positive enthalpy term dominates at room temperature, the reaction is nonspontaneous. The second term can dominate if we raise the temperature high enough. (c) Given: + Fe₂O₃(s) 2Fe(s) + 3H₂O(g) at 25 °C Find: and spontaneity. Can temperature be changed to make it spontaneous? Conceptual Plan: = - then = - then K then J/K kJ/K then 1kJ K = 273.15 + °C 1000J Solution: Reactant/Product (kJ/mol from Appendix IIB) H₂(₈) 0.0 Fe₂O₃(s) -824.2 Fe(s) 0.0 H₂O(₈) -241.8 Be sure to pull data for the correct formula and phase. = = [2(AHᵢ(Fe(s))) + - + = [2(0.0kJ) + - [3(0.0kJ) + 1(-824.2kJ)] = [-725.4kJ] [-824.2kJ] = kJ then Copyright © 2017 Pearson Education, Inc.