Energia Livre e Termodinâmica

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Chapter 18 Free Energy and Thermodynamics 407 K = e RT = = = 1.91 X 10⁴⁷ Check: The units (none) are correct. The free energy change was very negative, indicating a spontaneous reac- tion. This results in a very large K. (b) Given: 2H₂S(g) + at Find: K at 525 K Conceptual Plan: = - then = - then J/K kJ/K then T AG 1kJ then Solution: Reactant/Product from Appendix IIB) H₂S(g) -20.6 0.0 S₂(g) 128.6 Be sure to pull data for the correct formula and phase. = = + = [2(0.0kJ) + [2(-20.6kJ)] = [128.6kJ] [-41.2kJ] = +169.8kJ Reactant/Product K from Appendix IIB) 205.8 H₂(g) 130.7 S₂(g) 228.2 Be sure to pull data for the correct formula and phase. = = + = + 1(228.2J/K)] - = [489.6J/K] [411.6J/K] = +78.0J/K then = = = +169.8kJ = +128.85 kJ = +1.2885 X 10⁵ J then = K Rearrange to solve for K. K Check: The units (none) are correct. The free energy change is positive, indicating a nonspontaneous reaction. This results in a very small K. Copyright © 2017 Pearson Education, Inc.