General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 19 - Spontaenous Change: Entropy and Gibbs Energy - Exercises - (symbol) G and K as Functions of Temperature - Page 857: 57

Answer

1360 K

Work Step by Step

We find: $\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$ $\implies T=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{\Delta S^{\circ}}=\frac{(-24.8\,kJ)-(-45.5\,kJ)}{0.0152\,kJ/K}=1360\,K$
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