Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 15 - Chemical Thermodynamics - Exercises - Gibbs Free Energy Changes and Spontaneity - Page 605: 107

Answer

(a) 514 kJ (b) -1295 kJ (c) 34.9 kJ

Work Step by Step

We find: $\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$ $T= (273.15+45)K=318.15\,K$ (a) $\Delta G^{\circ}=(293000\,J)-(318.15\,K)(-695\,J/K)$ $=514000\,J=514\,kJ$ (b) $\Delta G^{\circ}=(-1137\,kJ)-(318.15\,K)(0.496\,kJ/K)$ $=-1295\,kJ$ (c) $\Delta G^{\circ}=(-86600\,J)-(318.15\,K)(-382\,J/K)$ $=34900\,J=34.9\,kJ$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.