Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 17 - Sections 17.1-17.9 - Exercises - Problems by Topic - Page 854: 63

Answer

-29.4 kJ

Work Step by Step

We obtain the required reaction by reversing the first reaction and then adding three times the second reaction with it. Therefore, $\Delta G^{\circ}_{rxn}=-[\Delta G^{\circ}_{rxn}(\text{first reaction})]+3[\Delta G^{\circ}_{rxn}(\text{second reaction})]$ $=-(-742.2\,kJ)+3(-257.2\,kJ)=-29.4\,kJ$
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.