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

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

Chapter 21 - Chemistry of the Main-Group Elements I: Groups 1, 2, 13, and 14 - 21-10 Concept Assessment - Page 955: 21-10


By using equation (7.21), $\Delta H^{\circ}$ for each reaction can be determined, and the only difference is replacing 1 mol $\mathrm{CO}_{2}$ in equation (21.29) by 1 mol $\mathrm{CO}$ in equation (21.30) . The difference in the heat of reaction is $$-110.5 \mathrm{kJ} / \mathrm{mol} \mathrm{CO}(\mathrm{g})-\left[-393.5 \mathrm{kJ} / \mathrm{mol} \mathrm{CO}_{2}(\mathrm{g})\right]=283 \mathrm{kJ} $$So, reaction (21.30) liberates 238 kJ less heat than reaction (21.29).

Work Step by Step

See answer.
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.