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

Published by Pearson Prentice Hal

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

#### Answer

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.

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