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.
