The heat of combustion of liquid benzene at constant volume is at . What is the heat of combustion o — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The combustion reaction is: $C_6H_6(l) + \frac{15}{2} O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$. The change in gaseous moles is $\Delta n_g = 6 - 7.5 = -1.5$. We use the relation $\Delta H = \Delta U + \Delta n_g RT$. Given $\Delta U = -3263.9 \text{ kJ mol}^{-1}$, $\Delta H = -3263.9 + (-1.5 \times 8.314 \times 10^{-3} \times 298.15) = -3263.9 - 3.71 = -3267.6 \text{ kJ mol}^{-1}$.