The standard heat of combustion of liquid benzene () at constant volume is measured as at . Calculat — Thermodynamics and Thermochemistry Chemistry Question
Question
The standard heat of combustion of liquid benzene ($C_6H_6$) at constant volume is measured as $-3263.9 \text{ kJ mol}^{-1}$ at $25^\circ\text{C}$. Calculate the absolute magnitude of the heat of combustion of benzene at constant pressure in $\text{kJ mol}^{-1}$ (rounded to one decimal place). ($R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$)
💡 Solution & Explanation
The correct combustion equation is: $C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$. The net change in gaseous stoichiometric moles is $\Delta n_g = 6 - 7.5 = -1.5$. The thermodynamic relationship is $\Delta H = \Delta U + \Delta n_g RT$. Substitute values: $\Delta H = -3263.9 + (-1.5 \times 8.314 \times 10^{-3} \text{ kJ K}^{-1}\text{mol}^{-1} \times 298.15 \text{ K}) = -3263.9 - 3.718 = -3267.618 \text{ kJ mol}^{-1}$. The absolute magnitude effectively rounds to $3267.6$.