The combustion of benzene releases CO_ 2 (g) and H_ 2 O(l). At 25^ circ C, the heat of combustion at constant volume (Delta U) is -3263.9text kJ mol ^ -1 . Calculate the heat of combustion at constant pressure (Delta H) in text kJ mol ^ -1 . (R = 8.314text J K ^ -1 text mol ^ -1 )

Home › MCQs › EEJ MAIN Chemistry › Question #7129

EEJ MAIN Chemistry QUESTION #7129

Question 1

The combustion of benzene releases $CO_{2}(g)$ and $H_{2}O(l)$. At $25^{\circ}C$, the heat of combustion at constant volume ($\Delta U$) is $-3263.9\text{ kJ mol}^{-1}$. Calculate the heat of combustion at constant pressure ($\Delta H$) in $\text{kJ mol}^{-1}$. ($R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$)

3260
-3267.6✔️
4152.6
-452.46

Correct Answer Explanation

The relationship is given by $\Delta H = \Delta U + \Delta n_g RT$.
Reaction: $C_{6}H_{6}(l) + 7.5O_{2}(g) \rightarrow 6CO_{2}(g) + 3H_{2}O(l)$.
$\Delta n_g = 6 - 7.5 = -1.5$.
$\Delta H = -3263.9 + (-1.5 \times 8.314 \times 10^{-3} \times 298) = -3263.9 - 3.7 = -3267.6\text{ kJ mol}^{-1}$.

More Options