JEE Main 2024 — 5 April, Shift 2 — Question 79

The combustion of 1 mol of benzene is represented as $\mathrm{C_6H_6(l) + \tfrac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)}$. The standard enthalpy of combustion of 2 mol of benzene is $-x$ kJ. Calculate the value of $x$, given that :

(1) Standard Enthalpy of formation of 1 mol of $\mathrm{C}_{6} \mathrm{H}_{6}(1)$, for the reaction 6 C (graphite) $+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(1)$ is $48.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

(2) Standard Enthalpy of formation of 1 mol of $\mathrm{CO}_{2}(\mathrm{~g})$, for the reaction $\mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)}$ is $-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

(3) Standard and Enthalpy of formation of 1 mol of $\mathrm{H}_{2} \mathrm{O}(1)$, for the reaction $\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1)$ is $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$.