JEE Main 2025 — 28 January, Evening Shift — Question 43

$\Delta \mathrm{H}_{\mathrm{f}}\left[\mathrm{CO}_{2}(\mathrm{~g})\right]=-393.5 \mathrm{~kJ} /$ mole $\Delta \mathrm{H}_{\mathrm{f}}\left[\mathrm{H}_{2} \mathrm{O}(\ell)\right]=-286.0 \mathrm{~kJ} /$ mole

$\Delta \mathrm{H}_{\mathrm{c}}\left[\mathrm{C}_{6} \mathrm{H}_{6}\right]=-3267.0 \mathrm{~kJ} / \mathrm{mole}$

$\Delta \mathrm{H}_{\mathrm{f}} \mathrm{C}_{6} \mathrm{H}_{6}=(?)$

$\mathrm{C}_{6} \mathrm{H}_{6}+\frac{15}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 6 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\ell)$

$\Delta \mathrm{H}_{\mathrm{R}}=\Delta \mathrm{H}_{\mathrm{C}}=\Sigma \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{P})-\Sigma \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{R})$

$-3267=6 \times(-393.5)+3(-286)-\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)$

$\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=48 \mathrm{~kJ} /$ mole