JEE MAIN - Chemistry (2025 - 28th January Evening Shift - No. 24)
Consider the following data :
Heat of formation of $\mathrm{CO}_2(\mathrm{g})=-393.5 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$
Heat of formation of $\mathrm{H}_2 \mathrm{O}(\mathrm{l})=-286.0 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$
Heat of combustion of benzene $=-3267.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The heat of formation of benzene is __________ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest integer)
Answer
48
Explanation
$$\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{f}}\left[\mathrm{CO}_2(\mathrm{~g})\right]=-393.5 \mathrm{~kJ} / \mathrm{mole} \\
& \Delta \mathrm{H}_{\mathrm{f}}\left[\mathrm{H}_2 \mathrm{O}(\ell)\right]=-286.0 \mathrm{~kJ} / \mathrm{mole} \\
& \Delta \mathrm{H}_{\mathrm{c}}\left[\mathrm{C}_6 \mathrm{H}_6\right]=-3267.0 \mathrm{~kJ} / \mathrm{mole} \\
& \Delta \mathrm{H}_{\mathrm{f}} \mathrm{C}_6 \mathrm{H}_6=(?) \\
& \mathrm{C}_6 \mathrm{H}_6+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\ell) \\
& \Delta \mathrm{H}_{\mathrm{R}}=\Delta \mathrm{H}_{\mathrm{C}}=\Sigma \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{P})-\Sigma \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{R}) \\
& -3267=6 \times(-393.5)+3(-286)-\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{C}_6 \mathrm{H}_6\right) \\
& \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{C}_6 \mathrm{H}_6\right)=48 \mathrm{~kJ} / \mathrm{mole}
\end{aligned}$$
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