For the combustion of 1 mole of liquid benzene at \(298 \mathrm{~K}\), the heat of reaction at constant pressure is \(-3268 \mathrm{~kJ} \mathrm{~mol}^{-1}\), what is heat of combustion at constant volume? \(\left(\mathrm{R}=8 \cdot 314 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right.\)

\(\begin{array}{l}
\mathrm{C}_{6} \mathrm{H}_{\mathrm{b}(\ell)}+\frac{15}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 6 \mathrm{CO}_{2(\mathrm{~g})}+3 \mathrm{H}_{2} \mathrm{O}_{(\ell)} \\
\mathrm{Q}_{\mathrm{p}}=-3268 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{n}_{(\mathrm{g})}=-1.5 \\
\mathrm{~T}=298 \mathrm{~K}, \quad \mathrm{R}=8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \\
\mathrm{Q}_{v}=?
\end{array}\)
Now,
\(\begin{aligned}
\mathrm{Q}_{\mathrm{p}} &=\mathrm{Q}_{\mathrm{v}}+\Delta \mathrm{n} \mathrm{RT} \\
\therefore \mathrm{Q}_{\mathrm{v}} &=\mathrm{Q}_{\mathrm{p}}-\Delta \mathrm{n} \mathrm{RT} \\
&=-3268-\left[(-1.5) \times 8.314 \times 10^{-3} \times 298\right] \\
&=-3268+3.716 \\
\therefore \mathrm{Q}_{\mathrm{v}} &=-3264.284 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}\)