For the combustion of 1 mole of liquid benzene at $298 \mathrm{~K}$, the heat of reaction at constant…

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? $\left(\mathrm{R}=8 \cdot 314 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right.$
  1. $-3264 \cdot 2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  2. $-1632 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  3. $-6728 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  4. $-672 \cdot 8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

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

Asked in: MHT CET 2020 (13 Oct Shift 2)

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