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JEE MAIN 2024
05-04-2024 S1
Question
The heat of combustion of solid benzoic acid at constant volume is -321.30 kJ at $27^{\circ} \mathrm{C}$. The heat of combustion at constant pressure is $(-321.30-x R) k J$, the value of $x$ is
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Solution
$\begin{aligned} & \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(\mathrm{S})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 7 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\ell) \\ & \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\ & =-321.30-\frac{1}{2} \frac{\mathrm{R}}{100} \times 300 \\ & =(-321.30-150 \mathrm{R}) \mathrm{kJ}\end{aligned}$
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