The heat of combustion of solid benzoic acid at constant volume is $-321.30 \mathrm{~kJ}$ at $27^{\circ} \mathrm{C}$. The heat of combustion at constant pressure is $(-321.30-x \mathrm{R}) \mathrm{~kJ}$, the value of $x$ is __________.
Answer (integer)
150
Solution
<p>$$\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(\mathrm{s})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 7 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{I})$$</p>
<p>$$\begin{aligned}
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_9 R T \\
& \Delta \mathrm{H}=-321.30-\frac{1}{2} \mathrm{R} \times 300
\end{aligned}$$</p>
<p>So, $x=150$</p>
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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