Q26 / 90 · JEE Main 2012 (12 May Online) · Chemistry
The difference between the reaction enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}\right)$ and reaction internal energy change $\left(\Delta_{\mathrm{r}} \mathrm{U}\right)$ for the reaction:
$$
2 \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+15 \mathrm{O}_2(\mathrm{~g}) \longrightarrow
$$
at $300 \mathrm{~K}$ is $\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$
- A$0 \mathrm{~J} \mathrm{~mol}^{-1}$
- B$2490 \mathrm{~J} \mathrm{~mol}^{-1}$
- C$-2490 \mathrm{~J} \mathrm{~mol}^{-1}$
- Correct answerD$-7482 \mathrm{~J} \mathrm{~mol}^{-1}$
Solution
$\Delta H=\Delta U+\Delta n_{\mathrm{g}} R T$
For the reaction $\Delta n_{\mathrm{g}}=12-15=-3$
$$
\begin{aligned}
\Delta H-\Delta U & =-3 \times 8.314 \times 300 \\
& =-7482 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.