The difference between the reaction enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}ight)$ and reaction…
$2 \mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l})+15 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 12 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
at $300 \mathrm{~K}$ is $\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}ight)$
- $0 \mathrm{~J} \mathrm{~mol}^{-1}$
- $2490 \mathrm{~J} \mathrm{~mol}^{-1}$
- $-2490 \mathrm{~J} \mathrm{~mol}^{-1}$
- $-7482 \mathrm{~J} \mathrm{~mol}^{-1}$
Solution
For the reaction $\Delta n_{\mathrm{g}}=12-15=-3$ $\Delta H-\Delta U=-3 \times 8.314 \times 300$
$=-7482 \mathrm{~J} \mathrm{~mol}^{-1}$
Asked in: JEE-TOPICTESTS-CHEMISTRY