$\begin{aligned} & \mathrm{S}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow…
& \mathrm{S}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+2 x \mathrm{kcal} \\
& \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+y \mathrm{kcal}
\end{aligned}$
The heat of formation of $\mathrm{SO}_2(\mathrm{~g})$ is given by :
- $x+y \mathrm{kcal}$
- $y-2 x \mathrm{kcal}$
- $\frac{2 x}{y} \mathrm{kcal}$
- $2 x+y \mathrm{kcal}$
Solution
(ii) $\mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+y$ kcal $\Delta \mathrm{H}_2$
(i) - (ii)
$\begin{aligned}
& \mathrm{S}(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_2(\mathrm{~g}) \Delta \mathrm{H} \\
& \Delta \mathrm{H}=\Delta \mathrm{H}_1-\Delta \mathrm{H}_2 \\
& =-2 x-(-y)=(y-2 x) \mathrm{kcal}
\end{aligned}$
Asked in: JEE Main 2025 (24 Jan Shift 2)