JEE MAIN - Chemistry (2025 - 24th January Evening Shift - No. 4)
$$\begin{aligned} & \mathrm{S}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+2 x \mathrm{kcal} \\ & \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+y \mathrm{kcal} \end{aligned}$$
The heat of formation of $\mathrm{SO}_2(\mathrm{~g})$ is given by :
Explanation
$\underset{(\mathrm{g})}{\mathrm{SO}_2}+\frac{1}{2} \underset{(\mathrm{~g})}{\mathrm{O}_2} \longrightarrow \underset{(\mathrm{~g})}{\mathrm{SO}_3} \quad \Delta \mathrm{H}=-\mathrm{y}$
$$\begin{aligned} &\begin{aligned} & \Delta \mathrm{H}_{\mathrm{r}}=\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_3}-\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} \\ & -\mathrm{y}=-2 \mathrm{x}-\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} \\ & \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2}=\mathrm{y}-2 \mathrm{x} \end{aligned}\\ &\text { Option (2) } \end{aligned}$$
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