$$\begin{aligned} & \mathrm{S}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+2 x \mathrm{kcal} \\ & \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+y \mathrm{kcal} \end{aligned}$$

The heat of formation of $\mathrm{SO}_2(\mathrm{~g})$ is given by :

<p>To determine the heat of formation for $\mathrm{SO}_2(\mathrm{~g})$, we need to consider the given reactions and their enthalpy changes.</p> <p>The reactions are:</p> <p><p>$\mathrm{S}(\mathrm{~g}) + \frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g}) + 2x \mathrm{kcal}$</p></p> <p><p>$\mathrm{SO}_2(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g}) + y \mathrm{kcal}$</p></p> <p>For reaction 2, we have:</p> <p>$ \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} = -y $</p> <p>Given that:</p> <p>$ -\mathrm{y} = -2 \mathrm{x} - \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} $</p> <p>Rearranging this equation gives:</p> <p>$ \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} = y - 2x $</p> <p>Thus, the heat of formation for $\mathrm{SO}_2(\mathrm{~g})$ is $y - 2x$ kcal.</p>