Given are the following data at 298 K. ${\mathrm{\Delta }}_{\mathrm{f}}{H}^{\circ }\left({\mathrm{SO}}_2,\mathrm{g}\right)=-296.8\mathrm{kJ}{\mathrm{mol}}^{-1}\text{ and }{\mathrm{\Delta }}_f{H}^{\circ }\left({\mathrm{SO}}_3,\mathrm{g}\right)=-395.7\mathrm{kJ}{\mathrm{mol}}^{-1}$ The value of Li/.!° for the reaction ${\mathrm{SO}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{SO}}_3\left(\mathrm{g}\right)\text{ is }$ 

$\text{ (i) }\mathrm{S}\left(\mathrm{s}\right)+{\mathrm{O}}_2\text{ (g) }⟶{\mathrm{SO}}_2\left(\mathrm{g}\right);\mathrm{\Delta }{H}^{\circ }=-296.8\mathrm{kJ}{\mathrm{mol}}^{-1}$

$\text{ (ii) }\mathrm{S}\left(\mathrm{s}\right)+\frac{3}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{SO}}_3\left(\mathrm{g}\right);\mathrm{\Delta }{H}^{\circ }=-395.7\mathrm{kJ}{\mathrm{mol}}^{-1}$

The required reaction ${\mathrm{SO}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{SO}}_3\left(\mathrm{g}\right)$ is obtained as Eq.(ii) - Eq. (i) Hence, ${\mathrm{\Delta }}_r{H}^{\circ }=[-395.7-(-296.8\left)\right]\mathrm{kJ}{\mathrm{mol}}^{-1}=-98.9\mathrm{kJ}{\mathrm{mol}}^{-1}$ For the given reaction ,$\mathrm{\Delta }{V}_{\mathrm{g}}=-1/2$. Hence $\begin{array}{r}{\mathrm{\Delta }}_{\mathrm{r}}{U}^{\circ }={\mathrm{\Delta }}_{\mathrm{r}}{H}^{\circ }-\left(\mathrm{\Delta }{v}_{\mathrm{g}}\right)RT=-98.9\mathrm{kJ}{\mathrm{mol}}^{-1}-(-1/2)\left(8.314\times {10}^{-3}{\mathrm{kJK}}^{-1}{\mathrm{mol}}^{-1}\left(298\mathrm{K}\right)\right.\\ =-(98.9+1.24){\mathrm{kJmol}}^{-1}=-97.66\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$