Answer:
[tex]\begin{array}{l}{\text { The value of } \Delta \mathrm{G}^{\circ} \text { at } 25^{\circ} \mathrm{C} \text { for the decomposition of gaseous sulfur trioxide to solid }} \\ {\text { elemental sulfur and gaseous oxygen is }+740.8 \mathrm{kJ} / \mathrm{mol}}\end{array}[/tex]
Option: A
Explanation:
[tex]\begin{array}{l}{\text { The value of } \Delta \mathrm{G}^{\circ} \text { at } 25^{\circ} \mathrm{C} \text { in the following reaction can be calculated as follows: }} \\ {2 \mathrm{SO}_{3}(\mathrm{g}) \rightarrow 2 \mathrm{S}(\mathrm{s}, \text { rhombic })+3 \mathrm{O}_{2}(\mathrm{g})}\end{array}[/tex]
[tex]\begin{array}{l}{\Delta \mathrm{G}^{\circ} \text { is Standard Gibbs free energy change which can be calculated from the standard free }} \\ {\text { energies of formation of the products and the reactants from the following equation: }}\end{array}[/tex]
[tex]\begin{array}{l}{\Delta \mathrm{G}^{\circ}=\Sigma \mathrm{G}_{\mathrm{f}(\text { products })}^{\circ}-\Sigma \mathrm{G}_{\text {creatants }}^{\circ}} \\ {\Delta \mathrm{G}^{\circ}=[\mathrm{Sum} \text { of standard free energies of formation of products }]-[\mathrm{Sum}\text { of standard } } \\ {\text { free energies bf formation of reactants] }}\end{array}[/tex]
[tex]\begin{array}{l}{\text { Now here standard values of } \Delta G^{\circ} f(k J / m o l) \text { for } S=0, O_{2}=0 \& S O_{3}=-370.4} \\ {\text { Hence these values can be substituted in above equation: }} \\ {\Delta G^{\circ}=\left[2 G_{f}^{\circ}(0)+3 G_{f}(0)\right]-[2(-370.4)]} \\ {\Delta G^{\circ}=[-0+0]-[-740.8]} \\ {\Delta G^{\circ}=+740.8 \mathrm{kJ} / \mathrm{mol}}\end{array}[/tex]
