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At what temperature does the reaction below change from favoring products to favoring reactants?
${SO}_{2} (g)$ $O_{2} (g)$ ${SO}_{3} (g)$
$Δ H_{f}^{\circ}, kJ \cdot {mol}^{- 1}$ -296.8 _ -395.7
$S^{\circ}, kJ \cdot {mol}^{- 1} \cdot K^{- 1}$ 0.248 0.205 0.257

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Detailed Solution

$\begin{array}{l} {SO}_{2} + \frac{1}{2} O_{2} ⟶ {SO}_{3} \\ Δ H^{\circ} = - 395.7 + 296.8 = - 98.9 \\ Δ S^{\circ} = 0.257 - 0.248 - \frac{0.205}{2} = - 0.0935 \\ Δ G^{\circ} = Δ H^{\circ} - T Δ S > 0 \\ \Rightarrow - 98.9 > T \times (- 0.0935) \end{array}$

$\Rightarrow T > \frac{98.9}{0.0935} = 1058 K$

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