For a reaction $\Delta \mathrm{H}=-30 \mathrm{~kJ}$ and $\Delta \mathrm{S}=-45 \mathrm{~J} \mathrm{~K}^{-1}$…
For a reaction $\Delta \mathrm{H}=-30 \mathrm{~kJ}$ and $\Delta \mathrm{S}=-45 \mathrm{~J} \mathrm{~K}^{-1}$, at what temperature reaction changes from spontaneous to non spontaneous?
$777 \cdot 0 \mathrm{~K}$
$675 \cdot 0 \mathrm{~K}$
$666 \cdot 6 \mathrm{~K}$
$375 \cdot 0 \mathrm{~K}$
Solution
$\begin{array}{l}
\Delta \mathrm{H}=-30 \mathrm{~kJ}, \Delta \mathrm{S}=-45 \mathrm{~J} \mathrm{~K}^{-1}=-0.045 \mathrm{~kJ} \mathrm{~K}^{-1} \\
\therefore \mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{-30 \mathrm{~kJ}}{-0.045 \mathrm{~kJ} \mathrm{~K}^{-1}}=666.67 \mathrm{~K}
\end{array}$
Since $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are both negative, the reaction is spontaneous at low temperature and nonspontaneous above $666.67 \mathrm{~K}$.