For the reaction at 298 K,
$2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}$
$\Delta H=400 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S=0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
From the expression,
$\Delta G=\Delta H-T \Delta S$
Assuming the reaction at equilibrium, $\Delta T$ for the reaction would be:
$T=(\Delta H-\Delta G) \frac{1}{\Delta S}$
$=\frac{\Delta H}{\Delta S}(\Delta G=0$ at equilibrium $)$
$=\frac{400 \mathrm{~kJ} \mathrm{~mol}^{-1}}{0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}$
T = 2000 K
For the reaction to be spontaneous, $\Delta G$ must be negative. Hence, for the given reaction to be spontaneous, $T$ should be greater than $2000 \mathrm{~K}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.