# For the reaction at 298 K,

Question:

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 Δand Δto be constant over the temperature range?

Solution:

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 $2000 \mathrm{~K}$.