Question

For the reaction at $298\ K$,
$2A+B→C$
$∆H = 400\ kJ mol^{-1}$ and $∆S = 0.2\ kJ K^{-1} mol^{-1}$
At what temperature will the reaction become spontaneous considering $∆H$ and $∆S$ to be constant over the temperature range.

Answer 1

From the expression,
$∆G = ∆H – T∆S$
Assuming the reaction at equilibrium, $∆T$ for the reaction would be:
$T = (△H-△G)\frac {1}{△S}$
$T= \frac {△H}{△S}$ ($∆G = 0$ at equilibrium)

$T=\frac { 400\ kJ mol^{-1}}{0.2\ kJK^{-1} mol^{-1}}$
$T = 2000\ K$
For the reaction to be spontaneous, $∆G$ must be negative. Hence, for the given reaction to be spontaneous, $T$ should be greater than $2000\ K$.