Question

A certain reaction is non-spontaneous at 300 K. The entropy change during the reaction is $120J{K}^{-1}$. Then the minimum value of $\Delta H$ for the reaction and the nature of the reaction will be:

• A. $36.0$ kJ; endothermic
• B. $36.0$ kJ; exothermic
• C. $2.8$ kJ; endothermic
• D. $-2.8$ kJ; exothermic

Answer 1

The correct option is A $36.0$ kJ; endothermic

For non-spontaneous reactions, $\Delta G$ is $+ve.$
$\Delta G=\Delta H-T\Delta S$; $\Delta S$ is $+120J{K}^{-1}$
To make $\Delta G$ as $+ve,\Delta H$ should be $+ve$, i.e., endothermic
For minimum value of entropy,
$\Delta H=T\Delta S$
$=300\times 120=36000J$
$=36.0kJ$