Question

A certain reaction is non-spontaneous at 298 K. The entropy change during the reaction is 121 J/K. What is the minimum value of
$△H$ (in $kJ$) for the reaction.

Answer 1

Given that $△G=+ve$ for non-spontaneous process.
As, $△G=△H-T△S$ and $△S=+121J/K$
$△H$ has to be positive, that is, the reaciton is endothermic.
To calculate the minimum value of $△H,△G=0.$
$\therefore △H=T△S$
$or△H=298\times 121J$
$△H=36.06kJ.$