Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However,…
- Both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are (-ve)
- $\Delta \mathrm{H}$ is (-ve) but $\Delta \mathrm{S}$ is (+ve)
- $\Delta \mathrm{H}$ is $(+\mathrm{ve})$ but $\Delta \mathrm{S}$ is $(-\mathrm{ve})$
- Both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are (+ve)
Solution
$\Delta G=\Delta H-T \Delta S$
For spontaneity,
$\Delta \mathrm{G} < 0$
Given,
$\Delta \mathrm{H}>0$
So, $\Delta \mathrm{S}$ should be +ve.
Asked in: JEE Main 2025 (24 Jan Shift 1)