JEE Main 2024 — Thermodynamics (C) Question with Solution

$∆\mathrm{G}$ can be expressed as:

$∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S}$

where $∆\mathrm{G}$ is the change in free energy

$∆\mathrm{H}$ is the change in enthalpy

$\mathrm{T}$ is the temperature

$∆\mathrm{S}$ is the change in entropy.

Conditions for spontaneous and non spontaneous reactions in terms of free energy change are as follows,

 $\mathrm{\Delta G}>0$for non-sponateneous reaction

$\mathrm{\Delta G}<0$for spontaneous reaction.

$\mathrm{\Delta G}=0$ for equilibrium reaction

$(\mathrm{\Delta G}{)}_{\mathrm{P},\mathrm{T}}=(+)\mathrm{ve}$ for non-spontaneous process

Hence option B is the answer.