JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Chemical Thermodynamics

A bomb calorimeter works at constant volume, so the heat released is the change in internal energy, $\Delta U=-1406\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The change in the number of gaseous moles is $2-(1+3)=-2$. Using $\Delta H=\Delta U+\Delta n_{g}RT$, the enthalpy change is $-1406+(-2)(8.3\times 10^{-3})(300)=-1410.98\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, which rounds to $-1411\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. At equilibrium the Gibbs energy change is zero, so $T\Delta S$ equals $\Delta H$. The minimum value of $T\Delta S$ is therefore $-1411\mathrm{kJ}$, whose magnitude is 1411.