JEE Main 2022 — Thermodynamics (C) Question with Solution

The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change $(∆\mathrm{S})$ of the reduction process is more on positive side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of ${∆}_{\mathrm{f}}\mathrm{G}$ becomes more on negative side and the reduction becomes easier.

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

$\because$ Entropy of liquid is more than solid

$\therefore$ on melting the entropy increases and $\mathrm{\Delta G}$ becomes more negative and hence it becomes easier to reduce metal.