JEE Advanced 2022 Paper 2 · Q10 · Thermodynamics
The correct option(s) about entropy (S) is(are) [R = gas constant, F = Faraday constant, T = Temperature]
Reveal answer + step-by-step solution
Correct answer:B, C, D
Solution
Evaluate each option.
(A) Thermodynamic identity: $\Delta S = nF \left(\frac{dE_{cell}}{dT}\right)$. With $n = 2$ and $dE_{cell}/dT = R/F$: $\Delta S = 2F \cdot (R/F) = 2R$, NOT $R$. (A) is WRONG.
(B) Concentration cell with H$^+$ going from 0.01 M (anode) to 0.1 M (cathode). $E^\circ_{cell} = 0$. The driving force is purely the entropy of mixing (ions moving from low to high concentration is the spontaneous direction here — actually the cell drives the spontaneous transfer of H$^+$ from 0.1 M to 0.01 M, increasing the entropy of the system). $\Delta G < 0$ and $\Delta H \approx 0$ (no bond/standard-state change), so the process is entropy-driven. (B) is CORRECT.
(C) Racemisation of a pure enantiomer yields a 50:50 racemic mixture — increases configurational entropy. $\Delta S > 0$. (C) is CORRECT.
(D) Chelate effect: [Ni(H$_2$O)$_6$]$^{2+}$ + 3 en $\rightarrow$ [Ni(en)$_3$]$^{2+}$ + 6 H$_2$O. Number of particles goes 1 + 3 = 4 on left to 1 + 6 = 7 on right. More particles ⇒ $\Delta S > 0$. (D) is CORRECT.
Correct options: B, C, D.
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