JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Chemical Thermodynamics

The correct option(s) about entropy (S) is(are)
[R = gas constant, F = Faraday constant, T = Temperature]

(A)For the reaction, $M (s) + 2 H^{+} (a q) \to H_{2} (g) + M^{2 +} (a q)$ , if $\frac{dE _{cell}}{dT} = \frac{R}{F}$ , then the entropy change of the reaction is R (assume that entropy and internal energy changes are temperature independent).
(B)The cell reaction, $Pt (s) ∣ H_{2} (g, 1$ bar $) ∣ H^{+} (a q, 0.01 M) ∥ H^{+} (a q, 0.1 M) ∣ H_{2} (g, 1$ bar $) ∣ Pt (s)$ , is an entropy driven process.✓ Correct
(C)For racemization of an optically active compound, $Δ S > 0$ .✓ Correct
(D) $Δ S > 0$ , for $[Ni (H_{2} O)_{6}]^{2 +} + 3$ en $\to [Ni (en)_{3}]^{2 +} + 6 H_{2} O$ (where en = ethylenediamine).✓ Correct

Answer: (B), (C), (D)

Solution

For option (A): Using

Δ S = n F (\frac{d E _{ce l l}}{d T})_{p}

and

\frac{d E _{ce l l}}{d T} = \frac{R}{F}

, we get

Δ S = n R

. For the given reaction, n = 2 (two electrons transferred), so

Δ S = 2 R

, not R. Incorrect.

For option (B): This is a concentration cell with

E_{ce l l} > 0

(positive), so

Δ G < 0

. Since

Δ H = 0

for concentration cells,

Δ G = - T Δ S < 0

implies

Δ S > 0

. The process is entropy-driven. Correct.

For option (C): Racemization is a spontaneous process converting one enantiomer to a racemic mixture. Since enantiomers have identical bond energies,

Δ H \approx 0

. For spontaneity,

Δ G < 0

, which requires

Δ S > 0

(increased disorder). Correct.

For option (D): The chelate effect makes

[Ni (en)_{3}]^{2 +}

more stable. The reaction is spontaneous with

Δ G < 0

. Going from 4 species (1 complex + 3 en) to 7 species (1 complex + 6 H₂O) increases disorder, so

Δ S > 0

. Correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · One or more correct options

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Chemical Thermodynamics

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