JEE Advanced 2022 — Paper 2 — CHEM-II solution

JEE Advanced2022Paper 2CHEM-II

The correct option(s) about entropy (S) is(are)

[R = gas constant, F = Faraday constant, T = Temperature]

A
For the reaction, M( $s$ ) + 2H $^{+}$ ( $a q$ ) $\to$ H $_{2}$ ( $g$ ) + M $^{2 +}$ ( $a q$ ), if $\frac{d E _{ce l l}}{d T} = \frac{R}{F}$ 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 $_{2}$ ( $g$ , 1bar) | H $^{+}$ ( $a q$ , 0.01M) || H $^{+}$ ( $a q$ , 0.1M) | H $_{2}$ ( $g$ , 1bar) | Pt( $s$ ), is an entropy driven process.
C
For racemization of an optically active compound, $Δ S > 0$
D
$Δ S > 0$ , for [Ni(H $_{2}$ O) $_{6}$ ] $^{2 +}$ + 3 en $\to$ [Ni(en) $_{3}$ ] $^{2 +}$ + 6H $_{2}$ O (where en = ethylenediamine).

Solution

(A) The Gibbs-Helmholtz relation for an electrochemical cell gives $Δ S = n F (\frac{d E _{ce l l}}{d T})$ . For the given reaction $n = 2$ , so $Δ S = 2 F \cdot \frac{R}{F} = 2 R$ , not $R$ . Incorrect.

(B) Concentration cell with $Δ H \approx 0$ . EMF $E = - \frac{2.303 R T}{F} lo g \frac{0.01}{0.1} = + \frac{2.303 R T}{F}$ , so $\frac{d E}{d T} = \frac{2.303 R}{F} > 0$ and $Δ S = n F \cdot \frac{d E}{d T} > 0$ . With $Δ H \approx 0$ , the cell operates purely because $- T Δ S < 0$ : entropy driven. Correct.

(C) Racemization converts a single enantiomer into a 50:50 mixture, increasing the number of accessible microstates. $Δ S > 0$ . Correct.

(D) The chelate effect: three bidentate en ligands replace six unidentate water molecules, with $Δ n = + 3$ (more particles on the product side). $Δ S > 0$ . Correct.

$∴$ (B), (C), (D).

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