JEE Advanced 2022 · Paper 2 · Chemistry Q10

ChemistryMultiple correct+4 / -2Question 46 of 54

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

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

For the reaction,

\ce{M(s) + 2H+(aq) -> H2(g) + M^{2+}(aq)}

, if

\dfrac{dE_{cell}}{dT} = \dfrac{R}{F}

, then the entropy change of the reaction is R (assume that entropy and internal energy changes are temperature independent).

The cell reaction,

\mathrm{Pt}(s)\,|\,\mathrm{H_2}(g, 1\,\text{bar})\,|\,\mathrm{H^+}(aq, 0.01\,\text{M})\,\|\,\mathrm{H^+}(aq, 0.1\,\text{M})\,|\,\mathrm{H_2}(g, 1\,\text{bar})\,|\,\mathrm{Pt}(s)

, is an entropy driven process.

For racemization of an optically active compound,

\Delta S > 0

\Delta S > 0

, for

\ce{[Ni(H2O)6]^{2+} + 3 en -> [Ni(en)3]^{2+} + 6H2O}

(where en = ethylenediamine).

Answer and worked solution

Official answer: B, C, D

Check each option.

(A) $\Delta S=nF\dfrac{dE}{dT}=2F\cdot\dfrac RF=2R$ , not R. False.
(B) This is a concentration cell: $\ce{H+}$ passes from 0.1 M to 0.01 M. $\Delta H=0$ and $\Delta G=-T\Delta S<0$ , so the process is driven by entropy. True.
(C) Racemisation turns a pure enantiomer into a 1:1 mixture. The entropy of mixing makes $\Delta S>0$ . True.
(D) Chelate effect: 4 particles on the left become 7 on the right, so $\Delta S>0$ . True.

(B), (C), (D).

JEE Advanced 2022 · Paper 2, Chemistry question 10