Question
The correct option(s) about entropy (S) is(are) [ R = gas constant, F = Faraday constant, T = Temperature]
JEE Advanced
Chemistry
Thermodynamics (C)
2022
JEE Advanced 2022 (Paper 2)
JEE Advanced Chemistry Question (2022) — Solution
Options
- A. For the reaction, M s + 2 H + aq → H 2 g + M 2 + aq , if dE cell   dT = R F then the entropy change of the reaction is R (assume that entropy and internal energy changes are temperature ind
- B. The cell reaction, Pt s ∣ H 2 g , 1   bar H + aq , 0 . 01 M ‖ H + aq , 0 . 1 M H 2 g , 1 bar ∣ Pt s , is an entropy driven process.
- C. For racemization of an optically active compound, Δ S > 0
- D. \( S >0\), for \( [ Ni ( H _2 O ) 6 ]^ 2+ +3 en [ Ni ( en )_ 3 ]^ 2+ +6 H _2 O \) (where en \(=\) ethylenediamine)
Answer
D. \( S >0\), for \( [ Ni ( H _2 O ) 6 ]^ 2+ +3 en [ Ni ( en )_ 3 ]^ 2+ +6 H _2 O \) (where en \(=\) ethylenediamine)
Step-by-step solution
(A) M s + 2 H + aq → H 2 g + M 2 + aq if dE cell   dT = R F Δ S = nF dE dT = 2   F R F = 2 R (B) E cell   = - 2 . 303 RT F log 0 . 01 0 . 1 = 2 . 303 RT F dE cell   dT = 2 . 303 R F ∴ Δ S = nF dE dT > 0 It is an entropy driven process. (C) It is correct Racemisation is a thermodynamically favourable method, and it proceeds spontaneously if a suitable pathway is accessible for the interconversion of the enantiomers. During racemisation of optically active compound, disorderness increases and hence, entropy increases. (D) For Ni H 2 O 6 2 + + 3 en → Ni en 3 + 3 + 6 H 2 O Entropy increases when bidentate ligands replace monodentate ligands due to increase in the number of molecules on the product side. Hence, (B, C, D) are correct.
Related: Chemistry — Thermodynamics (C) · All PYQ Banks