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

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

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

  1. For the reaction, Ms+2H+aq→H2g+M2+aq, if dEcell dT=RF then the entropy change of the reaction is R (assume that entropy and internal energy changes are temperature independent).
  2. The cell reaction, Pts∣H2g,1 barH+aq,0.01M‖H+aq,0.1MH2g,1bar∣Pts , is an entropy driven process.
  3. For racemization of an optically active compound, ΔS>0
  4. \(\Delta \mathrm{S}>0\), for \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right) 6\right]^{2+}+3 \mathrm{en} \rightarrow[\mathrm{Ni}(\mathrm{en})_{3}]^{2+}+6 \mathrm{H}_2 \mathrm{O}\) (where en \(=\) ethylenediamine)

Solution

(A) Ms+2H+aq→H2g+M2+aq

if dEcell dT=RF

ΔS=nFdEdT=2 FRF=2R

(B) Ecell =-2.303RTFlog0.010.1=2.303RTF

dEcell dT=2.303RF

∴ΔS=nFdEdT>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 NiH2O62++3en→Nien3+3+6H2O
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.

Asked in: JEE Advanced 2022 (Paper 2)

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