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JEE Advanced 2022Paper 2
Exam
JEE Advanced
Year
2022
Paper
Paper 2
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Advanced 2022 · Paper 2 · Chemistry · Thermodynamics (C)

The correct option(s) about entropy (S) is(are) [ R = gas constant,...

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

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

Options

  1. AFor the reaction, M(s)+2H+(aq)→H2(g)+M2+(aq)M\left(s\right) + 2{H}^{+}\left(aq\right) \to {H}_{2}\left(g\right) + {M}^{2 +}\left(aq\right), if dEcelldT=RF\dfrac{{dE}_{cell}}{dT} = \dfrac{R}{F} then the entropy change of the reaction is RR (assume that entropy and internal energy changes are temperature independent).
  2. BThe cell reaction, PtPt\left(s\right) ∣ {H}_{2}(g,1bar)∣H+(aq,0.01M)‖H+(aq,0.1M)∣H2(g,1bar)∣Pt(s)\left(g, 1bar\right)\left|{H}^{+}\left(aq, 0 . 01M\right) ‖ {H}^{+}\left(aq, 0 . 1M\right)\right|{H}_{2}\left(g, 1bar\right) ∣ Pt\left(s\right) , is an entropy driven process.
  3. CFor racemization of an optically active compound, \DeltaS>0\DeltaS > 0
  4. D(\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)

Correct answer

B. The cell reaction, PtPt\left(s\right) ∣ {H}_{2}(g,1bar)∣H+(aq,0.01M)‖H+(aq,0.1M)∣H2(g,1bar)∣Pt(s)\left(g, 1bar\right)\left|{H}^{+}\left(aq, 0 . 01M\right) ‖ {H}^{+}\left(aq, 0 . 1M\right)\right|{H}_{2}\left(g, 1bar\right) ∣ Pt\left(s\right) , is an entropy driven process.

Step-by-step solution

(A) M(s)+2H+(aq)→H2(g)+M2+(aq)M\left(s\right) + 2{H}^{+}\left(aq\right) \to {H}_{2}\left(g\right) + {M}^{2 +}\left(aq\right)

if dEcelldT=RF\dfrac{{dE}_{cell}}{dT} = \dfrac{R}{F}

\DeltaS=nFdEdT=2F(RF)=2R\DeltaS = nF\dfrac{dE}{dT} = 2F\left(\dfrac{R}{F}\right) = 2R

(B) Ecell=−2.303RTFlog0.010.1=2.303RTF{E}_{cell} = \dfrac{- 2 . 303RT}{F}log\dfrac{0 . 01}{0 . 1} = \dfrac{2 . 303RT}{F}

dEcelldT=2.303RF\dfrac{{dE}_{cell}}{dT} = \dfrac{2 . 303R}{F}

∴\DeltaS=nFdEdT>0∴ \DeltaS = nF\dfrac{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(H2O)6]2++3en→[Ni(en)3]+3+6H2O{\left[Ni{\left({H}_{2}O\right)}_{6}\right]}^{2 +} + 3en \to {\left[Ni{\left(en\right)}_{3}\right]}^{+ 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.

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