JEE Advance - Chemistry (2022 - Paper 2 Online - No. 10)
The correct option(s) about entropy (S) is(are)
$[\mathrm{R}=$ gas constant, $\mathrm{F}=$ Faraday constant, $\mathrm{T}=$ Temperature $]$
$[\mathrm{R}=$ gas constant, $\mathrm{F}=$ Faraday constant, $\mathrm{T}=$ Temperature $]$
For the reaction, $\mathrm{M}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{H}_{2}(g)+\mathrm{M}^{2+}(a q)$, if $\frac{d E_\text{cell}}{d T}=\frac{R}{F}$, then the entropy change of the reaction is $\mathrm{R}$ (assume that entropy and internal energy changes are temperature independent).
The cell reaction, $\operatorname{Pt}(s) \mid \mathrm{H}_{2}(g, 1$ bar $)\left|\mathrm{H}^{+}(a q, 0.01 \mathrm{M}) \| \mathrm{H}^{+}(a q, 0.1 \mathrm{M})\right| \mathrm{H}_{2}(g, 1 \mathrm{bar}) \mid \operatorname{Pt}(s)$, is an entropy driven process.
For racemization of an optically active compound, $\Delta \mathrm{S}>0$.
$\Delta \mathrm{S}>0$, for $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+3$ en $\rightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}+6 \mathrm{H}_{2} \mathrm{O}$ (where en $=$ ethylenediamine).
Explanation
(A) Given, Temperature coefficient $\frac{d \mathrm{E}_{\text {cell }}}{d \mathrm{~T}}=\frac{\mathrm{R}}{\mathrm{F}}$
$$ \mathrm{M}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{H}_2(\mathrm{~g})+\mathrm{M}^{2+}(\mathrm{aq}) $$
Here, number of electron $(n)=2$ transferred
Since,
$$ \begin{aligned} \Delta S & =\left(\frac{d \mathrm{E}_{\text {cell }}}{d \mathrm{~T}}\right) n \mathrm{~F} \\\\ \Delta S & =\frac{\mathrm{R}}{\mathrm{F}} \times 2 \times \mathrm{F} \\\\ \Delta \mathrm{S} & =2 \mathrm{R} \end{aligned} $$
Option (A) is incorrect.
(B)
$$ \begin{aligned} & \mathrm{E}_{\text {cell }}=\frac{-2.303 \mathrm{RT}}{\mathrm{F}} \log \frac{0.01}{0.1}=\frac{2.303 \mathrm{RT}}{\mathrm{F}} \\\\ & \frac{\mathrm{dE}_{\text {cell }}}{\mathrm{dT}}=\frac{2.303 \mathrm{R}}{\mathrm{F}} \\\\ & \therefore \Delta \mathrm{S}=\mathrm{nF} \frac{\mathrm{dE}}{\mathrm{dT}}>0 \end{aligned} $$
It is an entropy driven process.
(C) It is correct
During racemisation of optically active compound, disorder increases and hence, entropy increases.
(D) For $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}+3 \mathrm{en} \rightarrow\left[\mathrm{Ni}(\mathrm{en})_3\right]^{+3}+6 \mathrm{H}_2 \mathrm{O}$,
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.
$$ \mathrm{M}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{H}_2(\mathrm{~g})+\mathrm{M}^{2+}(\mathrm{aq}) $$
Here, number of electron $(n)=2$ transferred
Since,
$$ \begin{aligned} \Delta S & =\left(\frac{d \mathrm{E}_{\text {cell }}}{d \mathrm{~T}}\right) n \mathrm{~F} \\\\ \Delta S & =\frac{\mathrm{R}}{\mathrm{F}} \times 2 \times \mathrm{F} \\\\ \Delta \mathrm{S} & =2 \mathrm{R} \end{aligned} $$
Option (A) is incorrect.
(B)
$$ \begin{aligned} & \mathrm{E}_{\text {cell }}=\frac{-2.303 \mathrm{RT}}{\mathrm{F}} \log \frac{0.01}{0.1}=\frac{2.303 \mathrm{RT}}{\mathrm{F}} \\\\ & \frac{\mathrm{dE}_{\text {cell }}}{\mathrm{dT}}=\frac{2.303 \mathrm{R}}{\mathrm{F}} \\\\ & \therefore \Delta \mathrm{S}=\mathrm{nF} \frac{\mathrm{dE}}{\mathrm{dT}}>0 \end{aligned} $$
It is an entropy driven process.
(C) It is correct
During racemisation of optically active compound, disorder increases and hence, entropy increases.
(D) For $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}+3 \mathrm{en} \rightarrow\left[\mathrm{Ni}(\mathrm{en})_3\right]^{+3}+6 \mathrm{H}_2 \mathrm{O}$,
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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