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)
[ gas constant, Faraday constant, Temperature]
Options
- AFor the reaction, , if then the entropy change of the reaction is (assume that entropy and internal energy changes are temperature independent).
- BThe cell reaction, \left(s\right) ∣ {H}_{2} , is an entropy driven process.
- CFor racemization of an optically active compound,
- 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, \left(s\right) ∣ {H}_{2} , is an entropy driven process.
Step-by-step solution
(A)
if
(B)
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
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.