JEE Main On 16 April 2018 Question 5
Question: At 320K, a gas $ A _2 $ is 20% dissociated to $ A(g). $ The standard free energy change at 320K and 1 atm in $ J,mo{l^{-1}} $ approximately: $ (R=8.314,J{K^{-1}},mo{l^{-1}};\ln ,2=0.693;,\ln ,3=1.098) $ [JEE Main 16-4-2018]
Options:
A) 4763
B) 2068
C) 4281
D) 1844
Show Answer
Answer:
Correct Answer: C
Solution:
$ A _2\rightarrow 2A $ Initially, suppose $ [A _2]=1M $ and $ [A]=0M $
After 20 % dissociation, 80% of $ A _2 $ remains.
$ [A _2]=1\times \frac{80}{100}=0.8,M $ 20% of 1 M is
$ 1\times \frac{20}{100}=0.2.[A]=2\times 0.2=0.4M $
The equilibrium constant $ K=\frac{[A^{2}]}{[A _2]} $
$ K=\frac{{{[0.4]}^{2}}}{[0.8]}=0.2 $
$ \Delta G^{0}=-RT\ln k=-8.314J{K^{-1}} $ $ mo{l^{-1}}\times 320K\times \ln 0.2=4281,J/mol $
