A) 4763
B) 2068
C) 4281
D) 1844
Correct Answer: C
Solution :
\[{{A}_{2}}\rightleftharpoons 2A\] 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 \[K=\frac{[{{A}^{2}}]}{[{{A}_{2}}]}\] \[K=\frac{{{[0.4]}^{2}}}{[0.8]}=0.2\] \[\Delta {{G}^{0}}=-RT\ln k=-8.314J{{K}^{-1}}\] \[mo{{l}^{-1}}\times 320K\times \ln 0.2=4281\,J/mol\]You need to login to perform this action.
You will be redirected in
3 sec