A) 16
B) 60
C) 99
D) 132
Correct Answer: B
Solution :
[b] According to Arrhenius equation \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right)\] \[\log \frac{1.3\times {{10}^{-3}}}{1.3\times {{10}^{-4}}}=\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{1}{373}-\frac{1}{423} \right]\] \[I=\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{1}{373}-\frac{1}{423} \right]\] \[{{E}_{a}}=60kJ/mol\]You need to login to perform this action.
You will be redirected in
3 sec