A) \[52.2\,kJ\,mo{{l}^{-1}}\]
B) \[39.2\,kJ\,mo{{l}^{-1}}\]
C) \[52.9\,kJ\,mo{{l}^{-1}}\]
D) \[29.5\,kJ\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
Activation energy can be calculated from the equation. \[\frac{\log {{K}_{2}}}{\log {{K}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left( \frac{1}{{{T}_{2}}}-\frac{1}{{{T}_{1}}} \right)\] Given\[\frac{\log {{K}_{2}}}{\log {{K}_{1}}}=2{{T}_{2}}=308;{{T}_{1}}=208\] \[\therefore \]\[\log 2=\frac{-{{E}_{a}}}{2.303\times 8.314}\left( \frac{1}{308}-\frac{1}{208} \right)\] \[{{E}_{a}}=52.9kJ\,mo{{l}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
