The rate of a reaction at 400 K is 10 times than the rate of the reaction at 200 K . Calculate activation energy of the reaction.

Explanation

We know , `log""(K_(2))/(K_(1)) = (E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
Where `K_(1)` and `K_(2)` are the rate constants of the reaction at temperature `T_(1)` and `T_(2)(T_(2)gtT_(1))` and `E_(a)` is activation energy.
Given , `T_(1) = 200K , T_(2) = 400K "and" K_(2) = 10K_(1)`
`therefore log""(10K_(1))/(K_(1)) = (E_(a))/(2.303xx8.314)((400-200)/(200xx400))`
or, log`(10K_(1))/(K_(1))= 1.3xx10^(-4)E_(a)`
or, `E_(a) = 7658.8 J.mol^(-1)`