Rate constant of reaction at 300 K and 400 K are `0.0345 S^(-1)` and `0.1365 S^(-1)` respectively. Calculate the activation energy for the reaction.
[Given : `R = 8.314 J K^(-1) mol^(-1)`]

Explanation

`log.(K_(T_(2)))/(K_(T_(1))) = (E_(a))/(2.303 R)[(T_(2) - T_(1))/(T_(1) xx T_(2))]`
`E_(a) = (2.303 xx 8.314 xx 300 xx 400)/(10) log.(0.0345)/(0.1365)`
`= 13834 J mol^(-1)`
or `= 13.8 K J mol^(-1)`