The rate constant of a first order reaction at 300 K and 310 K are respectively `1.2 xx 10^(3) s^(-1)` and `2.4 xx 10^(3) s^(-1)`. Calculate the energy of activation. `(R = 8.314 J K^(-1) mol^(-1))`

Explanation

`log.(K_(2))/(K_(1)) = (E_(a))/(2.303 R) [(T_(2)-T_(1))/(T_(1)T_(2))]`
`log.(2.4 xx 10^(-3))/(1.2 xx 10^(-3)) = (E_(a))/(2.303 xx 8.314) [(310 - 300)/(300 xx 310)]`
`E_(a) = 53598 J` or `53.598 K J`.