At `27^(@)C` and `37^(@)C` , the rates of a reaction are given as `1.6 xx 10^(-2) mol L^(-1)s^(-1)` and `3.2 xx 10^(-2)` mol `L^(-1)s^(-1)` . Calculate the energy of activation for the given reaction .

Explanation

The ratio of specific rates at two different temperatuer are given as ,
`log = (K_(2))/(k_(1)) = (E_(a))/(2.303R) [(T_(2) - T_(1))/(T_(1)T_(2))]`
Substituting the values `K_(1),k_(2),T_(1),T_(2)` and molar gas constant R
` 0.301 = (E_(a) xx 1000)/(2.303 xx 8.314) xx (10)/(300 xx 310)`
Energy of activation ` = E_(a) = 53 KJ mol^(-1)`