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 temperatures 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)xx1000)/(2.303xx8.314)xx(10)/(300xx310)`
Energy of activation `=E_(a)=53kJmol^(-1)`