The rate constant k of a reaction increases three fold when temperature changes from `127^(@)` C to `45^(@)` C calculate the energy of activation for this reaction

Explanation

According to Arrhenius equation
`log(k_(2))/(k_(1))=(E_(a))/(2.303 R )[(1)/(T_(1))-(1)/(T_(2))]` ro log `(k_(2))/(k_(1))=(E_(a))/(2.303 R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`0.4771 =(E_(a))/(2.303xx8.314 j mol^(-1))xx(18)/(300xx318)`
`E_(a)=(0.4771xx2.303xx8.314xx300xx318)/(18) mol^(-)`
`=48416 j mol^(-1)`