The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature

Answer

`E_(a)=52.8 mol^(-1)`

Explanation

Let `k` at `293K=k_(1),k at 313K=k_(2)`
`:.k_(2)=4k_(1) or (k_(2))/(k_(1))=4`
Using Arrhenius equation,
`log``(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
`log4=(E_(a))/(2.303xx8.314J K^(-1)mol^(-1))xx((313-293))/(313xx293)`
`0.606=(E_(a))/(2.303xx8.314)xx(10)/(313xx293)`
Solvie for `E_(a):`
`:. E_(a)=52860J mol^(-1)=52.86mol^(-1)`