The rate constant for the first order reactior becomes three times when the temperatur is raised from `20^@`C to `50^@`C. Calculate the energy of activation for the reaction.

Explanation

`k_2//k_1=2,T_1=20^@C=(20+273)K=293K`
`T_2=50^@C=(50+273)K=323K,E_a=?`
`R=8.314 JK^(-1)mol^(-1)`
`log=(k_2)/(k_1)=(E_a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`
`log3=(E_a)/(2.303xx8.314JK^(-1)mol^(-1))[(323-293)/(323xx293k)]`
`0.4771=(E_a)/(19.15JK^(-1)mol^(-1))xx(30)/(323xx293k)`
`E_a=J mol^(-1)`
=28817.89 J `mol^(-1)`
=28.82 EJ `mol^(-1)`