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

`"log" (k_2)/(k_1) = (E_a)/(2.303R)[(T_2 - T_1)/(T_1 T_2)]`
Left the rule constant,
`k_1 = k`
Then, `k_2 = 3k`
`therefore "log" (3k)/(k) = (E_a)/(2.303 xx 8.314) [(323 - 293)/(94639)`
`0.4771 = (E_a)/(2.303 xx 8.314) [30/(94639)]`
`E_a = (0.4771 xx 2.303 xx 8.314 xx 94639)/(30)`
`E_a = 28.82 kJ "mol"^(-1)`