The rate constant for First order reaction become double when temperature is raised from 300 k to 400 k Find Activation Energy.

Explanation

`"log" (k_2)/(k_1) = (E_a)/(2.303 R)[(T_2 - T_1)/(T_1 - T_2)]`
Let `k_1 = k `then `k_2 = 2 k`
`"log" (2k)/(k) = (E_a)/(19.15) [(400 - 300)/(120000)]`
`therefore E_a = 0.301 xx 19.15 xx 120`
`= 6.91 kJ mol^(-1)`