The rate of constant of reaction at `300 K//320 K` are `5 xx 10^(-4) s^(-1) and 2.0 xx 10^(-3)s^(-1)` respectively. Calculate the value of activation energy of reaction.

Explanation

`"log " (k_2)/(k_1) = (E_a)/(19.17) [(T_2 - T_1)/(T_1 T_2)]`
`"log" (2.0 xx 10^(-3))/(5 xx 10^(-4)) [(320 - 300)/(96000)]`
`"log" 4 = (E_a)/(19.17) [(20)/(96000)]`
`E_a = (0.6021 xx 19.17 xx 9600)/(2)`
`E_a = 55.40 "kJ mol"^(-1)`