For a reaction, specific rate constant at 283 K is `2.25 xx 10^(-6)` L `mol^(-1)" "sec^(-1)` and at 293 K is `2.5 xx 10^(-5)` L `mol^(-1) " "s^(-1)`. Compute the energy of activation of the reaction.

Explanation

From Arrhenius equation,
`log(k_2)/(k_1)=(E_a)/(2.303R)[(1)/(T_1)-(1)/(T_2)]`
or, `E_a=(2.303RxxT_1T_2)/(T_2-T_1)xxlog(k_2)/(k_1)`
`=(2.303xx8.314xx283xx293)/(10)xxlog(2.50xx10^(-5))/(2.25xx10^(-6))`
=166030.929 J=166.030 KJ `mol^(-1)`