Question

The rate of reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_a) of the reaction assuming that it does not change with temperature. (R = 8.314 JK^-1 mol^-1)

Answer 1

According to Arrhenius equation:

Rate constant (k) = `"Ae"^(-"E"_"a"//"RT")` .....(1)

`log  "k"_2/"k"_1 = "E"_"a"/(2.303 "R") (1/"T"_1 - 1/"T"_2)` .....(2)

where, k₁ = rate constant at temperature T₁

k₂ = rate constant at temperature T₂

E_a = Activation energy

R = Gas constant

= 8.314 JK^-1 mol^-1

Given that, k₂ = 4k₁

T₁ = 293 K

T₂ = 313 K

Putting the values in equation (2),

`log  (4"k"_1)/"k"_1 = "E"_"a"/(2.303 xx 8.314) (1/293 - 1/313)` .....(∵ log 4 = 0.6021)

log 4 = `"E"_"a"/(2.303 xx 8.314) ((313 - 293)/(293 xx 313))`

E_a = `0.6021 xx 2.303 xx 8.314 xx ((293 xx 313)/20)`

= 52,863 J mol^-1

The energy of activation for the reaction will be E_a = 52.863 KJ mol^-1.