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 the Arrhenius equation:

Rate constant (k) = `A e^(-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 = 4 k₁

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.