Question

The rate constant of a reaction quadruples when the temperature changes from 300 K to 320 K. Calculate the activation energy for this reaction.

[log 2 = 0.30, log 4 = 0.60, 2.303 R = 19.15 JK⁻¹mol⁻¹]

Answer 1

Given: T₁ = 300 K, T₂ = 320 K, `k_2/k_1` = 4

R = 8.314 JK⁻¹ mol⁻¹, log 4 = 0.60

By Arrhenius's equation,

`log  k_2/k_1 = E_a/(2.303 xx R) (1/T_1 - 1/T_2)`

`log 4 = E_a/(2.303 xx 8.314)[1/300 - 1/320]`

`E_a = log 4 xx 2.303 xx 8.314 xx 320 xx 300/20`

`E_a = 0.60 xx 2.303 xx 8.314 xx 300 xx 320/20`

E_a = 55152 J/mol