Question

The rate constant of a reaction at 500K and 700K are 0.02 sec⁻¹ and 0.07 sec⁻¹ respectively. Calculate the value of E_a. (activation energy)

Answer 1

Given, k₁ = 0.02 s⁻¹; T₁ = 500 K

k₂ = 0.07 s⁻¹, T₂ = 700 K.

The equation is,

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

Substituting the values in the equation

`log  (0.07)/(0.02) = (E_a)/(8.314 xx 2.303("JK"^-1 "mol"^-1)700  xx  500K^2)` `((700  - 500)K)/(700 xx 500K^2)`

`=> log 3.5 = E_a/19.147 xx 200/(350000("mol"^-1))`

`=> E_a = (0.544 xx 19.147 xx 350000)/200 "J mol"^-1   ...[log 3.5 = 0.544]`

`therefore E_a = 18.228 " kJ  mol"^-1`

Thus, the value of activation energy (E_a) is 18.228 kJ mol⁻¹