Question

The rate constant of a first order reaction are 0.58 s^–1 at 313 K and 0.045 s^–1 at 293 K. What is the energy of activation for the reaction?

Answer 1

Given: 

Rate constant k₁ = 0.58 s^-1 ; Rate constant k₂ = 0.045 s^-1 

T₁ = 313 K; T₂ = 293 K 

R = 8.314 J K^-1 mol^-1 

To find: Activation energy (E_a)

Formula: log₁₀ \(\frac{k_2}{k_1}\) = \(\frac{E_a}{2.303\,R} [\frac{T_2\,-T_1}{T_1T_2}]\)

Calculation: From formula,

= 97455.34 J mol^-1 

= 97.45 kJ mol^-1