The rate constant of a reaction is 1.2 x 10^-3 sec^-1 at 30⁰ C and 2.1 x 10^-3 sec^-1 at 40⁰ C. calculate the energy of activation of the reaction.

We are given that:
k₁ = 1.2 x 10^-3 sec^-1
T₁ = 30 + 273 = 303 K
k₂ = 2.1 x 10^-3 sec^-1
T₂ = 40 + 273 = 313 K

Substituting these values the equation:
Log (k₂ / k₁)
= E_a / 2.303 R {(T₂ – T₁) / T₁ T₂}

We will get:
Log (2.1 x 10^-3 sec^-1/ 1.2 x 10^-3 sec^-1)
= E_a / 2.303 x 8.314 {(313 – 303) / 303 x 313}
Log (2.1 / 1.2) = E_a / 2.303 x 8.314 x (10 / 303 x 313)
E_a = 44126.3 J mol^-1
E_a = 44.13 kJ mol^-1
Hence, the energy of activation of the reaction is 44.13 kJ mol^-1