The rate of a particular reaction doubles when temperature changes from 27⁰ C to 37⁰ C. Calculate the energy of activation of the reaction.

We are given that:
When T₁ = 27 + 273 = 300 K
Let k₁ = k
When T₂ = 37 + 273 = 310 K
k₂ = 2 k
Substituting these values the equation:
Log (k₂ / k₁)
= E_a / 2.303 R {(T₂ – T₁) / T₁ T₂}

We will get:
Log (2 k / k) = E_a / 2.303 x 8.314 {(310 – 300) / 300 x 310}
Log 2 = E_a / 2.303 x 8.314 x (10 / 300 x 310)
E_a = 53598.6 J mol^-1
E_a = 53.6 kJ mol^-1
Hence, the energy of activation of the reaction is 53.6 kJ mol^-1