The rate constant is doubled when temperature increases from 27⁰C to 37⁰C. Activation energy in kJ is

We are given that:

When T1​=27+273=300K

Let k₁​=k

When T₂​=37+273=310K

k_2​=2k

Substituting these values the equation:

$\log \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=2.303 {\mathrm{E}}_{\mathrm{a}}\left(\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right)$

We will get:

$\log \frac{2\mathrm{K}}{\mathrm{K}}\left(\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\right)\times \frac{310-300}{310\times 300}$

$\log 2\left(\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\right)\times \frac{10}{300\times 310}$

E_a​=53598.6 Jmol⁻¹

E_a​=53.6 kJmol⁻¹

Hence, the energy of activation of the reaction is 53.6 kJmol⁻¹