We are given that:

k_{1} = 1.2 x 10^{-3} sec^{-1}

T_{1} = 30 + 273 = 303 K

k_{2} = 2.1 x 10^{-3} sec^{-1}

T_{2} = 40 + 273 = 313 K

Substituting these values the equation:

Log (k_{2} / k_{1})

= E_{a} / 2.303 R {(T_{2} – T_{1}) / T_{1} T_{2}}

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}