The rate of a reaction quadruples when the temperature changes from $293 \ K$ to $313 \ K$. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

(N/A) From the Arrhenius equation,we have:
$\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \ R} \left( \frac{T_{2} - T_{1}}{T_{1} \ T_{2}} \right)$
Given that $k_{2} = 4 \ k_{1}$,$T_{1} = 293 \ K$,$T_{2} = 313 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values:
$\log(4) = \frac{E_{a}}{2.303 \times 8.314} \left( \frac{313 - 293}{293 \times 313} \right)$
$0.6021 = \frac{E_{a}}{19.147} \left( \frac{20}{91709} \right)$
$E_{a} = \frac{0.6021 \times 19.147 \times 91709}{20}$
$E_{a} \approx 52863 \ J \ mol^{-1} = 52.86 \ kJ \ mol^{-1}$.
Thus,the activation energy is $52.86 \ kJ \ mol^{-1}$.