The rate of a reaction quadruples when the temperature changes from 293 K to 313 K.

Solution:

From Arrhenius equation, we obtain

$\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$

It is given that, $k_{2}=4 k_{1}$

$T_{1}=293 \mathrm{~K}$

$T_{2}=313 \mathrm{~K}$

Therefore, $\log \frac{4 k_{1}}{k_{2}}=\frac{E_{a}}{2.303 \times 8.314}\left(\frac{313-293}{293 \times 313}\right)$

$\Rightarrow 0.6021=\frac{20 \times E_{a}}{2.303 \times 8.314 \times 293 \times 313}$

$\Rightarrow 0.6021=\frac{20 \times E_{a}}{2.303 \times 8.314 \times 293 \times 313}$

 

$\Rightarrow E_{a}=\frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20}$

$=52863.33 \mathrm{~J} \mathrm{~mol}^{-1}$

$=52.86 \mathrm{~kJ} \mathrm{~mol}^{-1}$'

Hence, the required energy of activation is 52.86 kJmol−1.