Solution:
Solution
$$\log \left(\cfrac {k_2}{k_1}\right)=\cfrac {E_A}{2.303R} \left[\cfrac {1}{T_1}-\cfrac {1}{T_2}\right]$$
$$\Rightarrow \log (4)=\cfrac {E_A}{2.303\times 8.314}\times \left[\cfrac {1}{300}-\cfrac {1}{310}\right]$$
$$\Rightarrow 2\times 0.693= \cfrac {E_A}{2.303}\times \cfrac {0.0001075}{8.314}$$
$$\Rightarrow E_A=\cfrac {26.53793}{0.0001075}=2.46\times 10^{5}Jmol^{-1}$$