Question

The rate of most of the reaction double when their temperature is raised from $298K$ to $308K$. Calculate the activation energy of such a reaction.

Answer 1

Using Arrhenius Equation,

$k=A_o{e}^{\frac{-E_a}{RT}}$

$log\frac{K_2}{K_1}=\frac{E_a}{2.303}(\frac{1}{T_1}-\frac{1}{T_2})$

Given, rate of reaction doubles when temperature is raised from $208-308K$

$\therefore \frac{K_2}{K_1}=2$

$E_a=\frac{2.303\times 8.314\times 298\times 308\times 0.3014}{1000}$

$=52.89KJ/mol$