Question

The rate of a reaction doubles when its temperature changes from $300\, K$ to $310\,K$. Activation energy of such a reaction will be $(R = 8.314 JK^{-1} \,mol^{-1}$ and $log \,2 = 0.301)$

• A. $53.6\, kJ \,mol^{-1}$
• B. $48.6\, kJ \,mol^{-1}$
• C. $58.5\, kJ \,mol^{-1}$
• D. $60.5\, kJ \,mol^{-1}$

Answer 1

Answer: (A)

From Arrhenius equation, $\log \frac{k_2}{k_1}=\frac{-E_a}{2.303 R}\bigg(\frac{1}{T_2}-\frac{1}{t_1}\bigg)$
Given, $\frac{k_2}{k_1}=2T_2=310\,K$
$T_1=300\,K$
On putting values,
$\Rightarrow \log 2=\frac{-E_a}{2.303\times8.314}\bigg(\frac{1}{310}-\frac{1}{300}\bigg)$
$\Rightarrow E_a=53598.6 J \, = 53.6\, kJ \,mol^{-1}$