Question

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

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

Answer 1

Answer: (C)

$53.6$ $kJ$ ${mol}^{-1}$

$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$\log \left(2\right)=\frac{E_a}{2.303\times 8.314}[\frac{1}{300}-\frac{1}{310}]$

$E_a=\frac{0.301\times 2.303\times 8.314\times 300\times 310}{10}$

$=53598$ $J$ ${mol}^{-1}$
$=53.6$ $kJ$ ${mol}^{-1}$