Chemical Kinetics - Result Question 25
25. The rate of a reaction doubles when its temperature changes from $300 K$ to $310 K$. Activation energy of such a reaction will be $\left(R=8.314 JK^{-1} mol^{-1}\right.$ and $\left.\log 2=0.301\right)$
(a) $53.6 kJ mol^{-1}$
(b) $48.6 kJ mol^{-1}$
(c) $58.5 kJ mol^{-1}$
(d) $60.5 kJ mol^{-1}$
(2013 Main)
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Answer:
Correct Answer: 25. (a)
Solution:
- From Arrhenius equation, $\log \dfrac{k _2}{k _1}=\dfrac{-E _a}{2.303 R}\left(\dfrac{1}{T _2}-\dfrac{1}{T _1}\right)$
Given, $\quad \dfrac{k _2}{k _1}=2 T _2=310 K$
$ T _1=300 K $
On putting values,
$ \begin{array}{ccc} \Rightarrow & \log 2=\dfrac{-E _a}{2.303 \times 8.314}\left(\dfrac{1}{310}-\dfrac{1}{300}\right) \\ \Rightarrow & E _a=53.603 kJ / mol \end{array} $
BETA
