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)

Show Answer

Answer:

Correct Answer: 25. (a)

Solution:

  1. From Arrhenius equation, $\log \frac{k _2}{k _1}=\frac{-E _a}{2.303 R}\left(\frac{1}{T _2}-\frac{1}{T _1}\right)$

Given, $\quad \frac{k _2}{k _1}=2 T _2=310 K$

$$ T _1=300 K $$

On putting values,

$$ \begin{array}{ccc} \Rightarrow & \log 2=\frac{-E _a}{2.303 \times 8.314}\left(\frac{1}{310}-\frac{1}{300}\right) \ \Rightarrow & E _a=53.603 kJ / mol \end{array} $$



NCERT Chapter Video Solution

Dual Pane