SATHEE: Chemical Kinetics - Result Question 25

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 $(R = 8.314 J K^{- 1} m o l^{- 1}$ and $\log 2 = 0.301)$

(a) $53.6 k J m o l^{- 1}$

(b) $48.6 k J m o l^{- 1}$

(d) $60.5 k J m o l^{- 1}$

(2013 Main)

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Answer:

Correct Answer: 25. (a)

Solution:

From Arrhenius equation, $\log \frac{k_{2}}{k_{1}} = \frac{- E_{a}}{2.303 R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$

Given, $\frac{k_{2}}{k_{1}} = 2 T_{2} = 310 K$

$T_{1} = 300 K$

On putting values,

$\begin{array}{ccc} \Rightarrow & \log 2 = \frac{- E_{a}}{2.303 \times 8.314} (\frac{1}{310} - \frac{1}{300}) \Rightarrow & E_{a} = 53.603 k J / m o l \end{array}$

Chemical Kinetics - Result Question 25

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Chemical Kinetics - Result Question 25

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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