The rate of a reaction doubles when its temperature changes from 300 K to 310 K.

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 J K^{- 1} m o l^{- 1} a n d l o g 2 = 0.301)

(IIT-JEE-2013)

Options:

A . 48.6kJmol−1

B . 58.5kJmol−1

C . 60.5kJmol−1

D . 53.6kJmol−1

Answer: Option D
:
D
As per Arrhenius equation:

l o g \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.3 R} [\frac{T_{2} - T_{1}}{T_{2} T_{1}}]

2.3 l o g 2 = \frac{E_{a}}{8.314} [\frac{10}{300 \times 310}]

∴ E_{a} = 53.6 k J m o l^{- 1}

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