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

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 .

48.6 k J m o l^{- 1}

B .

58.5 k J m o l^{- 1}

C .

60.5 k J m o l^{- 1}

D .

53.6 k J m o l^{- 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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