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Q.

If rate of reaction becomes double when temperature is increased from 27° C to 37° C. Then find activation energy of reaction in calorie. (R = 2 cal/mol-K) On 2 = 0.7)
[Fill your answer after dividing by 10.]

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answer is 1302.

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Detailed Solution

It is known that,

$in \frac{k_{1}}{k_{2}} = \frac{E_{α}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$

Thus,

$\begin{array}{l} \ln 2 = \frac{E_{a}}{2} [\frac{1}{300} - \frac{1}{310}] \\ E_{a} = 13020 cal \end{array}$

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