Half life of a first order reaction:
$A (aq .) \to 2 B (aq .)$ is 40min at $27^{o} C$ and 20 min at $37^{o} C$ . The activation energy of reaction is $(\ln 2 = 0.7, R = 2.0 c a l m o l^{- 1} K^{- 1})$

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$19530 c a l / m o l$

$13020 c a l / m o l$

$26040 c a l / m o l$

$6510 c a l / m o l$

answer is B.

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Detailed Solution

$I n (\frac{K_{1}}{K_{2}}) = \frac{E a}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] \begin{array}{l} E a = \frac{0.7 \times 2 \times 310 \times 300}{10} \\ = 13020 c a l / m o l e \end{array}$

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