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First Order Reaction

A 1st order...

Question

A $1^{s t}$ order reaction is $50 %$ complete in $30$ minutes at $27^{o} C$ and in $10$ minutes at $47^{o} C$ . Calculate energy of activation for the reaction.
(Assume log $3$ = $0.48$ )

A

46.87 K J m o l^{- 1}

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B

42.21 K J m o l^{- 1}

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C

44.11 K J m o l^{- 1}

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D

49.20 K J m o l^{- 1}

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Solution

The correct option is C $44.11 K J m o l^{- 1}$
$k_{1} = \frac{0.693}{10}$

$k_{2} = \frac{0.693}{30}$

$l o g \frac{k_{2}}{k_{1}}$ = $\frac{E_{a}}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l o g 3 = \frac{E_{a}}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{320})$

$E_{a} = 44.11 K J {m o l}^{- 1}$

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