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Two reactions...

Question

Two reactions $R_{1}$ and $R_{2}$ have identical pre-exponential factors. Activation energy of $R_{1}$ exceeds that of $R_{2}$ by $10 k J m o l^{- 1}$ . If $k_{1}$ and $k_{2}$ are rate constants for reactions $R_{1}$ and $R_{2}$ respectively at 300 K, then $l n (\frac{k_{2}}{k_{1}})$ is equal to :
(R = 8.314 $J m o l^{- 1} K^{- 1}$ )

A

12

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B

6

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C

4

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D

8

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Solution

The correct option is C 4
$k_{1} = A e^{\frac{- E_{a_{1}}}{R T}}$

$k_{2} = A e^{\frac{- E_{a_{2}}}{R T}}$

$\frac{k_{2}}{k_{1}} = e^{- \frac{(E_{a_{2}} - E_{a_{1}})}{R T}}$

$\frac{k_{2}}{k_{1}} = e^{+ \frac{10 \times 10^{3}}{8.314 \times 300}}$

$l n \frac{k_{2}}{k_{1}} = 4$

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