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Rate Constant

Two reactions...

Question

Two reactions $R_{1}$ and $R_{2}$ have identical pre-exponential factors. Actiation energy of $R_{1}$ exceeds that of $R_{2}$ by $10 k J^{- 1}$ . If $K_{1}$ and $K_{2}$ are rate constants for reactions $R_{1}$ and $R_{1}$ respectively at 300 K, then In( $K_{1}$ / $K_{2}$ ) is equal to :-
$R = 8.314 J^{- 1} K^{- 1}$

A

8

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B

12

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C

6

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D

4

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Solution

The correct option is C 4
$K = A e^{- E_{a} / R T}$

$K_{1} = A e^{- E_{a (1)} / R T}$
$K_{2} = A e^{- E_{a (2)} / R T}$

Dividing both the rate equations , we get

$\frac{K_{1}}{K_{2}} = e^{E_{a (2)} - E_{a (1)} / R T}$

$l n \frac{K_{1}}{K_{2}} = \frac{10000}{8.314 \times 300}$

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

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