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Arrhenius Equation

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 constant for reactions $R_{1}$ and $R_{2}$ respectively at $300 K$ , then $l n (k_{2} / k_{1})$ 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
We know $K = A e^{- \frac{E_{a}}{R T}}$

$K_{1} = A e^{- \frac{E_{a 1}}{R T}}$

$K_{2} = A e^{- \frac{E_{a 2}}{R T}}$

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

$l n (\frac{K_{2}}{K_{1}}) = \frac{E_{a 1} - E_{a 2}}{R T}$

$=$ $\frac{10 \times 1000}{8.314 \times 300}$

$= 4.00$

The correct option is C.

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