# The activation energy of a reaction at a given temperature is found to be The ratio of rate constant to the Arrhenius factor is

1. A

0.01

2. B

0.1

3. C

0.02

4. D

0.001

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### Solution:

Let ${\mathrm{E}}_{\mathrm{a}}=$ activation energy, $k$ is rate constant and is Arrhenius factor. We know that:

$\mathrm{log}\frac{\mathrm{k}}{\mathrm{A}}=\frac{-{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{RT}};\mathrm{log}\frac{\mathrm{k}}{\mathrm{A}}=-\frac{2.300\mathrm{RT}}{2.303\mathrm{RT}}=-1$

$\frac{\mathrm{k}}{\mathrm{A}}=$ antilog $\left(-1\right)$ or $\frac{\mathrm{k}}{\mathrm{A}}=0.1$.

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