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The activation energy of a reaction at 1000 K is found to be 4.606 K cal mol. The ratio of rate constant to the Arrhenius factor constant is

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0.02

0.01

0.1

answer is D.

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Detailed Solution

$\log k = \log A - \frac{E_{0}}{2.303 RT} \Rightarrow \log k - \log A = - \frac{E_{0}}{2.303 RT} \log (\frac{k}{A}) = \frac{2.303 RT}{2.303 RT} = - 1 \Rightarrow \frac{k}{A} antilog (- 1.0) = 1 \times 10^{- 1} = 0.1$

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