ExamBro
ExamBro
KCET · Chemistry · Chemical Kinetics

The activation energy for a reaction at the temperature \(T \mathrm{~K}\) was found to be \(2.303 \mathrm{RT}\) \(\mathrm{J} \mathrm{mol}^{-1}\). The ratio of the rate constant to Arrhenius factor is

  1. A \(10^{-1}\)
  2. B \(10^{-2}\)
  3. C \(2 \times 10^{-3}\)
  4. D \(2 \times 10^{-2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(10^{-1}\)

Step-by-step Solution

Detailed explanation

Arrhenius equation is, rate constant,
\[
\begin{aligned}
&\mathrm{k}=A \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}} \\
&\mathrm{k}=\mathrm{Ae}^{-2.303 \mathrm{RT} / \mathrm{RT}} \\
&\frac{\mathrm{k}}{\mathrm{A}}=\mathrm{e}^{-2.303}
\end{aligned}
\]
On solving, we get
\[
\frac{\mathrm{k}}{\mathrm{A}}=10^{-1}
\]