Q29 / 180 · KCET 2010 · Chemistry

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

  • Correct answer
    A
    $10^{-1}$
  • B
    $10^{-2}$
  • C
    $2 \times 10^{-3}$
  • D
    $2 \times 10^{-2}$

Solution

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 $$ \frac{\mathrm{k}}{\mathrm{A}}=10^{-1} $$

Hence, the correct answer is option: 'A'.

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