Q35 / 180 · KCET 2011 · Chemistry
The activation energy of a reaction at a given temperature is found to be $2.303 R T \mathrm{~J} \mathrm{~mol}^{-1}$. The ratio of rate constant to the Arrhenius factor is
- A$0.01$
- Correct answerB$0.1$
- C$0.02$
- D$0.001$
Solution
$k=A e^{-E_{a} / R T}$
$$
\begin{aligned}
\frac{k}{A} &=e^{-E_{a} / R T} \\
\ln \left(\frac{k}{A}\right) &=\frac{-E_{a}}{R T} \\
2.303 \log \left(\frac{k}{A}\right) &=\frac{-E_{a}}{R T} \\
\log \left(\frac{k}{A}\right) &=\frac{-E_{a}}{2.303 R T} \\
\log \left(\frac{k}{A}\right) &=\frac{-2.303 R T}{2.303 R T} \\
\log \left(\frac{k}{A}\right) &=-1 \\
\frac{k}{A} &=0.1
\end{aligned}
$$
Hence, the correct answer is option: 'B'.