Question

Lindemann proposed the following detailed mechanism for gas phase isomerization, the reaction is

In the above scheme, $A^{\ast }$ is a molecule that has enough vibrational energy to decompose or isomerise. The Lindemann mechanism is followed by most of gas phase decomposition or isomerization reaction.
For the overall unimolecular isomerization of cyclopropane to propylene at , the values of versus concentration is as follows.
$\underset{k_{unit}}{\left|M\right|}\underset{\frac{1}{137.5}}{\left|1M\right|}\underset{\frac{1}{131.25}{\text{in s}}^{-1}}{\left|2M\right|}$

As per Lindemann mechanism $k_{unit}$ is

• A. $k_{unit}=\frac{k_1{k}_2}{k_{-1}}$
• B. $\frac{1}{k_{unit}}=\frac{k_{-1}}{k_1{k}_2}+\frac{k_1}{\left[M\right]}$
• C. $\frac{1}{k_{unit}}=\frac{k_{-1}}{k_1{k}_2}+\frac{\left[M\right]}{k_1}$
• D. $\frac{1}{k_{unit}}=\frac{k_{-1}}{k_1{k}_2}+\frac{1}{k_1}\times \frac{1}{\left[M\right]}$

Answer 1

The correct option is D $\frac{1}{k_{unit}}=\frac{k_{-1}}{k_1{k}_2}+\frac{1}{k_1}\times \frac{1}{\left[M\right]}$

Rate$=k_2\left[A^{\ast }\right]\left(i\right)$
On applying steady state approximation of $A^{\ast }$
$0=-k_1\left[A\right]\left[M\right]+k_{-1}\left[A^{\ast }\right]\left[M\right]+k_2\left[A^{\ast }\right]$
$\left[A^{\ast }\right]=\frac{k_1\left[A\right]\left[M\right]}{k_2+k_{-1}\left[M\right]}$
So rate of the reaction:
Rate $=k_2\times \frac{k_1\left[A\right]\left[M\right]}{k_2+k_{-1}\left[M\right]}$
$\frac{1}{k_{unit}}=\frac{k_{-1}}{k_1{k}_2}+\frac{1}{k_1}\times \frac{1}{\left[M\right]}$