Chapter 17, Problem 17F.1ST

Interpretation Introduction

Interpretation:

The rate constant for the activation step in the mechanism has to be calculated.

Answer to Problem 17F.1ST

The rate constant for the activation step in the mechanism was $24.69d{m}^{3}mol{s}^{-1}.$

Explanation of Solution

Effective rate constant (Lindemann-Hinshelwood mechanism):

  $\frac{\text{1}}{{\text{k}}_r}\text{=}\frac{\text{1}}{{\text{k}}_{\text{a}}\left[\text{A}\right]}-------\left(1\right)$.

Given data is as follows:

  $\begin{array}{l}{\text{k}}_{\text{r,1}}\text{=}\text{1}\text{.70}\text{×}{\text{10}}^{\text{-3}}{\text{s}}^{\text{-1}}\text{;}{\text{k}}_{\text{r,2}}\text{=}\text{2}\text{.20}\text{×}{\text{10}}^{\text{-4}}{\text{s}}^{\text{-1}}\\ \\ {\left[\text{A}\right]}_{\text{1}}\text{=}\text{4}\text{.37}\text{×}{\text{10}}^{\text{-4}}\text{mol}{\text{dm}}^{\text{-3}};{\left[\text{A}\right]}_2\text{=}\text{1}\text{.00}\text{×}{\text{10}}^{\text{-5}}\text{mol}{\text{dm}}^{\text{-3}}.\end{array}$

By substituting the given value into the equation (1) as follows,

  $\begin{array}{l}\frac{\text{1}}{{\text{k}}_{r,2}}-\frac{\text{1}}{{\text{k}}_{r,1}}\text{=}\frac{\text{1}}{{\text{k}}_{\text{a}}}\left[\frac{1}{{\left[\text{A}\right]}_2}-\frac{1}{{\left[\text{A}\right]}_1}\right]\\ \\ \frac{\text{1}}{\text{2}\text{.20}\text{×}{\text{10}}^{\text{-4}}{\text{s}}^{\text{-1}}}-\frac{\text{1}}{\text{1}\text{.70}\text{×}{\text{10}}^{\text{-3}}{\text{s}}^{\text{-1}}}\text{=}\frac{\text{1}}{{\text{k}}_{\text{a}}}\left[\frac{1}{\text{1}\text{.00}\text{×}{\text{10}}^{\text{-5}}\text{mol}{\text{dm}}^{\text{-3}}}-\frac{1}{\text{4}\text{.37}\text{×}{\text{10}}^{\text{-4}}\text{mol}{\text{dm}}^{\text{-3}}}\right]\\ \\ \frac{\left(\text{17}\text{.0}\text{×}{\text{10}}^{\text{-4}}{\text{s}}^{\text{-1}}\right)-\left(\text{2}\text{.20}\text{×}{\text{10}}^{\text{-4}}{\text{s}}^{\text{-1}}\right)}{\text{3}\text{.74}\text{×}{\text{10}}^{\text{-7}}{\text{s}}^{\text{-2}}}=\frac{\text{1}}{{\text{k}}_{\text{a}}}\left[\frac{\left(\text{4}\text{.37}\text{×}{\text{10}}^{\text{-4}}\text{mol}{\text{dm}}^{\text{-3}}\right)-\left(\text{1}\text{.00}\text{×}{\text{10}}^{\text{-5}}\text{mol}{\text{dm}}^{\text{-3}}\right)}{\text{4}\text{.37}\text{×}{\text{10}}^{\text{-9}}{\left(\text{mol}{\text{dm}}^{\text{-3}}\right)}^2}\right]\\ \\ 3.957\text{×}{\text{10}}^3\text{s}=\frac{\text{1}}{{\text{k}}_{\text{a}}}\left[\text{9}\text{.77}\text{×}{\text{10}}^4{\text{mol}}^{-1}{\text{dm}}^{\text{3}}\right]\\ \\ {\text{k}}_{\text{a}}=\frac{\text{9}\text{.77}\text{×}{\text{10}}^4{\text{mol}}^{-1}{\text{dm}}^{\text{3}}}{3.957\text{×}{\text{10}}^3\text{s}}=24.69d{m}^{3}mol{s}^{-1}.\end{array}$

Therefore, the rate constant for the activation step in the mechanism was $24.69d{m}^{3}mol{s}^{-1}.$