Chapter 17, Problem 17B.11P

Interpretation Introduction

Interpretation:

The order of the reaction with respect to A and its rate constant has to be calculated for the composition of a liquid phase reaction $\text{2A}\stackrel{}{\to }\text{B}$.

Explanation of Solution

The stoichiometry of the reaction relates product and reaction concentration as follows:

    $\begin{array}{l}\text{[A]}\text{=}{\text{[A]}}_{\text{0}}\text{-}\text{2[B]}\\ \end{array}$

When the reaction goes to completion, $\text{[B]}\text{=}{\text{[A]}}_{\text{0}}\text{/2;}\text{hence}{\text{[A]}}_{\text{0}}\text{=}\text{0}\text{.624}\text{mol}{\text{dm}}^{\text{-3}}$.  The half-life of A from its initial concentration is approximately $\text{1200}\text{s}$, and that its half-life from the concentration at $\text{1200}\text{s}\text{is}\text{also}\text{1200}\text{s}$.  This indicates a first-order reaction.  From the graph given below it is confirmed that it is a first order reaction.

    $\text{ln}\frac{{\text{[A]}}_{\text{0}}}{\text{[A]}}\text{=}{\text{k}}_{\text{A}}\text{t}$

Which follows from

$\begin{array}{l}\frac{\text{d[A]}}{\text{dt}}\text{=}{\text{-k}}_{\text{A}}\text{[A]}\\ \\ \text{t/s}\text{0}\text{600}\text{1200}\text{1800}\text{2400}\\ \text{[A]/(mol}{\text{dm}}^{\text{-3}}\text{)}\text{0}\text{.624}\text{0}\text{.446}\text{0}\text{.318}\text{0}\text{.224}\text{0}\text{.164}\\ \text{[B]/(mol}{\text{dm}}^{\text{-3}}\text{)}\text{0}\text{0}\text{.089}\text{0}\text{.153}\text{0}\text{.200}\text{0}\text{.230}\\ \text{ln}\frac{{\text{[A]}}_{\text{0}}}{\text{[A]}}\text{0}\text{0}\text{.34}\text{0}\text{.67}\text{1}\text{.02}\text{1}\text{.34}\end{array}$

Figure.1

The points lie on a straight line, which confirms first order kinetics.  Since the slope of the line is $\text{5}{\text{.6×10}}^{\text{-4}}{\text{s}}^{\text{-1}}$.  ${\text{k}}_{\text{A}}\text{=}\text{5}{\text{.6×10}}^{\text{-4}}{\text{s}}^{\text{-1}}$.  To express the rate law in the form

    $\begin{array}{l}\text{v}\text{=}\text{-}\frac{\text{1}}{\text{2}}\frac{\text{d[A]}}{\text{dt}}\text{=}\text{-}\left(\frac{\text{1}}{\text{2}}\right){\text{×(-k}}_{\text{A}}\text{[A])}\text{=}\frac{\text{1}}{\text{2}}{\text{k}}_{\text{A}}\text{[A]}\\ \\ \text{k}\text{=}\frac{\text{1}}{\text{2}}{\text{k}}_{\text{A}}\text{=2}{\text{.8×10}}^{\text{-4}}{\text{s}}^{\text{-1}}\end{array}$