Chapter 6, Problem 6.21P

Interpretation Introduction

Interpretation:

If the step 2 in the given mechanism is slow, then the rate is second order in $O_3$ and $-1$ order in $O_2$ has to be shown.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time. The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$ .

The equation that relates the reaction rate to the reactants concentrations that is raised to various powers is called as rate law.

Rate law can be determined by the slow step or otherwise called as rate-determining step.

Rate constant is defined as proportionality constant in the connection between rate and concentrations.

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$: A system is said to be in equilibrium when all the measurable properties of the system remains unchanged with the time.  Equilibrium constant is the ratio of the rate constants of the forward and reverse reactions at a given temperature. 

Consider the reaction where the reactant A is giving product B. 

    $\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging, $\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}={\text{K}}_{\text{c}}$

Where,

${\text{k}}_f$ is the rate constant of the forward reaction.

${\text{k}}_r$ is the rate constant of the reverse reaction.

$K_c$ is the equilibrium constant.

Explanation of Solution

Decomposition of ozone can be written as,

  $\begin{array}{l}2O_3\stackrel{}{\to }3O_2\\ 1.O_3\underset{k_1^{\text{'}}}{\overset{k_1}{\rightleftharpoons }}{O}_2+O---reverseequilibrium\\ 2.O+O_3\stackrel{k_2}{\to }2O_2---\text{Rate}\text{determinating}\end{array}$

First step is a fast equilibrium it is involving $O_3$ into $O_{2}andO$ is the oxygen atom.

Applying the steady state approximation to oxygen atom $\left(O\right)$

  $\begin{array}{c}\frac{d\left[O\right]}{dt}=k\left[O_3\right]-k\text{'}\left[O_2\right]\left[O\right]-k_2\left[O\right]\left[O_3\right]\\ \left[O\right]=\frac{k\left[O_3\right]}{\left(k\text{'}\left[O_2\right]+k_2\left[O_3\right]\right)}\end{array}$

Rate of decomposition of $O_3=-\frac{d\left[O_3\right]}{dt}$

  $\begin{array}{c}=k\left[O_3\right]-k\text{'}\left[O_2\right]\left[O\right]+k_2\left[O\right]\left[O_3\right]\\ =k\left[O_3\right]-k\text{'}\left[O_2\right]\times \frac{k\left[O_3\right]}{\left(k\text{'}\left[O_2\right]+k_2\left[O_3\right]\right)}+k_2\left[O_3\right]\times \frac{k\left[O_3\right]}{\left(k\text{'}\left[O_2\right]+k_2\left[O_3\right]\right)}\\ =\left(kk\text{'}\left[O_3\right]\left[O_2\right]+k{k}_2{\left[O_3\right]}^2-kk\text{'}\left[O_3\right]\left[O_2\right]+\frac{k{k}_2{\left[O_3\right]}^2}{k\text{'}\left[O_2\right]+k_2\left[O_3\right]}\right)\\ =\frac{2k{k}_2{\left[O_3\right]}^2}{\left(k\text{'}\left[O_2\right]+k_2\left[O_3\right]\right)}\end{array}$

Rate $=-\frac{d\left[O_3\right]}{dt}$

  $-\frac{d\left[O_3\right]}{dt}=\frac{2k{k}_2{\left[O_3\right]}^2}{\left(k\text{'}\left[O_2\right]+k_2\left[O_3\right]\right)}$

If second step is slow $k_2<<kandk\text{'}$

  $\left(k\text{'}\left[O_2\right]+k_2\left[O_3\right]\right)=k\text{'}\left[O_2\right]$

Rate (r)

  $\begin{array}{l}-\frac{d\left[O_3\right]}{dt}=\frac{2k{k}_2{\left[O_3\right]}^2}{k\text{'}\left[O_2\right]}\\ =\left(2kk/k\text{'}\right)\times \frac{{\left[O_3\right]}^2}{\left[O_2\right]}\\ =\left(2kk/k\text{'}\right)\times {\left[O_3\right]}^2\times {\left[O_2\right]}^{-1}\end{array}$

Therefore, the rate is second order in $\left(O_3\right)$ and $-1$ order in $O_2$.