Chapter 13, Problem 13.82QE

(a)

Interpretation Introduction

Interpretation:

The mechanism of the reaction of nitrogen dioxide with ozone to produce dinitrogen pentoxide has to be evaluated with the experimental results and also the intermediates have to be identified.

    $\begin{array}{l}{\text{2NO}}_{\text{2}}\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{Fast,}\text{reversible}\\ {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}+{\text{O}}_{\text{3}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{5}}+{\text{O}}_{\text{2}}Slow\end{array}$

Explanation of Solution

Reaction of nitrogen dioxide with ozone to produce dinitrogen pentoxide is given as follows;

    $\begin{array}{l}{\text{2NO}}_{\text{2}}\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{Fast,}\text{reversible}\\ {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}+{\text{O}}_{\text{3}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{5}}+{\text{O}}_{\text{2}}Slow\end{array}$

Overall stoichiometry:

The elementary steps have to be summed up in order to find the overall stoichiometry.  This is done as shown below;

    $\begin{array}{c}{\text{2NO}}_{\text{2}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\\ {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}+{\text{O}}_{\text{3}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{5}}+{\text{O}}_{\text{2}}\end{array}$

The species ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is common in both equation and hence it gets cancelled out.  Thus, the overall stoichiometry can be given as follows;

    ${\text{2NO}}_{\text{2}}+{\text{O}}_{\text{3}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{5}}+{\text{O}}_{\text{2}}$

This is consistent with the stoichiometry that is determined experimentally.

Rate Law:

Rate of the reaction is determined by the rate limiting step and it is the slowest step in the mechanism.  Therefore, the first step is the slow step and it is the rate-limiting step.  The rate law can be written as shown below considering the slow step;

    $\begin{array}{l}\text{Rate}\propto {[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}]}^1{[{\text{O}}_{\text{3}}]}^1\\ \text{Rate}=k_2{[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}]}^1{[{\text{O}}_{\text{3}}]}^1\end{array}$

Where,

    $[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}]$ is the concentration of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

    $[{\text{O}}_{\text{3}}]$ is the concentration of ${\text{O}}_{\text{3}}$.

    $k_2$ is the rate constant.

The species ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is the intermediate species that is formed and it exists for a very short time.

The fast and reversible step in the mechanism is given as follows;

    ${\text{2NO}}_{\text{2}}\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{Fast}$

The rate of formation of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is given as $k_1{[{\text{NO}}_{\text{2}}]}^2$.  The rate of decomposition of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is given as $k_{-1}[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}]+k_2[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}][{\text{O}}_{\text{3}}]$.  As the rate constant $k_{-1}$ will be larger than $k_2[{\text{O}}_{\text{3}}]$, it can be neglected.  Therefore, the rate of decomposition can be given as follows;

    $\text{Rate}\text{of}\text{decomposition}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}=k_{-1}[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}]$

The rate of formation and decomposition of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ expression can be equated as shown below;

    $\begin{array}{c}\text{Rate}\text{of}\text{formation}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}=\text{Rate}\text{of}\text{decomposition}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\\ k_1{[{\text{NO}}_{\text{2}}]}^2=k_{-1}{\text{[N}}_{\text{2}}{\text{O}}_{\text{4}}]\\ {\text{[N}}_{\text{2}}{\text{O}}_{\text{4}}]=\frac{k_1{[{\text{NO}}_{\text{2}}]}^2}{k_{-1}}\end{array}$

The concentration of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is substituted in the rate law equation as shown below;

    $\begin{array}{c}\text{Rate}=k_2{[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}]}^1{[{\text{O}}_{\text{3}}]}^1\\ =k_2\frac{k_1{[{\text{NO}}_{\text{2}}]}^2}{k_{-1}}{[{\text{O}}_{\text{3}}]}^1\\ Rate=k{[{\text{NO}}_{\text{2}}]}^2{[{\text{O}}_{\text{3}}]}^1\end{array}$

Thus the theoretical result is consistent with experimental data and the intermediate is ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

(b)

Interpretation Introduction

Interpretation:

The mechanism of the reaction of nitrogen dioxide with ozone to produce dinitrogen pentoxide has to be evaluated with the experimental results and also the intermediates have to be identified.

    $\begin{array}{c}{\text{NO}}_{\text{2}}+{\text{O}}_{\text{3}}\to {\text{NO}}_{\text{3}}+{\text{O}}_{\text{2}}\text{Slow}\\ {\text{NO}}_{\text{3}}+{\text{NO}}_{\text{2}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\text{Fast}\end{array}$

Explanation of Solution

Reaction of nitrogen dioxide with ozone to produce dinitrogen pentoxide is given as follows;

    $\begin{array}{c}{\text{NO}}_{\text{2}}+{\text{O}}_{\text{3}}\to {\text{NO}}_{\text{3}}+{\text{O}}_{\text{2}}\text{Slow}\\ {\text{NO}}_{\text{3}}+{\text{NO}}_{\text{2}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\text{Fast}\end{array}$

Overall stoichiometry:

The elementary steps have to be summed up in order to find the overall stoichiometry.  This is done as shown below;

    $\begin{array}{c}{\text{NO}}_{\text{2}}+{\text{O}}_{\text{3}}\to {\text{NO}}_{\text{3}}+{\text{O}}_{\text{2}}\\ {\text{NO}}_{\text{3}}+{\text{NO}}_{\text{2}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\end{array}$

The species ${\text{NO}}_{\text{3}}$ is common in both equation and hence it gets cancelled out.  Thus, the overall stoichiometry can be given as follows;

    ${\text{2NO}}_{\text{2}}+{\text{O}}_{\text{3}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{5}}+{\text{O}}_{\text{2}}$

This is consistent with the stoichiometry that is determined experimentally.

Rate Law:

Rate of the reaction is determined by the rate limiting step and it is the slowest step in the mechanism.  Therefore, the first step is the slow step and it is the rate-limiting step.  The rate law can be written as shown below considering the slow step;

    $\begin{array}{l}\text{Rate}\propto {[{\text{NO}}_{\text{2}}]}^1{[{\text{O}}_{\text{3}}]}^1\\ \text{Rate}=k_2{[{\text{NO}}_{\text{2}}]}^1{[{\text{O}}_{\text{3}}]}^1\end{array}$

Where,

    $[{\text{NO}}_{\text{2}}]$ is the concentration of ${\text{NO}}_{\text{2}}$.

    $[{\text{O}}_{\text{3}}]$ is the concentration of ${\text{O}}_{\text{3}}$.

    $k_2$ is the rate constant.

The species ${\text{NO}}_{\text{3}}$ is the intermediate species that is formed and it exists for a very short time.

Thus this is consistent with the experimental results.