Chapter 19.7, Problem 12PPA

Interpretation Introduction

Interpretation:

The reaction mechanism for the given chemical reaction should be showed that it is consistent with the given rate law of the chemical reaction.

Concept introduction:

In order to establish the plausibility of a mechanism, one must compare the rate law of the rate determining step to the experimentally determined rate law.

${\text{2NO + O}}_{\text{2}}{\text{ = 2NO}}_{\text{2}}\text{.}$

This reaction has been well established as a third-order homogeneous reaction.

Rate determining step:

In a chemical reaction the rate determining step is the slowest step in which the rate of the reaction depends on the rate of that slowest step.

Rate law:

It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

Answer to Problem 12PPA

At equilibrium the forward and reverse processes are occurring at the same rate, we can set their rates equal to each other and solve for the intermediate concentration.

Substituting the solution into the original rate law (rate=${\text{k}}_{\text{2}}{\text{[N}}_{\text{2}}{\text{O}}_{\text{2}}{\text{][O}}_{\text{2}}\text{]}\text{.}$) we will attain to the conclusion that the given reaction mechanism is consistent with the given rate law of the reaction.

Explanation of Solution

The rate law from the balanced equation of step 1 is as follows,

$\begin{array}{l}{\text{rate}}_{\text{forward}}=k_1\left[\text{NO}\right]\left[{\text{Br}}_{\text{2}}\right]\\ {\text{rate}}_{\text{reverse}}=k_{-1}\left[{\text{NOBr}}_{\text{2}}\right]\end{array}$

At equilibrium both the forward and the reverse reaction rate occurs at the same time hence we can equate both the rate and obtain the value for intermediate concentration.

$\begin{array}{l}{k}_1\left[\text{NO}\right]\left[{\text{Br}}_{\text{2}}\right]=k_{-1}\left[{\text{NOBr}}_{\text{2}}\right]\\ \left[{\text{NOBr}}_{\text{2}}\right]=\frac{k_1\left[\text{NO}\right]\left[{\text{Br}}_{\text{2}}\right]}{k_{-1}}\end{array}$

For step 2 the rate law is as follows,

$\text{rate}=k_2\left[{\text{NOBr}}_{\text{2}}\right]\left[\text{NO}\right]$

Substituting the obtained intermediated concentration value to the rate law obtained for the chemical equation in step 2 as below,

$\begin{array}{l}\text{rate}=k_2\left[{\text{NOBr}}_{\text{2}}\right]\left[\text{NO}\right]\\ =k_2\frac{k_1\left[\text{NO}\right]\left[{\text{Br}}_{\text{2}}\right]}{k_{-1}}\left[\text{NO}\right]\end{array}$

Now,

$k=k_2\frac{k_1}{k_{-1}}$

Therefore,

$\begin{array}{l}\text{rate}=k_2\left[{\text{NOBr}}_{\text{2}}\right]\left[\text{NO}\right]\\ =k\left[\text{NO}\right]\left[{\text{Br}}_{\text{2}}\right]\left[\text{NO}\right]\\ =k{\left[\text{NO}\right]}^2\left[{\text{Br}}_{\text{2}}\right]\end{array}$

The given rate law for the reaction is $\text{rate}=k{\left[\text{NO}\right]}^2\left[{\text{Br}}_{\text{2}}\right]$.

Hence it is proved that the given reaction mechanism is consistent with the given rate law of the chemical reaction.

Conclusion

The reaction mechanism for the given chemical reaction was showed that it is consistent with the given rate law of the chemical reaction.