Chapter 16, Problem 16.75P

(a)

Interpretation Introduction

Interpretation:

Among the given experiments the one that has valid mechanism has to be predicted.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right)\end{array}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Explanation of Solution

• Experiment (1):

The given reaction is one-step collision, hence the reaction equation becomes,

$2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)$ and the rate law of the reaction is, $Rate=k{\left[NO\right]}^2\left[O_2\right]$.

The given actual rate law, is $\text{Rate}\text{=}\text{k}{\left[\text{NO}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]$ which is as same as the obtained rate law.

Therefore, the rate law of the slow step is, $Rate=k\left[N_2{O}_5\right]$ as same as the given actual rate law.  So, it is valid mechanism.

• Experiment (2):

The given reaction involves multiple mechanism steps, by adding the entire individual steps gives rise to an overall reaction equation. Hence, the reaction equation as follows,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}2NO\left(g\right)\rightleftharpoons \cancel{N_2{O}_2\left(g\right)}\left[fast\right]\\ \cancel{N_2{O}_2\left(g\right)}+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)\left[slow\right]\end{array}}\\ 2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)Overallreaction\end{array}}$

The overall equation becomes,

$2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)$

In the reaction, the slow step is the rate determining step; and its rate law is the overall rate law.

Therefore, the rate law of the slow step is, $Rate=k^{\text{'}}\left[N_2{O}_2\right]\left[O_2\right]$, not as same as the given actual rate law.  So, it is not a valid mechanism.

The concentration $\left[N_2{O}_2\right]$ is the intermediate that cannot be involved in the overall rate law.  Hence, substitution is needed for the $\left[N_2{O}_2\right]$ intermediate.

The rate law for the given mechanism steps are,

The rate of mechanism (1), ${\text{Rate}}_{\text{1}}{\text{ = k}}_1{\left[\text{NO}\right]}^2\text{and}{\text{Rate}}_{\text{1}}{\text{ = k}}_{-1}\left[{\text{N}}_2{\text{O}}_2\right]$, thus the step is in equilibrium, the rate of forward and backward reaction rate are equal,

$\begin{array}{l}{\text{ k}}_1{\left[\text{NO}\right]}^2\text{ = }{\text{k}}_{-1}\left[{\text{N}}_2{\text{O}}_2\right]\\ \\ \left[{\text{N}}_2{\text{O}}_2\right]=\frac{{\text{k}}_1}{{\text{ k}}_{-1}}{\left[\text{NO}\right]}^2-----------------\left(2\right)\end{array}$

Thus, by substituting above relation into the equation (1), the rate law becomes,

$\begin{array}{l}Rate=k^{\text{'}}\left[N_2{O}_2\right]\left[O_2\right]\\ \\ =\frac{{\text{ k}}^{\text{'}}{\text{k}}_1}{{\text{ k}}_{-1}}{\left[\text{NO}\right]}^2\left[O_2\right]\\ \text{=}\text{k}{\left[\text{NO}\right]}^2\left[O_2\right]\text{,}\text{where}\frac{{\text{ k}}^{\text{'}}{\text{k}}_1}{{\text{ k}}_{-1}}\text{=}\text{k}\\ \\ \boxed{Rate=k{\left[NO\right]}^2\left[O_2\right]}\end{array}$

Therefore, the given rate law is consistent with the rate law of overall reaction obtained as above.

• Experiment (3):

The given reaction involves multiple mechanism steps, by adding the entire individual steps gives rise to an overall reaction equation. Hence, the reaction equation as follows,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}2NO\left(g\right)\rightleftharpoons \cancel{N_2\left(g\right)}+O_2\left(g\right)\left[fast\right]\\ \cancel{N_2\left(g\right)}+2O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)\left[slow\right]\end{array}}\\ 2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)Overallreaction\end{array}}$

The overall equation becomes,

$2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)$

In the reaction, the slow step is the rate determining step; and its rate law is the overall rate law.

Therefore, the rate law of the slow step is, $Rate=k^{\text{'}}\left[N_2\right]{\left[O_2\right]}^2$, not as same as the given actual rate law.

The concentration $\left[N_2\right]$ is the intermediate that cannot be involved in the overall rate law.  Hence, substitution is needed for the $\left[N_2\right]$ intermediate.

The rate law for the given mechanism steps are,

The rate of mechanism (1), ${\text{Rate}}_{\text{1}}{\text{ = k}}_1{\left[\text{NO}\right]}^2\text{and}{\text{Rate}}_{\text{1}}{\text{ = k}}_{-1}\left[{\text{N}}_2\right]\left[{\text{O}}_2\right]$, thus the step is in equilibrium, the rate of forward and backward reaction rate are equal,

$\begin{array}{l}{\text{k}}_1{\left[\text{NO}\right]}^2\text{ = }{\text{k}}_{-1}\left[{\text{N}}_2\right]\left[{\text{O}}_2\right]\\ \\ \left[{\text{N}}_2\right]=\frac{{\text{k}}_1}{{\text{ k}}_{-1}}\frac{{\left[\text{NO}\right]}^2}{\left[{\text{O}}_2\right]}-----------------\left(2\right)\end{array}$

Thus, by substituting above relation into the equation (1), the rate law becomes,

$\begin{array}{l}\text{Rate}\text{=}{\text{k}}^{\text{'}}\left[{\text{N}}_{\text{2}}\right]{\left[{\text{O}}_{\text{2}}\right]}^{\text{2}}\\ \\ \text{=}\frac{{\text{ k}}^{\text{'}}{\text{k}}_{\text{1}}}{{\text{ k}}_{\text{-1}}}\frac{{\left[\text{NO}\right]}^{\text{2}}}{\left[{\text{O}}_{\text{2}}\right]}{\left[{\text{O}}_{\text{2}}\right]}^{\text{2}}\\ \text{=}\text{k}{\left[\text{NO}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]\text{,}\text{where}\frac{{\text{ k}}^{\text{'}}{\text{k}}_{\text{1}}}{{\text{ k}}_{\text{-1}}}\text{=}\text{k}\\ \\ \boxed{Rate=k{\left[NO\right]}^2\left[O_2\right]}\end{array}$

Therefore, the given rate law is consistent with the rate law of overall reaction obtained as above.

Therefore, all the mechanisms are consistent with the rate law.

(b)

Interpretation Introduction

Interpretation:

Among the given experiments the one that is most reasonable has to be predicted.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right)\end{array}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Explanation of Solution

• Experiment (1):

The given reaction is one-step collision, hence the reaction equation becomes,

$2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)$

• Experiment (2):

The given reaction involves multiple mechanism steps, by adding the entire individual steps gives rise to an overall reaction equation.  Hence, the reaction equation as follows,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}2NO\left(g\right)\rightleftharpoons \cancel{N_2{O}_2\left(g\right)}\left[fast\right]\\ \cancel{N_2{O}_2\left(g\right)}+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)\left[slow\right]\end{array}}\\ 2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)Overallreaction\end{array}}$

The overall equation becomes,

$2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)$

• Experiment (3):

The given reaction involves multiple mechanism steps, by adding the entire individual steps gives rise to an overall reaction equation.  Hence, the reaction equation as follows,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}2NO\left(g\right)\rightleftharpoons \cancel{N_2\left(g\right)}+O_2\left(g\right)\left[fast\right]\\ \cancel{N_2\left(g\right)}+2O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)\left[slow\right]\end{array}}\\ 2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)Overallreaction\end{array}}$

The overall equation becomes,

$2NO\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2N{O}_2\left(g\right)$

Comparing all the experiments, the experiment (1) and (3) are termolecular molecularity; and the experiment (2) is bimolecular.  Since, the most reasonable mechanism is (2).