Chapter 16, Problem 16.110P

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

Reason for the correct option:

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,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}2N_2{O}_5\left(g\right)\stackrel{}{\to }\cancel{2N{O}_3\left(g\right)}+2N{O}_2\left(g\right)\left[Slow\right]\\ \cancel{2N{O}_3\left(g\right)}\stackrel{}{\to }2N{O}_2\left(g\right)+\cancel{2O}\left(g\right)\left[fast\right]\\ \cancel{2O}\left(g\right)\stackrel{}{\to }{O}_2\left(g\right)\left[fast\right]\end{array}}\\ 2N_2{O}_5\left(g\right)\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)Overallreaction\end{array}}$

The overall equation becomes,

$2N_2{O}_5\left(g\right)\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)$

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

Hence, the correct option is (II).

Reason for in-correct options:

• Experiment (1):

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

$2N_2{O}_5\left(g\right)\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)$ and the rate law of the reaction is, $Rate=k{\left[N_2{O}_5\right]}^2$.

The given actual rate law, is $\text{Rate}\text{=}\text{k}\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right]$ not equal to the obtained rate law.

• 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,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}{N}_2{O}_5\left(g\right)\rightleftharpoons \cancel{N{O}_3\left(g\right)}+\cancel{N{O}_2\left(g\right)}\left[fast\right]\\ \cancel{N{O}_2\left(g\right)}+N_2{O}_5\left(g\right)\stackrel{}{\to }3N{O}_2\left(g\right)+\cancel{O}\left(g\right)\left[slow\right]\\ \cancel{N{O}_3\left(g\right)}+\cancel{O\left(g\right)}\stackrel{}{\to }N{O}_2\left(g\right)+O_2\left(g\right)\left[fast\right]\end{array}}\\ 2N_2{O}_5\left(g\right)\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)Overallreaction\end{array}}$

The overall equation becomes,

$2N_2{O}_5\left(g\right)\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)$

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, $Rate=k\left[N_2{O}_5\right]\left[N{O}_2\right]$, not as same as the given actual rate law. So, it is not a valid mechanism.

• Experiment (4):

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,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}2N_2{O}_5\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)+\cancel{N_2{O}_3\left(g\right)}+\cancel{3O\left(g\right)}\left[fast\right]\\ \cancel{N_2{O}_3\left(g\right)}+\cancel{O\left(g\right)}\stackrel{}{\to }2N{O}_2\left(g\right)\left[slow\right]\\ 2\cancel{O\left(g\right)}\stackrel{}{\to }{O}_2\left(g\right)\left[fast\right]\end{array}}\\ 2N_2{O}_5\left(g\right)\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)Overallreaction\end{array}}$

The overall equation becomes,

$2N_2{O}_5\left(g\right)\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)$

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, $Rate=k\left[N_2{O}_3\right]\left[O\right]$, not as same as the given actual rate law. So, it is not a valid mechanism.

• Experiment (5):

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,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}2N_2{O}_5\left(g\right)\to \cancel{N_4{O}_{10}\left(g\right)}\left[Slow\right]\\ \cancel{N_4{O}_{10}\left(g\right)}\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)\left[fast\right]\end{array}}\\ 2N_2{O}_5\left(g\right)\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)Overallreaction\end{array}}$

The overall equation becomes,

$2N_2{O}_5\left(g\right)\stackrel{}{\to }4N{O}_2\left(g\right)+O_2\left(g\right)$

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, $Rate=k{\left[N_2{O}_5\right]}^2$, not as same as the given actual rate law. So, it is not a valid mechanism.

Hence, the in-correct options are (I), (III), (IV), and (V).