Chapter 11, Problem 71QRT

(a)

Interpretation Introduction

Interpretation:

The rate law for the given mechanism has to be determined.

The reaction:

  $2NO\left(g\right)+C{l}_2\left(g\right)\to 2NOC\mathrm{l}\left(g\right)$

The accepted mechanism:

  $\begin{array}{l}NO+C{l}_2\rightleftharpoons NOC{l}_{2}fast\\ \\ NOC{l}_2+NO\to 2NOClslow\end{array}$

Concept Introduction:

Steady-state approximation:

Once the steady state condition is reached, the rate of formation of the intermediate is equal to the rate of destruction of the intermediate.

Explanation of Solution

The reaction:

  $2NO\left(g\right)+C{l}_2\left(g\right)\to 2NOC\mathrm{l}\left(g\right)$

The accepted mechanism:

  $\begin{array}{l}NO+C{l}_2\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons }}NOC{l}_{2}fast\\ \\ NOC{l}_2+NO\stackrel{k_2}{\to }2NOClslow\end{array}$

The rate determining step is the slowest step in the mechanism.  Thus, the rate of the reaction is given below.

  $Rate=k_2\left[NOC{l}_2\right]\left[NO\right]$

Applying the steady-state approximation on the intermediate $\left(NOC{l}_2\right)$, the concentration of the intermediate can be obtained in terms of reactants.

At steady state condition, the rate of formation of the intermediate is equal to the rate of destruction of the intermediate.

$NOC{l}_2$ is created in the forward reaction of step $1$ and destroyed in the reverse reaction of step $1$ and step $2.$  Thus,

  $\begin{array}{l}Rat{e}_{forwardreaction1}=k_1\left[NO\right]\left[C{l}_2\right]\\ \\ Rat{e}_{reversereaction1}=k_{-1}\left[NOC{l}_2\right]\\ \\ Rat{e}_{reaction2}=k_2\left[NOC{l}_2\right]\left[NO\right]\end{array}$

Now,

  $\begin{array}{l}Rat{e}_{forwardreaction1}=Rat{e}_{reversereaction1}+Rat{e}_{reaction2}\\ \\ k_1\left[NO\right]\left[C{l}_2\right]=k_{-1}\left[NOC{l}_2\right]+k_2\left[NOC{l}_2\right]\left[NO\right]\end{array}$

Because the rate of step $2$ is presumed to be much smaller than the rate of step $1$, the second term is presumed to be negligibly small compared to the first term.

  $\begin{array}{l}Rat{e}_{forwardreaction1}\cong Rat{e}_{reversereaction1}\\ \\ k_1\left[NO\right]\left[C{l}_2\right]=k_{-1}\left[NOC{l}_2\right]\\ \\ \left[NOC{l}_2\right]=\frac{k_1}{k_{-1}}\left[NO\right]\left[C{l}_2\right]\end{array}$

Now, the rate law can be modified as given below.

  $\begin{array}{c}Rate=k_2\left(\frac{k_1}{k_{-1}}\left[NO\right]\left[C{l}_2\right]\right)\left[NO\right]\\ \\ Rate=k\text{'}{\left[NO\right]}^2\left[C{l}_2\right]\left[\because k\text{'}=k_2\frac{k_1}{k_{-1}}\right]\end{array}$

Therefore, the rate law of the reaction is $Rate=k\text{'}{\left[NO\right]}^2\left[C{l}_2\right].$

(b)

Interpretation Introduction

Interpretation:

Another mechanism that agrees with the same rate law has to be suggested.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Another mechanism that agrees with the same rate law is given below.

The imaginary mechanism:

  $\begin{array}{l}2N{O}_2\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons }}{N}_2{O}_{2}Fast\\ \\ N_2{O}_2+C{l}_2\stackrel{k_2}{\to }2NOC{l}_{2}Slow\end{array}$

The rate determining step is the slowest step in the mechanism.  Thus, the rate of the reaction is given below.

  $Rate=k_2\left[N_2{O}_2\right]\left[C{l}_2\right]$

Applying the steady-state approximation on the intermediate $\left(N_2{O}_2\right)$, the concentration of the intermediate can be obtained in terms of reactants.

At steady state condition, the rate of formation of the intermediate is equal to the rate of destruction of the intermediate.

$N_2{O}_2$ is created in the forward reaction of step $1$ and destroyed in the reverse reaction of step $1$ and step $2.$  Thus,

  $\begin{array}{l}Rat{e}_{forwardreaction1}=k_1{\left[N{O}_2\right]}^2\\ \\ Rat{e}_{reversereaction1}=k_{-1}\left[N_2{O}_2\right]\\ \\ Rat{e}_{reaction2}=k_2\left[N_2{O}_2\right]\left[C{l}_2\right]\end{array}$

Now,

  $\begin{array}{l}Rat{e}_{forwardreaction1}=Rat{e}_{reversereaction1}+Rat{e}_{reaction2}\\ \\ k_1{\left[NO\right]}^2=k_{-1}\left[N_2{O}_2\right]+k_2\left[N_2{O}_2\right]\left[C{l}_2\right]\end{array}$

Because the rate of step $2$ is presumed to be much smaller than the rate of step $1$, the second term is presumed to be negligibly small compared to the first term.

  $\begin{array}{l}Rat{e}_{forwardreaction1}\cong Rat{e}_{reversereaction1}\\ \\ k_1{\left[NO\right]}^2=k_{-1}\left[N_2{O}_2\right]\\ \\ \left[N_2{O}_2\right]=\frac{k_1}{k_{-1}}{\left[NO\right]}^2\end{array}$

Now, the rate law can be modified as given below.

  $\begin{array}{c}Rate=k_2\left(\frac{k_1}{k_{-1}}{\left[NO\right]}^2\right)\left[C{l}_2\right]\\ \\ Rate=k\text{'}{\left[NO\right]}^2\left[C{l}_2\right]\left[\because k\text{'}=k_2\frac{k_1}{k_{-1}}\right]\end{array}$

Therefore, the rate law of the reaction is $Rate=k\text{'}{\left[NO\right]}^2\left[C{l}_2\right].$

(c)

Interpretation Introduction

Interpretation:

Another mechanism that does not agree with the same rate law has to be suggested.

Explanation of Solution

Another mechanism that does not agree with the same rate law is given below.

The imaginary mechanism:

  $\begin{array}{l}NO+C{l}_2\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons }}NOCl+ClSlow\\ \\ NO+Cl\stackrel{k_2}{\to }NOClFast\end{array}$

The rate determining step is the slowest step in the mechanism.  Thus, the rate of the reaction is given below.

  $Rate=k_2\left[NO\right]\left[C{l}_2\right]$