Chapter 18, Problem 18.22P

Interpretation Introduction

Interpretation:

It has to be shown that the given mechanism also gives the observed rate law.  The observed rate law is $\text{Rate}\text{ = k }{\left[{\text{Cl}}_{\text{2}}\right]}^{3/2}\left[CO\right].$

Concept Introduction:

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

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

$\text{Reaction Rate}\text{ = k }{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}{\left[\text{C}\right]}^{\text{p}}\text{,where 'm, n and p' are orders of the reactants}\text{.}$

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

The observed rate of reaction is,

  $\text{Rate}\text{ = k }{\left[{\text{Cl}}_{\text{2}}\right]}^{3/2}\left[CO\right]$

Mechanism:

  $\begin{array}{l}\left(1\right)C{l}_2\left(g\right)\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}2Cl\left(g\right)\left(fastequilibrium\right)\\ \\ \left(2\right)Cl\left(g\right)+C{l}_2\left(g\right)\underset{k_{-2}}{\overset{k_2}{\rightleftarrows }}C{l}_3\left(g\right)\left(fastequilibrium\right)\\ \\ \left(3\right)C{l}_3\left(g\right)+CO\left(g\right)\stackrel{k_3}{\to }C{l}_{2}CO\left(g\right)+Cl\left(g\right)\left(slow\right)\end{array}$

As known, the overall reaction rate is same as the rate of the slowest reaction step.

The slowest step of the given mechanism is,

  $C{l}_3\left(g\right)+CO\left(g\right)\stackrel{k_3}{\to }C{l}_{2}CO\left(g\right)+Cl\left(g\right)$

The rate law for the above slowest reaction is, $\text{Rate}{\text{ = k}}_{\text{3}}\left[{\text{Cl}}_3\right]\left[\text{CO}\right]\mathrm{.......}\left(1\right)$

The concentration of the intermediate $\left(ClandC{l}_3\right)$ cannot be involved in the overall rate law. Hence, substitution is needed for the intermediate.

For equilibrium reaction $\left(1\right)$, ${\text{Cl}}_{\text{2}}\left(\text{g}\right)\underset{{\text{k}}_{\text{-1}}}{\overset{{\text{k}}_{\text{1}}}{\rightleftarrows }}\text{2Cl}\left(\text{g}\right)$

The rate of forward reaction, $\text{Rate}{\text{ = k}}_{\text{1}}\left[{\text{Cl}}_{\text{2}}\right]\mathrm{........}\left(2\right)$

The rate of reverse reaction, $\text{Rate}{\text{ = k}}_{\text{-1}}{\left[\text{Cl}\right]}^{\text{2}}\mathrm{........}\left(\text{3}\right)$

Equating both $\left(2and3\right)$, the value of $\left[\text{Cl}\right]$ can be obtained as given below.

  $\begin{array}{l}{\text{k}}_{\text{1}}\left[{\text{Cl}}_{\text{2}}\right]{\text{= k}}_{\text{-1}}{\left[\text{Cl}\right]}^{\text{2}}\\ \\ \left[\text{Cl}\right]={\left(\frac{{\text{k}}_{\text{1}}\left[{\text{Cl}}_{\text{2}}\right]}{{\text{k}}_{\text{-1}}}\right)}^{1/2}\mathrm{.....}\left(4\right)\end{array}$

For equilibrium reaction $\left(2\right)$, $C{l}_3\left(g\right)+CO\left(g\right)\stackrel{k_3}{\to }C{l}_{2}CO\left(g\right)+Cl\left(g\right)$

The rate of forward reaction, $\text{Rate}{\text{ = k}}_{\text{2}}\left[\text{Cl}\right]\left[{\text{Cl}}_2\right]\mathrm{........}\left(5\right)$

The rate of reverse reaction, $\text{Rate}{\text{ = k}}_{\text{-2}}\left[{\text{Cl}}_3\right]\mathrm{........}\left(6\right)$

Equating both $\left(\text{5 }and\text{ 6}\right)$, the value of $\left[{\text{Cl}}_3\right]$ can be obtained as given below.

  $\begin{array}{l}{\text{ k}}_{\text{2}}\left[\text{Cl}\right]\left[{\text{Cl}}_2\right]={\text{k}}_{\text{-2}}\left[{\text{Cl}}_3\right]\\ \\ \left[{\text{Cl}}_3\right]=\frac{{\text{ k}}_{\text{2}}\left[\text{Cl}\right]\left[{\text{Cl}}_2\right]}{{\text{k}}_{\text{-2}}}\mathrm{.....}\left(7\right)\end{array}$

Now, substituting equations $\left(4\right)$ into the equation $\left(7\right)$ results as,

  $\begin{array}{l}\left[{\text{Cl}}_3\right]=\frac{{\text{ k}}_{\text{2}}\left[\text{Cl}\right]\left[{\text{Cl}}_2\right]}{{\text{k}}_{\text{-2}}}\\ \\ \left[{\text{Cl}}_3\right]=\frac{{\text{ k}}_{\text{2}}\left[{\text{Cl}}_2\right]}{{\text{k}}_{\text{-2}}}{\left(\frac{{\text{k}}_{\text{1}}\left[{\text{Cl}}_{\text{2}}\right]}{{\text{k}}_{\text{-1}}}\right)}^{1/2}\\ \\ \left[{\text{Cl}}_3\right]=\frac{{\text{ k}}_{\text{2}}}{{\text{k}}_{\text{-2}}}{\left(\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\right)}^{1/2}{\left[{\text{Cl}}_2\right]}^{3/2}\mathrm{.......}\left(8\right)\end{array}$

Now, substituting equations $\left(8\right)$ into the equation $\left(1\right)$ results as,

  $\begin{array}{l}\text{Rate}{\text{ = k}}_{\text{3}}\left[{\text{Cl}}_3\right]\left[\text{CO}\right]\\ \\ \text{Rate}{\text{ = k}}_{\text{3}}\left[\text{CO}\right]\frac{{\text{ k}}_{\text{2}}}{{\text{k}}_{\text{-2}}}{\left(\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\right)}^{1/2}{\left[{\text{Cl}}_2\right]}^{3/2}\\ \\ \text{Rate}{\text{ = k}}_{\text{3}}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{-2}}}{\left(\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\right)}^{\text{1/2}}\left[\text{CO}\right]{\left[{\text{Cl}}_{\text{2}}\right]}^{3/2}\\ \\ \text{Rate}\text{ = k}\left[\text{CO}\right]{\left[{\text{Cl}}_{\text{2}}\right]}^{3/2},\text{Considering}{\text{k}}_{\text{3}}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{-2}}}{\left(\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\right)}^{\text{1/2}}\text{as}\text{k}\text{. }\end{array}$

Thus, the rate law of the overall reaction becomes,

  $\text{Rate}\text{ = k}\left[\text{CO}\right]{\left[{\text{Cl}}_{\text{2}}\right]}^{3/2}$

Therefore, the rate law of the given reaction mechanism is obtained as $\text{Rate}\text{ = k}\left[\text{CO}\right]{\left[{\text{Cl}}_{\text{2}}\right]}^{3/2}$ is consistent with the given rate law.

The mechanism having higher $k$ value should be taken into consideration.  In this mechanism, $C{l}_3\left(g\right)$ has been formed in second step, but $C{l}_3\left(g\right)$ does not exist.  Therefore, this mechanism is not the correct one.  The correct mechanism is given below.

  $\begin{array}{l}\left(1\right)C{l}_2\left(g\right)\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}2Cl\left(g\right)\left(fastequilibrium\right)\\ \\ \left(2\right)Cl\left(g\right)+CO\left(g\right)\underset{k_{-2}}{\overset{k_2}{\rightleftarrows }}ClCO\left(g\right)\left(fastequilibrium\right)\\ \\ \left(3\right)ClCO\left(g\right)+C{l}_2\left(g\right)\stackrel{k_3}{\to }C{l}_{2}CO\left(g\right)+Cl\left(g\right)\left(slow\right)\end{array}$