Chapter 18, Problem 18.18P

(a)

Interpretation Introduction

Interpretation:

The mechanisms which are consistent with the rate law have to be shown.

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

The given rate of reaction is,

  $\text{Rate}\text{ = k }\left[H_{\text{2}}\right]\left[I_2\right]$

Consider mechanism a:

  $H_2\left(g\right)+I_2\left(g\right)\stackrel{k_1}{\to }2HI\left(g\right)\left(one-stepreaction\right)$

The rate law for the above one – step reaction is, $\text{Rate}{\text{ = k}}_{\text{1}}\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]$

Therefore, the rate law of the given reaction mechanism is obtained as $\text{Rate}{\text{ = k}}_{\text{1}}\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]$ is consistent with the given rate law.

Consider mechanism b:

  $\begin{array}{l}\left(1\right)I_2\left(g\right)\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}2I\left(g\right)\left(Fastequilibrium\right)\\ \\ \left(2\right)H_2\left(g\right)+2I\left(g\right)\stackrel{k_2}{\to }2HI\left(g\right)\left(Slow\right)\end{array}$

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

The slowest step of the given mechanism is,

  $H_2\left(g\right)+2I\left(g\right)\stackrel{k_2}{\to }2HI\left(g\right)$

The rate law for the above slowest reaction is, $\text{Rate}{\text{ = k}}_2\left[{\text{H}}_{\text{2}}\right]{\left[\text{I}\right]}^2\mathrm{.......}\left(1\right)$

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

For equilibrium reaction $\left(1\right)$, $I_2\left(g\right)\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}2I\left(g\right)$

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

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

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

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

Now, substituting equations $\left(4\right)$ into the equation $\left(1\right)$ results as,

  $\begin{array}{l}\text{Rate}{\text{ = k}}_{\text{2}}\left[{\text{H}}_{\text{2}}\right]{\left[\text{I}\right]}^{\text{2}}\\ \\ \text{Rate}{\text{ = k}}_{\text{2}}\left[{\text{H}}_{\text{2}}\right]\left(\frac{{\text{k}}_{\text{1}}\left[{\text{I}}_{\text{2}}\right]}{{\text{k}}_{\text{-1}}}\right)\\ \\ \text{Rate}\text{ = }\frac{{\text{k}}_{\text{2}}{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]\\ \\ \text{Rate}\text{ = k}\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]\text{,Considering}\frac{{\text{k}}_{\text{2}}{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\text{as}\text{k}\text{.}\\ \end{array}$

Thus, the rate law of the overall reaction becomes,

  $\text{Rate}\text{ = k}\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]$

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

Consider mechanism c:

  $\begin{array}{l}\left(1\right)I_2\left(g\right)\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}2I\left(g\right)\left(Fastequilibrium\right)\\ \\ \left(2\right)I\left(g\right)+H_2\left(g\right)\underset{k_{-2}}{\overset{k_2}{\rightleftarrows }}{H}_{2}I\left(g\right)\left(Fastequilibrium\right)\\ \\ \left(3\right)H_{2}I\left(g\right)\stackrel{k_3}{\to }HI\left(g\right)+I\left(g\right)\left(Slow\right)\end{array}$

The rate law for the above slowest reaction is, $\text{Rate}{\text{ = k}}_{\text{3}}\left[{\text{H}}_{\text{2}}\text{I}\right]\mathrm{.......}\left(\text{1}\right)$

The concentration of the intermediates $\left(Iand{H}_{2}I\right)$ cannot be involved in the overall rate law. Hence, substitution is needed for the intermediate.

For equilibrium reaction $\left(1\right)$, $I_2\left(g\right)\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}2I\left(g\right)$

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

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

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

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

For equilibrium reaction $\left(2\right)$, $I\left(g\right)+H_2\left(g\right)\underset{k_{-2}}{\overset{k_2}{\rightleftarrows }}{H}_{2}I\left(g\right)$

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

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

Equating both $\left(\text{5 }and\text{ 6}\right)$, the value of $\left[{\text{H}}_{\text{2}}\text{I}\right]$ can be obtained as given below.

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

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

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

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

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

Thus, the rate law of the overall reaction becomes,

  $\text{Rate}\text{ = k}\left[{\text{H}}_{\text{2}}\right]{\left[{\text{I}}_{\text{2}}\right]}^{1/2}$

Therefore, the rate law of the given reaction mechanism is obtained as $\text{Rate}\text{ = k}\left[{\text{H}}_{\text{2}}\right]{\left[{\text{I}}_{\text{2}}\right]}^{1/2}$ is not consistent with the given rate law.

(b)

Interpretation Introduction

Interpretation:

The mechanisms which are consistent with both the rate law and the additional observation have to be shown.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The additional observation is that the reaction was catalyzed by light when the energy of the light was sufficient to break the $I-I$ bond in an $I_2$ molecule.  This additional observation can be seen in both mechanism b and c in the first step.  As mechanism c is not consistent with the observed rate law, only mechanism b is consistent with both the rate law and the additional observation.