Chapter 13, Problem 13.109QA

Interpretation Introduction

To explain:

Which of the following mechanisms is for thermal decomposition of NO₂ and which is for the photochemical decomposition of NO₂?

Answer to Problem 13.109QA

Solution:

Mechanism a is the photochemical decomposition of NO, while mechanism b and c are thermal decomposition of NO₂.

Explanation of Solution

1) Concept:

We are calculating the rate law for slow step for each given mechanism. If it is the same as the given rate of thermal decomposition of NO₂, then that mechanism is for thermal decomposition of NO₂. If it is same as the given rate of photochemical decomposition of NO₂, then that mechanism is for photochemical decomposition of NO₂.

2) Given:

$\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{O}}_2,\mathrm{ }\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}{\left[\mathrm{N}{\mathrm{O}}_2\right]}^2$

$\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{m}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{N}\mathrm{O}}_2,\mathrm{ }\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}\left[\mathrm{N}{\mathrm{O}}_2\right]$

$\mathbf{a}.\mathrm{ }\mathrm{ }\mathrm{N}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to \mathrm{N}{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{O}}_{\left(\mathrm{g}\right)}\dots \mathrm{s}\mathrm{l}\mathrm{o}\mathrm{w}$

${\mathrm{ }\mathrm{O}}_{\left(\mathrm{g}\right)}+\mathrm{N}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to \mathrm{N}{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\dots \mathrm{f}\mathrm{a}\mathrm{s}\mathrm{t}$

$\mathbf{b}.\mathrm{N}{\mathrm{O}}_{2\left(\mathrm{g}\right)}+\mathrm{N}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{N}}_2{{\mathrm{O}}_4}_{\left(\mathrm{g}\right)}\dots \mathrm{f}\mathrm{a}\mathrm{s}\mathrm{t}$

${\mathrm{N}}_2{{\mathrm{O}}_4}_{\left(\mathrm{g}\right)}\to \mathrm{N}{\mathrm{O}}_{\left(\mathrm{g}\right)}+\mathrm{N}{\mathrm{O}}_{3\left(\mathrm{g}\right)}\dots \mathrm{s}\mathrm{l}\mathrm{o}\mathrm{w}$

$\mathrm{N}{\mathrm{O}}_{3\left(\mathrm{g}\right)}\to \mathrm{ }\mathrm{N}{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\dots \mathrm{f}\mathrm{a}\mathrm{s}\mathrm{t}$

$\mathbf{c}.\mathrm{N}{\mathrm{O}}_{2\left(\mathrm{g}\right)}+\mathrm{N}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to \mathrm{N}{\mathrm{O}}_{\left(\mathrm{g}\right)}+\mathrm{N}{\mathrm{O}}_{3\left(\mathrm{g}\right)}\dots \mathrm{s}\mathrm{l}\mathrm{o}\mathrm{w}$

$\mathrm{N}{\mathrm{O}}_{3\left(\mathrm{g}\right)}\to \mathrm{ }\mathrm{N}{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\dots \mathrm{f}\mathrm{a}\mathrm{s}\mathrm{t}$

3. Calculations:

Rate law for mechanism a:

${\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}}_1={\mathrm{k}}_{\mathrm{a}}\left[\mathrm{N}{\mathrm{O}}_2\right]$

Here K_a is the rate constant for slow step of reaction a. This rate is the same as the given rate of photochemical decomposition of NO₂. Therefore, mechanism a is the mechanism of photochemical decomposition of NO₂.

Rate law for mechanism b:

$\mathrm{R}\mathrm{a}\mathrm{t}{\mathrm{e}}_2={\mathrm{k}}_{\mathrm{b}}\left[{\mathrm{N}}_2{\mathrm{O}}_4\right]$

Here, N₂O₄ is the intermediate formed in the step (1) (fast), so we need to write the rate law for both forward and backward reaction of step (1):

$\mathrm{R}\mathrm{a}\mathrm{t}{\mathrm{e}}_1={\mathrm{k}}_1\left[\mathrm{N}{\mathrm{O}}_2\right]\left[{\mathrm{N}\mathrm{O}}_2\right]={\mathrm{k}}_{-1}\left[{\mathrm{N}}_2{\mathrm{O}}_4\right]$

Here k₁ is the rate constant of forward reaction, and k_-1 is the rate constant of backward reaction.

Rearranging above rate equation for [N₂O₄], we get

$\left[{\mathrm{N}}_2{\mathrm{O}}_4\right]=\frac{{\mathrm{k}}_1\left[\mathrm{N}{\mathrm{O}}_2\right]\left[{\mathrm{N}\mathrm{O}}_2\right]}{{\mathrm{k}}_{-1}}=\frac{{\mathrm{k}}_1}{{\mathrm{k}}_{-1}}{\left[\mathrm{N}{\mathrm{O}}_2\right]}^2$

Plugging the value of [N₂O₄] in Rate₂ equation, we get

$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}=\frac{{\mathrm{k}}_{\mathrm{b}}.{\mathrm{k}}_1}{{\mathrm{k}}_{-1}}{\left[\mathrm{N}{\mathrm{O}}_2\right]}^2$

$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}{\left[\mathrm{N}{\mathrm{O}}_2\right]}^2$

$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e},\mathrm{ }\mathrm{k}=\frac{{\mathrm{k}}_{\mathrm{b}}.{\mathrm{k}}_1}{{\mathrm{k}}_{-1}}$

This rate of mechanism b is the same as the given rate of thermal decomposition of NO₂. Therefore, mechanism c is the mechanism of thermal decomposition of NO₂.

Rate law for mechanism c:

$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}={\mathrm{k}}_{\mathrm{c}}\left[\mathrm{N}{\mathrm{O}}_2\right]\left[{\mathrm{N}\mathrm{O}}_2\right]$

$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}={\mathrm{k}}_{\mathrm{c}}{\left[\mathrm{N}{\mathrm{O}}_2\right]}^2$

Here K_c is the rate constant for slow step of mechanism c. This rate is the same as the given rate of thermal decomposition of NO₂. Therefore, mechanism c is the mechanism of thermal decomposition of NO₂.

Conclusion:

Mechanism a is the photochemical decomposition of NO₂, while mechanism b and c are thermal decomposition of NO₂.