Question

The gaseous decomposition of ozone, \[2{{O}_{3}}\to 3{{O}_{2}}\], obeys the rate law r =
\[-\dfrac{d\left[ {{O}_{3}} \right]}{dt}=\dfrac{k{{\left[ {{O}_{3}} \right]}^{2}}}{\left[ {{O}_{2}} \right]}\].
Show that the following mechanism is consistent with the above rate law:
\[\begin{align}
  & {{O}_{3}}\overset{{{K}_{eq}}}{\mathop{\rightleftharpoons }}\,{{O}_{2}}+O(fast) \\
 & O+{{O}_{3}}\xrightarrow{{{k}_{2}}}2{{O}_{2}}(slow) \\
\end{align}\]

Answer 1

Hint: The reaction rate is expressed as a derivative of the concentration of reactant A or product C, with respect to time, t.
Consider the following reaction:
\[2A+B\to C\]
Reaction rate can be given as
Reaction rate= \[\dfrac{decrease\text{ }in\text{ }concentration\text{ }of\text{ }reactants}{time}\]=\[\dfrac{increase\text{ }in\text{ }concentration\text{ }of\text{ }products}{time}\]
= \[-\dfrac{1}{2}\dfrac{d[A]}{dt}=-\dfrac{d[B]}{dt}=\dfrac{d[C]}{dt}\]

Complete step by step solution: In the case of fast mechanism, since it is at equilibrium:
Forward - Rate of decomposition of \[{{O}_{3}}={{k}_{1}}\left[ {{O}_{3}} \right]\]
Reverse - Rate of formation of \[{{O}_{3}}={{k}_{1}}^{'}\left[ {{O}_{2}} \right]\left[ O \right]\]
In the case of slow mechanism,
Forward - Rate of consumption of \[{{O}_{3}}={{k}_{2}}\left[ O \right]\left[ {{O}_{3}} \right]\]
The net rate of decomposition of \[{{O}_{3}}\] is given by
Rate= \[-\dfrac{d\left[ {{O}_{3}} \right]}{dt}\]= \[{{k}_{1}}\left[ {{O}_{3}} \right]-{{k}_{1}}^{'}\left[ {{O}_{2}} \right]\left[ O \right]+{{k}_{2}}\left[ O \right]\left[ {{O}_{3}} \right]\] (i)
Apply the steady-state approximation to the intermediate (O atom) and after rearranging,

\[{{k}_{1}}\left[ {{O}_{3}} \right]-{{k}_{1}}^{'}\left[ {{O}_{2}} \right]\left[ O \right]-{{k}_{2}}\left[ O \right]\left[ {{O}_{3}} \right]=0\]
\[{{k}_{1}}\left[ {{O}_{3}} \right]={{k}_{1}}^{'}\left[ {{O}_{2}} \right]\left[ O \right]+{{k}_{2}}\left[ O \right]\left[ {{O}_{3}} \right]\]
\[\left[ O \right]=\dfrac{{{k}_{1}}\left[ {{O}_{3}} \right]}{{{k}_{1}}^{'}\left[ {{O}_{2}} \right]+{{k}_{2}}\left[ {{O}_{3}} \right]}\] (ii)
From (1) and (2), we have,
\[Rate={{k}_{1}}\left[ {{O}_{3}} \right]-{{k}_{1}}^{'}\left[ {{O}_{2}} \right]\dfrac{{{k}_{1}}\left[ {{O}_{3}} \right]}{{{k}_{1}}^{'}\left[ {{O}_{2}} \right]+{{k}_{2}}\left[ {{O}_{3}} \right]}+{{k}_{2}}\dfrac{{{k}_{1}}\left[ {{O}_{3}} \right]}{{{k}_{1}}^{'}\left[ {{O}_{2}} \right]+{{k}_{2}}\left[ {{O}_{3}} \right]}\left[ {{O}_{3}} \right]\]
        \[={{k}_{1}}\left[ {{O}_{3}} \right]-\dfrac{{{k}_{1}}^{'}\left[ {{O}_{2}} \right]{{k}_{1}}\left[ {{O}_{3}} \right]}{{{k}_{1}}^{'}\left[ {{O}_{2}} \right]+{{k}_{2}}\left[ {{O}_{3}} \right]}+\dfrac{{{k}_{2}}{{k}_{1}}{{\left[ {{O}_{3}} \right]}^{2}}}{{{k}_{1}}^{'}\left[ {{O}_{2}} \right]+{{k}_{2}}\left[ {{O}_{3}} \right]}\]
\[=\dfrac{2{{k}_{2}}{{k}_{1}}{{\left[ {{O}_{3}} \right]}^{2}}}{{{k}_{1}}^{'}\left[ {{O}_{2}} \right]+{{k}_{2}}\left[ {{O}_{3}} \right]}\]
It is provided in the question that the second step is relatively slower than the first step, therefore, we can make the approximation
\[{{k}_{2}}\left[ O \right]\left[ {{O}_{3}} \right]\ll {{k}_{1}}^{'}\left[ {{O}_{2}} \right]\left[ O \right],\,\,\,\,i.e.,\,\,\,\,{{k}_{2}}\left[ {{O}_{3}} \right]\ll {{k}_{1}}^{'}\left[ {{O}_{2}} \right]\]
Therefore, we get,
\[Rate=\dfrac{2{{k}_{2}}{{k}_{1}}{{\left[ {{O}_{3}} \right]}^{2}}}{{{k}_{1}}^{'}\left[ {{O}_{2}} \right]}=\dfrac{k{{\left[ {{O}_{3}} \right]}^{2}}}{\left[ {{O}_{2}} \right]}\]
          where, \[k=\dfrac{2{{k}_{1}}{{k}_{2}}}{{{k}_{1}}^{'}}\]
Hence, we have proved that the mechanism is consistent with the given rate law.


Note: The negative sign before the differentiation indicates that there will be a decrease in the concentration and positive means an increase in concentration. As it happens in a chemical reaction, the reactants get consumed to form the products. Therefore, the sign convention.