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Question: The gaseous decomposition of ozone, \[2{{O}_{3}}\to 3{{O}_{2}}\], obeys the rate law r = \[-\dfrac...

The gaseous decomposition of ozone, 2O3→3O22{{O}_{3}}\to 3{{O}_{2}}, obeys the rate law r =
−d[O3]dt=k[O3]2[O2]-\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:

& {{O}_{3}}\overset{{{K}_{eq}}}{\mathop{\rightleftharpoons }}\,{{O}_{2}}+O(fast) \\\ & O+{{O}_{3}}\xrightarrow{{{k}_{2}}}2{{O}_{2}}(slow) \\\ \end{aligned}$$
Explanation

Solution

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→C2A+B\to C
Reaction rate can be given as
Reaction rate= decrease in concentration of reactantstime\dfrac{decrease\text{ }in\text{ }concentration\text{ }of\text{ }reactants}{time}=increase in concentration of productstime\dfrac{increase\text{ }in\text{ }concentration\text{ }of\text{ }products}{time}
= −12d[A]dt=−d[B]dt=d[C]dt-\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 O3=k1[O3]{{O}_{3}}={{k}_{1}}\left[ {{O}_{3}} \right]
Reverse - Rate of formation of O3=k1′[O2][O]{{O}_{3}}={{k}_{1}}^{'}\left[ {{O}_{2}} \right]\left[ O \right]
In the case of slow mechanism,
Forward - Rate of consumption of O3=k2[O][O3]{{O}_{3}}={{k}_{2}}\left[ O \right]\left[ {{O}_{3}} \right]
The net rate of decomposition of O3{{O}_{3}} is given by
Rate= −d[O3]dt-\dfrac{d\left[ {{O}_{3}} \right]}{dt}= k1[O3]−k1′[O2][O]+k2[O][O3]{{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,

k1[O3]−k1′[O2][O]−k2[O][O3]=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]=0
k1[O3]=k1′[O2][O]+k2[O][O3]{{k}_{1}}\left[ {{O}_{3}} \right]={{k}_{1}}^{'}\left[ {{O}_{2}} \right]\left[ O \right]+{{k}_{2}}\left[ O \right]\left[ {{O}_{3}} \right]
[O]=k1[O3]k1′[O2]+k2[O3]\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=k1[O3]−k1′[O2]k1[O3]k1′[O2]+k2[O3]+k2k1[O3]k1′[O2]+k2[O3][O3]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]
=k1[O3]−k1′[O2]k1[O3]k1′[O2]+k2[O3]+k2k1[O3]2k1′[O2]+k2[O3]={{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]}
=2k2k1[O3]2k1′[O2]+k2[O3]=\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
k2[O][O3]≪k1′[O2][O],    i.e.,    k2[O3]≪k1′[O2]{{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=2k2k1[O3]2k1′[O2]=k[O3]2[O2]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=2k1k2k1′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.