Question
The chemical reaction, $$2{O_3} \to 3{O_2}$$ proceeds as
$$\eqalign{
& {O_3} \rightleftharpoons {O_2} + \left[ O \right]\left( {{\text{fast}}} \right) \cr
& \left[ O \right] + {O_3} \to 2{O_2}\left( {{\text{slow}}} \right) \cr} $$
The rate law expression will be
A.
$${\text{Rate}} = k\left[ O \right]\left[ {{O_3}} \right]$$
B.
$${\text{Rate}} = k{\left[ {{O_3}} \right]^2}{\left[ {{O_2}} \right]^{ - 1}}$$
C.
$${\text{Rate}} = k{\left[ {{O_3}} \right]^2}$$
D.
$${\text{Rate}} = k\left[ {{O_2}} \right]\left[ O \right]$$
Answer :
$${\text{Rate}} = k{\left[ {{O_3}} \right]^2}{\left[ {{O_2}} \right]^{ - 1}}$$
Solution :
\[\left[ O \right]+{{O}_{3}}\xrightarrow{{{k}_{2}}}2{{O}_{2}}\left( \text{slow} \right)\]
Rate of reaction is determined by slow step hence, $${\text{Rate}} = {k_2}\left[ O \right]\left[ {{O_3}} \right]$$
$$\left[ O \right]$$ is unstable intermediate so substitute the value of $$\left[ O \right]$$ in above equation.
Rate of forward reaction $$ = {k_1}\left[ {{O_3}} \right]$$
Rate of backward reaction $$ = {k_{ - 1}}\left[ {{O_2}} \right]\left[ O \right]$$
At equilibrium,
Rate of forward reaction = Rate of backward reaction
$${k_1}\left[ {{O_3}} \right] = {k_{ - 1}}\left[ {{O_2}} \right]\left[ O \right];$$ $$\left[ O \right] = \frac{{{k_1}\left[ {{O_3}} \right]}}{{{k_{ - 1}}\left[ {{O_2}} \right]}}$$
$${\text{Rate}} = {k_2}\left( {\frac{{{k_1}\left[ {{O_3}} \right]}}{{{k_{ - 1}}\left[ {{O_2}} \right]}}} \right)\left[ {{O_3}} \right]$$ $$ = \frac{{k{{\left[ {{O_3}} \right]}^2}}}{{\left[ {{O_2}} \right]}}$$