How is the 'rate law' and 'order' for the ozone decomposition reaction can be defined?

Question

How is the 'rate law' and 'order' for the following reaction can be defined?

$$\ce{2O3 -> 3O2}$$

"The expression that correlates the rate of reaction with concentration of reactants is known as rate law for that reaction." So the rate law only include the concentration of reactants, not products. But the rate law for this reaction is:

$$r = k[\ce{O3}]^2[\ce{O2}]^{-1}$$

where $k$ is rate constant.

"Order of a reaction is the number of reactant molecules whose concentration determines the rate expression."

Order for this reaction w.r.t. ozone is 2 and that w.r.t. oxygen is -1. But why we are considering oxygen, being a product of this irreversible reaction. This also means that the rate of this reaction will decrease with the increase in concentration of oxygen.

Now it seems that, this is a reversible reaction. But chemical kinetics is studied for only irreversible reaction?

Answer 1

Chemical kinetic describes both reversible and irreversible reactions.

The reaction scheme and rate expression do not refer to the same thing. The reaction scheme is an overall one whereas the rate equation refers to a particular multi-step reaction. Two examples are given below:

Ozone can be destroyed by UV light in the $240$ to $\pu{320 nm}$ region with the scheme $(\ce{M}$ is an inert gas needed to take away excess energy):

$$ \begin{align} \ce{O3 &->[$hν$] O2 + O} \\ \ce{O3 + O + M &-> O2 + O2 + M} \\ \hline \ce{2 O3 &-> 3 O2} \end{align} $$

Similarly, oxygen molecules can react with short wavelength UV as

$$ \begin{align} \ce{O2 &->[$hν$][$(<\pu{242 nm})$] O + O} & \\ \ce{O + O2 + M &-> O3 + M} &|\cdot 2 \\ \hline \ce{3 O2 &-> 2 O3} \end{align} $$

which also gives a similar net reaction. This is the Chapman mechanism and is very important in atmospheric chemistry.