Rate of reaction and dependence on concentrations Consider the decomposition of ozone as follows

(fast)
$$\ce{O3  &lt;=&gt; O2 + O}$$

(slow)
$$\ce{O + O3  -&gt; O2 }$$


The question comes here if the concentration of $\ce{O2}$ is increased , then comment on the rate of the reaction.

increases
decreases
constant
cannot be predicted

My try: I have a confusion that the rate law is not given so we cannot tell about the actual dependence of the rate on the concentration of $\ce{O2}$, I know that the rate law will depend upon the slow step as it is the rate determining step. Answer given is 2) decreases , can we apply Le Chatelier's principle here? Increase in $\ce{O2}$ , backward reaction so less rate?
If we carefully see them ozone formation is a complex ( not elementary) reaction. 
Yes the slowest step would be rate determining .
so for slowest step ;

Rate =$   k [O] [O _3] $

where $k$ would be rate constant. 

but  final reaction is :
$$\ce { 2 O_3 -&gt; 3O_2} $$
and so ,
equillibrium constant K(for 1st reaction)

$K= \frac{[O][O_2]}{[O_3]}$

thus substituting $[O] $from above equation to the rate equation .

We get 

Rate =$\frac {  k × K [O _3]^2 }{[O_2]}$

And hence if $[O_2]$ is increased it causes rate to decrease.