Consider the decomposition of ozone as follows

  1. (fast)

    $$\ce{O3 <=> O2 + O}$$

  2. (slow)

    $$\ce{O + O3 -> O2 }$$

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

  1. increases
  2. decreases
  3. constant
  4. 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?

  • $\begingroup$ For future reference, please don't use math markup $...$ to get italics. Use Markdown emphasis *...* instead. $\endgroup$
    – zwol
    Mar 12, 2017 at 14:15
  • $\begingroup$ @zwol I will take care of that in future questions. Thanks. $\endgroup$ Mar 12, 2017 at 14:26

3 Answers 3


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 : $$\ce { 2 O_3 -> 3O_2} $$ 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.


Here is an intuitive way to see that the rate of decomposition should decrease.

The second reaction only happens when an $\ce{O}$ radical collides with a ozone molecule. If you increase the concentration of $\ce{O2}$, the proportion of $\ce{O3}$ must go down (because the proportions of all species must add up to 1) and therefore the frequency of $\ce{O + O3}$ collisions must go down as well.


Intuitively, yes, increasing the concentration of $\ce{O2}$ will drive the (quasi-)equilibrium of the fast reaction $\ce{O3 <=> O2 + O}$ towards the left, decreasing the concentration of the intermediate species $\ce{O}$, and thus slowing down the rate-determining slow reaction $\ce{O + O3 -> 2 O2}$.

A minor complication here is that the rate of the slow reaction is also proportional to the concentration of $\ce{O3}$, which driving the fast reaction towards the left side will increase. Indeed, in principle, the rate of the slow reaction would be maximized when the fast quasi-equilibrium concentration of $\ce{O}$ was equal to that of $\ce{O3}$, with any deviation from this in either direction slowing down the reaction.

The missing piece of the puzzle is that the combined reaction is called "the decomposition of ozone", not "the dimerization of monatomic oxygen". It is thus reasonable to assume that $\ce{O3}$ is at least initially present in considerable excess of $\ce{O}$. Indeed, given the extreme reactivity of monatomic oxygen, it's never likely to be present at more than trace quantities under typical conditions. Thus, converting some $\ce{O}$ back into $\ce{O3}$ by reacting it with $\ce{O2}$ will reduce the (already very low) $\ce{O}$ concentration by a much larger factor than it will increase the (comparatively high) $\ce{O3}$ concentration, and will thus slow down the rate-determining step of the reaction.

In any case, even if we did somehow (e.g. by running the reaction in near vacuum) contrive a situation where the quasi-equilibrium of the fast reaction was shifted so far to the right that the concentration of $\ce{O}$ became comparable to that of $\ce{O3}$, then we would also need to consider the additional reaction $\ce{2 O -> O2}$ whose rate depends (quadratically!) only on the $\ce{O}$ concentration. I would expect that a more thorough analysis (which would, of course, have to take into account the relative rate constants of the different competing reactions) should turn up no situation where adding $\ce{O2}$ would actually speed up the overall decomposition of $\ce{O3}$. Of course, I could be wrong; these things are not always intuitive.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.