[wyldstylekenobi]

Cl + O₃ ClO + O₂ (fast) rate=k₁[Cl][O₃]
ClO + O Cl + O₂ (slow) rate=k₂[ClO][Ο]

so overall will be O₃ + O 2O₂ (feel confident about this much)
We use the slow rate law: rate=k₂[ClO][Ο]
but we have to get rid of the ClO because it is not one of the overall reaction's reactants (this is where my confidence diminishes) so the new rate law is: rate= k[Cl][O₃][Ο]

This is where everyone on the internet stops and I am so confused! Why can [Cl] be included in the rate law?! is it because it is a catalyst? Or is there, in fact, another step?

[chenbeier]

I think that is wrong, it has to be
Rate =k*[O₃]*[O ]

[AWK]

https://www.calstatela.edu/sites/default/files/dept/chem/07winter/102/chap-13-lecture.pdf

[Borek]

If the first step is fast enough and the second slow enough, concentration of [ClO] is directly proportional to the concentration of [Cl] (with some simplifying assumptions). Thus we can use either one (the difference will be just in the value of the k). [Cl] is in such a case more logical choice, as it is something that is present initially, regardless of the whole process.

In kinetics it is quite common do such mathematical tricks and approximations.