The chemical reaction, `2O_(3) to 3O_(2)` proceeds as
`O_(3) hArr O_(2) + [O]` (fast)
`[O] + O_(3) to 2O_(2)` (slow)
The rate law expression will be

Explanation

`2O_(3) overset(k_(1))rarr 3O_(2)`
Mechanism:
`O_(3) overset(k_(eq))hArr O_(2) + O` (fast)
`O + O_(3) overset(k)rarr2O_(2)` (slow)
Rate of the reaction form the slow step is
`r = k[O][O_(3)]` …(i)
Fast step is in equilibrium, the concentration of `[O]` is calculated form the step
`k_(eq) = ([O_(2)][O])/([O_(3)]) :. [O] = (k_(eq). [O_(3)])/([O_(2)])`
Substitute the value of `[O]` in Eq. (i)
`r = (k.k_(eq)[O_(3)][O_(3)])/(O_(2)) = (k_(1)[O_(3)]^(2))/([O_(2)]) = k_(1)[O_(3)]^(2)[O_(2)]^(-1)`