## Chemistry and Chemical Reactivity (9th Edition)

The slow step is rate-determining so: $r=k_S[O_3][O]$ we need to eliminate the intermediary; the back fast reaction has the rate: $r_{-F}=k_{-F}[O_2][O]$ since it's an equilibrium $r_{-F}=r_F=k_{F}[O_3]$ and $K_F=k_F/k_{-F}$ $r=k_S[O_3]\dfrac{r_F}{k_{-F}[O_2]}=(K_Fk_S)[O_3]^2/[O_2]$