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\begin{align} &\text{Step 1 (fast, reversible)} &\qquad \ce{O3 &<=>[k_1][k_{-1}] O2 + O} \\ &\text{Step 2 (slow)} &\qquad \ce{O + O3 &->[k_2] O2 + O2} \end{align} The rate is determined by the slowest step. $$r=k_2[\ce{O}][\ce{O3}]$$ If the concentration of $\ce{ O}$ which produced from the first step ...
You should write down the rate equations and set [O] as an intermediate so that $\ce{d[O]/dt} = 0$. This is the steady state approach and leads to $\displaystyle \frac{d\mathrm{[O_2]}} {dt}=2\frac{k_1k_2\mathrm{[O_3]}^2}{k_{-1}\mathrm{[O_2]}+k_2\mathrm{[O_3]}}$ which differs from your answer because when you assume a fast equilibrium you also assumed that \$...