Rate law for breakdown of ozone to oxygen For the following problem:


  Example 3: Write the rate law for $\ce{2 O3 -&gt; 3 O2}$ given that the f‌irst step is fast and reversible, and the second step is slow.
  
  $$
\begin{align}
&amp;\text{Step 1 (fast, reversible)} &amp;\qquad \ce{O3 &amp;&lt;=&gt;[$k_1$][$k_{-1}$] O2 + O} \\
&amp;\text{Step 2 (slow)} &amp;\qquad \ce{O + O3 &amp;-&gt;[$k_2$] O2 + O2}
\end{align}
$$
  
  The rate is determined by the slowest step.
  
  The rate of formation of $\ce{O2}$ is equal to $2$ times the rate of the slow step $(k_2[\ce{O}][\ce{O3}]),$ since two molecules of $\ce{O2}$ are formed.
  
  Thus, rate of formation of $\ce{O2} = 2k_2[\ce{O}][\ce{O3}],$ but $“\ce{O}”$ is an intermediate, solve for $“\ce{O}”$ in
  terms of products and reactants and rate constants.
  
  Since the first step is fast and reversible and the second step is slow, the f‌irst step is in equilibrium and we can write
  
  $$\frac{[\ce{O2}][\ce{O}]}{[\ce{O3}]} = \frac{k_1}{k_{-1}} = K_1 \quad\text{or}\quad [\ce{O}] = \frac{k_1[\ce{O3}]}{k_{-1}[\ce{O2}]}$$
  
  Substituting:
  
  $$
\begin{align}
\text{rate} &amp;= \frac{2k_2k_1[\ce{O3}]^2}{k_{-1}[\ce{O2}]} \\
\text{rate} &amp;= k_\mathrm{obs}\frac{[\ce{O3}]^2}{[\ce{O2}]}
\end{align}
$$


The solution indicates that the rate of formation $\ce{O2}$ is $2k_2[\ce{O}][\ce{O3}]$. Why isn't it $2k_2[\ce{O}][\ce{O3}] + k_1[\ce{O3}]$, because the first reaction also produces $\ce{O2}$?
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 $k_{-1} \gg k_1$. You should make this assumption after you have the full equation, if necessary, but not before you do the calculation. 

To answer your last point you still have the intermediate [O] which is an unknown and so has to be eliminated as shown by using the steady state approach used above.