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The mechanism for the reaction: \begin{align} \ce{NO_{(g)} + O_{2(g)} &<=>[k_{-1}][k_{1}] NO_{3(g)}} &\text{(fast)}\\ \ce{NO_{3(g)} + NO_{(g)} &<=>[K_{2}] 2\,NO_{2(g)}} &\text{(slow)}\\ \end{align}

The experimental rate law: $$Rate=-\dfrac{\Delta\ce{[NO]}}{\Delta t}=k\ce{[NO]}^2\ce{[O_{2}]}$$

At equilibrium, the forward and reverse rates are equal to each other. We then have $$k_{1}\ce{[NO]}\ce{[O_{2}]}=k_{-1}\ce{[NO_{3}]}$$

This can be re-arranged to $$\ce{[NO_{3}]}=\dfrac{K_{1}}{K_{-1}}\ce{[NO]}\ce{[O_{2}]}$$

Since $2\,\ce{NO}$ disappear in the overall reaction for every NO that reacts in the second step, we multiply the overall reaction rate by two and thus have:

$$2k_{2}\ce{[NO_{3}]}\ce{[NO]}$$

I understand everything up to this point. What I do not understand is how we get to this equation from the previous one.

$$Rate=2k_{2}\dfrac{k_{1}}{k_{-1}}\ce{[NO]}^2\ce{[O_{2}]}$$

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The final rate equation is generated from substituting $\ce{[NO3]}$ defined in the third to last equation:

$\ce{[NO3]} = \frac{k_1}{k_{–1}}\ce{[NO][O_2]}$

into the second to last equation:

$2k_2\ce{[NO_3][NO]}$

This will give you the final rate equation defined only in terms of the starting concentrations. The $\ce{[NO_3]}$ term is replaced, $\ce{[NO]}$ squared, and $\ce{[O_2]}$ remains. None of the equilibrium constants cancel out. The intermediate $\ce{[NO_3]}$ is not included in the final equation, as it is consumed in the overall reaction.

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