Calculating rate constant and order of a multi-step reaction

Question

If I’m given a multi-step reaction, say:

$$\ce{2 NO + O2 -> 2 NO2} \label{rxn:R1}\tag{R1}$$

with the following mechanism:

$$ \begin{align} \ce{NO + O2 &<=>[$K$] NO3} &\quad &\text{(fast)} \label{rxn:R1.1}\tag{R1.1}\\ \ce{NO3 + NO &->[$K_1$] NO2 + NO2} &\quad &\text{(slow)} \, , \label{rxn:R1.2}\tag{R1.2} \end{align} $$

then will the rate constant of the whole reaction be equal to the product of the equilibrium constant of the fast reaction and the rate constant of the slower reaction?

Because this is how they made us solve it at my institute:

$$K_\mathrm{eq} = \frac{[\ce{NO3}]}{[\ce{NO}][\ce{O2}]}$$

$$ \begin{align} \text{Rate of slow step \eqref{rxn:R1.2}} &= K[\ce{NO3}][\ce{NO}] \\ &= KK_\mathrm{eq}[\ce{NO}][\ce{O2}][\ce{NO}] \\ \Rightarrow\text{Rate of reaction} &= K'[\ce{NO}]^2[\ce{O2}] \end{align} $$

where $K' = KK_\mathrm{eq}$, and the order is $2 + 1 = 3$.

So, is $K'$ the rate constant for the overall reaction? And is the reaction a 3rd order reaction?

Answer 1

The method you've indicated in your solution is correct. Regarding:

Then will the rate constant of the whole reaction be equal to the product of the equilibrium constant of the fast reaction and the rate constant of the slower reaction?

In your question, the final $K'=K_1\cdot K_{eq}$, but this is not always true.

Remember that you've to adjust the final rate law expression in such a way such that it is free from all intermediates. In doing so, you might need to multiply the $K$ of another reaction, as you did above. But you may also need to divide or multiply the square/cube/etc. of another reaction.

Hence, $K_{final}$ is not always equal to $K_1\cdot K_2\cdot...\cdot K_n$. So, your assumption is incorrect.