# Calculating rate constant and order of a multi-step reaction

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?

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.

• Yeah, I get that. So whatever K’ I get in the final expression will be the rate constant of the overall reaction always? – Senthil Arihant Jan 4 '18 at 7:30
• Also, in the initial rate law expression (with the unstable NO3 concentration) will I always have to replace the most unstable species (i.e. NO3 here) to get be final rate law expression from which I’ll deduce the order of the reaction? Is that some kind of rule? – Senthil Arihant Jan 4 '18 at 7:32
• @SenthilArihant "Is that some kind of rule? " Yes, the rule is that the final rate law expression for a multi-step reaction should not contain any term for the intermediates in it. – Gaurang Tandon Jan 4 '18 at 8:38
• @SenthilArihant "I get in the final expression will be the rate constant of the overall reaction always?" Yes, of course – Gaurang Tandon Jan 4 '18 at 8:39