# How do you use the steady state approximation to determine the observed rate constant for an overall reaction, when knowing its mechanism and its individual rate constants?

Sep 29, 2015

The steady state approximation assumes that the change in the concentration of the intermediate is 0.

What this means is that the rate of the first step is slow enough that the intermediate is consumed as it is being produced, thereby making its concentration constant. So, let's take this reaction as an example.

Overall:

$2 {N}_{2} {O}_{5} \stackrel{{k}_{\text{obs}}}{\to} 4 N {O}_{2} + {O}_{2}$

Proposed Mechanism:

${N}_{2} {O}_{5} \stackrel{{k}_{1}}{r i g h t \le f t h a r p \infty n s} N {O}_{2} + N {O}_{3}$
${\textcolor{w h i t e}{\text{a"" "" }}}^{{k}_{- 1}}$
$N {O}_{2} + N {O}_{3} \stackrel{{k}_{2}}{\implies} N O + N {O}_{2} + {O}_{2}$
$N {O}_{3} + N O \stackrel{{k}_{3}}{\implies} 2 N {O}_{2}$

It's a bit tricky, but if you use the reactants in step 2, you can convince yourself that $N {O}_{2} + N {O}_{3}$ in step 2 is the same as ${N}_{2} {O}_{5}$, so this reaction does cancel out to be the overall reaction. Fun fact: this was on my exam on kinetics last semester.

Our goal is to get ${k}_{\text{obs}}$, the observed rate constant for the overall reaction, in terms of none of the intermediates.

So, using the steady state approximation, we can start by identifying the intermediates as $N O$ and $N {O}_{3}$ ($N {O}_{2}$ doesn't completely cancel out, so it doesn't count), and using that:

$\frac{d \left[N O\right]}{\mathrm{dt}} = 0 = {k}_{2} \left[N {O}_{2}\right] \left[N {O}_{3}\right] - {k}_{3} \left[N {O}_{3}\right] \left[N O\right]$

The first set of terms is positive because there, $N O$ is being produced. The second set of terms is subtracted because there, $N O$ is being consumed to make $2 N {O}_{2}$.

From here, we are trying to get the concentration of each intermediate in terms of the main reactants in the overall reaction:

${k}_{2} \left[N {O}_{2}\right] \cancel{\left[N {O}_{3}\right]} = {k}_{3} \cancel{\left[N {O}_{3}\right]} \left[N O\right]$

$\textcolor{g r e e n}{\left[N O\right] = {k}_{2} / {k}_{3} \left[N {O}_{2}\right]}$

Got the one for $N O$. Now for $N {O}_{3}$:

$\frac{d \left[N {O}_{3}\right]}{\mathrm{dt}} = 0 = {k}_{1} \left[{N}_{2} {O}_{5}\right] - {k}_{2} \left[N {O}_{2}\right] \left[N {O}_{3}\right] - {k}_{3} \left[N {O}_{3}\right] \left[N O\right] - {k}_{- 1} \left[N {O}_{2}\right] \left[N {O}_{3}\right]$

$N {O}_{3}$ was consumed in step 2, consumed in step 3, and would be consumed in the backwards reaction in step 1. Thus, the operations end up being that you subtract ${k}_{2}$, subtract ${k}_{3}$, and subtract ${k}_{- 1}$.

${k}_{1} \left[{N}_{2} {O}_{5}\right] = {k}_{2} \left[N {O}_{2}\right] \left[N {O}_{3}\right] + {k}_{3} \left[N {O}_{3}\right] \left[N O\right] + {k}_{- 1} \left[N {O}_{2}\right] \left[N {O}_{3}\right]$

${k}_{1} \left[{N}_{2} {O}_{5}\right] = \left[N {O}_{3}\right] \left({k}_{2} \left[N {O}_{2}\right] + {k}_{3} \left[N O\right] + {k}_{- 1} \left[N {O}_{2}\right]\right)$

$\textcolor{g r e e n}{\left[N {O}_{3}\right]} = \frac{{k}_{1} \left[{N}_{2} {O}_{5}\right]}{{k}_{2} \left[N {O}_{2}\right] + {k}_{3} \left[N O\right] + {k}_{- 1} \left[N {O}_{2}\right]}$

As you can see on this step, we needed $\left[N O\right]$ in order to figure out $\left[N {O}_{3}\right]$; how convenient!

$= \frac{{k}_{1} \left[{N}_{2} {O}_{5}\right]}{{k}_{2} \left[N {O}_{2}\right] + \cancel{{k}_{3}} \cdot {k}_{2} / \cancel{{k}_{3}} \left[N {O}_{2}\right] + {k}_{- 1} \left[N {O}_{2}\right]}$

$= \textcolor{g r e e n}{\frac{{k}_{1} \left[{N}_{2} {O}_{5}\right]}{2 {k}_{2} \left[N {O}_{2}\right] + {k}_{- 1} \left[N {O}_{2}\right]}}$

At this point, we don't want $\left[N {O}_{2}\right]$ in here because it's not a reactant in the overall reaction; it is both an intermediate and a product in some sense.

Since ${O}_{2}$ is a distinguishable product in the overall reaction, we can use that:

$\frac{d \left[{O}_{2}\right]}{\mathrm{dt}} = {k}_{2} \left[N {O}_{2}\right] \left[N {O}_{3}\right]$

Conveniently enough, after substitution, the $\left[N {O}_{2}\right]$ cancels out:

$= {k}_{2} \cancel{\left[N {O}_{2}\right]} \left(\frac{{k}_{1} \left[{N}_{2} {O}_{5}\right]}{2 {k}_{2} \cancel{\left[N {O}_{2}\right]} + {k}_{- 1} \cancel{\left[N {O}_{2}\right]}}\right)$

$= {k}_{2} \cdot \left(\frac{{k}_{1} \left[{N}_{2} {O}_{5}\right]}{2 {k}_{2} + {k}_{- 1}}\right)$

$= \frac{{k}_{1} {k}_{2}}{2 {k}_{2} + {k}_{- 1}} \left[{N}_{2} {O}_{5}\right]$

Therefore, comparing with the overall reaction:

$\frac{d \left[{O}_{2}\right]}{\mathrm{dt}} = {k}_{\text{obs}} \left[{N}_{2} {O}_{5}\right]$

$\frac{d \left[{O}_{2}\right]}{\mathrm{dt}} = \frac{{k}_{1} {k}_{2}}{2 {k}_{2} + {k}_{- 1}} \left[{N}_{2} {O}_{5}\right]$

...the observed rate constant is:

$\textcolor{b l u e}{{k}_{\text{obs}} = \frac{{k}_{1} {k}_{2}}{2 {k}_{2} + {k}_{- 1}}}$