# For the overall reaction shown below with a fast equilibrium, why is the true rate law NOT r(t) = k_2[A][B_2]? Also, why is the overall reaction order 3//2 and not 3?

## ${A}_{2} \stackrel{{k}_{1} \text{ }}{r i g h t \le f t h a r p \infty n s} A + A$ (fast) ${\textcolor{w h i t e}{a a a}}^{\textcolor{g r e e n}{{k}_{- 1}}}$ $A + {B}_{2} \stackrel{{k}_{2} \text{ }}{\implies} A B + B$ (slow) $\underline{A + B \stackrel{{k}_{3} \text{ }}{\implies} A B}$ (fast) ${A}_{2} + {B}_{2} \stackrel{{k}_{o b s} \text{ }}{\to} 2 A B$

Feb 12, 2016

I see what happened; what you wrote was a rate law that incorporated the intermediate $A$, when that wasn't part of the overall reaction. You actually wrote the rate law based on the slow step, but didn't get rid of the intermediate concentration.

OVERALL REACTION

$\setminus m a t h b f \left({A}_{2} + {B}_{2} \stackrel{{k}_{o b s} \text{ }}{\to} 2 A B\right)$

where ${k}_{o b s}$ is the observed rate constant, and may or may not contain more than one rate constant within it.

REACTION MECHANISM

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

where $\implies$ indicates an elementary step, i.e. a step with no implicit mechanism.

An elementary step has no intermediates and only one transition state. So, we know that that's exactly how that step happens and we can treat each reactant as first order.

(If there was $A + A \implies 2 A$, then $r \left(t\right) = k \left[A\right] \left[A\right] = k {\left[A\right]}^{2}$.)

THE FAST EQUILIBRIUM APPROXIMATION

Since the equilibrium is stated to be fast, apparently we are using the fast equilibrium approximation. We would not know this if we were not told the equilibrium was fast.

If we were told that the first step was slow, it could have been the steady-state approximation, which is not the same. That assumption says that $\Delta \left[I\right] \approx 0$ over time because all the steps are such that the concentration of the intermediate is approximately constant.

Under the fast equilibrium approximation, though:

• The reactant ${A}_{2}$ and intermediate $A$ (but not $B$) are assumed to be in equilibrium for the equilibrium step.
• ${r}_{1} \left(t\right)$ $\text{>>}$ ${r}_{2} \left(t\right)$, therefore ${k}_{1}$ $\text{>>}$ ${k}_{2}$.
• $\textcolor{b l u e}{{K}_{1} = \frac{{k}_{1}}{{k}_{- 1}} = \frac{{\left[A\right]}^{2}}{\left[{A}_{2}\right]}}$ because equilibrium is assumed to be already established.
• (${K}_{1}$ is evidently the equilibrium constant for step 1.)

So what's happening overall here is that the equilibrium is reached quickly, and then intermediate $A$ is being consumed little by little. When intermediate $B$ is produced, the reaction then rushes to finish because it made what it needed to get past a crucial barrier.

THE RATE OF THE RATE-DETERMINING STEP CAN BE USED TO REPRESENT THE OVERALL REACTION

When writing the rate law for this, we would say that the slow step (associated with ${k}_{2}$) is the rate-determining step because the rate of the other two steps are insignificant relative to step 2.

So, the rate law for that particular step is a good approximation to the actual rate law. In other words, ${r}_{2} \left(t\right) \approx r \left(t\right)$.

$r \left(t\right) = {k}_{o b s} \left[{A}_{2}\right] \left[{B}_{2}\right] \approx {r}_{2} \left(t\right) = {k}_{2} \left[A\right] \left[{B}_{2}\right]$

THE RATE LAW MUST CONTAIN ONLY REACTANTS PRESENT IN THE OVERALL REACTION

But we do have a problem: ${r}_{2} \left(t\right)$ involves an intermediate, which is NOT present in the original reaction, so we have to find a new expression for the intermediate $A$ that uses only reactants found in the overall reaction.

Otherwise, it would be a completely different reaction:

$\textcolor{red}{2} A + {B}_{2} \stackrel{{k}_{o b s} \text{ }}{\to} 2 A B$

DETERMINING THE OBSERVED RATE CONSTANT

From the fast-equilibrium approximation and bullet point 3, we get:

$\left[A\right] = {\left(\frac{{k}_{1} \left[{A}_{2}\right]}{{k}_{- 1}}\right)}^{\text{1/2}}$

So our new rate law is:

$r \left(t\right) = {k}_{2} {\left(\frac{{k}_{1} \left[{A}_{2}\right]}{{k}_{- 1}}\right)}^{\text{1/2}} \left[{B}_{2}\right]$

Simplifying this, we get:

color(blue)(r(t) = stackrel(k_(obs))overbrace(k_2((k_1)/(k_(-1)))^"1/2")[A_2]^"1/2"[B_2])

And remember that we got $\left[{B}_{2}\right]$ from an elementary step, so its contribution to the overall order is $1$.

Therefore, the reaction order overall is $\textcolor{b l u e}{\text{3/2}}$.