In a previous screencast, we looked at applying a steady-state approximation to this overall
reaction of 2A+B going to the product A2B. And we looked at it for a sequence of elementary
steps that are indicated here where A+B initially reacts to some reactive intermediate AB star, and
then that reactive intermediate subsequently reacts with A to make the product. And we
applied the steady-state approximation, we got the rate expression as shown here. And
the important thing to note is this is the rate expression at low concentrations of product.
And so now what we want to do is say suppose instead we applied the rate-determining step
approximation to this same sequence. The way we're going to apply it, we're going to assume
this second step, so we're going to assume this is the rate-determining step, which means
this must be, the first step must be in equilibrium, or we refer to it often as quasi-equilibrium.
This is an assumption that we're making, and then what we want to look at is what does
the rate expression look like, how does it compare to the steady-state approximation.
So we assume for step 1 that we're in equilibrium which means we have some equilibrium constant
that would be equal to concentration of AB star, concentration of A, over concentration of B -
product concentrations over reactant concentrations, raised to their stoichiometric coefficients,
which are 1. This equilibrium constant is the ratio of these two rate constants. The
rate-determining step approximation says the rate that we're making our product A2B is
going to be the rate of this second step since it's the rate-determining step. So that's
AB star times A minus k-2, concentration of product. Again, we're going to look at low concentrations
of our product because that's how we often run the experiments to make things simpler.
So we're going to set this term to zero. Then we're going to substitute for AB star into this
expression. So AB star is just the equilibrium constant times the concentration of A, concentration
of B. So if I make the substitution of the rate constant, equilibrium constant, concentration
of A squared, concentration of B - that's now the overall rate of reaction, low concentrations,
for the rate-determining step approximation. And so we can compare this now to the steady-state
approximation. What we see if we make the comparison is that the steady-state approximation
is the same as saying k-1 is greater than k2 times A. Or the other way of looking at it, if
I multiply both sides by AB star - and you'll see why I'm doing this - so what this is telling
us is that k1 times AB star - which is the reverse rate of the first step, the rate that AB star decomposes
in the first reaction - is greater than k2, concentration of A, the concentration of AB star, that's the rate
of this second step, the forward rate. So what the rate-determining step is saying is
that we make AB star, it's much more likely you decompose back to the reactants in the first
step than to react with A to make products. And so indeed, that's what the rate-determining
step approximation step is. The first step is in equilibrium, so made AB star is most likely to
go back to A and B and some small fraction will react with A to make product. And so
we can see that the rate-determining step is a subset of the steady-state approximation,
where we make some additional assumptions to get a rate expression that, if we have
some basis for making the assumption, may correspond better to what actual measurements
we made.