Physical Chemistry , 1st ed.

A→B (fast)

B→C (slow)

Here, B represents an intermediate product whose concentration is difficult to
determine experimentally. A and C represent normal chemical reactants whose
amounts can be measured. If the second step is the slow, rate-determining step,
then it will force the first step to be backed up like cars on a one-lane road.
However, we do know that no chemical reaction actually ever goes to 100%
completion. Rather, the reverse of the chemical reaction begins to occur and
eventually an equilibriumis established. Since the first step is backed up by the
second step, in many cases the reverse of the first step will begin to occur and
eventually the first step will establish an equilibrium between its products and
reactants. The two-step process is better represented as

A B (fast)

B→C (slow)

where the arrows are used to symbolize an equilibrium. If this first step is
truly in equilibrium, then the concentrations of A, B, and C are relatively
steady and unchanging. Because of this, this model is called the steady-state ap-
proximationof reaction mechanisms.
The steady-state approximation helps us relate the rate law determined
from the RDS with the rate law as determined from experiment. Recall that
the rate law of a mechanism is dictated by the stoichiometry of the RDS, but
how can we know if this rate law is consistent with the experimentalrate
law? We can use the fact that the preceding step(s) is/are at equilibrium to
derive a rate law in terms of the original reactants (whose amounts or con-
centrations we can measure). There are two ways to do this. First, we will
adopt a simplified approach. If the first step in the above two-step process
is in fact in equilibrium, then we can write an equilibrium constant expres-
sion for it:

K

[

[

A

B]

]

 (20.59)

where we are using molar concentrations as an arbitrary unit of amount.
(Thermodynamically, activities should be used, but using molarity still makes
our point.) If the RDS is the second step, then we can immediately write a rate
law for the reaction as

rate k[B]

But [B], the concentration of an intermediate species, may not be measurable!
No matter. We will use equation 20.59 to find an expression for [B]:

K[A][B]

[B]K[A]

If we substitute this expression for [B] into the rate law, we get

rate kK[A]

constants

Since kand Kare both constants, we will combine them into a new k that is
also some constant. The rate law becomes

rate k[A]

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20.8 The Steady-State Approximation 711

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