Page 104 - Mechanism and Theory in Organic Chemistry

Kinetics 93

Suppose th~t~he secadstq ismuch faster th2-n _the_ first. We would expect on
. ..
reac&-&&--
the basis- &a& &M&
. ..
--
/--
pend only on&+hzmx -c~~+,l,,,.r,----step
_
We
gQes instantly to C. _ -_ would say - u-nder hesccir~that& firs-step is
-
rate-determining.
The stationary-state approximation Kinetic analysis of Equations
2.37-2.38, first step rate-determining, takes the following form. Because B is
consumed as fast as it forms, its concentration is always very close to zero and
therefore approximately constant. We assume that

This assumption is known as the stationary-state approximation, and is valid for
highly reactive intermediates. We then write from the second step Equation 2.40

for the rate of product formation. But because B is a reactive intermediate, its
concentration will be difficult to measure; we require a rate equation expressed
in terms of measurable concentrations. We therefore write, from Equations 2.37
and 2.38,

d[Bl = k, [A] - k,[B]
dl
and, from Equation 2.39,

The stationary-state approximation thus allows us to equate k,[A] with k,[B].
The rates of formation and of disappearance of the reactive intermediate B are
equal. We can therefore write instead of 2.40 the final rate equation 2.43:

-= k, [A]
dt
Note that overall kinetic behavior is unaffected by events following a true rate-
determining step.
Suppose now that B is still a reactive intermediate, but the reverse of the first
step must be considered (Equations 2.44 and 2.45). There is now a competition

k.
B-C (2.45)
between two pathways: B may go on to C or return to A. Even though B still
does not accumulate, it is no longer true that every A reacting leads directly to C.
The first step is not strictly rate-determining, and the rate constant for the second
step enters into the rate equation (Problem 2).