Physical Chemistry , 1st ed.

Notice that this is the same overall rate law that we got when we used the equi-
librium constant of the first elementary process, so there is some consistency
between the two mathematical applications of the steady-state approximation.
The particular mathematical approach you might use depends on the infor-
mation available (that is, the value ofKor k 1 and k 1 ) as well as what you
might be trying to determine about a particular chemical reaction.
One kind of kinetics that uses the steady-state approximation is applied to
enzyme-catalyzed reactions. Because enzymes (which are proteins) are very
good catalysts, typically only a very small concentration is needed for a bio-
chemical reaction to occur, and determination of the reaction kinetics focuses
on following the change of concentration of the primary reactant, called the
substrate.
The first step in an enzyme-catalyzed process is the combination of the
proper enzyme, labeled E, with the substrate, labeled S. The second step, the
RDS, is the production of some product material P and the simultaneous re-
lease of the unchanged enzyme catalyst. The two elementary steps are repre-
sented as

k 1

E S ES (fast)

k 1

k 2

ES →E P (slow)

where ES represents the intermediate, an enzyme-substrate complex. Because
the second step is the RDS, the first elementary process reaches equilibrium,
and we can apply the steady-state approach to the intermediate ES and derive,
for the rate law,

rate 

k

k 2

1





k 1

k 2

[E][S] k[E][S] (20.64)

where kk 2 k 1 /(k 1 k 2 ). Applying equation 20.62, the amount of enzyme-
substrate complex ES is given by

[ES] 

k 2 

k 1

k 1

[E][S] (20.65)

We define [E 0 ] as the total amount of enzyme present in any form:

[E 0 ] [E] [ES] (20.66)

Using equation 20.65, we can rewrite the expression for [E 0 ] as

[E 0 ] [E] 

k 2 

k 1

k 1

[E][S] [E] 1 

k 2 

k 1

k 1

[S]

Solving for [E], we find that

[E] 

k 2

[E



0 ](

k

k



2

1





k

k



1

1

[

)

S]



In terms of [E 0 ], the rate of the equation is

rate

k

k 2

1





k 1

k 2

[E][S]



k 2

k



2 

k

k 1

1





[E

k

0

1

]

[S]

[S]

[E 0 ]



1 

k 2 

k 1

k 1

[S]

JQPJ

20.8 The Steady-State Approximation 713