Page 577 - Carrahers_Polymer_Chemistry,_Eighth_Edition

540 Carraher’s Polymer Chemistry

d[EM]
Rate of complex formation = = k [E][M] (16.26)
1
dt
The complex then either returns to form the initial reactants or forms the product(s) and the free
enzyme, E. In kinetic terms, the change, or rate of breakdown of the complex is described as

d[EM]
Rate of complex change = − = k [EM] + k [EM] (16.27)
2
−1
dt
The negative sign associated with the equation means that the terms are describing the rate of
decrease in complex concentration. The rate of complex formation is rapid, and fairly soon the rate
at which the complex is formed is equal to the rate at which it breaks down. This situation is called
a steady state. Mathematically, this is described by

d[EM] d[EM]
= − and (16.28)
dt dt
k [E][M] = k [EM] + k [EM] (16.29)
−1
2
1
The initial reaction rate, A to B, varies until the number of reactants clearly outnumbers the
number of reaction sites on the enzyme, at which time, C, the rate becomes zero order, independent
of the reactant concentration. Often it is difficult to directly measure the concentration of E as the

reaction progresses. Thus, the concentration of E is generally substituted for using the relationship

[E] = [E ] − [EM] (16.30)
o
where [E ] is the initial enzyme concentration.
o
Substitution of this description for [E] into Equation 16.29 gives

k ([E ][M] − [EM][M]) = k [EM] + k [EM] (16.31)
2
−1
o
1
Separating out [EM] on the right-hand side gives
k ([E ][M] − [EM][M]) = (k + k )[EM] (16.32)
1
−1
2
o
Bringing together all of the rate constants form both sides of the equation gives
[E ][M] − [EM][M] = k′[EM] (16.33)
o
Moving the [EM] containing terms to the right side gives

[E ][M] = k′[EM] + [EM][M] (16.34)
o
and separating out [EM] from the right side gives

[E ][M] = [EM] (k′ + [M]) (16.35)
o
Now division of both sides by k’ + [M] gives

[E ][M]
[EM] = 0
k + [M] (16.36)

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