Enzymes - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 378

or

C S K C C P
K C k B þK A k S þ k 1

C E0

A

r ¼ (8.128)

1þ ðK S K C k 1

3 þ k 1

þ k 1
3

K P ÞC S þðK C k 1 K S þk 1 K P þK S k c K P ÞC P

c 1 1

K C k 1

3 þK S k 1

þk 1
1

c

Based on Fig. 8.19 , one can think of the reaction network in analogy to the electric conduc-
tion: 1) the fluxes must be equal at any given point and 2) the total resistance is the summation
of all the resistors in series. If one were to look carefully, these statements could be detected as
both are embedded in Eqn (8.128) . Therefore, the rate expression not only is unique for PSSH
but also brings the reaction network to be directly in analog to electric circuit.
At this point, one may look back at the full solutions as illustrated in Figs 8.22 through 8.24 ,
especially with Figs 8.22b and d, 8.23c and d, and 8.25b , that the concentrations of interme-
diates (in this case, the concentrations of enzyme associated with S and with P) are hardly
constant or at “steady state” in any reasonably wide regions where we would like to have
the solutions meaningful. Therefore, the assumption is rather strong for the directly involved
species that we have eliminated from the rate expression. How do the solutions actually
measure up with the full solutions?
To apply the PSSH expression (8.127) or (8.128) , we must first ensure that the reaction
mixture in already in pseudo-steady state to minimize the error may cause in the solution.
This error can be negligible if the amount of catalyst is negligible or for steady flow reactors
where steady state is already reached.
Let us consider again the case where no P is present in the reaction mixture at the start of
the reaction. The concentrations of S and P charged into the batch reactor are C ST0 and
C ST0 ¼ 0. Since there is no P present in the initial reaction mixture, C P0 ¼ 0. Overall mole

balance at the onset of the reaction leads to
C ST 0 ¼ C S 0 þC SE 0 þC PE 0

(8.129)

where C SE0 and C PE0 satisfy Eqns (8.121) and (8.122) . Since C P0 ¼ 0, substituting Eqns (8.121)

and (8.122) into Eqn (8.129) , we obtain

ðK S K C k 1

3 þ k 1

þ k 1
3

K P ÞC S 0 C E 0

c

C ST 0 ¼ C S 0 þ (8.130)

K C k 1

3 þK S k 1

þ k 1 þðK S K C k 1 þ k 1

3 3 þ k 1

K P ÞC S 0

c c

This quadratic equation can be solved to give

q

ðC ST 0 C E 0 aÞ 2 þ4aC ST 0 C ST 0 C E 0 a þ

C S 0 ¼ (8.131)

2

where

K C k 1

3 þK S k 1

þ k 1
1

c

a ¼ (8.132)

K S K C k 1

3 þ k c þ k 1

K P

3