Biochemical Reactions and Enzyme Kinetics - Introduction to Biomedical Engineering - page 455

In general, it can be shown that the solution for q P in Eq. (8.14) is

0 1

e K 1 ðabÞ t b 1

@ A u ð t Þ

q P ¼ ð 8 : 21 Þ

b
a e K 1 ðabÞ t

1

where

s

2

Þ þ K

2
K 1

l ¼ ð q A ð 0 Þþ q B ð 0 Þþ 2 q P ð 0 Þ 4 ð q A ð 0 Þþ q P ð 0 Þ Þ q B ð ð 0 Þþ q P ð 0 Þ Þ

Þþ K 2
K

q A ð 0 Þþ q B ð 0 Þþ 2 q P ð 0

1 þ

l

a ¼

2

Þþ K 2
K

q A ð 0 Þþ q B ð 0 Þþ 2 q P ð 0

1

l

b ¼

2

Next, consider the case in which q B ð 0 Þ q A ð 0 Þ , where the change in q B is small and can

be treated as a constant q B ð 0 Þ: Thus, an approximation for Eq. (8.14) is

q P ¼ K

q B ð 0 Þ q A K 1 q P ð 8 : 22 Þ

1

Next, we remove q A in Eq. (8.22) by letting q A ¼ q P ð 0 Þþ q A ð 0 Þ q P , giving

q P ¼ K q B ð 0 Þ q A ð ð 0 Þþ q P ð 0 Þ Þ K ð q B ð 0 Þþ K 1 Þ q P ð 8 : 23 Þ

1 1

which can be straightforwardly solved as

ð K 1 q P ð 0 Þ K q B ð 0 Þ q A ð 0 Þ Þ

þ K

ð q B ð 0 Þ q A ð ð 0 Þþ q P ð 0 Þ Þ Þ

1 1

e K 1 q B ð0Þþ K 1

ð Þ t

q P ¼ u ð t Þ

ð 8 : 24 Þ

ð K q B ð 0 Þþ K 1 Þ ð K q B ð 0 Þþ K 1 Þ

1 1

8.1.4 Higher-Order Chemical Reactions and Sequential Reactions
Consider the higher-order chemical reaction in Eq. (8.25):

ð 8 : 25 Þ

in which a A of reactant A and b B of reactant B form product P, and P has a reverse reaction

to form a A þ b B. The law of mass action for Eq. (8.25) is

b
B K 1 q P

a

q P ¼ K 1 q

A q

b
B þ K 1 q P

a

q A ¼ K 1 q

A q

ð 8 : 26 Þ

b
B þ K 1

a

q B ¼ K

q

A q

q P

1