Biochemical Reactions and Enzyme Kinetics - Introduction to Biomedical Engineering - page 454

where

q P ¼18:3 ¼ q P ¼18:3 ¼

1 1

q P

B 1 ¼ q P ð 18 : 3 Þ 0 : 0989

ð q P 18 : 3 Þ q P ð 8 : 2 Þ ð 8 : 2 Þ

and

q P ¼8:2 ¼ q P ¼8:2 ¼

1 1

q P

B 2 ¼ q P

ð 8 : 2 Þ 0 : 0989

ð q P 18 : 3 Þ q P ð 8 : 2 Þ ð 18 : 3 Þ

Integrating Eq. (8.19) and rearranging yields

ð q P 8 : 2 Þ

ln 2 : 2 ¼ 20 : 2 t

ð q P 18 : 3 Þ

and after solving for q P , we have

u ð t Þ

e 20:2 t 8 : 21

q P ¼ ð 8 : 20 Þ

1 0 : 45 e 20:2 t

The solution for the reactants are found from q A ¼ q A ð 0 Þþ q P ð 0 Þ q P and

q B ¼ q B ð 0 Þþ q P ð 0 Þ q P , yielding

e 20:2 t

1 : 8059 þ 3 : 7

q A ¼ u ð t Þ

1 0 : 45

e 20:2 t

and

u ð t Þ

e 20:2 t 6 : 8059 þ 1 : 4798

q B ¼

1 0 : 4476 e 20:2 t

These results are shown in Figure 8.2. Note that conservation of mass is still maintained, since q P

contains both q A and q B ,so q A þ q B þ 2 q P ¼ 25 :

16

14

12

q B

10

8

q A

6

4

2

q P

0

0 0.05 0.1 0.15 0.2 0.25

Time

FIGURE 8.2 Illustration of the amount of product, q P ,inExample 8.1.