Page 172 - Modeling of Chemical Kinetics and Reactor Design

142 Modeling of Chemical Kinetics and Reactor Design

B (
d Ce ) = k C e ∫ kdt
2
kt 2
dt 1 A (3-135)
1 ∫
2
C e kt 2 = k C e ∫ kdt dt + Const. (3-136)
B A
where Const. = constant.

= k C AO ∫ e −kt 1 e • k t 2 dt + Const. (3-137)
1

= k C AO ∫ e (k 2 − )t dt + Const. (3-138)
k 1
1
( k 2 − )
k t
1
kC
e
Ce kt 2 = 1 AO 2 ( k 1 + Const. (3-139)
k − )
B
At time t = 0, the concentration of component B is C = 0. Therefore,
B
the constant Const. becomes

Const. =− kC AO (3-140)
1
k − )
( 2 k 1
Therefore, Equation 3-139 becomes

( k 2 − ) kC
k t
e
1
kC
Ce kt 2 = 1 AO − 1 AO (3-141)
k − k 1 k − k 1
B
2
2
and
1
1
C B = kC AO e − kt 1 − kC AO e − kt 2 (3-142)
k −
2
2 k 1 k − k 1
The final concentration of B is:
C = kC AO e ( − kt 1 − e − k t 2 ) (3-143)
1
k −
B
2 k 1