Page 174 - Modeling of Chemical Kinetics and Reactor Design

144 Modeling of Chemical Kinetics and Reactor Design

Substituting Equation 3-147 into Equation 3-149 gives

 ln( k 2 k ) 
1
k 
C − 2 
k
= e (3-150)
B max   k − 1  
2
C
AO
Taking the natural logarithm gives
C B  k   k 
ln max =  − 2  ln  2  (3-151)
C  k − k   k 
AO 2 1 1
Treating k /(k – k ) as an exponent and removing the natural
2
1
2
logarithm gives
 k 2 
k 
C  k    k − 1 
2
B max =  1  (3-152)
C AO  k 
2
This shows that C B max /C AO depends only on k and k , and k can
2
2
1
be evaluated from C /C at t .
B max AO max
From stoichiometry:


A →  → C
k 2
k 1
B
A B C
Amount at time t = 0 C AO 0 0
Amount at time t = t C C C
A B C
and from stiochiometry, (C – C ) = (C + C ). That is, C = (C
AO A B C C AO
– C ) – C . The concentration of C in terms of C AO , k , and k are:
A
B
2
1


C C =  1 − e − kt 1 −  k 1  e ( − kt 1 − e − k t 2 )
C AO    k − k   


1
2
 k k 
=  1 + 1 e − kt 2 − 2 e − k t 1  (3-153)
 k − k 1 k − k 1 
2
2