Page 156 - Modeling of Chemical Kinetics and Reactor Design

126 Modeling of Chemical Kinetics and Reactor Design

− ( r A )= −dC A = kC C B (3-59)
A
dt
From stoichiometry:

A B
Amount at time t = 0 C AO C BO
Amount at time t = t C C
A B
Amounts that have reacted (C AO –C ) C BO –C B
A
and from stoichiometry (C – C ) = 2(C – C ).
AO A BO B
Expressing the concentration of B in terms of A and the original
amounts C , C gives
AO BO

C = C BO − 1 C ( AO − C ) (3-60)
B
A
2
Substituting Equation 3-60 into Equation 3-59 gives

C A −dC t
∫
∫ A = dt

1
kC  − (C − C ) 0 (3-61)
 C
C AO A BO AO A
 2 
C A t
∫
− ∫ dC A = kdt
1
1
C   − C  + C  0 (3-62)

C AO A  C BO AO  A 
 2 2 
Taking the partial fractions of Equation 3-62 with α = [C BO –
(1/2)C AO ] gives the following:
1 ≡ p + q
 1  C 1 (3-63)
C  α + C  A α + C A
 2  2
A
A

1
1≡ p α + C  + qC (3-64)
 2 A  A
Solving for constants p and q involves equating the “constant” and
the coefficient of “C ” in Equation 3-64:
A