Page 183 - Modeling of Chemical Kinetics and Reactor Design

Reaction Rate Expression 153

For component C + ( r C ) = dC C = kC A (3-182)
2
dt
Rearranging and integrating Equation 3-180 between the limits gives

C A t
∫
− ∫ dC A = (k 1 + )dt
k
2
C (3-183)
C AO A 0
C
k
−ln A = (k 1 + )t (3-184)
2
C AO
The solution for the concentration of A is

C = C AO e −( 1 k ) t (3-185)
k + 2
A
Substituting Equation 3-185 into Equation 3-181, rearranging and
integrating gives

C B dC t
∫ B k ) t = k 1 ∫ dt (3-186)
k + 2
0 C AO e −( 1 0

Further rearrangement of Equation 3-186 gives

C B t
1 ∫ dC = k 1∫ e −( 1 k + k 2 t ) + Constant
C AO 0 B 0 (3-187)

C B =− k 1 e −( 1 k ) t + Constant
k + 2
C k + ) (3-188)
k
AO ( 1 2
At t = 0, C = 0
B
k
Constant = 1
(k + k )
1 2
The solution for concentration of B is