Page 182 - Modeling of Chemical Kinetics and Reactor Design

152 Modeling of Chemical Kinetics and Reactor Design

− ( r )=− dC A = k C A
A 1 ) (3-177)
dt (1 + kC A
2
Taking the reciprocals of Equation 3-177 gives

1 = 1+ kC A = 1 + k 2
2
− ( r A ) kC A kC A k 1 (3-178)
1
1
Plotting 1/(–r ) versus 1/C would give a straight line with the slope
A A
equal to 1/k and an intercept of k /k . Another analysis method
1
1
2
involves multiplying each side of Equation 3-178 by k /k and solving
2
1
for (–r ) to give
A
− ( r A )= k 1 − − ( r A ) (3-179)
k 2 kC A
2
Plotting (–r ) versus (–r )/C gives a straight line with the slope
A
A
A
equal to –1/k and the intercept equal to k /k .
2
1
2
SIMULTANEOUS IRREVERSIBLE SIDE REACTION
Consider the chemical reactions

A k 1 → B
↓
k 2

C

in which the reactions are first order, the initial concentration of A is
C , and both the concentrations of B and C are zero. The rate
AO
equations representing the reactions for a constant volume batch
system are:

For component A − ( r A )=− dC A = (k 1 + )C A (3-180)
k
2
dt
For component B + ( r B )= dC B = kC A (3-181)
1
dt