Page 123 - Introduction to chemical reaction engineering and kinetics

5.5 Series Reactions 105

This result may be used to eliminate cA in equation 5.53, to give a differential equation
from which en(t) may be obtained:

dc,ldt + k2cB = k,c,,exp(-k,t) (5.5-5)

This is a linear, first-order differential equation, the solution of which is

CB = [k,c,&,/(k2 - k,)] (6?-k1t - C?-k2’) (5.5-6)

Finally, co may be obtained from equation 5.5-4 together with equations 3.4-10 and
5.5-6:

cc = [cd@2 - kl)] w - e +w) - k,(l - p2’)] (5.5-7)
[

The features of the behavior of CA, cn, and co deduced qualitatively above are ihus-
trated quantitatively in Figure 5.4. Other features are explored in problem 5-10.
Values of the rate constants kI and b can be obtained from experimental measure-
at various times in a BR. The most sophisticated procedure is to
ments of cA and cn
0 use either equations 5.5-2 and -3 or equations 3.4-10 and 5.5-6 together in a nonlinear
V
parameter-estimation treatment (as provided by the E-Z Solve software; see Figure
7O-v
3.11). A simpler procedure is first to obtain k,
from equation 3.4-10, and second to ob-
tain h from k, and either of the coordinates of the maximum value of cB (t,,, or cn,max).
These coordinates can be related to k, and k2, as shown in the following example.

Obtain expressions relating t,,, and cn,,,ax in a BR to k, and k2 in reaction 5.5-la.

SOLUTION

Differentiating equation 5.5-6, we obtain
d(CdCAo) _ k~ -kzr _ kle-kit)
dt - k,-((k,e

Setting d(cn/c,,)/dt = 0 for t = tmax, we obtain

ln (k2/kl) (5.5-8)
t max =
k2 - k

From equation 5.5-3 with dcnldt = 0 at cB,max,

kl
= -CA&,& = &, ,-kltmax
CB,l?UZX
k2