Page 341 - Elements of Chemical Reaction Engineering Ebook

31 2 Multiple Reactions Chap. 6

Solution
First, we divide each equation through by the stoichiometric coefficient of the spe-
cies for which the rate law is given:

1: A + 1.25B C + 1.5D -rIA = kiACACi (E6-8.5)
2: A 0.75B 0.5E + 1.SD -r2~ = ~~ACACB (E6-8.6)

3: B +2C --+ 2F -r3~ = k,BC:CB (E6-8.7)

5
2
4: C+,A --+ ;E+D - r,, = k4, C, Ci’ (E6-8.8)
Stoichiometry. We will express the concentrations in terms of the molar flow
rates:

and then substitute for the concentration of each reaction species in the rate laws.
Writing the rate law for species A in reaction 1 in terms of the rate of formation,
rIA, and molar flow rates, FA and FB we obtain

I Thus

(E6-8.9)

Similarly for the other reactions,

FAFB
-
r2A = -k,C;,, (E6-8.10)
F,”
(E6-8.11)

-
r,, = -k&, SI3 FC F,?3 (E6-8.12)

Next, we determine the net rate of reaction for each species by using the appropriate
stoichiometric coefficients and then summing the rates of the individual reactions.

Net rates of formation:
2
Species A: ‘A = ‘lA+ r2A + 5 r4C (E6-8.13)