Page 174 - Introduction to chemical reaction engineering and kinetics

156 Chapter 7: Homogeneous Reaction Mechanisms and Rate Laws

If we simply add the three steps, we do not recapture (A). To get around this, we introduce
the stoichiometric numbel; S, for each step, as the number by which that step must be
multiplied so that addition of the steps results in (A):

sl(l) + d2) + +(3) = (4 (7.1-1)

where sl, s2, and s3 are the stoichiometric numbers for the three steps. To determine their
values systematically, we utilize the stoichiometric coefficients in the three steps for each
species in turn so as to correspond to the coefficient in (A):

N,O, : - Is, + OS, + OS, (from the three steps) = -2 (from (A))
NO,: 1st + (- 1 + l)s, + 2s3 = 4
0,: OS1 + lS2 + OS3 = 1
NO,: 1st - ls2 - ls3 = 0
NO: OS1 + 182 - ls3 = 0
This set of linear equations can be solved by inspection, or, more formally, by Gauss-Jordan
reduction of the augmented coefficient matrix:

with the result s1 = 2, s2 = 1, and s3 = 1. (Note that in this case the last two of the five
equations are redundant in obtaining values of the three stoichiometric numbers.) Thus,
the three steps are consistent with (A) if added as

2(1) + l(2) + l(3) = (A)
(b) From the mechanism, step (2),

rO* = k2 cNO, cNO, 09

We eliminate cNos (not allowed in the final rate law) by applying the stationary-state hy-
pothesis to NO,, rNo3 = 0 (and subsequently to NO):
rNO, = klCNz05 - k-1CN02CN03 - k2cN0,cN0, - k3cNOcN03 = o (0

rNO = k2CN02 cNO, - k3 cNOcNO, = o CD)
from (D),

CNO = (kdk3h02 (W

from (C) and (E),

kl cNzO,
cNO, = (F)
(k-l + 2k,k,02
from (B) and (F),

klk2 63
ro2 = kk, + 2kzCNZo5