Page 328 - Elements of Chemical Reaction Engineering Ebook

Sec. 6.3 Algorithm for Solution to Complex Reactions 299

Example 64 Stoichiometry and Rate Laws for Multiple Reactions

Consider the following set of reactions:
Rate Law#

Reaction 1: 4m3 + 6N0 5N2 + 6H,O -TINO = k,NocN,3cA:
+
Reaction 2: 2N0 - 0, r2N2 = k2N2ci0
N,
Reaction 3: N2 + 20, + 2N0, - r302 = k302CN2C:2
Write the rate law for each species in each reaction and then write the net rates of
formation of NO, 02, N, .
and

Solution
The rate laws for reactions 1, 2, and 3 are given in terms of species NO, N2 , and
O,, respectively. Consequently, to relate each reacting species in each reaction to its
rate law more clearly, we divide each reaction through by the stoichiometric ccleffi-
cient of the species for which the rate law is given.
2
1. NO+-NH, --+ 6 -IINO = klNOcNH3ck: (E6-4.1)
3
~N,+H,o
2. 2NO -9 N,+O, “2N2 = k2N2ci0 (E6-4.2)

-r302 = k302cN2ci2 (E6-4.3)
The corresponding rate laws are related by:

Reca,lling Eqn. (6-17)

Reaction 1: The rate law w.r.t. NO is

-riNo = kiN0cNH3c%
The relative rates are

rlNO
‘1H 0
‘IN
rl NH
-=3=2=2
(- 1) (-213) (5/6) (1)
Then the rate of disappearance of NH3 is
(E6-4.4)

From {prtusimetry data (1 1/2/19).