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3.3 Chemical Kinetics and Chemical Equilibrium 77

In the combustion products, for a mole of CO, there are b mole of NO, c mole of
CO 2 , d mole of O 2 and e mole of N 2 , therefore, the mole fraction of the species in
the combustion products can be determined as

a b c
y CO ¼ ; y NO ¼ ; y CO 2 ¼
a þ b þ c þ d þ e a þ b þ c þ d þ e a þ b þ c þ d þ e
d e
¼ ¼
y O 2 ; y N 2
a þ b þ c þ d þ e a þ b þ c þ d þ e
1
From the equilibrium chemical reaction CO 2 $ CO þ = 2O 2 , the chemical
equilibrium constant at 1 atm is

1=2 1=2
y co y ad
O 2 1=2
K P1 ¼ ¼ ð a þ b þ c þ d þ eÞ ¼ 0:3273 ð4Þ
c
y CO 2
Similarly, for the second equilibrium reaction ½ O 2 + ½ N 2 $ NO, its equi-
librium constant
b
y NO
K P2 ¼ 1=2 1=2 ¼ 1=2 1=2 ¼ 0:1222 ð5Þ
y y d e
O 2 N 2
From Eqs. (1)–(3), we can get

c ¼ 1 a
d ¼ 1=2ð1 þ a bÞ
e ¼ 1=2ð1 bÞ

Put them into (4) and (5), we can get

1=2
a 1 þ a b
0
¼ 0:3273 ð4 Þ
1 a 4 þ a
2b
¼ 0:1222 ð5 Þ
0
1=2
½ ð1 þ a bÞð1 bފ
Solving Eqs. (4′) and (5′) we get
a ¼ 0:3745 b ¼ 0:0675 c ¼ 0:6255 d ¼ 0:6535 e ¼ 0:4663