Page 295 - Advanced Thermodynamics for Engineers, Second Edition

284 CHAPTER 12 CHEMICAL EQUILIBRIUM AND DISSOCIATION

Rich mixture including dissociation is

C 8 H 18 þ 11:36ðO 2 þ 3:76N 2 Þ
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n R ¼55:09

5:7272 9
a 1 þ a 2 O 2 þ 42:71N 2
/ 5:7272ð1 a 1 ÞCO 2 þ 9ð1 a 2 ÞH 2 O þð2:2728 þ 5:7272a 1 ÞCO þ 9a 2 H 2 þ
2 2
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n P ¼59:73þ2:8636a 1 þ4:5a 2
(12.146)
It can be seen that this is a significantly more complex equation than for the weak mixture.
Application of a computer program gives the final temperature and pressure, without dissociation,as
2958.6 K and 64.15 bar respectively. The equilibrium constants are

p rCO 2
K p r1 ¼ 6:58152 ¼ 1=2
p rCO p
rO 2
(12.147)
p rCO p rH 2 O
K p r2 ¼ 6:82925 ¼
p
p rCO 2 rH 2
Substituting for the mole fractions gives
p; p; p
5:7272ð1 a 1 Þ ð2:2728 þ 5:7272a 1 Þ ð2:8636a 1 þ 4:5a 2 Þ
p CO 2 ¼ p CO ¼ p O 2 ¼
n P n P n P
(12.148)
Hence
2 2
K 2 5:7272 ð1 a 1 Þ n P p 0 (12.149)
ð2:2728 þ 5:7272a 1 Þ ð2:8636a 1 þ 4:5a 2 Þ p
p r1 ¼ 43:3164 ¼ 2
Similarly

9a 2
9ð1 a 2 Þ
p (12.150)
p; p H 2
p H 2 O ¼ ¼
n P n P
and
1 a 2
(12.151)
ð2:2728 þ 5:7272a 1 Þ 9 ð1 a 2 Þ ð2:2728 þ 5:7272a 1 Þ
K p r2 ¼ ¼
a 2
9a 2 5:7272ð1 a 1 Þ 5:7272ð1 a 1 Þ
equations (Eqns (12.149) and (12.151)) are again nonlinear equations in a 1 and
The K p r1 and K p r2
a 2 , and they can be solved in the same manner as before.
The solution to this problem, allowing for dissociation is a final temperature of 2891 K and a final
pressure of 62.99 bar. This results in the following degrees of dissociation: a 1 ¼ 0:0096; a 2 ¼
0:05540: It is left to the reader to prove these values are correct.
EXAMPLE 6: THE FORMATION OF NITRIC OXIDE
When a mixture of oxygen and nitrogen is heated above about 2000 K, these two elements will join
together to form nitric oxide. The elements combine in this way because the energy of formation of