Page 288 - Advanced Thermodynamics for Engineers, Second Edition

12.11 DISSOCIATION PROBLEMS 277

From Eqn (12.109), the partial pressures of the constituents are

8ð1 a 1 Þ p
p rCO 2 ¼
n P p 0
8a 1 p 9ð1 a 2 Þ p
p rCO ¼ p rH 2 O ¼
n P p 0 n P p 0
ð1:389 þ 4a 1 þ 4:5a 2 Þ p 9a 2 p
p rO 2 ¼ p rH 2 ¼
n P p 0 n P p 0
Hence

2
K 2 ½8ð1 a 1 ފ n P p 0 (12.111)
p r1 ¼ 43:3164 ¼ 2 ð1:389 þ 4a 1 þ 4:5a 2 Þ p
½8a 1 Š
and

8a 1 a 1
9ð1 a 2 Þ ð1 a 2 Þ
(12.112)
K p r2 ¼ ¼
9a 2 8ð1 a 1 Þ a 2 ð1 a 1 Þ
equation (Eqn (12.111)) contains a pressure term, but this can be replaced by the tem-
The K p r
perature of the products by applying the perfect gas law to the mixture. Then
p P V P ¼ n P <T P
(12.113)
p R V R ¼ n R <T R
Thus

n P n R <T R n R T R
(12.114)
¼ ¼
p P p R <T P p R T P
Substituting gives
2
2 ð1 a 1 Þ n R T R
K p 0 (12.115)
p r1 ¼ 43:3164 ¼ 2
a 1:389 þ 4a 1 þ 4:5a 2 p R T P
1
Inserting values for these parameters gives
2
43:3164 10 2800 ð1 a 1 Þ
(12.116)
¼ 2
67:111 500 a 1:389 þ 4a 1 þ 4:5a 2
1
which can be expanded to give

3
2

144:58a þ a 49:2052 þ 162:652a 2 þ 2a 1 1 ¼ 0 (12.117)
1
1
This equation contains both a 1 and a 2 . If it is assumed that a 1 >> a 2 then Eqn (12.117) is a cubic
equation in a 1 . Based on this assumption the value of a 1 is approximately 0.11. Substituting this value
in Eqn (12.112) gives a 2 ¼ 0.017776, which vindicates the original assumption. Recalculation around
this loop gives