Page 274 - Advanced Thermodynamics for Engineers, Second Edition

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12.7 EXAMPLES OF SIGNIFICANCE OF K p

Dissociation of CO 2 (Eqn (12.29))
a
aCO 2 5 aCO þ O 2
2
Total combustion equation with dissociation

1 a
1:1CO þ ðO 2 þ 3:76N 2 Þ / ð1 aÞCO 2 þð0:1 þ aÞCO þ O 2 þ 1:88N 2 (12.77)
2 2
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n R ¼3:48 n P ¼2:98þa=2
, is given by Eqn (12.70):
The equilibrium constant, K p r

p p 0
CO 2
K p r ¼
1
2
p p 0 p p 0
CO O 2
The values of partial pressures in the products are
p 2 1 a p 2 p CO p 2 0:1 þ a p 2 p 2 a p 2
p CO 2 p O 2
; ; : (12.78)
¼ x CO 2 ¼ ¼ x CO ¼ ¼ x O 2 ¼
p 0 p 0 n P p 0 p 0 p 0 n P p 0 p 0 p 0 2n P p 0
Also, from the ideal gas law
p 2 V 2 ¼ n 2 <T 2 ; and p 1 V 1 ¼ n 1 <T 1 : (12.79)
Thus

n P n 1 T 1 245:65
: (12.80)
¼ ¼
p 2 p 1 T 2 T 2
Hence, substituting these values in Eqn (12.70) gives
1
2
2 2 245:65
2n P p 0
ð1 aÞ 2 ð1 aÞ
K p r ¼ ; giving K ¼ p 0 (12.81)
p r 2
a p 2 aT 2
ð0:1 þ aÞ
ð0:1 þ aÞ
2
This is an implicit equation in the products temperature, T 2 , because K p ¼ f (T 2 ). Writing K p as X gives
2 2 2
Xð0:1 þ aÞ aT 2 ¼ð1 aÞ 2 245:65p 0 ¼ð1 aÞ 2 245:65 1:01325 (12.82)
Expanding Eqn (12.82) gives
2 3
0 ¼ 497:77 að995:54 þ 0:01XT 2 Þþ a ð497:77 0:2XT 2 Þ a XT 2 (12.83)
2
The term XT 2 can be evaluated for various temperatures because XT 2 ¼ K T 2
p r
T 2 (K) K p XT 2
2800 6.582 121303
2900 4.392 55940
3000 3.013 27234