Page 278 - Advanced Thermodynamics for Engineers, Second Edition

12.8 THE VAN’T HOFF RELATIONSHIP 267

The ratios of the amounts of substance in the equilibrium products are defined by the equilibrium
constant,
1=2 1=2
p p 0 an p
CO 2 P 0 :
K p r ¼ (12.89)
1=2 1=2 1=2
¼
p p 0 p p 0 bc p
CO O 2 P
Hence,
2
2
2
b c p P K ¼ a ; (12.90)
p r
p 0 n P
and, from the atomic balances, this can be written in terms of a as
1
2 p P 2 2
K ¼ a (12.91)
2 p 0 n P
ð1:1 aÞ ð1 aÞ p r
The previous calculations showed that the temperature of the products which satisfies the gov-
¼ 13:1488. These values will be used to
erning equations is T P ¼ 2950 K; which gives a value of K p r
demonstrate this example. Then
2 2 3 p P 13:1488 2
1:21 2:2a þ a 1:21a 2:2a þ a ¼ a : (12.92)
p 0 n P 2
It is possible to evaluate the ratio p P =p 0 n P from the perfect gas relationship, giving
p P p R T P 8:5 2950
¼ 12:009
¼ ¼
p 0 n P n R T R 3:48 600
This enables a cubic equation in a to be obtained
2
3
95:532 269:23a þ 251:65a 78:953a ¼ 0 (12.93)
The solution to this equation is a ¼ 0.8506, and hence the chemical equation becomes
1
1:1CO þ ðO 2 þ 3:76N 2 Þ/0:8506CO 2 þ 0:2494CO þ 0:0747O 2 þ 1:88N 2 (12.94)
2
which is the same as that obtained previously. The advantage of this approach is that it is possible to
derive a set of simultaneous equations which define the equilibrium state, and these can be easily
solved by a computer program. The disadvantage is that it is not possible to use the intuition that most
engineers can adopt to simplify the solution technique. It must be recognised that the full range of
iteration was not used in this demonstration, and the solution obtained from the original method was
simply used for the first ‘iteration’.

12.8 THE VAN’T HOFF RELATIONSHIP BETWEEN EQUILIBRIUM
CONSTANT AND HEAT OF REACTION
It has been shown that, Eqn (12.54)

1 0 0 0 0
y a m þ y b m y c m y d m : (12.95)
<T
ln K p r ¼ a b c d