Page 294 - Advanced thermodynamics for engineers

282 CHAPTER 12 CHEMICAL EQUILIBRIUM AND DISSOCIATION

and the total amount of substance in the products is
n P ¼ a þ b þ c þ d þ e þ f (12.135)
Considering the atom balances gives
Carbon: 8 ¼ a þ c giving c ¼ 8 a (12.136a)

Hydrogen: 18 ¼ 2b þ 2d giving b ¼ 9 d (12.136b)
Oxygen: 27:778 ¼ 2a þ b þ c þ 2e
giving e ¼ 13:889 a ðb þ cÞ=2 (12.136c)
Nitrogen: f ¼ 52:223 (12.136d)
From the perfect gas law

n P n R T R
¼ ; (12.137)
p P p R T P
giving Eqn (12.133) as
2
2 a n R T R
K ¼ 43:3164 ¼ p 0 (12.138)
p r1 2
c e p R T P
Equation (12.138) is a member of a nonlinear set of equations in the coefficients of the chemical
equation. One way to solve this is to make an assumption about the relative magnitude of d and the
other parameters. If it is assumed that d is small compared to the other parameters in Eqn (12.131),
then Eqn (12.138) becomes a cubic equation in a. This is equivalent to assuming that a 2 is small
compared to a 1 , as was assumed in the previous calculation. If Eqns (12.136(a)–(c)) are substituted
into Eqn (12.138), with the assumption that d ¼ 0, then Eqn (12.138) becomes

2 a 2 n R T R a 2 67:111 500 2
ð8 aÞ ð5:389 a=2Þ¼ 2 p 0 ¼ 1 ¼ 0:027667a ; (12.139)
K p R T P 43:3164 10 2800
p r1
giving
2
344:896 118:224a þ 13:3613a a 3 2 ¼ 0 (12.140)
The solution to Eqn (12.140) relevant to this problem is a ¼ 7:125; giving c ¼ 8 a ¼
0:875 and b ¼ 9. It is now possible to check if this result satisfies the equilibrium condition for the
water gas reaction, viz. Equations (12.134), which gives
cb 0:875 9
d ¼ ¼ ¼ 0:16184: (12.141)
aK p r2 7:125 6:89295
The calculation should now be performed again based on the new value of d given in Eqn (12.141).
This produces a new cubic equation in a, similar to that in Eqn (12.140) but with the following
coefficients.
2 3
350:07 119:519a þ 13:442a a 2 ¼ 0: