Page 289 - Advanced Thermodynamics for Engineers, Second Edition

278 CHAPTER 12 CHEMICAL EQUILIBRIUM AND DISSOCIATION

a 1 ¼ 0:108
a 2 ¼ 0:01742
These values are close enough solutions for the dissociation coefficients and for the chemical
equation (Eqn (12.109) which becomes

C 8 H 18 þ 13:889ðO 2 þ 3:76N 2 Þ
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
n R ¼67:111
(12.118)
/ 7:136CO 2 þ 8:84322H 2 O þ 0:846CO þ 0:15678H 2 þ 1:8994O 2 þ 52:222N 2
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
n P ¼71:121
This equation satisfies the equilibrium constraints and the chemistry of the problem, but it still has
to be checked to see if it meets the First Law. The First Law is defined by the equation

U P T P U P T s ¼ ðQ v Þ þ U R T R U R T s (12.119)
s
Note two things about this equation. First, it is more convenient to evaluate the difference between
the internal energies of the products because now the composition of the products is also a function of
temperature and hence U P (T s ) is not a constant. Second, with dissociation, (Q v ) s is not the full value of
the internal energy of reaction of octane because not all the octane has been oxidised fully to CO 2 and
water. The value of (Q v ) s in this case is given by
~
Q
s
v s ¼ðQ v Þ 0:108 8 ð 283000Þ 0:01742 9 ð 241827Þ
¼ 5116320 þ 244512 þ 37913:6 (12.120)
¼ 4833894 kJ=kmol octane

ALTERNATIVE METHOD FOR CALCULATING THE ENERGY RELEASED
BY COMBUSTION
The energy released from partial combustion can be calculated using Hess’ law. In this case, from
Table 9.4,
X X
~
DU 0 ¼ DU 0 DU 0
Products Reactants
¼ 7:136ðDU 0 Þ þ 8:84322ðDU 0 Þ H 2 O þ 0:846ðDU 0 Þ CO ðDU 0 Þ
CO 2 C 8 H 18

¼ 7:136 ð 393405Þþ 8:84322 ð 239082Þþ 0:846 ð 113882Þ ð 74897Þ
¼ 4943040 kJ=kmol C 8 H 18 burned
(12.121)
This value can be related to the energy released at the standard temperature of 25 C by the

following equation (*note, the fuel properties have been based on methane because those of octane
were not available).