Page 275 - Advanced Thermodynamics for Engineers, Second Edition

264 CHAPTER 12 CHEMICAL EQUILIBRIUM AND DISSOCIATION

Solving for a at each temperature gives

a
T P
2800 0.090625
2900 0.1218125
3000 0.171875

These values of a all obey the chemical equation, i.e. dissociation, but they do not all obey the
energy equation. It is necessary to consider the energy equation now to check which value of T 2
balances an equation of the form

s
U P ðT P Þ¼ ðQ v Þ þ½U R ðT R Þ U R ðT s ފ þ U P ðT s Þ
This energy equation, based on the internal energy of reaction at T ¼ 0, may be rewritten
(12.84)
0 ¼ DU 0 U P ðT P Þþ U R ðT R Þ
Now U R ðT R Þ is constant and is given by

Constituent CO O 2 N 2
12626.2 12939.6 12571.7
u 600
n 1.1 0.5 1.88
U R ðT R Þ¼ 43993:4kJ

Hence,
U P ðT P Þ¼ð1 aÞ 279523 þ 43993 (12.85)
It can be seen that the value of U P is a function of a: the reason for this is because the com-
bustion of the fuel (CO) is not complete when dissociation occurs. In a simple, single degree of
freedom reaction like this the reduction in energy released is directly related to the progress of the
reaction.

a U P (T P )
T P
2800 0.090625 298184
2900 0.1218125 289467
3000 0.171875 275473