Page 266 - Advanced Thermodynamics for Engineers, Second Edition

12.5 CALCULATION OF CHEMICAL EQUILIBRIUM 255

12.5 CALCULATION OF CHEMICAL EQUILIBRIUM
AND THE LAW OF MASS ACTION

1
General relationships will be derived, and the particular case of the CO þ O 2 reaction will be shown
2
in brackets { }.
It was previously shown that for a system at constant pressure and temperature to be in an equi-
librium state, it must have a minimum value of Gibbs energy, i.e. dG) p,T ¼ 0.
But, by definition, for a system at constant pressure and temperature
¼ m dm 1 þ m dm 2 þ .m dm n : (12.37)
dgÞ p;T 1 2 n
where m 1 , m 2 ,.m n are the masses (or amounts) of the possible constituents of the mixture.
Only four constituents will be considered during this discussion, two reactants and two products,
but the theory can be extended to any number of constituents. The equilibrium equation for the
complete reaction is:

y a A þ y b B 5 y c C þ y d D

1 (12.38)
CO þ O 2 5 CO 2
2
At some intermediate stage in the reaction the state may be represented as:
y a A þ y b B/ð1 εÞy a A þð1 εÞy b B þ εy c C þ εy d D

1 1 (12.39)
e:g: CO þ O 2 /ð1 εÞCO þð1 εÞ O 2 þ ε CO 2
2 2
where stoichiometric coefficients equal to unity are implicit.
ε is known as the fraction of reaction and is an instantaneous value during the reaction as opposed
to a, the degree of dissociation, which is a final equilibrium value. The use of ε allows the changes in
Gibbs energy to be considered as the reaction progresses.
For the purposes of evaluating the dissociation phenomena it is possible to consider the reaction to
occur at constant temperature and pressure (equal to values obtained by other means or calculated by an
implicit iterative technique). The Gibbs energy of the system may be described by the following equation.
G ¼ð1 εÞy a m þð1 εÞy b m þ εy c m þ εy d m d
a
c
b

1 (12.40)
G ¼ð1 εÞm CO þ ð1 εÞm þ εm :
2 O 2 CO 2
To find the equilibrium condition while maintaining p and T constant, this function has to be
minimised with respect to ε, i.e. it is necessary to locate when

vG
¼ 0:
vε
p;T
Now
X
m dm i (12.41)
dG ¼ Vdp SdT þ i