Page 227 - Physical Chemistry

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Chapter 7 the reacting species. Each independent chemical reaction provides one relation between
One-Component Phase Equilibrium the chemical potentials, and, like relations (7.3) to (7.6), each such relation can be used
and Surfaces
to eliminate one variable from T, P, and the mole fractions. If the number of indepen-
dent chemical reactions is r, then the number of independent intensive variables is re-
duced by r and the phase rule (7.7) becomes
f c p 2 r (7.8)
By independent chemical reactions, we mean that no reaction can be written as a com-
bination of the others (Sec. 6.5).
In addition to reaction-equilibrium relations, there may be other restrictions on the
intensive variables of the system. For example, suppose we have a gas-phase system
containing only NH ; we then add a catalyst to establish the equilibrium 2NH 3 ∆
3
N 3H ; further, we refrain from introducing any N or H from outside. Since all
2
2
2
2
the N and H comes from the dissociation of NH , we must have n 3n and x
3
2
2
H 2
H 2
N 2
3x . This stoichiometry condition is an additional relation between the intensive
N 2
variables besides the equilibrium relation 2m m 3m . In ionic solutions, the
NH 3 N 2 H 2
condition of electrical neutrality provides such an additional relation.
If, besides the r reaction-equilibrium conditions of the form n m 0, there are
i
i
i
a additional restrictions on the mole fractions arising from stoichiometric and elec-
troneutrality conditions, then the number of degrees of freedom f is reduced by a and
the phase rule (7.8) becomes
f c p 2 r a (7.9)*
where c is the number of chemical species, p is the number of phases, r is the number
of independent chemical reactions, and a is the number of additional restrictions.
We can preserve the simple form (7.7) for the phase rule by defining the number
of independent components c ind as
c ind c r a (7.10)
Equation (7.9) then reads
f c ind p 2 (7.11)*
Many books call c ind simply the number of components.
EXAMPLE 7.2 The phase rule

For an aqueous solution of the weak acid HCN, write the reaction equilibrium
conditions and find f and c .
ind

The system has the five chemical species H O, HCN, H , OH , and CN ,
2

so c 5. The two independent chemical reactions H O ∆ H OH and
2

HCN ∆ H CN give two equilibrium conditions: m H 2 O m H m OH and
m HCN m H m CN . The system has r 2. In addition, there is the electroneutral-
ity condition n H n CN n OH ;division by n gives the mole-fraction relation
tot
x H x CN x OH . (See also Prob. 7.6.) Thus, a 1. The phase rule (7.9) gives
f c p 2 r a 5 1 2 2 1 3
c ind c r a 5 2 1 2
The result f 3 makes sense, since once the three intensive variables T, P, and
the HCN mole fraction are specified, all the remaining mole fractions can be cal-
culated using the H O and HCN dissociation equilibrium constants. The two
2
independent components are most conveniently considered to be H O and HCN.
2