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Section 7.1
Exercise The Phase Rule
Find f and c for (a) an aqueous solution of HCN and KCN; (b) an aqueous
ind
solution of HCN and KCl; (c) an aqueous solution of the weak diprotic acid
H SO . [Answers: (a) 4, 3; (b) 4, 3; (c) 3, 2.]
2 3

EXAMPLE 7.3 The phase rule

Find f in a system consisting of CaCO (s), CaO(s), and CO (g), where all the
3 2
CaO and CO come from the reaction CaCO (s) ∆ CaO(s) CO (g).
2 3 2
A phase is a homogeneous portion of a system, and this system has three
phases: CaCO (s), CaO(s), and CO (g). The system has three chemical species.
3 2
There is one reaction-equilibrium condition, m m m , so r
CaCO 3 (s) CaO(s) CO 2 (g)
1. Are there any additional restrictions on the mole fractions? It is true that the
number of moles of CaO(s) must equal the number of CO moles: n
2 CaO(s)
n . However, this equation cannot be converted into a relation between the
CO 2 (g)
mole fractions in each phase, and it does not provide an additional relation be-
tween intensive variables. Hence
c ind c r a 3 1 0 2
f c ind p 2 2 3 2 1
The value f 1 makes sense, since once T is fixed the pressure of CO gas in
2
equilibrium with the CaCO is fixed by the reaction-equilibrium condition, and
3
so P of the system is fixed.
Exercise

Find c and f for a gas-phase mixture of O , O, O , and e , in which all the O
ind 2

comes from the dissociation of O and all the O and e come from the ioniza-
2
tion of O. Give the most reasonable choice of independent intensive variables.
(Answer: 1, 2; T and P.)
In doubtful cases, rather than applying (7.9) or (7.11), it is often best to first list
the intensive variables and then list all the independent restrictive relations between
them. Subtraction gives f. For example, for the CaCO –CaO–CO example just given,
3
2
the intensive variables are T, P, and the mole fractions in each phase. Since each phase
is pure, we know that in each phase, the mole fraction of each of CaCO , CaO, and
3
CO is either 0 or 1; hence the mole fractions are fixed and are not variables. There
2
is one independent relation between intensive variables, namely, the already stated
reaction-equilibrium condition. Therefore f 2 1 1. Knowing f, we can then cal-
culate c ind from (7.11) if c ind is wanted.
In dG SdT VdP m dn , the Gibbs equation (4.78) for a phase, the
i
i
i
sum is over all the actual chemical species in the phase. Provided the phase is in reac-
tion equilibrium, it is possible to show that this equation remains valid if the sum is
taken over only the independent components of the phase; see Prob. 7.70. This is a
useful result, because one often does not know the nature or the amounts of some of
the chemical species actually present in the phase. For example, in a solution, the
solute might be solvated by an unknown number of solvent molecules, and the solvent
might be dissociated or associated to an unknown extent. Despite these reactions that
produce new species, we need only extend the sum m dn over the two independent
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i