Hillert M. Phase Equilibria, Phase Diagrams and Phase Transformations: Their Thermodynamic Basis
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20.7 Dilute solution approximations 455
after replacing x
A
with 1 − x
B
− x
C
,
G
A
=
o
G
A
+ RT ln x
A
+
x
2
B
+ x
B
x
C
L
AB
+
x
B
x
C
+ x
2
C
L
AC
− x
B
x
C
L
BC
(20.61)
G
B
=
o
G
B
+ RT ln x
B
+
1 − 2x
B
+ x
2
B
− x
C
+ x
B
x
C
L
AB
+
−x
C
+ x
B
x
C
+ x
2
C
L
AC
+ (x
C
− x
B
x
C
)L
BC
(20.62)
G
C
=
o
G
C
+ RT ln x
C
+
−x
B
+ x
2
B
+ x
B
x
C
L
AB
+
1 − x
B
+ x
B
x
C
− 2x
C
+ x
2
C
L
AC
+ (x
B
− x
B
x
C
)L
BC
. (20.63)
The two formalisms are again identical if the second-order terms are neglected and the
following relations are obtained by comparing Eqs (20.59) and (20.60) with Eqs (20.62)
and (20.63).
ε
B
B
=−2L
AB
/RT (20.64)
ε
C
B
= (L
BC
− L
AB
− L
AC
)/RT = ε
B
C
(20.65)
ε
C
C
=−2L
AC
/RT. (20.66)
Equation (20.55)isstill valid. We note that ε
C
B
and ε
B
C
must be equal in order to make
the two formalisms identical.
Equation (20.61) demonstrated that one needs second-order terms in order to model
any deviation from Raoult’s law for the solvent, A. There are no such terms in the ε
formalism and a related feature is that the expressions for G
A
, G
B
and G
C
obtained from
the ε formalism do not satisfy the Gibbs–Duhem relation. Pelton and Bale [41] showed
that it can be satisfied by adding a particular term, K,tothe expressions for all three G
A
,
G
B
and G
C
. Inspection of Eqs (20.62) and (20.63)reveals that the set of second-order
terms is the same in all cases and it actually represents the K term,
KRT =
x
2
B
+ x
B
x
C
L
AB
+
x
2
C
+ x
B
x
C
L
AC
− x
B
x
C
L
BC
. (20.67)
Using the relations (20.64)to(20.66)itcould as well be expressed in the form proposed
by Pelton and Bale.
− K =
1
/
2
ε
B
B
x
2
B
+ ε
C
B
x
B
x
C
+
1
/
2
ε
C
C
x
2
C
. (20.68)
Darken [42] noticed that the regular solution model for binary solutions is often very
inadequate and showed that it sometimes works well in the A-rich part by using a hypo-
thetical state of reference for the solute. This is called the quadratic formalism and simply
means that one substitutes
o
G
B
+ M
B
for
o
G
B
,which represents the real Gibbs energy
of pure B in the same phase. This can be done directly in Eq. (20.62) and similarly
for pure C in Eq. (20.63). The values of M
A
and M
B
are given by a comparison of
Eqs (20.59) and (20.60).
ln
o
f
B
= (M
B
+ L
AB
)/RT = M
B
/RT −0.5ε
B
B
(20.69)
ln
o
f
C
= (M
C
+ L
AC
)/RT = M
C
/RT −0.5ε
C
C
. (20.70)
456 Mathematical modelling of solution phases
The corrected ε formalism is completely equivalent to the quadratic formalism and thus
offers a convenient way of evaluating the M and L coefficients to fit the experimental
information.
Exercise 20.10
Derive the expression for K in Eq. (20.70).
Hint
Apply the Gibbs–Duhem relation in the x
B
direction and in the x
C
direction, respectively,
in a ternary system, after first expressing x
A
in terms of x
B
and x
C
.
Solution
0 =x
A
dlna
A
+x
B
dlna
B
+x
C
dlna
C
=x
A
d(ln x
A
+ K ) + x
B
d(ln x
B
+ ε
B
B
x
B
+ ε
C
B
x
C
+
K ) + x
C
d(ln x
C
+ ε
B
C
x
B
+ ε
C
C
x
C
+ K ).∂/∂x
B
of this gives since x
A
= 1 − x
B
− x
C
:0= x
A
(−1/x
A
+ ∂ K /∂ x
B
) + x
B
(1/x
B
+ ε
B
B
+ ∂ K /∂ x
B
) + x
C
(ε
B
C
+ ∂ K /∂ x
B
);
∂ K /∂ x
B
=−x
B
ε
B
B
− x
C
ε
B
C
.Inthe same way we find ∂ K /∂ x
C
=−x
B
ε
C
B
− x
C
ε
C
C
. These
conditions are satisfied with K =−(ε
B
B
x
2
B
+ 2ε
C
B
x
B
x
C
+ ε
C
C
x
2
C
)/2.
20.8 Predictions for solutions in higher-order systems
Taking all binary interactions into account we obtain the following expression for the
excess Gibbs energy in a multicomponent solution phase.
E
G
m
=
i
j>i
x
i
x
j
I
ij
. (20.71)
If this model applies, all the I
ij
coefficients can be determined experimentally on the
respective binary systems and the properties of the higher-order system can be predicted
by combination. On the other hand, if a composition-dependent I is required in order
to represent the experimental information on a binary system, then there is no simple
physical model predicting the properties of the higher-order system. It may be stated
that a composition-dependent I implies that the interaction energy is not determined
completely by the pair-wise interaction of atoms. If I in a binary system is equal to
0
L + (x
A
− x
B
)
1
L, then the excess Gibbs energy is described by
E
G
m
= x
A
x
0
B
L + x
A
x
A
x
1
B
L − x
A
x
B
x
1
B
L. (20.72)
The last two terms seem to originate from interactions within groups of three atoms. If
that is the case, one should expect similar effects in the higher-order system, for instance
an interaction between an A,aBand a C atom given by x
A
x
B
x
C
I
ABC
.Ofcourse, there
20.8 Predictions for solutions in higher-order systems 457
is no way by which this interaction can be predicted from the binary systems where that
group of atoms does not occur.
When the binary interaction energies depend upon the composition, it should be
advisable to introduce I
ABC
in the description of the ternary system. However, it must
be realized that there is no unique way of defining such a parameter because there is
no unique way of predicting how the properties of a binary system contribute to the
properties of the higher-order system unless I
AB
is a constant. For variable I
AB
several
expressions are used
I
AB
=
0
Lx
2
A
+
1
Lx
A
x
B
+
2
Lx
2
B
(20.73)
I
AB
=
0
L +
1
Lx
B
+
2
Lx
2
B
(20.74)
I
AB
=
0
L +
1
L(2x
B
− 1) +
2
L
6x
2
B
− 6x
B
+ 1
(20.75)
I
AB
=
0
L +
1
L(x
A
− x
B
) +
2
L
x
2
A
− 4x
A
x
B
+ x
2
B
(20.76)
I
AB
=
0
L +
1
L(x
A
− x
B
) +
2
L(x
A
− x
B
)
2
. (20.77)
These expressions are quite equivalent when applied to a binary system because they
can be transformed into each other by the use of x
A
+ x
B
= 1. When applied to a higher-
order system the expressions give different results because x
A
+ x
B
is no longer unity.
For practical reasons it may be important to select a particular expression and at present
there is a strong preference for the last one. It was first suggested by Redlich and Kister
[39] that this particular form, initially intended for binary systems, should be used for
representing the binary contributions in a multicomponent system.
The same kind of problem appears when one wants to predict the properties of a
quaternary solution from the four ternaries. A general method may be based upon the
observation that all the expressions, listed for the binary interaction energy I
AB
, become
identical if x
A
is replaced by x
A
+ (1 − x
A
− x
B
)/2 and x
B
by x
B
+ (1 − x
A
− x
B
)/2. In
fact, with this method all second-order expressions can be transformed into expressions
identical to the Redlich–Kister polynomial. Generalizing this method we find that x
A
should be replaced by x
A
+ (1 − x
A
− x
B
− x
C
)/3, x
B
by x
B
+ (1 − x
A
− x
B
− x
C
)/3
and x
C
by x
C
+ (1 − x
A
− x
B
− x
C
)/3inthe ternary interaction energy I
ABC
. This
method can easily be extended to higher-order terms [43].
Exercise 20.11
Show that Eqs (20.74) and (20.77) become identical for a binary system if x
B
is replaced
by x
B
+ (1 − x
A
− x
B
)/2. Show how the second set of L parameters can be evaluated
from the first one.
Hint
The expression by which we shall replace x
B
is equal to 1/2 − (x
A
− x
B
)/2.
458 Mathematical modelling of solution phases
. . .
Kohler Colinet Muggianu
AA
BCBC
A
BC
Figure 20.7 Selection of binary alloys (black points) according to three symmetric methods of
predicting ternary properties.
Solution
I
AB
(in Eq. (20.74)) =
0
L +
1
L/2 −
1
L/2 · (x
A
− x
B
) +
2
L/4 +
2
L/4(x
A
− x
B
)
2
−
2
2
L/4 · (x
A
− x
B
).
Comparison shows that:
0
L (in Eq. (20.77)) =
0
L +
1
L/2 +
2
L/4;
1
L(in Eq. (20.77))
=−(
1
L +
2
L)/2;
2
L(in Eq. (20.77)) =
2
L/4.
20.9 Numerical methods of predictions for higher-order solutions
As an alternative to the analytical methods described in the preceding section, several
numerical methods have been suggested. They allow the properties of a ternary solution
phase to be estimated from binary solutions without first assessing all the information
available in the binary systems. However, it should be emphasized that they are never-
theless based upon some assumptions regarding the properties. The expression for the
Gibbs energy of a ternary alloy of the composition x
A
, x
B
, x
C
must contain the terms
x
A
o
G
A
+ x
B
o
G
B
+ x
C
o
G
C
and it is common also to include the ideal entropy of mixing.
This leaves only the excess Gibbs energy and the following three symmetric methods
have been proposed.
Kohler [44]:
E
G
m
=
i
j>i
(x
i
+ x
j
)
2
·
E
G
ij
m
x
i
x
i
+ x
j
;
x
j
x
i
+ x
j
. (20.78)
Colinet [45]:
E
G
m
=
i
j>i
x
j
/2
1 − x
i
·
E
G
ij
m
(x
i
;1− x
i
) +
x
i
/2
1 − x
j
·
E
G
ij
m
(1 − x
j
; x
j
)
.
(20.79)
Muggianu [46]:
E
G
m
=
i
j>i
4x
i
x
j
(1 + x
i
− x
j
)(1 + x
j
− x
i
)
×
E
G
ij
m
((1 + x
i
− x
j
)/2; (1 + x
j
− x
i
)/2). (20.80)
All the weighting factors have been selected in such a way that the methods correctly
reproduce the term x
i
x
j
0
L
ij
. Colinet and Muggianu also reproduce the terms x
i
x
j
·
l
L
ij
(x
i
− x
j
). In addition, Muggianu reproduces all the higher-power terms if they are
written in the form x
i
x
j
·
k
L
ij
(x
i
− x
j
)
k
.Asaconsequence, the numerical method by
20.9 Numerical methods of predictions for higher-order solutions 459
Muggianu and the analytical method based upon the Redlich–Kister polynomial give the
same result and may be recommended for general use.
These numerical methods were initially defined geometrically as illustrated in
Fig. 20.7.Itexplains how one selected the binary alloys, the Gibbs energies of which
were used to estimate the ternary properties. It should be emphasized that these methods
only apply to integral excess quantities.
Exercise 20.12
Refer to Fig. 20.7 and show that the binary A–B alloy used in Muggianu’s method has
an A content of (1 + x
A
− x
B
)/2.
Hint
Draw lines through the ternary alloy, parallel to sides A–C and B–C. The intercepts on
the A–B side have lengths equal to x
B
, x
C
and x
A
.
Solution
The distance of the binary alloy from the B corner is x
C
/2 + x
A
. Using the relation
x
C
= 1 − x
A
− x
B
one can transform this expression into 1/2 + x
A
/2 − x
B
/2.
21
Solution phases with sublattices
21.1 Sublattice solution phases
In the substitutional solutions discussed in Section 20.4 all lattice sites were equivalent
and a solution was formed from a pure substance by substituting new kinds of atoms
for the initial one. However, relatively few crystalline phases belong to this class. The
great majority have different kinds of lattice sites and can be described by using two
or more sublattices. Examples of such phases will be discussed in this chapter. It will
be demonstrated that a great variety of such phases can be modelled in a very direct
way using an approach often called the compound energy model or formalism. It is a
crude model in the sense that it assumes random mixing within each sublattice. The
expression for the entropy of such phases is simple and was presented in Section 19.8,
‘Restricted random mixtures’, but the excess Gibbs energy can easily become very
complicated. However,it should be realized that actual calculations of equilibria, and even
of whole phase diagrams, can now be carried out with sophisticated computer programs
which only require that the expression for the molar Gibbs energy of each phase is
defined.
Section 19.8 gave the expression for the entropy assuming random mixing of all the
components present in each sublattice. The result was expressed in terms of the site
fraction variable, y
i
, and in Section 19.10 it was then applied to interstitial solutions,
which are a special case of solution phases with sublattices. We shall first consider the
rather simple case where there is only one component, M, in one sublattice, and a number
of components, i, j,...,inanother sublattice. It is then convenient to consider a formula
unit with one mole of atoms in the second sublattice (M)
b
(i, j,...)
1
.For 1 mole of such
formula units we get
S
ideal
m
=−Ry
i
ln y
i
. (21.1)
The deviation from ideal solution behaviour may be represented by the interactions
between the components in the second sublattice. Using the Redlich–Kister type of
power series we have, for the interaction between components i and j,when the first
sublattice is filled with M,
E
G
m
= y
i
y
j
n
k=0
k
L
M
ij
(y
i
− y
j
)
k
. (21.2)
21.1 Sublattice solution phases 461
For y
i
= 1 the phase is identical to a compound M
b
i and the whole expression of the
molar Gibbs energy for 1 mole of formula units will be
G
m
=
i
y
i
o
G
M
b
i
+ RT
i
y
i
ln y
i
+
i
j>i
y
i
y
j
n
k=0
k
L
M
ij
(y
i
− y
j
)
k
. (21.3)
It should be noticed that y
j
is at the same time the site fraction of j in the second sublattice
and the mole fraction of the compound M
b
j among all M
b
i compounds. From G
m
we
may thus derive expressions for the partial Gibbs energies of the compounds, using
Eq. (4.6)but replacing x fractions with y fractions.
G
M
b
l
=
o
G
M
b
l
+ RT ln y
l
+
i=l
$
0
L
M
li
y
i
(1 − y
l
) +
n
k=1
k
L
M
li
y
i
(y
l
− y
i
)
k−1
·[(k + 1)(1 − y
l
)(y
l
− y
i
) + ky
i
]
%
−
i=l
j=l,>i
y
i
y
j
$
0
L
M
ij
+
n
k=1
k
L
M
ij
(y
i
− y
j
)
k
(k + 1)
%
. (21.4)
The expression is thus analogous to the one for a substitutional solution in Eq. (20.34). A
ternary phase of this type would behave as a binary, substitutional phase. It is sometimes
called a quasi-binary or pseudo-binary solution. For the same reason one may call a
quaternary phase of this type a quasi-ternary phase.
As already shown in Eq. (4.56) one can generalize Eq. (4.6)toasolution phase with
several sublattices, t, u and υ, and with several constituents on the sublattices, e.g.,
G
A
a
B
b
C
c
= G
m
+ ∂G
m
/∂y
t
A
+ ∂G
m
/∂y
u
B
+ ∂G
m
/∂y
υ
C
− y
s
i
∂G
m
/∂y
s
i
. (21.5)
The last term is a summation over all the constituents, i,inall the sublattices, s. Further-
more,
G
A
a
B
b
C
c
≡ µ
A
a
B
b
C
c
= aµ
A
+ bµ
B
+ cµ
C
. (21.6)
Exercise 21.1
High-temperature measurements have shown complete miscibility in the solid phase of
Al
2
O
3
− Cr
2
O
3
. Information from lower temperatures is less certain but there is some
report of a miscibility gap with a maximum at about T
crit
= 2000 K. Model this solution
phase.
Hint
Introduce a constant regular solution parameter. Express the ideal entropy contribution
with regard to the size of the formula unit chosen.
462 Solution phases with sublattices
Solution
Define the unit as M
1
O
1.5.
Then we get G
m
= y
Al
o
G
AlO
1.5
+ y
Cr
o
G
CrO
1.5
+ RT(y
Al
ln y
Al
+
y
Cr
ln y
Cr
) + y
Al
y
Cr
L. Because we have chosen a symmetric description, the miscibil-
ity gap will be modelled as symmetric. Its maximum will be found at y
Al
= y
Cr
=
0.5which yields 0 = d
2
G
m
/dy
2
Cr
= RT/y
Al
y
Cr
− 2L; L = 2RT
crit
= 2R · 2000 =
33 000 J/mol.
21.2 Interstitial solutions
In Section 19.10 we discussed the ideal entropy of mixing in an interstitial solution. In
fact, it may be regarded as the special case of a solution with two sublattices obtained by
allowing vacant sites on one of the sublattices. For 1 mole of sites in that sublattice we
can write the formula as (M)
b
(Va, i, j, ...)
1
. All the equations derived for the phase with
two sublattices in the preceding section can be applied if the vacancies are included as
a component. For a binary M–C system we write the interstitial solution as M
b
(Va, C)
1
and Eq. (21.3) yields,
G
m
= y
C
o
G
M
b
C
+ y
Va
o
G
M
b
Va
+ RT(y
C
ln y
C
+ y
Va
ln y
Va
)
+ y
C
y
Va
n
k=0
k
L
CVa
(y
C
− y
Va
)
k
(21.7)
G
M
b
C
=
o
G
M
b
C
+ RT ln y
C
+ y
2
Va
$
0
L
CVa
+
n
k=1
k
L
CVa
(y
C
− y
Va
)
k−1
[(2k + 1)y
C
− y
Va
]
%
(21.8)
G
M
b
Va
=
o
G
M
b
Va
+ RT ln y
Va
+ y
2
C
$
0
L
CVa
+
n
k=1
k
L
CVa
(y
C
− y
Va
)
k−1
[y
C
− (2k + 1)y
Va
]
%
. (21.9)
Since all the sites in the second sublattice are vacant in the compound M
b
Va,itisidentical
to b moles of pure M. G
M
is thus obtained by dividing the last equation by b. G
C
can be
obtained by taking the difference between the two equations because
G
M
b
C
− G
M
b
Va
= bG
M
+ G
C
− bG
M
− G
Va
= G
C
− G
Va
, (21.10)
and it may be assumed that the chemical potential of vacancies is zero at equilibrium.
We thus obtain
G
C
= G
M
b
C
− G
M
b
Va
=
o
G
M
b
C
− b
o
G
M
+ RT ln(y
C
/y
Va
) +
0
L
CVa
(y
Va
− y
C
)
+
n
k=1
k
L
CVa
(y
C
− y
Va
)
k−1
[2ky
C
y
Va
− (y
C
− y
Va
)
2
]. (21.11)
21.2 Interstitial solutions 463
Site fraction, y
C
1.0
0
0 1.0
Activity
a
C
a
M
a
M
b
C
a
M
b
Va
Figure 21.1 Variation of different activities in an ideal interstitial solution of C in M with the
formula M
1/3
(Va, C)
1
.
Actually, we could have obtained G
M
b
C
− G
M
b
Va
and thus G
C
directly from G
m
using
Eq. (4.8),
G
C
= G
M
b
C
− G
M
b
Va
=
∂G
m
∂y
C
−
∂G
m
∂y
Va
(21.12)
It is important to notice that pure C cannot exist in the interstitial sublattice without M
in the other sublattice. As a consequence, the reference state for C must be chosen using
another phase. If pure C in a form α is chosen and y
Va
is replaced by 1 − y
C
, then we
obtain
RT ln a
C
= G
C
−
o
G
α
C
=
o
G
M
b
C
−
o
G
M
b
Va
−
o
G
α
C
+ RT ln[y
C
/(1 − y
C
)]
+
0
L
CVa
· (1 −2y
C
) +
1
L
CVa
·
−1 + 6y
C
− 6y
2
C
+
2
L
CVa
·
1 − 10y
C
+ 24y
2
C
− 16y
3
C
+···. (21.13)
One formula unit of M
b
Va is identical to b mole of M. The activity coefficient for very
dilute solutions would be given by
o
f
C
= exp
(
o
G
M
b
C
− b
o
G
M
−
o
G
α
C
+
0
L
CVa
−
1
L
CVa
+
2
L
CVa
−···)/RT
. (21.14)
The deviation from Henry’s law for less dilute solutions depends strongly upon the
choice of composition variable. In order to get a constant activity coefficient for C over
the whole system, when all the L parameters are zero, one must use y
C
/(1 − y
C
)asthe
composition variable. If y
C
is used one will find that a
C
goes to infinity at y
C
= 1. This
is demonstrated in Fig. 21.1.Onthe other hand, in this type of diagram the activities of
M
b
Va and M
b
C show an ideal behaviour when all the parameters are zero. The diagram
was drawn for b = 1/3 and all L parameters equal to zero.
464 Solution phases with sublattices
Exercise 21.2
In the plot of a
M
versus y
C
for an interstitial solution of C in M it may seem surprising
in view of Raoult’s law that the curve for a
M
does not approach the diagonal close to
y
C
= 0. Find the reason and calculate the initial slope according to Raoult’s law.
Hint
Raoult’s law says that a
M
should approach x
M
, i.e.1 − x
C
, not 1 − y
C
.Evaluate x
M
from
y
C
using the formula M
b
(Va, C)
1
with b = 1/3 (the value for bcc).
Solution
Forabinary interstitial solution we have x
M
= b/(b + y
C
) = 1/(1 + y
C
/b)
∼
=
1 −
y
C
/b = 1 − 3y
C
. The initial slope should be –3 and not –1.
21.3 Reciprocal solution phases
With a stoichiometric phase one usually means a phase with a constant composition. This
may, for instance, be caused by a crystalline structure which is composed of different
sublattices, one for each component. An example is cementite, Fe
3
C. Such a phase is
also described as a compound. When a further component is added, it may go into one
of the existing sublattices, an example being manganese-alloyed cementite, (Fe,Mn)
3
C.
The composition of such a phase may thus vary along a line in the ternary phase diagram
and it is sometimes described as a quasi-binary phase. Such phases were described
in Section 21.1.Ifstill another component is added, two alternatives result, examples
being (Fe,Mn,Cr)
3
C and (Fe,Mn)
3
(C,N)
1
. The latter type of phase is sometimes called a
reciprocal solution phase because the central alloy can be regarded as a solution between
either Fe
3
C and Mn
3
NorFe
3
N and Mn
3
C. Both kinds of phases have the restriction
to the variation in composition which we have called stoichiometric constraint. The
composition of a reciprocal solution phase is represented by two sets of site fractions,
one set for each sublattice, y
A
+ y
B
= 1 and y
C
+ y
D
= 1.
Accepting the stoichiometric constraint it is logical to consider the binary compounds
as the components of the system. They were introduced in Section 4.8 and are called
component compounds. Let us discuss the reciprocal case represented by (A,B)
b
(C,D)
c
.
This is a quaternary phase but its composition can only be varied with two degrees of
freedom instead of three due to the stoichiometric constraint. All possible compositions
can thus be represented on a plane just like a ternary system. As shown in Fig. 13.12,a
square diagram is now the natural shape and each corner is an end-member and represents
a component compound. Perpendicular to that plane we may plot the Gibbs energy (see
Fig. 21.2).
The diagram demonstrates that it is, in general, impossible to place a plane through
the four points representing the Gibbs energy values of the four component compounds.
The question is then what surface of reference one should use when giving the Gibbs