Hillert M. Phase Equilibria, Phase Diagrams and Phase Transformations: Their Thermodynamic Basis

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21.3 Reciprocal solution phases 465

AC AD

Figure 21.2 Surface of reference for a reciprocal phase.

energy value for a composition inside the system. The most natural choice seems to

be the curved surface shown in the diagram and for a simple case that choice may be

supported by the random mixing version of the nearest-neighbour bond energy model

to be described in the next chapter. This surface of reference is accepted as the basis for

the compound energy formalism [47] and it yields the following expression for the

Gibbs energy.

= y

+ y

+ RT[by

ln y

+ by

ln y

+ cy

ln y

+ cy

ln y

] +

. (21.15)

This expression may be regarded merely as a deﬁnition of the excess term

but it

becomes very important if one, as a ﬁrst approximation, neglects the excess term. G

refers to 1 mole of formula units M

. The simplest type of power series representation

of the excess term makes use of the remaining two second-power terms, y

and y

However, in order to allow different behaviour on two opposite sides of the composition

square we shall not use them but go directly to third-power terms,

= y

AB:C

+ y

AB:D

+ y

A:CD

+ y

B:CD

. (21.16)

The colon in the subscript is used to separate sublattices. For any side of the system this

expression reduces to the expression discussed in the preceding section and all these

interaction energies can thus be evaluated from the properties of the side systems. As an

example, for the A

–B

side of the system we have y

= 0, y

= 1 and the excess

Gibbs energy expression reduces to y

AB:C

For more complicated cases one may express each one of the parameters I

AB:C

, etc.,

with a power series of the Redlich–Kister type using site fractions. We have already done

this for the simple case of a phase with sublattices considered previously. In this way we

can introduce a large number of higher-power terms but it should be noticed that for each

such term all the y

except for one come from one sublattice and all those coefﬁcients

can be evaluated from the side systems. In order to adjust a description to information

from inside the composition square we need a term like y

AB:CD

466 Solution phases with sublattices

In our discussion on constituents in Section 4.8 we saw that the chemical potentials

of the compounds can be evaluated from Eq. (4.56)which here yields

= bG

+ cG

= G

+ ∂G

/∂

+ ∂G

/∂

− y

∂G

/∂y

(21.17)

We thus obtain, for instance,

+ y

· 

AD+BC

+ bRT ln y

+ cRT ln y

(21.18)

− y

· 

AD+BC

+ bRT ln y

+ cRT ln y

. (21.19)

where the quantity



AD+BC

≡

−

, (21.20)

is the Gibbs energy for the reciprocal reaction, A

+ B

→ A

+ B

.For

constant interaction energies, to be denoted by L,weﬁnd

= y

+ y

AB:C

+ y

A:CD

+ y

− y

B:CD

+ y

− y

AB:D

(21.21)

= y

+ y

AB:C

+ y

B:CD

+ y

− y

A:CD

+ y

− y

AB:D

. (21.22)

It should be noticed that the quantity 

AD+BC

is evaluated from information on the

four pure component compounds and does not even concern the quasi-binary sides. It

often has a dominating inﬂuence on the properties of alloys inside the quaternary system.

One may regard 

AD+BC

as a representation of the difference in interaction between

nearest neighbours, i.e. usually between atoms in different sublattices. The L values, on

the other hand, which enter in the excess Gibbs energy and control the behaviour of the

quasi-binary sides, represent the interactions between atoms in the same sublattice, i.e.

next-nearest neighbours in most cases, and may thus be of secondary importance.

It is worth noting that the partial Gibbs energies of the four component compounds

are not independent of each other. From Eq. (4.49)itisevident that

+ G

− G

= bG

+ cG

+ bG

+ cG

−bG

− cG

− bG

− cG

= 0. (21.23)

If one, for some reason, wants to consider the partial Gibbs energies of the elements,

then it can be done relative to the value of one of them, e.g. G

= G

− bG

(21.24)

= G

− bG

(21.25)

= G

− G

+ bG

= G

− G

+ bG

. (21.26)

However, G

is indeterminate unless a second phase is present. The same phenomenon

is illustrated for binary and ternary systems in Figs. 7.8 and 7.9 but it cannot be easily

illustrated for a phase with four components.

One can introduce Redlich–Kister polynomials to describe the composition depen-

dence of the interaction energies. When calculating

we must then evaluate several

21.3 Reciprocal solution phases 467

kinds of contributions if there are many components. We ﬁnd contributions of the fol-

lowing forms from interactions on the second sublattice



M:ij

+ y

− 2y

) +



k=1

M:ij

− y

)

k−1

{(y

− y

)

× [y

(1 + k)(1 − y

) + y

− y

] + ky

}

(21.27)



N:ij

(1 − 2y

) +



k=1

N:ij

− y

)

k−1

{(y

− y

)

× [(1 +k)(1 − y

) − y

] + ky

} (21.28)



M:lj

(1 − 2y

) +



k=1

M:lj

− y

)

(1 − 2y

− ky

)

(21.29)



N:lj

(−2) +



k=1

N:lj

− y

)

(−2 − k).

(21.30)

Equivalent terms would come from the interactions on the ﬁrst sublattice, L

MN:i

MN: j

, L

KN:i

and L

KN: j

The model for reciprocal phases, which has been discussed here, can be generalized

to several sublattices and several components on each one. For three sublattices we ﬁnd

= y

+ RT(ay

ln y

+ by

ln y

+ cy

ln y

) +

(21.31)

It should be repeated that one can make computer calculations by simply deﬁning

the G

expression. The complicated expressions for partial excess quantities given here

are seldom needed.

Exercise 21.3

Examine how G

varies along the A

− B

diagonal if all the L parameters are

zero. Compare with a binary substitutional solution A–B.

Hint

On the diagonal y

= y

and y

= y

Solution

The model gives G

+ y



G + bRT ln y

+ cRT ln y



G + (b + c)RT ln y

. This holds for one mole of atoms if b +c = 1 and it then

resembles the regular solution model if

L = 

G ≡

−

468 Solution phases with sublattices

21.4 Combination of interstitial and substitutional solution

The compound energy model, used to describe a reciprocal phase, can also be used for

the case where there are two substitutionally mixed elements and one interstitial element.

If we use C to represent the interstitial element then D represents the vacant interstitial

sites. A and B represent the two substitutional elements. The compound A

will thus

simply represent b atoms of A and B

represents b atoms of B. The difference between

and A

or the difference between B

and B

can be used to represent c

atoms of C. By the methods used in our preceding discussion on interstitial solutions we

now obtain

− y



AD+BC

/b + RT ln y

+ (c/b)RT ln(1 − y

) +

(21.32)

+ y



AD+BC

/b + RT ln y

+ (c/b)RT ln(1 − y

) +

(21.33)

= y



−



/c + y



−



+ RT ln[y

/(1 − y

)] +

, (21.34)

where

= y

AB:D

− L

AB:C

− L

B:CD

+ L

A:CD

) + y

AB:D

+ y

A:CD

+2y

AB:C

− L

AB:D

) + 2y

B:CD

− L

A:CD

) (21.35)

= y

AB:D

− L

AB:C

+ L

B:CD

− L

A:CD

) + y

AB:D

+ y

B:CD

+2y

AB:C

− L

AB:D

) + 2y

A:CD

− L

B:CD

) (21.36)

= y

AB:C

− L

AB:D

) + (1 − 2y

)(y

A:CD

− y

B:CD

). (21.37)

The expression for G

has been made symmetric for A and B. Alternatively, we can

modify the expression by considering A as the base metal.

Exercise 21.4

By considering the Fe–Mn–S melt as a reciprocal solution (Fe,Mn)

(Va,S)

, estimate the

parameter at 1900 K from the following binary information,

◦

MnS

−

◦

−

◦

−139.4kJ/mol and

◦

FeS

−

◦

−

◦

=−62.4kJ/mol.

Hint

The model is actually the combination of a substitutional and interstitial model. Neglect-

ing binary interaction energies we obtain for S, which in the model plays the role of

an interstitial element, G

◦

FeS

−

◦

+ RTln[y

/(1 − y

)] + y



◦

G. The ε

parameter is deﬁned from Eq. (20.6): (G

−

)/RT = ln

+ ln x

+ ε

but for low Mn and in particular low S contents we may approximate x by y

and neglect the last term.

21.5 Phases with variable order 469

Solution

By comparing the two expressions we ﬁnd RTε

∼



G ≡

MnS

−

FeS

=−139.4 + 62.4 =−77.0kJ/mol; ε

=−77000/1900R =−4.9.

21.5 Phases with variable order

So far we have discussed reciprocal phases where each element can go into one sublattice,

only. However, there are many cases where an element can go into two or more sublattices

although it energetically prefers a particular one. The distribution on various sublattices

will then vary with temperature, with the highest degree of order found at the lowest

temperature at which the atoms are still able to move. To illustrate the case we shall

consider a phase with the formula (A,B)

(B,A)

where A prefers the ﬁrst sublattice

and B the second one. For simplicity we shall choose b + c = 1. The four component

compounds will be A

, A

, B

and B

.Wecan apply all the equations already

derived for a reciprocal phase but now the site fractions are not ﬁxed by the composition.

However, there is a relation between composition and site fractions and we can write it

in two different ways,

= by



+ cy



= y





+ by





+ cy





(21.38)

= by



+ cy



= y





+ by





+ cy





. (21.39)

These relations can be used to simplify the G

expression obtained by applying Eq.

(21.15)tothe present case.

= y





+ y





+ y





+ y





+ RT(by



ln y



+ by



ln y



+ cy



ln y



+ cy



ln y



)

= x

+ x

+ y





(

− b

−c

)

+ y





(

− b

− c

)

+ RT(by



ln y



+ by



ln y



+ cy



ln y



+ cy



ln y



), (21.40)

where

and

are the Gibbs energy of 1 mole of A or B in this structure and

may be denoted by

and

.Wecan introduce the following two quantities, which

represent the Gibbs energy of formation of the compounds A

and B



− b

− c

− b

− c

(21.41)



− b

− c

− b

− c

. (21.42)

The sum of the two quantities is identical to the Gibbs energy of the reciprocal reaction

introduced in Section 21.3,



+ 

−

≡ 

AB+BA

(21.43)

470 Solution phases with sublattices

We now obtain

= x

+ x

+ y







+ y







+ RT(by



ln y



+ by



ln y



+ cy



ln y



+ cy



ln y



). (21.44)

There are only two independent y

parameters and after ﬁxing the composition there will

be only one independent parameter, e.g. y



yielding dy



=−dy



and dy



=−dy



(b/c)dy



. The equilibrium value of y



under constant composition and any T is obtained

by minimizing G

c · ∂ G

/∂y



= (cy



+ by



)

− (cy



+ by



)

+bc · RT(1 + ln y



− 1 −ln y



− 1 −ln y



+ 1 +ln y



) = 0.

(21.45)

It is easiest to solve for T at any chosen value of y



bcRT =

(cy



+ by



)

− (cy



+ by



)

ln(y









)

. (21.46)

When the two sublattices are equivalent, b = c = 0.5 and



= 

= 0.5

AB+BA

. (21.47)

and the equilibrium condition is

−

AB+BA



− y



− y



+ y



ln(y









)

, (21.48)

where 

AB+BA

is deﬁned by Eq. (21.20). This may be called the symmetric case.

The disordered state, y



= y



= x

,isasolution to the equation at all temperatures and

compositions but in order to examine where it is a stable or unstable equilibrium, we

must study the second-order derivative,

c · ∂

/∂y

2

= (b + b + b + b) · 0.5

AB+BA

+ bRT(c/y



+ c/y



+ b/y



+ b/y



)

= 2b

AB+BA

+ bRT/x

. (21.49)

For positive values of 

AB+BA

the second derivative is always positive and the disor-

dered state is always the stable state. There is no ordering tendency. For negative values

of 

AB+BA

it is positive at higher temperatures and negative at lower and there is

a transition temperature where it is zero, corresponding to the criterion g

ξξ

= 0inEq.

(15.18). We ﬁnd

−

AB+BA

= 2x

. (21.50)

The disordered state is thus a state of stable equilibrium above this critical temperature

but an unstable equilibrium below. Two new solutions to the equation appear there,

representing stable ordered states. This is demonstrated for a composition of x

= 0.6

in Fig. 21.3(b) and the dashed line represents the unstable, disordered state below the

transition point. The phase diagram is shown in Fig. 21.3(a). The whole curve is there

21.5 Phases with variable order 471

1.2

(a) (b)

1.0

0.8

0.6

T/T

1.2

1.0

0.8

0.6

0.5 1.0

y ′

Figure 21.3 Second-order transition in a binary system. (a) Phase diagram. (b) Variation of

degree of order with temperature for x

= 0.6.

drawn as a dashed line because it is not a phase boundary but a transition line, the

transition being of second order. The site fraction y



approaches the x

value without

any jump, as demonstrated in Fig. 21.3(b).Wehave thus managed again to model the

second-order type of ordering transition but this time we have used a model containing

parameters of some physical signiﬁcance. In Section 15.2 we followed Landau’s approach

and simply worked with coefﬁcients in a power series expansion.

When 

and 

are not equal, the result will be quite different. This

is demonstrated in Fig. 21.4(a) for b = c = 0.5, x

= 0.5, and 

− 

−0.1

AB+BA

. This may be called the asymmetric case. As soon as the two compound

energies differ, the completely disordered state will never be stable and the ordered region

in the phase diagram does not show an order–disorder transition. The variation of order

with composition is shown at three temperatures in Fig. 21.4(b). The results for x

= 0.5

can be read on the diagonal between the upper left and lower right corners.

Exercise 21.5

Where in Fig. 21.4(a) would a line for 

− 

=+0.1

AB+BA

fall?

Hint

Do not try to solve this problem by looking at the equations. The answer should be based

upon a more basic consideration.

Solution

A change of sign of 

◦

AB+BA

means that the other sublattice will be preferred by the

A atoms but otherwise the effects will be the same as before. The effect of 

◦

AB+BA

is to accentuate the order, independent of what sublattice the A atoms prefer. We will

472 Solution phases with sublattices

2.0

1.5

1.0

0.5

0.50 0.75 1.00

asymmetric

symmetric

T/T

y ′

y ′′

1.0

0.8

0.6

0.4

0.2

0 0.5 1.0

1000 K

1500 K

500 K

(a) (b)

Figure 21.4 Model calculation of an ordered phase without a transition to a completely

disordered state at any temperature. (a) Result for x

= 0.5 compared with the result for a

symmetric case showing a transition (dashed line). (b) Variation with composition.

thus get the same shape of curve as in Fig. 21.4(a) but starting from y



= 0atT = 0

and approaching y



= 0.5 from the left.

21.6 Ionic solid solutions

In this chapter we have modelled various types of phases with sublattices without actually

discussing the nature of the atoms (or ‘species’ to be more general). The compound

energy model also applies to ionic substances but there are some complications due to

the requirement of electroneutrality which should now be discussed.

Let us ﬁrst consider solid solutions between NaCl, KCl, NaBr and KBr which all

have the same crystalline structure. All the elements are ionized and we could give the

formula as (Na

, K

)

(Cl

−1

, Br

−1

)

. One can vary the composition freely within the

composition square because all the ions are univalent and the condition of electroneu-

trality is automatically fulﬁlled over the whole square by the fact that the two sublattices

have the same number of sites. The compound energy formalism can be applied with no

additional complications in this case.

Let us next consider the solution of CaCl

in NaCl. A complication is then caused by Ca

being divalent and in order to compensate for this some of the cation sites will be vacant.

The formula would thus be (Na

, Ca

, Va

)

(Cl

−1

)

. This seems to resemble a ternary

system with Na

, Ca

and Va

as the components. However, electroneutrality

requires that y

= y

and one can only vary the composition along a straight line in the

composition triangle. In order to model the properties of solutions on this line we shall

apply the ordinary expression for the compound energy model but with the additional

21.6 Ionic solid solutions 473

condition of electroneutrality. For 1 mole of formula units we get by neglecting the excess

Gibbs energy,

= y

NaCl

+ y

CaCl

+ y

Va Cl

+ RT



ln y

. (21.51)

Here we have two quantities which cannot be studied experimentally,

CaCl

and

Va Cl

because they are deﬁned for charged compounds. However, due to the auxiliary condi-

tion y

= y

, they always appear in the neutral combination, (

CaCl

Va Cl

). The

properties of this combination can be studied by studying neutral solutions. Instead of

introducing this combination in the equation, it may be recommended to keep the original

form and apply the condition of electroneutrality in the ﬁnal expression one wants to

use. When listing the parameter values for an ionic system one could select one of the

charged compounds and give all the other charged compounds relative to that one. As

an example, one may like to express the chemical potential of CaCl

in the solution. It

is obtained as

CaCl

≡ G

CaCl

+ G

Va Cl

CaCl

+ RT ln y

Va Cl

+ RT ln y

CaCl

Va Cl

+ 2RT ln y

, (21.52)

since y

= y

. The neutral combination (

CaCl

Va Cl

) can be given a numerical

value.

Let us now consider the opposite case, the solution of NaCl in CaCl

. Electroneu-

trality may there be satisﬁed by the formation of vacant sites on the anion sublat-

tice, (Ca

, Na

)

(Cl

−1

, Va

)

, and the condition of electroneutrality is now 1 · y

2 · y

. This is a reciprocal system and the compound energy formalism yields

= y

CaCl

+ y

CaVa

+ y

NaCl

+ y

NaVa

+ RTy

ln y

. (21.53)

The chemical potential of NaCl is obtained by applying Eq. (21.19)

2µ

NaCl

= G

NaCl

+ G

NaVa

NaCl

− y



G + RT ln y

+2RT ln y

NaCl

NaVa

+ y



+ RT ln y

+2RT ln y

NaVa

NaCl

NaVa

+(1 − y

)



+2RT ln



0.5y

(1 − 0.5y

)



NaCl

NaVa

, (21.54)

where

NaCl

NaVa

and 

◦

G (which is identical to

CaCl

NaVa

−

CaVa

−

NaCl

and may be written as 

CaCl +NaVa

) represent neutral combinations and can

be given numerical values. Figure 21.5 illustrates the neutral line in the two cases.

How the model may describe the properties outside these lines is of no practical

consequence.

Many ionic compounds are non-stoichiometric, an example being CeO

.Itmaybe

modelled by assuming that some of the Ce ions are only trivalent, yielding the formula

(Ce

, Ce

)

−2

, Va

)

= y

+ y

+ RT2y

ln y

. (21.55)

474 Solution phases with sublattices

VaCI

NaCI CaCI

CaCI

NaCI

NaVa

CaVa

Figure 21.5 Neutral lines for two cases of ionic solutions, the solution of CaCl

in NaCl and of

NaCl in CaCl

The deviation from the stoichiometric composition depends upon the oxygen potential

in the surroundings. In order to derive an expression for the oxygen potential one must

ﬁnd a reaction formula for the formation of oxygen which balances atoms as well as

charges.

4Ce

+ 2O

−2

→ 4Ce

+ O

. (21.56)

Then the reaction formula must be expressed in terms of the component compounds in

the model. Instead of 2O

−2

we thus write Ce

–Ce

.With all charges shown we

write the reaction as

5(Ce

)

−2

)

→ (Ce

)

(Va

)

+ 4(Ce

)

−2

)

+ O

. (21.57)

Applying Eq. (3.19)wethus obtain

2µ

= µ

= 5G

− G

− 4G

= 5

+ 5y



Va +Ce

+5RT ln y

+ 10RT ln y

+ 5

−

+ y



Va +Ce

− RT ln y

−2RT ln y

−

− 4

+ 4y



Va +Ce

−4RT ln y

− 8RT ln y

− 4

= 5

−

−4

+ RT ln





+(y

+ 4y

)

Va +Ce

+ 5

−4

−

, (21.58)

where the factor y

+ 4y

can be written as 8y

because the condition of electroneu-

trality for (Ce

, Ce

)

−2

, Va

)

1 · [4(1 − y

) + 3y

] = 2 · 2(1 − y

)ory

= 4y

. (21.59)