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

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20.4 Empirical approach to substitutional solutions 445

0.3

0.2

0.1

1.0

−0.1

k = 0

–x

)

Figure 20.2 Properties of Redlich–Kister terms.

20.4 Empirical approach to substitutional solutions

We shall now discuss an empirical approach to the modelling of the excess Gibbs energy

of solution phases. We have already seen that the difference between the properties of

a real solution and an ideal solution may be represented by a quantity called the excess

Gibbs energy,

.For a substitutional solution we can immediately see that it must go

to zero for the pure components and a convenient way of representing its composition

dependence in a binary system is to introduce the factor x

which goes to zero for

pure A as well as pure B. We shall thus express

as x

I.Weshall allow I to

be a function of composition. It may for instance be represented by a power series,

often called Redlich–Kister polynomial [39]. Figure 20.2 illustrates the properties of the

various terms in

I =

L +

L(x

− x

) +

L(x

− x

)

+ ··· (20.21)

= x



k=0

L(x

− x

)

. (20.22)

The parameter I may be regarded as a representation of the interaction between the

two components. It is convenient to give the two components as an index to I and write

it as I

for an A–B solution. When I is a constant, independent of composition as well

as temperature, one talks about a regular solution and

L may thus be called the regular

solution parameter. When it appears alone, it is usually denoted by L.

L may be called

the subregular solution parameter and

L the subsubregular solution parameter. However,

it should be mentioned that a regular solution is sometimes deﬁned as a solution where

is independent of temperature but may have any composition dependence. On the

other hand, in recent years there is a tendency to call a solution regular as soon as I is

independent of composition, whether or not I is independent of temperature.

Positive values of the regular solution parameter result in a tendency of demixing. If

not interrupted by other reactions a miscibility gap will form when the temperature is

446 Mathematical modelling of solution phases

lowered. It is easy to calculate its spinodal from the condition for the stability limit. For

a constant I we get

= dG

/dx

= (x

− x

)

L + RT(x

+ ln x

− x

− ln x

) (20.23)

= d



= (−1 − 1)

L + RT(1/x

+ 1/x

) = 0 (20.24)

= x

· 2

L/R (20.25)

= RT

L. (20.26)

The spinodal will thus be a parabola in this simple case and its maximum will fall at

= x

= 0.5 and be a consolute point.

cons

L/2R and x

= T

/4T

cons

. (20.27)

A second equation is required in order to calculate the consolute point if I is not a constant

and it is obtained through the condition of a critical point,



= 0. (20.28)

For further reference it is convenient here to derive expressions for the partial

Gibbs energies from the Redlich–Kister polynomial. Using the standard method pre-

sented in Section 4.1,wecan evaluate the last two terms in the following general

expression

− T ·

ideal

. (20.29)

With the power series representation of

we ﬁrst obtain a general expression for the

effect of an interaction between two components, l and i,

(1 − x

) +



k=1

− x

)

k−1

[(k + 1)(1 − x

)(x

− x

) + kx

(20.30)

Forabinary A–B system we can replace 1 − x

by x

and 1 − x

by x

and obtain

= x



k=1

− x

)

k−1

[(2k + 1)x

− x

]

(20.31)

= x



k=1

− x

)

k−1

− (2k + 1)x

]

. (20.32)

It should be noticed that in our notation the sign of

changes for odd k if the order

between A and B is reversed. It is also interesting to note that the expression for the partial

excess Gibbs energy of one component contains the square of the molar content of the

other. The partial excess Gibbs energy of a component will thus go to zero asymptotically

as the pure component is approached. In this respect the very dilute solutions are ideal,

a result which is usually formulated in Raoult’s law. It will be discussed in the next

section.

Forasolution with more than two components we should consider interactions within

each combination of two components and possibly also interactions between more than

20.4 Empirical approach to substitutional solutions 447

two components, so-called ternary, quaternary, etc., interactions. By limiting the present

discussion to binary interactions we obtain



+ RT



ln x



j>i

, (20.33)

and

will be the sum of contributions from all I

but also from other I

.Weobtain



i=l

(1 − x

) +



k=1

− x

)

k−1

[(k + 1)

×(1 − x

)(x

− x

) + kx

]

−



i=l



j=l,>i





k=1

− x

)

(k + 1)



. (20.34)

Foraternary system with constant interaction energies (i.e. for the regular solution

model) we can write the result as follows by omitting the superscript 0 in

+ RT ln x

+ x

− x

+ x

(20.35)

+ RT ln x

+ x

− x

(20.36)

+ RT ln x

− x

+ x

(20.37)

Exercise 20.4

Apply the expressions for

and

,givenby Eqs (20.31) and (20.32), to a binary

system and show that G

= x

Hint

We know that for an ideal solution G

= x

.Now it is thus sufﬁcient to show that

= x

Solution

x

= x

+ 

− x

)

k−1

[(2k + 1)x

− x

]



+ x

+ 

− x

)

k−1

− (2k + 1)x

]



= x

+ x

) + 

− x

)

k−1

× [x

(2k + 1)x

− x

+ x

− x

(2k + 1)x

]



= x



+ 

− x

)



448 Mathematical modelling of solution phases

Exercise 20.5

Mo has the same structure as α–Fe(bcc) and it is thus possible to combine them into

the same model, covering the whole range of composition of bcc in the Fe–Mo phase

diagram. Determine the G

expression for this model from the mutual solubilities at

1300

◦

Cwhich are 0.16 Mo in Fe and 0.075 Fe in Mo.

Hint

Since two pieces of information are given, we can determine two parameters. The excess

term will thus be written as x

[

L +

L(x

− x

)] where

L and

L will be consid-

ered as independent of T.Atequilibrium G

−

, i.e. RT ln x

would have

the same value in both phases and so would RT ln x

.For a binary solution

we ﬁnd

= x

[

L +

L(3x

− x

)] and

= x

[

L +

L(3x

− x

)].

Solution

RT ln x

= RT ln 0.84 + (0.16)

[

L +

L(3 · 0.84 − 0.16)] = RT ln 0.075 +

(0.925)

[

L +

L(3 · 0.075 − 0.925)]; RT ln x

= RT ln 0.16 + (0.84)

[

L +

L(0.84 − 3 · 0.16)] = RT ln 0.925 + (0.075)

[

L +

L(0.075 − 3 · 0.925)].

The numerical result is

L = 34 500 and

L =−4500 J/mol.

20.5 Real solutions

As discussed already in Section 7.1, one often represents the properties of real solutions

with the activity a

and activity coefﬁcient f

, deﬁned through

= µ

REF

+ RT ln a

= µ

REF

+ RT ln x

+ RT ln f

. (20.38)

It is common to choose as reference the state of pure i at the same temperature and

pressure as the state under consideration, usually denoted

The activity coefﬁcient is often intended for use in dilute solutions only. It is interesting

to examine its variation with composition by applying the regular solution model for a

binary substitutional solution. Figure 20.3 shows three curves for the activity obtained

with

L/RT = 2, 0or− 2. For high B contents x

is small and the regular solution

model yields

= x

exp



/RT



∼

; f

∼

1. (20.39)

As a consequence, all the curves approach asymptotically the diagonal in the diagram.

This is Raoult’s law and its validity is very general. We have already seen that

has

the factor x

independent of what power series has been chosen. This factor originates

from the factor x

, introduced in order to make

go to zero for pure B and for pure

A. For low B contents where x

may be approximated by 1 we obtain

= x

exp



/RT



∼

exp(

L/RT); f

∼

exp(

L/RT). (20.40)

20.5 Real solutions 449

Henry's law

Henry's

law

Raoult's law

L = 2RT

L = 0

L =−2RT

1.0

0.8

0.6

0.4

0.2

Figure 20.3 The properties of a binary solution according to the regular solution model with three

different values of the regular solution parameter.

This gives the slope of the tangent to the curve at the origin. It is self-evident that at low

B contents the alloys fall close to this tangent. This is often formulated as Henry’s law.

It must be emphasized that this law says nothing about the slope of the tangent, contrary

to Raoult’s law. It is interesting to mention that Raoult’s law for component A can be

derived by applying Henry’s law to component B.

A convenient method of studying the properties of a binary solution is based upon the

presence of the factor x

in the expression for the partial excess Gibbs energy of B in

the empirical model. We obtain



k=1

− x

)

k−1

− (2k + 1)x

] =

= RT ln (a

)/x

= RT ln f

. (20.41)

If a plot of experimental values of RT ln f

versus x

can be represented by a straight

line then the properties can be represented by two parameters,

and

.Ifexper-

imental data are available for both components, then the two-parameter representation

requires that one ﬁnds two straight lines, one for each component and such that they

yield the same set of

L and

L values. An example for liquid Fe–Ni at 1852 K is given in

Fig. 20.4.InFig.20.4(b) the two lines have the same slope, representing

Fe,Ni

, and

the same value at x

= 0.25 for the Ni line and x

= 0.25 for the Fe line, representing

Fe,Ni

. The experimental scatter is considerable in Fig. 20.4(b), especially at low Ni

contents for Fe and at low Fe contents for Ni because of the very small values then taken

by the factors x

and x

in the denominators.

It should be emphasized that there are many binary systems where the power series

representation is not very convenient. Figure 20.5 shows an example from liquid Bi–Mg

at 973 K where a very large number of power terms would be needed for a satisfactory

representation of the data. It is evident that some particular physical effect occurs at

the centre of the system and it would be difﬁcult to represent such data mathematically

450 Mathematical modelling of solution phases

1.0

0.8

0.6

0.4

0.2

1.0

RT Inf

(kJ/mol)

−10

−20

activity

Figure 20.4 The properties of liquid Fe–Ni solutions at 1852 K according to direct measurements

and a subregular solution model with

Fe,Ni

=−10 and

Fe,Ni

= 5kJ/mol.

RTInf

(kJ/mol)

1.0

−10

−20

−30

−40

Figure 20.5 Experimental data from liquid Bi–Mg alloys at 973 K. This behaviour cannot be well

represented with a power series.

without identifying that effect and representing it with an adequate model. Such models

will be described later on.

Exercise 20.6

Consider the gaseous mixture (solution) of H

and O

at 1 bar and a temperature high

enough for the reaction 2H

+ O

→ 2H

Otogotoequilibrium but still low enough to

make the formation of H

O practically complete. Examine how the activity of O

would

vary across the binary H

–O

system as a function of x

Hint

The binary system has two components and it is evident that they are deﬁned as H

and

and that x

is thus deﬁned with no regard for the formation of H

O. Suppose that the

20.5 Real solutions 451

1.0

0.8

0.6

0.4

0.2

0 1.0

Figure 20.6 Solution to Exercise 20.6.

gas is actually an ideal gas mixture of H

and H

O. At constant T and P we then have

RT ln x

= µ

−

= RT ln y

where the deﬁnition of y

takes into account the

presence of H

Solution

Between x

= 0 and 1/3 there are practically no molecules of O

. All oxygen goes into

O. Between x

= 1/3 and 1 there are, for the same reason, practically no H

molecules

and y

+ y

= 1 and x

= (y

+ 0.5y

)/(y

+ 1.5y

) and thus a

= y

(3x

− 1)/(x

+ 1). It is evident that it is impossible to describe the deviation of this

solution from the ordinary ideal solution with a power series. Any reasonable model

must recognize that the interaction between the two components is here so strong that

large quantities of a new kind of molecule form (Fig. 20.6).

Exercise 20.7

It is common to deﬁne the activity of a component i in a solution with reference to pure

i at the same temperature. The chemical potential of i in the solution is then written as

(T ) + RT ln a

. Suppose that one would instead like to deﬁne an activity by

referring to pure i at absolute zero as reference. Derive a relation between the two kinds

of activity, a

and a

Hint

The chemical potential of i in the solution, µ

,isawell deﬁned quantity even though its

value can be given only relative to a reference, i.e., as µ

−

Solution

is the same quantity in the two equations: µ

(T ) + RT ln a

and

(0) + RT ln a

. Let us take the difference: 0 =

(T ) −

(0) +

452 Mathematical modelling of solution phases

RT ln(a

); a

= a

exp[(

(T ) −

(0))/RT]. Thus a

< a

since

(0) >

(T ).

20.6 Applications of the Gibbs–Duhem relation

At constant T and P the Gibbs–Duhem relation, Eq. (3.34), reduces to

x

= 0orx

dlna

= 0. (20.42)

We have here chosen the notation G

instead of µ

since we shall consider a single solution

phase. By introducing partial excess Gibbs energies or activity coefﬁcients through

+ RT ln x

+ RT ln f

, (20.43)

we get

x

= 0orx

dln f

= 0, (20.44)

because

x

dlnx

= x

= dx

= 0. (20.45)

In a binary system it is thus possible to evaluate f

from measurements of f

by integration

from pure 1.

ln f

=−



dln f

. (20.46)

Forgraphical or numerical integration Wagner [40] suggested that this equation should

ﬁrst be transformed by integration by parts, yielding

ln f



ln f

−

ln f

. (20.47)

One can also evaluate

from the information on f

. The last equation gives

ln f

+ x

ln f

= x



ln f

, (20.48)

and this is identical to

/RT.

As an introduction to a discussion of ternary systems, it may be useful to repeat the last

derivation by starting from the well-known expression for a binary system in Section 7.1,

when x

is regarded as a dependent variable.

+ (1 − x

/dx

. (20.49)

This can be rearranged into

= (1 − x

)



/(1 − x

)



/dx

. (20.50)

20.6 Applications of the Gibbs–Duhem relation 453

Integration from x

= 0where

= 0, yields

= (1 − x

)







= 0) + (1 − x

)





/(1 − x

)









= (1 − x

)





/(1 − x

)



, (20.51)

which is actually identical to the result of Eq. (20.48).

Foraternary system one may derive the same equations if the ratio x

is regarded

as the second independent variable. However, when integrating from x

= 0 one now

starts from a binary 1–3 alloy and its

is not zero. We should thus write the

result as

= (1 − x

)







= 0) +





/(1 − x

)









. (20.52)

This equation can be useful if one has measured

(i.e. RT ln f

)insections of constant

Exercise 20.8

Show that G

= G

+ (1 − x

) · (∂ G

/∂x

)

in a ternary system by starting from

G = NG

, x

Hint

Since x

= N

we ﬁnd (∂(x

)/∂ N

)

= 0.

Solution

= N

/(N

+ N

); (∂ x

/∂ N

)

= (N − N

)/N

= (1 − x

)/N; G

≡

(∂G/∂ N

)

= G

+ N · (∂ G

/∂x

)

· (∂ x

/∂ N

)

+ N · (∂G

/∂(x

))

(∂(x

)/∂ N

)

+N · (∂G

/∂x

)

· (1 − x

)/N + N · (∂ G

/∂(x

))

0 = G

+ (1 − x

) · (∂ G

/∂x

)

Exercise 20.9

By studying the ratio of vapour pressures over A–B alloys one can measure how the ratio

of activities a

varies with composition. Show how one can evaluate a

and a

from

such information.

454 Mathematical modelling of solution phases

Hint

Use the Gibbs–Duhem relation in the form x

dlna

+ x

dlna

= 0. Replace ln a

ln(a

) + ln a

Solution

0 = x

dlna

+ x

d ln(a

) + x

dlna

= (x

+ x

)d ln a

+ x

d ln(a

) =

dlna

+ x

d ln(a

); ln a

=−



d ln(a

)

20.7 Dilute solution approximations

When discussing dilute solutions Wagner [40] suggested that one should consider the

composition dependence of the activity coefﬁcient. For a dilute solution of B in A he

wrote

−

)/RT = ln a

= ln f

+ ln x

= ln

+ ln x

+ ε

(20.53)

This may be compared with the expression we obtained for the regular solution model,

i.e., Eq. (20.32) with a constant parameter.

+ RT ln x



1 − 2x

+ x



(20.54)

The two formalisms are identical if the second-order term can be neglected, i.e. for dilute

solutions. We identify parameters as follows if the same reference is used in both cases.

= L

/RT (20.55)

=−2L

/RT (20.56)

However, a different reference state is often used in the ε formalism. As an example,

the inﬁnite-dilution reference state is deﬁned in such a way that it makes ln

= 0 and

gives the relation

inf.dil.

+ L

(20.57)

inf.dil.

+ RT ln x



−2x

+ x



(20.58)

Foraternary system, where small amounts of B and C are dissolved in A, Wagner

introduced an interaction coefﬁcient between the two solutes, ε

= ε

, and gave the

expressions

(

−

)

/RT = ln a

= ln f

+ ln x

= ln

+ ln x

+ ε

(20.59)

−

)/RT = ln a

= ln f

+ ln x

= ln

+ ln x

+ ε

. (20.60)

This may be compared with the regular solution model yielding Eqs (20.35)to(20.37)