principles of thermodynamic modeling of igneous processes

8/2/2019 Principles of Thermodynamic Modeling of Igneous Processes

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PART 3:Principles of Thermodynamic Modelingof Igneous Processes

Geology 531: Igneous Petrology Notes


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GLGY 431 PRINCIPLES OF THERMODYNAMIC MODELING OF IGNEOUS PROCESSES

Principles of Thermodynamic Modeling of Igneous Processes

Introduction to the Concept of Chemical Equilibrium

Thermodynamic systems can be described in two different ways. There is, first,

the real collection of atoms, molecules, phases, and energy that constitutes the real world.

Then there is the mathematical structure that represents the real world. The literature

contains several definitions of equilibrium in thermodynamic systems. Some definitions

are phrased in terms of the behavior of the real collection. Other definitions are

mathematical statements. Prigogine and Defay (1954, p. 69, Equation 6.23) use the value

of a linear combination of chemical potentials to describe the equilibrium condition:

(1) =iii 0

where i is a stoichiometric coefficient and i is the chemical potential of the chemical

entity i. Equation (1) represents the equilibrium for a chemical transformation that can be

written:

(2) =i

ii 0x

where xi is a chemical formula and Equation (2) represents a balanced chemical

transformation.

Denbigh (1981) describes thermal equilibrium with his zeroth law:

..if bodies A and B are each in thermal equilibrium with a third body, they

are also in thermal equilibrium with each other.

In other words, bodies in equilibrium, at least bodies in thermal equilibrium, have the

same temperature. Nordstrom and Munoz (1985) state that equilibrium occurs when the

equality holds in the Clausius equation:

0 (3)dSTdQ

where Q is the heat gained by the system, T is the temperature in Kelvin, and S is

entropy. Equality holds for reversible processes. In other words, equilibrium and

reversibility are synonymous with this definition. Anderson and Crerar (1993) define

equilibrium with two statements:

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A system at equilibrium has none of its properties changing with time, no

matter how long it is observed.

A system at equilibrium will return to that state after being disturbed, that

is, after having one or more of its parameters slightly changed, [they will]then change back to the original values.

Other statements of equilibrium postulate a minimum for thermodynamic potentials that

depend on a particular set of variables. For example, equilibrium obtains if the Gibbs

energy for a system is a minimum at some fixed values of pressure and temperature.

All of these statements are correct in that, if the criteria hold, then the system is at

equilibrium. Except for the definition by Prigogine and Defay (1954), however, the

statements are sufficient rather than necessary criteria for equilibrium in chemical

systems. In some systems, the other statements will be incorrect whereas the statement by

Prigogine and Defay (1954) (hereafter termed the First Statement) will still be correct. If

the other statements obtain then so does the First Statement. The converse is not true. The

First Statement can obtain, although one or more of the others do not obtain.

In summary the sufficient statements are:

1. Equilibrium systems do not change, no matter how long they are observed.

2. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.

3. When infinitesimally perturbed, equilibrium systems return to their original

equilibrium state.

4. A minimum in the value of a thermodynamic potential occurs at equilibrium.

5. Equilibrium systems are at constant temperature.

Although many equilibrium thermodynamic systems cannot be realized in nature,

they often provide adequate representations of reality. If even part of a natural system can

be adequately approximated by an equilibrium system then significant, mathematical

simplification over a dynamic model usually follows. Consequently, a precise definition

of equilibrium leads to precise descriptions of representations of reality and a better

understanding of processes that affect chemical systems.

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The trivial case of one phase: Suppose two grains of kyanite, a and b. Then

05252

= b SiOAla

SiOAl

Equilibrium Systems Do Not Change, No Matter How Long They Are Observed

As a counter example to this definition of equilibrium, consider a glass of water

and ice in an environment at room temperature. The temperature in the glass is 0C. In

fact, the glass contains an assemblage of phases that defines a point on the

thermodynamic temperature scale. Yet, obviously the system is changing with time as the

ice melts. Purists will protest that the temperature is not really 0C but it is slightly

greater than that. However, as a representation of reality, the mathematical formulation is

adequate. A temperature scale based on phase assemblages out of equilibrium would be

inconvenient at best and unusable normally. Consequently, for the practical purpose of

the temperature scale, a mixture of ice and water is an equilibrium system and the

appropriate and adequate expression is:

0 (4)22

= WaterOHIce

OH

an example of the First Statement.

At Equilibrium, the Rate of the Forward Reaction Equals the Rate of the Reverse

Reaction

Again, the mixture of ice and water example provides a counter example. The rate

of melting is faster then freezing, yet, as shown above, the system can be represented as

an equilibrium system.

When Infinitesimally Perturbed, Equilibrium Systems Return to Their Original

Equilibrium State

Consider a large volume of water and ice into which a drop of hot water is added.

The temperature quickly returns to 0C but the small amount of ice that melted to cool

the drop does not freeze again. Unless the equilibrium state is defined by temperature

alone, then the system of ice and water does not return to its original state when

infinitesimally perturbed.

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A Minimum in the Value of a Thermodynamic Potential Occurs at Equilibrium

A counter example to this condition is more difficult to contrive than the others.

Consider a solid solution that can have a varying degree of order, for example, an

ilmenite-hematite solid solution. Ghiorso (1990) derives an expression for the chemical

potential of the hematite component:

ysHm s

Gsy

GyG

= (5)

where y is the mole fraction of Fe2O3 in solid solution and s describes the degree of

ordering of Fe and Ti on the metal sites of the structure. Disordered FeTiO3 is

characterized by s equal to zero whereas fully ordered FeTiO3 has a value ofs equal to

one. G is the Gibbs potential for the solution. In general, s is evaluated by setting the

partial derivative ofG with respect to s equal to zero (see, for example, Sack, 1980).

This exercise is equivalent to minimizing the Gibbs potential for the solid solution with

respect to diffusion. However, it may happen that a solid solution encounters an

environment where it will react with another phase under conditions where rates of

internal diffusion in the solid solution are too slow to alter the ordering state. Under such

conditions, the chemical potential of Fe2O3 still exists [Equation. (5)] and could be used

to describe an equilibrium change in a geochemical system with an expression of the

form of Equation (1). In such a system, the Gibbs potential for the system will not be a

minimum because the Gibbs potential for the solid solution will not be a minimum.

Equilibrium Systems are at Constant Temperature

A counter example to this statement can be constructed by recourse to that

magical device, an adiabatic semi-permeable membrane. Suppose a solution on one side

of the wall and a vapor consisting of the pure solvent on the other. The wall is assumed

permeable to the solvent but impermeable to the solute. As an example, suppose the

solvent is water and that the solution is ideal. The chemical potential for H2O in the

solution is:

(6)xlnRTT712.16683172 WW

Water

OH =

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R is the gas constant and TW is the temperature of the solution. The chemical potential of

H2O as steam, assuming ideal gas behavior is:

(7)PlnRTT11.4557952 SS

Steam

OH +=

TSis the temperature of the vapor. If the chemical potentials in Equations (6) and (7) are

set equal, an equations in four unknowns results, TW, TS, x, and P. It remains to show that

the equation can be satisfied by values of the unknowns that are physically realizable:

0 < TW< TW< TS< (8)

0 < P < 0


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Representations of equilibrium by linear combinations chemical potentials equal

to zero do in fact, provide convenient methods for defining equilibrium systems.

Equilibrium systems are those parts of reality that can be adequately and appropriately

represented by such linear combinations. The non-equilibrium parts must be represented

by linear combinations of chemical potentials different from zero. Describing systems in

terms of equality and inequality constraints puts the initial focus on whether or not that

part of reality can be described in terms of equilibrium thermodynamics or whether the

description must be a dynamic one. The causes for the differences can then be interpreted

in terms of geometric walls, membranes and barriers to the transfer of matter and energy.

As Reiss (1964) emphasizes, the behavior of thermodynamic systems is usually dictated

by the nature of the constraints (wall, membranes, and barriers) on the thermodynamic

system. Just as thermodynamics is independent of our understanding of the atomic nature

of matter, the mathematical structure that represents reality can be constructed without

recourse to the precise nature of the real parts of the universe.

Often equilibrium systems are equated with parts of reality characterized by

reversible reactions or changes. Although systems containing reversible processes are

systems at equilibrium, the converse need not be true. Some systems containing

irreversible processes can be analyzed and described with equilibrium statements. One

such set of systems is the set that contains isenthalpic changes (see page 34). Examples

include expansion of gases into a vacuum (the Joule-Thompson effect, see Lewis and

Randall, 1961, p. 48), adiabatic decompression during uplift (Waldbaum, 1971), and

rapid ascent of nephelinites (Nicholls, 1990; Trupia and Nicholls, 1996). Such systems

should be distinguished from systems described by what are called irreversible

thermodynamics (e.g. the Soret effect). In these systems, linear combinations of chemical

potentials differ from zero. The study of such systems is better described as non-

equilibrium thermodynamics rather than irreversible thermodynamics.

Phase Rules and Duhem's Theorem

This chapter is an outline of the thermodynamic basis for modeling crystal-melt

equilibria. Because the purpose of the chapter is to illustrate the principles of

thermodynamic modeling, rather than solve the problems associated with specific

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applications, numerical analysis will be kept to a minimum. The main consequence of

keeping numerical analysis to a minimum is that all solutions will be treated as ideal, thus

eliminating the effects of activity coefficients. Once the features of ideal systems are

mastered, the principles involved in modeling real systems are easily followed.

The first question to answer is: How much information must we bring to a

problem in order to model a thermodynamic system? The answer is surprisingly simple,

if we know the composition of the system, then we need only two other pieces of

information. Duhem's theorem, a little known form of a phase rule can be stated:

Given the composition of a chemical system, then two and only two independent

variables need be fixed to determine the equilibrium state of the system.

The more commonly known phase rule, the Gibbs phase rule, can be written:

F = c + 2 (9)

where F is the degrees of freedom or number of variables that must be fixed in order to

define the equilibrium state of the system, c is the number of components in the system

and is the number of phases in the system. The import of Duhem's theorem is that F is

equal to 2 if we know the composition of the system.

In chemical systems of interest to petrologists, the intensive variables include P, T

and the compositions of all the phases present. In addition to P, T and phase

compositions, igneous petrologists have traditionally been preoccupied with the modal

abundances of the phases in the system. Modal amounts are not considered intensive

variables in derivations of the Gibbs phase rule (e.g. Prigogine and Defay, 1954, p. 174-

175). They are, however, intensive variables because they are measured in relative units

(e.g. volume percent, weight percent, etc.). The intensive variables describing the phase

compositions are related to properties of the phases, such as volumes, by partial molar

quantities:

v = ivi xi (10)where v

is the molar volume of phase , vi

is the partial molar volume of component i

in phase , and xi is the mole fraction of component i in phase . The summation is over

index i = 1..N, N being the number of components in phase .

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Whereas phase properties are related to partial molar quantities and compositions,

properties of the chemical system are related to the phase properties and the modal

amounts:

V = z

v

(11)

where V is the molar volume of the chemical system, z is the molar concentration of

phase in the system and v is the molar volume of phase . In Equation (11), the sum is

over index = 1..M, M being the number of phases in the system. Consequently, we see

that the relationship between the molar volume of the system and the molar phase

volumes parallels that between the molar phase volumes and the partial molar volumes.

Both the molar system properties defined in this way and the molar phase properties are

intensive variables.

A typical statement of Duhem's theorem is (Prigogine and Defay, 1954, p. 188):

"Whatever the number of phases, of components or of chemical reactions, the

equilibrium state of a closed system, for which we know the initial masses [of the

constituents] is completely determined by two independent variables."

Compared to the statement given earlier, this is a much more restrictive version.

First, the initial masses can be more difficult to estimate than the composition of the

system. For example, even if a particular pluton crystallized as a closed system, a

questionable assumption, estimating the masses of the elements in the rock body might be

impossible because of the inaccessibility of parts of the pluton. Second, the system need

not remain closed after we calculate its equilibrium state; to apply Duhem's theorem, all

we need is its composition in an equilibrium state.

The two different statements can be reconciled (mass vs concentration) by

realizing that for calculating the properties of the phases or of the system, the size of an

equilibrium system can be arbitrarily set: Two systems differing in size but with the same

composition and in the same equilibrium state will differ neither in P, T, phase

compositions, nor in modal concentration of the phases. The difference in size can be

accounted for by extensive variables that are related by a single scalar multiple.

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The equilibrium state of an igneous system is defined when numerical values for

P, T, the phase compositions and the mode are available. Consequently, the variables are

P, T, M values of z

where M is the number of phases in the system, and M N values of

xi

where N is the number of components in the system. The total number of variables in

a system of known composition is 2 + M(1 + N). With this many variables, only two of

which can be arbitrarily changed, there must be some equations relating them.

Given the composition of the system, we can always write down N mass balance

constraints. Generally, the mass balance constraints are written in terms of extensive

amounts of the phases and constituents:

m

x1

= n1 (12)

m

x2

= n2.

.

m

xN

= nN

where the summation index, , is over the number of phases in the system; = 1..M. The

mand ni are number of moles in the system. These extensive variables can be

converted to intensive ones by dividing by the size of the system, represented by:

Sn = ini (13)

After the division, the mass balance equations are:

z

x1

= y1 (14)

z

x2

= y2.

.

z

xN

= yN

where the z

= m/Sn and yi = ni/Sn are the mole fractions of the phases in the system

and the mole fractions of the components in the system, respectively.

Subtracting the number of mass balance equations, N, from the number of

variables leaves 2 + M + N(M 1). Duhem's theorem says only two of these can be

independently varied. Consequently, there must be an additional M + N(M 1) equations

we can write down. The additional equations are thermodynamic ones, reflecting the

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equilibrium conditions. The simplest and most general equations (see page 1) to follow

are equations relating chemical potentials:

i(P,T,xi

) = i

(P,T,xi

) (15)

Generally, petrologic modeling has been done with P and T fixed arbitrarily; the set ofnonlinear equations are then solved for the remaining variables.

Example of the Application of Duhem's Theorem

As a simple example of the application of Duhem's theorem, consider a ternary

chemical system consisting of a binary solid solution and a melt (Figure 2). The

composition of the system is known and is expressed as the mole fractions of two

components, C and B. The concentration of the third component, A, in the system is

given by:

yA = 1 yB yC (16)

The variables describing the equilibrium state of the system are T, P, the

composition of the solid, x, the composition of the melt, yB and yC, and the fractions of

each phase, solid and melt, in the system, zS

and zL, a total of seven.

Relating these variables are three mass or materials balance equations:

(1 x) zS

+ (1 yB yC) zL

= 1 yB yC (17)x z

S+ yB z

L= yB (18)

yC zL

= yC (19)

and two thermodynamic ones:

AS(P,T,x) = A

L(P,T,yB,yC) (20)

BS(P,T,x) = B

L(P,T,yB,yC) (21)

for a total of five equations in seven variables. As Duhem's theorem states, we must fix

two variables in order to calculate the others.

The two variables that are most commonly fixed in thermodynamic modeling of

igneous processes are P and T. Any two of the seven listed variables could be fixed,

however, and the remaining ones calculated. In addition, variables that are functions of

the seven listed could also be fixed. For example, the volume of the system is given by

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the appropriate modification of Equation (11):

V = vS

zS

+ vL

zL

(22)

If the molar volume of the system, V, is known, this last equation provides a

relationship between P, T, and the compositional variables because vS

and vL

are

functions of P, T and phase composition. Consequently, with V fixed, P or T and the

compositional variables can be calculated from the sets of Equations (17)-(19), (20)-(21)

and (22)).

The simplest and crudest approximation to the thermodynamic equations

[Equations (20) and (21)] is to use distributions coefficients and assume they are

approximately constant:

x/yB = Kd1 (23)

(1 x)/(1 yB yC) = Kd2 (24)

The distribution coefficients are either experimentally determined (e.g. Roeder

and Emslie, 1970) or derived from thermodynamic data sets that are, in turn, constructed

from experimental data (Ghiorso, et al., 1983; Nielsen and Dungan, 1983; Berman,

1988). Distribution coefficients are seldom, if ever, constants and to make reasonably

accurate models one must use a thermodynamic data set to evaluate the thermodynamic

equations.

The sets of equations that must be solved for the remaining variables, after fixing

two, are nonlinear. Nonlinear equations always create numerical and calculation

problems. Consequently, there is a considerable literature on how to solve the equilibrium

problem. The most efficient way is not to solve the set of nonlinear equations; rather, the

preferred method is to minimize the Gibbs energy, G, or some other thermodynamic

potential for the system (see, for example, Ghiorso, 1987). The numerical consequences

of the different approaches to this problem are significant (Press et al., 1986, p. 269-272).

The preferred method, however, does not guarantee that all physical-chemical

configurations can be modeled in this way.

If the degrees of freedom calculated with the Gibbs phase rule are zero or one ( i.e.

if the system is invariant or univariant) then the two independent variables cannot be

arbitrarily chosen; rather, the independent variables must be system variables. Both must

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be system variables if the system is invariant; one must be a system variable if the system

is univariant. Under any set of constraints, invariant, univariant, bivariant, or more, the

independent variables may be system variables. The restriction is on the number of non-

system variables, ones that are commonly called intensive variables.

For example, suppose the system consists of two polymorphs, A and B. Then

there are no chemical unknowns, both phases have the same composition, that of the

system itself. The unknowns are P, T, and the fractions of the two phases, zA

and zB. One

equation relating the variables is an expression of thermodynamic equilibrium between

the two phases:

G(P,T) = 0 (25)

where G(P,T) is the Gibbs energy change for the transformation of A to B. This

equation is a function of P and T only because the phases have the same composition and

is independent of the concentrations of the phases in the system, zA

and zB. Consequently,

it provides no information about the amounts of the phases in the system. Given P we can

calculate T and vice-versa. Consequently, we cannot arbitrarily fix both P and T and we

cannot calculate zA

and zB.

In order to calculate zA

and zB

we require a relationship that depends on how

much of each phase is present. Variables which depend on the amounts of the phases

present are system variables such as density, heat content and heat capacity. Others are

refractive index, compressibility, thermal expansion and functions of these variables such

as the difference between the sound velocities in the system:

Vp2

- (4/3) Vs2

= cp/[( cp - T v 2)] (26)

where Vp and Vs are the acoustic (seismic) wave velocities, cp is the molar heat capacity

of the system at constant pressure, is the density of the system, v is the specific volume

of the system, and and are the thermal expansion and compressibility of the system:

= (v/T)P/v (27)

= (v/P)T/v (28)

Suppose we select density as an independent variable and that we want to know

the state of the system at some fixed density. The reciprocal of the density is the specific

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volume of the system. The coefficient of thermal expansion for a phase, k, in the system

is given by:

k= (vk/T)P/vk (29)

where vk

is the specific volume of phase kand has units of m

3

/kg. Coefficients of thermalexpansion are often approximately constant over temperature ranges of a few hundred

degrees. Over a limited temperature range then, we can integrate Equation (29) to get:

vk= vokexp(kT) (30)

where vokis the specific volume of phase kat T = 0 and at P = 0. vkis then the specific

volume at any T in the limited temperature range and at P = 0.

The compressibility of a phase, k, in the system is defined as:

k= (vk/P)T/vk (31)

Like the coefficient of thermal expansion, compressibility is approximately constant over

a limited range of the controlling variable, pressure. If compressibility is approximately

constant, then Equation (31) can be integrated:

vk= C(T) exp(kP) (32)

where C(T) is the volume of phase k at zero pressure. Combining with Equation (30)

gives the specific volume of phase kat any P and T:

vk= vokexp(kT kP) (33)

The specific volume of the system will be a linear combination of the phase volumes (see

page 7, Equation (10):

v = zA

voA

exp(AT AP) + (1 zA) voB

exp(BT BP) (34)

where zkis the weight fraction of phase k in the system.

This last equation is a function of P, T, and the concentration of phase A in the

system. The concentration of B is one minus the concentration of A. By fixing either P or

T, we can calculate the other with the equilibrium relation [Equation (25)]. By fixing

density, we fix the specific volume of the system. Consequently, the weight fractions of

the phases can be calculated with Equation (34).

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Mathematical and Thermodynamic Preliminaries

Reactions between solid and melt are written in the following pages as

crystallization reactions. If component A enters the solid from the melt then the reaction

is written:

Am

= As

The equilibrium condition for this reaction is:

AsA

m= A = 0 (35)

where the subscripts designate the components and the superscripts designate the phase.

The standard states are the pure components in the solid or melt state. The

thermodynamic condition for the crystallization of the pure component is written:

AA* = A* (36)

where the superscripts and * indicate standard states for the solid and melt of

component A, respectively. We assume the Gibbs energies for the standard state

crystallization reactions are linear functions of temperature. This assumption is equivalent

to assuming there is no heat capacity change for the crystallization reaction.

Thermodynamics requires that:

i* = Hi* T Si* (37)

Because of our assumption, the heats of crystallization (Hi*) and entropies of

crystallization (Si*) must be constants. A melt is more disordered than a solid of the

same composition and, therefore, has a larger entropy (remember entropies are positive).

Consequently, the sign ofSi* for the crystallization reaction will be negative. At the

fusion temperature, i* is zero and, because T is always positive, we see that Hi* must

be negative also. This conclusion is consistent with the fact that heat is liberated when

substances freeze (latent heat of crystallization).

If the solid and melt solutions are ideal, the equilibrium condition for component

A can be written:

A + RT ln xAA* RT ln yA = 0 (38)

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where xA and yA are the mole fractions of component A in the solid and melt,

respectively. We will consistently use xi as a label for the concentration of a component

in a solid and yi as a label for the concentration of a component in the melt. The particular

component will be indicated by the subscript i. Equation (38) can be rearranged into a

frequently used form:

xA = yA exp(A*/RT) (39)

We will often use the abbreviation:

fA = exp(A*/RT) (40)

The derivative of Equation (40) is required when analyzing crystallization paths

and zoning profiles expected during crystallization. Its derivative is:

dfA/dT = fAHA*/RT2

(41)

We will use the abbreviation:

hA = HA*/RT2

(42)

The reader should note that Equation (41) is completely general in that it does not

depend upon i* being a linear function of temperature [See a thermodynamics text, for

example, Denbigh, 1981;, p. 90]. Hence, even in real systems, the derivative of fi is easily

calculated if the enthalpy of crystallization is known.

Writing Equation (40) in the form:

fi = exp(Hi*/RT + Si*/R) (43)

provides a convenient way to examine its properties. The function, fi, is always positive

because the exponential function is positive and the slope of Equation (43) is always

negative because the sign of dfi/dT [Equation (41)] is determined by the sign ofHi*,

which is negative.

At high temperatures, the argument of the exponential function in Equation (43) is

dominated by Si*/R. Because Si* is less than zero, the function becomes asymptotic to

a horizontal line which has an intercept less than one. The argument to the exponential

function is zero at the fusion temperature, T if, of component i and there the value of fi is

equal to one. At temperatures less than Tif the value of fi is greater than one and the

steepness of the slope is proportional to Hi*.

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A schematic diagram illustrating these properties is shown onFigure 3.

Liquidus Surfaces and Fractionation Curves

This section illustrates the calculation of isotherms, cotectic curves, and

fractionation curves in a simple ternary system. Suppose a ternary system, ABC. C

crystallizes in the system only as a pure solid whereas A and B form a complete series of

solid solutions. No other solid phases form in the system. Both the melt and the solid

solutions are ideal. The standard state Gibbs energies for the crystallization reactions of

the pure end members are:

A: A* = 45000 + 45 TB: B* = 60000 + 50 T

C: C* = 30000 + 20 TThe units are Joules, moles and Kelvins. Plots of the Gibbs energies of crystallization are

shown onFigure 4.

The problem is to construct a phase diagram for this system showing the

following features:

1. The cotectic between melt, solid C and AxB(1 x) solid solutions.

2. Isotherms on the liquidus surface.

In addition, we want to extract as much information as possible about

crystallization paths. Consequently, we illustrate the following:

3. Equilibrium crystallization paths for the following bulk compositions:

1. 40% A, 45% B, 15% C.2. 10% A, 85% B, 5% C.

4. Perfect fractionation curves that pass through the following bulk compositions:

3. 40% A, 45% B, 15% C.4. 10% A, 85% B, 5% C.

Construction of the Phase Diagram

The first step in the construction of a phase diagram is to find the fusion

temperatures for the end members of the solid phases that form in the chemical system

represented on the phase diagram. Equilibria between pure solid and pure liquid end

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member compositions occur at the fusion temperatures. At the fusion temperature, the

Gibbs energy for the crystallization reaction must be zero; hence, i* is zero for the

crystallization reaction at the fusion temperature for each end member:

A: 45000 + 45 T = 0; TA,f = 1000 KB: 60000 + 50 T = 0; TB,f = 1200 KC: 30000 + 20 T = 0; TC,f = 1500 K

Next, calculate liquidus temperatures for the A-C and B-C binary edges of the

diagram. Because no solid solution occurs in compositions on these edges, they can be

represented by simple eutectic diagrams. Along the A-C edge there are the following

exchange reactions:

A = A

Melt SolidC = C

Melt Solid

Because the solids are pure in our binary eutectic system, iS

= i. The melt, on

the other hand, is an ideal solution. Consequently:

iL

= i* + RT ln yi (44)

where yi is the mole fraction of component i in the melt. By setting iSi

L= 0, we get

the following equilibrium relations representing the crystallization reactions for the two

end members, A and C [See Equation (39), Page15].

yA = exp(A*/RT) (45)

yA = 1 exp(C*/RT) (46)

where yA is the mole fraction of component A in the melt.

The easiest way to find the liquidus compositions is to enter several values of T

into the Equations (45) and (46) and calculate the corresponding values of y A and yC.

Some results are shown in Table 1 and the A-C binary diagram is shown on Figure 5.

Notice that at 950 K the mole fraction of component A in the melt is the same for

both equations (0.752). This is the eutectic composition and 950 K is the eutectic

temperature on the A-C binary side of the ternary system.

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An alternative method of solving Equations (45) and (46) is to substitute the

expression for yA from Equation (45) into Equation (46) and rearrange it in the form:

1 exp(A*/RT) exp(C*/RT) = 0 (47)

This form is an implicit function of a single variable, T. It is, however, nonlinear andspecial techniques are required to solve it.

Table 1: Liquidus melt compositions on the A-C side of the ternary phase diagram.

T K yA Equation (45) yA Equation (46)

1500 0.000

1400 0.158

1300 0.3091200 0.452

1100 0.583

1000 1.000 0.700950 0.752 0.752

900 0.550 0.800

800 0.260 0.880

A similar procedure with the B-C binary returns a eutectic temperature of 1101 K and a

mole fraction of B in the eutectic melt of 0.582.

Slope of the Liquidus Curve

Not often considered in the experimental and theoretical construction of a phase

diagram are thermodynamic constraints on the slope of the liquidus curve or surface intemperature-composition space. Van Laar (1936, see Prigogine and Defay, 1954, p. 361)

showed that in a binary system with a compound showing no solid solution, the liquidus

surface near the compound's composition can have either a positive or negative curvature.

To repeat Van Laar's demonstration we start with Equation (45):

y = exp(*/RT) (48)

where we have temporarily dropped the subscript A in favor of a concise notation.

To study the nature of the curvature of the liquidus curve on the binary

temperature-composition diagram we need the slope (dT/dy) and the curvature (d2T/dy

2).

Solving Equation (45) for T, assuming H* and S* are constants, gives:

T = (H*/R)/(ln y + S*/R) (49)

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By differentiating Equation (49) and then back substituting from Equation (49) we

get:

dT/dy = - T/[y(ln y + S*/R)] (50)

d

2

T/dy

2

= [R

2

T

3

/ (y H*)2

](ln y + S*/R + 2) (51)The sign of the curvature will obviously depend on the sign of the last term in

Equation (51) because the first term, in square brackets, is positive. We are considering a

crystallization reaction, consequently, both H* and S* must be negative. Also, we are

interested in the slope and curvature at y = 1 where the temperature is the fusion

temperature for the pure component. At y = 1, then:

IfS*/R + 2 > 0, the curvature is positive.IfS*/R + 2 < 0, the curvature is negative.

Table 2: Liquidus slopes near the melting points of some silicate phases. The -quartz

melting point is metastable with respect to the melting point of cristobalite.

Phase S* Tf y' T'Cristobalite - 4.091 1996 0.221 491

-quartz - 28.934 1740 MetastableFayalite - 61.860 1490

Forsterite - 78.668 2163

Consequently, in order for the slope of the liquidus curve to be concave upwards

on a temperature-composition plot near the composition of the pure phase, S* must liebetween 0 and -16.6286 J/mole K (- 2 R). In addition, by setting the curvature, d

2T/dy

2, to

zero we can find the coordinates of the inflection point where the curvature changes from

negative to positive with increasing y:

y' = exp(- S*/R - 2) (52)

T' = - H*/(2 R) (53)

where T' and y' are the temperature-composition coordinates of the inflection point.

Some results for end members of petrologic interest are shown in Table 2 and

Figure 6. Cristobalite obviously has a small enough entropy of crystallization for a

concave upward liquidus curve. Other phases have entropies of crystallization of the

order of the olivine end members. Consequently, we expect most phase diagrams to have

liquidus curves with negative curvature, as commonly drawn. However, phase diagrams

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with cristobalite on the liquidus should have concave upward liquidus curves. This

feature is present on diagrams for Na2OSiO2 and K2OSiO2 [See Morey, 1964].

Calculation of the Binary Loop

The next problem is to calculate the binary loop for the A-B edge of the ternary

system. If we let:

xA = mole fraction of A in the solid

yA = mole fraction of A in the liquid

then the equilibrium relations for the components A and B are:

yA = xA exp(A*/RT) (54)

1 yA = (1 xA) exp(B*/RT) (55)

Table 3: Temperature - composition coordinates of the binary loop along the A-B side of

the ternary phase diagram.

T K yA xA

1200 0.000 0.000

1150 0.371 0.1831100 0.652 0.399

1050 0.857 0.662

1000 1.000 1.000

Solving for xA:

xA = [1 exp(B*/RT)]/[exp(A*/RT) exp(B*/RT)] (56)

Equation (56) is easiest to evaluate if xA is calculated at several temperatures

between the fusion temperatures of the end members. The mole fraction of A in the melt

is then calculated with Equation (54) or Equation (55). The results are shown onFigure 7

and in Table 3.

Liquidus Slopes on Binary Loops

The slope of the liquidus curve on a binary loop can be extracted from Equations

(54) and (55) by eliminating x and solving for yA:

yA = (1 f2)/(f1 f2) (57)

where:

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f1 = exp(A*/RT)f2 = exp(B*/RT)

Table 4: Fusion properties of some common rock-forming mineral end members. Units

are K, Joules and moles.

Tf H* S* ReferenceAnorthite 1830 135562 74.077 Weill, et al. (1980)Anorthite 1830 133000 72.678 Richet & Bottinga (1984b)Albite 1373 62800 45.739 Stebbins, et al. (1983)Albite 1373 64300 46.832 Richet & Bottinga (1984a)Sanidine 1473 57700 39.172 Stebbins, et al. (1984)Sanidine 1500 54000 36.000 Richet & Bottinga (1984a)Diopside 1670 137700 82.455 Richet & Bottinga (1984b)Enstatite 1834 77400 42.203 Stebbins, et al. (1984)Enstatite 1834 73200 39.913 Richet & Bottinga (1986)

Titanite 1670 123805 77.134 King, et al. (1954)Fayalite 1490 89300 59.933 Stebbins & Carmichael (1983)Forsterite 2163 114000 52.705 Navrotsky, et al. (1989)

Notice that on the binary AB edge of the ternary diagram, yA is equal to (1 yB).

Differentiation of Equation (57) gives:

dT/dyA = (f1 f2)2/[f2 h2(1 f1) f1 h1(1 f2)] (58)

where:

hi = Hi*/RT2

The typical phase diagram has composition plotted on the x-axis and temperature

on the y-axis. We look first for conditions where the liquidus is flat or vertical; in other

words, where dT/dyA is zero or infinite. The slope is zero if the numerator of Equation

(58) is zero or if f1= f2. Generally, the fi for two compositions become equal only outside

the temperature range of the binary T-x loop (Figure 8). Consequently, there is no

likelihood of having a horizontal liquidus on a binary T-x loop.

The second special case, a vertical slope, can occur if the denominator of

Equation (58) is zero. The denominator can be written:

(df2/dT)(1 f1) = (df1/dT)(1 f2) (59)

(See Equation (41),page 15).

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Our system was labeled such that component A melts at a lower temperature than

component B. The temperature range accessible to liquidus conditions of such a binary

loop is (TAf< T < TBf). Within this temperature range, (1 f2) is negative whereas (1 f1)

is positive (Figure 9). Consequently, the distribution of signs on the factors in Equation

(59) are:

[()(+)] =? [()()] (60)

Clearly, this is an impossibility: the left-hand side will be negative whereas the

right-hand side will be positive. A positive number cannot equal a negative number,

hence we have shown there cannot be a vertical liquidus in a binary loop.

A similar treatment of the solidus equation [Equation (56)] will show that it also

cannot be flat or vertical in the temperature interval between the fusion temperatures ofthe end members.

The liquidus curves of a binary loop can be either convex or concave upwards.

Examples are shown on Figure 8. To determine the conditions for a convex or concave

upward liquidus curve analytically is difficult because the second derivative with respect

to yA is required for Equation (58). Equation (58) is an implicit representation of the

liquidus curve; it is a function of T, not yA. Consequently, the chain rule would have to be

used and the resulting expression is complex enough to almost defy analysis. Fortunately,

there is a graphical method that works to some extent.

The Gibbs energies of crystallization for the end members are linear functions of

temperature in our simplified treatment. Consequently, graphs of the molar Gibbs

energies of crystallization for the two end members of a binary loop can have only one of

three configurations: The lines representing the changes on crystallization can intersect at

temperatures greater than the fusion temperature of the more refractory end member; they

can be parallel lines; or they can intersect at temperatures below the fusion temperature of

the less refractory end member(Figure 8).

Each configuration produces a unique shape for the liquidus and solidus. If the

intersection of the curves representing the Gibbs energies of crystallization occurs at a

temperature greater than the fusion temperature of the more refractory end member, then

the liquidus curve is convex upwards. It is also true for the case where the entropies of

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crystallization are equal and the curves are parallel. On the other hand, the liquidus curve

is concave upwards for the case where the lines representing the Gibbs energies of

crystallization intersect below the lower fusion temperature (Figure 8).

The heats of crystallization of several important end members are shown in Table4. Solid solutions between albite and sanidine could have a configuration for a binary

loop like the third case. However, the nonideal behavior required by the azeotrope and

the incongruent melting of sanidine may make such a configuration difficult to detect.

The olivines could also have a configuration like that predicted by case three. Again,

there may be complications because the intersection of the lines representing the Gibbs

energies of crystallization is at negative temperatures (K). Consequently, the linear

approximation for the Gibbs energies of crystallization is invalid and one would have to

investigate the effects of heat capacity terms on the slopes of the liquidus and solidus.

Location of the Ternary Cotectic

The location of the ternary cotectic (the line representing compositions in

equilibrium with pure C and A-B solid solutions) can be calculated next. There are now

three simultaneous equations, one for each component:

A: yA = xA exp(A*/RT) (61)

B: yB = (1 xA) exp(B*/RT) (62)C: yC = exp(C*/RT) (63)

In addition, we require that the sum of the mole fractions in the melt equal one:

yA + yB + yC = 1 (64)

Solving Equations (61)(64) for xA gives:

xA = [1 exp(B*/RT) exp(C*/RT)]/[exp(A*/RT) exp(B*/RT)] (65)

The ternary cotectic curve runs between the binary eutectic temperatures

calculated earlier. The easiest method for locating the curve is to calculate xA at a series

of temperatures between the binary eutectic temperatures with Equation (65). Then use

these values of xA and the chosen temperatures to calculate yA and yB with Equations (61)

and (62). The last mole fraction, yC, can be calculated with Equation (64). Some results

are shown in Table 5.

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Calculation of the Isotherms

Equations representing the isotherms in this system are particularly easy to find.

In the primary phase field of component C, the equilibrium relation can be written:

T = 30000/(20 R ln yC) (66)At constant values of the mole fraction in the melt of component C, the right hand side of

Equation (66) is a constant. Consequently, in the primary phase field of C, the isotherms

are parallel to lines of constant mole fraction.

To find the equations representing isotherms in the primary phase field of the A-B

solid solutions, Equations (61), (62), and (64) can be combined in the form:

yC = [(f1 f2)/f2] yA + (f2 1)/f2 (67)

where:

f1 = exp(A*/RT)f2 = exp(B*/RT)

If the temperature is constant, as it is on an isotherm, then Equation (67) has the

form of a straight line, yC = a yA + b. Consequently, isotherms in the primary phase field

of A-B solid solutions will be straight lines. These equations complete the description of

the phase diagram. We know the compositions and temperatures of the eutectics, the

liquidus and solidus of the binary solid solution diagram, the temperature andcomposition coordinates of the cotectic and the equations for the isotherms. The

completed phase diagram is shown on Figure 9.

Table 5: Composition and temperature coordinates of the ternary cotectic.

T K xA yA yC

1100 0.0000 0.0000 0.41691075 0.1213 0.1770 0.3864

1050 0.2525 0.3267 0.3567

1025 0.4008 0.4574 0.3280

1000 0.5707 0.5707 0.3004975 0.7676 0.6682 0.2738

950 1.0000 0.7514 0.2485

Linear isotherms are a consequence of the ideality assumption we are making and

the absence of intermediate compounds in the system. They do not result from the linear

form of the standard state Gibbs energies of crystallization. Rather they occur because T

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and composition do not occur as products or quotients in the equations describing ideal

solutions of compounds that are also components. Because the isotherms are linear in

simple ideal systems, the liquidus surfaces are cylindrical surfaces with generators

parallel to the isotherms.

Calculation of Crystallization Paths.

A crystallization path on a ternary diagram consists of a set of compositional

coordinates. This is true whether the path is one of perfect fractional crystallization or

perfect equilibrium crystallization. However, there are also the conditions, pressure and

temperature, to contend with. Generally, phase diagrams are isobaric, polythermal ones

when applied to petrological problems. Consequently, we actually need to construct a

crystallization path in temperature-composition space. The first step, then, is to find the

saturation temperature of the crystallizing assemblage.

At this point, it is an advantage to take stock of the number of equations and the

number of unknowns in those equations that we can put together. We assume the

composition of the initial melt on the crystallization curve is known. Consequently, the

mole fractions of the components in the melt are known. However, the composition of the

saturating solid and the temperature of saturation are both unknown. If the solid is pure,

then there is one equilibrium relation [Equation (63), for example]. The only remaining

unknown is temperature, which can be calculated with the aid of Equation (63). If the

solid is binary, then there are two equilibrium relations, [for example Equations (61) and

(62)]. There is only one compositional variable in a binary solution, however, because the

mole fractions of the two components must sum to one. Consequently, we can use the

two equilibrium relations to calculate temperature and the composition of the solid

solution.

It is easy to demonstrate that for any solid solution of N components we can write

equilibrium relations for each component. But the number of independent compositional

variables in the solid solution is N 1 because the sum of the mole fractions must equal

one. Consequently, we can always calculate the saturation temperature and composition

of a phase, at a given pressure and melt composition.

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The discussion in the last paragraph provides us with a means for determining in

which primary phase field a given composition falls:

Calculate the saturation temperature for all possible phases. The primary phase

field belongs to the solid with the highest saturation temperature.

For example, composition A with 15% C saturates with C at approximately 839 K

[Equation (63)] whereas the solid solution saturates at approximately 1104 K. If you have

constructed the phase diagram as outlined above, you can determine from the isotherms

that this last temperature is approximately correct.

The determination of the saturation temperature for a solid solution requires some

numerical analysis techniques because the equations are nonlinear. In our example, the

equilibrium relations for components A and B can be written:

xA = yA exp(A*/RT) (68)

xA = 1 yB exp(B*/RT) (69)

Equations (68) and (69) are a simultaneous pair in two unknowns, T and xA

(remember yA and yB are known). It is relatively easy to eliminate xA by setting the two

equal to each other. However, solving the equations for T (or eliminating T, which is the

same thing) is difficult because T occurs as part of the argument to the exponential

function. Consequently, iterative methods of solutions of systems of nonlinear equationsmust be used to find xA and T.

One can use these techniques to construct liquidus diagrams as functions of

pressure and temperature. At several pressures, one calculates the saturation temperatures

of the phases in the system. The points are plotted and connected with smooth curves.

These are the saturation curves for the phases in P-T space. The curve that lies at the

highest temperature at any pressure is the liquidus phase at that pressure. The metastable

portions of the saturation curves can be plotted also. Where the curves are separated bysmall ranges in temperature, at the same pressure, the phase with the curve at a lower

temperature than the saturation temperature of the liquidus phase will usually be the

second phase to saturate at the pressure of interest. An example is shown inFigure 11.

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Perfect Equilibrium Paths

The calculation of perfect equilibrium crystallization paths can be efficiently

organized by invoking Duhem's theorem. Remember that this theorem states that if we

know the bulk composition of an equilibrium system plus two other variables then all the

equilibrium properties of the system are fixed (i.e. they can be calculated).

The required compositional values for composition #1 are the percentages of A,

B, and C (40%, 45%, and 15%) in the system. Let's call these values NA, NB and NC. If

NM and NS represent the amounts of melt and solid solution in the system at some fixed

pressure and temperature (the other two variables we must know to satisfy the conditions

of Duhem's theorem) then conservation of mass requires that:

xA NS + yA NM = NA (70)(1 xA) NS + yB NM = NB (71)yC NM = NC (72)

In addition to the equations representing conservation of matter, there are the

equilibrium relations between components of the melt and solid solution. At constant

pressure and temperature these can be written in the following form:

yA = xA g1 (73)

yB = (1 xA) g2 (74)

where:

g1 = exp(A*/RT)g2 = exp(B*/RT)

In addition, we require that the sum of the mole fractions of the melt components equal 1:

yA + yB + yC = 1 (75)

Equations (70)-(75) constitute a system of six equations in six unknowns, NS, NM,

xA, yA, yB and yC. Some algebra can be used to eliminate all except xA to give:

xA2 [(g1 g2)(NA + NB + NC)] xA [(NA + NC)(g1 g2) + (NA + NC)(1 g2)] +

[NA(1 g2)] = 0 (76)

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Table 6: Perfect equilibrium path for composition #1. Calculated with Equations (73),

(74), (75) and (76).

T K xA yA yC

1100 0.249 0.408 0.158

1080 0.292 0.436 0.2011070 0.312 0.444 0.225

1060 0.329 0.447 0.2501050 0.345 0.446 0.276

1040 0.359 0.442 0.304

1030 0.371 0.435 0.332

Equation (76) is a quadratic equation in xA which will provide two values on solution.

Examination of the phase diagram (Figure 10) will show that the smaller value is correct.

It is then a straightforward matter to back substitute into Equations (70)-(75) to calculate

the remaining variables. Some results for composition #1 are listed in Table 6. The

perfect equilibrium crystallization path for composition A is plotted on Figure 10, curve

E.

Perfect Fractionation Paths

Although perfect fractionation can be approximated by a series of perfect

equilibrium crystallization calculations, such a procedure does not provide a means for

extrapolating to precursors of the present melt composition, assuming it was derived by

perfect fractionation. The use of fractionation curves for determining more primitive melt

compositions was described several years ago (Irvine, 1977; Pearce, 1978). At the time,

the lack of knowledge of the temperature and compositional dependence of the

distribution coefficients of components between crystal and melt precluded accurate

calculations.

Our first objective is to derive equations from which we can calculate the

composition of the liquid and solids at any stage of the fractionation process. The

fractionation curves of ternary phase diagrams are the geometric representation of the

equations that provide the compositions of the melts. We will examine the case where the

fractionating solid is a binary solution in components one and two. The mole fraction of

component one in the solid will be labeled x. Consequently, the mole fraction of

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component two in the solid will be (1 x). Components in the melts, whether ternary or

higher, will be labeled yi, i = 1,N.

The mole fractions in the melt are related to the numbers of moles of the

components in the melt by:

y1 = n1/(n1 + n2 + sn) (77)

y2 = n2/(n1 + n2 + sn) (78)

where the ni are the number of moles of component i in the melt and sn is the sum:

sn = ini, i = 3, N (79)N is the number of components in the chemical system. If the system is ternary, then sn is

simply equal to n3. We will define:

Sn = ini, i = 1, N (80)The two sums, Sn and sn, are related by:

Sn = n1 + n2 + sn (81)

Note that because there is only one fractionating phase containing components

one and two, sn is a constant. Therefore, the differentials of y1 and y2 are given by:

dy1 = (1 y1) dn1/Sn y1 dn2/Sn (82)dy2 = y2 dn1/Sn + (1 y2) dn2/Sn (83)

The definition of perfect fractionation requires that the change in melt

composition equal that of the crystallizing solids. In our problem this means that the

composition of the binary solution must be a function of the instantaneous changes of n1

and n2 in the melt:

x = dn1/(dn1 + dn2) (84)

The next step is to solve Equation (84) for dn2, substitute the result into Equations (82)

and (83), and eliminate dn1 from the resulting two equations:

dy2/dy1 = (1 x y2)/(x y1) (85)

Equation (85) is our fundamental equation. It describes the effect of perfect

fractionation of a binary solid solution on the composition of a melt. Notice that the

expression is independent of the mole fractions of all components except those that enter

the solid solution (i.e. the expression is independent of y3, y4,.....yN).

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Integration of Equation (85) will provide y2 as a function of y1. Consequently, if

we know the initial composition of the fractionating system, we can plot the fractionation

curve through the initial composition at all other values of y1. The integration cannot be

carried out, however, until the relationship between x and y1 is specified. This

relationship is necessarily a thermodynamic one and its exact nature will depend on the

degree of non-ideality of the solid and liquid solutions.

Equation (85) can also be derived from geometric arguments if the system is

ternary. Such a derivation provides an illustration of the compositional meaning of the

mathematical notation in Equation (85). A fundamental postulate of phase diagrams is

that liquid compositions move directly away from the instantaneous composition of the

crystallizing solid. The situation is shown on Figure 12.

The solid composition lies on the line between pure component one and pure

component two. Suppose the solid crystallizing has the composition of point S at

coordinates (x,y2). However, the binary nature of the solid solution requires that:

y2 = x + 1 (86)

The liquid composition, L, with coordinates (y1, y2), lies on the fractionation

curve (labeled F onFigure 10). The straight line connecting L and S is, by definition (see

above), tangent to the fractionation curve. The derivative of y2 with respect to y1 gives the

instantaneous slope of the fractionation curve. We need only equate the derivative with

the slope of the straight line between L and S to get:

dy2/dy1 = (y2 y2)/(x y1) (87)

Substitution of the value of y2 from Equation (86) gives the result obtained

earlier:

dy2/dy1 = (1 x y2)/(x y1) (88)

The second equation that describes the state of a system during a perfectfractionation process is one from which we can calculate the amount of uncrystallized

melt. As in most petrological calculations, we are not after the absolute amount of melt

but rather the fraction of the original melt that has not crystallized. This fraction, F, is

defined by:

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F = Sn/Sn (89)

where Sn is the number of moles of components 1, 2,....,N originally present in the

magma before fractionation.

Equation (89) is a definition rather than a practical recipe for calculating F.However, the definitions of y1, y2, sn and Sn [Equations (77) (80)] can be combined to

give:

F = (1 y1 y2)/(1 y1 y2) (90)

where the superscript, , indicates the values of the mole fractions in the melt prior to

fractionation.

Equations (85) and (90) are our main weapons for calculating more primitive melt

compositions. Before delving into the calculations a description of the petrologic

situation is in order. Suppose we have a lava or, better yet, a glass composition and

associated with this lava or glass is an olivine that is too Mgrich in its core to be in

equilibrium with the melt. If we call Mg2SiO4 component A (or 1) and Fe2SiO4

component B (or 2) then one way we can recognize this fact is to calculate an equilibrium

olivine composition with Equations (68) and (69). If the calculated value of the

equilibrium olivine differs significantly from the composition of the cores of the olivine

phenocrysts then presumably the crystals and melt were not in equilibrium.Consequently, the olivine is perhaps a cognate xenocryst that formed during an earlier

stage in the evolution of the lava which then fractionated olivine to achieve its present

state. What we want to do is calculate the composition of a melt that satisfies two criteria:

(1) The calculated melt is in equilibrium with the Mg-rich olivine.

(2) The observed composition of the lava or glass lies on a fractionation curve

that passes through the calculated melt composition.

Integral Form of Fractionation Curves

Our desired position is an integrated form of Equation (85). Only under special

circumstances, however, can an integrated form of Equation (85) be found with the result

expressed in elementary functions. In order to integrate Equation (85), xA has to be

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eliminated, say with Equation (68). This elimination introduces T which can, in theory be

eliminated with Equation (69). In other words, we would have to eliminate T between

Equations (68) and (69), solve the result explicitly for xA, and substitute into Equation

(85). If the distributions of components A and B were independent of T so that:

xA = Kd1 yA (91)

1 xA = Kd2 yB (92)Then under these circumstances Equation (88) becomes:

dyB/dyA = (Kd2 1) yB/[(Kd1 1) yA] (93)

Integration gives:

yB/yB' = (yA/yA')K (94)

where K = (Kd2 1)/(Kd1 1).

We know, however, that Kdi is equal to:

Kdi = exp(Gi*/RT) = exp[(Hf* T Sf*)/RT] (95)

In order for Kdi to be constant, the argument in the exponential function must be constant.

The only way this can happen is if the temperature is constant, an impossibility if

crystallization is constrained to fall on a fractionation curve. Consequently, this integrated

form can, at best, be only an approximation over a small temperature range and we might

as well go to numerical integration methods directly.

Numerical Integration of Fractionation Slopes

Our starting point is with Equations (68) and (69). These two equations contain

the composition of the olivine, x, temperature, T, and the mole fractions of components

one and two in the melt, y1 and y2, as potential unknowns. Because the composition of a

melt on the fractionation curve is, by hypothesis, the composition of the rock or glass, we

can determine y1 and y2. Equations (68) and (69) can be solved for x and T, the remaining

variables. The next step is to calculate the composition of a melt slightly displaced in

composition from the original but up in temperature along the fractionation curve. To do

this we slightly increase the value of y1 (the Mg-component of the melt) and call this

increment dy1. Because the Mg-component has a very much higher fusion temperature

than the Fe-component, increasing y1 will usually guarantee we are moving up the

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fractionation curve (for an exception to this statement see Ghiorso, et al., 1983, p. 125).

We next calculate the increment in y2, dy2, with Equation (88). The mole fractions of y1'

and y2' in the new melt are then given by:

y1'= y1 + dy1 (96)y2' = y2 + dy2 (97)

The remaining mole fractions in the new melt are easily calculated with the aid of

the definition of the fractionation stage F [Equation (89)]. The mole fraction of

component j, a component that remains in the melt during fractionation is given by:

yj' = nj/Sn (98)

Substitution for Sn from the defining equation for F gives:

yj' = nj/(F Sn) (99)

However, the number of moles of component j is the same in the new melt as in the

original, consequently, the original mole fraction of component j is given by:

yj = nj/Sn (100)

Combining the last two equations gives the expression for the mole fraction of a

component that remains in the melt and is not fractionated into the solids:

yj' = yj/F, j = 3,...,N (101)

The mole fractions of all the components in the melt on the fractionation curve,which is the precursor to the melt represented by the lava or glass, can be calculated with

Equations (96), (97), and (101). One then begins the process anew: first, calculate an

olivine composition and equilibrium temperature for the new melt with Equations (68)

and (69) then find another melt composition on the fractionation curve by incrementally

increasing y1 and y2. The process is repeated until the calculated composition of the

olivine in equilibrium with the primitive melt matches that of the Mg-rich core of the

olivine found in the rock. The values of y1 and y2 that result in this olivine composition

also determine the primitive melt composition when substituted into Equations (90) and

(101) and the calculations are complete.

To apply this discussion to our current problem, we first make a small change in

notation to Equation (88):

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dyB/dyA = (1 xA yB)/(xA yA) (102)

The first step of the calculations has already been done. From the calculation of the

saturation temperature and composition of the solid solution we know xA. The original

composition of the system provides yA and yB. Substitution into Equation (102) gives us

the slope of the fractionation curve at the composition of the original melt

(dyB/dyA = 1.94).

Next pick a fractionation step by choosing a value for dyA. This is possibly the

most critical step. If you choose too large a step, then the fractionation path will be

imprecisely located; if the step is too small, at best you will spend a lot of time and effort

calculating the path. In addition, if the calculation is part of a computer program, round

off error may cause the calculations to be wrong. Because the purpose of the problem is

to provide an idea of the concepts of calculating phase diagrams, let's choose a fairly

large step, say 0.1, for the change in yA. Then dyB can be obtained from Equation (102) as

0.194. Adjusting the values of the mole fractions and ensuring that the sum of the mole

fractions of the melt components is one gives:

yA = 0.500

yB = 0.256

yC = 0.244

When the new melt composition is plotted on Figure 10, it falls very close to the

1050 K isotherm. This value can be substituted into Equation (68) or Equation (69) and a

new value of xA calculated. The slope is again evaluated and another step along the

fractionation path is calculated. The result is shown onFigure 10 (Curve F). One can also

change the sign of the fractionation step, dyA, and calculate the fractionation path to

higher (that is, more primitive compositions). This part of the curve is labeled F'Figure

10.

Polybaric Fractionation Curves

To calculate isobaric fractionation curves, we need only vary one melt component

to step along the curve. Given an initial composition, there is only one fractionation path

that can be calculated. If a polybaric fractionation curve is required however, the system

of equations will have two independently variable compositional components and the

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path will not be unique for a given starting composition. If another thermodynamic

property of the melt can be set, then the pressure need not be fixed and there will be only

one independently variable compositional component. A unique polybaric fractionation

path can then be calculated which will reflect the constraint provided by the additional

thermodynamic variable. A tractable path to calculate is one in which there is no heat lost

from the melt; the only heat extracted from the magma would be that carried by the

fractionating solids.

Only the melt is on the isenthalpic path; fractionated solids are removed from the

system on crystallization. The derivation is for a system fractionating a single binary

solid solution.

The differentials of H and S as functions of P, T, V, S and number of moles in the

melt are given by:

dH = V dP + T dS + i dni (103)

and:

dS = (Cp/T) dT + (S/P)T,n dP + (S/ni)T,P dni (104)

where the summation index, i, runs from 1 to N.

In order to find dH as a function of P and T, we need to substitute for dS (as a

function of P, T and the ni) in Equation (103). In order to do this, we need to find an

expression for the coefficient of dP in Equation (104) in terms of melt properties (V, Cp,

etc.). The differential of the Gibbs function is given by:

dG = V dP S dT + i dni (105)

Applying Maxwell's relations gives:

(S/P)T,n = (V/T)P,n (106)

The coefficient of thermal expansion, , is related to the partial derivative of the volume

with respect to temperature by:

(V/T)P,n = V (107)

Substituting into Equation (104) gives:

dS = (Cp /T) dT V dP + (S/ni)T,P dni (108)

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A further substitution of this last equation into Equation (103) gives:

dH = V(1 T) dP + Cp dT + T (S/ni)T,P dni + i dni (109)

The coefficients of dni can be combined by noting that the partial derivatives of S

with respect to ni are partial molar entropies and are related to the partial molarenthalpies, hi, by:

i + T (S/ni)T,P = hi (110)

The general equation for changes in H due to changes in P, T and ni is given by the

following equation:

dH = V(1 T) dP + Cp dT + hi dni (111)

The next step is to introduce the constraints generated by a single, fractionating, binary

solid. The mole fraction of component 1 in the solid will be represented by x and the

corresponding mole fraction in the melt will be represented by y1. The mole fraction of

component 2 in the solid is (1 x) and is y2 in the melt. The size of the system is the sum

of the moles of each component and is given by:

Sn = ni (112)

Divide Equation (111) by Sn:

dH/Sn = (V/Sn)(1 T) dP + (Cp /Sn) dT + hi dni/Sn (113)

Introduce molar quantities:

v = V/Sn; cp = Cp /Sn (114)

to get:

dH/Sn = v(1 T) dP + cp dT + hi dni/Sn (115)

If only components 1 and 2 are crystallizing as a solid solution then Equation (115)

becomes:

dH/Sn = v(1 T) dP + cp

dT + h1

dn1/Sn + h

2dn

2/Sn (116)

The composition of the fractionating solid is given by:

x = dn1/(dn1 + dn2) (117)

Substituting this last equation into the penultimate one gives:

dH/Sn = v(1 T) dP + cp dT + [x h1 + (1 x) h2](dn1 + dn2)/Sn (118)

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If only a binary solid composed of components 1 and 2 is fractionating, then the change

in the size of the system is given by:

dSn = dn1 + dn2 (119)

The mole fraction of component 1 in the melt is given by:

y1 = n1/Sn (120)

Its differential is given by:

dy1 = dn1/Sn y1 dSn/Sn (121)

An analogous expression can be written for y2:

dy2 = dn2/Sn y2 dSn/Sn (122)

Adding Equations (121) and (122) gives:

(dn1 + dn2)/Sn = (dy1 + dy2)/(1 y1 y2) (123)

Substituting the result into Equation (118) gives:

dH/Sn = v(1 T) dP + cp dT + [x h1 + (1 x) h2](dy1 + dy2)/(1 y1 y2) (124)

Under conditions of constant H, Equation (124) can be written:

(1 y1 y2)[v(1 T) dP/dT + cp] = [x h1 + (1 x) h2](dy1/dT + dy2/dT) (125)

In a fractionating system, the derivatives of the mole fractions in the melt are not

independent, but are related by:

dy2/dT = [(1 x y2)/(x y1)] dy1/dT (126)

Combining these last two equations gives:

(x y1)[v(1 T) dP/dT + cp] = [x h1 + (1 x) h2] dy1/dT (127)

An example may make the use of Equation (127) easier to follow.

To minimize numerical problems, and cp will be set to zero in the following

illustration. Generally, is of the order 1 10-9 and T is of the order 1000. Consequently,

T is considerably less than one and the assumption of equal to zero is a reasonable

one. The molar heat capacity, cp, however is not zero and would need to be included in a

real calculation. With these stipulations, Equation (127) becomes:

dP = - [x h1 + (1 - x) h2] dy1/[v(x - y1)] (128)

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where hi is the standard state enthalpy of component i. The standard state molar

enthalpies can be used in place of the partial molar enthalpies because of the assumption

of ideal melt solutions.

Given an initial melt composition and an initial pressure, the calculation of thefractionation path begins with solving the pair of equations:

1s(T, P, x) = 1

m(P, T, yi) (129)

2s(T, P, x) = 2

m(P, T, yi) (130)

for x and T. A value for dy1 is selected and the change in concentration of component 2 is

calculated with:

dy2 = (1 - x - y2) dy1/(x - y1) (131)

The pressure change is calculated from Equation (128) and the cycle of calculations

repeated.

An example is shown on Figure 13. As expected, the polybaric, isenthalpic path

diverges from the isobaric path. As a result, the solid compositions fractionating from

derivative melts will be different along the two paths, even though the initial melt

composition and conditions were identical. The isobaric perfect fractionation path is

reversible. By changing the sign of dy1 at any stage of fractionation one reproduces points

on the path above the fractionation stage where the sign change was made. In contrast,the polybaric, isenthalpic path is not reversible. The isenthalpic path shown onFigure 13

was calculated from high pressure to low starting with a composition where the two

paths, isobaric and isenthalpic, diverge on the left-hand diagram (40% A, 45% B). Again

change the sign of dy1 and starting at the low pressure end of the original isenthalpic path

(32.6% A, 5.5% B) calculate a path under isenthalpic conditions; it does not exactly

coincide with the original. This is more apparent on the plot of melt mole fraction y1vs

solid mole fraction, x, [Figure 13(b)] where the compositions near the maximum on the

fractionation paths are distinct. Isenthalpic paths are not reversible paths [See, for

example, Waldbaum, 1971; Lewis and Randall, 1961, p. 46-51) whereas isentropic paths

are.

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Example from Yukon Territory

The lava flows from Volcano Mountain form one of the few reported occurrences

of Recent nephelinites in the Canadian Cordillera (Francis and Ludden, 1990; Fiesinger

and Nicholls, 1977; Nicholls, et al. 1982). They contain olivine phenocrysts,

clinopyroxene microphenocrysts, and olivine, pyroxene, microlitic nepheline and leucite

in the groundmass, and host ultramafic xenoliths (spinel lherzolites) and olivine

xenocrysts.

The curve labeled olivine on Figure 14is the saturation curve for olivine in a melt

with the composition of a Yukon nephelinite. An isenthalpic fractionation curve (Nicholls

1990) starting at the olivine saturation temperature at 2 GPa is shown with a dashed line.

The curve labeled clinopyroxene connects points where olivine and clinopyroxene

coexist with melts that formed by perfect fractionation of olivine under isobaric

conditions. The numbers on the points designated with plus signs are the calculated

olivine compositions that fractionated along the isenthalpic curve on ascent to the surface.

The numbers in italics on the points shown with open squares are the calculated olivine

compositions that coexist with clinopyroxene and melts derived by isobaric fractionation.

Observed olivine compositions in the phenocryst cores and groundmass are indicated

with bold numbers.

If the magma that formed this nephelinite had ascended along the isenthalpic path

shown with a dashed line on Figure 14, the expected product, according to the model,

would be a rock with a monomineralic olivine phenocryst assemblage and a glassy or

finely crystalline groundmass, as the isenthalpic decompression path never crosses the

clinopyroxene saturation curve. Introducing a small amount of heat loss, (i.e., deviating

slightly from a perfectly isenthalpic crystallization path) would produce

microphenocrysts of clinopyroxene.

The observed textures in the nephelinites show a monomineralic phenocryst

assemblage with small modal amounts of pyroxene microphenocrysts. Furthermore, the

expected compositions of the phenocrysts under isobaric crystallization conditions at the

pressures and temperatures modeled do not match the observed values but they nearly

match those predicted to form on an isenthalpic fractionation path (Figure 14). These

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observations suggest that the modeled nephelinite in particular and the other nephelinites

from Volcano Mountain crystallized under polybaric conditions and that loss of heat from

the magma during ascent was small. This interpretation of the crystallization path is

consistent the fact that lava flows usually erupt rapidly with little heat loss during

transport.

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x

P(bars)0 5 10 15 200.0

0.2

0.4

0.6

0.8

1.0

25

T = 300 Kw

T = 500 Ks

Figure 1: Diagram representing the equality of chemical potentials of H O in steam and2an ideal solution. The two phases are at different temperatures.


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L

SystemComposition

SA B

dS

dL

C

Mole Fraction, x

Sz = d / (d + d )S S L

Lz = d /(d + d )L S L

ThermodynamicEquations

S Lm = mA A

Unknown Variables - Number of Equations = 2

Variables1. x: Mole fraction component B in binary solid.2. y : Mole fraction component B in melt.B3. y : Mole fraction component C in melt.C

L4. z : Fraction of melt in system.

S5. z : Fraction of solid in system.6. T: Temperature.7. P: Pressure.

Known Values

y : Mole fraction component C in system.Cy : Mole fraction component B in system.B

Mass BalanceEquations

S(1 - x) z + (1 - y - y )B CLz = 1 - y - yB C

Figure 2: Example of the application of Duhem's theorem. Given the initial composition of thechemical system, y and y , and the values of two other variables, T and P, there are fiveB Cequations to be solved for the remaining variables; three mass balance equations and twothermodynamic ones. The remaining variables are the composition of the melt (y , y ), theB C

S Lcomposition of the solid (x, 0) and the fractions of each phase, z and z .


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fi

f1

f2

f 01

f > 12

(1 - f ) < 02

5

4

3

2

1

500 1000 1500 2000TK

TA,f TB,f

2df /dT = fD H /RT < 02 2 B

2df /dT = fD H /RT < 01 1 A

Figure 3: Diagram showing the properties and relationships for two curves of fvs Ti[f = exp(m */RT)]. In one instance, the parameters are: H * = 45000, S * = 45. Fori i A Athe other curve, the parameters are: H * = 60000, S * = 50. Slopes are shown at TB Bequal 1100.


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T800 900 1000 1100 1200 1300 1400 1500

25

20

15

10

5

0

-5-10

-15

-20

A

B

C

D m *i

Figure 4: Plot ofD m * for the three end members of the ternary system discussed in the text. Theicrystallization (or fusion) temperatures of the end members are given by the intersection of thecurves and the line where D m * = 0 (1000 for A, 1200 for B, and 1500 for C).i


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1500

1300

1100

900

0.0 0.2 0.4 0.6 0.8 1.0

yA

T

Te

ye

C A

Figure 5: The phase diagram for the AC binary side of the ternary diagram discussed in thetext. The eutectic temperature, Te, and composition, ye, are marked. The liquidus curves are

dashed where metastable below the eutectic temperature.


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0.0 0.2 0.4 0.6 0.8 1.00

500

1000

1500

2000

b -QuartzCristobalite

Mole Fraction SiO in Ideal Melt2

TK

Inflection point

Figure 6: Binary liquidus curves for cristobalite and b - quartz, assumingideal solution behavior in the melt and D c = 0 for the fusion reaction. Theplarge circle marks the inflection point on the liquidus curve for cristobalite.


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0.0 0.2 0.4 0.6 0.8 1.0

y , xA A

1200

1100

1000

TK

B A

F(x, T)

F(y, T)

Figure 7: The binary loop along the A-B side of the ternary diagram.


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-40

-20

-60

0

20

40

G

(kJ)

Dxl

TfA

TfB

TfA

principles of thermodynamic modeling of igneous processes

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