Thermodynamics For Geologists 2012

1. Introduction and definitions

The mineralogies of rocks sampled at the Earth's surface reflect the pressures and temperatures at

which the rocks were formed and their chemical compositions. The methods of chemical

thermodynaics can be used to predict the most stable, or equilibrium, mineral assemblage of a rock

under and given pressure-temperature conditions.

It has long been apparent that chemical equilibirum is approached at some time during the

crytallization history of many metamorphic (e.g., Eskola, 1920) and slowly cooled igneous rocks, and

even during rapid cooling of volcanic rocks (e.g., Carmichael, 1967). The application of

thermodynamic principles to these rocks can therefore provide valuable insight into their histories.

In such cases it may be valid to assume equilibrium and to apply thermodynamic considerations in

order to determine physical conditions (pressure, temperature, etc.) of crystallization. It is apparent that

the volume of rock which is in equilibrium may be small because of the preservation of original

compositional difference on the cm scale even during high-grade metamorphism. Given the

fundamental assumption of chemical equilibrium we shall begin by defining the important

thermodynamic variables.

1.1 System

A system is simply a group of atoms, minerals, or rocks which are under consideration. The exact

position of the boundary of a system may be fixed at will. For example, it may be convenient to choose

as the system a whole outcrop; or a hand specimen, or a single mineral or even a fixed volume of ocean

water at a given depth. The boundary of a system is generally fixed in such a way that the minerals,

fluids (or gases—more on this distinction later) and melts within it may be regarded as having the

potential to be at equilibrium.

Changes that take place within a system may or may not involve interaction with surrounding material.

An isolated system is one which cannot exchange energy or matter with its surroundings. A closed

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system can exchange energy but not mass with its surroundings, e.g., isochemical metamophism. An

open system can exchange both energy and mass with its surroundings, e.g. if the system is sediments

at the base of a water column matter can move between the water and the sediments.

A system is made up of one or more phases. A phase is a restricted part of the system with distinct

physical and chemical properties. When considering rocks on often identifies several mineral phases,

e.g. all the olivine crystals constitute the olivine phase, all of the plagioclase crystals constitute the

plagioclase phase and so on. In addition there may be a molten phase and a fluid phase containing

H2O, CO2, S, F, Cl and CO.

1.2 Components

Each phase in a system may be considered to be composed of one or more components. For any

particular phase the components may be defined in a number of different ways. For example, in an

(Mg,Fe)2SiO4 olivine solid solution some of the possible sets of components are: (a) Mg2SiO4 and

Fe2SiO4, (b) MgO, FeO, SiO2, and (c) Mg2+, Fe2+, Si4+ and O2-. The choice of components is arbitrary

and depends upon the type of thermodynamic problem under consideration. Generally, if is most

convenient to take the end-members of solid solution series as the components of complex mineral

phases:

Olivine: Mg2SiO4, Fe2SiO4

Plagioclase: CaAl2Si2O8, NaAlSi3O8, KalSi3O8

Clinopyroxene: CaMgSi2O6, Mg2Si2O6, Fe2Si2O6, CaFeSi2O6, NaAlSi2O6

The reason behind the choice of these end-member formula units is that thermodynamic data are not

available for all possible choices of components. For example, the thermodynamic properties of pure

Mg2SiO4 olivine and NaAlSi3O8 plagioclase are well known, but those of Mg2+ in olivine and Na+ in

plagioclase are not; this is because the ions cannot be studied in isolation.

1.3 Chemical potential, µ

In order to determine the directions of chemical change in rocks it is necessary to introduce the concept

of the chemical potentials, µs, of the components present in phases. A chemical potential is analogous

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to an electrical or gravitational potential in that by measuring the µs of components in different parts of

the system, the tendency (or opposite) of material to flow or react can be deduced. The direction of

flow of components is always from regions of high chemical potential to those of low chemical

potential so that the system as a whole approaches its lowest possible energy state.

Consider a system consisting of two polymorphs of CaCO3, aragonite and calcite. The chemical

potential of the CaCO3 component in the calcite phase (µcalciteCaCO3) relative to that in the aragonite phase

(µaragoniteCaCO3) at different pressures and temperatures may be deduced from the calcite-aragonite phase

diagram (Fig. 1.1)

At the point X, in the calcite stability field, µcalciteCaCO3 is less than µaragonite

CaCO3 and material will flow

from aragonite to calcite until all of the aragonite disappears. At Y the reverse will occur, and at Z the

values of µcalciteCaCO3 and µcaragonite

CaCO3 are equal and both phases are stable and in equilibrium with each

other. In fact, this is a requirement of chemical equilibrium as we shall see.

To express the relationship between calcite and aragonite it is customary to write a balanced chemical

reaction in terms of the component (or components) of interest and to consider the relative values of µ

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in the manner discussed above:

CaCO3aragonite⇔CaCO3

calcite

Importantly, note how this chemical reaction is written. The reaction is written using components and

the phases in which the components are found are the superscripts. Chemical reactions are never

written as reactions between phases. Also note that the reaction is mass-balanced—there are equal

numbers of atoms on both sides of the reaction. In general, the components on the left hand side (lhs)

of reactions are considered to be the reactants (or the initial state) and those on the right hand side (rhs)

are the products (or final state). Following this convention the difference in µCaCO3 between products

and reactions can be calculated as

CaCO3=CaCO3products−CaCO3

reactants

or

CaCO3=CaCO3calcite

−CaCO3aragonite

Therefore in Figure 1 it can be seen that

at X, ∆µCaCO3 < 0

at Y, ∆µCaCO3 > 0

at Z, ∆µCaCO3 = 0

The last of these simple equations yields the fundamental condition for chemical equilibrium.

The chemical potentials of components are expressed in joules per mol (j mol-1). Because chemical

potential is a molar property it is independent of the actual number of moles in the system, this type of

thermodynamic variable, or property, is labeled intensive. Other examples of intensive properties are

temperature, pressure, and density.

The chemical potential is function of the state of the system. This means that the chemical potential of

any component i, µi, has a unique value if pressure, temperature and the bulk composition of the system

are fixed (which defines the state of the system). For example, consider Figure 1.1. If the system is

pure CaCO3, then at point X µcalciteCaCO3 and µaragonite

CaCO3 each have unique values. These values are

independent of the path the system followed to reach the temperature and pressure corresponding to

point X. The only way to change µcalciteCaCO3 and µaragonite

CaCO3 is to change the state of the system, i.e., by

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changing pressure or temperature or adding other components.

1.4 Gibbs Free Energy and chemical potential

In the general case of phases and systems consisting of more than one component it often becomes

necessary to consider the thermodynamic properties of the phase (or system) as a whole rather than the

properties of it constituent components. The sum of the chemical potentials of each component times

the number of moles of each component for all components in a phase or system is defined as the

Gibbs Free Energy:

Gtotal=∑ii ni ,

where ni is the number of moles of component i in the phase of interest and the summation extends

over all of the components in the phase. For an olivine solid solution consisting of n1 moles of

Mg2SiO4 and n2 moles of Fe2SiO4 the free energy of the olivine phase is given by

Gtotalolivine

=n1Mg2 SiO2

olivinen2Fe2 SiO2

olivine

It is apparent that G is a function of the total number of moles of components present in the phase or

system of interest. Properties, or variables, that depend upon the amount of material present are

extensive (e.g., mass, volume, heat content) .

Since the free energy of a multicomponent phase is a function of its composition, the chemical

potential of any component i in a phase may be obtained by partial differentiation of G with respect to

ni for constant amounts of all other components, nj's

∂Gtotal

∂n i

P , T , nj

=i

where the subscripted variables on the right: Pressure, P, Temperature, T, and the nj's are all held

constant during the differentiation (more on this topic later).

For a pure, one-component phase the following relationship is evident

G=n=G /n

So for a pure phase the chemical potential of its constituent component is equal to the molar Gibbs Free

Energy of the phase. Note that from the above equations the units for G are Joules.

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For most thermodynamic calculations it is convenient to work with molar properties of phases or of

components of phases (although there are some notable exceptions). For this reason Ga will refer to the

Gibbs Free Energy per mole of a phase a and µia will refer to the chemical potential of component i in

phase a. If phase a is composed purely of i then

Ga=ia .

Absolute values of Gibbs Free Energies of phases, or of components, cannot be determined. But,

thankfully this is not necessary. The Gibbs Free Energies of phases can be measured relative to an

arbitrary set of standard substances, usually either elements or oxides, and from this a scale of relative

Gibbs Free Energies or chemical potentials can be constructed. Since the equilibrium condition for

reactions is that the chemical potential difference is zero, it is only necessary to know the difference

between the µs of reactant and product components in order to determine whether the reaction proceeds

from right to left, or vice versa.

1.5 Equilibria involving more than one component

Equilibria in system containing more than one component can be considered in a similar manner that

that already discussed for the aragonite-calcite reaction. At high pressures, and moderate temperatures

anorthite decomposes to grossular, aluminosilicate and SiO2:

3 CaAl2Si2O8plagioclase ⇔ Ca3Al2Si3O12

garnet + 2 Al2SiO5kyanite + SiO2

quartz

Following the principles elucidated above ∆G can be defined:

G=Ca3 Al2 Si3 O12

garnet 2Al2 Si O5

kyanite SiO2

quartz−3CaAl2 Si2 O8

plagioclase

and at equilibrium where all components in the phases are stable

G=0 .

If ∆G is less than 0 the products are stable and if it is greater than 0 the reactants are stable. This is the

key to predicting thermodynamic equilibrium. Note in the above calculation of ∆G the chemical

potential of the individual components must be multiplied by the number of moles of the components

in the reaction.

However, in rocks the garnets are never pure grossular (Ca3Al2Si3O12) and the plagioclases are never

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pure anorthite (CaAl2Si2O8). In both cases these multicomponent, solid solutions are of widely variable

compositions; the composition of the phase affects the chemical potential of the component.

Nevertheless, whenever there is equilibrium between the plagioclase solid-solution, the garnet solid-

solution, kyanite (almost always nearly pure Al2SiO5) and quartz (also almost always nearly pure SiO2)

then the ∆G for the reaction is zero. This condition applies regardless of the complexity of the phases

in the rocks and the type and number of other components and other phases in the rocks.

Although the chemical potentials of components in solid solutions are not the same as in pure phases

(more on this later), the reactions and equilibrium conditions may be considered in terms of simple

end-member components and the equilibrium condition. The consideration of equilibrium relations

between the various simple components in a multicomponent natural rock is an extremely powerful

tool when applied to geological problems because often multiple reactions between multiple

components can be written and used to constrain the conditions at which the rock formed.

1.6 Determination of µ and G

In order to determine the equilibrium conditions for a reaction between components the numerical

values of µ and G must be known for the components and phases of interest. Although the absolute

values of these quantities cannot be determined, the relative values can be obtained by specifying an

arbitrary reference. The fundamental equation that specifies the Gibbs Free Energy of a system or a

phase is:

G=H−TS ,

where H is the enthalpy, or heat content of the system or phase, T is the absolute temperature (on the

Kelvin scale) and S is its entropy.

It is important to recognize the difference between the heat content and temperature, which are often

used interchangeably in everyday language. The heat content, that is enthalpy, is a measure of the

internal energy, E (sometimes U is also used) plus the pressure, P, multiplied by the volume, V, of the

material:

H=EPV .

Its units are in J or J mol-1. On the other hand, temperature is a measure of the hotness or coldness of

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an object. One of the more rigorous relationships defining temperature from statistical mechanics

relates it to the average kinetic energy of a gas divided by Boltzmann's constant, k:

T=m v2

3 k,

where m is the mass of a gas molecule and v is its velocity. Classical thermodynamics provides other

definitions of temperature that need not be dealt with at this time.

The enthalpy of the system is usually expressed relative to the enthalpies of the constituent elements,

which by convention are taken to be zero at 1 bar and the temperature of interest (i.e., any temperature).

Thus a scale of relative enthalpies of compounds of elements can be established (and looked up in

tables). Consider the enthalpy of pure forsterite, Mg2SiO4. Because the heat contents of Mg (crystal),

Si (crystal) and O2 (gas) at 1 bar are defined to be zero, the enthalpy of Mg2SiO4 is obtained from the

reaction

2 Mg + Si + 2 O2 ⇔ Mg2SiO4

and calculated by subtracting the values of the reactants from those of the products:

∆Hreaction = Hforsterite – (2 HMg + Hsi+ 2HO2)

= Hforsterite – 0 = -2175.68 kJ at 1 bar, 298 K

But how is Hforsterite determined? Conceptually the measurement of Hforsterite is simple: Dissolve one

mole of Mg2SiO4 in acid, or a solvent composed of an oxide melt, at a specific temperature and

measure the heat generated; that heat is Hforsterite. However, the experimental procedure is vastly more

complex than this sentence implies. Thankfully, in many, if not most, cases the necessary values for

thermodynamic calculations have already been measured. Note that the negative sign for Hforsterite in the

reaction indicates that heat is given off during the formation of forsterite from its consituent oxides; this

type of reaction in which heat is liberated is an exothermic reaction, a reaction in which heat is

absorbed is an endothermic reaction.

The entropy of a system (or phase), S, may be regarded as a measure of the degree of randomness or

disorder in the system. The greater the degree of disorder the higher the value of the entropy. As an

example let us consider a system of a large volume containing a small amount of a gaseous species.

The probability of finding a molecule of the gas within any given volume of the system is very small

because a few molecules are dispersed throughout a large volume. The degree of randomness of the

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system is therefore large and the molar entropy of the gas is also large. As the volume of the system is

decreased the pressure of the gas increases, pressure being inversely proportional to volume, and

intermolecular distances decrease. By decreasing the total volume of the system the probability of

finding a molecule in any given volume of gas is increased and entropy or randomness is therefore

decreased. If pressure is increased sufficiently, the gas may liquefy or even solidify. Using this

argument, it is apparent that molecules in the liquid state with low volume and high density should

have lower molar entropy than those in the gaseous state, and the the molar entropy in the solid state

will, in general, be lower still. In addition to the increased molecular density in liquid and solid states

relative to the gaseous state, the translational and rotational degrees of freedom of molecules (also

related to randomness) are much lower in the liquid and solid states than in the gaseous state because of

the existence of stronger intermolecular forces.

As a rule of thumb, the following relative entropy scale may be constructed:

(Sgas)low P > (Sgas)high P > Sliquid > Ssolid

Since G is an extensive property both H and S are extensive properties. However, because it is

intended to use molar values of G throughout the rest of this book, Ha and Sa will henceforward be

taken to refer to the enthalpy per mole and entropy per mole of phase a. The units of Ha are J mol-1

and of Sa are J mol-1 K-1. The analogous expression to that for G, but in this case for µ is

ia=H

_

ia−T S

_

ia ,

where H_

ia is the partial molar enthalpy of i in phase a and S

_

ia is the partial molar entropy of i in

phase a. The partial molar enthalpy of component i is the enthalpy per mole of this component in the

phase of composition a. If phase a only consists of i, as with G and µ, the molar enthalpy of a is equal

to the partial molar enthalpy of i in a. But, if phase a is a mixture of components, this equality does not

apply (see Chapter 3). The same definition and relations also exist for the partial molar entropy.

Consider the reaction of forsterite and quartz to form enstatite (all phases are pure):

Mg2SiO4forsterite + SiO2

quartz ⇔ Mg2Si2O6enstatite

The change in enthalpy of the system produced by reacting 1 mole of forsterite with 1 mole of quartz is

given by:

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∆H = Henstatite – Hquartz – Hforsterite

For example at 900 K, 1 bar:

∆H = -3 091 930 – (905 422 + 2 172 360) = - 14 142 J

Note the negative sign indicating an exothermic reaction that gives off heat (which remember is energy

). ∆H is the heat evolved (or absorbed) as a reaction takes place at constant pressure. In a similar

manner the entropy change for the reaction is:

∆S = Senstatite – Squartz – Sorsterite,

and at 900 K, 1 bar:

∆S = 360.33 – (109.16 + 258.91) = -7.74 J K-1

Note that the results of the calculations do not have mols in the units. This is because when each of the

values of H and S in the above equations are multiplied by the stoichiometric coefficients (number of

mols) of the components involved in the reactions. Thus for a general reaction of the type:

x A + y B ⇔ z C + a D

∆H = zHC + aHD – (xHA + yHB)

∆S = zSC + aSD – (xSA + ySB),

∆G = zGC + aGD – (xGA + yGB)

Note the general rule of “Products – Reactants” in these and all thermodynamic calculations.

In order to determine whether or not forsterite would react with quartz to form enstatite or vice versa at

a given set of P, T conditions it is necessary to know ∆G at those chosen conditions:

G=Mg2 Si2 O6

enstatite−Mg2 Si O4

forsterite−Si O2

quartz

and because in this example the phases are all pure:

G=Genstatite−GforsteriteGquartz =H−T S

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and by analogy with the case of calcite and aragonite:

∆G < 0 reaction proceed to right

∆G > 0 reaction proceed to left

∆G = 0 reaction at equilibrium

For this specific example at 900 K, 1 bar:

∆G = - 7 176 J

and the reaction proceeds to the right and enstatite is produced from forsterite and quartz.

Hess's law

A useful property of thermodynamic calculations is that they can be mathematically manipulated. Such

manipulations allow us to calculate thermodynamic values that cannot be directly measured. A good

example is the formation of diamond from graphite because of the high temperatures and pressures

necessary for:

Cgraphite ⇔ Cdiamond ∆H1

However, we can calculate ∆H1 from other reactions involving the reactants and products. Consider the

two simple-to-perform reactions:

Cgraphite + O2, gas ⇔ CO2, gas ∆H2

Cdiamond + O2, gas ⇔ CO2, gas ∆H3 .

We can reverse the second reaction

CO2, gas ⇔ Cdiamond + O2, gas -∆H3

and then add the graphite combustion reaction with the reversed diamond combustion reaction to make

Cgraphite + CO2, gas + O2, gas ⇔ Cdiamond + O2, gas + CO2, gas

∆H = ∆H2 – ∆H3

But each side of the reaction contains CO2, gas + O2, gas and so, similarly to mathematical equations, they

cancel each other out to produce the final reaction:

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Cgraphite ⇔ Cdiamond ∆H1 = ∆H2 – ∆H3.

This is know as Hess's law, and the concept is frequently used to calculate values that cannot be

directly measured. All extensive variables (G, S, H, and V, or volume) are additive and they can all be

determined in analogous indirect ways.

1.7 Temperature dependence of H and S, µ and G

If a substance is heated at constant pressure, then its heat content, or enthalpy, increases proportionally

to temperature. The relationship between the temperature increase and enthalpy increase is different

for different substances and is expressed by a characteristic constant of proportionality, the heat

capacity, Cp (strictly speaking this is the heat capacity at constant pressure; Cv, the heat capacity at

constant volume also exists but we shall not be using it, much . . . ). If the substance is heated through

an infinitesimally small temperature interval of dT then the infinitesimally small increase in enthalpy,

dH is rigorously defined by:

dH = Cp dT.

Because enthalpies of reactants and products change with temperature, the enthapy change of a reaction

(e.g., A ⇔ Β) is also temperature dependent:

dHA = (Cp)A dT; dHB = (Cp)B dT;

d∆H = (Cp)B dT - (Cp)A dT = ∆Cp dT

where ∆H = ΣΗproducts - ΣHreactants and ∆Cp = Σ(Cp)products – Σ(Cp)reactants. In tables of thermodynamic data it

is common to give enthalpies of materials at a reference temperatures, usually, although not always, at

298.15 K. In order to obtain the enthalpy at higher temperatures the above expression for dH must be

integrated from the reference temperature to the temperature of interest:

∫T

298dH= ∫

T

298C p dT or HT=H298 ∫

T

298C p dT .

If Cp is constant and therefore independent of temperature (which it rarely is) it can be removed from

the intregral and

HT=H298Cp T−298 .

However, in most cases Cp is a complex function that goes to 0 at 0 K:

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To express such a function are typically expressed as a polynomial. A simple example (e.g., Kelley

1960) would be:

Cp=abTc

T 2,

where the a, b and c are experimentally determined constants for the phase (usually pure) of interest

and T is the temperature in K (as usual for thermodynamics). Thus to determine the enthalpy of most

substances the polynomial must be integrated:

HT=H298 ∫T

298abT

c

T2dT ,

which when integrated yields

HT=H298[a12

bT 2−

cT ]

298

T

Example/Exercise 1. Determine the enthalpy of (low) albite at 900 K and 1 bar given the value at 298

K and 1 bar as well as the other data in the following table.

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Table from W. White Geochemistry textbook

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Increasing the heat content of a phase also leads to an increase in the entropy. If the amount of heat,

dH is supplied at constant pressure then:

dS=dHT

and because dH = CP dT we find

dS=Cp dT

T

which yields

ST=S298 ∫T

298

C p

TdT

or, using the previous temperature dependent expression for Cp

ST=S298 ∫T

298

aTb

c

T 3dT

which when integrated produces

ST=S298[a lnTbT−c

2T 2 ]298

T

.

Example/Exercise 2 Calculate ∆H800, ∆S800 and ∆G800 for the following reaction at 1 bar:

NaAlSi3O8albite ⇔ NaAlSi2O6

jadeite + SiO2quartz

using the information in the table above. In which direction will the reaction proceed? Compare the

calculated ∆G800 from what you would calculate if ∆Cp = 0. (Note that this does not mean that the

invividual Cp's of the reactants and products are zero, but that their sum is equal to zero. Also, the

results you find for this example are applicable to virtually all solid-solid reactions, i.e., those only

involving minerals and not involving gases, fluids, or melts.)

1.8 Pressure dependence of µ and G

At constant temperature T, the effect of pressure on the free energy of phase a (Ga) is given by

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∂Ga

∂ P

T

=V a .

Similarly the change in chemical potential of component i in phase a is equal to the partial molar

volume of component i in phase a:

∂i

a

∂P

T

= V ia .

For a reaction involving a volume change, ∆V, where

V=V products−V reactants

the analogous expression is

∂G∂P

T

=V

Integration of these expressions from a reference pressure (almost always 1 bar) at a given temperature

high pressure allows the calculation of Ga, µi and ∆G at the temperature and pressure of interest:

GP ,T=G1 bar ,T ∫P

1 barV dP .

1.9 Temperature and pressure dependence of µ and G

The Gibbs Free Energy of a reaction at any temperature and pressure may be calculated by combining

the equations for the temperature dependence with that for the pressure dependence:

GP ,T=H 1 bar, T ∫T

298C p dT−T S1 bar, T ∫

T

298

Cp

TdT ∫

P

1 barV dP

Note that this expression implicitly assumes that ∆Cp is pressure independent.

For solid-solid reactions the simplifying assumption that is commonly made (and is often an excellent

approximation) is that the ∆V is independent of temperature and pressure:

∫P

1 barV dP=P−1V

However, situations exist in which the effects of isobaric thermal expansion, α, and isothermal

compressibility, β, on the reactants and products must be taken into account; the isobaric thermal

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expansion is defined as:

=1VdVdT

P

and the isothermal compressibility as:

=−1V

dVdP

T

In these cases

dV =V dTdP

and

V=V 1 bar, 298T−298P−1

Unfortunately, α and β are not constants and the integration of the above equation can be complicated.

1.10 Calculation of a reaction boundary

Consider the reaction

NaAlSi3O8albite ⇔ NaAlSi2O6

jadeite + SiO2quartz .

At equilibrium we know that :

G=NaAlSi 2 O6

jd SiO2

qz −NaAlSi 3O 8

ab =0

which, if all phases are pure (more about impure phases later), is equivalent to

G=G jdGqz−Gab=0

Using the table of data given above we calculate:

∆H1 bar, 298 = -1.569 kJ and ∆S1 bar, 298 = -35.23 J K-1

Note that enthalpies are almost always in kJ and entropies in J. It is recommended that enthalpies be

converted to J in order to avoid errors in subsequent calculations. Furthermore, numerical values

should not be rounded off until the final calculation is performed.

We can now determine which side of the reaction is stable at 1 bar, 298 K, but we want to calculate the

equilibrium of this reaction at various temperatures and pressures. Thus, we need to perform the

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integrations discussed above to convert from these values at 1 bar, 298 K to high pressures and

temperatures.

If we were interested in the equilibrium at 298 K and high pressure the appropriate equation is:

GP , 298=H 1 bar, 298−298S1 bar, 298 ∫P

1 barV dP=0

We can compare this expression to that for a point that is infinitesimally higher it temperature, T + dT,

and in pressure, P + dP :

GP+dP,T+dT=H 1 bar, 298C pdT−298dT S1 bar, 298C p

298dTdT ∫

P+dP

1 barV dP=0

Equating these last two equations (because they are infinitesimally different in P and T) yields

− S dTV dP=0

or, the famous Clausius-Clapeyron equation:

dPdT

=SV

which gives us the slope of the reaction in P-T space at any specified temperature and pressure.

A simplifying assumption that ∆Cp =0 implies that ∆H and ∆S are constant for all temperatures and

allows the simple calculation of the approximate reaction boundary. Thus, at equilibrium and 1 bar:

G=H 1 bar−T SH 1 bar=T S

T=−1569 J

−35.23 J K−1=44.5 K

At higher pressure the boundary can be calculated by:

G=0=H 1 bar−T SP−1V

or, the P,T point at 1 bar (T = 44.5 K) can be extrapolated to higher pressure using the Clausius-

Clapeyron equation and assuming no change in ∆S or ∆V with pressure and temperature, which for this

particular example reaction results in:

P=1dPdT

T−44.5 .

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This assumption is often good for solid solid reactions, but bad for reactions involving melts and

terrible for reactions involving gases or fluids. For these reactions, and if we want the most precise

calculations for solid-solid reactions the complete equation, with all of its integrals must be solved, i.e.,

GP , T=H 1 bar, T ∫T

298C p dT−T S1 bar, T ∫

T

298

C p

TdT ∫

P

1 barV dP

at various temperature and pressure conditions. Thankfully, this is now easily done by computer.

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