14 - Thermodynamics and Metamorphism

8/7/2019 14 - Thermodynamics and Metamorphism

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Thermodynamics and Metamorphism

Geology 212 Petrology Prof. Stephen A. Ne


is document last updated on NaN-Dec-NaN

Equilibrium and Thermodynamics

lthough the stability relationships between various phases can be worked out using the experimental

ethod, thermodynamics gives us a qualitative means of calculating the stabilities of various compounds

ombinations of compounds (mineral assemblages). We here give an introductory lesson in thermodyna

help us better understand the relationships depicted on phase diagrams.

heFirst Law of Thermodynamics states that "the internal energy, E, of an isolated system is constant".osed system, there cannot be a loss or gain of mass, but there can be a change in energy, dE. This chan

energy will be the difference between the heat, Q, gained or lost, and the work , W done by the system

o,

dE = dQ - dW (1)

Work, W, is defined as force x distance. Since Pressure, P, is defined as Force/surface area, Force = P x

urface area, and thus W = P x surface area x distance = P x V, where V is volume. If the work is done a

onstant pressure, then W = PdV. Substitution of this relationship into (1) yields:

dE = dQ - PdV (2)

his is a restatement of the first law of thermodynamics.

he Second Law of Thermodynamics states that the change in heat energy of the system is related to the

mount of disorder in the system. Entropy is a measure of disorder, and so at constant Temperature and

ressure:

dQ = TdS

hus, substituting into (2) we get:

dE = TdS - PdV (3)

he Gibbs Free Energy, G, is defined as the energy in excess of the internal energy as follows:

ttp://www.tulane.edu/~sanelson/geol212/thermodynamics&metamorphism.htm (1 of 17) [22/12/2004 01:09:35 p.m.]


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G = E + PV - TS (4)

ifferentiating this we get:

dG = dE +VdP + PdV - TdS - SdT

ubstituting (3) into this equation then gives:

dG = TdS - PdV + VdP + PdV - SdT - TdS

r

dG = VdP - SdT (5)

or a system in equilibrium at constant P and T, dG = 0.

we differentiate equation (5) with respect to P at constant T, the result is:

(6)

nd if we differentiate equation (5) with respect to T at constant P we get:

(7)

quation (6) tells us that phases with small volume are favored at higher pressure, and equation (7) tells u

at phases with high entropy (high disorder) are favored at higher temperature.

quation (5) tells us that the Gibbs Free Energy is a function of P and T. We can see this with reference

e diagram below, which shows diagrammatically how G, T, and P are related in a system that contains

ossible phases, A and B.



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n the diagram, phase A has a steeply sloping free energy

urface. Phase B has a more gently sloping surface. Where the

wo surfaces intersect, Phase A is in equilibrium with phase B,

nd GA = GB.

ext we look at 3 cross-sections through this figure. In the first,e look a section of G versus T at constant P, such as along the

ont face of the figure to the right.

n this section at constant P we see that at temperatures below TE phase A

as the lowest Gibbs Free Energy, G. At these temperatures, phase A is

able and phase B is not stable because it has a higher free energy. Noteat at TE the free energy of the phase A, GA, is the same as the free energy

f phase B, GB. At temperatures greater than TE phase B has a lower free

nergy than phase A, and thus phase B is stable.

ext, we look at a section of G versus P at constant T. In this section, GA

GB at PE. At pressures greater than PE phase B is stable because it has a

wer G than phase A. At pressures less than PE phase A is stable because

has a lower G than phase B.



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nally, we look at a cross-section across the bottom of the first figure.

ere we project the line of intersection of the Free Energy surfaces onto

e P - T plane. Along the line of intersection of the surfaces GA = GB.

he line separates two fields, one at low P in which A is the stable phase

nd one at higher P in which phase B is stable. This is a classic P-T phase

agram. The line represents all values of P and T where Phase A and

hase B are in equilibrium or where GA = GB. If the chemical reaction that

ccurs on this line is:

A B

hen we can write:

G = GB - GA = 0

ote that G is defined as Gproducts - Greactants for the chemical reaction as written above.

aking a similar approach, we can rewrite equation (5), above, as:

dG = VdP - SdT (8)

Where in general,

G = the change in Gibbs Free Energy of the reaction = Gproducts - Greactants

S = the change in entropy of the reaction = Sproducts - Sreactants

nd V = the change in volume of the reaction = Vproducts - Vreactants

t equilibrium, as we have just seen, G = 0, so from equation (8)

0 = VdP - SdT

earranging this equation yields

( 9)

his relation is known as the Clausius - Clapeyron Equation. It is important because it tells us the slop

e equilibrium boundary or reaction boundary on a Pressure versus Temperature phase diagram.



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We next look at two cases of chemical reactions. In the first case, the chemical reaction is between only

olid phases. In the second case a fluid or gas is involved as one of the products of the reaction.

olid - Solid Reactions

solid-solid reaction only involves the solid phases as both reactants and products, with no fluid phases

owing up in the chemical reaction. Most solid-solid reactions appear as straight lines on Pressure-

emperature diagrams. The reason for this comes from the Clausius-Clapeyron equation.

dP/dT = S/V

where S = the change in entropy of the reaction = Sproducts - Sreactants

and V = the change in volume of the reacti0on = Vproducts - Vreactants

general, both the entropy, S, and the Volume, V of any phase varies with temperature and pressure. Amperature increases, both S and V tend to increase (things become more disorganized at high temperatu

creasing the entropy and molecules vibrate more at high temperature, increasing the volume). Similarly

oth S and V tend to decrease with increasing pressure (less room to vibrate means better organization an

wer volume). In addition, the change in volume and entropy at any given temperature and pressure tend

e small. The net result of this is that for solid - solid reactions the effects of increasing temperature tend

e offset by the effects of increasing pressure, and thus dP/dT remains more or less constant. A curve wh

ope is constant is a straight line.

et's use these principles to analyze some solid - solid reactions, such as those

the Al2SiO

5phase diagram. Note that for the solid-solid reaction

ndalusite Kyanite, dP/dT is positive. This is what we expect, because

e product kyanite, occurs on the low T side of the reaction boundary and

ould have a lower entropy, making S positive. Increasing the pressure

uses a decrease in volume, so Kyanite should have lower volume than

ndalusite and thus V is also positive. With both S and V positive, the

ope of the boundary curve dP/dT is positive.

or the reaction Kyanite Sillimanite, the product Sillimanite occurs on the high T side of the bounda

nd thus SKyanite < SSillimanite, so S is negative. Since Kyanite occurs on the high P side of the boundar

urve, VKyanite < VSillimanite, so V is also negative. Thus, dP/dT is also positive.

ut, note that the reaction boundary for Andalusite Sillimanite has a negative slope on the diagram.

he product of the reaction, Sillimanite, has a smaller volume than the reactant Andalusite. So, V is

egative. But, Sillimanite occurs on the high temperature side of the reaction and thus has a higher entro

an andalusite. Thus, since the reactant, andalusite, has a lower entropy then the product, sillimanite, S

egative, making dP/dT negative.



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or the reaction Kyanite Sillimanite, the product Sillimanite occurs on the high T side of the bounda

nd thus SKyanite < SSillimanite, so S is negative. Since Kyanite occurs on the high P side of the boundar

urve, VKyanite < VSillimanite, so V is also negative. Thus, dP/dT is also positive.

ut, note that the reaction boundary for Andalusite Sillimanite has a negative slope on the diagram.

he product of the reaction, Sillimanite, has a smaller volume than the reactant Andalusite. So, V is

egative. But, Sillimanite occurs on the high temperature side of the reaction and thus has a higher entro

an andalusite. Thus, since the reactant, andalusite, has a lower entropy then the product, sillimanite, S

egative, making dP/dT negative.

evolatization Reactions

nlike sold-solid reactions, devolatilization reactions appear as curves on Pressure - Temperature diagram

see why this is true let's analyze the simple dehydration reaction:

A B + H2O.

or this reaction we can write:

S = SB + SH2O - SA = Ssolids +SH2O

nd

V = VB + VH2O - VA = Vsolids + VH2O

ncreasing temperature will generally cause S to be

ositive, especially for this reaction in which a gas or fluid

hase is produced, because gases always have a higher

ntropy (randomness) than solids. At low pressure

Vsolids will generally be negative. But at low pressure

H2O will be very large, because a fluid or gas will expand

fill space available. Thus, dP/dT will be positive. As

e pressure increases, the fluid or gas will be more

ompressible than the solids, so the total

V will becomecreasingly smaller. Thus, dP/dT will increase.

ventually because of the compressibility of the gas or fluid phase, Vsolids becomes equal to VH2O and

V of the reaction becomes 0, making dP/dT infinite. At pressures above this point, the fluid becomes s

ompressed that VH2O gets smaller and smaller, making V for the reaction increasingly negative. Thu

P/dT becomes increasingly negative.



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Calculation of Reaction Boundaries

nother relationship that is useful is:

G = H - TS

here G is the Gibbs Free Energy, H is the enthalpy, T is the absolute temperature in Kelvin, and S is thentropy.

or a chemical reaction, we can rewrite this as:

G = H - TS (10)

here again:

G = the change in Free Energy of the reaction = Gproducts - Greactants

H = the change in Enthalpy of the reaction = Hproducts - Hreactants

S = the change in Entropy of the reaction = Sproducts - Sreactants

n general G, H, S, and V are dependent of Pressure and Temperature, but at any given T & P:

G < 0 (negative) the chemical reaction will be spontaneous and run to the right,

G = 0 the reactants are in equilibrium with products,

nd ifG > 0 (positive) the reaction will run from right to left.

emperature Dependence of G, H, and S

s stated above, G, H, and S depend on Temperature and Pressure. But, because G depends on H and S,

sually more convenient to consider the temperature dependence of H and S, so that if we know H and S

ny given temperature, we can calculate G.



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where Cp is the heat capacity at constant pressure. The heat capacity is the amoun

f heat necessary to raise the temperature of the substance by 1o K.

hus:

r

f Cp is not a function of temperature, then further integration results in:

Note that in general, Cp is a function of temperature, and the known function of

emperature could be inserted before integration, but this introduces complication

hat are beyond the scope of this course).

ables of thermodynamic data are usually tabulated at some known reference

emperature and pressure, most commonly at a Temperature of 298 K, and Pressu

f 1 bar ( = 0.1 MPa ~ 1 atm). Thus, we if we need to know H at some temperatu

, other than 298 K, we can use the above equation to determine H at the new

emperature:

or a reaction, the above equation can be rewritten as:

(11)

he temperature dependence of entropy, S, is given by:



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r

Again, if Cp is not a function of T, then integration results in:

Or, since data are usually available at 298 K and 0.1MPa, for a reaction, this can b

written as:

(12)

quation 10 can then be combined with equations 11 and 12 to give the dependence ofG on temperatur

We can simplify this even further if we assume that for a reaction, Cp = 0:

(13)

hus, using the assumptions above, we can now calculate G at our reference pressure, P1 at any

mperature, if we know H and S at our reference temperature of 298 K.



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ressure Dependence of G andG

rom equation 6, above:

r, for a reaction:

ntegrating this, results in:

f we assume that V is not a function of pressure (in general this is a good

ssumption for solids, but not for liquids or gases, as we have seen above), then

ntegration results in:

(14)

Notice that the second term on the left in equation 14 is the same equation (13),

bove. Thus, we can substitute equation 13 into equation 14, and rearrange to get

ur final expression for G as a function of Pressure and Temperature:

(15)

gain, however, we must remember the assumptions involved in using this equation. The assumptions a

atCP = 0 and V is not a function of pressure.

alculation of Reaction Boundaries

we have thermodynamic data for minerals involved in a chemical reaction, then we can use equation 1

alculate reaction boundaries on a Pressure - Temperature diagram. Let's consider the following chemica

action:



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NaAlSi3O8 NaAlSi2O6 + SiO2

albite jadeite quartz

PhaseH298, 0.1Mpa

(joules/mole)

S298, 0.1Mpa

(joules/Kmole)

V298, 0.1Mpa

(cm3/mole)

albite -3921618.201 224.12 100.83

jadeite -3025118.24 133.574 60.034

quartz -908626.77 44.207 23.7

(Data from Berman, 1988)

or the reaction, at equilibrium, we can write:

H298, 0.1Mpa = Hjadeite + Hquartz - Halbite= -3025118.24 + (-908626.77) - (-3921618.201)

S298, 0.1Mpa = Sjadeite + Squartz - Salbite= 133.574 + 44.207 - 224.12

V298, 0.1Mpa = Vjadeite + Vquartz - Valbite = 60.034 + 23.7 - 100.83

hen, since at equilibrium, G = 0, we can write:

o be consistent we should have V in units of m3/mole, in which case P will have units of Pa (Pascals)

ut since our initial atmospheric pressure is 0.1 MPa, if we leaveV in units of cm3/mole, the results wi

ome out with pressure units of MPa (note that 103 MPa = 1GPa).

ugging in the values ofH, S, and V, as shown above, we can then plug in a value of T (in Kelvin)nd solve for P (in MPa).



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he results should look

omething like the graph

hown here, where Pressure

as been converted to GPa and

mperature is shown in oC.

he dashed line in the graph

hows the same boundaryalculated using a more

gorous approach, that is

king into account the

ariation ofH with

mperature and V with

ressure.

ote that the reaction boundary is still a straight line, as we would expect from our analysis of solid-solid

actions and the Clapeyron equation, as discussed above.

We next consider the more complicated case of a dehydration reaction. The reaction we will consider is:

KAl3Si3O10(OH)2 + SiO2 KAlSi3O8 + Al2SiO5 + H2O

muscovite quartz kspar sillimanite fluid

n this reaction, the Al2SiO5 mineral is shown as sillimanite, whereas in reality we would have to take int

onsideration that the reaction boundary may cross the stability fields of andalusite, sillimanite, and kyanor simplicity sake, however, we will just consider sillimanite.

ecause this reaction involves a fluid or gas phase, our assumption about V being a constant will more

an likely not be true, since we know that fluids and gases are easily compressed with increasing pressu

nd expand with increasing temperature, much more so than do solid phases.

s a first approximation, we note that for a gas phase, the ideal gas law states that:

PV = nRT

or

V = nRT/P

gain, showing that the volume of the gas is a function of temperature and pressure. Thus, substituting V

om the ideal gas law into equation 6, above, results in:



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ntegrating this from P1 to P2 results in:

r

or the gas phase we can let P1 be the reference pressure (0.1 MPa), and the value

f G can be at the pressure, P, and temperature, T, of the reaction, so that we can

ewrite this as:

(16)

f, as we did before, we calculate G for this reaction, including all phases includ

he gas as part of the H and S terms, including only the solids in the V term, a

xpressing the variation of the volume of gas as a function of pressure, as shown

bove, then:

H298, 0.1Mpa = Hkspar + Hsillimanite + HH2O - (Hmuscovite + Hquartz)

S298, 0.1Mpa = Skspar + Ssillimanite + SH2O - (Smuscovite + Squartz)

V298, 0.1Mpa = Vkspar + Vsillimanite - (Vmuscovite + Vquartz) =Vsolids

hen, substituting these values along with equation (16) into equation (15),



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f we assume that PH2O = P, then we can substitute to get:

(17)

his equation is difficult to separate so that T can solved for P or P solved for T.

With a spreadsheet, however, one can set up the equation so that for each T on can

uess at a value of P until one is found that makes G = 0. One can then make a

lot of P vs. T to show the reaction curve.

uch a plot is shown here. Again, we

hould review the assumptions

ehind this solution. We are

suming that Cp = 0, that V is

ot a function of pressure, that

llimanite is the Al2SiO5 mineral

roduced in the reaction, that

uscovite, kspar, quartz, and

llimanite are pure phases (not solid

olutions), and that H2O behaves as

n ideal gas and is the only gas

resent in the fluid phase.

ote how the reaction boundary is

urved and how it asymptotically

pproaches the temperature axis at

w temperature. This is typical of

ehydration and decarbonationactions.



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he assumptions about the lack of solid solution in the solids is pretty good to for these solids to a first

pproximation (with the possible exception of k-spar), but in the earth, it is likely that the fluid phase is n

ure H2O, and is more likely a mixture of H2O and CO2.

We can take this into account by substituting the partial pressure, pH2O of H2O in equation (17).

pH2O = XH2OP

here XH2O = the mole fraction of H2O in the fluid or gas phase. Equation (17) then becomes:

efore looking at the results of this equation, let's reason out what effects we shou

ee. If we look at the reaction:



e should be able to see that reducing the amount of H2O on the right-hand side of the equation should m

e reaction run from left to right, thus decreasing the stability of muscovite + quartz.

we now solve the equation, again

y setting up a spreadsheet so that

e can take guesses at the value of P

ecessary to reduce G to 0 at any

ven T, we get the graph shown

ere. It is notable that our reasoning

bout decreasing the stability of

uscovite + quartz was correct.

ecall that we said that most of the

hases with the exception of the

uid phase and possibly kspar in this

action are pure. Let's now

onsider what the effect would be on

e reaction curve if kspar was

ctually a solid solution between

ure KAlSi3O8 and NaAlSi3O8,



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hich of course it is in nature.

gain, looking at the reaction:



e should be able to see that reducing the concentration of KAlSi3O8 in the kspar solid solution should h

similar effect as reducing the concentration of H2O in the fluid phase. That is the stability curve wouldove to lower temperature at any given pressure, thus reducing the stability of muscovite + quartz.

y similar reasoning, if one of the reactants, say muscovite were not pure muscovite, but instead was a so

olution, the reaction would begin to run from right to left, thus increasing the stability of the products an

oving the reaction boundary to higher temperature at any given pressure.

nfortunately, we do not have time in this course to explore how impure phases are taken care of in the

ermodynamic treatment.

he basic principals of the thermodynamic treatment presented here allow metamorphic petrologists to w

ut reactions that can serve as geothermometers (to determine temperature) and geobarometers (to determ

ressure), and thus determine where in the earth metamorphism took place in a given suite of metamorph

cks.



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Reaction Rates (Kinetics)

hermodynamics can tell us what mineral phases are in equilibrium at a specific temperature and pressur

ut does not tell us anything about the rates at which chemical equilibrium is achieved. During prograde

etamorphism, temperatures are generally increasing, and rates of chemical reactions tend to increase w

creasing temperature. Thus, during prograde metamorphism there is usually ample time for equilibriu

ccur. But, as these minerals are brought back to the pressure temperature conditions present at the Earthurface, temperature will be decreasing. The rates of the reactions will thus be much slower, and phases

able at high pressure and temperature will often be preserved when we find the rocks exposed at the

urface. Thus, it is common to find metastable phases, i.e. phases that are not stable at the surface, but

present equilibrium at some higher temperature and pressure.

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14 - Thermodynamics and Metamorphism