Thermodynamics: The Science of Energy Relations

Thermodynamics

: the branch of 

physical science 

that

deals with 

the relations between 

heat and other forms

of energy 

(such as 

mechanical, electrical, or chemical

energy), and, by extension, of 

the relationships and

interconvertibility

 of all forms of energy.



A 

system

 is some part of the universe that you want to study

and understand

•

The 

surroundings

 are everything else in the universe that is not

in our system

•

The 

system

 can be 

open

or 

close

d

 to 

(

isolated

 from)

 the

surroundings in terms of both matter and energy

Natural systems tend toward states of

minimum energy

System/components definitions 1

Some definitions



The 

Phase 

is any 

mechanically separable 

and 

chemically

homogenous 

portion of the system



Component

 is minimum number of constituents required to

describe system.

System/components definitions 1

Some definitions

Some definitions

•

 In our case, a 

system

 is likely to be a mineral or a rock

•

 In this case, it is comprised of chemical 

components

 that 

describe chemical

variability in that mineral or a rock

•

 Typical 

components

 might be FeO, MgO and SiO

2

 used to describe olivine

•

 A 

phase

 is any mechanically separable and chemically 

homogenous portion of

the system, e.g. a melt, a fluid, or a mineral in a rock

•

 A 

reaction

 is anything that changes the nature of the 

phases within a system

System/components definitions 2

Some definitions

•

 Thermodynamics is primarily concerned with macroscopic energies of microscopic

processes that we might or might not fully understand.

•

 It is convenient to group all of the variables required into two classes:

Intensive

variables

 are independent of the amount of 

material present:

e.g. Pressure (

P

)

& Temperature

 (T)

Extensive

variables

 are dependent on the amount of 

material present:

e.g. Volume (

V

) & Entropy (

S

)

System/components definitions 3

The 

First Law of Thermodynamics states that "the internal energy, E, of an

isolated system is 

c

onstant"

.

 In a 

closed system

, there cannot be a loss or gain of mass, but there can be a

change in energy, dE.

dE = dQ - dW                 (1)

Q=heat

W=work done by the system

W=force x distance.

Pressure, P= Force/surface area,

Force = P x surface area,

Therefore, W = P x surface area x distance = P x V

V is volume.

If the work is done at constant pressure, then dW = PdV.  Substitution of this

relationship into (1) yields:

dE = dQ - PdV                 (2)

This is a restatement of the first law of thermodynamics.

The 

Second Law of Thermodynamics states that the change in heat energy of the

system is 

related to the amount of disorder in the system.

Entropy is a measure of disorder,

and so at constant Temperature and Pressure:

dQ = TdS

Thus, substituting into (2) we get:

dE = TdS - PdV              (3)

The 

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

follows:

G = E + PV - TS                         (4)

Differentiating this we get:

dG = dE +VdP + PdV - TdS - SdT

Substituting (3) into this equation then gives:

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

or

dG = VdP - SdT     (5)

For a system 

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

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

                                                                        (6)

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

                          (7)

Equation (6) tells us that phases with small volume are favored at higher pressure,

and equation

(7) tells us that phases with high entropy (high disorder) are favored at higher

temperature.

Equation (5), dG = VdP – SdT, tells us that the Gibbs

Free Energy is a function of P and T.

In the diagram, phase A has a steeply sloping free

energy

surface.  Phase B has a more gently sloping surface.

Where the two surfaces intersect, Phase A is in

equilibrium with phase B, and

G

A

 = G

B

.

At constant P

below T

E

 phase A has the lowest Gibbs Free

Energy, G.  At these

temperatures, phase A is stable and phase B is

not stable because

it has a higher free energy.

Note that at T

E

 the free energy of the

phase A and B

are same.

At

temperatures greater than T

 phase B has a lower free energy

than phase A, and thus phase B is stable.

Next, we look at a section of G versus P at constant T.

G

A

= G

B 

at P

E

.  At pressures greater than P

 phase B is

stable because it has a lower G than phase A.  At pressures

less than P

E

 phase A is stable because it has a lower G than phase B.

Finally, we look at a cross-section across the

bottom of the first

figure.  Here we project the line of intersection

of the Free Energy

surfaces onto the P - T plane.  Along the line of

intersection of the

surfaces G

A

= G

B

.  The line separates two fields, one at low P in

which A is the stable phase and one at higher P

in which phase B

B

is stable.  This is a classic P-T phase diagram.

The line represents

all values of P and T where

For reaction, A ↔B

d

Δ

G = 

Δ

VdP - 

Δ

SdT            (8)

ΔG = the change in Gibbs Free Energy of the reaction

ΔS = the change in entropy of the reaction

and  ΔV = the change in volume of the reaction

Δ

S=∑ S

Pr

- ∑ S

re

Δ

V= ∑ V

Pr

- ∑ V

Re

At equilibrium, as we have just seen, ΔG = 0, so from equation (8)

0 = Δ

VdP - 

Δ

SdT

Rearranging this equation yields

                                                                                   ( 9)

This relation is known as the 

Clausius - Clapeyron Equation.  It is important

because it tells 

us the slope of the equilibrium boundary or reaction

boundary on a Pressure versus Temperature phase diagram.

We next look at two cases of chemical reactions.  In the first case, the

chemical reaction is between only solid phases.  In the second case a fluid or

gas is involved as one of the products of the reaction.

Solid - Solid Reactions

Only involves 

solid phases

,  with 

no  fluid 

phases.

Most solid-solid reactions appear as 

straight lines 

on P-T diagrams.

Why?????

The reason for this comes from the Clausius-Clapeyronequation.

dP/dT = 

Δ

S/

Δ

V

As T increases, both S and V increase, disorganized at high temperature, high

entropy, molecule vibrates more, high V

Similarly, V tend to decrease with increasing pressure (less room to vibrate

means better organization and lower volume).

The  change in volume and entropy at any given temperature and pressure

tends to be small. A curve whose slope is constant is a straight line.

Let's use these principles to analyze some solid - solid reactions,

such as those in the Al2SiO5

  phase diagram.  Note that for the

solid-solid reaction Andalusite <=> Kyanite, dP/dT is positive.

This is what we expect, because the product kyanite, occurs on the

low T side of the reaction boundary

Entropy for Ky low or high?

ΔS negative

Increasing the pressure causes a decrease in volume,

Volume for Ky low or high?

 ΔV is also negative

With both ΔS and ΔV megative, the slope

of the boundary curve dP/dT is positive.

For the reaction 

Kyanite <=> Sillimanite

,

ΔS is positive.-RIGHT???

 ΔV is also positive-?????

is also positive.

Note that the reaction boundary for  

Andalusite <=> Sillimanite

has a negative slope

.

ΔV is negative or postive?

ΔS is positive or negative?

dP/dT negative.

Devolatization Reactions

Devolatilization reactions appear as curves on Pressure Temperature

diagrams.

A <=> B + H2O.

For this reaction we can write:

and

Δ

S = S

B

 + S

H2O

- S

A

 = Δ

S

solids

+S

H2O

Δ

V = V

B 

+ V

H2O

- V

A

 = Δ

V

solids

+ V

H2O

Increasing temperature will generally cause S to

be positive

 especially for this reaction in which a

gas or fluid phase is produced, because gases

always have a higher entropy (randomness) than

solids.

 At low pressure ΔV positive.

At low pressure fluid or gas will expand to fill

space available

dP/dT will be positive.

As the pressure increases, the fluid or gas will be

more

compressible than the solids, so the total ΔV will

become increasingly smaller.  Thus, dP/dT(= ΔS/

ΔV) will  increase.

Δ

S = S

B

 + S

H2O

- S

A

 = Δ

S

solids

+S

H2O

Δ

V = V

B 

+ V

H2O

- V

A

 = Δ

V

solids

+ V

H2O

Another relationship that is useful is:

G = H - TS

where G is the Gibbs Free Energy, H is the enthalpy, T is the absolute temperature in

Kelvin,

and S is the entropy.

For a chemical reaction, we can rewrite this as:

Δ

G = 

Δ

H - T

Δ

S  (10)

For a reaction Reactant=Product

ΔG = ∑ G

Pr

- ∑ G

re

Δ

S=∑ S

Pr

- ∑ S

re

Δ

H= ∑ H

Pr

- ∑ H

re

In general ΔG, ΔH, ΔS, and ΔV are dependent of Pressure and Temperature, but at any

given T & P:

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

If ΔG = 0 the reactants are in equilibrium with products,

and if ΔG > 0 (positive) the reaction will run from right to left.

u

G is a measure of relative chemical stability for a phase

u

We can determine G for any phase by measuring H and S for the reaction

creating the phase from the elements

u

We can then determine G at any T and P mathematically

F

Most accurate if know how V and S vary with P and T

•

dV/dP is the coefficient of isothermal compressibility

•

dS/dT is the heat capacity (Cp)

Temperature Dependence of G, H, and S

As stated above, G, H, and S depend on Temperature and Pressure.  But,

because G depends on

H and S, it is usually more convenient to consider the temperature dependence

of H and S, so

that if we know H and S at any given temperature, we can calculate G.

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

amount of heat necessary to raise the temperature of the substance by 1

o

K.

Tables of thermodynamic data are usually tabulated at some known reference

temperature and pressure, most commonly at a Temperature of 298 K, and

Pressure of 1 bar ( = 0.1 MPa ~ 1atm).

Thus, we if we need to know H at some temperature, T, other than 298 K, we can

use the above equation to determine H at the new temperature:

(T1 is 298K)

C

P

 can be expressed in terms of power series f T

C

P

=a+bT-+c/T

2

............

(a+bT+c/T

2

) dT.............(13a)

=

a(T

2

 - T

1

) + b(T

2

2

 - T

1

2

)/2 - c/(T

2

-1

 - T

1

-1

)

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

(a+bT+c/T

2

) dT.............(13b)

a(

ln

 T

2

 - 

ln

 T

1

) + b(T

2

 - T

1

) - c(T

2

-2

 -

T

1

-2

)/2

NOT USED

NOT USED

•

Exothermic vs. Endothermic

•

If 



r

H° < 0 the reaction produces a reduction in

enthalpy and is exothermic (heat is given up by the

rock and gained by the surroundings). If 



r

H° > 0 the

reaction produces an increase in enthalpy and is

endothermic (heat from the surroundings is consumed

by the rock). An easy way to remember this is that

spontaneous reactions produce a decrease in internal

energy, and because we know that

E

P

∞ H

P

a decrease in H

P

 is also a decrease in E

P

.

•

Determining Enthalpies

Thus, if we want to measure how the internal energy E of a crystal

changes 



E with increasing temperature at constant pressure, we

want to know 



H, and we can get that by integrating the heat

capacity C

P

 over the temperature range of interest.

There's another way to measure 



H, though: calorimetry.

•

By dissolving a mineral in acid and measuring the heat produced by

the dissolution, we get a heat of dissolution (usually positive). The

enthalpy of "formation"



f

H° of the mineral is then just the

opposite of the heat of dissolution (usually negative). Exceptions to

the "usually positive/negative" rule include CN, HCN, Cu

2+

, Hg

2+

,

NO, Ag

+

, and S

2-

. Enthalpies of formation appear in tables of

thermodynamic data and are usually referenced to 298 K and 1

atm.

•

Enthalpy of Reaction

To get an enthalpy of reaction 

r

H° we can measure the enthalpies of

formation of the reactants and products 

f

H° and then take the

difference between them as 



r

H° = 



f

H°

products

- 



f

H°

reactants

•

For example, we can compute the enthalpy of the reaction

anhydrite + water = gypsum:

•

CaSO

4

 + 2H

2

O = CaSO

4

2H

2

O

•

from Ca + S + 2O

2

 = CaSO

4       



f

H° = -1434.11 kJ/mol

•

H

2

 + 0.5O

2

 = H

2

O                      



f

H° = -285.830 kJ/mol

•

Ca + S + 3O

2

 + 2H

2

 = CaSO

4

2H

2

O      



f

H° = -2022.63 kJ/mol

•

Thus, 



r

H° =



f

H°gypsum - 



f

H°anhydrite - 



f

H°water = -16.86

kJ/mol.

•

Calculating 

f

H° at Temperatures Other Than 298 K

•

So far we know how to calculate the change in

enthalpy caused by heating and we know that we can

get enthalpies of formation from tables. What if we

want to know the enthalpy of formation of a mineral at

a temperature other than 298 K?

We do this by calculating 



r

C

P

 for the reaction that

forms the mineral of interest: 



r

C

P

 = 



r

C

Pproducts

 - 



r

C

Preactants

•

and then integrating. Thus, for example if we want to

know 



f

H° for quartz at 

1000 K

, we get coefficients for

the heat capacities of Si, O

2

 and SiO

2

:

•

http://www.geol.ucsb.edu/faculty/hacker/geo

124T/lecture.html

NaAlSi

3

O

8

 = NaAlSi

2

O

6

 + SiO

2

P - T phase diagram of the equilibrium curve

How do you know which side has which phases?

Figure 27.1. 

Temperature-pressure

phase diagram for the reaction:

Albite = Jadeite + Quartz calculated

using the program TWQ of Berman

(1988, 1990, 1991).

 Winter (2010)

An Introduction to Igneous and

Metamorphic Petrology. Prentice

Hall.

pick any two points on the equilibrium curve

d



G = 0 = 



VdP - 



SdT

Figure 27.1. 

Temperature-pressure

phase diagram for the reaction:

Albite = Jadeite + Quartz calculated

using the program TWQ of Berman

(1988, 1990, 1991).

 Winter (2010)

An Introduction to Igneous and

Metamorphic Petrology. Prentice

Hall.

Return to dG = VdP - SdT, for an isothermal process:

Gas Phases

Gas Phases

For solids it was fine to ignore V as f(P)

For gases this assumption is shitty

You can imagine how a gas compresses as P increases

How can we define the relationship between V and P for a gas?

Gas Pressure-Volume Relationships

Ideal Gas

–

As P increases V decreases

–

PV=nRT

Ideal Gas Law

•

P = pressure

•

V = volume

•

T = temperature

•

n = # of moles of gas

•

R = gas 

constant

 = 8.3144 J mol

-1

 K

-1

P x V is a constant at constant T

Figure 5.5.

Piston-and-cylinder apparatus to

compress a gas.   

Winter (2010) An Introduction to

Igneous and Metamorphic Petrology. Prentice Hall.

Gas Pressure-Volume Relationships

Since

we can substitute RT/P for V (for a single mole of gas), thus:

and, since R and T are certainly independent of P:

z

Gas Pressure-Volume Relationships

And since

G

P2

 - G

P1

 = RT 

ln 

P

2

 - 

ln 

P

1

 = RT

 ln

 (P

2

/P

1

)

Thus the free energy of a gas phase at a specific P and T, when

referenced to a standard atate of 0.1 MPa becomes:

G

P, T

 - G

T

 = RT

 ln

 (P/P

o

)

G of a gas at some P and T = G in the reference state (same T and 0.1 MPa)

+ a pressure term

o

Gas Pressure-Volume Relationships

The  form of this equation is very useful

G

P, T

 - G

T

 = RT 

ln

 (P/P

o

)

For a 

non-ideal gas

 (more geologically appropriate) the same

form is used, but we substitute 

fugacity ( 

f 

)

 for P

where

f 

= 



P



 is the fugacity coefficient

Tables of fugacity coefficients for common gases are available

At low pressures most gases are ideal, but at high P they are not

o

Dehydration Reactions

•

Mu + Q = Kspar + Sillimanite + H

2

O

•

We can treat the solids and gases separately

G

P, T

 - G

T

 = 



V

solids

 (

P

 - 0.1) + RT

 ln

 (

P

/0.1)

 (isothermal)

•

The treatment is then quite similar to solid-solid reactions, but

you have to solve for the equilibrium P by iteration

Dehydration Reactions

(qualitative analysis)

Figure 27.2. 

Pressure-temperature

phase diagram for the reaction

muscovite + quartz = Al

2

SiO

5

 + K-

feldspar + H

2

O, calculated using SUPCRT

(Helgeson 

et al

., 1978). 

Winter (2010)

An Introduction to Igneous and

Metamorphic Petrology. Prentice Hall.

Gibbs energy

An introduction to (or some revision on) thermodynamics

Gibbs energy

Ignoring (for a while) heat capacity, thermal expansion and isothermal

compression…

•

If a chemically closed 

system

 has two possible 

states

 (configurations of

phases

), the one with the lowest absolute 

G

at any 

PT

 should be more

stable.

•

If both have the same absolute 

G

(the

∆

G

of moving from one state to

the other

= 0), they have the same relative stability and a 

reaction

between them is stable.

If only Fayalite

If ‘N’=∑n=n

i

+n

j

+n

k

+.....

G



/N=

∑(n

i/

N)µ

i

=

∑X

i

µ

i  

 (X=Mole fracion)

G



/N= Molar Gibb’s free energy

Intensive or extensive variable?

For pure phase X

i

=1 as (n

i

/n

i

),

Therefore, Molar Gibb’s free energy=

µ

i

o 

 for

pure phase

dG



=∑µ

i

dn

i 

       at constant P, T

G



=∑µ

i

n

i

Diffusion

•

At equilibrium- µ

i 

of component i must be

same in every coexisting phases that contain

it.

•

If µ

i 

is lower in one phase that the other, free

energy of system could be lowered by

migration of the component from a phase

with higher µ

i 

to a phase with lower µ

i

•

Difference in µi drives diffusion.

•

Jd+Qtz=Ab

•

If pure phases and no solid solution, we can

write



G=G

Ab

-G

Qtz

-G

Jd

•

G



/N=

∑(n

i/

N)µ

i

=∑X

i

µ

i,

•

Therefore for this equation G



=µ

i

 and we can

write it as 



G=µ

Pl

Ab

-µ

Qtz

-µ

Cpx

Jd

=0 (at

eqilibrium)

•

In reality, no phases behave ideally and we

write

µ

i



 - µ°

i

 = RT 

ln

a

i



µ of the almandine (Fe

3

Al

2

Si

3

O

12

) component

of a garnet solid solution ((Fe, Mg, Ca,

Mn)

3

Al

2

Si

3

O

12

) is µ

alm

 = µ °

alm

 + RT 

ln

a

alm

µ° is chemical potential of component in its

pure reference state he activity forms a bridge

between idealized behavior and real behavior

Activity

•

a = (



X)

Z

where a = activity of a compound, ‘Z’ is the

"site occupancy coefficient" (e.g., = 2 for Mg in

Mg

2

SiO

4

), and 



 is the 

activity coefficient

 that

describes the non-ideal behavior.

•

For pure compounds a=1 as X=1.

For ideal solution 



=1. 

Activity Models (Activity-Composition

Relations) for Crystalline Solutions

•

a

prp

 = 



Mg

3

X

Mg

3

•

a

alm

 = 



Fe

3

X

Fe

3

•

a

grs

 =



Ca

3

X

Ca

3

•

a

sps

 = 



Mn

3

X

Mn

3

If we assume ideal behavior (



= 1) in garnet

•

a

alm

 = X

alm

3

 = [Fe/(Fe + Mg + Ca + Mn)]

3

•

a

prp

 = X

prp

3

 = [Mg/(Fe + Mg+ Ca + Mn)]

3

......................................................

Solutions: T-X relationships

Example: orthopyroxenes  (Fe, Mg)SiO

3

–

Real vs. Ideal Solution Models

Figure 27.3. 

Activity-composition relationships for the enstatite-ferrosilite mixture in orthopyroxene at 600

o

C and 800

o

C. Circles are data

from Saxena and Ghose (1971); curves are model for sites as simple mixtures (from Saxena, 1973) 

Thermodynamics of Rock-Forming

Crystalline Solutions

. 

Winter (2010) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

•

For solid:

a = (



X)

Z

•

For ideal gas:

a

i



=P

i

/P

i

ref

•

For real gas:

a

i



=f

i

/f

i

ref

The Equilibrium Constant

•

At equilibrium the sum of the Gibbs free energies of

the reactants equals the sum of the Gibbs free energies

of the products.

•

Equally, the sum of the partial molar Gibbs free

energies (chemical potentials) of the reactants equals

the sum of the partial molar Gibbs free energies (µ) of

the products.

•

cC + dD = aA + bB

At equilibrium 



r

µ = 0 = cµ

C

 + dµ

D

 - aµ

A

 - bµ

B

•

If we then remember that  µ - µ ° = RT 

ln

a

•

and rewrite it as µ = µ ° + RT 

ln

a

•

we can reformat the earlier equation as 



r

µ = 0 = c(µ

C°

+ RT 

ln

a

C

) + d(µ

D°

 + RT 

ln

a

D

) - a(µ

A°

 + RT 

ln

a

A

) - b(µ

B°

 + RT

ln

a

B

)

or, 



r

µ = 



r

µ ° + RT 

ln

 (a

C

c

 a

D

d

/a

A

a

 a

B

b

)

or. 



r

µ = 



r

µ ° + RT 

ln

 Q

 and Q is the 

activity product ratio

. The activities in the

Q term change as the reaction progresses toward

equilibrium, 



r

µ=0

•

At equilibrium, the product and reactant activities

have adjusted themselves such that 



r

µ = 0.

We write this (with K instead of Q, to signify

equilibrium) as 0 = 



r

G° = -RT 

ln

 K

•

The utility of K is that it tells us for any reaction

and any pressure and temperature, what the

activity ratios of the phases will be at equilibrium

Solutions: T-X relationships

Back to our reaction:

Simplify for now by ignoring dP and dT

For a reaction such as:

aA + bB = cC + dD

At a constant P and T:

where:

Compositional variations

Effect of adding Ca to albite = jadeite + quartz

plagioclase = Al-rich Cpx + Q





G

T, P

 = 



G

o

T, P

+ RT

l

n

K

Let’s say 



G

o

T, P

was the value that we calculated for

equilibrium in the pure Na-system (= 0 at some P and T)



G

o

T, P

  = 



G

298, 0.1

 + 



V (P - 0.1) - 



S (T-298)  = 0

By adding Ca we will shift the equilibrium by 

RT

l

n

K

We could assume ideal solution and

All coefficients = 1

Solutions: T-X relationships

Ab = Jd + Q was calculated for 

pure

 phases

When solid solution results in impure phases

the activity of each phase is reduced

Compositional variations

Effect of adding Ca to albite = jadeite + quartz





G

P, T

 = 



G

o

P, T

 + RT

l

n

K

numbers are values for K

Figure 27.4. 

P-T phase diagram for the reaction Jadeite + Quartz =  Albite for various values of K. The equilibrium curve for K = 1.0 is

P-T phase diagram for the reaction Jadeite + Quartz =  Albite for various values of K. The equilibrium curve for K = 1.0 is

the reaction for pure end-member minerals (Figure 27.1). Data from SUPCRT (Helgeson 

the reaction for pure end-member minerals (Figure 27.1). Data from SUPCRT (Helgeson 

et al

et al

., 1978). 

., 1978). 

Winter (2010) An Introduction to

Winter (2010) An Introduction to

Igneous and Metamorphic Petrology. Prentice Hall.

Igneous and Metamorphic Petrology. Prentice Hall.

Geothermobarometry

Use measured distribution of elements in coexisting

phases from experiments at known P and T to estimate

P and T of equilibrium in natural samples

Thermometry

Exchange Reactions

•

Many thermometers are based on 

exchange

reactions

, which are reactions that exchange

elements but preserve reactant and product

phases.

•

For example: Fe

3

Al

2

Si

3

O

12

+ KMg

3

AlSi

3

O

10

(OH)

2

 =

Mg

3

Al

2

Si

3

O

12

 + KFe

3

AlSi

3

O

10

(OH)

2

•

almandine + phlogopite = pyrope + annite

We can reduce this reaction to a simple exchange

vector: (FeMg)

gar

+1

 = (FeMg)

bio

-1

•

Popular thermometers include garnet-biotite (GARB),

garnet-clinopyroxene, garnet-hornblende, and

clinopyroxene-orthopyroxene; all of these are based on

the exchange of Fe and Mg, and are excellent

thermometers because 



r

V

 is small, such that

dP/dT= 



r

S/ 



r

V 

is large (i.e., the reactions have steep

slopes and are little influenced by pressure).

Let's write the equilibrium constant for this exchange

reaction K = (a

prp

a

ann

)/(a

alm

a

phl

)

•

thus 



r

G = -RT 

ln

 (a

prp

a

ann

)/(a

alm

a

phl

)

•

If we assume idealbehaviour, The equilibrium constant is

K = (X

Mg

gar

 X

Fe

bio

)/(X

Fe

gar

 X

Mg

bio

)

•

When discussing element partitioning it is common to define a

distribution coefficient

 K

D

, which is just the equilibrium constant

without the exponent (this just describes the partitioning of

elements and not the partitioning of chemical potential):

K

D

 = (X

Mg

gar

 X

Fe

bio

)/(X

Fe

gar

 X

Mg

bio

) = (Mg/Fe)

gar

 /(Mg/Fe)

bio

 = K

1/3

•

John Ferry and Frank Spear in 1978 measured experimentally the

distribution of Fe and Mg between biotite and garnet at 2 kbar and

found the following relationship:

Geothermobarometry

The Garnet - Biotite geothermometer

•

If you compare their empirical equation

ln

 K

D

 = -2109 / T + 0.782

If we compare this relation with

ln

 K = - (

r

G° / RT) = -(

r

H / RT) - (P

r

V /

RT) + (

r

S / R)

for this reaction 

r

S = 3*0.782*R =

19.51 J/mol K

(the three comes from the site

occupancy coefficient; i.e., K = K

D

3

)

and -(

r

H / R) - (P

r

V / R) = -2109

or 

r

H = 3*2109*R -2070*

r

V

•

Molar volume measurements show that for

this exchange reaction 

r

V = 0.238 J/bar, thus 

r

H

= 52.11 kJ/mol

•

The full equation is then 52,110 - 19.51*T(K) +

0.238*P(bar) + 3RT 

ln

 K

D

 = 0

•

To plot the K

D

 lines in PT space

Geothermobarometry

The Garnet - Biotite geothermometer

Figure 27.5.

Graph of 

l

n

K vs. 1/T (in Kelvins) for the Ferry and Spear (1978) garnet-biotite exchange equilibrium at 0.2 GPa from Table

27.2.  

Winter (2010) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

Geothermobarometry

The Garnet - Biotite geothermometer

Figure 27.6. 

AFM projections showing the relative distribution of Fe and Mg in garnet vs. biotite at approximately 500

o

C

(a)

and 800

o

C 

(b)

.

From Spear (1993) 

Metamorphic Phase Equilibria and Pressure-Temperature-Time Paths

. Mineral. Soc. Amer. Monograph 1.

•

Net-Transfer Reactions

•

Net-transfer reactions are those that cause

phases to appear or disappear. Geobarometers

are often based on net-transfer reactions because



r

V is large and relatively insensitive to

temperature. A popular one is GASP: 3CaAl

2

Si

2

O

8

= Ca3Al

2

Si

3

O

12

 + 2Al

2

SiO

5

 + SiO

2

 anorthite =

grossular + kyanite + quartz

which describes the high-pressure breakdown of

anorthite.

For this reaction

•

r

G = -RT 

ln

 [(a

qtz

a

ky

2

a

grs

) / a

an

3

] = -RT 

ln

 a

grs

 / a

an

3

•

(the activities of quartz and kyanite are one because they are pure

phases). A best fit through the experimental data for this reaction

by Andrea Koziol and Bob Newton yields P(bar) = 22.80 T(K) - 7317

•

for 



r

V = -6.608 J/bar. Again, if we use 

ln

 K = -(



r

H / RT) - (P



r

V /

RT) + (



r

S / R)

•

and set 

ln

 K = 0 to calculate values at equilibrium, we can rewrite

the above as (P



r

V / R) = -(



r

H / R) + (T



r

S / R)

•

or P= T



r

S /



r

V - 



r

H / 



r

V

•

if T



r

S /



r

V = 22.8 then



r

S = -150.66 J/mol K

if 



r

H / 



r

V = 7317 then 



r

H = -48.357 kJ/mol

So, we can write the whole shmear as 0 = -48,357 + 150.66 T(K) -

6.608 P (bar) + RT 

ln

 K

•

Contours of 

ln

 K on a PT diagram for GASP look like this:

•

To plot the K

D

 lines in PT space

•

Figure 27.14. 

Chemically zoned plagioclase and poikiloblastic garnet from meta-pelitic sample 3, Wopmay Orogen, Canada. a. Chemical

profiles across a garnet (rim 



 rim). 

b.

An-content of plagioclase inclusions in garnet and corresponding zonation in neighboring plagioclase.

After St-Onge (1987) 

J. Petrol.

 28, 1-22 .

Geothermobarometry

P-T-t Paths

Figure 27.15. 

The results of applying the garnet-biotite geothermometer of Hodges and Spear (1982) and the GASP geobarometer of Koziol

(1988, in Spear 1993) to the core, interior, and rim composition data of St-Onge (1987). The three intersection points yield P-T estimates which

define a P-T-t path for the growing minerals showing near-isothermal decompression. After Spear (1993).

Geothermobarometry

P-T-t Paths

•

Dsc xx

What is a P-T-t path

Metamorphism is a

dynamic 

process

,

involving changes in

temperature ±

pressure through

time. The pressure

(P) - temperature (T) -

time (t) 

path

 of a

metamorphic rock is

the set of all P-T

conditions

experienced by a rock

during its

metamorphic history

A common pressure-temperature path for regional metamorphism. The

rate of prograde metamorphism (heating) and rate of retrograde

metamorphism (cooling) may not be the same. The duration of the path

from start (onset of metamorphism) to finish (exposure of the rock at the

Earth's surface) will vary from rock to rock depending on the tectonic

history

What are some common P-T paths? 

•

P-T paths are commonly described as 'clockwise' or

'anticlockwise' (a.k.a. 'counterclockwise').

•

Paths can be clockwise or anticlockwise, and can also vary in

terms of

`

(1) how different the prograde and retrograde segments of

the path are – very similar or very different (Figure 2b), and

(2) how different the maximum pressure and maximum

temperature are from each other (Figure 2c). 

Common P-T paths, including (a) Clockwise versus counterclockwise

paths, (b) Paths with similar vs different prograde and retrograde

segments, and (c) Paths with coincident maximum P and T conditions vs

paths with very different maximum P and T conditions. Note that the T

maximum is known as the 'peak' of metamorphism. 

How are P-T paths determined? 

How are P-T paths determined? 

•

For some rocks, the only part of the P-T path recorded in the 

mineral assemblage

and texture of the rock is the 'peak' of metamorphism 

– the conditions of the

thermal maximum (i.e., the maximum temperature and the pressure at the

maximum temperature).

•

At the peak of metamorphism, the mineral assemblage presumably equilibrated,

and no (or little) further reaction took place as the rock cooled and decompressed

en route to the Earth's surface.

•

Some rocks may record more of their P-T paths. If a rock contains a partial record

of its P-T path, this is both good and bad. This is 

good

 because we 

want as much P-

T path information 

as possible, so as to be able to interpret the thermal/tectonic

processes and history as well as possible. This is 

bad

 because the mineralogical

and textural evidence for P-T path segments 

other

 than the conditions of the peak

of metamorphism represent 

disequilibrium

. In most cases, however, the evidence

for disequilibrium can be very useful because it can be used to reconstruct P-T

path segments.

•

A few common methods for inferring P-T path

segments are: 

1. Mineral inclusions 

Some minerals contain 

inclusions

 of other minerals. For

example, garnets commonly contain inclusions of

minerals that were present in the rock matrix as the

garnet grew, but that were not completely eliminated

by metamorphic reactions during progressive

metamorphism. The growing garnets surrounded these

relict minerals as the garnets grew, and the relict

minerals are preserved as mineralogical evidence of an

earlier stage in the metamorphic history of the rock.

Figure 3. Left: Photomicrograph (plane light) of kyanite inclusions in garnet; field of view = 2 mm. Right: Photomicrograph (crossed polars) of sillimanite in the matrix; field of view = 4 mm. 

Figure 3. Left: Photomicrograph (plane light) of kyanite inclusions in garnet; field

of view = 2 mm. Right: Photomicrograph (crossed polars) of sillimanite in the

matrix; field of view = 4 mm. 

•

Some inclusions don't contain a lot of information

about P-T conditions because the minerals are 

stable

over such a wide range of conditions

 (for example:

quartz).

•

Other mineral inclusions are very useful for inferring

P-T conditions and path segments, especially if these

minerals no longer exist in the matrix of the rock (that

is, outside the garnet).

•

In Figure 3, kyanite inclusions occur in garnet in a rock

that has only sillimanite in the matrix, indicating that

the rock was previously in the kyanite stability field but

that P-T conditions changed. When the matrix of the

rock (including the garnet rim) equilibrated, the rock

was in the silliminate stability field (Fig. 4).

•

Figure 4. P-T diagram showing

stability fields of the Al

2

SiO

5

polymorphs: andalusite,

kyanite, and sillimanite (after

Holdaway, 1971). Two

possible P-T paths are shown

to illustrate different ways

that kyanite can be replaced

by sillimanite: one path (a)

involves a decrease in

pressure (decompression);

the other (b) involves an

increase in temperature

(prograde metamorphism). 

Even though it may not be possible to determine the

trajectory of the P-T path from inclusions alone,

inclusions may be used with other chemical and

textural information to better define the path. In

addition, the composition of inclusions may be used in

thermobarometry to determine P-T conditions along

the path, provided the inclusions have not chemically

reacted with their host mineral (e.g., Whitney, 1991).

2. Element zoning 

•

Metamorphic (and igneous) minerals may change composition in response

to changing chemical and physical conditions, such as changes in pressure-

temperature conditions, deformation variables, or chemical factors (e.g.,

presence of fluids). The chemical response of a mineral to these changes

may be recorded in minerals that have crystal chemical and structural

characteristics that allow compositions acquired early in the mineral's

growth history to be preserved during later stages of growth at different

conditions. Crystals that have different regions with different compositions

are 

zoned 

.

•

A very common zoning pattern involves a difference in composition of a

mineral's center (core) compared to its rim, and concentric rings of

different compositions arranged between the core and the rim (Fig. 5).

Minerals that typically show this type of zoning in the major or trace

elements are garnet, plagioclase, zircon, and tourmaline (Fig. 5). Of these,

garnet and plagioclase are relevant to studies of metamorphic P-T paths,

and zircon is relevant to determination of the timing of petrologic events. 

Figure 27.14. 

Chemically zoned plagioclase and poikiloblastic garnet from meta-pelitic sample 3, Wopmay Orogen, Canada. a. Chemical

profiles across a garnet (rim 



 rim). 

b.

An-content of plagioclase inclusions in garnet and corresponding zonation in neighboring plagioclase.

After St-Onge (1987) 

J. Petrol.

 28, 1-22 .

Geothermobarometry

P-T-t Paths

Figure 5. Common zoned minerals. (a) False color X-ray map showing Mn distribution

in a garnet from Iran (Sepahi et al., 2003). The garnet core contains more Mn than

the garnet rim (or the matrix); this is typical growth zoning. The garnet is 1.3 mm in

diameter; (b) Photomicrograph (crossed polarized light) showing zoning in

plagioclase in a metamorphosed igneous rock; oscillatory zoning is common in

igneous plagioclase, but metamorphic plagioclase may also be zoned in the anorthite

(Ca) and albite (Na) components; field of view = 2 mm; (c) Photomicrograph (plane

polarized light) of a zoned tourmaline crystal in a kyanite schist; field of view = 2 mm;

(d) Cathodoluminescence image of isotopically zoned zircon from the Nigde Massif,

Turkey, with the U-Pb age of the core and rim labeled in millions of years (Whitney et

al., 2003). 

•

Using zoning information to reconstruct the part of the P-T path

experienced by the zoned mineral is not simple, but a few general aspects

of the relationship of zoning to P-T path may easily be inferred:

(a) Garnets with Mn-rich cores and Mn-poorer rims record 

growth zoning

that represents the change from the lower-T conditions at which the

garnet core grew to the higher-T conditions at which the garnet rim grew

(i.e., prograde metamorphism involving increasing temperature and

pressure). Mn is preferentially partitioned into garnet relative to most

other common minerals, so Mn is sequestered in early-formed garnet,

depleting the local environment of the growing garnet in Mn.

(b) Minerals that show major element growth zoning probably did not

experience very high metamorphic temperatures. 

At high temperature (>

700 C) and sufficient duration, zoning may be homogenized as

intracrystalline diffusion becomes more effective at eliminating

compositional variation

. An unzoned mineral that is typically zoned at

low-medium metamorphic grades has either experienced high

temperature conditions or was never zoned (owing to a simple reaction

history at limited P-T or to growth entirely at high-T). 

3. Reaction textures 

•

Some metamorphic rocks contain evidence for incomplete reactions

or other textural evidence for part of the P-T path. A simple

example is the partial replacement of andalusite by sillimanite (Fig.

6a) by the polymorphic transformation of andalusite to sillimanite.

Textural evidence may also be useful in cases where the reactants

have been completely consumed if the shape of one or more

reactants are preserved; for example, 

pseudomorphs

(Figs. 6b-d).

An example of a more complex reaction texture involves the

formation of 

coronas 

, which consist of one or more shells (rims,

moats) of a mineral or minerals around a central (reactant) phase

(Fig. 6e). In many cases, coronas also involve the fine-scale

intergrowth of minerals in a texture known as 

symplectite

(Figs. 6e-

f).

Figure 6. Images of reaction textures. (a) Photomicrograph (plane light) showing the partial replacement of andalusite by

sillimanite in a schist from Iran. Note: The crystallization sequence (sillimanite after andalusite) can't be inferred only from this

photo; (b) Photomicrograph (plane light) showing the complete replacement of kyanite by sillimanite in a sample of gneiss from

the Thor-Odin dome, British Columbia. The former presence of kyanite is known because some pseudomorphs (not shown)

contain relict kyanite. Without these relics, and based only on the tabular shape of the pseudomorph, it would be difficult to

infer whether the original mineral was kyanite or andalusite; (c) Crossed polar view of (b) showing the randomly oriented

sillimanite in the pseudomorph; (d) Photomicrograph (plane light) showing a partial pseudomorph of chlorite after garnet from a

retrograded eclogite, Turkey. Garnet relics are present, but the former presence of garnet is clear also from the shape of the

pseudomorph; (e) Photomicrograph (plane light) showing a corona texture from a Thor-Odin dome gneiss. The central Al2SiO5

phase (sillimanite after kyanite) is rimmed by an inner shell of spinel + cordierite symplectite and an outer shell of cordierit; (f)

Backscattered electron image of spinel (brightest phase) + cordierite (darkest gray) + anorthite (medium gray) from a Thor-Odin

symplectite. 

How are P-T-t paths interpreted? 

•

An important part of using 

P-T paths or P-T-t paths 

to understand

metamorphic and tectonic processes

 is to relate the P-T conditions,

path shape, and (if age information is available) duration and rate of

P-T path segments to the driving forces of metamorphism.

•

P-T path shape by itself does not provide a unique interpretation of

tectonic process or metamorphic driving forces. For example,

clockwise paths can form in continental collision belts of subduction

zones. Similarly, some subduction zone rocks record clockwise paths

and some record counterclockwise paths. However, the integration

of P-T path characteristics, time/rate information, structural data,

and other petrologic information can provide significant

information about metamorphic and tectonic processes. Therefore,

although subduction zone rocks can follow various paths during

subduction and exhumation, determining the specific path that a

particular exhumed subduction zone rock followed is important for

understanding subduction dynamics.

Integrating deformation into P-T-t

histories: 

P-T-t-d

 paths 

•

The idealized view of P-T paths is that mineral assemblages equilibrate at

every stage of the path from the onset of metamorphism to the peak of

metamorphism, at which the final assemblage is locked in.

•

It is important to recognize the influence of deformation on metamorphic

reactions, and the extent to which deformation (strain energy) may assist

reactions. It is possible that two rocks of the same bulk composition that

follow the same P-T path but that have different deformation histories

(e.g., one is pervasively deformed and the other is not, perhaps because

strain is localized in weaker rocks nearby) will contain different mineral

assemblages. The deformed rock may contain the predicted equilibrium

assemblage for the P-T conditions attained by the rock, whereas the

undeformed or less deformed rock may contain more metastable phases.

P-T conditions and paths should therefore be considered in their structural

context, and, if possible, a P-T-t-d path (Pressure-Temperature-time-

deformation) constructed.