Gibbs Phase Rule

• The number of variables which are required to describe the state of a system:

• p+f=c+2 f=c-p+2– Where p=# of phases, c= # of components,

f= degrees of freedom– The degrees of freedom correspond to the

number of intensive variables that can be changed without changing the number of phases in the system

Variance and f• f=c-p+2• Consider a one

component (unary) diagram

• If considering presence of 1 phase (the liquid, solid, OR gas) it is divariant

• 2 phases = univariant

• 3 phases = invariant

Free Energy• Gibbs realized that for a reaction, a certain

amount of energy goes to an increase in entropy of a system.

• G = H –TS or G0R = H0

R – TS0R

• Gibbs Free Energy (G) is a state variable, measured in KJ/mol

• Tabulated values of G0R are in Appendix

)reactants()( 000i

iii

iiR GnproductsGnG

• Now, how does free energy change with T and P?• From G=H-TS:

• T and P changes affect free energy and can drive reactions!!

2

1

2

1

2

)1(2

1

)1(11222)( 121,1,,

T

T

P

PT

PT

TPTPTPTP dPVdT

T

CTdTCTTSGG P

P

Volume Changes (Equation of State)

VdPdG

T

PTV

V

1

TPV

V

1

Volume is related to energy changes:

Mineral volume changes as a function of T: , coefficient of thermal expansion

Mineral volume changes as a function of P: , coefficient of isothermal expansion

For Minerals:

Volume Changes (Equation of State)

• Gases and liquids undergo significant volume changes with T and P changes

• Number of empirically based EOS solns..• For metamorphic environments:

– Redlich and Kwong equation:

• V-bar denotes a molar quatity, aRw and bRK are constants

)(2/1RK

Rw

RK bVVTa

bVRTP

Phase Relations• Rule: At equilibrium, reactants and products have

the same Gibbs Energy– For 2+ things at equilibrium, can investigate the P-T

relationships different minerals change with T-P differently…

• For GR = SRdT + VRdP, at equilibrium, Grearranging:

R

R

G VS

TP

0Clausius-Clapeyron equation

Remember that a line on a phase diagram describes equilibrium, GR=0!!

V for solids stays nearly constant as P, T change, V for liquids and gases DOES NOT

• Solid-solid reactions linear S and V nearly constant, S/V constant + slope in diagram

• For metamorphic reactions involving liquids or gases, volume changes are significant, V terms large and a function of T and P (and often complex functions) – slope is not linear and can change sign (change slope + to –)

R

R

G VS

TP

0

P

R

TR

R

TV

VS

TC

TS P

P

R

SR change with T or P?

V = Vº(1-P)

21

22

00

00

(2

1

2

1

2

PPPVS

VdPSdPPSSS

P

PT

P

PP

R

R

G VS

TP

0

Example – Diamond-graphite• To get C from

graphite to diamond at 25ºC requires 1600 MPa of pressure, let’s calculate what P it requires at 1000ºC:

graphite diamond

(K-1)

1.05E-05 7.50E-06

(MPa-1)

3.08E-05 2.27E-06

Sº(J/mol K)

5.74 2.38

Vº(cm3/mol)

5.2982 3.417

Clausius-Clapyron Example

Phase diagram• Need to represent how mineral reactions

at equilibrium vary with P and T

R

R

G VS

TP

0P

R

R

R

TV

VS

TC

TS P

P

R