1. THERMODYNAMICS
WHO NEEDS IT?

2. WHY THERMODYNAMICS?
At what conditions does one mineral transform into another?
Al2Si2O5(OH)4 = Al2SiO5 + SiO2 +2H2O
Kaolinite
Andalusite
At what conditions does a magma crystallize a particular mineral?
Mg2+ + SiO44- = Mg2SiO4
Magma
Olivine
The conditions of interest are P, T and the activities or
Thermodynamic concentrations of chemical species

3. THE GIBBS FUNCTION w Recall the fundamental equation: dU = dq + dw
or dU = TdS – PdV
and (U/S)v = T; (U/V)s = – P q
As U is a function of S, V we can transform the fundamental equation and define Gibbs energy as :
G = U – S(U/S)v – V(U/V)s
= U –TS +PV
dG = dU– TdS – SdT + PdV + VdP
= TdS – PdV – TdS – SdT + PdV + VdP
dG= -SdT + VdP

4. IMPLICATIONS FOR EQUILIBRIUM
Consider the reaction Kyanite = Andalusite
GAndalusite
GKyanite G
G/P = V
1 Bar
Peq
GAndalusite
GKyanite G
G/T = -S
298 K
T eq

5. STABLE AND METASTABLE EQUILIBRIUM
dGP,T = 0
Metastable equilibrium
Stable equilibrium
GT,P

6. THE GIBBS FUNCTION (Continued)
Since dG = (G/T)pdT + (G/P)TdP
At equilibrium G Andalustite = G Kyanite
dGT,P = 0
Whereas for the Gibbs function we transformed the fundamental equation on both S and V we can define enthalpy by transforming it only on S and thus
H = U + V(U/V)s
or H = U + PV
but
G = U –TS +PV or
U = G + TS – PV
and thus G = H –TS
or GT,P = H - TS

7. APPLICATION OF THE GIBBS FUNCTION
We cannot measure energy parameters of the phase of interest in absolute terms
We need a standard set of conditions to which the parameters can be related
 We need to relate energies of phases to those of their elements and define a standard state, e.g., 298 K and 1 bar
Consider the reaction H2O(V) = H2O (L) at 25 C, (298 K, 1 bar)
220 1 25
374
Temperature (oC)
Pressure (bar)
Water
Steam
Ice
C.P.
fGoH2O = GoH2O(L) – GoH GoO2
fGoH2O = GoH2O(V) – GoH GoO2
rGoH2O = GoH2O(L) – GoH2O(V)

8. HEAT CAPACITY AND PHASE EQUILIBRIA
Consider the reaction Kyanite = Andalusite
At equilibrium, GKyanite = GAndalusite and rG = 0
rG = rH - TrS and therefore rS = rH/T
HT – H298
J mol-1
298
500
1000
Temperature (K)
2000
6000
10000
dH/dT = Cp
= a + bT –cT-2
Kyanite
rHT = rH298 +  rCpdT
298 T
Andalusite
rST = rS298 +  (rCp/T)dT
298 T
rGP,T = rH298 +  rCpdT – T(rS298 +  (rCp/T)dT) +  rV
298 T 1 P

9. CLAPEYRON EQUATION Consider the reaction Kyanite = Andalusite
At equilibrium, GKyanite = GAndalusite and rG = 0
but dG = -SdT + VdP and dG =0
 VdP = SdT
i.e., dP/dT = rS/rV
Calculate the equilibrium temperature at 1 bar
(rG/T)p = -rS and
rGT - rG298 =  - rSdT
298 T
If rST = rS298 then
rGT - rG298 = - rS (T – )

10. APPLY THE CLAPEYRON EQUATION
i.e., dP/dT = rS/rV
rGT - rG298 = - rS (T – )
Pressure (Kbar)
Temperature (oC)
dP/dT
Kyanite
Andalusite

11. CLAPEYRON EQUATION (Continued)
rG/T)p = - rS T
rGT - rG298 =  - rS
298
rGT - rG298 = - rS (T – )
0 – 1220 = (T ) = K or C
dP/dT = rS/rV
= 9.41/0.744 (1 cm3 = 0.01Jbar-1) = bar/ K
fGo So Vo
kJ mol-1 J mol-1K-1 cm3 mol-1
Kyanite
Andalusite
Sillimanite

12. ALUMINOSILICATE STABILITY
Kyanite = Andalusite
1 bar = dP/dT = bar K-1
Andalusite = Sillimanite
1 bar = C; dP/dT = bar K-1 10
Andalusite = Sillimanite 8
dP/dT = bar K-1 6
Pressure (Kbar)
Kyanite
Triple point 4
Sillimanite
416 C; 3310 bar 2
Andalusite
100
200
300
400
500
600
700
Temperature (oC)

13. REACTIONS INVOLVING A GAS
An assumption implicit in evaluating equilibria involving solids is that volume does not change with P and T
This is not true for gases, e.g., PV = RT
Consider the reaction Calcite + Quartz = Wollastonite + CO2
rGP,T = rH298 +  rCpdT – T(rS298 +  (rCp/T)dT) +  rVdP
298 T 1 P
= rH298 +  rCpdT – T(rS298 +  (Cp/T)dT) + Vs(P-1)
298 T 1 P
+  r(RT/P)dPg T T
= rH298 +  rCpdT – T(rS298 +  (rCp/T)dT) + rVs(P-1)
298
298
+ RTlnPg

14. REACTIONS INVOLVING A GAS (Continued)
Calcite + Quartz = Wollastonite + CO2
CaCO3 + SiO2 = CaSiO3 + CO2
300
400
500
600
700
1000
1500
2000
Calcite +
Quartz
Wollastonite
CO2
Temperature (oC)
Pressure (bar)
Wollastonite
Calcite
dp/dt
large
dp/dt
small
Quartz
CO2
298
= rH298 +  rCpdT – T(rS298 +  (rCp/T)dT) + rVs(P-1) T
+ RTlnf CO2 T

15. CHEMICAL POTENTIAL  nidμi +  μidni = -SdT + PdV +  μidni
If the system is open then we need to consider the contribution
of components that act as intensive parameters
The partial G of component i, Gi = (G/ni)P,T,ni (I = 1,2…) = μi -
dG =-SdT + VdP +  μi d ni i
G =  niμi and dG =  nidμi +  μidni i
 nidμi +  μidni = -SdT + PdV +  μidni i
Rearrange
0 = -SdT + PdV +  nidμi i
At constant P and T  nidμi = 0 i

16. PRACTICAL SIGNIFICANCE OF CHEMICAL POTENTIAL
Consider the reaction Aragonite = Calcite (CaCO3) P T Z Y X
Aragonite
Calcite μ
CaCO3
Calcite
Aragonite X Y Z =

17. GIBBS DUHEM EQUATION At constant P and T  nidμi = 0
Consider two components to which the system is open
nadμa + nbdμb = 0
dμa / dμb= -nb/na
Consider phase relations in the system Fe-S-O
Magnetite
Slope = 3/2
FeS + 1/2S2 = FeS2
μO2
3FeS +2O2 = Fe3O4 + 3/2S2
Slope = 3/4
Pyrite
Fe3O4 + 3S2 = 3FeS2 +2O2
Pyrrhotite
μS2

18. FUGACITY AND ACTIVITY Consider the change in G of an ideal gas with P
 dG = GP – Go =  VdP =  (RT/P)dP P 1
G – Go = RT lnP
Most gases are not ideal. We therefore define an effective thermodynamic pressure, fugacity, which is related to pressure as follows f = P
G – Go = RT ln f
Therefore μi – μio = RT ln fi/fio for component i in a gas mixture
Extending the concept to solutions generally we define:
ai = fi/fio and ai = Xii
μi – μio = RT ln ai

19. THE EFFECT OF MIXING Ideal mixing Non-ideal mixing A B XB
VMixing
VAo
VBo VA VB
V =XAVAo+XBVBo A B XB
V ideal > Vnon-ideal
μAo
μBo μB μA
XAμAo + XBμAo
GMixing A B XB
No change of V or H with Ideal mixing. S increases with both ideal and non ideal mixing.
S = k ln 

20. EFFECT OF MIXING ON ACTIVITY
Ideal solid solutions
For a pure phase a = 1 but for a solid solution ai = (Xii)n where i is a component and n is the number of crystallographic sites of mixing
If mixing ideal i = 1 and ai = (Xi)n
Consider olivine comprising forsterite (Mg2SiO4) + fayalite (Fe2SiO4) in reaction Mg2SiO4 + 2H2O = 2Mg(OH)2 + SiO2
Olivine Brucite Quartz
If Olivine contains 70 mol% forsterite then aForsterite = (0.7)2 = 0.49
and μForsterite = μForsteriteo + RT ln(0.49)

21. LAW OF MASS ACTION
Rate of reaction is directly proportional to the concentration of each reacting substance
Consider reaction aA + bB  cC + dD
k1[A]a[B]b
At equilibrium aA + bB = cC + dD
k1[A]a[B]b = k2[C]c[D]d or k1/k2 = K
The equilibrium constant, K = [C]c[D]d/[A]a[B]b
ln K =cln[C] + dln[D] – aln[A] – bln[B]
More generally
Ln K = ni  lnXi

22. THE EQUILIBRIUM CONSTANT
Consider reaction aA + bB = cC + dD
rG = rμi = cμC + dμD – aμA - bμB
but μi – μio = RT ln ai
rG = rμa-d = rμa-do + RTln (aCcaDd/aAaaBb)
rG = r μo +RT ln K
Ln K = ni  ln ai
At equilibrium rG = 0 and r μo = -RT ln K
rG = - RT ln K and ln K = - rG/RT