1. Thermodynamics Internal Energy (E) of a mineral = Sum of energy stored
A system, e.g. a mineral, is an isolated portion of the Universe.
Internal Energy (E) of a mineral = Sum of energy stored
in the bonding and in the kinetic energy of the atomic vibration
Adding heat to a mineral dQ
increasing kinetic energy
increasing internal energy (dE)
dE = dQ - pdV
pdV = work at constant pressure, volume expansion
This is a statement of the
First Law of Thermodynamics

2. Thermodynamics Added heat (Q) to a system versus its Entropy (S)
At constant temperature: dQ/T = dS
Second law of thermodynamics:
In any reversible process the change in the
Entropy of the system (dS) is equal to the
heat received by the system (dQ) divided by the absolute temperature.
Adding heat to ice = increase in entropy through
Vibration
Breaking hydrogen bonds between (H2O) groups
(Partial Melting)

3. Thermodynamics Combinations of first and second law of thermodynamics:
E = Q – PV, Q = H at constant P
H = E + PV
dQ = TdS
dE = TdS - PdV
G = H – TS
G = E + PV – TS
dG = dE + PdV + VdP – TdS – SdT
dG = TdS – PdV + PdV + VdP – TdS – SdT
To integrate properly we must know how V and S vary with P and T (hence the calculus), but we shall simplify the math and assume V and S are constant
This simplifies our math considerably (but may lead to some errors)
We can check the validity of our assumptions by comparing to experiments or more rigorous calculations (performed by computer)

4. Free Energy vs. Temperature
Slope of GLiq > Gsol since
Ssolid < Sliquid
A: Solid more stable than liquid (low T)
B: Liquid more stable than solid (high T)
Slope dG/dT = -S
Slope S < Slope L
Equilibrium at Teq
GLiq = GSol
dG = VdP – SdT

5. Free Energy vs. Pressure
Slope of GLiq > Gsol since
Vsolid < Vliquid
A: Solid more stable than liquid (high P)
B: Liquid more stable than solid (low P)
Slope dG/dP = V
Slope S < Slope L
Equilibrium at Peq
GLiq = GSol
dG = VdP – SdT

6. Thermodynamics dGB = VBdP – SBdT dG = VdP – SdT
In a phase diagram with the environmental variables temperatures and pressures we can relate the change of the free energy of a phase A as
dGA = VAdP – SAdT
and of the phase B
dGB = VBdP – SBdT
To integrate properly we must know how V and S vary with P and T (hence the calculus), but we shall simplify the math and assume V and S are constant
This simplifies our math considerably (but may lead to some errors)
We can check the validity of our assumptions by comparing to experiments or more rigorous calculations (performed by computer)

7. Thermodynamics VBdP – SBdT = VAdP – SAdT At equilibrium or =
dGB = dGA or
VBdP – SBdT = VAdP – SAdT =
(VA- VB)dP = (SA– SB)dT
dP/dT = Δ S/ Δ V Clapeyron equation
To integrate properly we must know how V and S vary with P and T (hence the calculus), but we shall simplify the math and assume V and S are constant
This simplifies our math considerably (but may lead to some errors)
We can check the validity of our assumptions by comparing to experiments or more rigorous calculations (performed by computer)

8. Thermodynamics dG = VdP – SdT, at constant T: dG = VdP
Ideal gas law: V = RT/P
dG = RT(dP/P)
Integrating from the standard pressure P°
to a pressure P :
Gp – G° = RT(lnP - lnP°)
(n G°R)products – S(nG°R)reactants = RT(S(nlnP)products – S(nlnP)reactants)

9. Thermodynamics RT(S(nlnP)products – S(nlnP)reactants) = RT ln K
- DG°R = RT ln K
R = 1.987cal/deg mol
T = K
ln K = log K
DG°R = log K
K = 10-DG°R/1.364
To integrate properly we must know how V and S vary with P and T (hence the calculus), but we shall simplify the math and assume V and S are constant
This simplifies our math considerably (but may lead to some errors)
We can check the validity of our assumptions by comparing to experiments or more rigorous calculations (performed by computer)
Note: R = J/deg mol

10. The Equilibrium Constant
For a one-component system, the Second Law of Thermodynamics is
dG = -SdT + VdP
but dG = ndG = ndµ
where µ = (∂G/∂n)
At constant T and for simplicity an ideal gas,
dG = VdP
From the gas law, V = nRT/P
Integrating,
µ P
n∫µ◦ dµ = nRT∫P◦ dP
µ - µº = RT ln P/P◦
If P◦ is the standard state of the ideal gas at 1 atm pressure and T, P◦=1
µ = µº + RT ln P
(Note that P is unitless as it is a ratio as defined above.)

11. The Equilibrium Constant
For a multi-component gas mixture at equilibrium:
vj+1Aj vi-1Ai-1+ viAi = v1A1 + v2A vjAj
where v refers to stoichiometric coefficients and A to the gas species.
The subscripts j+1 to I refer to the reactants and 1 to j refer to the
products.

12. The Equilibrium Constant
µi = µi° + RT ln Pi.
where Pi = P (ni /n) = PXi
∆Gr = ∑viµi
= ∑vi (µiº + RT ln Pi)
= ∑viµiº + ∑viRT ln Pi
= ∆G◦ + RT ∑ln Pivi
= ∆G◦ + RT ln K
where
P1v1P2v Pj v j
K = Pj+1vj+1Pj+2vj Pivi the equilibrium constant
(note that the value of K is dependent on the quanitites used in the expression and the standard states selected for them. In other words, K is not a universal constant for a reaction, but only for the conditions considered in deriving it.)

13. Thermodynamics DG°R = RT ln K DG°R = DH°R – TDS°R ln K = - DG°R / RT
lnK = -(DH°R/T – DS°R)/ R
dlnK/dT = DH°R / RT2
dln K = DH°R/R (dT/T2)
dln K = DH°R/R (dT/T2)

14. TEMPERATURE DEPENDENCE OF THE EQUILIBRIUM CONSTANT
Van’t Hoff Equation

15. EXAMPLE VAN’T HOFF PLOT

16. Application: Solubility of silica
SiO2(am) + 2H2O(l)  H4SiO40
G°f(H4SiO40) = kcal
G°f(SiO2(am)) = kcal
nG°f(2H2O(l)) = 2( ) kcal
DG°R = – ( ( ) = kcal
K =
DH°R = kcal, endothermic reaction
Log KT = – [3.47x103(cal) / ( x 1.987) x (1/T – 1/298.15)
Log KT = – 758.4(1/T – )

17. Application: Solubility of silica
Log KT = – 758.4(1/T – )
With increasing T, the term (1/T – ) becomes negative, the term – 758.4(1/T – ) becomes positive and log KT > -2.96
If the enthalpy change is negative (exothermic), the last term will be negative with increasing T and
log KT < log K(298.15)
Endothermic reaction: K increases with T
Exothermic reactions: K decreases with T

18. Application: Solubility of silica
Endothermic reaction: K increases with T
Exothermic reactions: K decreases with T
Endothermic dissolution reaction = higher ratio of products versus reactants = increasing solubility with increasing T
Exothermic dissolution reactions = lower ratio of products versus reactants = decreasing solubility with increasing T

19. Second Example: dissolution of feldspar
2KAlSi3O8 + 9H2O + 2H+ 
Al2Si2O5(OH)4 + 2K+ + 4H4SiO4
Equilibrium constant K at 25 and 100°C???
Compound
G°f (kcal)
H°f (kcal)
KAlSi3O8
-894.9
-948.7
H2O
Al2Si2O5(OH)4
-983.5 K+
-67.70
-60.32
H4SiO4
-348.3

20. DH °R = S (n H°f)products - S(n H°f)reactants=
( (-60.32) + 4(-348.3)) –
(2(-948.7) + 9( )) = – =
kcal
Compound
G°f (kcal)
H°f (kcal)
KAlSi3O8
-894.9
-948.7
H2O
Al2Si2O5(OH)4
-983.5 K+
-67.70
-60.32
H4SiO4
-348.3

21. Log KT = -5.207 – 3275.36(-0.000674) Log KT = -5.207 – -2.208
DG°R = kcal
DH °R = kcal
K = 10-DG°R/1.364 =
Log K = at K
Log K at K?
R = 1.987cal/deg mol
Log KT = – ( )
Log KT = –
Log KT =

22. DG°R = kcal
DH °R = kcal
K = 10-DG°R/1.364 =
Log K = at K
Log K at K =
Endothermic dissolution reaction =
higher ratio of products versus reactants at higher T=
increasing solubility with increasing T