1. Natural systems tend toward states of minimum energy
Thermodynamics
a system:
Some portion of the universe that you wish to study
The surroundings:
The adjacent part of the universe outside the system
Changes in a system are associated with the transfer of energy
a system:
Some portion of the universe that you wish to study
a glass of water
plagioclase
the mantle
Changes in a system are associated with the transfer of energy
lift an object: stored chemical  potential
drop an object: potential  kinetic
pump up a bicycle tire: chemical  mechanical  heat (friction + compression)
add an acid to a base: chemical  heat
Natural systems tend toward states of minimum energy

2. Energy States Unstable: falling or rolling
Stable: at rest in lowest energy state
Metastable: in low-energy perch
Figure 5.1. Stability states. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

3. Gibbs Free Energy Gibbs free energy is a measure of chemical energy
All chemical systems tend naturally toward states of minimum Gibbs free energy
G = H - TS
Where:
G = Gibbs Free Energy
H = Enthalpy (heat content)
T = Temperature in Kelvins
S = Entropy (can think of as randomness)

4. Thermodynamics a Phase: a mechanically separable portion of a system
Mineral
Liquid
Vapor
a Reaction: some change in the nature or types of phases in a system
reactions are written in the form:
reactants = products

5. Thermodynamics
The change in some property, such as G for a reaction of the type:
2 A B = C + 4 D
DG = S (n G)products - S(n G)reactants
= GC + 4GD - 2GA - 3GB
The side of the reaction with lower G will be more stable
How do we go about determining this for a reaction?
First we must be able to determine G for the phases in the reaction at any P and T

6. Thermodynamics
For a phase we can determine V, T, P, etc., but not G or H
We can only determine changes in G or H as we change some other parameters of the system
Example: measure DH for a reaction by calorimetry - the heat given off or absorbed as a reaction proceeds
Arbitrary reference state and assign an equally arbitrary value of H to it:
Choose K and 0.1 MPa (lab conditions)
...and assign H = 0 for pure elements (in their natural state - gas, liquid, solid) at that reference

7. Si (metal) + O2 (gas) = SiO2 DH = -910,648 J/mol
Thermodynamics
In our calorimeter we can then determine DH for the reaction:
Si (metal) + O2 (gas) = SiO DH = -910,648 J/mol
= molar enthalpy of formation of quartz (at 298, 0.1)
It serves quite well for a standard value of H for the phase
Entropy has a more universal reference state: entropy of every substance = 0 at 0K, so we use that (and adjust for temperature)
Then we can use G = H - TS to determine G of quartz
= -856,288 J/mol

8. z Thermodynamics G VdP SdT - =
For other temperatures and pressures we can use the equation:
dG = VdP – SdT (ignoring DX for now)
where V = volume and S = entropy (both molar)
We can use this equation to calculate G for any phase at any T and P by integrating z G
VdP
SdT T P 2 1 - =
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)

9. Thermodynamics If V and S are constants, our equation reduces to:
GT2 P2 - GT1 P1 = V(P2 - P1) - S (T2 - T1)
which ain’t bad!
The P and T corrections can be done separately in any order, since we are concerned only with the initial and final states, not the path the system followed to get from one to the other

10. Thermodynamics In Worked Example 1 we used
GT2 P2 - GT1 P1 = V(P2 - P1) - S (T2 - T1)
and G298, 0.1 = -856,288 J/mol to calculate G for quartz at several temperatures and pressures
Low quartz
Eq. 1
SUPCRT
P (MPa)
T (C)
G (J) eq. 1
G(J)
V (cm3)
S (J/K)
0.1 25
-856,288
-856,648
22.69
41.36
500
-844,946
-845,362
22.44
40.73
-875,982
-890,601
23.26
96.99
-864,640
-879,014
23.07
96.36
Agreement is quite good
(< 2% for change of 500o and 500 MPa or 17 km)

11. Thermodynamics Summary thus far:
G is a measure of relative chemical stability for a phase
We can determine G for any phase by measuring H and S for the reaction creating the phase from the elements
We can then determine G at any T and P mathematically
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)
Use?
If we know G for various phases, we can determine which is most stable
Why is melt more stable than solids at high T?
Is diamond or graphite stable at 150 km depth?
What will be the effect of increased P on melting?
All of these questions depend on reactions comparing 2 or more phases

12. Does the liquid or solid have the larger volume?
High pressure favors low volume, so which phase should be stable at high P?
Does liquid or solid have a higher entropy?
Consider the phase diagram (as we anticipate igneous petology)
We can apply thermodynamics qualitatively to assess these diagrams
Figure 5.2. Schematic P-T phase diagram of a melting reaction. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.
High temperature favors randomness, so which phase should be stable at higher T?
We can thus predict that the slope of solid-liquid equilibrium should be positive and that increased pressure raises the melting point.

13. Does the liquid or solid have the lowest G at point A?
What about at point B?
Figure 5-2. Schematic P-T phase diagram of a melting reaction. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.
The phase assemblage with the lowest G under a specific set of conditions is the most stable

14. Free Energy vs. Temperature
dG = VdP - SdT at constant pressure: dG/dT = -S
Because S must be (+) G for a phase decreases as T increases
Figure 5.3. Relationship between Gibbs free energy and temperature for a solid at constant pressure. Teq is the equilibrium temperature. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.
From dG = VdP - SdT we can deduce that at constant pressure: dG = -SdT
And since S must be (+) G for a phase decreases as T increases
Would the slope for the liquid be steeper or shallower than that for the solid?

15. 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 dP/dT = -S
Slope S < Slope L
Equilibrium at Teq
GLiq = GSol
Figure 5.3. Relationship between Gibbs free energy and temperature for the solid and liquid forms of a substance at constant pressure. Teq is the equilibrium temperature. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

16. Now consider a reaction, we can then use the equation:
dDG = DVdP - DSdT (again ignoring DX)
For a reaction of melting (like ice  water)
DV is the volume change involved in the reaction (Vwater - Vice)
similarly DS and DG are the entropy and free energy changes
dDG is then the change in DG as T and P are varied
DG is (+) for S  L at point A (GS < GL)
DG is (-) for S  L at point B (GS > GL)
DG = 0 for S  L at point x (GS = GL)
DG for any reaction = 0 at equilibrium

17. dP dT DS = DV dDG = 0 = DVdP - DSdT Y X
Pick any two points on the equilibrium curve
DG = ? at each
Therefore dDG from point X to point Y = = 0 dP dT DS = DV
dDG = 0 = DVdP - DSdT

18. Figures I don’t use in class
Figure 5.4. Relationship between Gibbs free energy and pressure for the solid and liquid forms of a substance at constant temperature. Peq is the equilibrium pressure. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

19. Figures I don’t use in class
Figure 5.5. Piston-and-cylinder apparatus to compress a gas. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.