ThermodynamicsThermodynamics 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

Natural systems tend toward states of minimum energy

Energy StatesEnergy 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.

Gibbs Free EnergyGibbs Free EnergyGibbs free energy is a measure of chemical energy

All chemical systems tend naturally toward states of minimum Gibbs free energy

G = H - TSWhere:

G = Gibbs Free EnergyH = Enthalpy (heat content)T = Temperature in KelvinsS = Entropy (can think of as randomness)

ThermodynamicsThermodynamicsa 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

ThermodynamicsThermodynamicsThe change in some property, such as G for a reaction of the type:

2 A + 3 B = C + 4 DG = (n G)products - (n G)reactants

= GC + 4GD - 2GA - 3GB

ThermodynamicsThermodynamicsFor 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 systemExample: measure H 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 298.15 K and 0.1 MPa (lab conditions)

...and assign H = 0 for pure elements (in their natural state - gas, liquid, solid) at that reference

ThermodynamicsThermodynamicsIn our calorimeter we can then determine H for the reaction:

Si (metal) + O2 (gas) = SiO2 H = -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 phaseEntropy 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

ThermodynamicsThermodynamicsFor other temperatures and pressures we can use the equation:

dG = VdP – SdT (ignoring X 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 zzG G VdP SdTT P T P T

T

P

P

2 1 11

2

1

2

2

ThermodynamicsThermodynamicsIf V and S are constants, our equation reduces to:

GT2 P2 - GT1 P1 = V(P2 - P1) - S (T2 - T1)

which ain’t bad!

ThermodynamicsThermodynamicsIn 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. 1G(J) V (cm3) S (J/K)0.1 25 -856,288 -856,648 22.69 41.36500 25 -844,946 -845,362 22.44 40.730.1 500 -875,982 -890,601 23.26 96.99500 500 -864,640 -879,014 23.07 96.36

Agreement is quite good (< 2% for change of 500o and 500 MPa or 17 km)

ThermodynamicsThermodynamicsSummary 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?

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?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.

Figure 5.2. Schematic P-T phase diagram of a melting reaction. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

Does the liquid or solid have the lowest G at point A?

What about at point B?

The phase assemblage with the lowest G under a specific set of conditions is the most stable

Figure 5-2. Schematic P-T phase diagram of a melting reaction. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

Free Energy vs. Free Energy vs. TemperatureTemperature

dG = VdP - SdT at constant pressure: dG/dT = -S

Because 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?

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.

Free Energy vs. Free Energy vs. TemperatureTemperature

Slope of GLiq > Gsol

since Ssolid < Sliquid

A: Solid more stable than liquid (low T)

B: Liquid more stable than solid (high T) Slope P/T = -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.

Now consider a reaction, we can then use the equation:

dG = VdP - SdT(again ignoring X)For a reaction of melting (like ice water)

V is the volume change involved in the reaction (Vwater - Vice)

similarly S and G are the entropy and free energy changesdG is then the change in G as T and P are

varied G is (+) for S L at point A (GS < GL) G is (-) for S L at point B (GS > GL) G = 0 for S L at point x (GS = GL)

G for any reaction = 0 at equilibrium

Pick any two points on the equilibrium curve G = ? at each Therefore dG from point X to point Y = 0 - 0 = 0

dG = 0 = VdP - SdT

X

Y

dPdT

SV

Figures I don’t use in Figures I don’t use in classclass

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.

Figures I don’t use in Figures I don’t use in classclass

Figure 5.5. Piston-and-cylinder apparatus to compress a gas. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.