Thermodynamicsa system:Some portion of the universe that you wish to studyThe surroundings:The adjacent part of the universe outside the systemChanges in a system are associated with the transfer of energyNatural systems tend toward states of minimum energy

Energy StatesUnstable: falling or rollingStable: at rest in lowest energy stateMetastable: in low-energy perchFigure 5-1. Stability states. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

Gibbs Free EnergyGibbs free energy is a measure of chemical energyAll chemical systems tend naturally toward states of minimum Gibbs free energyG = H - TSWhere:G = Gibbs Free EnergyH = Enthalpy (heat content)T = Temperature in KelvinsS = Entropy (can think of as randomness)

Thermodynamicsa Phase: a mechanically separable portion of a systemMineralLiquidVapora Reaction: some change in the nature or types of phases in a systemreactions are written in the form:reactants = products

ThermodynamicsThe change in some property, such as G for a reaction of the type:2 A + 3 B = C + 4 DDG = S (n G)products - S(n G)reactants = GC + 4GD - 2GA - 3GB

ThermodynamicsFor a phase we can determine V, T, P, etc., but not G or HWe can only determine changes in G or H as we change some other parameters of the systemExample: measure DH for a reaction by calorimetry - the heat given off or absorbed as a reaction proceedsArbitrary 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

ThermodynamicsIn our calorimeter we can then determine DH for the reaction:

Si (metal) + O2 (gas) = SiO2 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 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

ThermodynamicsFor other temperatures and pressures we can use the equation:dG = VdP SdT (ignoring DX for now)where V = volume and S = entropy (both molar)

ThermodynamicsIf V and S are constants, our equation reduces to:GT2 P2 - GT1 P1 = V(P2 - P1) - S (T2 - T1)

which aint bad!

ThermodynamicsIn Worked Example 1 we usedGT2 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 pressuresAgreement is quite good (< 2% for change of 500o and 500 MPa or 17 km)

ThermodynamicsSummary thus far:G is a measure of relative chemical stability for a phaseWe can determine G for any phase by measuring H and S for the reaction creating the phase from the elementsWe can then determine G at any T and P mathematicallyMost accurate if know how V and S vary with P and TdV/dP is the coefficient of isothermal compressibilitydS/dT is the heat capacity (Cp)Use?If we know G for various phases, we can determine which is most stableWhy 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 stableFigure 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. TemperaturedG = VdP - SdT at constant pressure: dG/dT = -SBecause S must be (+) G for a phase decreases as T increasesWould 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. TemperatureSlope of GLiq > Gsol since Ssolid < SliquidA: Solid more stable than liquid (low T)B: Liquid more stable than solid (high T)Slope dP/dT = -SSlope S < Slope LEquilibrium at TeqGLiq = GSolFigure 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: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 changesdDG is then the change in DG as T and P are variedDG 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

Figures I dont use in classFigure 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 dont use in classFigure 5-5. Piston-and-cylinder apparatus to compress a gas. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

a system:Some portion of the universe that you wish to studya glass of waterplagioclasethe mantleChanges in a system are associated with the transfer of energylift an object: stored chemical potentialdrop an object: potential kineticpump up a bicycle tire: chemical mechanical heat (friction + compression)add an acid to a base: chemical heat

The side of the reaction with lower G will be more stableHow 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

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 constantThis 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)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

All of these questions depend on reactions comparing 2 or more phases

Consider the phase diagram (as we anticipate igneous petology)We can apply thermodynamics qualitatively to assess these diagrams

From dG = VdP - SdT we can deduce that at constant pressure: dG = -SdTAnd since S must be (+) G for a phase decreases as T increases