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

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.

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)

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

Thermodynamics The change in some property, such as G for a reaction of the type: 2 A + 3 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

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 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

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) = 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 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

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)

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

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)

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

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.

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

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?

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.

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

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.

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.