Part 1 - Thermodynamics

1. Part 1 - Thermodynamics We will begin with a brief review of the Thermodynamics and Phase Diagram construction. We discussed these topics in Mineralogy

2. Mineral Stability/Equilibrium Phase Stability is defined by the state (solid, liquid, gas) and internal structure of a compositionally homogeneous substance under particular external conditions of pressure and temperature A Mineral of constant composition is considered a solid phase Mineral stability is commonly portrayed on a Phase Diagram

4. Stability of a phase (or mineral) is related to its internal energy, which strives to be as low as possible under the external conditions. Metastability exists in a phase when its energy is higher than P-T conditions indicate it should be. (1) Activation Energy is the energy necessary to push a phase from its metastable state to its stable state. (2-1) Equilibrium exists when the phase is at its lowest energy level for the current P-T conditions. (3) (Two minerals that are reactive with one another, may be found to be in equilibrium at particular P-T conditions which on phase diagrams are recognized as phase boundaries) Note: most metamorphic and igneous minerals at the earth�s surface are now metastable, but often were in equilibrium when formed

5. Energy States Unstable: falling or rolling

6. Components and Phases Components are the chemical entities necessary to define all the potential phases in a system of interest

7. HOWEVER!

8. Degrees of Freedom f by Examples If T and P can change without changing the mineral assemblage, the system has two degrees of freedom f = 2 If neither T or P can change without changing the mineral assemblage, the system has zero degrees of freedom f = 0 If T and P must change together to maintain the same mineral, the system has one degree of freedom f = 1

9. The Phase Rule The number of minerals that may stably coexist is limited by the number of chemical components p + f = c + 2 where P is the number of mineral phases, c the number of chemical components, and f is the number of degrees of freedom.

10. Basic Thermodynamics The theoretical basis of phase equilibrium Three Laws of Thermodynamics 1. First Law: Change in Internal Energy dE = dQ � dW Q � heat energy W � work = F * distance (notice distance is a length) Pressure has units Force/Area = (F/dist2) so F* dist = P * area * dist = P * V At constant pressure the change in work dW = PdV So if Pressure is constant: (1) dE = dQ � PdV where dV is thermal expansion or contraction

11. Second and Third Laws of Thermodynamics 2. All substances move to the greatest state of disorder (highest Entropy �S�) for a particular T and P. (2) dQ/T = dS so dQ =TdS �The state of greatest order [lowest S] is at the lowest temperature. With increasing temperature, disorder becomes more prevalent.� Minerals with simple atomic structure and simple chemistry have lower entropy. (1) dE = dQ � PdV = TdS �PdV first law rewritten 3. At absolute zero (0�K), Entropy is zero Does this suggest how to measure S? Can we measure changes in S by keeping track of temperature as we add heat to a system? �The entropy of a pure crystalline substance can therefore be obtained directly from heat-capacity measurements by assuming that S0 (at 0o K) is zero� Wood and Fraser (1978) p38

12. Gibbs Free Energy Define G � the energy of a system in excess of its internal energy E. This is the energy necessary for a reaction to proceed For a single phase define G = E + PV � TS Differentiating dG = dE +PdV +VdP -TdS � SdT (product rule) Also dE = dQ - PdV = TdS � PdV (1) first law rewritten so dG = TdS � PdV + PdV + VdP -TdS � SdT = VdP �SdT This is equation (3) dG = VdP �SdT Stable changes have negative DG. If T = constant dT = 0, then dG = VdP, if V decreases, P can increase without increasing G (2) at constant T (dG/dP)T = V (dense (low V) phases are favored at high P) If P = constant dP = 0, then dG = -SdT, if T increases then S can increase without increasing G (3) at constant P (dG/dT)P = -S (disordered phases (high S) are favored at high T)

13. Enthalpy Earlier we saw G(T,p) = E + pV - TS But the Enthalpy H(S,p) = E + pV So DG = DH �TDS DH can be measured in the laboratory with a calorimeter. So can DS. You measured DH in chemistry lab.

14. Gibbs Free Energy Gibbs free energy is a measure of chemical energy

15. Thermodynamics

17. Constructing Phase Diagrams with for systems with reactions DG = DH �TDS DG 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

18. To get an equilibrium curve for a phase diagram, could use dDG = DVdP - DSdT and G, S, V values for albite, jadeite and quartz to calculate the conditions for which DG of the reaction: Ab = Jd + Q is equal to 0

19. NaAlSi3O8 = NaAlSi2O6 + SiO2 Albite = Jadeite + Quartz P - T phase diagram of the equilibrium curve How do you know which side has which phases? Calculate DG for products and reactant for pairs of P and T, spontaneous reaction direction at that T P will have negative DG When DG < 0 the product is stable Where on this diagram is Albite stable? Jadeite? In what direction would be the Ab + Qtz = Ne reaction? Why?Where on this diagram is Albite stable? Jadeite? In what direction would be the Ab + Qtz = Ne reaction? Why?

20. Clausius -Clapeyron Equation Defines the state of equilibrium between reactants and products in terms of S and V From Eqn.3, if dG =0, dP/dT = ?S / ?V (eqn.4) The slope of the equilibrium curve will be positive if S and V both decrease or increase with increased T and P

21. To get the slope, at a boundary DG is 0 dDG = 0 = DVdP - DSdT

22. Return to dG = VdP � SdT. For an isothermal process dT is zero, so:

23. Ideal Gas As P increases V decreases PV=nRT Ideal Gas Law P = pressure V = volume T = temperature n = # of moles of gas R = gas constant = 8.3144 J mol-1 K-1 Gas Pressure-Volume Relationships

24. Gas Pressure-Volume Relationships Since we can substitute RT/P for V (for a single mole of gas), thus: and, since R and T are certainly independent of P:

25. Gas Pressure-Volume Relationships And since GP2 - GP1 = RT (ln P2 - ln P1) = RT ln (P2/P1) Thus the free energy of a gas phase at a specific P and T, when referenced to a standard state of 0.1 MPa becomes: GP, T - GT = RT ln (P/Po) G of a gas at some P and T is equal to G in the reference state (same T and 0.1 MPa) + a pressure term

26. Gas Pressure-Volume Relationships The form of this equation is very useful GP, T - GT = RT ln (P/Po) For a non-ideal gas (more geologically appropriate) the same form is used, but we substitute fugacity ( f ) for P where f = gP g is the fugacity coefficient Tables of fugacity coefficients for common gases are available At low pressures most gases are ideal, but at high P they are not

27. Dehydration Reactions Ms + Qtz = Kspar + Sillimanite + H2O We can treat the solids and gases separately GP, T - GT = DVsolids (P - 0.1) + RT ln (P/0.1) (isothermal) The treatment is then quite similar to solid-solid reactions, but you have to solve for the equilibrium P by iteration. iterative methods are those which are used to produce approximate numerical solutions to problems. Newton's method is an example of an iterative method.

28. Dehydration Reactions

29. Solid Solutions: T-X relationships Ab = Jd + Q was calculated for pure phases When solid solution results in impure phases the activity of each phase is reduced Use the same form as for gases (RT ln P or ln f) Instead of fugacity f, we use activity a Ideal solution: ai = Xi y = # of crystallographic sites in which mixing takes place Non-ideal: ai = gi Xi where gamma gi is the activity coefficient

30. Solutions: T-X relationships Example: orthopyroxenes (Fe, Mg)SiO3 Real vs. Ideal Solution Models, activity vs composition

31. Solutions: T-X relationships Back to our reaction: Simplify for now by ignoring dP and dT For a reaction such as: aA + bB = cC + dD At a constant P and T: 0 = where:

32. Compositional variations Effect of adding Ca to albite = jadeite + quartz plagioclase = Al-rich Cpx + Q DGT, P = DGoT, P + RTl n K Let�s say DGoT, P was the value that we calculated for equilibrium in the pure Na-system (= 0 at some P and T) DGoT, P = DG298, 0.1 + DV (P - 0.1) - DS (T-298) = 0 By adding Ca we will shift the equilibrium by RTl n K We could assume ideal solution and

33. Compositional variations So now we have: DGT, P = DGoT, P + RTln since Q is pure X quartz = 1 DGoT, P = 0 as calculated for the pure system at P and T DGT, P is the shifted DG due to the Ca added (no longer 0) Thus we could calculate a DV(P-Peq) that would bring DGT, P back to 0, solving for the new Peq

34. Effect of adding Ca to albite = jadeite + quartz DGP, T = DGoP, T + RTln K Compositional variations

35. The Payoff Our experiments and calculations allow us to construct the 3-D plot in (a), and to project the mineral with the lowest free energy at each PT onto the graph in (b).