1. Ch 27 more Gibbs Free Energy Gibbs free energy is a measure of chemical energy Gibbs free energy for a phase: G = E + PV – TS => G = H - TS Where: G = Gibbs Free Energy E = Internal Energy H = Enthalpy (heat content) = E + PV T = Temperature in degrees Kelvin o K P = Pressure, V = Volume S = Entropy (randomness, disorder)

2. Changes Thermodynamics treats changes Thermodynamics treats changes Regardless of path G = E + PV – TS Regardless of path G = E + PV – TS We should rewrite the equation for Gibbs Free Energy in terms of changes,  G We should rewrite the equation for Gibbs Free Energy in terms of changes,  G  G = E + P  V – T  S for P, T constant  G = E + P  V – T  S for P, T constant  G =  H – T  S  G =  H – T  S  pronounced “delta” means “the change in”  H can be measured in the laboratory with a calorimeter.  S can also be measured with heat capacity measurements. Values are tabulated in books. The change in Gibbs free energy, ΔG, in a reaction is a very useful parameter. It can be thought of as the maximum amount of work obtainable from a reaction.

3. Thermodynamics For a reaction at other temperatures and pressures The change in Gibbs Free Energy is d  G =  VdP -  SdT We can use this equation to calculate G for any phase at any T and P by integrating the above equation. If V and S are ~constants, If V and S are ~constants, dG = V dP – S dT our equation reduces to: G T2 P2 - G T1 P1 = V(P 2 - P 1 ) - S (T 2 - T 1 ) FOR A SOLID_SOLID REACTION

4.  Suppose 3A + 2B = 2C +1D reactants = products  G = 2G C +1G D -3G A – 2G B  Gibbs Free Energy (G) is measured in KJ/mol or Kcal/mol  One small calorie cal ~ 4.2 Joules J Gibbs for a chemical reaction Hess’s Law applied to Gibbs for a reaction 298.15K, 0.1 MPa Same procedure for  H,  S,  V

5. Which direction will the reaction go?  G for a reaction of the type: 2 A + 3 B = C + 4 D  G =  (n G) products -  (n G) reactants = G C + 4G D - 2G A - 3G B = G C + 4G D - 2G A - 3G B The reaction with negative  G will be more stable, i.e. if  G is negative for the reaction as written, the reaction will go to the right “For chemical reactions, we say that a reaction proceeds to the right when  G is negative and the reaction proceeds to the left when  G is positive.” Brown, LeMay and Bursten (2006) Virtual Chemistry p 163 Same procedure for  H,  S,  V

6. Since G = E + PV – TS And we saw the slope of a sum is the sum of the slopes Differentiating dG = dE +PdV +VdP -TdS – SdT What is dE? dE = dQ – dW First Law, and dQ =TdS 2 nd law So dE = dQ - PdV => dE = TdS – PdV Most of these terms cancel, so dG = VdP –SdT And if we need the changes when moving to a new T,P d  G =  VdP -  SdT

7. To get an equilibrium curve for a phase diagram, could use d  G =  VdP -  SdT and G, S, V values for Albite, Jadeite and Quartz to calculate the conditions for which  G of the reaction: Ab = Jd + Q is equal to 0 is equal to 0 From G values for each phase at 298K and 0.1 MPa list  G 298, 0.1 for the reaction, do the same for  V and  S From G values for each phase at 298K and 0.1 MPa list  G 298, 0.1 for the reaction, do the same for  V and  S  G at equilibrium = 0, so we can calculate an isobaric change in T that would be required to bring  G 298, 0.1 to 0  G at equilibrium = 0, so we can calculate an isobaric change in T that would be required to bring  G 298, 0.1 to 0 0 -  G 298, 0.1 = -  S (T eq - 298)(at constant P) Similarly we could calculate an isothermal change Similarly we could calculate an isothermal change 0 -  G 298, 0.1 = -  V (P eq - 0.1)(at constant T) Method:

8. NaAlSi 3 O 8 = NaAlSi 2 O 6 + SiO 2 Albite = Jadeite + Quartz P - T phase diagram of the equilibrium curve How do you know which side has which phases? Calculate  G for products and reactant for pairs of P and T, spontaneous reaction direction at that T P will have negative  G When  G < 0 the product is stable Figure 27-1. Temperature-pressure phase diagram for the reaction: Albite = Jadeite + Quartz calculated using the program TWQ of Berman (1988, 1990, 1991).

9. 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 = ΔS / ΔV (eqn.4) slope The slope of the equilibrium curve will be positive if S and V both decrease or increase with increased T and P dG = VdP –SdT

10. To get the slope, at a boundary  G is 0 d  G = 0 =  VdP -  SdT solve dP dT S V    Figure 27-1. Temperature- pressure phase diagram for the reaction: Albite = Jadeite + Quartz calculated using the program TWQ of Berman (1988, 1990, 1991). Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall. gives us the slope

11. End of review

12. Return to dG = VdP – SdT. For an isothermal process dT is zero, so: GGVdP PP P P 21 1 2  Gas Phases For solids it was fine to assume V stays ~ constant For gases this assumption is wrong A gas compresses as P increases How can we define the relationship between V and P for a gas?

13. Gas Laws 1600’s to 1800’s Combined as ideal gas law: n= # moles, and R is the universal gas constant R = 8.314472 N·m·K −1 ·mol −1 Pressure times Volume is a constant Increase Temp, Volume increases Increase Temp, Pressure increases Increase moles of gas, Volume increases

14. 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 = 8.3144 J mol -1 K -1 So P x V is a constant at constant T Gas Pressure-Volume Relationships Figure 5-5. Piston-and-cylinder apparatus to compress a gas. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

15. Gas Pressure-Volume Relationships Since we can substitute RT/P for V (for a single mole of gas), thus: we can substitute RT/P for V (for a single mole of gas), thus: and, since R and T are certainly independent of P: GG VdP PP P P 21 1 2  GG RT P dP PP P P 21 1 2  GGRT P dP PP P P 21 1 2  1

16. Logarithms Logarithms (Logs) are just exponents Logarithms (Logs) are just exponents if b y = x then y = log b x if b y = x then y = log b x log 10 (100) = 2 because 10 2 = 100 log 10 (100) = 2 because 10 2 = 100 Natural logs (ln) use e = 2.718 as a base Natural logs (ln) use e = 2.718 as a base For example ln(1) = log e (1) = 0 For example ln(1) = log e (1) = 0 because e 0 = (2.718) 0 = 1 because e 0 = (2.718) 0 = 1 Anything to the zero power is one. Anything to the zero power is one. b x /b y = b x-y so log b x - log b y = log b (x/y) b x /b y = b x-y so log b x - log b y = log b (x/y)

17. Early on we looked at slopes and areas, and defined derivatives and integrals. We can just look these up in tables. Here is another slope d ln u = 1 du dx u dx The area under the curve is the reverse operation

18. Gas Pressure-Volume Relationships b x /b y = b x-y so log b x - log b y = log b (x/y)

19. Gas Pressure-Volume Relationships The form of this equation is very useful G P, T - G T = RT ln (P/P o ) G P, T - G T = RT ln (P/P o ) For a non-ideal gas (more geologically appropriate) the same form is used, but we substitute fugacity ( f ) for P where f =  P  is the fugacity coefficient G P, T - G o T = RT ln (f /P o ) so G P, T - G o T = RT ln (f /P o ) so  H2O ranges 0.1 – 1.5,  CO2 ranges 2 – 50 At low pressures most gases are ideal, but at high P they are not o

20. 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 RT ln f ) Instead of fugacity f, we can use activity a Ideal solution: a i = X i y = # of crystallographic sites in which mixing takes place Non-ideal: a i =  i X i where gamma  i is the activity coefficient y y

21. Dehydration Reactions Ms + Qtz = Kspar + Sillimanite + H 2 O Ms + Qtz = Kspar + Sillimanite + H 2 O We can treat the solids and gases separately We can treat the solids and gases separately G P, T - G T =  V solids (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 pressure P by iteration. The treatment is then quite similar to solid-solid reactions, but you have to solve for the equilibrium pressure 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. Iterative methods are those which are used to produce approximate numerical solutions to problems. Newton's method is an example of an iterative method. Iterative methodsNewton's method Iterative methodsNewton's method o

22. Newton’s Method

23. Dehydration Reactions dP dT S V    Figure 27-2. Pressure-temperature phase diagram for the reaction muscovite + quartz = Al 2 SiO 5 + K-feldspar + H 2 O, calculated using SUPCRT (Helgeson et al., 1978). Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall. Muscovite is unstable at High T while Qtz present, dehydrates by reacting w Qtz, forms K-spar and Al- silicate + water.  V high at low P so high  V gas ->  S/  V low (gentle slope)  V low at high P (already near limit of compressibility) so ->  S/  V high (steep slope) Result: Characteristic concave shape; decarbonation and other devolitilazation reactions are similar

24. Ch 27b Geothermobarometry For any reaction with one or more variable components, at any given P,T,we can solve for the equilibrium curve using For any reaction with one or more variable components, at any given P,T,we can solve for the equilibrium curve using  G=0=  G 0 + RT ln K (27-17)  G=0=  G 0 + RT ln K (27-17) So ln K = -  G 0 /RT So ln K = -  G 0 /RT

25. Equilibrium Constant K G P, T = G o T + RT ln (P/0.1 MPa ) G P, T = G o T + RT ln (P/0.1 MPa ) At equilibrium the ratio in the parentheses, regardless of how it is expressed (Pressures, chemical potentials, activities), is a constant, called the equilibrium constant, K At equilibrium the ratio in the parentheses, regardless of how it is expressed (Pressures, chemical potentials, activities), is a constant, called the equilibrium constant, K G P, T = G o T + RT ln (K) G P, T = G o T + RT ln (K)

26. Calculating an Equilibrium Constant for a Reaction The units M (molar) are moles per liter A mixture of gasses in an inclusion was allowed to reach equilibrium. 0.10 M NO, 0.10 M H 2, 0.05 M N 2 and 0.10 M H 2 O was measured. Calculate the Equilibrium Constant for the equation:

27. K for an example reaction For a reaction 2A + 3B = C + 4D For a reaction 2A + 3B = C + 4D K = X C X 4 D.  C  4 D K = X C X 4 D.  C  4 D X 2 A X 3 B.   A  3 B X 2 A X 3 B.   A  3 B where X i is the mole fraction and  i is the correction, i.e. the activity coefficient, so i.e. K = K D. K  We will define the Distribution Coefficient, K D, again below. We saw it earlier in Chapter 9.

28. G P, T - G o T = RT ln (K) and at equilibrium G P, T = 0 G P, T - G o T = RT ln (K) and at equilibrium G P, T = 0 ln K = -  G 0 /RT ln K = -  G 0 /RT but  G o =  H o –T  S o +  V dP but  G o =  H o –T  S o +  V dPSo ln K = -  H o /RT +  S o /R - (  V/RT) dP (27-26) ln K = -  H o /RT +  S o /R - (  V/RT) dP (27-26)

29. A Garnet-Biotite Reaction Below is the stoichiometric equation for the Fe-Mg exchange in the reaction between the biotites and Ca-free garnets: Below is the stoichiometric equation for the Fe-Mg exchange in the reaction between the biotites and Ca-free garnets: Fe 3 Al 2 Si 3 O 12 + KMg 3 Si 3 AlO 10 (OH) 2 = Mg 3 Al 2 Si 3 O 12 + KFe 3 Si 3 AlO 10 (OH) 2 Fe 3 Al 2 Si 3 O 12 + KMg 3 Si 3 AlO 10 (OH) 2 = Mg 3 Al 2 Si 3 O 12 + KFe 3 Si 3 AlO 10 (OH) 2 Almandine + Phlogopite = Pyrope + Annite Almandine + Phlogopite = Pyrope + Annite This false color image of a garnet crystal in equilibrium with biotites. The garnet passed from an initial composition of Magnesium-rich Pyrope in its core to Fe-rich Almandine on its rim. Phlogopite is the magnesium end-member of the biotite solid solution series Annite is the iron end-member of the biotite solid solution series ln K = -  H/RT +  S/R - (  V/RT) dP

30. Garnet-Biotite Geothermometer

31. The Distribution Coefficient K D

32. The Garnet - Biotite Fe –Mg exchange reaction Figure 27-5. Graph of l nK vs. 1/T (in Kelvins) for the Ferry and Spear (1978) garnet-biotite exchange equilibrium at 0.2 GPa from Table 27-2. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall. Application to  H and  S determination lnK = -  H/RT +  S/R - (  V/RT) dP This is a line! From (27-26) we can extract  H from the slope and  S from the intercept! y = slope. x + b

33. The GASP geobarometer Garnet-aluminosilicate-silica-plagioclase Figure 27-8. P-T phase diagram showing the experimental results of Koziol and Newton (1988), and the equilibrium curve for reaction (27-37). Open triangles (yellow) indicate runs in which An grew, closed triangles (red) indicate runs in which Grs + Ky + Qtz grew, and half- filled triangles (yellow/red) indicate no significant reaction. The univariant equilibrium curve is a best-fit regression of the data brackets. The line at 650 o C is Koziol and Newton’s estimate of the reaction location based on reactions involving zoisite. The shaded area is the uncertainty envelope. After Koziol and Newton (1988) Amer. Mineral., 73, 216- 233Geothermobarometry

34. Assessment of reaction textures Identify which minerals are early, which are late, and which are part of a stable assemblage. Identify which minerals are early, which are late, and which are part of a stable assemblage. Early minerals are likely to be inclusions or broken. Early minerals are likely to be inclusions or broken. Late minerals may be in cracks or strain shadows. Late minerals may be in cracks or strain shadows. Minerals that are in textural equilibrium should not be separated by reaction zones. Minerals that are in textural equilibrium should not be separated by reaction zones.

35. The GASP geobarometer 3CaAl 2 Si 2 O 8 = Ca 3 Al 2 Si 3 O 12 + 2Al 2 SiO 5 + SiO 2 3CaAl 2 Si 2 O 8 = Ca 3 Al 2 Si 3 O 12 + 2Al 2 SiO 5 + SiO 2 3 Anorthite = Grossular + 2 Al 2 SiO 5 + Quartz 3 Anorthite = Grossular + 2 Al 2 SiO 5 + Quartz Garnet-aluminosilicate-silica-plagioclase These Grossular garnets (in association with SiO 2 and Al 2 SiO 5 ) have Anorthite plagioclase rims. They tell us only that the rock passed somewhere through this equilibrium line.

36. However … if we have another mineral equilibrium, we may get a crossing line on our PT diagram if we have another mineral equilibrium, we may get a crossing line on our PT diagram Pyrophyllite is Al 2 Si 4 O 10 (OH) 2

38. Determining P-T-t History Zoning in Pl gives successive stages in P-T history; Zoning in Pl gives successive stages in P-T history; if we can date these different stages, then we can if we can date these different stages, then we can get P-T-t path. get P-T-t path. How is this done?

39. GASP Gar-Bt Spear’s Classic Paper 1 bar = 100000 pascal1 mb [mbar, millibar] = 100 Pascals 1 atmosphere [atm, standard] = 1.01 bar Spears 15-47

40. Garnet-Biotite line Calculate K D then draw in a Garnet-Biotite line Calculate Pressures in Kilobars for 400 and 700C 1000 bar = 1 kilobar GASP Line Draw in the GASP Line Crossing Point gives the P-T conditions You have a thick section of a metamorphic rock containing Plagioclase, Biotites, Garnets and aluminosilicates (Al 2 SiO 5 ), so you run electron microprobe scans across interesting areas. In a scan where garnet contacts biotite, you find X Mg = 0.310, X Fe = 0.690 for Garnets; and X Mg = 0.606, X Fe = 0.324 for Biotite. Find the Pressure and Temperature