Ch 27 more Gibbs Free Energy Gibbs free

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)

Changes n Thermodynamics treats changes n Regardless of path G = E + PV – TS n We should rewrite the equation for Gibbs Free Energy in terms of changes, D G n DG = E + P DV – T DS for P, T constant n DG = DH – T DS D = pronounced “delta” means “the change in” 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. DH can be measured in the laboratory with a calorimeter. DS can also be measured with heat capacity measurements. Values are tabulated in books.

Thermodynamics For a reaction at other temperatures and pressures The change in Gibbs Free Energy is d. DG = DVd. P - DSd. T We can use this equation to calculate G for any phase at any T and P by integrating the above equation. FOR A SOLID_SOLID REACTION If V and S are ~constants, our equation reduces to: d. G = V d. P – S d. T GT 2 P 2 - GT 1 P 1 = V(P 2 - P 1) - S (T 2 - T 1)

Gibbs for a chemical reaction Hess’s Law applied to Gibbs for a reaction o Suppose 3 A + 2 B = 2 C +1 D reactants = products o DG = 2 GC +1 GD -3 GA – 2 GB 298. 15 K, 0. 1 MPa Same procedure for DH, DS, DV o Gibbs Free Energy (G) is measured in KJ/mol or Kcal/mol o One small calorie cal ~ 4. 2 Joules J

Which direction will the reaction go? DG for a reaction of the type: 2 A + 3 B =C +4 D DG = S (n G)products - S(n G)reactants = GC + 4 GD - 2 GA - 3 GB Same procedure for DH, DS, DV The reaction with negative DG will be more stable, i. e. if DG 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 DG is negative and the reaction proceeds to the left when DG is positive. ” Brown, Le. May and Bursten (2006) Virtual Chemistry p 163

Since G = E + PV – TS And we saw the slope of a sum is the sum of the slopes Differentiating d. G = d. E +Pd. V +Vd. P -Td. S – Sd. T What is d. E? d. E = d. Q – d. W First Law, and d. Q =Td. S 2 nd law So d. E = d. Q - Pd. V => d. E = Td. S – Pd. V Most of these terms cancel, so d. G = Vd. P –Sd. T And if we need the changes when moving to a new T, P d. DG = DVd. P - DSd. T

To get an equilibrium curve for a phase diagram, could use d. DG = DVd. P - DSd. T 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 Method: From G values for each phase at 298 K and 0. 1 MPa list DG 298, 0. 1 for the reaction, do the same for DV and DS n DG at equilibrium = 0, so we can calculate an isobaric change in T that would be required to bring DG 298, 0. 1 to 0 0 - DG 298, 0. 1 = -DS (Teq - 298) (at constant P) n Similarly we could calculate an isothermal change 0 - DG 298, 0. 1 = -DV (Peq - 0. 1) (at constant T) n

Na. Al. Si 3 O 8 = Na. Al. Si 2 O 6 + Si. O 2 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 Figure 27 -1. Temperature-pressure phase diagram for the reaction: Albite = Jadeite + Quartz calculated using the program TWQ of Berman (1988, 1990, 1991).

Clausius -Clapeyron Equation d. G = Vd. P –Sd. T • Defines the state of equilibrium between reactants and products in terms of S and V From Eqn. 3, if d. G =0, d. P/d. T = ΔS / ΔV (eqn. 4) The slope of the equilibrium curve will be positive if S and V both decrease or increase with increased T and P

To get the slope, at a boundary DG is 0 d. DG = 0 = DVd. P - DSd. T d. P DS = solve d. T DV Figure 27 -1. Temperaturepressure 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

End of review

Gas Phases Return to d. G = Vd. P – Sd. T. For an isothermal process d. T is zero, so: P GP - GP = 2 1 2 P 1 Vd. P 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?

Gas Laws • 1600’s to 1800’s Pressure times Volume is a constant Increase Temp, Volume increases Increase Temp, Pressure increases Increase moles of gas, Volume increases • Combined as ideal gas law: • n= # moles, and R is the universal gas constant • R = 8. 314472 N·m·K− 1·mol− 1

Gas Pressure-Volume Relationships Ideal Gas – As P increases V decreases – PV=n. RT Ideal Gas Law § P = pressure § V = volume § T = temperature § n = # of moles of gas § R = gas constant = 8. 3144 J mol-1 K-1 So P x V is a constant at constant T Figure 5 -5. Piston-and-cylinder apparatus to compress a gas. Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.

Gas Pressure-Volume Relationships Since GP - GP = 2 1 P 2 P 1 Vd. P we can substitute RT/P for V (for a single mole of gas), thus: P GP - GP = 2 1 2 P 1 RT d. P P and, since R and T are certainly independent of P: G P - G P = RT 2 1 P 2 P 1 1 d. P P

Logarithms n Logarithms (Logs) are just exponents if by = x then y = logb x log 10 (100) = 2 because 102 = 100 n Natural logs (ln) use e = 2. 718 as a base For example ln(1) = loge(1) = 0 because e 0 = (2. 718)0 = 1 Anything to the zero power is one. n bx /by = bx-y so logbx - logb y = logb(x/y)

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 The area under the curve is the reverse operation

Gas Pressure-Volume Relationships bx /by = bx-y so logbx - logb y = logb(x/y)

Gas Pressure-Volume Relationships The form of this equation is very useful o 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 = g. P g is the fugacity coefficient GP, T - Go. T = RT ln (f /Po) so g H 2 O ranges 0. 1 – 1. 5, g CO 2 ranges 2 – 50 At low pressures most gases are ideal, but at high P they are not

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: ai = Xi Non-ideal: ai = gi Xi y y y = # of crystallographic sites in which mixing takes place where gamma gi is the activity coefficient

Dehydration Reactions n Ms + Qtz = Kspar + Sillimanite + H 2 O n We can treat the solids and gases separately o GP, T - GT = DVsolids (P - 0. 1) + RT ln (P/0. 1) (isothermal) n The treatment is then quite similar to solid-solid reactions, but you have to solve for the equilibrium pressure P by iteration. n Iterative methods are those which are used to produce approximate numerical solutions to problems. Newton's method is an example of an iterative method.

Newton’s Method

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

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

Equilibrium Constant K = Go. T + RT ln (P/0. 1 MPa) n 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 n GP, T = Go. T + RT ln (K)

The units M (molar) are moles per liter Calculating an Equilibrium Constant for a Reaction 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:

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

n GP, T n ln - Go. T = RT ln (K) and at equilibrium GP, T = 0 K = - DG 0/RT n but DGo = DHo –TDSo + DV d. P So n ln K = - DHo/RT +DSo/R - (DV/RT) d. P (27 -26)

Phlogopite is the magnesium end-member of the biotite solid solution series Annite is the iron end-member of the biotite solid solution series A Garnet-Biotite Reaction n Below is the stoichiometric equation for the Fe-Mg exchange in the reaction between the biotites and Ca-free garnets: n Fe 3 Al 2 Si 3 O 12 n Almandine + Phlogopite = Pyrope + + KMg 3 Si 3 Al. O 10(OH)2 = Mg 3 Al 2 Si 3 O 12 + KFe 3 Si 3 Al. O 10(OH)2 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 Ferich Almandine on its rim. ln K = - DH/RT +DS/R - (DV/RT) d. P Annite

Garnet-Biotite Geothermometer

The Distribution Coefficient KD

Application to DH and DS determination The Garnet - Biotite Fe –Mg exchange reaction ln. K = - DH/RT +DS/R - (DV/RT) d. P y = slope. x +b This is a line! From (27 -26) we can extract DH from the slope and DS from the intercept! Figure 27 -5. Graph of ln. K 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.

Geothermobarometry 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 halffilled 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, 216233

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

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

However … n 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

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

GASP 1 bar = 100000 pascal 1 mb [mbar, millibar] = 100 Pascals Spears 15 -47 1 atmosphere [atm, standard] = 1. 01 bar Spear’s Classic Paper Gar-Bt

You have a thick section of a metamorphic rock containing Plagioclase, Biotites, Garnets and aluminosilicates (Al 2 Si. O 5) , so you run electron microprobe scans across interesting areas. In a scan where garnet contacts biotite, you find X Mg = 0. 310, XFe = 0. 690 for Garnets; and XMg = 0. 606, XFe = 0. 324 for Biotite. Find the Pressure and Temperature Calculate KD then draw in a Garnet-Biotite line Calculate Pressures in Kilobars for 400 and 700 C 1000 bar = 1 kilobar Draw in the GASP Line Crossing Point gives the P-T conditions