Mineral Stability DiagramsandChemical Weathering of FeldsparsAlbite Jadeite + Quartz dΔG = ΔVdP - ΔSdTand G, S, V values for albite, jadeite and quartz to calculate theconditions for which ΔG of the reaction: Ab = Jd + Q ΔG = 0• from G values for each phase at 298K and 0.1 MPa calculate ΔG298, 0.1 for thereaction, do the same for ΔV and ΔS• ΔG at equilibrium = 0, so we can calculate an isobaric change in T that would berequired to bring ΔG298, 0.1 to 00 - ΔG298, 0.1 = -ΔS (Teq - 298) (at constant P)• Similarly we could calculate an isothermal change0 - ΔG298, 0.1 = -ΔV (Peq - 0.1) (at constant T)Mineral S(J) G (J) V (cm3/mol) Low Albite 207.25 -3,710,085 100.07 Jadeite 133.53 -2,844,157 60.04 Quartz 41.36 -856,648 22.688From Helgeson et al. (1978).Table 27-1. Thermodynamic Data at 298K and0.1 MPa from the SUPCRT DatabaseMethod:NaAlSi3O8 = NaAlSi2O6 + SiO2P - T phase diagram of the equilibrium curveHow do you know which side has which phases?Figure 27-1. Temperature-pressure phase diagram for thereaction: Albite = Jadeite +Quartz calculated using theprogram TWQ of Berman (1988,1990, 1991). Winter (2001) AnIntroduction to Igneous andMetamorphic Petrology. PrenticeHall.Pick any two points on the equilibrium curvedΔG = 0 = ΔVdP - ΔSdTThusdPdTSV=ΔΔFigure 27-1. Temperature-pressure phase diagram for thereaction: Albite = Jadeite +Quartz calculated using theprogram TWQ of Berman (1988,1990, 1991). Winter (2001) AnIntroduction to Igneous andMetamorphic Petrology. PrenticeHall.Does the liquid orsolid have the largervolume?High pressure favorslow volume, so whichphase should be stableat high P?Does liquid or solid have ahigher entropy?High temperature favorsrandomness, so whichphase should be stable athigher T?We can thus predict that the slopeof solid-liquid equilibrium shouldbe positive and that increasedpressure raises the melting point….Recall decompression melting?Figure 5-2. Schematic P-T phase diagram of a melting reaction.Winter (2001) An Introduction to Igneous and MetamorphicPetrology. Prentice Hall.Does the liquid or solidhave the lowest G atpoint A?What about at point B?The phase assemblage with the lowest G under a specific set ofconditions is the most stableFigure 5-2. Schematic P-T phase diagram of a melting reaction.Winter (2001) An Introduction to Igneous and MetamorphicPetrology. Prentice Hall.dΔG = ΔVdP - ΔSdTFree 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 theliquid be steeper orshallower than that forthe solid?Figure 5-3. Relationship between Gibbs free energy andtemperature for a solid at constant pressure. Teq is the equilibriumtemperature. Winter (2001) An Introduction to Igneous andMetamorphic Petrology. Prentice Hall.Slope of GLiquid > GSolidsince SS < SLA: Solid more stable thanliquid (low T)B: Liquid more stable thansolid (high T)– Slope δP/δT = -S– Slope S < Slope LEquilibrium at Teq– GL = GSFree Energy vs. TemperatureFigure 5-3. Relationship between Gibbs free energy andtemperature for a solid at constant pressure. Teq is the equilibriumtemperature. Winter (2001) An Introduction to Igneous andMetamorphic Petrology. Prentice Hall.Now consider a reaction, we can then use the equation:dΔG = ΔVdP - ΔSdT (again ignoring ΔX)For a reaction of melting (like ice → water)• ΔV is the volume change involved in the reaction (Vwater - Vice)– similarly ΔS and ΔG are the entropy and free energy changesdΔG is then the change in ΔG as T and P are varied• ΔG is (+) for S → L at point A (GS < GL)• ΔG is (-) for S → L at point B (GS > GL)• ΔG = 0 for S → L at point x (GS = GL)ΔG for any reaction = 0 at equilibriumFigure 5-4. Relationship between Gibbs free energy and pressure forthe solid and liquid forms of a substance at constant temperature.Peq is the equilibrium pressure. Winter (2001) An Introduction toIgneous and Metamorphic Petrology. Prentice Hall.Free Energy vs. PressureWhich species are in solution when microcline weathers to kaolinite?444522283SiO4H 2K (OH)OSiAl H O9H O2KAlSi++↔++++This is a chemical weathering process that can occuron Earth’s surface and in groundwater….STEP 1: calculate ΔGoR (appendix B)=[(-906.84)+2(-67.70)+4(-312.66)] - [2(-894.9)+9(-56.687)] = +7.103 kcalSTEP 2: calculate K364.110oRGKΔ−=21.5364.1103.71010−−==K444522283SiO4H 2K (OH)OSiAl H O9H O2KAlSi++↔++++STEP 3: set up the equilibrium expression21.52244410][][][−++=HKSiOHSTEP 4: take the log of the equilibrium expression21.5]log[2]log[2]log[444−=−+++HKSiOHSTEP 5: collect terms21.5]log[]log[2]log[444−=−+++HKSiOHSTEP 6: re-arrange and divide by 260.2]log[2][][log44−−=++SiOHHKEquation of a liney=mx+b()• Any values off this line are not in equilibrium• If log(H4SiO4) > equilibrium reaction moves left so kaolinite becomes microcline. Microcline is stable and kaolinite is unstable• If log(H4SiO4) < required for equilibrium then kaolinite is stable and microcline is unstable444522283SiO4H 2K (OH)OSiAl H O9H O2KAlSi++↔++++60.2]log[2][][log44−−=++SiOHHKFaure ch. 12, fig. 12.1, p. 174]log[ . ][][log44SiOHvsHK++• What if the system is closed with a very low water:rock ratio?]log[ and ][][log44SiOHHK++May change as reaction proceeds until equilibrium is reached then kaolinite and microcline co-exist in aqueous solution• What if the system is open and the water:rock ratio is high?Then the reaction will have a tough time getting to equilibrium sothe reaction continues until either kaolinite or microcline is used upand the reaction goes to completion444522283SiO4H 2K (OH)OSiAl H O9H O2KAlSi++↔++++ΔH°R = + 14.895The reaction is endothermic and the reaction consumes heat in the forward directionAn increase in temperature favors the conversion ofmicrocline to kaolinite and causes the equilibrium constant to increase…−Δ−=15.298113025.2loglogTRHKKoRoTTUsing the Van’t Hoff Equation:Where is the equilibrium line at 35°C?Compare KT° to KT:KT° = 10-5.21Plugging already calculated numbers into the Van’t Hoff eqn:KT = 10-4.85Since 10-4.85 > 10-5.21, equilibrium shifts to the right in favor of kaolinite with increasing TFaure ch. 12, fig. 12.1, p. 174]log[ . ][][log44SiOHvsHK++Faure ch. 12, fig. 12.1, p. 174]log[ . ][][log44SiOHvsHK++Stability field for most natural watersStability field for seawaterClay