Presentation on theme: "Mineral Stability Diagrams and Chemical Weathering of Feldspars"— Presentation transcript:
1Mineral Stability Diagrams and Chemical Weathering of Feldspars
2Ab = Jd + Q Method: Albite Jadeite + Quartz 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 + QDG = 0Method:from G values for each phase at 298K and 0.1 MPa calculate DG298, 0.1 for the reaction, do the same for DV and DS• DG at equilibrium = 0, so we can calculate an isobaric change in T that would be required to bring DG298, 0.1 to 00 - DG298, 0.1 = -DS (Teq - 298) (at constant P)Similarly we could calculate an isothermal change0 - DG298, 0.1 = -DV (Peq - 0.1) (at constant T)
3NaAlSi3O8 = NaAlSi2O6 + SiO2P - T phase diagram of the equilibrium curveHow do you know which side has which phases?Where on this diagram is Albite stable? Jadeite?In what direction would be the Ab + Qtz = Ne reaction? Why?Figure 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.
4Pick any two points on the equilibrium curve dDG = 0 = DVdP - DSdTThusdPdTSV=DFigure 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.
5Does 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?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?Consider the phase diagram (as we anticipate igneous petology)We can apply thermodynamics qualitatively to assess these diagramsWe can thus predict that the slope of solid-liquid equilibrium should be positive and that increased pressure raises the melting point….Recall decompression melting?
6dDG = DVdP - DSdTDoes 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
7Free Energy vs. Temperature dG = VdP - SdT; at constant pressure: dG/dT = -SBecause S must be (+), G for a phase decreases as T increasesWould the slope for the liquid be steeper or shallower than that for the solid?From dG = VdP - SdT we can deduce that at constant pressure: dG = -SdTAnd since S must be (+) G for a phase decreases as T increasesFigure 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.
8Free Energy vs. Temperature Slope of GLiquid > GSolidsince SS < SLA: Solid more stable than liquid (low T)B: Liquid more stable than solid (high T)Slope dP/dT = -SSlope S < Slope LEquilibrium at TeqGL = GSFigure 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.
9Now 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 changesdDG 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
10Free Energy vs. Pressure 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.
11Which species are in solution when microcline weathers to kaolinite? This is a chemical weathering process that can occuron Earth’s surface and in groundwater….
12STEP 1: calculate DGoR (appendix B) =[( )+2(-67.70)+4( )] - [2(-894.9)+9( )]= kcalSTEP 2: calculate K
13( ) STEP 3: set up the equilibrium expression STEP 4: take the log of the equilibrium expressionSTEP 5: collect terms()Equation of a liney=mx+bSTEP 6: re-arrange and divide by 2
14Any values off this line are not in equilibrium If log(H4SiO4) > equilibrium reaction moves left so kaolinite becomes microcline. Microcline is stable andkaolinite is unstableIf log(H4SiO4) < required for equilibrium then kaolinite isstable and microcline is unstable
16What if the system is closed with a very low water:rock ratio? May change as reaction proceedsuntil equilibrium is reached thenkaolinite and microcline co-existin aqueous solutionWhat 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 completion
17DH°R =The reaction is endothermic and the reaction consumesheat in the forward directionAn increase in temperature favors the conversion ofmicrocline to kaolinite and causes the equilibriumconstant to increase…
18Using the Van’t Hoff Equation: Where is the equilibrium line at 35°C?Compare KT° to KT:KT° =Plugging already calculated numbers into the Van’t Hoff eqn:KT =Since > , equilibrium shifts to the right in favorof kaolinite with increasing T
20Stability field for seawater Stability field for most natural waters Clay minerals(e.g., kaolinite)deposited inthe ocean tendto take up K+and are convertedto illite (musc.)or low-T feldspar(with enough time)Stabilityfield formostnaturalwatersFaure ch. 12, fig. 12.1, p. 174
21Phases involved in microcline weathering to kaolinite SiO2 — Al2O3 — K2O — HCl — H2OThe system isn’t just kaolinite and microclineHow do you know which phases are stable during the reaction?You have to be able to identify ALL of the phases present:gibbsite, muscovite, smectite, pyrophyllite, and amorphous silica…Understanding Figure 12.1 and log[H4SiO4]Measuring [K+], [H+] and [H4SiO4] allows stability field to be identified
23Complementary mineral stability diagrams Add in Na+ and Ca2+ to look at thestability of albiteand anorthite…Ab=albiteAn=anorthiteC=calciteG=gibbsiteK=kaolinitePa=paragonitePy=pyrophylliteSi=amorphous silica