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Exsolution and Phase Diagrams Lecture 11 Alkali Feldspar Exsolution ‘Microcline’ - an alkali feldspar in which Na- and K-rich bands have formed perpendicular to the twinning direction. This leads to this cross-hatched or fabric-like texture under crossed polarizers. G-bar–X and Exsolution • We can use G-bar–X diagrams to predict when exsolution will occur. • Our rule is that the stable configuration is the one with the lowest free energy. • A solution is stable so long as its free energy is lower than that of a physical mixture. • Gets tricky because the phases in the mixture can be solutions themselves. Inflection Points • At 800˚C, ∆Greal defines a continuously concave upward path, while at lower temperatures, such as 600˚C (Figure 4.1), inflections occur and there is a region where ∆Greal is concave downward. All this suggests we can use calculus to predict exsolution. • Inflection points occur when curves go from convex to concave (and visa versa). • What property does a function have at these points? • Second derivative is 0. Albite-Orthoclase Inflection Points • Second derivative is: æ ¶2 G ö RT æ ¶2 Gex ö çè ¶X 2 ÷ø = X X + çè ¶X 2 ÷ø 2 1 2 2 • First term on r.h.s. is always positive (concave up). • Inflection will occur when æ ¶2 Gex ö RT £ çè ¶X 2 ÷ø XX 2 1 2 Spinodal Actual solubility gap can be less than predicted because an increase is free energy is required to begin the exsolution process. Phase Diagrams • Phase diagrams illustrate stability of phases or assemblages of phases as a function of two or more thermodynamic variables (such as P, T, X, V). • Lines mark boundaries where one assemblage reacts to form the other (∆Gr=0). Thermodynamics of Melting • • • • Melting occurs when free energy of melting, ∆Gm, is 0 (and only when it is 0). This occurs when: ∆Gm = ∆Hm –T∆Sm Hence: DH m TM = ∆ Sm Assuming ∆S and ∆H are independent of T: Ti,m T Rln ai, = 1- i,m T ∆ H i,m • T-X phase diagram for the system anorthite-diopside. where Ti,m is the freezing point of pure i, T is the freezing point of the solution, and the activity is the activity of i in the liquid phase. Computing an Approximate Phase Diagram We assume the liquid is an ideal solution (ai = Xi) and compute over the range of Xi T= ∆ H i,m ∆ H i,m - R ln Xi Ti,m Constructing T-X phase diagrams from G-bar–X diagrams We can use thermodynamic data to predict phase stability, in this case as a function of temperature and composition Phase Rule and Phase Diagrams • Phase rule: ƒ = c – ϕ + 2; c = 2 for a binary system. • Accordingly, we have ƒ = 4 – ϕ and: • Univariant • • Divariant Trivariant Phases Free compositional variables ϕ = 3; 2 solids + liquid, 2 liquids + solid 3 solids or liquids 0 ϕ = 2; 1 solid + 1 liquid, 2 solids, 2 liquids 0 ϕ = 1; 1 solid or 1 liquid 1 Trivariant System G-bar-X diagram for a trivariant, one-phase system exhibiting complete solid solution. Need to specify P, T, and X to completely describe the system. Divariant Systems • • We need to specify both T and P (G-bar–X relevant only to that T and P). Two phases coexist on a plane in T–P–X space. G-bar-X diagrams for different divariant systems o o o o • • (a) Liquid solution plus pure solid (b) Liquid solution plus solid solution (c) Two pure solids (d) Limited solid solution (limited liquid solution would be the same) The free energy of the system as a whole is that of a mechanical mixture of phases – described by straight line through or tangent to free energies of individual phases. We deduce compositions of solutions by drawing tangents between curves (or points) for phases. Univariant Systems • One degree of freedom. o We specify only P or T. o Three phases in binary system can coexist along a line (not a plane) in P-T-X space. o only at one T, once we specify P (and visa versa). • Compositions of solutions are determined by drawing tangents. Plagioclase Solution • Unlike alkali feldspar, Na-Ca feldspar (plagioclase) forms a complete solid (and liquid) solution. • Let’s construct the melting phase diagram from thermodynamics. • For simplicity, we assume both liquid and solid solutions are ideal. Plagioclase Solution • Condition for equilibrium: µia = µib o e.g. s µAb = µAb • Chemical potential is µia = µioa + RT ln Xia • Combining these: s æ X Ab ö µ - µ = RT ln ç è X ÷ø o Ab os Ab Ab s æ X An ö µ - µ = RT ln ç è X ÷ø o An os An An o standard states are the pure end member solids and liquids. Plagioclase Solution s æ X An ö µ - µ = RT ln ç è X ÷ø o An os An An • The l.h.s. is simply ∆Gm for the pure component: ∆G Ab m s æ X Ab ö = RT ln ç è X ÷ø Ab • rearranging Ab s X Ab = X Ab e∆ G m /RT • Since XAn = 1 - XAb An ∆ G m /RT (1- X Ab )e error in book: Ab on lhs should be An = 1- X Ab e Ab ∆ G m /RT Plagioclase Solution • From: • Solving for mole fraction of Ab in the liquid: An e∆ G m /RT -1 X Ab = Ab An e∆ G m /RT + e∆ G m /RT The mole fraction of any component of any phase in this system can be predicted from the thermodynamic properties of the end-members. In the ideal case, as here, it simply depends on ∆Gm and T. In a non-ideal case, it would depend on Gexcess as well. Computing the equation above (and a similar one for the solid), we can compute the phase diagram. • • • • An ∆ G m /RT (1- X Ab )e = 1- X Ab e Ab ∆ G m /RT