Phase Equilibrium At a constant pressure simple compounds (like ice) melt at a single temperature More complex compounds (like silicate magmas) have very.

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Presentation transcript:

Phase Equilibrium At a constant pressure simple compounds (like ice) melt at a single temperature More complex compounds (like silicate magmas) have very different melting behavior

Makaopuhi Lava Lake Magma samples recovered from various depths beneath solid crust From Wright and Okamura, (1977) USGS Prof. Paper, 1004.

Makaopuhi Lava Lake Thermocouple attached to sampler to determine temperature Olivine decreases Color of glass changes- change composition From Wright and Okamura, (1977) USGS Prof. Paper, 1004.

Makaopuhi Lava Lake Temperature of sample vs. Percent Glass 100 90 70 60 50 40 30 20 10 Percent Glass 900 950 1000 1050 1100 1150 1200 1250 Temperature oc 80 > 200 oC range for liquid -> solid Fig. 6.1. From Wright and Okamura, (1977) USGS Prof. Paper, 1004.

olivine decreases below 1175oC Makaopuhi Lava Lake Minerals that form during crystallization 1250 1200 1150 1100 1050 1000 950 10 20 30 40 50 Liquidus Melt Crust Solidus Olivine Clinopyroxene Plagioclase Opaque Temperature oC olivine decreases below 1175oC Fig. 6.2. From Wright and Okamura, (1977) USGS Prof. Paper, 1004.

Makaopuhi Lava Lake Mineral composition during crystallization 100 Olivine Augite Plagioclase 90 80 Weight % Glass 70 60 50 .9 .8 .7 .9 .8 .7 .6 80 70 60 Mg / (Mg + Fe) Mg / (Mg + Fe) An Fig. 6.3. From Wright and Okamura, (1977) USGS Prof. Paper, 1004.

Crystallization Behavior of Melts 1. Cooling melts crystallize from a liquid to a solid over a range of temperatures (and pressures)

Crystallization Behavior of Melts 1. Cooling melts crystallize from a liquid to a solid over a range of temperatures (and pressures) 2. Several minerals crystallize over this T range, and the number of minerals generally increases as T decreases

Crystallization Behavior of Melts 1. Cooling melts crystallize from a liquid to a solid over a range of temperatures (and pressures) 2. Several minerals crystallize over this T range, and the number of minerals increases as T decreases 3. Minerals that form do so sequentially, with considerable overlap

Crystallization Behavior of Melts 1. Cooling melts crystallize from a liquid to a solid over a range of temperatures (and pressures) 2. Several minerals crystallize over this T range, and the number of minerals increases as T decreases 3. Minerals that form do so sequentially, with considerable overlap 4. Minerals that involve solid solution change composition as cooling progresses

Crystallization Behavior of Melts 1. Cooling melts crystallize from a liquid to a solid over a range of temperatures (and pressures) 2. Several minerals crystallize over this T range, and the number of minerals increases as T decreases 3. The minerals that form do so sequentially, with consideral overlap 4. Minerals that involve solid solution change composition as cooling progresses 5. The melt composition also changes during crystallization

Crystallization Behavior of Melts 1. Cooling melts crystallize from a liquid to a solid over a range of temperatures (and pressures) 2. Several minerals crystallize over this T range, and the number of minerals increases as T decreases 3. The minerals that form do so sequentially, with consideral overlap 4. Minerals that involve solid solution change composition as cooling progresses 5. The melt composition also changes during crystallization 6. The minerals that crystallize (as well as the sequence) depend on T and X of the melt

Crystallization Behavior of Melts 1. Cooling melts crystallize from a liquid to a solid over a range of temperatures (and pressures) 2. Several minerals crystallize over this T range, and the number of minerals increases as T decreases 3. The minerals that form do so sequentially, with consideral overlap 4. Minerals that involve solid solution change composition as cooling progresses 5. The melt composition also changes during crystallization 6. The minerals that crystallize (as well as the sequence) depend on T and X of the melt 7. Pressure can affect the types of minerals that form and the sequence

Crystallization Behavior of Melts 1. Cooling melts crystallize from a liquid to a solid over a range of temperatures (and pressures) 2. Several minerals crystallize over this T range, and the number of minerals increases as T decreases 3. The minerals that form do so sequentially, with consideral overlap 4. Minerals that involve solid solution change composition as cooling progresses 5. The melt composition also changes during crystallization 6. The minerals that crystallize (as well as the sequence) depend on T and X of the melt 7. Pressure can affect the types of minerals that form and the sequence 8. The nature and pressure of the volatiles can also affect the minerals and their sequence

The Phase Rule F = C - f + 2 F = # degrees of freedom The number of intensive parameters that must be specified in order to completely determine the system

The Phase Rule F = C - f + 2 F = # degrees of freedom f = # of phases The number of intensive parameters that must be specified in order to completely determine the system f = # of phases phases are mechanically separable constituents

The Phase Rule F = C - f + 2 F = # degrees of freedom The number of intensive parameters that must be specified in order to completely determine the system f = # of phases phases are mechanically separable constituents C = minimum # of components (chemical constituents that must be specified in order to define all phases)

The Phase Rule F = C - f + 2 F = # degrees of freedom f = # of phases The number of intensive parameters that must be specified in order to completely determine the system f = # of phases phases are mechanically separable constituents C = minimum # of components (chemical constituents that must be specified in order to define all phases) 2 = 2 intensive parameters Usually = temperature and pressure for us geologists

High Pressure Experimental Furnace Cross section: sample in red Carbide Pressure Vessle 1 cm Furnace Assembly Talc SAMPLE Graphite Furnace 800 Ton Ram the sample! Fig. 6.5. After Boyd and England (1960), J. Geophys. Res., 65, 741-748. AGU

1 - C Systems 1. The system SiO2 Fig. 6.6. After Swamy and Saxena (1994), J. Geophys. Res., 99, 11,787-11,794. AGU

1 - C Systems 2. The system H2O Fig. 6.7. After Bridgman (1911) Proc. Amer. Acad. Arts and Sci., 5, 441-513; (1936) J. Chem. Phys., 3, 597-605; (1937) J. Chem. Phys., 5, 964-966.

2 - C Systems A. Systems with Complete Solid Solution 1. Plagioclase (Ab-An, NaAlSi3O8 - CaAl2Si2O8) Fig. 6.8. Isobaric T-X phase diagram at atmospheric pressure. After Bowen (1913) Amer. J. Sci., 35, 577-599. T - X diagram at constant P = 0.1 MPa F = C - phi + 1 since P is constant: lose 1 F

Bulk composition a = An60 Point a: at 1560oC = 60 g An + 40 g Ab XAn = 60/(60+40) = 0.60 Example of cooling with a specific bulk composition Point a: at 1560oC phi = ? (1) F = ? = C - phi + 1 = 2 - 1 + 1 = 2

= a divariant situation F = 2 1. Must specify 2 independent intensive variables in order to completely determine the system = a divariant situation 2. Can vary 2 intensive variables independently without changing f, the number of phases same as: Intensive variables can be any ones (P, T, X, density, molar G-V-S, etc. as far as the Phase Rule is concerned Geologically, it’s best to choose T and X (P is constant) Thus F = T and X(An)Liq is a good choice Because X(Ab)Liq = 1 - X(An)Liq they are not independent, so may pick either but not both

Now cool to 1475oC (point b) ... what happens? Get new phase joining liquid: first crystals of plagioclase: = 0.87 (point c) F = ? X An plag Must specify T and or can vary these without changing the number of phases X An liq Now cool to 1475oC (point b) ... what happens? Now cool to 1475oC (point b) ... what happens?

X X F = 2 - 2 + 1 = 1 (“univariant”) Must specify only one variable from among: T X An liq Ab plag (P constant) Considering an isobarically cooling magma, and are dependent upon T X An liq plag The slope of the solidus and liquidus are the expressions of this relationship a “tie-line” connects coexisting phases b and c As long as have two phases coexisting at equilibrium, specifying any one intensive variable is sufficient to determine the system

At 1450oC, liquid d and plagioclase f coexist at equilibrium A continuous reaction of the type: liquidB + solidC = liquidD + solidF As cool, Xliq follows the liquidus and Xsolid follows the solidus, in accordance with F = 1 At any T, X(An)Liq and X(An)Plag are dependent upon T We can use the lever principle to determine the proportions of liquid and solid at any T

The lever principle: = D Amount of liquid Amount of solid ef = Amount of solid de where d = the liquid composition, f = the solid composition and e = the bulk composition d e f D liquidus de ef solidus

When Xplag ® h, then Xplag = Xbulk and, according to the lever principle, the amount of liquid ® 0 Thus g is the composition of the last liquid to crystallize at 1340oC for bulk X = 0.60

X Final plagioclase to form is i when = 0.60 An plag Final plagioclase to form is i when = 0.60 Now f = 1 so F = 2 - 1 + 1 = 2

1. The melt crystallized over a T range of 135oC * Note the following: 1. The melt crystallized over a T range of 135oC * 4. The composition of the liquid changed from b to g 5. The composition of the solid changed from c to h Numbers refer to the “behavior of melts” observations * The actual temperatures and the range depend on the bulk composition

Equilibrium melting is exactly the opposite Heat An60 and the first melt is g at An20 and 1340oC Continue heating: both melt and plagioclase change X Last plagioclase to melt is c (An87) at 1475oC

Fractional crystallization: Remove crystals as they form so they can’t undergo a continuous reaction with the melt At any T Xbulk = Xliq due to the removal of the crystals Thus liquid and solid continue to follow liquidus and solidus toward low-T (albite) end of the system Get quite different final liquid (and hence mineral) composition

Remove first melt as forms Melt Xbulk = 0.60 first liquid = g Partial Melting: Remove first melt as forms Melt Xbulk = 0.60 first liquid = g remove and cool bulk = g ® final plagioclase = i Fractional crystallization and partial melting are important processes in that they can cause significant changes in the final rock that crystallizes beginning with the same source rock.

Note the difference between the two types of fields The blue fields are one phase fields Any point in these fields represents a true phase composition The blank field is a two phase field Any point in this field represents a bulk composition composed of two phases at the edge of the blue fields and connected by a horizontal tie-line Plagioclase Liquid plus

2. The Olivine System Fo - Fa (Mg2SiO4 - Fe2SiO4) also a solid-solution series Fig. 6.10. Isobaric T-X phase diagram at atmospheric pressure After Bowen and Shairer (1932), Amer. J. Sci. 5th Ser., 24, 177-213.

2-C Eutectic Systems Example: Diopside - Anorthite No solid solution Fig. 6.11. Isobaric T-X phase diagram at atmospheric pressure. After Bowen (1915), Amer. J. Sci. 40, 161-185.

Cool composition a: bulk composition = An70 At a: phi = 1 F = C - phi + 1 = 2 - 1 + 1 = 2 Good choice is T and X(An)Liq

What happens at b? Pure anorthite forms (point c) phi = 2 Cool to 1455oC (point b) What happens at b? Pure anorthite forms (point c) phi = 2 F = 2 - 2 + 1 = 1 composition of all phases determined by T Xliq is really the only compositional variable in this particular system

Continue cooling as Xliq varies along the liquidus Continuous reaction: liqA ® anorthite + liqB Lever rule shows less liquid and more anorthite as cooling progresses

at 1274oC f = 3 so F = 2 - 3 + 1 = 0 invariant (P) T and the composition of all phases is fixed Must remain at 1274oC as a discontinuous reaction proceeds until a phase is lost What happens at 1274oC? get new phase Pure diopside g joins liquid d and anorthite h d is called the eutectic point: MUST reach it in all cases

Discontinuous Reaction: all at a single T Use geometry to determine d between g and h, so reaction must be: diopside + anorthite = liquid must run right -> left as cool (toward low entropy) therefore first phase lost is liquid Below 1274oC have diopside + anorthite phi = 2 and F = 2 - 2 + 1 = 1 Only logical variable in this case is T (since the composition of both solids is fixed)

Left of the eutectic get a similar situation Cool An20 1350oC (e) and diopside forms first at f Continue cooling: continuous reaction with F = 1 as liqA -> liqB + diopside At 1274oC anorthite (h) joins diopside (g) + liquid (d) Stay at 1274oC until liquid consumed by discontinuous reaction: liquid -> diopside + anorthite

1. The melt crystallizes over a T range up to ~280oC Note the following: 1. The melt crystallizes over a T range up to ~280oC 2. A sequence of minerals forms over this interval - And the number of minerals increases as T drops 6. The minerals that crystallize depend upon T - The sequence changes with the bulk composition #s are listed points in text

Augite forms before plagioclase Gabbro of the Stillwater Complex, Montana This forms on the left side of the eutectic

Plagioclase forms before augite Ophitic texture Diabase dike This forms on the right side of the eutectic

Also note: The last melt to crystallize in any binary eutectic mixture is the eutectic composition Equilibrium melting is the opposite of equilibrium crystallization Thus the first melt of any mixture of Di and An must be the eutectic composition as well

Fractional crystallization: Since solids are not reactants in eutectic-type continuous reactions, the liquid path is not changed Only the final rock will differ: = eutectic X, and not bulk X Fig. 6.11. Isobaric T-X phase diagram at atmospheric pressure. After Bowen (1915), Amer. J. Sci. 40, 161-185.

Partial Melting: if remove liquid perfectly as soon as it forms: melt Di + An to 1274oC discontinuous reaction: Di + An -> eutectic liquid d consume either Di or An first depending on bulk X then melting solid = pure Di or An and jump to 1- C system Thus heat 118 or 279oC before next melt

C. Binary Peritectic Systems Three phases enstatite = forsterite + SiO2 Figure 6.12. Isobaric T-X phase diagram of the system Fo-Silica at 0.1 MPa. After Bowen and Anderson (1914) and Grieg (1927). Amer. J. Sci. Cool bulk composition a (42 %) At a phi = 1 (liquid) F = 2 - 1 + 1 = 2 Cool to 1625oC Cristobalite forms at b phi = 2 F = 2 - 2 + 1 = 1 Xliq = f(T)

C. Binary Peritectic Systems Figure 6.12. Isobaric T-X phase diagram of the system Fo-Silica at 0.1 MPa. After Bowen and Anderson (1914) and Grieg (1927). Amer. J. Sci. As T lowered Xliq follows path to c: the eutectic At 1543oC, enstatite forms: d Now f = 3 and F = 2 - 3 + 1 = 0 invariant Discontinuous reaction: liq = En + Crst colinear Stay at this T until liq is consumed Then have En + Crst phi = 2 F = 2 - 2 + 1 = 1 univariant At 1470oC get polymorphic transition Crst -> Trid Another invariant discontinuous rxn

At 1557oC get opx (En) forming phi = 3 F = 2 - 2 + 1 = 0 invariant Figure 6.12. Isobaric T-X phase diagram of the system Fo-Silica at 0.1 MPa. After Bowen and Anderson (1914) and Grieg (1927). Amer. J. Sci. Next cool f = 13 wt. % at 1800oC get olivine (Fo) forming phi = 2 F = 2 - 2 + 1 = 1 univariant Xliq = f(T) At 1557oC get opx (En) forming phi = 3 F = 2 - 2 + 1 = 0 invariant

i m k d c i = “peritectic” point 1557oC have colinear Fo-En-liq geometry indicates a reaction: Fo + liq = En consumes olivine (and liquid) ® resorbed textures c d i k m Fo En 1557 When is the reaction finished? Expanded view of the Fo-SiO2 diagram: When is the reaction finished?? When a phase is used up Which phase will it be? Since the bulk composition lies between En and Fo, liq must be used up first Then have En + Fo: f = 2 F = 1 Bulk X

Olivine with orthopyroxene rims, Mt McLoughlin, OR

Same photo with olivine at extinction to make Opx more visible

1543 c d i k m Fo En 1557 bulk X x y Cr Now begin with bulk composition shown, hi T: F = 2 olivine y forms at 1590oC liq = x phi = 2 F = 1 1557oC En joins Fo and liq phi = 3 F = 0 again discontinuous reaction: Fo + liq = En this time olivine consumed before liquid olivine appears then disappears Xliq follows liquidus -> c below m: En + liq phi = 2 F = 1 At 1543oC get En + Crist + liq Eutectic invariant liq -> En + Cr

Incongruent Melting of Enstatite Melt of En does not ® melt of same composition Rather En ® Fo + Liq i at the peritectic Partial Melting of Fo + En (harzburgite) mantle En + Fo also ® first liq = i Remove i and cool Result = ? 1543 c d i Fo En 1557 Cr

Immiscible Liquids Cool X = n At 1960oC hit solvus exsolution ® 2 liquids o and p f = 2 F = 1 both liquids follow solvus Mafic-rich liquid Silica-rich Crst 1695 Reaction? At 1695oC get Crst also

Pressure Effects Different phases have different compressibilities Thus P will change Gibbs Free Energy differentially Raises melting point Shift eutectic position (and thus X of first melt, etc.) Figure 6.15. The system Fo-SiO2 at atmospheric pressure and 1.2 GPa. After Bowen and Schairer (1935), Am. J. Sci., Chen and Presnall (1975) Am. Min. Note effect on Fo-SiO2 system: Raises m.p. Hi P -> eutectic

D. Solid Solution with Eutectic: Ab-Or (the alkali feldspars) Eutectic liquidus minimum Cool composition a first solid at b: 1090oC last liquid at e: 1000oC don’t reach eutectic point final solid = d 780oC intersect solvus -> 2 solid phases: exsolution (perthite) phi = 2 so F = 1 Xsolids = f(T) Mineral-pair geothermometry Figure 6.16. T-X phase diagram of the system albite-orthoclase at 0.2 GPa H2O pressure. After Bowen and Tuttle (1950). J. Geology.

Fractional crystallization -> liquids approach eutectic Cool composition i first solid at j: 1020oC last liquid at k: 970oC now liq evolves -> more Or don’t reach eutectic point final solid = Xi also intersect solvus -> exsolution (antiperthite) Fractional crystallization -> liquids approach eutectic Partial melting liquids closer to eutectic composition than solid source

Effect of PH O on Ab-Or 2 Increasing PH2O progressively lowers melting point (liquidus) with little effect on the solvus (LeChatelier) At 500 MPa (about 15 km depth) the solidus intersects the solvus, and solid solution becomes limited Figure 6.17. The Albite-K-feldspar system at various H2O pressures. (a) and (b) after Bowen and Tuttle (1950), J. Geol, (c) after Morse (1970) J. Petrol.