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GEO 4860 Advanced petrology Part 1: Thermodynamics and phase diagrams, mostly igneous petrology January 24 – March 16 Evaluation: Written exercises (20%)

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Presentation on theme: "GEO 4860 Advanced petrology Part 1: Thermodynamics and phase diagrams, mostly igneous petrology January 24 – March 16 Evaluation: Written exercises (20%)"— Presentation transcript:

1 GEO 4860 Advanced petrology Part 1: Thermodynamics and phase diagrams, mostly igneous petrology January 24 – March 16 Evaluation: Written exercises (20%) and mid-term exam (30%) Reidar Trønnes, Natural History Museum, UiO www. nhm.uio.no/om-museet/seksjonene/forskning-samlinger/ansatte/rtronnes/index-eng.xml Part 2: Mostly metamorphic petrology, petrological and structural field relations, training in petrological research March 20 – June 14 Evaluation: Project report (20%) and final exam (30%) Håkon Austrheim, Dept. of Geosciences / Centre for Physics of Geological Processes, UiO

2 Lectures and tutorials Tuesday and Friday and in GEO 114 (Skolestua) Examinations Mid-term exam: Friday, March 16, Final exam: Wednesday, June 14, – Textbook Winter (2010): Principles of Igneous and Metamorphic Petrology. 2. ed. Prentice Hall / Pearson. ppt-presentasjoner for hvert av kapitlene på:

3 List of participants: 12 Advantage for student – instructor communication Presentation of student status, M.Sc. thesis project, supervisor, etc. Completed courses (or courses in progress) Optical mineralogy and petrography ? Isotope geochemistry ? Structural geology ?

4 Mineral- and rock compositions Weight% versus atom% (or mol%) Very basic features - Silicate minerals and common rocks are nearly "fully oxidized", with O as the most important and often the only anion BUT: Fe occurs as a combination of Fe 2+ (as FeO) and Fe 3+ (as Fe 2 O 3 / FeO 1.5 ) - The contents of halogenides and sulphides are very low in common silicate rocks - Oxide-, carbonate-, phosphate-, sulphate- and tungstate-minerals are also "fully oxidized" Chemical analyses of minerals and rocks are normally presented as oxides BUT: useful to recalculate to cation basis, especially for minerals (i.e. relatively simple stoichiometric compounds)

5 Basic information: Valency / oxidation state for major and trace elements - The great majority of major and trace elements has one dominating valency in silicate minerals and rocks in planetary mantles and crusts. - Important exception: Fe Important question Choice of stoichiometry / formula for the oxides SiO 2 Al 2 O 3 or AlO 1.5 FeO Fe 2 O 3 or FeO 1.5 CaO Na 2 O or NaO 0.5 K 2 O or KO 0.5

6 Is 19 wt% Al 2 O 3 identical to 19 wt% AlO 1.5 ? Yes ! even if Al 2 O 3 has 2 cations + 3 anions, whereas AlO 1.5 has 1 cation O-ions Calculation of mineral formulas based on wt%: I prefer to use one-cation-oxides

7 Mineral formula, normalized to 5 cations = cp * 5/18635 Si Al Fe Ca Na K  cations   charge SiO 2 AlO 1.5 FeO 1.5 CaO NaO 0.5 KO 0.5 wt% mol. wt mw cation prop. cp = 10 4 * wt%/mw  Calculation of mineral formula (structural formula) for a feldspar analysis Generell feltspat-formel: (K,Na,Ca)(Al,Si) 4 O 8, i.e. 5 cations + 8 O-ions Mineral formula, normalized to 8 O-ions = op * 8/(O * 29769) Si Al Fe Ca Na K  cations   charge oxygen-prop. op = 10 4 * O * wt%/mw  Mineral formula normalized to 5 cations: K 0.35 Na 0.62 Ca 0.03 (Fe 0.04 Al 1.02 Si 2.94 ) O 7.99 Mineral formula normalized to 8 O-ions: K 0.35 Na 0.62 Ca 0.03 (Fe 0.04 Al 1.03 Si 2.95 ) O 8.00

8 Mol% FeO: 1Fe + 1O = 2, 1/2 = 50% Fe Fe 2 O 3 : 2Fe + 3O = 5, 2/5 = 40% Fe Fe 3 O 4 : 3Fe + 4O = 7, 3/7 = 43% Fe Graphic presentation of mineral compositions Along a simple axis, system O-Fe: 3 phases (minerals): FeO (wüstite), Fe 2 O 3 (hematite) and FeO ˑ Fe 2 O 3 (Fe 3 O 4, magnetitt) Weight% FeO: Fe O = 71.85, 78% Fe Fe 2 O 3 : 2 ˑ Fe + 3 ˑ O = , 70% Fe Fe 3 O 4 : 3 ˑ Fe + 4 ˑ O = , 72% Fe

9 Fo Mol% Mg 2 SiO 4 : 2MgO + 1SiO 2, 1/3 = 33.3% SiO 2 MgSiO 3 : 1MgO + 1SiO 2 = 2, 1/2 = 50.0% SiO 2 SiO 2 : 100% SiO 2 MgAl 2 O 4 : 1MgO + 1Al 2 O 3 = 2, 1/2 = 50.0% Al 2 O 3 Mg 3 Al 2 Si 3 O 12 : 3MgO + 1Al 2 O 3 + 3SiO 2 = 7, 3/7 = 42.9 % MgO 3/7 = 42.9% SiO 2 1/7 = 14.3% Al 2 O 3 In triangular diagram, system MgO-SiO 2 -Al 2 O 3. Mineral endmembers (components) forsterite (Mg 2 SiO 4 ), enstatite (MgSiO 3 ), silica (SiO 2 ), spinel (MgAl 2 O 4 ) og pyrope (Mg 3 Al 2 Si 3 O 12 ) Graphic presentation of mineral compositions MgO Al 2 O 3 SiO 2 Weight% Mg 2 SiO 4 : 2 ˑ = , 42.7% SiO 2 MgSiO 3 : = , 59.9% SiO 2 SiO 2 : 100% SiO 2 MgAl 2 O 4 : = , 71.7% Al 2 O 3 Mg 3 Al 2 Si 3 O 12 : 3 ˑ ˑ = , 30.0% MgO 44.7% SiO % Al 2 O 3 Py En Fo Sp

10 Definition of the terms phase, system and components Phase: chemically homogenous substance Examples: water, ice, steam, kyanite, sillimanite, quarts, granitic melt System components: chemical entities necessary for the characterization of a system (e.g. elements, oxides or more complex formula units) System: a collection of phases under consideration We decide the boundaries of our system. Try to find the most convenient system in order to describe and understand the chemical and thermodynamic equilibria. Examples: - hand speciment of a basalt - a 10 m long by 2 m tall road cut exposing lenses of eclogite in gneis - an experimental sample capsule with 50 mol% MgSiO mol% Mg 3 Al 2 Si 3 O 12 - the entire Earth’s mantle + core Phase components: chemical entities necessary for the characterization of a phase (e.g. elements, oxides or more complex formula units)

11 Considerations w.r.t. choice of components Al-silicate system containing the phases kyanite, sillimanite, andalusite. Polymorphs with chemical composition: Al 2 SiO 5 3 elements: Al, Si, O 2 ooxides: Al 2 O 3, SiO 2 How many components ? 1: Al 2 SiO 5

12 The ”olivine system” of solid solution between forsteritt (Mg 2 SiO 4 ) og fayalitt (Fe 2 SiO 4 ) 4 elements: Mg, Fe, Si, O 3 oxides: MgO, FeO, SiO 2 How many components? How many phases? 2: Mg 2 SiO 4 and Fe 2 SiO 4 1: olivine

13 The phase rule For a system in equilibrium: P + F = C + m F: Number of degrees of freedom (variance): number of variables that can change independently of each other (e.g. p-T-X, pressure-temperature-composition) C: The smallest number of chemical components required to characterize all of the phases P: Number of phases present in equilibrium at any given location m: Number of external variables that influence the system, commonly m=2 (p and T)

14 Phase rule: P + F = C + 2 System: Al 2 SiO 5 C = 1 (component Al 2 SiO 5 is common to all 3 phases ) P = 3  3 + F = F = 0 (invariant point) P = 2  2 + F = 3 F = 1 (univariant line) P = 1  1 + F = 3 F = 2 (divariant field)

15 Olivine group Solidus Liquidus Forsterite: very high melting point System forsterite-fayalite: full solid solution and liquid solution the very first melt olivine Bulk composition

16 This is a T-X-diagram at constant (fixed) pressure (1 bar)  Phase rule: System: (Mg,Fe) 2 SiO 5 or Mg 2 SiO 4 – Fe 2 SiO 4 C = 2 both of the phases comprise different proportions of the components forsterite and fayalite P = 1  F = 2 - divariant fields, can vary both T and X - i.e. for a fixed T (e.g C) can olivine vary from Fo 100 to Fo 0 P = 2  F = 1 - univariant curves. If T is fixed, the compositions of the two phases are also fixed Three phases are never present in this diagram. Therefore, there are no binary invariant points. BUT the melting points for pure forsterite and pure fayalite are unary invariant points (in the two one-component systems Fo and Fa. P + F = C + 1 P + F = = 3 F = 3 – P - changing T  changing X sol and X melt

17 The lever rule: Measuring the mass proportion of two phases Bulk composition Fo 20 Equilibrium at 1660 C 99.99% olivine (Fo 20 ) % melt (Fo 51 )

18 Vektstang-regelen: Måling av mengdeforholdet mellom to faser 41 7 Bulk composition Fo 20 Equilibrium at 1700 C olivine (Fo 84 ) + melt (Fo 56 ) 100 * 41/48 = 85.4% 100 * 7/48 = 14.6%

19 Vektstang-regelen: Måling av mengdeforholdet mellom to faser 22 5 Bulk composition Fo 20 Equilibrium at 1800 C olivine (Fo 93 ) + melt (Fo 77 ) 100 * 5/27 = 18.5% 100 * 22/27 = 81.5%

20 Teaching Phase Equilibria

21 One-component system Two-component system Eutectic point, 23% NaCl

22 –2ºC –18ºC Simple experiment i freezer(s) with adjustable T Make a salt solution with 2-3% NaCl Put plastic bags with the solution in the freezer(s) at two different T (–2 and –18 ºC) - take out the next day - pick out the pieces of ice rinse in clean water - taste the ice - taste the solutions - estimate the mass proportion solid/liquid What will you observe: –2ºC ? –18 ºC ?

23 Phase relations: basalt – peridotite, binary systems

24 Melting relations for natural rocks – Multicomponent systems Eutektic point Not a simple binary system: A ternary system is much better, but still only a simplified model peridotite basalt

25 First law: (internal energy of an isolated system is constant) dE = dQ – dW = TdS – pdV Third law: S crystal = 0 at T = 0 K Second law of thermodynamics: Change of internal energy (heat content) and entropy dQ = TdS (for a reversibel process) dE 1 Contribution 1: added energy (dE 1 ) may increase the internal energy (by heating, dQ) dQ goes into the system  therefore positive Contribution 2: added energy (dE 2 ) enable the system to perform work on the surroundings ( = volume increase against constant pressure) dW goes out of the system  therefore negative dE 2 =  dW =  pdV Work: W = pV W = Fs (Nm = J) p = F/s 2 (N/m 2 = Pa) W = ps 3 = pV (m 3 N/m 2 = Nm) dE 1 = dQ = TdS Our system: e.g. a crystal dE 2 volume increase heating

26 First law: The internal energy in an isolated system is constant dE = dQ – dW = TdS – pdV Gibbs free energy: energy in addition to the internal energy Definition: G = E – (TS – pV) = E + pV – TS = H – TS At chemical equilibrium:  G = G prod – G react = 0

27 Gibbs free energy : energy in addition to the internal energy Definition: G = E – (TS – pV) = E + pV – TS = H – TS At chemical equilibrium:  G = G prod – G react = 0 Simple melting reaction: SiO 2 = SiO 2 solid melt (e.g. tridymite) Lowest energy level: greatest stability

28 The aluminium silicates: Al 2 SiO 5 Kyanite: Al [6] Al [6] Si O 5 Triclinic Sillimanite: Al [6] Al [4] Si O 5 Orthorombic Andalusite: Al [6] Al [5] Si O 5 Orthorombic T p Al [6] Al [5] Al [4] c c c

29  G =  E + p 1  V – T 1  S = 0  G =  E + (p 1 +dp)  V – (T 1 +dT)  S = 0 Reaction Al 2 SiO 5 = Al 2 SiO 5 andalusite kyanite andalusite kyanite p T p1p1 T1T1 dp dT  V = V ky – V and positive? or negative? negative  V and  S gives positive dp/dT-slope Subtraction of upper equation from the lower one: Consider two points on the phase boundary (equilibrium between kyanite and andalusite)  G = 0 p 1,T 1 p 1 +dp, T 1 +dT dp  V = dT  S dp/dT =  S/  V Clayperon-equation  S = S ky – S and < 0 < 0

30 Aluminium silicates: Al 2 SiO 5 Kyanite: Al [6] Al [6] Si O 5 Triclinic, D: 3600 kg/m 3 Sillimanite: Al [6] Al [4] Si O 5 Orthorombic, 3250 kg/m 3 Andalusite: Al [6] Al [5] Si O 5 Orthorombic, 3180 kg/m 3 T p Al 2 SiO 5 = Al 2 SiO 5 andalusite sillimannite  V = V sill – V and  S = S sill – S and positive? or negative? dp/dT =  S/  V positive? or negative? < 0 > 0

31 General features regarding melting and crystallisation What is melting ? Solid/ordered crystal lattice breaks down (”dissolves”) What is required? (Reaction: crystal → melt) - heating to T m - heat of fusion,  H m =  E + p  V > 0 First law (Internal energy in isolated system is constant) dE = dQ – dW = TdS – pdV Gibbs free energy: energy in addition to the internal energy Definition: G = E – (TS – pV) = E + pV – TS = H – TS At equilibrium:  G = G prod – G react = 0 Equilibrium:  G sm =  H ̶̶ T  S = 0, d.v.s.  S > 0 always positive  S m What about  V? Mostly:  V > 0 possible exception at very high p High p →  V may be negative, but  E is always positive

32 Basic question (try to reason only intuitively): Assuming that  V m >0, how does T m change with increasing p? T m increases with increasing p because the volume increase due to melting (  V m > 0) becomes more unfavourable as p increases. Increasing T m is required to compensate for the unfavourable p-V-effect. Why ?

33  G =  E + p 1  V – T 1  S = 0  G =  E + (p 1 +dp)  V – (T 1 +dT)  S = 0 Reaction Mg 2 SiO 4 = Mg 2 SiO 4 forsterite melt Forsterite p T T1T1 p1p1 dT dp  V = V sm – V fast > 0  S > 0 positive  V and  S: positive dp/dT-slope Clapeyron-equation for melting dp  V = dT  S dp/dT =  S/  V Melt Forsterite p T Melt

34 Older and somewhat wrong phase diagram for SiO 2 from Perkins (2011, Mineralogy) Correct melting curve based on more recent experiments (Kanzaki 1990; Zhang et al 1993) p-T melting curves are not linear, but have mostly increasing dp/dT with increasing p and T. Why? How can we answer this question? Extremely high T, difficult experiments

35 Consider the Clapeyron-slope of the melting curve: dp/dT =  S m /  V m Melting reaction: SiO 2 (solid) = SiO 2 (melt)   S m = S melt  S solid  V m = V melt  V solid Remember:  S m is always postitive  for simplicity we can assume that it is nearly constant What about  V m ? What is most compressible: mineral or melt ?  V m → 0 in the highest p-range of  -quartz and coesite dp/dT =  S m /  V m → ∞


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