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Metamorphic Petrology GLY 262Compatibility diagrams

Chemographic Diagrams3-C mineral compositions are plotted on a triangularchemo-graphic diagram.

They may form the hypothetical mineralsx, y, z, xz, xyz, and x2 z

Note that this subdivides the chemographic diagram into 5 sub-triangles, labeled (A)-(E)

x-xy-x2z = (A)xy-xyz-x2z = (B)xy-xyz-y = (C)xyz-z-x2z = (D)y-z-xyz = (E)

Each of these assemblages forms in rocks of different bulk composition,and they are called compatible assemblages for a specific P-TAny point within the diagram represents a rock of specific bulk composition.

Equilibrium mineral assemblages are indicated by tie-lines which connect co-existing minerals.

Chemographic DiagramsValid compatibility diagram must be referenced to a specific range of P-T conditions, such as a zone in some metamorphic terrane, because the stability of the minerals and their groupings vary as P and T vary

Previous diagram refers to a P-T range in which the fictitious minerals x, y, z, xy, xyz, and x2z are all stable and occur in the groups shownAt different grades the diagrams change

Other minerals become stableDifferent arrangements of the same minerals (different tie-lines connect different coexisting phases)

A + B = C + D

Can be used to deduce reactions.

At the isograd

Above the isograd

Below the isograd

Compatibility diagrams explain clearly why rocks, even though equilibrated at the same metamorphic grade, often develop different mineral assemblages

A shift in bulk composition, even if only slightly, can vary the mineral assemblage If move from (E) to (D) a rock will contain the mineral x2 z (along with z and

xyz) and no longer the mineral y that occurred in (E) Only rocks with bulk compositions corresponding to sub-triangle (A), for

example, develop the mineral x, because they are rich in the x component Suppose some mineral, such as mineral xyz, is an important index mineral

(e.g. garnet). The diagram can readily explain why all pelitic rocks need not contain garnet, even if they are within the garnet zone

Any rock with a bulk composition in sub-triangle (A) will be garnet-free

Any point within the trianglecontains 3 phases and as it is a 3-C systemPhi = C therefore it consistent withthe mineralogical phase rule

What happens if you pick a composition that falls directly on a tie-line, such as point (f)?

In this case the mineral assemblage consists of xyz and z only (the ends of the tie-line), since by adding these two phases together in the proper proportion you can produce the bulk composition (f)

In such a situation phi = 2

But how many components do you need for this two phase system?

2 - Therefore the mineralogical phaserule still holds

In the unlikely event that the bulk composition equals that of a single mineral, such as xyz, then

= 1, but C = 1 as well

compositionally degenerate

A diagram in which some minerals exhibit solid solution

Solid solutionin (y,z)

Solid solutionIn x,y,z

Suppose a bulk rock composition is in the shaded field of the mineral (xyz)ss

phi = 1, but the system is not degenerate

Due to the variable nature of the composition of the phase, C must still equal 3 and the phase rule tells us that:F = C - phi + 2 = 4Thus P, T, and any 2 of the 3 components in the phase are independently variable

The shaded area of mineral (xyz)ss is thus an area (compositionally divariant), and our single-phase rock can have any composition within the shaded solid solution limits

If Xbulk on a tie-line The composition of the two minerals that correspond to bulk rock composition (f) are indicated by the two shaded dots at the ends of the tie-line through (f)

phi = 2 and C is still 3, so F = 3 - 2 + 2 = 3

Since P and T are independently variable, that means the composition of each phase is univariant, and must vary along the lines where the bundles of tie-lines end

However, if you fix the composition of one you automatically fix the other

Xbulk in 3-phase triangles F = 2 (P & T) so Xmin fixed

In such situations phi = 3, and C = 3, as predicted by the mineralogical phase rule

Since F = 2 and corresponds to P and T the phase rule tells us that all of the compositional variables for each phase are fixed

Chemographic Diagrams for Metamorphic Rocks

Most common natural rocks contain the major elements: SiO2 , Al2 O3 , K2 O, CaO, Na2 Oof , FeO, MgO, MnO and H2 O such that C = 9

Its actually 11 if you include TiO2 and Fe2 O3 (we will briefly demonstrate how you take into account everything with advanced phase diagram modelling)

Three components is the maximum number that we can easily deal with in two dimensions

What is the right choice of components? Some simplifying methods:

1) Simply ignore componentsTrace elementsElements that enter only a single phase (we can drop both the component and the phase without violating the phase rule)Perfectly mobile components or saturated components e.g. H2O

2) Combine componentsComponents that substitute for one another in a solid solution: (Fe + Mg)Warning: sometimes you may want to see how Fe-Mg varies e.g. pelites

3) Limit the types of rocks to be shownOnly deal with a sub-set of rock types for which a simplified system works

4) Use projections

The phase rule and compatibility diagrams are rigorously correct when applied to complete systems

A triangular diagram thus applies rigorously only to true (but rare) 3-component systemsIf you drop components and phases, combine components, or project from phases, we may miss out on some important information. Lose a rigorous correlation between the behavior of the simplified system and realityHowever, being able to graphically display the simplified system, many aspects of the systems behavior become apparent. Need to understand a simple system before moving onto more complex ones. Compatibility diagrams are really becoming outdated in modern metamorphic petrology. However they do serve a purpose for understanding some basic principles and chemographic relationships.

The ACF DiagramIllustrate metamorphic mineral assemblages in maficrocks on a simplified 3-C triangular diagramConcentrate only on the minerals that appeared or disappeared during metamorphism, thus acting as indicators of metamorphic grade

Figure 24-4. After Ehlers and Blatt (1982). Petrology. Freeman. And Miyashiro (1994) Metamorphic Petrology. Oxford.

The ACF DiagramThe three pseudo-components are all calculated on an molecular basis:

A = Al2 O3 + Fe2 O3 - Na2 O - K2 OC = CaO - 3.3 P2 O5F = FeO + MgO + MnO

The ACF DiagramA = Al2 O3 + Fe2 O3 - Na2 O - K2 O

Why the subtraction?Na and K in the average mafic rock are typically combined with Al to produce Kfs and AlbiteIn the ACF diagram, we are interested only in the other K-bearing metamorphic minerals, and thus only the amount of Al2O3 that occurs in excess of that combined with Na2O and K2O (in albite and K-feldspar)Because the ratio of Al2O3 to Na2O or K2O in feldspars is 1:1, we subtract from Al2O3 an amount equivalent to Na2O and K2O in the same 1:1 ratio

The ACF DiagramC = CaO - 3.3 P2 O5

C is formulated in a similar way to AAll the P2 O5 in most rocks is combined with CaO (in the ratio 1:3.3) to create apatite. Apatite is an ubiquitous accessory mineral, and in the ACF diagram we are interested only in the CaO that exists in excess of that captured by P2 O5 to create apatite

Thus we subtract from CaO an amount equal to 3.3 times the amount of P2 O5 to eliminate P2 O5 as a component and apatite as a phase without altering CaO in the rest of the system

F is a combined pseudo-component based on the common exchangeability of Fe, Mg, and Mn in solid solution on the octahedral sites of most mafic minerals. If they behave in such a similar fashion, it makes some sense to treat them as a single component

The ACF Diagram

Water is omitted under the assumption that it is perfectly mobileNote that SiO2 is simply ignored

We shall see that this is equivalent to projecting from quartz (in excess)

By creating these three pseudoBy creating these three pseudo--components, components, EskolaEskola reduced reduced the number of components in mafic rocks from 8 to 3the number of components in mafic rocks from 8 to 3

The ACF Diagram

Anorthite CaAl2Si2O8A = 1 + 0 - 0 - 0 = 1, C = 1 - 0 = 1, and F = 0Provisional values sum to 2. Normalize to 100 % or 1A = 0.5 or 50%C = 0.5 or 50%F = 0

An example: minerals are easy you read off the An example: minerals are easy you read off the chemical formula. Bulkchemical formula. Bulk--compositions will require compositions will require the subtractions the subtractions

Figure 24-4. After Ehlers and Blatt (1982). Petrology. Freeman. And Miyashiro (1994) Metamorphic Petrology. Oxford.

A typical ACF compatibility diagram, referring to a specific range of P and T (the kyanite zone in the Scottish Highlands)

Figure 24-5. After Turner (1981). Metamorphic Petrology. McGraw Hill.

The AKF Diagram

In the AKF diagram, the pseudo-components are:

A = Al2 O3 + Fe2 O3 - Na2 O - K2 O - CaOK = K2 OF = FeO + MgO + MnO

Because pelitic sediments are high in Al2 O3 and K2 O, and low in CaO, Eskola proposed a different diagram that included K2 O to depict the mineral assemblages that develop in them

AKF compatibility diagram (Eskola, 1915) illustrating paragenesis of pelitic hornfelses, Orijrvi region Fin

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