metamorphic petrology gly 262 - weebly

Metamorphic Petrology GLY 262The phase rule - revisited

The Phase Rule in Metamorphic Systems

•

Phase rule, as applied to systems at equilibrium:

F = C -

φ

+ 2 The phase ruleφ

= the number of phases in the system

C

= the number of components: the minimum number of chemical constituents required to specify every phase in the system

F

= the number of degrees of freedom: the number of independently variable intensive parameters of state (such as temperature, pressure, the composition of each phase, etc.)

Stable Mineral Assemblages in Metamorphic Rocks

•

Equilibrium Mineral Assemblages•

At equilibrium, the mineralogy (and the composition of each mineral) can be entirely fixed by the T, P, and X of the system

•

Relict minerals or later alteration products are excluded unless specifically stated

Use the phase rule to calculate F for fields 1-3 and thered line

Note ALL fieldsAlso containpl + qtz

+ ilm

1.

2. 3.


F ≥

2 is the most common situationF = C -

φ

+ 2 = ≥

2

φ ≤ C

This additional rule is known as Goldschmidt’s mineralogical phase rule, or simplythe mineralogical phase rule

N.B. Do not confuse the phase rule and the mineralogical phase rule!


Suppose we have determined C for a rockConsider the following three scenarios:

a) φ

= C and therefore F = 2The standard divariant situation

The rock probably represents an equilibrium mineral assemblage from within a metamorphic zone


Consider the following three scenarios:

C = 1φ = 1 commonφ = 2 uncommonφ = 3 v.rare - only at the specific P-T conditions of the invariant point

The P-T phase diagram for the system Al2 SiO5 calculated using the program TWQ (Berman, 1988, 1990, 1991). Winter (2001) An Introduction to Igneous and Metamorphic Petrology. Prentice Hall.


b) φ

< C and therefore F = ≥

3

Common with mineral systems that exhibit solid solution

PlagioclasePlagioclase

Liquid

LiquidLiquid

plus

Plagioclase

The phase rule in metamorphic systems

bi liq

sill

bi g liq

sill

g liqsill

cd

g liq

sill

cd

g liq

bi cd

liq

sill

bi cd

liq

*All fields contain quartz*

1.

2.

High variance (F ≥

3)mineral

assemblages predominate in

most metamorphic

rocks

What is the variance (F)of the white fields?

What is the variance (F)of the grey-blue fields?

F = C –

Φ

+ 2

3 = 6 –

5 + 2

4 = 6 –

4 + 2

KFMASH


c) φ

> CA more interesting situation, and at least one of three situations must be responsible:

1)

F < 2The sample is collected from a location right on a univariant reaction curve (isograd) or invariant point

Situations a)

and b)

together cover nearly all rocks, hence the formulation of Goldschimdt’s

mineralogical

phase rule (F = C –

Φ

+ 2 = ≥

2)


2)

Equilibrium has not been attainedThe phase rule applies only to systems at equilibrium, and there could be any number of minerals coexisting if equilibrium is not attained


3)

We didn’t choose the # of components correctly

Some guidelines for an appropriate choice of CBegin with a 1-component system, such as CaAl2Si2O8(anorthite), there are 3 common types of major/minor components that we can add

a)

Components that generate a new phaseAdding a component such as CaMgSi2O6 (diopside), results in an additional phase: in the binary Di-An system diopside coexists with anorthite below the solidus


3)


b) Components that substitute for other components

Adding a component such as NaAlSi3O8 (albite) to the 1-C anorthite system would dissolve in the anorthitestructure, resulting in a single solid-solution mineral(plagioclase) below the solidusFe and Mn commonly substitute for MgAl may substitute for SiNa may substitute for K


3)


c) “Perfectly mobile”

componentsMobile components are either a freely mobile fluid component or a component that dissolves readily in a fluid phase and can be transported easilyThe chemical activity of such components is commonly controlled by factors external to the local rock systemThey are commonly ignored in deriving C for metamorphic systems


Consider the very simple metamorphic system, MgO- H2

OPossible natural phases in this system are periclase (MgO), aqueous fluid (H2O), and brucite(Mg(OH)2)How we deal with H2O depends upon whether water is perfectly mobile or notA reaction can occur between the potential phases in this system:

MgO

+ H2

O → Mg(OH)2

Per + Fluid = Bru

1) Suppose H2 O is mobile and we ignore it as a componentBegin with periclase above the reaction temperature: phi = 1 (water doesn’t count) and C = 1 (MgO). It is divariant

AND the

mineralogical phase rule holds

As we cool to the temperature of the reaction curve, periclase reacts with water to form brucite: MgO

+ H2

O → Mg(OH)2Reaction: periclase coexists with brucite: phi = 2 F = 1 mineralogical phase ruledoes not hold. Phi > C

To leave the curve, all the periclase must be consumed by the reaction, and brucite is the solitary remaining Phase, so phi = 1 and C = 1 again so phi = C

Imagine the diagram to be a field area, and the reaction an isograd. For any common sample phi = 1 (since we ignore H2

O and don’t really see the fluid in the rock), and phi = C in accordance with the mineralogical phase ruleOnly an uncommon sample, collected exactly on the isograd, will have phi > C

2) If water is not perfectly mobile, we must treat the system differentlyWater is not a freely permeating phase, and it may become a limiting factor in the reaction

A reaction involving more than one reactant will proceed until any one of the reactants is consumed. As the system cools, the reaction will proceed until either periclase or water is consumed. When one runs out, the other has nothing more to react with. Periclase can react only with the quantity of water that can diffuse into the system. If water is not perfectly mobile, and is limited in quantity,it will be consumed first What is a consequence of

this?

Once the water is gone, the excess periclase remains stable as conditions change into the brucite stability field

As a general rule, reactions such as this do indeed represent the absolute stability boundary of a phase such as brucite, because it is the only reactant in the prograde reaction (brucite → periclase + water)However,

the reaction is not the absolute

stability boundary of periclase, or of water, because either can be stable across the boundary if the other reactant is absent This is true for ANY reaction

Involving multiple phases

A) Isograds caused by discontinuous reactions

e.g. staurolite isograd

grt + chl + mu ----> staur + bt + qtz + H2 O

Phase Rule: Φ

+ F = C + 2

C = 6

Φ

= 7

∴

F = 1

discontinuous reaction forms a line on P-T graph

and a line in the field. Can be deduced by the breaking of a tie-line on a Δ diagram

The phase rule and reactions

B) Isograds caused by continuous reactions

e.g. garnet isograd

chl1 + mu + qtz ----> chl2 + grt + bt + H2 OFe-rich Mg-rich

Phase Rule: Φ

+ F = C + 2

C = 6

Φ

= 6

∴

F = 2

A continuous reaction forms an area on a P-T grid and area in the field.

In this example tie-lines migrate to more Mg-rich comps.

Summary

•

The phase rule can help:1)

Determine the type of reaction occurring

2)

Determine whether the mineral assemblage is in equilibrium

3)

Provide information about mobile components

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