### Lecture 11

```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
```