1. Binary Phase Diagrams
GLY 4200
Fall, 2016

2. Binary Diagrams Binary diagrams have two components
We therefore usually choose to plot both T (temperature) and X (composition) with pressure held constant
P-X (T fixed) or P-T (X fixed) are also possible

3. Binary System Examples
Binary solid solution - olivine, plagioclase feldspar
Binary eutectic with congruent melting - potassium feldspar - silica
Binary peritectic and eutectic with solid to solid conversion - leucite - Potassium feldspar - silica
Binary minimum melting point - potassium feldspar - albite
Binary minimum melting point with solvus - potassium feldspar - albite

4. Phase Rule for Binary Systems
f = c - p + 2 = 2 - p + 2 = 4 - p
If two phases are present, there are two degrees of freedom (both T and X)
If three phases are present, there is one degree of freedom (either T or X)

5. Binary Solid Solution
Plagioclase Feldspar series, 39% of the earth’s crust
Diagram show the plagioclase feldspar series. Plot the liquidus in green, the solidus in brown.
Starting at An60 and 1600̊C trace the behavior of the melt in red. Answer questions:
At what temperature does the first crystal appear?
What is the composition of the first crystal?
Trace the behavior of the solid in blue.
What is the composition of the solid at 1350̊C?
What is the composition of the liquid at 1350̊C?
At what temperature does the last liquid disappear?
What is the composition of the last liquid?
What is the composition of the solid corresponding to the last liquid?
Starting at An40 and 1000̊C trace the behavior of the solid in blue.
Answer questions:
At what temperature does the first liquid appear?
What is the composition of the first liquid?
Trace the behavior of the melt in red.
What is the composition of the solid at 1250̊C?
What is the composition of the liquid at 1250̊C?
At what temperature does the last solid disappear?
What is the composition of the last solid?
What is the composition of the liquid corresponding to the last solid?

6. Phase Rule Application
How many degrees of freedom are present at points 1,2, and 3?
How many degrees of freedom are present at points 1,2, or 3?

7. Binary Eutectic Diagram Conventions
Two immiscible components: such as CaAl2Si2O8 (calcic plagioclase) and FeMgSiO4 (olivine) plotted along the horizontal axis, OR olivine (isolated tetrahedra) and pyroxene (single chain tetrahedra), which are immiscible because they have different crystal structures
One variable, temperature, plotted along the vertical axis.
Pressure is held constant at 1 atmosphere.
Three phases, crystal A, crystal B, and melt.
Complete miscibility of the melt (magma)
The binary eutectic phase diagram explains the chemical behavior of two immiscible (unmixable) crystals from a completely miscible (mixable) melt, such as olivine and pyroxene, or pyroxene and Ca plagioclase.
We can generalize the system to two minerals, A and B. The conventions for the phase diagram include the following:

8. Binary Eutectic Diagram Assumptions
The system remains in equilibrium throughout its history, so that all reactions can take place and everything can come to stability
Everything in the original melt remains in communication throughout the crystallization process

9. Binary Eutectic Diagram
Source:
Plot the liquidus in green, the solidus in brown. Note that the liquidus line is continuous, but the solidus line is not. In addition, the solidus line is vertical. The line marked “solidus” in the diagram is INCORRECT!
Comments:
1. The liquidus line separates the all melt phase from the melt+crystal phase.
2. The solidus line separates the melt+crystal phase from the all crystal phase.
NOTE: The solidus and liquidus lines are experimental, they have been determined by melting and cooling many melts at different percent compositions.
3. The eutectic is the point at which all three phases can exist simultaneously, A, B, and melt. The eutectic here is 50% B, but can be any percent depending on the minerals involved.
4. For pure A (far left of diagram) the melting (crystalizing) temperature is TA about 1380̊C.
5. For pure B (far right of diagram) the melting (crystalizing) temperature is TB about 1485̊C.

10. Binary Eutectic Diagram – Intermediate Compositions
Source:
1. If we add some B to a melt of A (say 20% B; red arrow) the temperature of melting (crystallization) is lowered to about 1360̊C. The more B we add the lower the melting temperature becomes; that is, it moves down the liquidus line toward the eutectic.
2. Any mixture of A and B lowers the melting (crystallizing) temperature. For example, a mix of 60% B (blue arrow) crystallizes at about 1320̊C where pure B crystallizes (melts) at about 1480̊C.

11. Binary Eutectic Crystallization
Source:
1. Assume a melt composition of 70% B and 30% A.
2. Cool melt to liquidus line along red arrow. Only B crystals form at about 1380̊C (B is immiscible with A). Only B crystallizes because we are on the Melt+B liquidus line; no A can crystallize until the eutectic is reached.
3. Removing (crystallizing) B changes the melt composition making in richer in A. Therefore the melt composition begins to migrate to the left, but down the liquidus line toward the eutectic point. The system must stay on the liquidus line since going above it would raise the temperature high enough to melt everything
4. At the eutectic point, and only at the eutectic point, can A finally begin to crystallize out of the melt, and A and B now crystallize out together at a ratio of 50/50 until all the melt is gone
5. Finally after all the melt is gone the two crystals A+B can leave the eutectic. Since the original composition of 70% B has not changed we therefore shift the path right to the 70% point, and continue to drop the temperature straight down.
This path is the same any time the composition of B is greater than the eutectic value. If the composition is less than the eutectic, the path is similar, but a mirror image (see blue arrows for a 20% B).

12. Binary Eutectic Melting
Source:
We can reverse the process and begin with a rock, heating it slowly until it melts. In this case the diagram is read the reverse of the crystallization steps.
Comments:
1. Start with a rock containing 70 % B and 30% A.
2. It is slowly heated until it reaches the solidus line. At the solidus line the system shifts laterally to the eutectic point.
3. At the eutectic both components A and B begin melting at the same time in the ratio of the eutectic point (here 50% A/ 50% B). Melting is always at the ratio of the eutectic, regardless of the starting composition. Thus, even though the rock is 70% B and 30% A, the first melt is 50% A and 50% B.
Melting continues at a 50/50 ratio until all of the least abundant is melted (here A at 30%) and all the remaining unmelted rock is mineral B (which since all of A is melted is now 100% B.)
4. Now, imagine the process stops just at the point the last crystal of A has melted, and all the melt is removed (fractionated) from the system. A rock with a 50% B/50% A composition will solidify from this melt.
5. The unmelted rock contains all the remaining mineral B and is thus 100% B and its composition now shifts all the way to the right side, and its melting temperature now becomes that of pure B.
6. As the temperature continues to rise, B will remain unmelted until the TB of about 1485̊C is reached.
By this process the original rock has been split into two fractions. There is a melt fraction of 50% A/50% B, and an unmelted (residue) fraction of 100% B. Of course, the separation could occur at any time in the process, before all A has melted, or after A has all melted and some of B as well. The resulting two fractions will differ depending on the circumstances.

13. Melting A Rock With A 30/70 Eutectic
Source:
1. Imagine a rock containing 20 % A and 80% B.
2. It is slowly heated until it reaches the solidus line. At the solidus line the system shifts laterally to the 30/70 eutectic point.
3. At the eutectic both components A and B begin melting at the same time in the ratio of 30% B and 70% A. Melting at the eutectic is always at the ratio of the eutectic, regardless of the starting composition. Thus, even though the rock is 80% B and 20% A, the first melt is 30% B and 70% A. Melting continues at a 30/70 ratio until all of the A is melted, and all the remaining unmelted rock is mineral B (which since all of A is melted is now 100% B.)
4. Once all of B is gone then the system can leave the eutectic. Because B is the only crystal left, as the temperature rises the system will climb up the Melt+B liquidus.
5. It climbs up the Melt+B liquidus until it reaches the melting (crystallization) temperature for a 80% B/20% A system, at which point the last crystal of B melts.
6. With further rise in temperature the systems just follows the red line up to the top of the diagram.

14. Fractionation
Source:
Binary eutectic systems can undergo fractionation.
Comments:
1. Imagine a rock containing 60 % B and 40% A.
2. It is slowly heated until it reaches the solidus line. At the solidus line the system shifts laterally to the 30/70 eutectic point.
3. At the eutectic both components A and B begin melting at the same time in the ratio of 30% B and 70% A. Melting at the eutectic is always at the ratio of the eutectic, regardless of the starting composition. Thus, even though the rock is 60% B and 40% A, the first melt is 30% B and 70% A.
Melting continues at a 30/70 ratio until all of the A is melted, and all the remaining unmelted rock is mineral B (which since all of A is melted is now 100% B.) (Ratios get hard to calculate in a case like this, 60/40 start and 30/70 melting; we will just go along as if it were simple.)
4. Now, imagine the process stops just at the point the last crystal of A has melted, and all the melt is removed (fractionated) from the system. A rock with a 30% B/70% A composition will solidify from this melt.
5. The unmelted rock contains all the remaining mineral B and is thus 100% B and its composition now shifts all the way to the right side and its melting temperature now becomes that of pure B. As the temperature continues to rise it will remain unmelted until the TB of about 1485̊C is reached.
By this process the original rock has been split into two fractions. There is a melt fraction of 30% B/70% A, and an unmelted (residue) fraction of B. Of course, the separation could occur at any time in the process, before all B has melted, or after A has all melted and some of B as well. The resulting two fractions will differ depending on the circumstances.

15. Congruent Melting The previous case is an example of congruent melting
Congruent melting means melting of a substance directly to a liquid that is of the same composition as the solid

16. Incongruent Melting
Melting accompanied by decomposition or by reaction with the liquid, so that one solid phase is converted into another
Melting to give a liquid different in composition from the original solid
One example occurs in the forsterite-quartz system

17. Binary Eutectic with Incongruent Melting
L = Liquid
Fo is fosterite
En is enstatite
Qtz is quartz
Diagram: fo-qtz6.gif
These type of abbreviations for important rock-forming minerals are common, especially on diagrams where space is limited. One rule is always followed. The abbreviation cannot be a chemical symbol. Thus, we use Ab for albite, NOT Al.

18. Reaction MgSiO3 + SiO2 = Mg2SiO4 En + Qtz = Fo
Note that the forsterite - quartz system is binary, and thus has two components. We see three minerals in the diagram, so this may appear contradictory. However, we can write a chemical reaction relating the three minerals: (as shown)
This removes one possible component

19. Fo- Qtz
Plot the liquidus in green, the solidus in brown. Note that the liquidus line is continuous, but the solidus line is not. In addition, the solidus line is vertical.
Comments, Case 1:
1. Imagine melt with composition Qtz20 at 1900̊C. F = 3-1 = 2 (since composition is specified)
2. The melt cools until the liquidus line is encountered. At that point, pure Fo begins to crystallize. F = = 1
3. The melt continues to crystallize, and is gradually enriched in Qtz by the removal of Fo.
4. At 1557̊C, we encounter the “peritectic” point, marked P.
5. At the peritectic, Fo reacts with the melt to produce En. F = 3 -3 = 0. The temperature remains constant.
6. Eventually, all of the L is used up. The system is now En + Fo, so F = 3-2 =1. The temperature is free to change, and the system cools with a mix of Fo and En.

20. Forsterite – Quartz at Higher T
Two liquid phases are present on the right side of the diagram
Diagram: fo-qtz.gif
A more complete, but less detailed diagram for the Fo-Qtz system shows some new features. The stable form of silica present at lower temperatures in this diagram is tridymite (Tr). At 1470̊C, the tridymite is converted to cristabalite (Cr). At a composition of Qtz70. Starting at 1350̊C, the mix of Tr and En would heat until 1470̊C is reached. Tridymite would begin to convert to Cr, and three phases (Tr, Cr, and En) would be present. That means that T is invariant. Once the Tr was gone, the system would heat again.
At 1695̊C, the Cr + L mixture would separate into two liquids. One liquid has a composition of Qtz55 and the other Qtz93. The cristabolite would melt. Once it is gone, the system would heat, with the two liquids starting to dissolve in each other. Above approximately 1980̊C, the liquids would form a single melt phase.

21. Binary Minimum Melting Point
Diagram: phase21.gif from
Another possible form of behavior is the minimum melting point system, This resembles the binary solid solution in that both types having curving solidus lines.
Plot the liquidus in green, the solidus in brown. Note that both the liquidus and solidus linse are continuous.
Comments, Case 1:
1. Imagine a liquid composed of K20 at 100̊C. It will cool without change until it hits the liquidus. F = (for fixed composition).
2. When the cooling reaches the liquid, a solid begins to form. This will be a solid solution composed of approximately K12 initially. As cooling proceeds, the composition of the liquid will follow the liquidus, and the composition of the solid will follow the solidus F = 2-2 = 0.
3. When the solid reaches K20, all of the liquid is exhausted, and cooling of the single solid solution continues. F =
Comments, Case 2:
1. Imagine a solid composed of K80 at -100̊C. It will heat without change until it hits the solidus. F = (for fixed composition).
2. When the cooling reaches the solidus, a liquid begins to form. The composition of the liquid will be approximately K64, initially. The temperature will be approximately -20̊C. As heating proceeds, the composition of the liquid will follow the liquidus, and the composition of the solid will follow the solidus F = 2-2 = 0.
3. When the liquid reaches K80, all of the solid is exhausted, and heating of the single liquid continues. F =

22. Albite - Orthoclase Diagram: phdifig4.gif
This figure show a simplified diagram of the albite-orthoclase system. It is similar to the previous example, except that it adds a solvus line in the subsolidus region. Solvus lines separate regions of a single solid phase from regions of two solid phases.
Comments:
1. Suppose a liquid of composition Or70 is present at 1300̊C. It will cool until it hits the liquidus at approximately 1180̊C. A solid of composition approximately Or85 will form.
2. As cooling proceeds, the solid will gradually be enriched in albite, until its composition is Or70. At that point, the liquid is exhausted. The solid begins to cool.
3. Cooling proceeds until the solidus line is reached at point A in the diagram. The solid then exsolves into a phase enriched in albite (approximately Or32) shown at point B.
4. As cooling proceeds, assuming equilibrium is maintained, the two solids phases will exchange atoms, with the Or-poor phase giving up K and receiving Na from the Or-rich phase. At 300̊C, the compositions will be approximately Or5 (point C) and Or95 (Point D). This mixture is often called a perthite.

23. Lever
Adapted from:
The Lever Rule (or Law of Moments) can be used to estimate the relative percentage of solid or liquid in a mixture of solid plus melt. The principle of the lever rule is quite simple, as Figure 1 illustrates:
Assume distance LF is 33 mm, while SF is 89mm. The letters L and S can be assumed to represent liquid and solid, respectively. F represents the fulcrum. LF + SF = 122mm, and represents the entire composition. The percentage of liquid and solid are therefore as shown
The computation is quite easy. Two difficulties remain. The first is where to place the fulcrum. The second is which side of the fulcrum represents liquid and which side represents solid?

24. Application of Lever Rule
Adapted from:
At point 1 the alloy is completely liquid, with a composition C.
C = 65 weight% β

25. Point 2
Source:
At point 2 the alloy has cooled as far as the liquidus, and solid phase B starts to form.
Phase B first forms with a composition of 96 weight% β.
The green dashed line below is an example of a tie-line. A tie-line is a horizontal (i.e., constant-temperature) line through the chosen point, which intersects the phase boundary lines on either side.
The composition is still essentially 100% liquid.
Solid phase B starts to form with a composition of 96 weight% β

26. Point 3 C1 = 58 weight% β and C2 = 92 weight% β
Source:
C1 = 58 weight% β and C2 = 92 weight% β

27. Point 3 Calculations
Fraction of solid b = ( ) / ( ) = 20 weight%
Fraction of liquid = ( ) / ( ) = 80 weight%
A tie-line is drawn through the point, and the lever rule is applied to identify the proportions of phases present.
Intersection of the lines gives compositions C1 and C2 as shown.

28. Point 4 C3 = 48 weight% β C4 = 87 weight% β
Source:

29. Point 4 Calculations
Fraction of solid with composition C4 = ( ) / ( ) = 44 weight%.
Fraction of liquid at eutectic = 56 weight%

30. Point 5 C5 = 9 weight% β and C6 = 91 weight% β
Source:
C5 = 9 weight% β and C6 = 91 weight% β

31. Point 5 Calculations
Fraction of solid with composition C6 = (65 - 9) / (91 - 9) = 68 weight%
Fraction of solid with composition C5 = ( ) / (91 - 9) = 32 weight%

32. Effects of Pressure
Source:
The diagrams we have been looking at assume that pressure is fixed. In geology, we often need to understand systems at higher pressures. This can be done by examining a series of diagrams of the same system at different pressures.
The diagram shows the forsterite-quartz system previously examined. As pressure increases, the following changes are noted:
With increased pressure, from 1 atm to 7 kbar, En changes from incongruent to congruent melting behavior
The field of liquid immiscibility, the 2L field on the 1 atm diagram, disappears at slightly elevated pressures and is no longer evident in the 3 kbar diagram
With increased pressure, from 1 atm to 7 kbar, En changes from incongruent to congruent melting behavior
The field of liquid immiscibility, the 2L field on the 1 atm diagram, disappears at slightly elevated pressures and is no longer evident in the 3 kbar diagram.

33. Orthoclase – Albite at Low P
Source:
Another example of increasing pressure is shown in the orthoclase - albite system. Orthoclase is K-spar (KAlSi3O8) and albite is sodium plagioclase, NaAlSi3O8. Since K is much larger than Na, Goldschmidt’s rules tell us immiscibility is limited. Another mineral is also present. This is leucite, KAlSi2O6, which cannot be formed from ab + or, so this system is only pseudobinary. Observed changes in the diagram as pressure increases:
1. Disappearance of leucite from the binary diagram. Leucite is unstable at P > 2 kbars. Leucite is not found in plutonic rocks which have crystallized at depths corresponding to pressures in excess of 2 kbars. Leucite is most commonly found in volcanic rocks, which are undersaturated with respect to silica.
2. Increasing P also lowers the melting point of both end member components, Ab and Or.

34. Orthoclase – Albite at 5 kbars
Source:
Examination of a more detailed view of the 2 kbar diagram shows that crystallization and melting paths in this system are comparable to those examined in the Ab-An system. As liquids cool between the liquidus and solidus the solid phase is constantly changing composition, along the solidus, to maintain equilibrium with the liquid, which also is changing composition, along the liquidus.
This diagram also shows the presence of the solvus, below which the feldspar solid solution exsolves to form two separate feldspars, one rich in Ab and one rich in Or. This exsolution results in the formation of a variety of textures which can be recognized by examining the resulting rock under the microscope.
1. If the feldspar grain consists of an Or-rich host and exsolution lamellae of Ab, then the grain is a perthite and exhibits a perthitic texture.
2. If the feldspar grain consists of an Ab-rich host and lamellae of Or, the grain is an antiperthite, resulting in an antiperthitic texture.
The Lever Rule can be used to calculate the percent of each phase present at a given temperature. Using the red dot at any point on or below the solvus line, we can measure the tie line, and the size of each fragment of the tie line, and use that data to calculate the percent of each phase present. Note that the percentages will change as temperature decreases. The composition of the phases will also more closely approach that of the end members. Note also that Or is more easily able to accommodate the smaller Na, whereas albite tolerates only very small amounts of K.

35. Reading a Ternary Diagram
Image: tri.gif
Three component systems may be plotted on a ternary composition diagram. Since three components are being plotted, there is no axis available for either temperature or pressure. Thus both pressure and temperature are regarded as fixed. Either one of these may be plotted as overlays on the composition plot.
The graphic shows a ternary diagram. Three components, labeled A, B, and C are shown. The method of reading the diagram is shown (40%A, 50%B, 10%C).

36. Points on Diagram Edge Points on the edge have only two components
Image: tri2.gif
Points on the edge have only two components

37. Reading Sample Points
Note that the scales are plotted increasing in a CCW directions
Image: ternread-d.gif
This diagram is plotted in terms of generic components A, B, and C. The scales are plotted so they increase in a counterclockwise (CCW) fashion. We can read the composition of the points:
%A, 20%B, 20% C
Note the sum must equal 100. You only need to read two components, but it is a good idea to read all three, and check the sum.
%A, 40%B, 40% C
%A, 70%B, 20% C
%A, 20%B, 80% C

38. Ternary Rock Composition Diagram
Ternary diagrams can be used to plot any three things
Image: ternread-f.gif
Image: ternread-d.gif
Ternary diagrams can be used to total any three things whose sum is 100. The diagram shows a frequent use of a ternary diagram, to show sedimentary rock composition.
We can read off the composition of the points plotted:
% sandstone
% shale
% limestone
% sandstone, 10% shale, 30% limestone
% sandstone, 60% shale, 10% limestone
% sandstone, 10% shale, 80% limestone

39. Texture Diagram
Another use of ternary diagrams
Image: textsand.gif

40. Ternary System Sample ternary diagram
A blank ternary diagram is shown. We can plot our own diagram.
A - Nepheline, NaAlSiO4 (lower left)
B - Kalsilite KAlSiO4 (lower right)
C - Silica SiO2 (top)
Note that Nepheline + Silica = Soda feldspar (albite, NaAlSi3O8)
Kalsilite + Silica = Potassium feldspar (KAlSi3O8)

41. Silica % Silica ↑ % Kalsilite ↓ Nepheline Kalsilite ← % Nepheline 100
100 90 10 80 20 70 30 60 40
% Silica ↑ 50
% Kalsilite ↓ 50 60 40 70 30
Plot the axes:
0-100% Nepheline from top to bottom on the bottom
0-100% Kalsilite from top to bottom on the right
0-100% Silica from bottom to top on the left
Plot points:
A - 75% silica, 0% nepheline, 25% kalsilite
B - 30% silica, 50% nepheline. 20% kalsilite
C - 20% silica, 35% nepheline, 45% kalsilite 80 20 90 10
100 90 80 70 60 50 40 30 20 10
100
Nepheline
Kalsilite
← % Nepheline

42. Real Plot Image: System_Na_K_AL_Si_O.gif
The diagram shows the same system we just plotted. In addition to the composition information, temperature overlays are shown. The lines with arrows are cotectic lines. Any composition on the diagram will move downhill toward the nearest cotectic as it cools. Once on a cotectic, the system will cool toward the ternary eutectic (1020̊C).
Note that any side of this diagram is a binary plot. Where the cotectic line intersects a side, it represents a binary eutectic point.