1. Phase Diagrams Homework Exercises (a-only): 8.3,4, 6, 13, 15
Problems 8.2 & 8.4 1

2. Definitions
Phase(P) -“State matter which is uniform throughout not only in chemical composition but also in physical state” J. Willard Gibbs
Solid
Various phases [e.g. crystal structures (diamond; graphite) or compositions (UC;UC2)]
Alloys (sometimes its difficult to tell this - microscopic examination may be necessary {dispersions uniform on macroscopic scale})
Miscible one phase (P=1)
Immisible multiple phases (P>1)
Liquid
Miscible liquids are one phase
Immiscible liquids are multiple phases (P>1)
Gas
Systems consisting of gases can have only one phase
Shape or degree of subdivision irrelevant 2

3. Definitions Heterogeneous and homogeneous systems
Systems with one phase are homogeneous
Systems with more than one phase are heterogeneous
Constituent- a chemical species (ion or molecule which is present
Component (C) - chemically independent constituents of a system
C = #of independent chemical constituents - # of distinct chemical reactions
#of independent chemical constituents = total # of constituents minus the number of any restrictive conditions (charge neutrality, material balance etc.) 3

4. Counting Components Components: Example: CaCO3(s)  CaO(s) + CO2(g)
Phases: P = 2 solid + 1 gas = 3
Component: C = 3 consitiuents - 1 reaction = 2
Example: Consider the system NaCl, KBr and H2O. Suppose you can also isolate the following from it KCl(s), NaBr(s), NaBr. H2O(s), KBr. H2O(s), and NaCl. H2O(s)
Phases: P = 5 solid
Components:
There are 8 constituents
You can write the following reactions four:
NaCl + KBr  KCl + NaBr; NaCl + H2O  NaCl. H2O
KBr + H2O  KBr. H2O; NaBr + H2O  NaBr. H2O
Conditions: 1 -Material balance moles of KCl = moles of NaBr + NaBr. H2O
Components: C = (8 constituents -1 condition) - 4 reactions = 3 4

5. Counting Components (continued)
Remember you must count reactions which actually occur not those which could occur
Consider a system in which we has the following : O2(g), H2(g), H2O(g)
If conditions are such O2(g) + H2(g)  H2O(g) does not occur, C=3
If conditions are such (T, catalyst) O2(g), H2(g)  H2O(g) occurs, then C=3components -1reaction =2
If conditions are such (T, catalyst) O2(g), H2(g)  H2O(g) occurs and you impose the condition that all the hydrogen and oxygen come from dissociation of water, then C=3 constituents - 1 condition -1 reaction = 1
Identity of components is a matter of some choice, number isn’t
Chose components that cannot be converted into one another by reactions
E.g. CaCO3(s)  CaO(s) + CO2(g) 5

6. Phase Rule
In discussing phase equilbria, you need only consider intensity factors (temperature, pressure, and concentration)
Only certain of these can be varied independently
Some are fixed by the values chosen for the independent variables and by the requirements of thermodynamic equilibrium
e.g. if you chose  and T for a system P is fixed
Number of variables which can be varied independently without changing the number of phases is called the degrees of freedom of the system
The number of degrees of freedom or variance of a system, F, is related to the number of components(C) and number of phases(P)
F = C - P + 2 6

7. Phase Rule Proof Only C-1 mole fractions are needed since the xJ = 1
In any system the number of intensive variables are: pressure, temperature plus the mole fractions of each component of each phase.
Only C-1 mole fractions are needed since the xJ = 1
Thus for P phase, the number of intensive variables = P(C-1) + 2
At equilibrium the chemical potential of phase must be equal,
i.e. µJP1= µJP2= µJP3= µJP4= µJP5….{there are P-1 such equations}
Since there are C components, equilibrium requires that there are C(P-1) equations linking the chemical potentials in all the phases of all the components
F = total required variables - total restraining conditions
F = P(C-1) C(P-1) = PC - P + 2 -CP + C = C- P + 2 7

8. One Component Systems
Phase rule says that you can have at most 3 phases
F = C- P +2; C=1 so F=3-P
If P=3, F=0 system is invariant
Specified by temperature and pressure and occurs at 1 point (called the triple point)
If one phase is present, F = 2 that is P and T can be varied independently
This defines an area in a P,T diagram which only one phase is present
If two phases are present, F = 1 so only P or T can be varied independently.
This defines a line in a P, T diagram 8

9. Single Component Phase Diagram
Point a: only vapor present
Point b: Liquid boils - 1 atm= Tb but vapor and liquid co-exist along line(vapor pressure curve)
Point C : Liquid freezes
Point D: Only solid
Triple point: 3 phases in equilibrium
Line Below triple point: Vapor pressure above solid
Note: You don’t necessarily have to have 3 phases and they don’t have to be solid liquid and gas 9

10. Cooling Curve
You can generate a cooling constant pressure (isobar) from previous phase diagram
Halts occur during 1st order phase transitions (e.g. freezing) 10

11. Experimental Measurements
Phase changes can be measured by performing DTA (differential thermal analysis) on samples
In DTA sample is heated vs. a reference
1st order transitions can be measured even when they can’t be observed
They will occur as peaks in DTA
High Pressures can be achieved with diamond anvil cells
See text descriptions 11

12. Two Component Systems For two component systems, F = 2-P+2 = 4-P
If P or T is held constant, F’ = 3-P (’ indicates something is constant)
Maximum value for F’ is 2
If T is constant one degree of freedom is pressure and the other is mole fraction
Phase diagram (Constant T) is map of pressure and compositions at which each phase is stable
If P is constant one degree of freedom is temperature and the other is mole fraction
Phase diagram (Constant P) is map of temperature and compositions at which each phase is stable
Both Useful 12

13. Vapor Pressure Diagrams
By Raoult’s law (pA = xApA * & pB = xBpB * ), the total pressure p is
p = pA + pB = xApA * + xBpB *
But xB = (1-xA)so xApA * + xBpB * = xApA * + (1-xA) pB * = pB * + xA(pA * -pB *)
@ Constant T, total vapor pressure is proportional to xA (or xB )
The composition of the vapor is given by Raoult’s law so the mole fraction in the gas phase, yA and yB is
yA = pA/p and yB = pB/p {also yA = 1-yA)
From above yA = xApA * /[pB * + xA(pA * -pA *)]
If pA * /pB *= pA/B then yA = (xA * pA/B) /(1+ (xA * pA/B) - xA )
Or yA = (xA * pA/B) /(1+ (xA *( pA/B - 1) ) 13

14. Effect of Ratio of Vapor Pressure on Mole Fraction in Vapor
This shows the vapor is richer in the more volatile component
If B is non volatile then yB = 0 14

15. Pressure Composition Diagrams
Assume the composition on the x axis is the overall composition, zA (as mole fraction)
In Liquid region zA = xA
In vapor Region region zA = yA
In between two phases present
F’=1 so at given pressure compositions are fixed by tie lines 15

16. Isopleth Compositions in Each Phase
A vertical line represents a line of constant composition or isopleth
Until pressure = p1sample is liquid vapor phase composition is a1
At p1, vapor composition is given by tie line to vapor curve
At p2 vapor composition is a’2, liquid composition is a2 and overall composition is a
At p3 vitually all the liquid is vapor and trace of liquid has composition given by tie line to liquid 16

17. Determining proportions of Phases (lever rule)
The composition of each phase is given by the each end of the tie line
The relative proportion of each phase is given by the length of the tie line
nl = nl or n= nl / l 17

18. Temperature-Compositions
Assume A more volatile than B
Region between two curves is 2-phase region
F’=1 (pressure is fixed)
At given temperature compositions are fixed by tie lines
Region outside lines composition & temperature variable
Heat liquid with composition a1
Hits boiling curve, vapor has composition a2’, liquid a2 (=a1)
vapor is richer in more volatile component
Distillation
Vapor condensed (a2’-a3)
New concentration a3’ (richer still)
New etc until nearly pure liquid obtained

19. Distillation/Theoretical Plates
A theoretical plate is a vaporization-condensation step
Previous example has 3 theortetical plates
If the two curves move closer together, more theoretical plates are required to achieve same degree of separation
Curves more together if components have similar vapor pressures

20. Non-Ideal T-C Diagrams - High Boiling Azeotropes
Maximum in phase diagram occurs when interactions in liquid between A & B stabilize the liquid
GE is more negative
If such a liquid is boiled, as vapor is removed, composition of liquid is richer in B (less A)
As vapor is removed you move to right up the curve until you reach point b
At b liquid boils with constant composition
Called an azeotrope (unchanging Gr.)
Example HCl-water boils @80 wt % water at 108.6°C

21. Non-Ideal T-C Diagrams - Low Boiling Azeotropes
Minimum in phase diagram occurs when interactions in liquid between A & B destabilize the liquid
GE is more positive
If such a liquid is boiled, & vapor is condensed , composition of vapor is richer in B (less A)
As vapor is removed you move to right down the curve until you reach point b
At b liquid boils with constant composition
Example ethanol-water boils at constant water content of 4 78°C

22. Non-Ideal T-C Diagrams - Immisicble Liquids
If two liquids immiscible and in intimate contact then p is nearly the sum of vapor pressures of pure components (p = pA* +pB*)
mixture will boil when p = atmospheric pressure
intimate contact (& trace level saturation maintained)
If two liquids immiscible and not in intimate contact then p for each is the vapor pressures of pure components (p = pA* and p = pB*)
Each will boil separately when respective pA* = atmospheric pressure and/or pB* = atmospheric pressure

23. Liquid-Solid Phase Diagrams
Liquids miscible & solids immiscible
Consider Cooling along isopleth from a1
At a2 pure B starts to come out of solution
At a3 solution is mixture of B + Liquid with composition b3 (ratio by lever rule)
At a4 liquid has composition “e” and freezes
In solid region there are two phases pure A and pure B
Composition given by tie line, ratio by lever rule
“e” is called a eutectic

24. Examples of Simple Eutectic Systems

25. Eutectics
In previous diagram, the eutectic (easily melted, Gr.) point is a temperature at which a mixture freezes without first depositing pure A or B
Like a melting point in that it it is a definite temperature
That’s because, since C=2 and P=3, by Phase rule, F’=0
A cooling (or heating) curve will have a halt at the eutectic temperature
If pure A and pure B are in contact a liquid will form at the eutectic temperature
Examples
solder lead/tin (67/33) melting point 183°C
NaCl and water (23/77) melting point -21.1°C

26. Liquid-Solid Phase Diagrams - Reacting Systems
Some Binary systems react to produce one (or more) compounds
Definite composition
Unique melting point
Congruent melting point, I.e. melts to a liquid of identical composition
Maximum in phase diagram
Phase diagram interpreted as before except now there are additional regions

27. PARTICLE BED REACTORS (PBRs)
High Power Density Nuclear Sources for Space Power & Propulsion
Performance superior to chemical rockets (H2/O2)
Enabling technology for Mars mission
Multiple coolants possible
He for power applications
H2, NH3 for propulsion applications
High power densities (10’s MW/liter)
Superior performance to NERVA system (70’s era nuclear propulsion system)
Typical operating temperatures >2500 K for propulsion 2

28. PBR - Schematic

29. Materials Requirements for PBR Hot Components
Hot frit, nozzle, etc.
Withstand H2 Environment
~2800 K
70 atmospheres
Large Temperature Gradients
12K to Tmax over a few cm
Multiple Thermal Cycles
Long Exposure Times
10’s minutes
Launch Stresses
Withstand Radiation Fields
Not Affect Reactor Criticality
Fuel
Same as general components
Provide for adequate reactivity
Optimize coating thickness and type to maintain criticality Maintain (Keff)
No HfC coatings
Maintain coolable geometry
No large gaps between layers
No particle clumping
No reaction with other components, e.g. hot frit
Minimum F.P. and U release
Criticality and safety criterion 3

30. Hot Component Material Selection
Reactor Components (Hot Frit)
Rhenium (monolithic & coatings)
Large neutronic penalty for monolithic Re
High radiation heating for monolithic Re
Extensive alloying of Re with fuel T > 2760 K
Pyrolytic BN
High cost 11BN required
Unacceptable thermal decomposition (3 wt % in K
Reaction with baseline ZrC fuel coating
Carbide-Coated Carbon (graphite & carbon-carbon)
Potential for CTE mismatch between coatings and substrate
Fuels
Baseline fuel HTGR-type with ZrC coating
Conclusion: Materials development program focused on carbide coatings of carbon materials 4

31. “HTGR” Type Fuel UC Outer carbide shell Pyrocarbon layer(s)
HTGR - SiC
PBR - ZrC
Pyrocarbon layer(s)
“Spongy” layer
Dense layer
Inner kernel
HTGR - UO2
PBR - UC
UC melting point 2525°C
UC °C UC

32. Carbide Phase Diagrams
Tantalum-Carbon
Zirconium-Carbon

33. Liquid-Solid Phase Diagrams - Reacting Systems (Incongruent Melting)
If the compound is not stable as a liquid incongruent melting occurs
Compound melts into components
Called the peritectic melting point
One solid phase “melts around” the other
Isopleths
“a”
a1-> a2 liquid phase with A + B
a2 solid B precipitates
a3->a4 Solid B + compound
“b”
b2-> b3 liquid phase with A + B
b3 B reacts to form compound
b3->b4 Solidcompound + liquid
b5 solid A precipitates with compound

34. Zone Refining
Ultra high purity can be obtained by moving a small molten zone across a sample.
Impurities more soluble in liquid than solid so they continually move down the liquid front
One end becomes purer while other end is dirtier
Multiple passes can be used to achieve high purity

35. Ternary Phase Diagrams
Each composition must be defined by two compositions or mole fractions
Composition diagrams are therefore two dimensional
Triangle with each edge one line of binary phase diagram
Pressure or temperature add third dimension
Usually temperature
Phase diagrams are usually given as a succession of surfaces at constant temperature
To examine temperature variation you hold composition constant

36. Ternary Phase Diagrams