Química Física

1. Chapter 8: Phase Diagrams
Homework
Exercises (a-only): 8.3,4, 6, 13, 15
Problems 8.2 & 8.4

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

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.)

4. Counting 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

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)

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

7. Phase Rule Proof
 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) + 2 - C(P-1) = PC - P + 2 -CP + C = C- P + 2

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

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

10. Cooling Curve
 You can generate a cooling
A Gas Cooling
curve @ constant pressure
B
Gas Condensing (isobar) from previous phase
100°C
diagram
Liquid Cooling
Temp.
 Halts occur during 1st order
Solid f reezing
phase transitions (e.g. freezing)
C
0°C
D
Solid Coolin g
Time

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

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

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) )

14. Effect of Ratio of Vapor Pressure
on Mole Fraction in Vapor
1.2
p*/p*=1000
1 A B
p*/p*=10
0.8 p*/p*=50 A B
A B
0.6 p*/p*=4
A B
p*/p*=2 p*/p*=1
A B
A B
0.4
0.2
0
0 0.2 0.4 0.6 0.8 1
Mole Fraction of A in Liquid
 This shows the vapor is richer in the more volatile component
 If B is non volatile then yB = 0

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

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

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

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 vapor @ concentration a3’ (richer still)
» New condensate @a4 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
wt% @ 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
Componen t Tm of A Componen t Tm of Eutectic Eutectic
A (K) B B Mol %B Tm B (K)
(K)
Picric Acid 395 TNT 353 64 333
Sb 903 Pb 599 81 519
Cd 594 Bi 444 55 417
KCl 1063 AgCl 724 69 579
Si 1685 Al 930 89 851

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

28. PBR - Schematic

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

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 coatings @ T > 2760 K
» Pyrolytic BN
 High cost 11BN required
 Unacceptable thermal decomposition (3 wt % in 10 min @ 2700 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

31. “HTGR” Type Fuel
 Outer carbide shell
» HTGR - SiC
UC » PBR - ZrC
 Pyrocarbon layer(s)
» “Spongy” layer
» Dense layer
 Inner kernel
» HTGR - UO2
» PBR - UC
 UC melting point 2525°C
 UC2 2350-2400°C

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
C
2.6 cm
cm 4.2
2
4.2
cm cm
4.2
1
5.7 cm
4.2 cm
A B
100 %
Component "A"
To obtain relative amounts d rop a perpendicular to each axis
Ratio of the lengths of lines is the ratio of compone nets
Ratio A:B:C
"1 " = 1:1:1 or 33% o f each component
"2 " = 1.6:1:2.2 or 33 .6%A, 20.8%B, 45 .6%C