Lecture 36. the Phase Rule

Lecture 36. The Phase Rule

P = number of phases C = number of components (chemically independent constituents) F = number of degrees of freedom xC,P = the mole fraction of component C in phase P

The variables used to describe a system in equilibrium:

x11, x21, x31,...,xC −1,1 phase 1

x12 , x22 , x32 ,..., xC−1,2 phase 2

x1P , x2P , x3P ,...,xC−1,P phase P T,P

Total number of variables = P(C-1) + 2

Constraints on the system: m11 = m12 = m13 =…= m1,P P - 1 relations m21 = m22 = m23 =…= m2,P P - 1 relations mC,1 = mC,2 = mC,3 =…= mC,P P - 1 relations 1 Total number of constraints = C(P - 1)

Degrees of freedom = variables - constraints

F=P(C- 1) + 2 - C(P - 1) F=C- P+2

Single Component Systems: F = 3 - P

In single phase regions, F = 2. Both T and P may vary.

At the equilibrium between two phases, F = 1. Changing T requires a change in P, and vice versa.

At the triple point, F = 0. Tt and Pt are unique.

2 Four phases cannot be in equilibrium (for a single component.)

Two Component Systems: F = 4 - P

The possible phases are the vapor, two immiscible (or partially miscible) liquid phases, and two solid phases. (Of course, they don’t have to all exist. The liquids might turn out to be miscible for all compositions.)

3 Liquid-Vapor Equilibrium

Possible degrees of freedom: T, P, mole fraction of A xA = mole fraction of A in the liquid yA = mole fraction of A in the vapor zA = overall mole fraction of A (for the entire system)

We can plot either T vs zA holding P constant, or P vs zA holding T constant.

Let A be the more volatile substance:

* * PA >PB and Tb,A

Pressure-composition diagrams Fix the temperature at some value, T. Assume Raoult’s Law:

* * P=PA xA +PB xB

* * * * * P=PA xA +PB (1 - xA)=PB + (PA - PB )xA

4 Composition of the vapor

P x P* y = A = A A > x A * + ()* − * A P PB PA PB x A

P x P* y = B = B B < x B * + ()* − * B P PB PA PB xA

Point a: One phase, F = 3 (T, P, xA)

Point b: Liquid starts to vaporize, F = 2 (T,P; xA not free.) xA = zb,yA = yb” Vapor is rich in A. Point c: Liquid has lost so much A that its composition is xA = xc’. The vapor is now poorer in A, yA = yc”

Ratio of moles in the two phases is given by the lever rule: n ′′ liq = cc ′ nvap cc

5 Pointd:Liquidisalmostallgone,xA = xd’,yA = yd =zA

For points below d, only the vapor is present (F=3).

Numerical example: benzene-toluene at 20 C Exercise 8.4b. Let B = toluene and A = benzene.

* * Given: PA = 74 Torr, PB = 22 Torr, zA =0.5

Q. At what pressure does the mixture begin to boil? A. At point b, P = 0.5x22 + 0.5x74 = 48 Torr.

Q. What is the composition of the vapor at this point? A. PB = 0.5x22 = 11, PA = 0.5x74 = 37, yA = 37/48 = 0.77

Q. What is the composition and vapor pressure of the liquid when the last few drops of liquid boil? A. Point d.

74x y = z = 0.5 = A A A + − 74xA 22(1 xA )

xA = 0.229, xB =0.771

P = 0.229x74 + 0.771x22= 33.9 Torr

6 Lecture 37. Construction of the Temperature- Composition Diagram

Fix the total pressure at some chosen value, P. Boiling does not occur at a unique temperature, but rather over a range of temperatures. (Contrast with a pure substance.)

The vapor pressure curve is determined by the liquid composition:

P=PA(T) + PB(T)

PA(T) is the vapor pressure of pure A at temperature T.

7 Assume Raoult’s Law:

= * + * P xA PA (T) xB PB (T) = * + − * xAPA (T) (1 xA )PB (T) = * + * − * PB (T ) xA (PA (T ) PB (T ))

The vapor pressure pure liquid A at any temperature T is given by the Clapeyron-Clausius equation:

 *  ∆H    PA (T)  = − vap,A  1 − 1  ln *     PA (T0 )  R T T0 

* PA (T0 ) =1atm,T0 = normal boiling point of A. The same reasoning applies to substance B.

* * Once we have PA (T ) and PB (T ) , Raoult’s Law gives us xA as a unique function of T and P. This allows us to construct the liquid curve.

8 Composition of the vapor:

P x P* (T) y = A = A A A P P ∆H  1 1  * − vap  −  y P (T ) RT  T T  A = A = e  b, A  xA P

This result is equivalent to

µ * + = µ * + A,vap RT ln y A A,liq RT ln xA

This last result is the starting point for the boiling point elevation formula, except that there we assume yA =1.

9 Recipe for construction of the phase diagram:

1. Calculate Tb,A and Tb,B at the pressure of the diagram.

2. Choose a temperature Tb,A

* * 3. Calculate PA (T ) and PB (T ).

P − P* (T ) x = B 4. A * − * PA (T ) PB (T )

= * 5. y A x APA (T)/P.

10 Cooling of a vapor mixture

Point a. Pure vapor, yA =zA

Point b. Vapor begins to condense at xA =zb’,yA =zA =zb

Point c. Comparable amounts of the two phases are n ′ vapor = cc present. Lever rule: ′′ nliquid cc

Point d. Vapor is almost all gone; xA =zd’ =zA,yA =zd”

Point e. Only liquid is present.

11 Distillation

Point a. Mixture starts to boil, with xA =zA,yA =zb

Points b-c. Vapor is condensed to form a liquid with xA=zb =zc

Point c. The liquid that was collected in the previous step is boiled to form a vapor with xA=zd

Condensation of the last bit of vapor produces a liquid very rich in either A (if Tb,A Tb,B on the right).

12 Non-ideal solutions

Left: Impossible phase diagram, because at Pmax,where the liquid of this composition just starts to boil, there is no corresponding point on the vapor curve. (There is no tie line.)

The vapor should always lie below the liquid in a pressure-composition diagram. At an extremum they must touch.

Center: Vapor pressure reaches a maximum because of repulsion between A and B. This is an azeotrope, where the liquid and vapor have the same composition. (Note an error in the drawing: The vapor curve does not have a cusp, but rather is tangent to the liquid curve.)

Right: Vapor pressure reaches a minimum because of attraction between A and B. This is also an azeotrope.

13 Distillation of non-ideal solutions

Left diagram: low boiling azeotrope. A solution of composition za first boils at Ta. The vapor comes off with composition zb. It is condensed and then boils at Tc

The final vapor to come off has the azeotropic composition, and the remaining liquid is enriched in B (or A, depending on which side of the azeotrope the process started).

Right diagram: high boiling azeotrope. As before, za first boils at Ta

The final vapor to come off is enriched in B (or A), and the remaining liquid has the azeotropic composition.

14 Equilibrium between immiscible liquids

Upper Critical Lower Critical Double Critical Temperature Temperature Temperature

The two phase region is always described by a tie line. It is a “no man’s land.”

15 Boiling of immiscible liquids

Melting of immiscible solids

Mixture of non-reactive Solids A and B react to solids. form compound AB.

16 Lecture 38. Equilibrium Constants

Consider the following reaction:

H2 +Cl2 Ø 2HCl

Does it occur spontaneously? Let’s calculate DG for one mole of reaction.

∆ 0 = G f (H 2 ) 0 ∆ 0 = G f (Cl2 ) 0 ∆ 0 = − G f (HCl) 95.30kJ / mol ∆ 0 = − Gr 190.6

It seems that the reaction occurs spontaneously. But what if we start with pure HCl. Shouldn’t some of it react to form H2 and Cl2?

17 To determine when the reaction occurs spontaneously, we must take the partial pressures into account:

 P  ∆G (Cl ) = 0 + RT ln Cl2  f 2  P o   Cl2   P  ∆ = − +  HCl  G f (HCl) 95.30 RT ln o   PHCl   P 2  P   P  ∆ = − + HCl −  H 2  −  Cl2  Gr 190.6 RT ln  RT ln  RT ln   P0   Po   Po 

At equilibrium, DGr =0.Thatis,

 2   PHCl   (P o ) 2   P 2  190.6 = −∆G o = RT ln = RT ln HCl  r  P P    H 2 Cl2  PH PCl    2 2  P o P o 

= −∆ 0 RT ln K P Gr

ln KP = 76.93 at 298 K

33 KP =2.57x10

18 So what happens if we start with pure HCl?

P P = P = 0 Suppose that initially HCl =1barand H2 Cl2

P = 1− 2z P = P = z At equilibrium, HCl and H 2 Cl2

()1− 2z 2 = 2.57x1033 z2

1 ≈ 2.57x1033 z2

z=1.97 x 10-17

This reaction goes nearly to completion. But this won’t be the case at higher T.

19 Another example:

N2O4 F 2NO2

∆ 0 = G f 97.89 and 51.31 kJ/mol

∆ 0 = − = Gr 2x51.31 97.89 47.30

ln KP = -4.730/RT = 1.909

 2 PNO  2   Po  P2 = = = NO2 K P 0.148 o PN O P P 2 4 N2O4 Po

What is the composition of 2 bar of this material?

Let x be the mole fraction of NO2.

()xP 2 K = P ()1− x PPo

Noting that Po =1bar,

20 2 2 x P =KPP - KPxP

2 2 P x +KPPx - KPP=0

− K P ± K 2 P2 + 4K P3 x = P P P 2P2

For P = 2 bar, x = 0.238. (Only the + sign is physically meaningful.)ForP=0.01bar,x=0.940

What is the composition of a very low pressure gas?

− K P + K P 1+ 4P / K x = P P P 2P2

− + ()+ − 2 2 ≈ K P P KP P 1 2P / KP P / 2K P = − P x 2 1 2P 2K P

(Here we used the series (1+e)1/2 =1+e/2 -e2/8 + …)

For P = 0.01 bar bar, x ≈ 0.966

21 Suppose you somehow arranged to start with pure N2O4. What is the composition of an equilibrium mixture?

N2O4 NO2 Initial P0 0 Final P0-z 2z

(2z)2 K = P − P0 z

For KP = 0.148 and P0 =1bar,z=0.175 P = 1− 0.175 = 0.825 N 2O4 P = 2x0.175 = 0.349 NO2

Ptot =1.1744 x = 0.349 /1.1744 = 0.297 NO2

22 The book defines the fraction of dissociation, a.

N2O4 NO2 Initial no. N0 moles Final no. (1-a)n 2an moles −α −α α Final mole 1 = 1 2 fraction 1−α + 2α 1+α 1+α

P2 x2 P2 4α 2P K = NO2 = NO2 = P P x P 1−α 2 N 2O4 N 2O4

But P = (1+ a)P0

Substituting this result gives the same equation as before. This method is useful if the total pressure is specified.

How does KP change if P is increased?

How does the composition change if P is increased?

23 More complex stochiometry:

2H2S+CH4 F 4H2 +CS2

4  P   P  CS  H 2   2   o   o   P   P  K = P 2  P   P  CH  H 2 S   4   o   o   P   P  Set Po =1bar,sothat

P4 P K = H 2 CS 2 P P P H 2 S CH 4

Problem: Given that T = 700 C and P = 762 Torr Initial Final H2S 11.02 mmol CH4 5.48 H2 0 CS2 00.711

o Find KP and DGr

24 Solution:

2H2S+CH4 F 4H2 +CS2

4    P  PCS  H2   2   o   o  4  P   P  x x P2 K = = H2 CS2 P  2  P  x2 x (P0 )2 P CH H2S CH4  H2S   4   o   o   P   P  Set Po =1bar,sothat

= 2 KP KxP

Problem: Given that T = 700 C and P = 762 Torr = 1.012 bar Initial Final x H2S 11.02 mmol 9.598 0.536 CH4 5.48 4.769 0.266 H2 0 2.844 0.159 CS2 00.7110.040

o Find KP and DGr -4 Kx = 3.34x10 -4 KP = 3.43 x10 0 DGr = -RT lnKP = 94.5 kJ/mol

25 Lecture 39. Equilibrium Constants II

Derivation of KP for H2 +Cl2 F 2HCl

G = µ n + µ n + 2µ n m H 2 H 2 Cl2 Cl2 HCl HCl dG = µ dn + µ dn + 2µ dn m H 2 H 2 Cl2 Cl2 HCl HCl

Let x indicate the “extent” of reaction.Itisadevicefor handling the stochiometry of the reaction. dn = −dξ H 2 dn = −dξ Cl2 = ξ dnHCl 2d

The chemical potentials are  P  µ = ∆ o +  H 2  H G f ,m (H 2 ) RT ln  2  P o   P  µ = ∆ o +  Cl2  Cl G f ,m (Cl2 ) RT ln  2  Po   P  µ = ∆Go (HCl) + RT ln HCl  HCl f ,m  Po 

26 Putting all the pieces together:

 P   P  = −∆ o −  H 2  − ∆ o −  Cl2  dGm { G f ,m (H 2 ) RT ln  G f ,m (Cl2 ) RT ln   P o   P o   P  + 2∆G o (HCl) + 2RT ln HCl }dξ f ,m  P o  dG  P 2  m = ∆G o + RT ln HCl  dξ r  P P   H 2 Cl2 

At equilibrium, dG m = 0 dξ = −∆ o RT ln K P Gr

 P2  K =  HCl  P  P P   H2 Cl2 

27 Note that KP is independent of pressure (for ideal gases). This does not mean that the equilibrium ratios of mole fractions or concentrations are independent of P.

Let’s use 2H2S+CH4 as an example.

P = x P CS 2 CS 2 etc.

4 4  P   P  x   x   2 H2 o CS2 o   =  P   P  = P K P K x    P 2  P   Po  x2   x   H2S  Po  CH4  Po 

In general, ∆ν  P  K = K   P x  Po 

28 For example, for N2O4 F 2N2O  P  K = K   P x  Po  2 −∆ o x K e Gr ,m / RT K = N2O = P = − x x P P N2O4

= lim xN O 0 P→0 2 4

= lim xNO 0 P→∞ 2

Increasing the pressure reduces the total number of moles, in accord with le Chatlier’s principle.

29 Temperature dependence of KP

∆Go ln K = − r,m P RT

d(∆G o / T ) d ln K P = − 1 r,m dT R dT

Gibbs-Helmholtz equation:

∆H o d ln K P = 1 r,m dT R T 2

∆H o dT d ln K = r,m P R T 2

Van’t Hoff equation:

  ∆ o   K (T ) H r,m 1 1 ln P 2  = −  −   K P (T1 )  R  T2 T1 

30 ∆ o < Suppose that the reaction is exothermic, so that H r 0. For T2 >T1,

 K (T )  ln P 2  < 0  K P (T1 ) 

In other words, increasing T shifts the reaction to the left. Otherwise, the reaction would “run away,” in accord with le Chatlier.

31 Example of H2 +Cl2 F 2HCl

∆ o = − H r,m 184.62 kJ/mol

At 5000 K (assuming that DHisconstant),

184,620  1 1  ln K (5000K) = 76.93 +  −  = 6.86 P 8.31451 5000 298 

KP = 952

Nowifwestartoffwith1barofpureHCl,the equilibrium mixture contains 0.032 bar of H2 and 0.032 bar of Cl2.

32 How do real gases and condensed phases behave?

µ = µ o + i i RT ln ai

o = lim f i Pi Gases:ai =fi/P , P→0

lim a = x Solvent: a = g x , → A A A A A x A 1

lim a = x Solute:a =P /K = g x , → B B B B B B B xB 0

= γ ′ We can also write aB BbB

Note: n x = B B + nA nB

= nB bB nAM A

The activity of a pure solid or liquid is 1. Why? Because its activity doesn’t change, and no “correction term” is needed.

33 Example: Liquid-vapor equilibrium:

CCl4(liq) F CCl4(vapor) at 298 K

a P / P o K = vapor = vapor = P vapor in bar aliq 1

o DG vap,m(298) = 4.46 kJ/mol

∆Go 4460 ln K = − vap,m = − = −1.80 RT 8.314x298

K = 0.165 ﬂ vapor pressure = 124 Torr

Note: From the van’t Hoff equation,

∆G 0 (T )   vap = ∆ 0  1 − 1  H vap   RT  T Tb 

This is the same result that we get from the Clapeyron- Clausius equation.

34 Lecture 40. Equilibrium of Elecytrolytes

Solubility Product

For a saturated solution,

AgCl(s) Ø Ag+(aq) + Cl-(aq)

γ +b + γ −b − K = Ag Cl SP (bo )2

-10 b + = b − Given that K=1.8x10 and Ag Cl , and assuming that g+g- =1,

b + Ag = K bo SP −5 + = bAg 1.35x10 mol/Kg

35 Debye-Híckel limiting law:

γ = − log`10 ± | z+ z− | A I

I = ionic strength of the solution (dimensionless) A = 0.50926 for water at 25 C.

= 1 2 o I ∑ zi (bi / b ) 2 i zi = charge number of ion i

-5 Here, z+ =1,z- = -1, I = b @ 1.35x10 log10 g≤ = -0.00187 g≤ = 0.9957

−5 + = bAg 1.35x10 mol/Kg

36 Example: Determine the solubility of Ni3(PO4)2,given -32 that KSP = 4.74x10 . Assume first that g≤ =1.

One mole of electrolyte produces 3 moles of Ni+ and 2+ two moles of PO4 . Let b = the molality of the Ni3(PO4)2 that gets dissolved. The activity of the 2+ dissolved Ni is 3g+b, and the activity of the 2- dissolved PO4 is 3g-b.

3 2 5   b    b   b  K = 3γ +   2γ −   =108  SP   bo    bo   bo 

b = 2.13x10-7 mol/kg

Now calculate the ionic strength. 2 2 -6 z+ =2,z- = 3I = 0.5{3b(2) + 2b(3) } = 3.196x10

logγ ± = −0.50926(2)(3) I = −0.005463

γ 5 = γ 3γ 2 ± + − g≤ =0.9875  5 5  b  K = 108γ ±   SP  bo 

b = 2.16x10-7 mol/kg

37 Ionization of water

Water: + - 2H2O F H3O +OH

+ − a(H O )a(OH ) + − + = 3 = = 2 Kw 2 [H3O ][OH ] [H3O ] [a(H 2O)]

Note: pH meters are calibrated for activity.

+ = [H3O ] KW

0 -13.997 -14 KW(25 C) = 10 @ 10

+ -7 [H3O ]=10 M

+ pH = -log10[H3O ]=7

0 What is Kw at 0 C?

We need DG(00C) for acid dissociation. Must correct for DCP.

 ∆  ∆  G  = − H d 2 dT  T  T

38 ∆ = ∆ + ∆ ()− H (T ) H (T0 ) CP T T0

     ∆ = T ∆ + ∆ − T + ∆ − −  T  G(T ) G(T0 ) H (T0 )1  CP T T0 T ln  T0  T0    T0 

DG(250C) = 79.868 kj/mol

DH(250C) = 56.563 kj/mol

0 DCP(25 C) = 19.7 j/mol/deg

DG(00C) = 78.127 kj/mol

-15 Kw = 1.198x10

pH = 7.47

39 Lecture 42. Acid-Base Equilibria

Strong acid: + - HA + H2O Ø H3O +A

+ -7 [H3O ]=[HA]0 >> 10

pH = - log[HA]0

Strong base: - + [OH ]>>[H3O ]

+ - [H3O ]=KW/[OH ]

pH = pKW +log[B]

Weak acid: + - HA + H2O F H3O +A

+ − + 2 = [H3O ][A ] ≈ [H3O ] Ka [HA] [HA]0

+ = >> −7 [H3O ] Ka[HA]0 10

pH = ½ pKa - ½ logA0

40 Weak base:

- + - A +H2O Ø H3O +OH

[HA][OH − ] [OH − ]2 K = ≈ b [A− ] [A− ]

KaKb =Kw

− = = [OH ] KbB0 B0Kw / Ka

+ = − = [H3O ] Kw /[OH ] KwKa / B0

pH = ½ pKw +½pKa - log B0

41 Problem 9.16: Acid rain. Calculate the pH of rainwater caused by dissolved CO2,given:

P = 3.6x10−4 atm CO2 6 KH =1.25x10 Torr pKa =6.37

The reaction involved is

- + H2CO3(aq) + H2O(liq) F H2CO3 (aq) + H3O (aq)

− + + 2 K = [H 2CO3 ][H3O ] ≈ [H3O ] a [H 2CO3 ] [H 2CO3 ]

pKa = 2pH + log[H2CO3]

Assume that [H2CO3]=[CO2] in the liquid phase.

n n [CO ] [CO ] x = CO2 ≈ CO2 = 2 = 2 CO2 n + n n [H O] 1000/18.015 CO2 H 2O H 2O 2

[CO ] = x / 0.018015 2 CO2

42 x We calculate CO2 using Henry’s law.

P = K x CO2 H CO2

−4 P 3.6x10 x760 − x = CO2 = = 2.2x10 7 CO2 6 K H 1.25x10

−7 2.2x10 − [CO ] = =1.2x10 5 M 2 0.018015

43 Exact solution: 4 equations with 4 unknowns

+ - - x=[H3O ], y = [OH ], z = [A ], A = [AH]

Kw =xy

Ka =xz/A

x=y+z

A=A0 - z

Solution: y=x- z

Kw =x(x-z) ﬂ z=x- Kw/x

x 2 − K x 3 − K x K = w = w a K A x − x 2 + K K A − x − w 0 a w 0 x

3 − = − 2 + x K w x K a A0 x K a x K a K w

pH= ½ pKa - ½log(1.2x10-5)=5.645

44 Titration of a Weak Acid with a Strong Base

Chemical equations:

BOH Ø B+ +OH-

- - OH +HAØ H2O+A

+ - HA + H2O F H3O +A

- - A +H2O F HA + OH

+ - + - Unknowns: HA, H3O ,A ,B,OH

Algebraic equations: [H O + ][A− ] K = 3 a [HA]

[HA][OH − ] K = b [A− ]

(note: KaKb =Kw)

+ + - - H3O +B =A +OH

- [HA] = [HA]0 +[A ]

45 + [B ] = [BOH]0 =S

Initial point: pH = ½ pKa - ½logA0

Before the equivalence point:

- - OH +HAØ H2O+A

+ - HA + H2O F H3O +A

[A-]=S

[HA] = A0 - S

(S is the concentration of salt produced by + - BOH + HA Ø H2O+BA )

+ − [H O ][A ] + S K = 3 = [H O ] a 3 − [HA] A0 S

pH = pKa - log((A0 - S)/S)

46 At the equivalence point:

HA has been all converted, and the relevant chemical - - equation is A +H2O F HA + OH

[HA][OH − ] [OH − ]2 K = = b [A− ] S

− = = 1/ 2 [OH ] SKb [S(K w / K a )]

pOH = ½ pKw - ½pKa - ½logS=pKw - pH

pH = ½ pKa +½pKw +½logS

(Exercise: Work out the corresponding equation for titration of a weak base by a strong acid.)

End point:

+ - [H3O ]=Kw/[OH ]=Kw/(B0-A0)

pH = pKw +log(B0-A0)

47 Lecture 43. Standard States

The choice of standard state is arbitrary. It affects the value of K, but not what you actually observe in the lab.

Gas: 1 bar Liquid or solid: the pure material Solute, including H+:1M H+ in biochemical reactions: 10-7 M(i.e.,pH7)

Suppose you lived on a planet where the atmospheric pressure is l bar. You my choose to use as your standard a pressure of l bar. How does this affect m and KP?  λPo  µ′o = µ o + RT ln  = µ o + RT lnλ  Po 

48 For the conditions of your experiment,  P  µ = µ o + RT ln   Po   P λPo  = µ′o − RT ln λ + RT ln   λPo Po   P  = µ′o − RT ln λ + RT ln  + RT ln λ  λPo   P  = µ′o + RT ln   P′o 

For the reaction aA + bB F cC + dD ,

∆ ′o = ∆ o + + − − λ = ∆ o + ∆ν λ Gr,m Gr,m (c d a b)RT ln Gr,m RT ln

 P c  P d  C   D  λ o λ o ′ =  P   P  = λ−∆ν KP K P  P a  P b  A   B   λPo   λPo  A reaction that is at equilibrium at Po will not be at equilibrium at P′o. \ This idea applies also to ionic equilibria. Normally we choose for the standard state co =1M. But for biochemical systems, we choose for H+

c′o = 10−7 M

49 + + +  []H  µ(H ) = µ o (H ) + RT ln a = 0 + RT ln  = −2.302RT ⋅ pH  co 

 []+  µ′ + =  H  = µ + − (H ) RT ln −  (H ) 2.302RT 10 7 co 

For a reaction A + nH+ Ø P

∆ ′o = ∆ o + ν ⋅ Gr,m Gr,m 7 2.303RT

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3.091 – Introduction to Solid State Chemistry

Lecture Notes No. 10

PHASE EQUILIBRIA AND PHASE DIAGRAMS

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Sources for Further Reading: 1. Campbell, J.A., Why Do Chemical Reactions Occur?, Prentice-Hall, Englewood Cliffs, NJ, 1965. (Paperback) 2. Barrow, G.M., Physical Chemistry, McGraw-Hill, New York, 1973. 3. Hägg, G., General and Inorganic Chemistry, Wiley, 1969. 4. Henish, H., Roy, R., and Cross, L.E., Phase Transitions, Pergamon, 1973. 5. Reisman, A., Phase Equilibria, Academic Press, 1970. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

PART A: PHASE EQUILIBRIA AND PHASE DIAGRAMS

Phase diagrams are one of the most important sources of information concerning the behavior of elements, compounds and solutions. They provide us with the knowledge of phase composition and phase stability as a function of temperature (T), pressure (P) and composition (C). Furthermore, they permit us to study and control important processes such as phase separation, solidification, sintering, purification, growth and doping of single crystals for technological and other applications. Although phase diagrams provide information about systems at equilibrium, they can also assist in predicting phase relations, compositional changes and structures in systems not at equilibrium.

1. GASES, LIQUIDS AND SOLIDS

Any material (elemental or compound) can exist as a gas, a liquid or a solid, depending on the relative magnitude of the attractive interatomic or intermolecular forces vs the disruptive thermal forces. It is thus clear that the stability (existence) of the different

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states of aggregation, which are referred to as phases, is a function of temperature and pressure (since with increased pressure the atoms, for exampled of a gas phase, are closer spaced and thus subject to increased interatomic attraction).

In general terms, a “phase” is a homogeneous, physically distinct, mechanically separable portion of a material with a given chemical composition. To illustrate this definition, let us look at a few examples of common multi-phase systems. Ice cubes in water constitute a two-phase system (ice and liquid water), unless we include the vapor above the glass in our system, which would make it a three-phase system. A mixture of oil and water would also be a two-phase system. Just as oil and water represent two distinct liquid phases, two regions of a solid with distinctly different composition or structure separated by boundaries represent two solid phases.

If we look at a one-component system, such as liquid water, we recognize that because of the energy distribution of the water molecules, some water molecules will always

possess sufficient energy to overcome the attractive forces on the surface of H2O and enter into the gas phase. If thermal energy is continuously supplied to a liquid in an open container, the supply of high energy molecules (which leave the liquid phase) is replenished and the temperature remains constant – otherwise the loss of high energy molecules will lower the temperature of the system. The total quantity of heat necessary to completely “vaporize” one mole of a liquid at its boiling point is called its

molar heat of vaporization, designated by ΔHV. Similarly, the heat required to completely melt one mole of a solid (the heat required to break the bonds established in the solid phase) is called the (latent) heat of fusion (ΔHV).

Visualize a liquid in a sealed container with some space above the liquid surface. Again, some of the most energetic liquid molecules will leave the liquid phase and form a “gas phase” above the liquid. Since gas molecules will thus accumulate in the gas phase (at a constant temperature), it is inevitable that as a result of collisions in the gas

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phase some molecules will re-enter the liquid phase and a situation will be established whereby the rate of evaporation will equal the rate of condensation – i.e., a dynamic equilibrium between the liquid and gas phase will exist. The established pressure in the gas phase is referred to as the equilibrium vapor pressure, which is normally significantly less for solids than for liquids.

For obvious reasons it is desirable to know for any given material the conditions (P, T) under which the solid state, the liquid state and the gaseous state are stable, as well as the conditions under which the solid and liquid phases may coexist. These conditions

are graphically presented in equilibrium phase diagrams, which can be experimentally determined.

2. THE ONE-COMPONENT PHASE DIAGRAM

Figure 1 illustrates the temperatures and pressures at which water can exist as a solid, liquid or vapor. The curves represent the points at which two of the phases coexist in equilibrium. At the point Tt vapor, liquid and solid coexist in equilibrium. In the fields of the diagram (phase fields) only one phase exists. Although a diagram of this kind delineates the boundaries of the phase fields, it does not indicate the quantity of any phase present.

It is of interest to consider the slope of the liquid/solid phase line of the H2O phase diagram. It can readily be seen that if ice – say at –2°C – is subjected to high

pressures, it will transform to liquid H2O. (An ice skater will skate not on ice, but on water.) This particular pressure sensitivity (reflected in the slope of the solid/liquid phase line) is characteristic for materials which have a higher coordination number in

the liquid than in the solid phase (H2O, Bi, Si, Ge). Metals, for example have an opposite slope of the solid/liquid phase line, and the liquid phase will condense under pressure to a solid phase.

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Phase Diagram of Water

Supercritical fluid 217.7 Critical point ) m t a (

r Normal melting point u s s Solid Liquid e r

P 1 Normal boiling point

-3 6.0 x 10 Gas Triple point (Tt)

0 0.0098 100 374.4 Temperature (oC)

Image by MIT OpenCourseWare. Fig. 1 Pressure/Temperature Diagram for Water. (Not drawn to scale.)

3. PHASE RULE AND EQUILIBRIUM

The phase rule, also known as the Gibbs phase rule, relates the number of components and the number of degrees of freedom in a system at equilibrium by the formula F = C – P + 2 [1] where F equals the number of degrees of freedom or the number of independent variables, C equals the number of components in a system in equilibrium and P equals the number of phases. The digit 2 stands for the two variables, temperature and pressure.

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The number of degrees of freedom of a system is the number of variables that may be changed independently without causing the appearance of a new phase or disappearance of an existing phase. The number of chemical constituents that must be specified in order to describe the composition of each phase present. For example, in the reaction involving the decomposition of calcium carbonate on heating, there are three phases – two solid phases and one gaseous phase.

CaCO3 (s) � CaO (s) + CO2 (g) [2]

There are also three different chemical constituents, but the number of components is only two because any two constituents completely define the system in equilibrium. Any third constituent may be determined if the concentration of the other two is known. Substituting into the phase rule (eq. [1]) we can see that the system is univariant, since F = C – P + 2 = 2 – 3 + 2 = 1. Therefore only one variable, either temperature or pressure, can be changed independently. (The number of components is not always easy to determine at first glance, and it may require careful examination of the physical conditions of the system at equilibrium.)

The phase rule applies to dynamic and reversible processes where a system is heterogeneous and in equilibrium and where the only external variables are temperature, pressure and concentration. For one-component systems the maximum number of variables to be considered is two – pressure and temperature. Such systems can easily be represented graphically by ordinary rectangular coordinates. For two-component (or binary) systems the maximum number of variables is three – pressure, temperature and concentration. Only one concentration is required to define the composition since the second component is found by subtracting from unity. A graphical representation of such a system requires a three-dimensional diagram. This, however, is not well suited to illustration and consequently separate two-coordinate diagrams, such as pressure vs temperature, pressure vs composition and temperature

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vs composition, are mostly used. Solid/liquid systems are usually investigated at constant pressure, and thus only two variables need to be considered – the vapor pressure for such systems can be neglected. This is called a condensed system and finds considerable application in studying phase equilibria in various engineering materials. A condensed system will be represented by the following modified phase rule equation: F = C – P + 1 [3] where all symbols are the same as before, but (because of a constant pressure) the digit 2 is replaced by the digit 1, which stands for temperature as variable. The graphical representation of a solid/liquid binary system can be simplified by representing it on ordinary rectangular coordinates: temperature vs concentration or composition.

4. H vs T PHASE DIAGRAM

With the aid of a suitable calorimeter and energy reservoir, it is possible to measure the heat required to melt and evaporate a pure substance like ice. The experimental data obtainable for a mole of ice is shown schematically in fig. 2. As heat is added to the

solid, the temperature rises along line “a” until the temperature of fusion (Tf) is reached. The amount of heat absorbed per mole during melting is represented by the length of

line “b”, or ΔHF. The amount of heat absorbed per mole during evaporation at the boiling point is represented by line “d”. The reciprocal of the slope of line “a”, (dH/dT), is the heat required to change the temperature of one mole of substance (at constant

pressure) by 1°CF. (dH/dT) is the molar heat capacity of a material, referred to as “Cp”.

As the reciprocal of line “a” is Cp (solid), the reciprocals of lines “c” and “e” are

Cp (liquid) and Cp (vapor) respectively.

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Fig. 2 H vs T Diagram for Pure H2O. (Not to scale.)

From a thermodynamic standpoint, it is important to realize that fig. 2 illustrates the energy changes that occur in the system during heating. Actual quantitative measurements show that 5.98 kJ of heat are absorbed at the melting point (latent heat of fusion) and 40.5 kJ per mole (latent heat of evaporation) at the boiling point. The latent heats of fusion and evaporation are unique characteristics of all pure substances.

Substances like Fe, Co, Ti and others, which are allotropic (exhibit different structures at different temperatures), also exhibit latent heats of transformation as they change from one solid state crystal modification to another.

5. ENERGY CHANGES

When heat is added from the surroundings to a material system (as described above), the energy of the system changes. Likewise, if work is done on the surroundings by the material system, its energy changes. The difference in energy (ΔE) that the system

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experiences must be the difference between the heat absorbed (Q) by the system and the work (W) done on the surroundings. The energy change may therefore be written as: ΔE = Q – W [4]

If heat is liberated by the system, the sign of Q is negative and work done is positive. Q and W depend on the direction of change, but ΔE does not. The above relation is one way of representing the First Law of Thermodynamics which states that the energy of a system and its surroundings is always conserved while a change in energy of the system takes place. The energy change, ΔE, for a process is independent of the path taken in going from the initial to the final state.

In the laboratory most reactions and phase changes are studied at constant pressure. The work is then done solely by the pressure (P), acting through the volume change, ΔV. W = PΔV and ΔP = 0 [5] Hence: Q = ΔE + PΔV [6] Since the heat content of a system, or the enthalpy H, is defined by: H = E + PV [7] ΔH = ΔE + PΔV [8] so that: ΔH = Q – W + PΔV [9] or ΔH = Q [10]

Reactions in which ΔH is negative are called exothermic since they liberate heat, whereas endothermic reactions absorb heat. Fusion is an endothermic process, but the reverse reaction, crystallization, is an exothermic one.

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6. ENTROPY AND FREE ENERGY

When a gas condenses to form a liquid and a liquid freezes to form a crystalline solid, the degree of internal order increases. Likewise, atomic vibrations decrease to zero when a perfect crystal is cooled to 0°K. Since the term entropy, designated by S, is considered a measure of the degree of disorder of a system, a perfect crystal at 0°K has zero entropy.

The product of the absolute temperature, T, and the change in entropy, ΔS, is called the entropy factor, TΔS. This product has the same units (Joules/mole) as the change in enthalpy, ΔH, of a system. At constant pressure, P, the two energy changes are related to one another by the Gibbs free energy relation: ΔF = ΔH – TΔS[ 11] where F = H – TS [12]

The natural tendency exhibited by all materials systems is to change from one of higher to one of lower free energy. Materials systems also tend to assume a state of greater disorder whereby the entropy factor TΔS is increased. The free energy change, ΔF, expresses the balance between the two opposing tendencies, the change in heat content (ΔH) and the change in the entropy factor (TΔS).

If a system at constant pressure is in an equilibrium state, such as ice and water at 0°C, for example, at atmospheric pressure it cannot reach a lower energy state. At equilibrium in the ice-water system, the opposing tendencies, ΔH and TΔS, equal one another so that ΔF = 0. At the fusion temperature, TF: �H � � F SF [13] TF Similarly, at the boiling point:

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�H � � V SV [14] TV Thus melting or evaporation only proceed if energy is supplied to the system from the surroundings.

The entropy of a pure substance at constant pressure increases with temperature according to the expression:

C �T �S � p (since : �H � C �T) [15] T p where Cp is the heat capacity at constant pressure, ΔCp, ΔH, T and ΔT are all measurable quantities from which ΔS and ΔF can be calculated.

7. F vs T

Any system can change spontaneously if the accompanying free energy change is negative. This may be shown graphically by making use of F vs T curves such as those shown in fig. 3.

Fig. 3 Free energy is a function of temperature for ice and water.

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The general decrease in free energy of all the phases with increasing temperature is the result of the increasing dominance of the temperature-entropy term. The increasingly negative slope for phases which are stable at increasingly higher temperatures is the result of the greater entropy of these phases.

PART B: PHASE DIAGRAMS (TWO-COMPONENT SYSTEMS)

1. SOLID SOLUTIONS

A solution can be defined as a homogeneous mixture in which the atoms or molecules of one substance are dispersed at random into another substance. If this definition is applied to solids, we have a solid solution. The term “solid solution” is used just as “liquid solution” is used because the solute and solvent atoms (applying the term solvent to the element in excess) are arranged at random. The properties and composition of a solid solution are, however, uniform as long as it is not examined at the atomic or molecular level.

Solid solutions in alloy systems may be of two kinds: substitutional and interstitial. A substitutional solid solution results when the solute atoms take up the positions of the solvent metal in the crystal lattice. Solid solubility is governed by the comparative size of the atoms of the two elements, their structure and the difference in electronegativity.

If the atomic radii of a solvent and solute differ by more than 15% of the radius of the solvent, the range of solubility is very small. When the atomic radii of two elements are equal or differ by less than 15% in size and when they have the same number of valency electrons, substitution of one kind of atom for another may occur with no distortion or negligible distortion of the crystal lattice, resulting in a series of homogeneous solid solutions. For an unlimited solubility in the solid state, the radii of the two elements must not differ by more than 8% and both the solute and the solvent elements must have the same crystal structure.

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In addition to the atomic size factor, the solid solution is also greatly affected by the electronegativity of elements and by the relative valency factor. The greater the difference between electronegativities, the greater is the tendency to form compounds and the smaller is the solid solubility. Regarding valency effect, a metal of lower valency is more likely to dissolve a metal of higher valency. Solubility usually increases with increasing temperature and decreases with decreasing temperature. This causes precipitation within a homogeneous solid solution phase, resulting in hardening effect of an alloy. When ionic solids are considered, the valency of ions is a very important factor.

2. CONSTRUCTION OF EQUILIBRIUM PHASE DIAGRAMS OF TWO-COMPONENT SYSTEMS

To construct an equilibrium phase diagram of a binary system, it is a necessary and sufficient condition that the boundaries of one-phase regions be known. In other words, the equilibrium diagram is a plot of solubility relations between components of the system. It shows the number and composition of phases present in any system under equilibrium conditions at any given temperature. Construction of the diagram is often based on solubility limits determined by thermal analysis – i.e., using cooling curves. Changes in volume, electrical conductivity, crystal structure and dimensions can also be used in constructing phase diagrams.

The solubility of two-component (or binary) systems can range from essential insolubility to complete solubility in both liquid and solid states, as mentioned above. Water and oil, for example, are substantially insoluble in each other while water and

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alcohol are completely intersoluble. Let us visualize an experiment on the water-ether system in which a series of mixtures of water and ether in various proportions is placed in test tubes. After shaking the test tubes vigorously and allowing the mixtures to settle, we find present in them only one phase of a few percent of ether in water or water in ether, whereas for fairly large percentages of either one in the other there are two phases. These two phases separate into layers, the upper layer being ether saturated with water and the lower layers being water saturated with ether. After sufficiently increasing the temperature, we find, regardless of the proportions of ether and water, that the two phases become one. If we plot solubility limit with temperature as ordinate and composition as abscissa, we have an isobaric [constant pressure (atmospheric in this case)] phase diagram, as shown in fig. 4. This system exhibits a solubility gap.

One phase Temperature Two phases

0 20 40 60 80 100 Composition percent, water 100 80 60 40 20 0 Composition percent, ether

Image by MIT OpenCourseWare.

Fig. 4 Schematic representation of the solubilities of ether and water in each other.

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3. COOLING CURVES

Fig. 5 Cooling curves: (a) pure compound; (b) binary solid solution; (c) binary eutectic system.

Source: Jastrzebski, Z. The Nature and Properties of Engineering Materials. 2nd edition. New York, NY: John Wiley & Sons, 1976. Courtesy of John Wiley & Sons. Used with permission.

4. SOLID SOLUTION EQUILIBRIUM DIAGRAMS

Read in Jastrzebski: First two paragraphs and Figure 3-4 in Chapter 3-8, "Solid Solutions Equilibrium Diagrams," pp. 91-92.

Read in Smith, C. O. The Science of Engineering Materials. 3rd ed. Englewood Cliffs, NJ: Prentice-Hall, 1986. ISBN: 9780137948840. Last two paragraphs and Figure 7-8 in Chapter 7-3-1, "Construction of a Simple Equilibrium Diagram," pp. 247-248.

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Fig. 6 Plotting equilibrium diagrams from cooling curves for Cu-Ni solid solution alloys. (a) Cooling curves; (b) equilibrium diagram. Source: Jastrzebski, Z. The Nature and Properties of Engineering Materials. 2nd edition. New York, NY: John Wiley & Sons, 1976. Courtesy of John Wiley & Sons. Used with permission.

Fig. 7 15 LN–10

5. INTERPRETATION OF PHASE DIAGRAMS

From the above discussion we can draw two useful conclusions which are the only rules necessary for interpreting equilibrium diagrams of binary systems.

Rule 1 - Phase composition: To determine the composition of phases which are stable at a given temperature we draw a horizontal line at the given temperature. The projections (upon the abscissa) of the intersections of the isothermal line with the liquidus and the solidus give the compositions of the liquid and solid, respectively, which coexist in equilibrium at that temperature. For example, draw a horizontal temperature

line through temperature Te in fig. 7. The Te line intersects the solidus at f and the liquidus at g, indicating solid composition of f% of B and (100–f)% of A. The liquid composition at this temperature is g% of B and (100–g)% of A. This line in a two-phase region is known as a tie line because it connects or “ties” together lines of one-fold saturation – i.e., the solid is saturated with respect to B and the liquid is saturated with respect to A.

Rule 2 - The Lever Rule: To determine the relative amounts of the two phases, erect an ordinate at a point on the composition scale which gives the total or overall composition of the alloy. The intersection of this composition vertical and a given isothermal line is the fulcrum of a simple lever system. The relative lengths of the lever arms multiplied by the amounts of the phase present must balance. As an illustration, consider alloy � in fig. 7. The composition vertical is erected at alloy � with a composition of e% of B and (100–e)% of A. This composition vertical intersects the

temperature horizontal (Te) at point e. The length of the line “f–e–g” indicates the total amount of the two phases present. The length of line “e–g” indicates the amount of

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solid. In other words: eg x 100 � % of solid present fg fg x 100 � % of liquid present fg

These two rules give both the composition and the relative quantity of each phase present in a two-phase region in any binary system in equilibrium regardless of physical form of the two phases. The two rules apply only to two-phase regions.

6. ISOMORPHOUS SYSTEMS

An isomorphous system is one in which there is complete intersolubility between the two components in the vapor, liquid and solid phases, as shown in fig. 9. The Cu-Ni system is both a classical and a practical example since the monels, which enjoy extensive commercial use, are Cu-Ni alloys. Many practical materials systems are isomorphous.

Fig. 9 Schematic phase diagram for a binary system, A-B, showing complete intersolubility (isomorphism) in all phases.

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7. INCOMPLETE SOLUBILITY

Read in Smith, C. O. The Science of Engineering Materials. 3rd ed. Englewood Cliffs, NJ: Prentice-Hall, 1986. ISBN: 9780137948840. Chapter 7-3-5, "Incomplete Solubility," pp. 252-253.

8. EUTECTIC SYSTEMS

Read in Smith: Chapter 7-3-6, "Eutectic," pp. 253-256.

9. EQUILIBRIUM DIAGRAMS WITH INTERMEDIATE COMPOUNDS

Read in Jastrzebski, Z. D. The Nature and Properties of Engineering Materials. 2nd ed. New York, NY: John Wiley & Sons, 1976. ISBN: 9780471440895.

Chapter 3-11,"Equilibrium Diagrams with Intermediate Compounds," pp. 102-103.

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Fig. 15 Binary system showing an intermediate compound. C is the melting point (maximum) of the compound AB having the composition C’. E is the eutectic of solid A and solid AB. E’ is the eutectic of solid AB and solid B.

Source: Jastrzebski, Z. The Nature and Properties of Engineering Materials. 2nd edition. New York, NY: John Wiley & Sons, 1976. Courtesy of John Wiley & Sons. Used with permission.

19 MIT OpenCourseWare http://ocw.mit.edu

3.091SC Introduction to Solid State Chemistry Fall 2009

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. TERNARY PHASE DIAGRAMS Credit forscanning thephasediagrams: Credit forPhase DiagramDrawings: San JoseStateUniversity Guna Selvaduray An Introduction Richard Brindos Richard rne Croom Brenden

G. Selvaduray - SJSU - Oct 2004 Utility ofTernary PhaseDiagrams ¾ ¾ ¾ ¾ ¾ ¾ Solder metallurgy Stainless steels Aluminum alloys Refractories Glass compositions Several other applications

G. Selvaduray - SJSU - Oct 2004 .Rimn PhaseEquilibria A. Reisman, .N hns PhaseDiagrams inMetallurgy F. N.Rhines, G. Masing,TernarySystems A. Prince,AlloyPhaseEquilibria C. G. Bergeron andS.H.Ri C. G.Bergeron .F eadandD.R.Wilder,Fundamentals ofPhaseEquilibria M. F.Berard D. R.F.West,TernaryEquilibrium Diagrams York, 1956 New York,1982 in CeramicSystems inCeramics Equilibria New York,1944 1984 New York,1966 References on Ternary Phase , R.A.N.Publishers, Ohio,1990 Diagrams , TheAmerican CeramicSociety,Ohio, sbud, IntroductiontoPhase , ReinholdPublishingCompany, , Academic Press,1970 , ElsevierPublishingCompany, , McGraw-Hill, New , Chapman and Hall, , ChapmanandHall,

G. Selvaduray - SJSU - Oct 2004 What areTernary PhaseDiagrams? Normally, pressure isnotaviable variablein Diagrams thatrepresent theequilibrium therefore held constantat1atm. ternary phase diagramconstruction, andis temperature. between threecomponents, asafunctionof between thevarious phases thatareformed

G. Selvaduray - SJSU - Oct 2004 Components are “independent components” P =4when C=3andF0 For C=3, the maximumnumber ofphaseswill P F =C+1- For isobaricsystems: P F =C+2- co-exist when F=0 The GibbsPhaseRulefor 3-Component Systems

G. Selvaduray - SJSU - Oct 2004 ¾ ¾ ¾ ¾ ¾ Solidification sequence Amount ofeachphase Chemical composition ofindividualphases Number ofphases Overall composition Some Important Terms

G. Selvaduray - SJSU - Oct 2004 The Gibbs Triangle:Anequilateral triangleon The Gibbs Triangleisalwaysused todetermine Sum oftheconcentration ofthethree or Can beexpressedas either“wt.%” The concentrationof eachofthethree by each corner which thepure components arerepresented the overall composition components mustaddupto100% components “molar %” vrl opsto 1 Overall Composition -

G. Selvaduray - SJSU - Oct 2004 G. Selvaduray - SJSU - Oct 2004 The concentrations ofAand C,inX, canbe intersectsthe side ABtells Where the line A’C’ Draw lines passingthroughX, and parallelto Let theoverallcomposition berepresentedby Refer toFiguresOC1 andOC2 Method 1 There arethreeways ofdeterminingtheoverall determined inan identical manner us theconcentration ofcomponent BinX each ofthe sides the pointX composition vrl opsto 2 Overall Composition -

G. Selvaduray - SJSU - Oct 2004 G. Selvaduray - SJSU - Oct 2004 This method can besomewhat confusing, and Concentration ofA=B’B Concentration ofC=A’B’ Concentration ofB=AA’ intersectsABatB’ B’C” intersectsABatA’ A’C’ Draw linesthroughX, paralleltothesidesof Method Two is not recommended Gibbs Triangle vrl opsto 3 Overall Composition - :

G. Selvaduray - SJSU - Oct 2004 G. Selvaduray - SJSU - Oct 2004 %A =AX Draw straightlinesfrom eachcorner,throughX Application oftheInverse LeverRule Method 3 Always determine theconcentration ofthe Important Note: adding themup toobtain100% components independently, thencheck by MB CL BN AM vrl opsto 4 Overall Composition - %B =BX %C =CX

G. Selvaduray - SJSU - Oct 2004 G. Selvaduray - SJSU - Oct 2004 h iudsSurface The Liquidus System Isomorphous h oiu Surface The Solidus solid solubility made upofthreebinaries thatexhibittotal components. Theternary systemistherefore components aretotally solubleintheother case) thathasonlyone solidphase.All composition solid phase forms forany givenoverall temperatures below which a(homogeneous) liquid forms foranygivenoverall composition temperatures abovewhichahomogeneous enr smrhu System Ternary Isomorphous : Aplotof the : Aplotof the : Asystem(ternaryin this

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System iudsSurface Liquidus

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System oiu Surface Solidus

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System Identify temperature of interest,T Refer toFiguresBT1A, BT1BandBT1C Isothermal Section Connect points 3&4with curvature reflecting Connect points 1&2withcurvature reflecting Draw thehorizontal, intersecting theliquidus temperature through thespacediagram ataspecified ternary phasediagram obtainedbycutting h oiu surface the solidus surface the liquidus surfacesatpoints 1,2,3&4 and solidus enr smrhu System Ternary Isomorphous A“oiotl sectionofa : A“horizontal” 1 here

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System ra1234 topaergo liquid +solid Area 1-2-3-4: twophaseregion - Area C-3-4: homogeneoussolid phase Area A-B-1-2: homogeneousliquid phase The lineconnectingpoints 3&4representsthe The lineconnectingpoints 1&2representsthe surface intersection oftheisotherm withthesolidus surface intersection oftheisotherm withtheliquidus enr smrhu System Ternary Isomorphous

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System Temperature =T Isothermal Section Area 1-2-3-4: two phase region - liquid +solid liquid Area 1-2-3-4: twophaseregion - Area B-C-4-3:homogeneous solidphase Area A-1-2:homogeneous liquidphase B, butabovemelting pointofC enr smrhu System Ternary Isomorphous 2 , belowmeltingpoints ofA& continued -

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System when theoverall compositionis inatwophase ofeachphasepresent Amount (b) compositionofphases present Chemical (a) Determination of: region enr smrhu System Ternary Isomorphous

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System 4. Thechemical compositionof thesolidphase 3. Thechemicalcomposition oftheliquidphase 2. Drawtie-linepassing throughX,tointersect 1. Locateoverallcomposition usingtheGibbs the Gibbs Triangle is givenby thelocationofpoint Zwithin the Gibbs Triangle is givenby thelocationofpoint Ywithin the phaseboundaries atYandZ triangle enr smrhu System Ternary Isomorphous

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System 6. Fraction ofliquid=ZX/YZ 5. Fraction ofsolid=YX/YZ Amount ofeachphase presentisdeterminedby Tie line using theInverseLever Rule compositions enr smrhu System Ternary Isomorphous : Astraightlinejoining anytwoternary

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System

G. Selvaduray - SJSU - Oct 2004 Ternary Isomorphous System 3. Exceptfor thetwobounding tie-lines,they 2. Theymust runbetweentwo one-phase 1. Thedirectionsoftie linesvarygraduallyfrom Drawing tie-linesintwo phaseregions corners ofthecompositional triangle are notnecessarily pointedtoward the regions other, withoutcrossing eachother that ofoneboundary tielinetothatofthe enr smrhu System Ternary Isomorphous

G. Selvaduray - SJSU - Oct 2004 Ternary Eutectic System – No Solid Solubility L = The TernaryEutectic Reaction: Made upof threebinaryeutectic systems,allof A liquidphase solidifiesintothree separate which exhibitno solidsolubility solid phases α Ternary Eutectic System + βγ (No SolidSolubility)