lecture 36. the phase rule · 2013. 9. 21. · the phase rule p = number of phases c = number of...
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1
Lecture 36. The Phase Rule
P = number of phasesC = number of components (chemically independentconstituents)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:
1,1312111 ,...,,, −Cxxxx phase 1
2,1322212 ,...,,, −Cxxxx phase 2
PCPPP xxxx ,1321 ,...,,, − phase PT,P
Total number of variables = P(C-1) + 2
Constraints on the system:
m11 =m12 =m13 =…=m1,P P - 1 relationsm21 =m22 =m23 =…=m2,P P - 1 relations
mC,1 =mC,2 =mC,3 =…=mC,P P - 1 relations
2
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. ChangingT requires a change in P, and vice versa.
At the triple point, F = 0. Tt and Pt are unique.
3
Four phases cannot be in equilibrium (for a singlecomponent.)
Two Component Systems: F = 4 - P
The possible phases are the vapor, two immiscible (orpartially miscible) liquid phases, and two solid phases.(Of course, they don’t have to all exist. The liquidsmight turn out to be miscible for all compositions.)
4
Liquid-Vapor Equilibrium
Possible degrees of freedom: T, P, mole fraction of A
xA = mole fraction of A in the liquidyA = mole fraction of A in the vaporzA = 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 < Tb,B
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
5
Composition of the vapor
( ) AABAB
AAAA x
xPPP
Px
P
Py >
−+==
***
*
( ) BABAB
BBBB x
xPPP
Px
P
Py <
−+==
***
*
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 isxA = xc’. The vapor is now poorer in A, yA = yc”
Ratio of moles in the two phases is given by the leverrule:
cc
cc
n
n
vap
liq
′′′
=
6
Point d: Liquid is almost all gone, 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 theliquid when the last few drops of liquid boil?A. Point d.
)1(2274
740.5zy AA
AA
A
xx
x
−+===
xA = 0.229, xB = 0.771
P = 0.229x74 + 0.771x22= 33.9 Torr
7
Lecture 37. Construction of the Temperature-Composition Diagram
Fix the total pressure at some chosen value, P. Boilingdoes not occur at a unique temperature, but rather over arange of temperatures. (Contrast with a pure substance.)
The vapor pressure curve is determined by the liquidcomposition:
P = PA(T) + PB(T)
PA(T) is the vapor pressure of pure A at temperature T.
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Assume Raoult’s Law:
)()( ** TPxTPxP BBAA +=)()1()( ** TPxTPx BAAA −+=))()(()( *** TPTPxTP BAAB −+=
The vapor pressure pure liquid A at any temperature T isgiven by the Clapeyron-Clausius equation:
−
∆−=
0
,
0*
* 11
)(
)(ln
TTR
H
TP
TP Avap
A
A
)( 0* TPA = 1 atm, T0 = normal boiling point of A.
The same reasoning applies to substance B.
Once we have )(* TPA and )(* TPB , Raoult’s Law gives usxA as a unique function of T and P. This allows us toconstruct the liquid curve.
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Composition of the vapor:
P
TPx
P
Py AAA
A
)(*
==
−
∆−
== Ab
vap
TTRT
H
A
A
A eP
TP
x
y ,
11* )(
This result is equivalent to
AliqAAvapA xRTyRT lnln *,
*, +=+ µµ
This last result is the starting point for the boiling pointelevation formula, except that there we assume yA = 1.
10
Recipe for construction of the phase diagram:
1. Calculate Tb,A and Tb,B at the pressure of thediagram.
2. Choose a temperature Tb,A < T < Tb,B
3. Calculate )(* TPA and ).(* TPB
4. )()(
)(**
*
TPTP
TPPx
BA
BA −
−=
5. (T)/P.P*AAA xy =
11
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
present. Lever rule: cc
cc
n
n
liquid
vapor
′′′
=
Point d. Vapor is almost all gone; xA = zd’ = zA, yA = zd”
Point e. Only liquid is present.
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Distillation
Point a. Mixture starts to boil, with xA = zA, yA = zb
Points b-c. Vapor is condensed to form a liquid withxA= zb = zc
Point c. The liquid that was collected in the previousstep is boiled to form a vapor with xA=zd
Condensation of the last bit of vapor produces a liquidvery rich in either A (if Tb,A < Tb,B on the left) or B (ifTb,A > Tb,B on the right).
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Non-ideal solutions
Left: Impossible phase diagram, because at Pmax, wherethe liquid of this composition just starts to boil, there is nocorresponding point on the vapor curve. (There is no tieline.)
The vapor should always lie below the liquid in apressure-composition diagram. At an extremum theymust touch.
Center: Vapor pressure reaches a maximum because ofrepulsion between A and B. This is an azeotrope, wherethe liquid and vapor have the same composition. (Note anerror in the drawing: The vapor curve does not have acusp, but rather is tangent to the liquid curve.)
Right: Vapor pressure reaches a minimum because ofattraction between A and B. This is also an azeotrope.
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Distillation of non-ideal solutions
Left diagram: low boiling azeotrope. A solution ofcomposition za first boils at Ta. The vapor comes off withcomposition zb. It is condensed and then boils at Tc
The final vapor to come off has the azeotropiccomposition, and the remaining liquid is enriched in B (orA, depending on which side of the azeotrope the processstarted).
Right diagram: high boiling azeotrope. As before, za firstboils at Ta
The final vapor to come off is enriched in B (or A), andthe remaining liquid has the azeotropic composition.
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Equilibrium between immiscible liquids
Upper CriticalTemperature
Lower CriticalTemperature
Double CriticalTemperature
The two phase region is always described by a tie line. Itis a “no man’s land.”
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Boiling of immiscible liquids
Melting of immiscible solids
Mixture of non-reactivesolids.
Solids A and B react toform compound AB.
17
Lecture 38. Equilibrium Constants
Consider the following reaction:
H2 + Cl2 Ø 2HCl
Does it occur spontaneously? Let’s calculate DG for onemole of reaction.
0)( 20 =∆ HG f
0)( 20 =∆ ClG f
molkJHClG f /30.95)(0 −=∆
6.1900 −=∆ rG
It seems that the reaction occurs spontaneously.But what if we start with pure HCl. Shouldn’t some of itreact to form H2 and Cl2?
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To determine when the reaction occurs spontaneously, wemust take the partial pressures into account:
+=∆
oCl
Clf P
PRTClG
2
2ln0)( 2
+−=∆
oHCl
HClf P
PRTHClG ln30.95)(
−
−
+−=∆o
Cl
o
HHClr P
PRT
P
PRT
P
PRTG 22 lnlnln6.190
2
0
At equilibrium, DGr = 0. That is,
=
=∆−=2222
22
2
ln)(
ln6.190ClH
HCl
o
Cl
o
H
oHCl
or PP
PRT
P
P
P
PP
P
RTG
0ln rP GKRT ∆−=
ln KP = 76.93 at 298 K
KP = 2.57 x 1033
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So what happens if we start with pure HCl?
Suppose that initially HClP = 1 bar and 022
== ClH PP
At equilibrium, zPHCl 21−= and zPP ClH ==22
( ) 332
2
1057.221
xz
z =−
332 1057.2
1x
z≈
z=1.97 x 10-17
This reaction goes nearly to completion. But this won’tbe the case at higher T.
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Another example:
N2O4 F 2NO2
=∆ 0fG 97.89 and 51.31 kJ/mol
30.4789.9731.5120 =−=∆ xGr
ln KP = -4.730/RT = 1.909
oON
NO
o
ON
o
NO
P PP
P
P
PP
P
K42
2
42
2
2
2
148.0 =
==
What is the composition of 2 bar of this material?
Let x be the mole fraction of NO2.
( )( ) oP PPx
xPK
−=
1
2
Noting that Po = 1 bar,
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x2P2 = KPP - KPxP
P2 x2 + KPPx - KPP = 0
2
322
2
4
P
PKPKPKx PPP +±−
=
For P = 2 bar, x = 0.238. (Only the + sign is physicallymeaningful.) For P = 0.01 bar, x = 0.940
What is the composition of a very low pressure gas?
22
/41
P
KPPKPKx PPP ++−
=
( )P
PPPP
K
P
P
KPKPPKPKx
21
22//21
2
22
−=−++−≈
(Here we used the series (1+e)1/2 = 1 + e/2 -e2/8 + …)
For P = 0.01 bar bar, 966.0≈x
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Suppose you somehow arranged to start with pure N2O4.What is the composition of an equilibrium mixture?
N2O4 NO2
Initial P0 0Final P0-z 2z
zP
zKP −
=0
2)2(
For KP = 0.148 and P0 = 1 bar, z = 0.175825.0175.01
42=−=ONP
349.0175.022
== xPNO
Ptot = 1.1744297.01744.1/349.0
2==NOx
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The book defines the fraction of dissociation, a.
N2O4 NO2
Initial no.moles
N 0
Final no.moles
(1-a)n 2an
Final molefraction α
ααα
α+−=
+−−
1
1
21
1
αα
+1
2
2
2222
1
4
42
2
42
2
αα−
=== P
Px
Px
P
PK
ON
NO
ON
NOP
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?
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More complex stochiometry:
2H2S + CH4 F 4H2 + CS2
=
o
CH
o
SH
o
CS
o
H
P
P
P
P
P
P
P
P
P
K42
22
2
4
Set Po = 1 bar, so that
42
22
4
CHSH
CSHP PP
PPK =
Problem: Given that T = 700 C and P = 762 TorrInitial Final
H2S 11.02 mmol
CH4 5.48H2 0CS2 0 0.711
Find KP and DGro
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Solution:
2H2S + CH4 F 4H2 + CS2
20
2
2
4
2
4
)(42
22
42
22
P
P
xx
xx
P
P
P
P
P
P
P
P
KCHSH
CSH
o
CH
o
SH
o
CS
o
H
P =
=
Set Po = 1 bar, so that
2PKK xP =
Problem: Given that T = 700 C and P = 762 Torr = 1.012 bar
Initial Final xH2S 11.02 mmol 9.598 0.536CH4 5.48 4.769 0.266H2 0 2.844 0.159CS2 0 0.711 0.040
Find KP and DGro
Kx = 3.34x10-4
KP = 3.43 x10-4
DGr0 = -RT lnKP = 94.5 kJ/mol
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Lecture 39. Equilibrium Constants II
Derivation of KP for H2 + Cl2 F 2HCl
HClHClClClHHm nnnG µµµ 22222++=
HClHClClClHHm dndndndG µµµ 22222++=
Let x indicate the “extent” of reaction. It is a device forhandling the stochiometry of the reaction.
ξddnH −=2
ξddnCl −=2
ξddnHCl 2=
The chemical potentials are
+∆=
o
HomfH P
PRTHG 2
2ln)( 2,µ
+∆= o
ClomfCl P
PRTClG 2
2ln)( 2,µ
+∆=o
HClomfHCl P
PRTHClG ln)(,µ
27
Putting all the pieces together:
ξdP
PRTHClG
P
PRTClG
P
PRTHGdG
oHClo
mf
o
Clomfo
Homfm
}ln2)(2
ln)(ln)({
,
2,2,22
+∆+
−∆−
−∆−=
+∆=
22
2
lnClH
HClor
m
PP
PRTG
d
dG
ξ
At equilibrium,
0=ξd
dGm
orP GKRT ∆−=ln
=
22
2
ClH
HClP PP
PK
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Note that KP is independent of pressure (for ideal gases).This does not mean that the equilibrium ratios of molefractions or concentrations are independent of P.
Let’s use 2H2S + CH4 as an example.
PxP CSCS 22= etc.
2
22
44
42
22
=
= ox
oCHoSH
oCSoH
P P
PK
P
Px
P
Px
P
Px
P
Px
K
In general,ν∆
=oxP P
PKK
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For example, for N2O4F 2N2O
=oxP P
PKK
P
e
P
K
x
xK
RTGP
ON
ONx
omr /2 ,
42
2
∆−
−===
0lim420
=→ ON
Px
0lim2
=∞→ NO
Px
Increasing the pressure reduces the total number of moles,in accord with le Chatlier’s principle.
30
Temperature dependence of KP
RT
GK
omr
P,ln
∆−=
dT
TGd
RdT
Kd omrP
)/(1ln ,∆−=
Gibbs-Helmholtz equation:
2
,1ln
T
H
RdT
Kd omrP
∆=
2
,lnT
dT
R
HKd
omr
P
∆=
Van’t Hoff equation:
−
∆−=
12
,
1
2 11
)(
)(ln
TTR
H
TK
TK omr
P
P
31
Suppose that the reaction is exothermic, so that .0<∆ orH
For T2 > T1,
0)(
)(ln
1
2 <
TK
TK
P
P
In other words, increasing T shifts the reaction to the left.Otherwise, the reaction would “run away,” in accord withle Chatlier.
32
Example of H2 + Cl2 F 2HCl
62.184, −=∆ omrH kJ/mol
At 5000 K (assuming that DH is constant),
86.6298
1
5000
1
31451.8
620,18493.76)5000(ln =
−+=KK P
KP = 952
Now if we start off with 1 bar of pure HCl, theequilibrium mixture contains 0.032 bar of H2 and 0.032bar of Cl2.
33
How do real gases and condensed phases behave?
ioii aRT ln+= µµ
Gases: ai = fi/Po, ii
PPf =
→0lim
Solvent: aA = gA xA, AAx
xaA
=→1
lim
Solute: aB = PB/KB = gB xB, BBx
xaB
=→0
lim
We can also write BBB ba γ′=
Note:
BA
BB nn
nx
+=
AA
BB Mn
nb =
The activity of a pure solid or liquid is 1. Why? Becauseits activity doesn’t change, and no “correction term” isneeded.
34
Example: Liquid-vapor equilibrium:
CCl4(liq) F CCl4(vapor) at 298 K
vapor
ovapor
liq
vapor PPP
a
aK ===
1
/in bar
DGovap,m(298) = 4.46 kJ/mol
80.1298314.8
4460ln , −=−=
∆−=
xRT
GK
omvap
K = 0.165 fl vapor pressure = 124 Torr
Note: From the van’t Hoff equation,
−∆=
∆
bvap
vap
TTH
RT
TG 11)(0
0
This is the same result that we get from the Clapeyron-Clausius equation.
35
Lecture 40. Equilibrium of Elecytrolytes
Solubility Product
For a saturated solution,
AgCl(s) Ø Ag+(aq) + Cl-(aq)
2)( o
ClAgSP b
bbK
−+ −+=γγ
Given that K=1.8x10-10 and −+ =ClAg
bb , and assuming thatg+g- = 1,
SPo
Ag Kb
b=
+
51035.1 −=+ xbAg mol/Kg
36
Debye-H�ckel limiting law:
IAzz ||log 10` −+± −=γ
I = ionic strength of the solution (dimensionless)A = 0.50926 for water at 25 C.
∑=i
oii bbzI )/(
2
1 2
zi = charge number of ion i
Here, z+ = 1, z-
= -1, I = b @ 1.35x10-5
log10 g≤ = -0.00187
g≤
= 0.9957
51035.1 −=+ xbAg mol/Kg
37
Example: Determine the solubility of Ni3(PO4)2, giventhat KSP = 4.74x10-32. Assume first that g
≤= 1.
One mole of electrolyte produces 3 moles of Ni+ andtwo moles of PO4
2+. Let b = the molality of theNi3(PO4)2 that gets dissolved. The activity of thedissolved Ni2+ is 3g+b, and the activity of thedissolved PO4
2- is 3g-b.
523
10823
=
= −+ oooSP b
b
b
b
b
bK γγ
b = 2.13x10-7 mol/kg
Now calculate the ionic strength.z+ = 2, z
-= 3I = 0.5{3b(2)2 + 2b(3)2} = 3.196x10-6
005463.0)3)(2(50926.0log −=−=± Iγ
235−+± = γγγ g
≤= 0.9875
5
5108
= ± oSP b
bK γ
b = 2.16x10-7 mol/kg
38
Ionization of water
Water:2H2O F H3O
+ + OH-
Note: pH meters are calibrated for activity.
WKOH =+][ 3
KW(250C) = 10-13.997@ 10-14
[H3O+] = 10-7 M
pH = -log10[H3O+] = 7
What is Kw at 00C?
We need DG(00C) for acid dissociation. Must correct forDCP.
dTT
H
T
Gd
2
∆−=
∆
2332
2
3 ][]][[)]([
)()( +−+−+
=== OHOHOHOHa
OHaOHaKw
39
( )00 )()( TTCTHTH P −∆+∆=∆
−−∆+
−∆+∆=∆
00
000
0
ln1)()()(T
TTTTC
T
TTHTG
T
TTG P
DG(250C) = 79.868 kj/mol
DH(250C) = 56.563 kj/mol
DCP(250C) = 19.7 j/mol/deg
DG(00C) = 78.127 kj/mol
Kw = 1.198x10-15
pH = 7.47
40
Lecture 42. Acid-Base Equilibria
Strong acid:HA + H2O Ø H3O
+ + A-
[H3O+] = [HA]0 >> 10-7
pH = - log[HA]0
Strong base:[OH-] >> [H3O
+]
[H3O+] = KW/[OH-]
pH = pKW + log[B]
Weak acid:HA + H2O F H3O
+ + A-
0
233
][
][
][
]][[
HA
OH
HA
AOHKa
+−+
≈=
703 10][][ −+ >>= HAKOH a
pH = ½ pKa - ½ logA0
41
Weak base:
A- + H2O Ø H3O+ + OH-
][
][
][
]][[ 2
−
−
−
−
≈=A
OH
A
OHHAKb
KaKb = Kw
awb KKBBKOH /][ 00 ==−
03 /]/[][ BKKOHKOH aww == −+
pH = ½ pKw + ½ pKa - log B0
42
Problem 9.16: Acid rain. Calculate the pH of rainwatercaused by dissolved CO2, given:
atmxPCO4106.3
2
−=KH = 1.25 x 106 Torr
pKa =6.37
The reaction involved is
H2CO3(aq) + H2O(liq) F H2CO3-(aq) + H3O
+(aq)
][][
][]][[
32
23
32
332
COHOH
COHOHCOH
aK++−
≈=
pKa = 2pH + log[H2CO3]
Assume that [H2CO3] = [CO2] in the liquid phase.
015.18/1000
][
][
][ 2
2
2
2
2
22
2
2
CO
OH
CO
n
n
nn
nx
OH
CO
OHCO
COCO ==≈
+=
018015.0/][22 COxCO =
43
We calculate 2COx using Henry’s law.
22 COHCO xKP =
76
4
102.21025.1
760106.32
2
−−
=== xx
xx
K
Px
H
COCO
Mxx
CO 57
2 102.1018015.0
102.2][ −
−
==
44
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
wa
w
w
wa KKxxA
xKx
x
KxA
KxK
+−−
=−−
−=
20
3
0
2
waaaw KKxKxAKxKx +−=− 20
3
pH= ½ pKa - ½ log(1.2x10-5) = 5.645
45
Titration of a Weak Acid with a Strong Base
Chemical equations:
BOH Ø B+ + OH-
OH- + HA Ø H2O + A-
HA + H2O F H3O+ + A-
A- + H2O F HA + OH-
Unknowns: HA, H3O+, A-, B+, OH-
Algebraic equations:
][
]][[ 3
HA
AOHK a
−+
=
][]][[
−
−
=A
OHHAKb
(note: KaKb = Kw)
H3O+ + B+ = A- + OH-
[HA] = [HA]0 + [A-]
46
[B+] = [BOH]0 = S
Initial point:pH = ½ pKa - ½ log A0
Before the equivalence point:
OH- + HA Ø H2O + A-
HA + H2O F H3O+ + A-
[A-] = S
[HA] = A0 - S
(S is the concentration of salt produced byBOH + HAØ H2O + B+A-)
SA
SOH
HA
AOHKa −
== +−+
03
3 ][][
]][[
pH = pKa - log((A0 - S)/S)
47
At the equivalence point:
HA has been all converted, and the relevant chemicalequation is A- + H2O F HA + OH-
S
OH
A
OHHAKb
2][][
]][[ −
−
−
==
2/1)]/([][ awb KKSSKOH ==−
pOH = ½ pKw - ½ pKa - ½ log S = pKw - pH
pH = ½ pKa + ½ pKw + ½ log S
(Exercise: Work out the corresponding equation fortitration of a weak base by a strong acid.)
End point:
[H3O+] = Kw/[OH-] = Kw/(B0-A0)
pH = pKw + log(B0-A0)
48
Lecture 43. Standard States
The choice of standard state is arbitrary. It affects thevalue of K, but not what you actually observe in the lab.
Gas: 1 barLiquid or solid: the pure materialSolute, including H+: 1 MH+ in biochemical reactions: 10-7 M (i.e., pH 7)
Suppose you lived on a planet where the atmosphericpressure is l bar. You my choose to use as your standarda pressure of l bar. How does this affect m and KP?
λµλµµ lnln RTP
PRT o
o
ooo +=
+=′
49
For the conditions of your experiment,
′
+′=
+
+−′=
+−′=
+=
oo
oo
o
o
oo
oo
P
PRT
RTP
PRTRT
P
P
P
PRTRT
P
PRT
ln
lnlnln
lnln
ln
µ
λλ
λµ
λλ
λµ
µµ
For the reaction aA + bB F cC + dD ,
λνλ lnln)( ,,, RTGRTbadcGG omr
omr
omr ∆+∆=−−++∆=′∆
νλ
λλ
λλ ∆−=
=′ Pb
oB
a
oA
d
oD
c
oC
P K
P
P
P
P
P
P
P
P
K
A reaction that is at equilibrium at Po will not be atequilibrium at .oP′\
This idea applies also to ionic equilibria.Normally we choose for the standard state co = 1M.But for biochemical systems, we choose for H+
Mc o 710−=′
50
[ ]pHRT
c
HRTaRTHH o
o ⋅−=
+=+=
+++ 302.2ln0ln)()( µµ
[ ]RTH
c
HRTH
o302.2)(
10ln)(
7−=
=′ +
−
++ µµ
For a reaction A + nH+Ø P
RTGG omr
omr 303.27,, ⋅+∆=′∆ ν
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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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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.
4
Normal melting point
Supercriticalfluid
Criticalpoint
LiquidSolid
Gas
Normalboilingpoint
Triple point (Tt)6.0 x 10-3
1
217.7
Pres
sure
(atm
)
Temperature (oC)
0 0.0098 100 374.4
Phase Diagram of Water
Image by MIT OpenCourseWare.
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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.
6
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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
7
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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.
8
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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:
�HF�SF � [13]
TF
Similarly, at the boiling point:
9
�H�S �
VV [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:
Cp�T �S � (since : �H � Cp�T) [15]
T
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
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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.
10
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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.
Fig. 4 Schematic representation of the solubilities of ether and water in each other.
13
0 20 40 60 80 100Composition percent, water
100 80 60 40 20 0Composition percent, ether
Two phases
Tem
pera
ture
One phase
Image by MIT OpenCourseWare.
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3. COOLING CURVES
14
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 SolutionsEquilibrium 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 inChapter 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.
15
Fig. 7
Source: Jastrzebski, Z. The Nature and Properties of EngineeringMaterials. 2nd edition. New York, NY: John Wiley & Sons, 1976.Courtesy of John Wiley & Sons. Used with permission.
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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.
17
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7. INCOMPLETE SOLUBILITY
18
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.
19
Source: Jastrzebski, Z. The Nature and Properties of EngineeringMaterials. 2nd edition. New York, NY: John Wiley & Sons, 1976.Courtesy of John Wiley & Sons. Used with permission.
MIT OpenCourseWarehttp://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.
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TERNARY PHASE DIAGRAMS
An Introduction
Guna SelvaduraySan Jose State University
Credit for Phase Diagram Drawings:Richard Brindos
Credit for scanning the phase diagrams:Brenden Croom
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Utility of Ternary Phase Diagrams
Glass compositionsRefractoriesAluminum alloysStainless steelsSolder metallurgySeveral other applications
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References on Ternary Phase Diagrams
A. Prince, Alloy Phase Equilibria, Elsevier Publishing Company, New York, 1966
D. R. F. West, Ternary Equilibrium Diagrams, Chapman and Hall, New York, 1982
G. Masing, Ternary Systems, Reinhold Publishing Company, New York, 1944
C. G. Bergeron and S. H. Risbud, Introduction to Phase Equilibria in Ceramics, The American Ceramic Society, Ohio, 1984
M. F. Berard and D. R. Wilder, Fundamentals of Phase Equilibriain Ceramic Systems, R.A.N. Publishers, Ohio, 1990
F. N. Rhines, Phase Diagrams in Metallurgy, McGraw-Hill, New York, 1956
A. Reisman, Phase Equilibria, Academic Press, 1970
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What are Ternary Phase Diagrams?
Diagrams that represent the equilibrium between the various phases that are formed between three components, as a function of temperature.
Normally, pressure is not a viable variable in ternary phase diagram construction, and is therefore held constant at 1 atm.
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The Gibbs Phase Rule for3-Component Systems
F = C + 2 - PFor isobaric systems:F = C + 1 - P
For C = 3, the maximum number of phases will co-exist when F = 0
P = 4 when C = 3 and F = 0
Components are “independent components”
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Some Important Terms
Overall compositionNumber of phasesChemical composition of individual phasesAmount of each phaseSolidification sequence
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Overall Composition - 1
The concentration of each of the three components
Can be expressed as either “wt. %” or “molar %”
Sum of the concentration of the three components must add up to 100%
The Gibbs Triangle is always used to determine the overall composition
The Gibbs Triangle: An equilateral triangle on which the pure components are represented by each corner
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Overall Composition - 2
There are three ways of determining the overall composition
Method 1Refer to Figures OC1 and OC2Let the overall composition be represented by
the point XDraw lines passing through X, and parallel to
each of the sidesWhere the line A’C’ intersects the side AB tells
us the concentration of component B in XThe concentrations of A and C, in X, can be
determined in an identical manner
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Method Two:Draw lines through X, parallel to the sides of the
Gibbs TriangleA’C’ intersects AB at A’B’C” intersects AB at B’
Concentration of B = AA’Concentration of C = A’B’Concentration of A = B’B
This method can be somewhat confusing, and is not recommended
Overall Composition - 3
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Overall Composition - 4
Method 3
Application of the Inverse Lever RuleDraw straight lines from each corner, through X%A = AX %B = BX %C = CX
AM BN CL
Important Note:Always determine the concentration of the
components independently, then check by adding them up to obtain 100%
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Ternary Isomorphous System
Isomorphous System: A system (ternary in this case) that has only one solid phase. All components are totally soluble in the other components. The ternary system is therefore made up of three binaries that exhibit total solid solubility
The Liquidus Surface: A plot of the temperatures above which a homogeneous liquid forms for any given overall composition
The Solidus Surface: A plot of the temperatures below which a (homogeneous) solid phase forms for any given overall composition
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dura
y -S
JSU
-O
ct 2
004
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Liquidus Surface
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Solidus Surface
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Isomorphous System
Isothermal Section: A “horizontal” section of a ternary phase diagram obtained by cutting through the space diagram at a specified temperature
Refer to Figures BT1A, BT1B and BT1CIdentify temperature of interest, T1 hereDraw the horizontal, intersecting the liquidus
and solidus surfaces at points 1, 2, 3 & 4Connect points 1 & 2 with curvature reflecting
the liquidus surfaceConnect points 3 & 4 with curvature reflecting
the solidus surface
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Isomorphous System
The line connecting points 1 & 2 represents the intersection of the isotherm with the liquidussurface
The line connecting points 3 & 4 represents the intersection of the isotherm with the solidussurface
Area A-B-1-2: homogeneous liquid phaseArea C-3-4: homogeneous solid phaseArea 1-2-3-4: two phase region - liquid + solid
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Isomorphous System
Isothermal Section - continuedTemperature = T2, below melting points of A &
B, but above melting point of CArea A-1-2: homogeneous liquid phaseArea B-C-4-3: homogeneous solid phaseArea 1-2-3-4: two phase region - liquid + solid
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Isomorphous System
Determination of:
(a) Chemical composition of phases present
(b) Amount of each phase present
when the overall composition is in a two phase region
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Isomorphous System
1. Locate overall composition using the Gibbs triangle
2. Draw tie-line passing through X, to intersect the phase boundaries at Y and Z
3. The chemical composition of the liquid phase is given by the location of the point Y within the Gibbs Triangle
4. The chemical composition of the solid phase is given by the location of the point Z within the Gibbs Triangle
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Isomorphous System
Tie line: A straight line joining any two ternary compositions
Amount of each phase present is determined by using the Inverse Lever Rule
5. Fraction of solid = YX/YZ
6. Fraction of liquid = ZX/YZ
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Isomorphous System
Drawing tie-lines in two phase regions
1. The directions of tie lines vary gradually from that of one boundary tie line to that of the other, without crossing each other
2. They must run between two one-phase regions
3. Except for the two bounding tie-lines, they are not necessarily pointed toward the corners of the compositional triangle
Tern
ary
Isom
orph
ous
Syst
em
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
The Ternary Eutectic Reaction:
L = α + β + γ
A liquid phase solidifies into three separate solid phases
Made up of three binary eutectic systems, all of which exhibit no solid solubility
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Phase regions:
Homogeneous liquid phaseLiquid + one solid phaseLiquid + two solid phasesThree solid phases
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
The Liquidus Surface
The liquidus surface “dips down” somewhere in the middle, to the ternary eutectic point, which would be at a temperature lower than all three binary eutectic temperatures
All points in space, above the liquidus surface, represent the existence of a homogeneous liquid phase
All points in space, below the liquidus surface, represent the existence of two or more phases (more on this later)
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
The Liquidus Surface (continued)
The Liquidus Projection - a projection of the liquidus surface onto a plane, with indications of isotherms and phase regions
The liquidus surface also represents the boundary between the single phase liquid region and the (liquid + one solid phase) regions
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Boundaries between 2 phase regions & 3 phase regions
With reference to Figure EB1:The surface P-E-F-J represents the boundary
between the liquid region and the (liquid + one solid phase) region
The surfaces P-E-G-I-P and P-J-H-I-J-P represent the boundary between the (liquid + one solid phase) region and the (liquid + two solid phases) regions
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Isothermal Section at T1, below melting point of A, but above melting points of B and C
We have two regions: a region of “liquid” and a region of “liquid + A”
The boundary between these two regions is a line, the curvature of which is in accordance with the liquidus surface
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Isothermal Section at T2, below the melting points of A and C, but above the melting point of B, and above the eutectic temperature of the system A-C
This isothermal section has three regions - L, L+A, and L + C
The boundary between L and L+A is determined by where the isothermal plane cuts the liquidus surface
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Isothermal section at temperature T3, below the eutectic temperature of the system A-C, but still above the melting point of B
The isothermal section now has four regions: L, L+A, L + C, L + A + C
Note that points 2 and 4 have converged and moved into the Gibbs Triangle; this represents the path that connects the A-C eutectic point to the ternary eutectic point
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
The boundaries between the one-phase and two-phase regions, and the two-phase and three-phase regions are lines, but the one-phase region and the three-phase region meet at a point
The three-phase region is a triangle - called atie-triangle
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Law of Adjoining Phases(for isothermal sections of ternary phase diagrams)
R1 = R - D- - D+ >= 0R1: dimension of boundary between
neighboring phase regions; 0 for point contact, 1 for line contact and 2 for surface contact
R: dimension of the concerned diagram or section of a diagram; for an isothermal section of a ternary, R = 2
D-: number of phases that disappear when crossing the boundary from one phase region to another
D+: number of phases that appear when crossing the boundary from one phase region to another
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Law of Adjoining Phases(continued)
1-phase region with 2-phase region: line1-phase region with 3-phase region: point2-phase region with 3-phase region: tie-line2-phase region with 2-phase region: point1-phase region with 1-phase region: point
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Isothermal section at temperature T4, below the melting points of A, B and C, and below the eutectic temperature of the A-C system, but above the A-B and B-C eutectic temperatures
The isothermal section now has five regions:L, L+A, L+B, L+C, L+A+C
Point 2 has moved further into the Gibbs Triangle, towards the ternary eutectic point
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Isothermal section at temperature T5, above the eutectic temperature of the B-C system, but below all other melting points and eutectic points
The isothermal section now has six regions:L, L+A, L+B, L+C, L+A+B, L+A+C
In addition to Point 2, Point 1 has also moved into the Gibbs Triangle, towards the ternary eutectic
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Isothermal section at temperature T6, above the ternary eutectic temperature, but below all other melting points and eutectic points
The isothermal section now has seven regions:L, L+A, L+B, L+C, L+A+B, L+A+C, L+B+C
In addition to Points 1 & 2, Point 3 has also moved into the Gibbs Triangle, towards the ternary eutectic
Note that the liquid regions is slowly converging towards the ternary eutectic point
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Solidification Sequence
Given an overall composition, determine the sequence of solidification, assuming equilibrium conditions
Let the overall composition be given by the point X
Imagine a line, orthogonal to the plane of the liquidus projection, passing through X
Let this line intersect the liquidus surface at a temperature T1
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Solidification Sequence (continued)
For all temperatures T > T1, there is one homogeneous liquid phase
Solidification begins when T = T1
The first solid to appear is: A
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Solidification Sequence (continued)
When T < T1, then precipitation of A occursAs the temperature drops, the composition of
the liquid phase “travels” along the line XY, on the liquidus surface, towards Y.
Let the temperature at Y be T2
At temperatures of T2 < T < T1, there are two phases in equilibrium - A and L
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Solidification Sequence (continued)In order to determine the amount of each phase
present, we need to fix the temperature first.Let T = T’, being at point ZWe use the Inverse Lever RuleFraction of L = AX/AZFraction of A = XZ/AZChemical composition of the liquid phase is
determined by the composition of point Z within the Gibbs Triangle
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Solidification Sequence (continued)
At point Y, where T = T2, the second solid phase, B, begins to precipitate
Over the temperature range of T2 > T > TE:- The solid phases A and B exist in equilibrium
with L- Both solid phases, A and B, coprecipitate as
the temperature is lowered
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Solidification Sequence (continued)
Analysis when T = T”, i.e., at point MComposition of L is given by the composition of
M within the Gibbs Triangle
How do we determine the amounts of A, B and L?
Let temperature T” correspond to the point M
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Solidification Sequence (continued)
Construct the triangle A-B-MThis triangle is a ternary system in which the
overall composition X can be represented in terms of the three constituents
Fraction of A = PX/PAFraction of B = QX/QBFraction of L = RX/RM
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Phase analysis at a given temperature
Isothermal Section at T = T4 will be used as reference
For all overall compositions that fall within the region marked L, the chemical composition of the liquid phase is the same as the overall composition
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Phase analysis at a given temperature
If the overall composition falls within a two phase region, e.g., L + C, then:
1. Locate the position of the overall composition, X, within the Gibbs Triangle
2. Draw tie lines, in this case connecting point C to line 2-3, and passing through X, and intersecting line 2-3 at Y
3. Use the Inverse Lever Rule to determine the amounts of L and CFraction of L = CX/CYFraction of C = YX/CY
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Phase analysis at a given temperature
The chemical composition of C in this case is 100% C
The chemical composition of L is given by determining the composition the point Y represents within the Gibbs Triangle
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Phase analysis at a given temperature
If the overall composition falls within a three phase region, e.g., L + A + C
1. Locate the position of the overall composition, Y, within the Gibbs Triangle
2. Construct the following straight lines:A-Y-Q, M-Y-R and C-Y-P
3. Use the Inverse Lever Rule to determine the amounts of L, A and C
Fraction of A = QY/QAFraction of C = PY/PCFraction of L = YR/MR
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System(No Solid Solubility)
Phase analysis at a given temperature
The chemical composition of A in this case is 100% A
The chemical composition of C in this case is 100% C
The chemical composition of L is given by determining the composition the point Y represents within the Gibbs Triangle
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Some Useful Rules regarding Phase Diagrams
The Boundary Rule: Any p-phase region can be bounded only by regions containing p +/- 1 phases, where p denotes the number of phases.
The Boundary Curvature Rule: Boundaries of one-phase regions must meet with curvatures such that the boundaries extrapolate into the adjacent two-phase regions.
The Solubility Rule: All components are soluble to some degree in all phases
Tern
ary
Eut
ectic
Sys
tem
–N
o S
olid
Sol
ubilit
y
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Ternary Eutectic System
(With Solid Solubility)
G. S
elva
dura
y -S
JSU
-O
ct 2
004
Tern
ary
Eut
ectic
Sys
tem
–W
ith S
olid
Sol
ubilit
y
G. S
elva
dura
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Alamade Lines
A join connecting the composition of the primary crystals of two areas havinga common boundary line.Alkamade Theorem: The intersection of a boundaryline with its Corresponding Alkamade linerepresents a temperature maximum on thatboundary line and a temperature minimum onthe Alkamade Line.
Source: Bergeron & Risbud
Alkamade lines never cross one another
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Alamade LinesA
B CBC
Alkamade Lines:A-B B-C C-AB-BC BC-CA-BC
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Colligative Properties- Page 1 Lecture 4: Colligative Properties
• By definition a colligative property is a solution property (a property of mixtures) for which it is the amount of solute dissolved in the solvent matters but the kind of solute does not matter.
• Coming to grips with this concept should immediately remind you of kinetic molecular theory of gases—in that case we
treated gas molecules as indistinguishable hard spheres and (ideally) it was the number of them, rather than the type of molecules, that determined gas properties.
• This means that when considering the impact of solute on a colligative property,
1 mole of sugar ≡ 1 mole Na + ≡ 1 mole O −2 ≡ 1 mole urea ≡ 1 mole pickles do exactly the same thing Listed below are the four colligative properties we will examine during this lecture—each is kind of fun because it is associated with fairly famous physical phenomena that you might like to explain to a friend . Colligative Properties For each of these properties you will be introduced to the physical phenomenon behind the property and learn how to perform simple calculations to determine the magnitude the change in solution state function associated with a colligative property.
♦Vapor Pressure Lowering—explains the value of putting antifreeze in your radiator to keep a car from overheating ♦Boiling Point Raising—explains how you can cook spaghetti faster in salt water ♦Freezing Point Lowering explains why salt is placed on roads to keep ice from forming ♦Osmotic Pressure explains why your little brother killed the family fish when he placed them in pure water while cleaning the tank.
Colligative Properties- Page 2 Concentration and Colligative Properties If colligative properties depend on the amount of the solute in the solvent, then the equations defining them must include a concentration term, and sure enough, they do. Over the next few pages you will be introduced to the equations in the context of the specific properties, but for now, simply note the similarities in structure for the equations: each equation includes a colligative property on the left side of the equation that is set equal to a concentration term and a solvent constant. ΔP vapor pressure ΔT temperature raising or lowering π osmotic pressure ΔT f = -mK f
ΔP = χP 0 ΔT f = -mK f π = MRT Three properties set equal to three different concentrations terms times a solvent constant Let’s practice performing concentration calculations. Two of these, molarity, M, and mole fraction, X, should be familiar to you. A third, molality, m, may be new. But all are useful ways to define the amount of stuff in solution—the more stuff, the larger the concentration. Let’s start by imagining that we are placing 50 g (0.146 mole) of the solute, sugar, in 117 g (6.5 mole) of the solvent, H 2 O. ♦What is χ (mole fraction) of 50g of sugar in 117 g of water? χ =
molesBmolesAmolesA
+ =
OHsugarsugar
25.6146.0146.0
+ = 0.022 mole fraction sugar
♦What is m (molality) of 50g of sugar in 117 g of water? Note that the moloaity calculation is similar to a molarity calculation except that we divide by the mass of the solvent in kg rather than the liters of solution.
m = solventkg
molesA,
= OHkg
sugarmole
2,117.0,146.0 = 1.25 molal
Colligative Properties- Page 3
♦What is M (molarity) of 50g of sugar in 117 g of water? First we need to find the volume of solution from a density calculation.
V solution = (mass)(density) = (50 g + 117 g)(g
mL34.1
1 ) = 125 mL
M = solutionV
molesA = L
moles125.0
146.0 = 1.17 Molar
So we have 3 ways to describe 50g of sugar in 117 g of water, each of which is used in a colligative property calculation.
0.022 mole fraction ≡ 1.25 m ≡ 1.17 M
Colligative Properties- Page 4 Time out for the Van’t Hoff equation. Every test on colligative properties includes a question that employs the Van’t Hoff equation. Text books make this seem a lot harder than it is. Very simply, Van’t Hoff corrects for the fact that the number of particles you thrown into solution is not always the number of particles that determine the magnitude of the property. For example, think about what happens when you put the following one mole quantities into a liter of water. Which one raises the boiling point the most? 1. 1 mole NaCl 2. 1 mole Na 2 S 3. 1 mole CaS 4. 1 mole sugar You might think 1 mole is 1 mole is 1 mole, and they are all the same. But 1 mole NaCl 1 mole Na 2 S 1 mole CaS 1 mole sugar In fact each produces a different number of dissolved particles in solution.
• 1mole of NaCl is 2 moles of particles in the solvent • 1 mole of Na2S is 3 moles of particles in the solvent • 1 mole of insoluble CaS is 0 moles of particles in the solvent • 1 mole of sugar is 1 mole of particles in solution
What we need is a correction factor for each compound, i, the Van’t Hoff factor, which is i = 2, 3, 0 and 1, respectively for the four solutions. i is simply inserted into every colligative property equation to make the correction. Oh, and the answer to the original question about the change in boiling point for the four one mole samples?
1 mole Na 2 S > 1 mole NaCl > 1 mole sugar > 1 mole CaS
1 mole Na + and 1 mole Cl −
2 moles Na + 1 mole S −2
0 moles CaS insoluble
1 mole sugar
Colligative Properties- Page 5 So how big of an effect does a solute concentration have on a colligative property? Now time for some math with the four different equations for colligative properties. Suppose we wanted to measure just how much 50 grams of sugar in 117 grams of water changed the magnitude of a state function. Colligative property 1: Vapor pressure depression ΔP = P 0 χ mole fraction which is the amount of solute added depression constant which is the vapor pressure of pure solvent at a given T. of pure solvent in vapor pressure For H 2 O at 25 o C the pure vapor pressure is 23.8 torr So the vapor pressure depression in ΔP = 23.8 torr (0.022) = 0.524 torr And the new vapor pressure is now about 23.3 torr. By the way, this equation is referred to as Raoult’s Law which says simply that the vapor pressure above a solution is proportional to the mole fraction of the solute.
Colligative Properties- Page 6 Time out for a famous vapor pressure calculation. Calculating the total vapor pressure of a binary mixture of two volatile solvents. Note that in the problem above, we determined the reduction in vapor pressure of the solvent. It decreased from 23.8 torr to 23.3 torr. But what if two volatile compounds were mixed together. Each would contribute to the other having a vapor pressure lowering, but the overall vapor pressure would have to be determined by adding together the individual vapor pressures of the two compounds. Sample problem. Suppose you combine two compounds, A and B, in a mole ratio of 0.25 to 0.75. At a given temperature, the pure vapor pressure of compound A is 100 torr and the pure vapor pressure of compounds B is 50 torr. What are the new vapor pressures for A and B combined. Calculation. First, determine the vapor pressure for each compound: For compound A-- Pa = Pa
0 χ = 100 torr x 0.25 = 25 torr contribution for A. For compound B—Pb = Pb
0 χ = 50 torr x 0.75 = 37.5 torr contribution for B. Total vapor pressure above the solution (assuming gas ideality and using Dalton’s Law of partial Pressures) Ptot = Pa + Pb = 25 + 37.5 torr = 62.5 torr. Note this problem can be made more complicated by making you work to calculate a mole fraction (I started by having a mole fraction.)
Colligative Properties- Page 7 Colligative Property 2: boiling point elevation
ΔT b = K b m molality which is the amount of solute added temperature constant which is the boiling point elevation constant for water increase because of solute
For water, Kb is 0.512 o C/molal
So the boiling point elevation is ΔT b = K b m = (0.512)(1.25 m) = 0.64 o And the new b.p. of water with a heck of a lot of sugar in it is 100.64 o C Colligative Property 3: freezing point depression
ΔTf = Kfm molality which is the amount of solute added temperature constant which is the boiling point elevation constant for water decrease because of solute For water, Kb 1.86 o C/molal
So the boiling point elevation is ΔT f = -K f m = -(1.86)(1.25) = 2.32 o C
And the new f.p. of water with a heck of a lot of sugar in it is -2.32 o C
Colligative Properties- Page 8 Colligative Property 4: osmotic pressure
π = MRT molarity which is the amount of solute added Osmotic pressure constant, RT, which seems to pop up everywhere increase because of solute The constants R and T are the ideal gas law constant and the system temperature
So the osmotic pressure change is π = MRT = (1.17 M)(0.082)(298 K) = 28.6 atm
Colligative Properties- Page 9 Comparison of magnitudes of colligative property changes. Let’s look at how much the colligative property change was for the same solute/solvent combination—note that what determines the magnitude of the overall change is the amplification by the constant term—in other words, the bigger the constant, the bigger the change, and the clear winner is the RT term for osmotic pressure. Sample values for freezing and boiling points constants for different solvents also give you an idea of what kind of changes to expect for different solvents.
• ΔP lowers water vapor pressure from 23.8 to 23.3 torr • ΔT b raises boiling point of water from 100 o C to 100.64 o C • ΔT f lowers freezing point of water from 0 o C to -2.32 o C • π raises osmotic pressure from 1 atm to 28.6 atm
While the temperature was changing by a degree or so for boiling point elevation or freezing point depression, there was a near 30-fold increase in osmotic pressure—no wonder fish are so sensitive. Not that even with small changes, significant impacts can occur in the real world, as we will see:
• Vapor pressure lowering explains why we add ethylene glycol to water. • Boiling point elevation explains how adding salt to water speeds up cooking pasta. • Freezing point depression explains how putting salt on icy roads melts ice. • Osmotic pressure change explains lysing of cells or why you can’t put salt water fish in fresh water.
Colligative Properties- Page 10 A bit of theory with bad pictures to explain colligative properties Colligative property 1: V.P depression and B.P elevation. What happens when y ou add to pure ? Look at the drawing below to see a simple explanation. Mixture 75 % to 25% compared to Pure solution of 100% Surface molecules ΔP = χP 0 ΔP = χP 0 25% pure solvent 0% pure solvent Since we know that vapor pressure is a surface phenomenon, we see that one obvious reason for the reduction in vapor pressure is that there are fewer surface locations to put a molecule. So the solvent will have to have a reduction in vapor pressure because fewer molecules are present to leave the surface. The plot of mole fraction versus vapor pressure shows this obvious consequence. A second reason for the reduction in vapor pressure is an entropic effect. Recall that highly ordered systems are not favored, and so for example, pure liquids will have a larger driving force from the change in entropy compared to mixtures which have a smaller entropy change. Finally, note that the same argument that explains vapor pressure lowering also explains boiling point elevation. If the definition of boiling is that the vapor pressure exceeds atmospheric pressure, and the vapor pressure is lowered, then you need to raise temperature to achieve boiling.
Colligative Properties- Page 11 Examples of vapor pressure lowering in action—your car radiator While you may not know a lot about the car daddy bought you, one thing that would make sense is that you want the liquid in your radiator to remain a liquid. Turning into a solid at really cold temperature, or turning into a vapor at high temperature, probably isn’t the best way to get the cooling action of recirculating fluid in a radiator going. So what happens if you decide to use pure water as your radiator coolant: f.p. b.p. H 2 O 0 o 100 Or maybe you gould use a different solvent like pure ethylene glycol—that green stuff you buy in stores. f.p. b.p. ethylene -12 o 135 o glycol Note that this isn’t the greatest temperature range for people who live in Canada and vacation in Florida. Your car radiator can freeze or overheat pretty easily with these pure liquids. But what happens if we take advantage of vapor pressure lowering by creating a mixture of two solvents. f.p. b.p. 50-50 H 2 O, E.G. -34 o 265 o 30-70 H 2 O, E.G. -84 o 276 o Wow a Δ of >350 o A three order of magnitude increase in temperature range!! A couple of points:
• That 50-50 mixture they tell you to use isn’t as good as a 30-70 ratio—try impressing your Dad by telling him he can get an extra 50oC of liquid range (especially useful in cold climates) by using a little less water.
• The incredible temperature range achieved isn’t explained by anything we have mentioned but rather on interesting
non-ideal effects that occur in mixtures, something you don’t learn about unless you take upper division physical chemistry.
Colligative Properties- Page 12 Colligative Property 2—freezing point depression is explained Freezing point depression is again thermodynamic effect. In order to make ice, which is a pure crystal, extra work must be done to separate the solute from the solvent—NaCl from water, for example. This means you need to reduce the freezing point to thermodynamically drive the reaction. This has the advantage of allowing for the deicing of streets by requiring that a temperature lower than 0oC be reached for ice to form. By the way, a lot of different kinds of compounds can be used to achieve this lower of the freezing point of ice. Salt is not the best candidate not only because it can cause only about a temperature drop of -15oC (making it useless when it gets colder) but also because it is so corrosive for automobiles. But it is cheap. liquid creating ice in a mixture frozen ` liquid creating ice in pure solvent Frozen -
Colligative Properties- Page 13 Colligative property 3: Osmotic pressure. Osmotic pressure in action with good pictures. The movement of solvent across a semi-permeable membrane to establish equal concentration. Note that whenever you give solutes a chance, they will distribute evenly across a solvent—for example, a few drops of food coloring to water and over time, a homogeneity of color results from equal concentration of the dye in the solvent. Now imagine that you create a barrier so that there is a selective ability to pass material throughout the solution. Semi-
permeable membranes do this—they selectively allow one kind of molecule or another to cross the membrane. In the example shown, the membrane is cellulose acetate that selectively allows water to pass through to create the pressure differential. Scientists are very creative at creating materials that allow one compound or another to selectively pass and so as you move along in science you will see many examples of selective mobility like this. Of course the most famous example of a semipermeable membrane is the membrane in cells. There is an elaborate process by which water and various electrolytes cross
the cell membrane barrier—(I have tried to forget all this because I couldn’t get into medical school and no longer care about electrolyte imbalance. But all of you in biology probably know this and could explain it to me if I cared.)
Colligative Properties- Page 14 So in the picture of red blood cells below, note that just like the porridge in Goldilocks and the Three Bears, the solution
around the cells be just the right concentration making for nice healthy cells (first picture), too high a concentration in which case the water flows out of the cell and they shrivel up (second picture), or too low a concentration in which case water flows into the cells and they lyse or rupture (third picture.)
Osmotic pressure in action with bad pictures.. In Figure A, note the concentration imbalance inside the baggie formed by the semi-permeable membrane. The concentration gradient that results drives selectively into the baggie across the membrane (the cannot penetrate the baggie and leave. Consequently the large influx of to achieve a concentration balance results in an increase in pressure (osmotic pressure) inside the baggie. Now imagine this happening in a cell, or heaven forbid, a fish. Your pet guppie, Toto, is happily swimming around in a dirty container of liquid in Figure C. Feeling badly for the fish and his dirty confines, you decide to place the fish in some clean water in Figure D. The water selectively enters the fish to ease the concentration gradient, and the increase in osmotic pressure, while a cleansing sort of thing in one respect, causes the fish to explode and die. So now you know, it was you that killed your favorite pet.
Colligative Properties- Page 15 Figure A— Figure C—dirty water Figure D-clean water Figure E—R.I.P. Figure B
Chem 467 Supplement to Lectures 33
Phase Equilibrium
Chemical Potential Revisited
We introduced the chemical potential as the conjugate variable to amount. Briefly reviewing, the total
Gibbs energy of a system consisting of multiple chemical species and/or multiple phases is given by
G=G T , p , n1, n2,. . .
The total differential of the Gibbs energy is written as
dG=−S dTV dp∑ii dni
where we have used the Gibbs equation for the Gibbs energy and the definition of chemical potential
i=∂G∂n i T , p ,n j≠i
As we will see in the coming days, the chemical potential is probably the single most powerful quantity
we will use in our discussion of thermodynamics. We will use the chemical potential to describe such
things as:
Single component phase transitions
Simple mixtures and binary phase diagrams
Chemical equilibrium
Electrochemistry
Other areas we won't discuss in as much detail, but also depend on the chemical potential, include:
Separation science and retention
Adhesion and wetting phenomena
Biochemical processes, such as enzymes and metabolism
Fuel cells and alternative fuel sources
Chemical Potential of a Single Component System
Now, let's consider a single component system for a while. We can write the following relation for the
total Gibbs energy in terms of the molar Gibbs energy at some particular T and p:
G=n G
If we take the derivative of this expression, we can write dG=G dn or dGdn
=G
If we look at the definition of the chemical potential, we can see that
= ∂G∂ n
T , p=G
For a pure material, the chemical potential is equivalent to the molar Gibbs energy. If we continue with
a pure material, we can write
d =d G=−S dTV dp
We also conclude that
∂∂T p=−S ∂∂ p T
=V
Because we have criteria for spontaneity in terms of the Gibbs energy, we can write that a spontaneous
process is one that reduces the chemical potential.
d GT , p=d 0
With this information, we are prepared to discuss the phase equilibrium of a pure substance.
Phase Transitions of a Pure Component
You are all familiar with phase equilibrium. You have seen ice
melt and liquid water boil. You should also be familiar with phase
diagrams, that indicate the temperature and pressure conditions
under which different phases are stable. A generic phase diagram
is given in the figure at right. These types of diagrams are
generally based on empirical data. We will now develop the
underlying thermodynamics that cause these phase transitions to
take place.
Let's take a closed system of one component at constant T and p, but allow for two different phases,
which we'll call and . We would then write the differential of the total Gibbs energy as
dGT , p= dn dn
Now, because this is a pure material, any increase in the moles of phase must have an
accompanying decrease in the moles of phase . Thus, we can write
dn=−dn and dGT , p=−dn
Recall our criterion for a spontaneous process, that dGT, p < 0. This means that if dn0 then
−0 or
This tells us that we will have spontaneous conversion from phase to phase if this inequality
holds. Thus the thermodynamically stable phase is the one with the lower chemical potential. If the
chemical potentials are equal, then dGT, p = 0 and the system is at equilibrium.
Changes with T
The result we just derived is general; at any given temperature and pressure, the stable phase is the one
with the lower chemical potential. If the chemical potentials of the two phases are equal, the two
phases exist in equilibrium. When we look at a phase diagram, we see a boundary between the two
phases. How does that boundary depend on changes in conditions? We will first look at how the
equilibrium boundary depends on temperature.
We already have the general result ∂∂T p=−S
Since entropy is always a positive quantity, the chemical potential must decrease as the temperature
increases. Let's consider the chemical potential of the solid, liquid and gas phases of a pure substance.
We know that the entropy of a condensed phase is less than for the gaseous phase, and the entropy of a
solid is less than a liquid. Or, mathematically
Sgas > Sliq > Ssolid
Therefore, the chemical potential of a gas will drop more
rapidly than that of a liquid, which drops faster than the
chemical potential of a solid. Assuming that the entropy of the
individual phase is constant with respect to T, we would plot the
temperature dependence of the chemical potential (or the Gibbs
energy) as seen in the figure.
According to the figure, what happens in terms of phase
transitions as the temperature increases?
How would this plot be different for CO2 at 1 atm?
The temperature at which the transition occurs from liquid to solid is known at the melting point. If
the pressure is our standard pressure, then this is the standard melting point. Likewise, the
temperature where a liquid transitions to a gas at the standard pressure is the standard boiling point.
It is possible, depending on the chemical potentials of the different species, to go directly from solid to
vapor, and if that happens, the temperature at which it occurs is the sublimation point.
Response of Melting to Applied Pressure
Let's consider what happens to our melting point as we change the pressure. We know that at a given
temperature, the change in chemical potential with changes in pressure is given by
∂∂ p T=V
Since the molar volume is always a positive quantity, the chemical potential will increase with
increasing pressure. If we do this at each temperature, we will see our line of µ vs T move to higher µ.
To a first approximation, S won't change much with these changes in pressure, so the slope doesn't
change. But, we also have to consider the fact that the molar volume of the different phases can be
different. If the molar volume of the liquid is greater than that of the solid, or the density of the liquid
is less than the solid, the liquid line will move up more than the solid line. This means the temperature
at which they intersect, the melting point, will increase with increasing pressure. However, for water
we have the reverse case. The density of the solid is less than the density of the liquid, meaning the
molar volume of the solid is greater than that of the liquid and so the solid curve will move up more
than the liquid curve. This means that the melting point will decrease as the pressure increases.
Response of Boiling to Applied Pressure
Using the same logic as we did for melting, what happens to the boiling point as the pressure changes?
Phase Boundaries
Now that we have qualitatively discussed how phase transitions depend on the temperature and
pressure, we can develop a more quantitative treatment. Let's consider the case of two phases, and
present in equilibrium at some particular T and p. In order to be at equilibrium, the chemical
potential of the two phases must be equal.
p , T = p ,T
It is also necessary that T, p and dµ be the same for both phases if we are staying on the phase
boundary. We can write for each phase
d =−S dT V dp
This let's us write
− S dT V dp=− S dT V dp
We can collect the pressure and temperature terms and write
V − V dp= S − SdT
We now define the entropy and volume changes of the phase transition as
trs S= S − S trs V = V − V
This let's us write
dpdT
=trs Strs V
This is known as the Clapeyron equation. This expression is exact and applies to any phase boundary
of a pure substance. We have not made any assumptions in deriving this relation. Keep in mind that
we are allowing temperature and pressure to vary and in principle the volume and the entropy of the
transition will vary with the conditions. Assuming we know these quantities, this expression gives us
the slope of the phase boundary at that point. Now let's look a little more at this equation for particular
cases.
Solid-Liquid Phase Boundary – A melting (fusion) process is accompanied by an enthalpy change ∆fusH
at any particular temperature. If we know this quantity per mole of material, we know the heat at the
pressure of the phase transition. Thus we can write for the entropy
trs S=qtrs
T=trs H
T
We can put this into our expression to write dpdT
=trs H
T trs V
If we have data for the enthalpy and the volume of the transition as a function of temperature, we can
determine a formula for the phase boundary, which will be a curve on our p-T plot.
∫p∗
pdp=∫T ∗
T fus HT fus V
dT≈ fus H fus V
∫T ∗
T dTT
Here we have assumed that the enthalpy and volume changes are constant over our temperature range.
We then write the approximate phase boundary as
p≈ p∗ fus H fus V
ln TT ∗
If the temperature range is narrow, we can approximate the ln term as
ln TT∗=ln 1T−T ∗
T∗ ≈T−T ∗
T ∗
This let's us write
p≈ p∗ fus H
T∗ fus VT −T∗
So if we start at some temperature T *, the phase boundary is a steep straight line on a p-T plot. What
about the slope of this line? ∆fusH > 0 is always true; it always takes heat to make a solid melt.
However, ∆fusV can be positive or negative, depending on the relative density of the liquid and solid.
For most materials, the solid is more dense than the liquid, so the volume increases upon melting. Thus
the p-T line will have a positive slope. With water, however, the liquid density is greater than that of
the solid, so this line has a negative slope. The completely agrees with our result from above, as it
should.
Liquid-Vapor Boundary – Let's go back to our earlier statement of the Clapeyron equation, but now
write it for a liquid-vapor phase boundary.
dpdT
=vap H
T vap V
Because the molar volume of a gas is much larger than that of the liquid, we can treat the volume
change as simply the volume of the gas; vap V ≈V g . If we then treat the gas as ideal, we can write
dpdT
=vap H
T RT / por 1
pdpdT
=d ln pdT
=vap HRT 2
This is known as the Clausius – Clapeyron equation and is an approximation for the liquid-vapor phase
boundary. How would we get a better expression for this phase boundary?
As we did before, we can determine a formula for the phase boundary by integrating this expression
∫ln p∗
ln pd ln p=ln p
p∗=vap H
R ∫T∗
T dTT 2 =
−vap HR 1
T− 1
T ∗ ln p
p∗=−vap H
R 1T− 1
T ∗ =vap H T −T ∗R T T ∗
This is only an approximation, based on the assumption that we can treat ∆vapH as a constant. If we
want to evaluate this boundary over a wider temperature range, we have to account for the temperature
dependence on the enthalpy of vaporization. It should be noted that ∆vapH approaches zero as T
approaches the critical temperature. Thus, there is no phase boundary above Tc.
Solid-Vapor Phase Boundary – The Clausius – Clapeyron equation also applies to the boundary
between the solid and vapor phases, again because the volume of the vapor is much greater than the
volume of the solid. Also, recall that ∆subH = ∆fusıH + ∆vapH, so we would write
ln pp∗=
subl H T −T∗R T T∗
and the solid-vapor line should be steeper than the liquid-vapor line at similar temperatures.
Note on standard states – It should be noted that all pressures should be defined relative to our
standard pressure. This makes sense from a consideration of the units of our expressions. In deriving
the Clausius – Clapeyron equation, we switched from p to ln p, but you may be concerned about the
units when we do this. Because we are writing a differential equation, we can always add a constant
factor and not change the derivative. Thus, in the step when we used
1p
dp=d ln p
we should technically write
1p
dp=d ln p−d ln po=d ln ppo
where p° is our standard pressure. Because p° is a fixed point, its derivative is zero, so we can add it,
or the derivative of its ln, without changing our equation. Writing the expression in this way keeps our
units constant. Because p° is usually taken to be 1 bar, it is often dropped in the expression of the
Clausius – Clapeyron equation. When we integrate the equation, the standard state is removed from the
expression. So, the way we wrote it works and is a simpler notation, but you need to keep in mind that
all pressures are with respect to the standard state pressure.
Vapor Pressure and Liquid/Gas Coexistence
We know from our own observations that when we have a sample of liquid there exists some amount of
vapor of that material. For example, if we have a closed container of a particular liquid and we remove
all the gas from it, there will be some pressure still in the container, from the vapor of the substance.
This amount of pressure is known as the vapor pressure. We can use the chemical potential to
determine the vapor pressure of a pure material at conditions other than the equilibrium phase
boundary. We define the standard boiling point as the temperature at which a liquid boils at 1 bar
pressure. If we know these points, we can approximately determine the vapor pressure at any other
temperature by evaluating the Clausius – Clapeyron equation for the temperature of interest. An
alternate form of this equation is
d ln pd 1/T
=−vap H
R
(You will derive this form of the equation in the homework.) This equation tells us that the ln of the
vapor pressure should decrease linearly with the reciprocal temperature. This is what is experimentally
observed sufficiently far from the critical point. As with many properties, vapor pressures for many
liquids are tabulated.
Degrees of Freedom and the Triple Point
For a pure material, keeping amount fixed constant, we have three variables, or degrees of freedom, T,
p and Vm. However, because of the equation of state, we can only independently vary two of them. If
we require that we keep two phases in equilibrium, that brings in an additional constraint. Now, we can
only vary one parameter, such as the pressure. The temperature is fixed by the Clapeyron equation and
the volume is fixed by the equation of state. Each constraint removes a degree of freedom. If we want
to have three phases in equilibrium with each other, that adds another constraint, meaning we can't
change any of the state variables. This is why phase diagrams have triple points, which are invariant.
This is also why you can't have a point in a single component phase diagram where more than three
phase are in equilibrium; that would require negative degrees of freedom, which doesn't make physical
sense.
PHASE EQUILIBRIUM
� Component: is either pure metal and/or compound of which an alloy is composed. They refer to the independent chemical species that comprise the system.
� Solid Solution: It consists of atoms of at least two different types where solute atoms occupy either substitutional or interstitial positions in the solvent lattice and the crystal structure of the solvent is maintained.
� Solubility limit: for almost all alloy systems, at specific temperature, a maximum of solute atoms can dissolve in solvent phase to form a solid solution. The limit is known as solubility limit.
� Phase: defined as a homogenous portion of a system that has uniform physical and chemical characteristics i.e. it is a physically distinct from other phases, chemically homogenous and mechanically separable portion of a system.
� Variable of a system: theses include two external variables namely temperature and pressure along with internal variable such as composition © and number of phases (P). Number of independent variable among these gives the degree of freedom (F).
� They are given by Gibbs Phase rule
P + F = C + 2
EQUILIBRIUM PHASE DIAGRAM
� A diagram that depicts existence of different phases of
a system under equilibrium is termed as phase
diagram. Equilibrium phase diagrams represent the
relationships between temperature and the
compositions and the quantities of phases at
equilibrium. In general practice it is sufficient to
consider only solid and liquid phases, thus pressure is consider only solid and liquid phases, thus pressure is
assumed to be constant (1 atm.) in most applications.
CONTD…
� Important information obtainable from a phase diagram
can be summarized as follows:
� To show phases are present at different compositions and
temperatures under slow cooling (equilibrium) conditions.
� To indicate equilibrium solid solubility of one
element/compound in another. element/compound in another.
� To indicate temperature at which an alloy starts to solidify
and the range of solidification.
� To indicate the temperature at which different phases start
to melt.
� Amount of each phase in a two-phase mixture can be
obtained.
� A phase diagram is actually a collection of solubility
limit curves. Set of solubility curves that represents
locus of temperatures above which all compositions are
liquid are called liquidus, while solidus represents set
of solubility curves that denotes the locus of
temperatures below which all compositions are solid.
Every phase diagram for two or more components
must show a liquidus and a solidus. Phase diagrams
are classified based on the number of components in are classified based on the number of components in
the system. Single component systems have unary
diagrams, two-component systems have binary
diagrams, three-component systems are represented
by ternary diagrams, and so on. When more than two
components are present, phase diagrams become
extremely complicated and difficult to represent.
UNARY PHASE DIAGRAM
� In these systems there is no composition change
(C=1), thus only variables are temperature and
pressure. Thus in region of single phase two
variables (temperature and pressure) can be
varied independently. If two phases coexist then,
according to Phase rule, either temperature or
pressure can be varied independently, but not pressure can be varied independently, but not
both. At triple points, three phases can coexist at a
particular set of temperature and pressure. At
these points, neither temperature nor the pressure
can be changed without disrupting the
equilibrium i.e. one of the phases may disappear.
BINARY PHASE DIAGRAM
� Binary phase diagrams are maps that represent the
relationships between temperature and the
compositions and quantities of phases at equilibrium,
which influence the microstructure of an alloy. Many
microstructures develop from phase transformations,
the changes that occur when the temperature is altered
(ordinarily upon cooling).This may involve the (ordinarily upon cooling).This may involve the
transition from one phase to another, or the appearance
or disappearance of a phase. Binary phase diagrams are
helpful in predicting phase transformations and the
resulting microstructures, which may have equilibrium
or non-equilibrium character.
DETERMINATION OF PHASE AMOUNTS
� The tie line must be utilized in conjunction with a
procedure that is often called the lever rule,
which is applied as follows:
� 1. The tie line is constructed across the two-phase
region at the temperature of the alloy.
� 2. The overall alloy composition is located on the
tie line.
� 3. The fraction of one phase is computed by � 3. The fraction of one phase is computed by
taking the length of tie line from the overall alloy
composition to the phase boundary for the other
phase, and dividing by the total tie line length.
� 4. The fraction of the other phase determined in
the same manner.
� The tie line has been constructed that was used for the
determination of and L phase compositions. Let the
overall alloy composition be located along the tie line
and denoted as and mass fractions be represented by
WL and Wa for the respective phases. From the lever
rule, WL may be computed according to
DEVELOPMENT OFMICROSTRUCTURE IN ISOMORPHOUS ALLOYS
BINARY EUTECTIC SYSTEMS
� A number of features of this phase diagram are
important: First, three single-phase regions are found
on the diagram: @, b, and liquid. The @ phase is a solid
solution rich in copper; it has silver as the solute
component and an FCC crystal structure. The b -phase
solid solution also has an FCC structure, but copper is
the solute. Pure copper and pure silver are also the solute. Pure copper and pure silver are also
considered to be @ and b phases, respectively.
� Thus, the solubility in each of these solid phases is
limited, in that at any temperature below line BEG
only a limited concentration of silver will dissolve in
Copper and similarly for copper in silver.
CU-AG BINARY EUTECTIC SYSTEM
PB-SN BINARY EUTECTIC SYSTEM
IRON-CARBON SYSTEM
� Of all binary alloy systems, the one that is possibly the
most important is that for iron and carbon. Both steels
and cast irons, primary structural materials in every
technologically advanced culture, are essentially iron–
carbon alloys. Pure iron, upon heating, experiences two
changes in crystal structure before it melts. At room
temperature the stable form, called ferrite, or @ iron,
has a BCC crystal structure. Ferrite experiences a has a BCC crystal structure. Ferrite experiences a
polymorphic transformation to FCC austenite, or γ
iron, at 912°C or (1674F). This austenite persists to
1394°C (2541F), at which temperature the FCC
austenite reverts back to a BCC phase known as δ
ferrite, which finally melts at 1538°C. All these changes
are apparent along the left vertical axis of the phase
diagram.
INVARIANT REACTION
