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Review for Mid-Term 3, 2015
Based on material in lecture notes 6-8 inclusive.
Mid-term exam will be in the recitation period,
November 23rd, 7-8:20 pm.
Split into two rooms as usual:
Engineering 143 (A-M), Frey 119 (N-Z)
Bring ID, calculator, pencil/pen, etc.
2
Phase Transition
Aphase transition is the spontaneous transformation of one
phase into another at a characteristic T & p.
How do we know the phase of a substance at a given p and T?
Using the Gibbs energy: spontaneous processes have negative
changes in Gibbs free energy. For example, if you want to know the
most stable phase of water at a given pressure and temperature, you
can calculate the Gibbs energies for each of the phases, compare
them and pick the lowest one
Chemical potentials are equivalent to molarGibbs energies for pure
substances, so the phase with the lowest chemical potential will be the
most stable phase
At the transition temperature and pressure the two
phases have the same chemical potential.
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Phase Transition, pure substance
Heat exchange without temperature change
Constant pressure cooling curve for
a substance. Calorimetry can spot a
phase change even if there are no
outward visible changes.
A discontinuity indicates a phase
change. Heat is going towards
changing internal structure, entropy.
4
Temperature Dependence of a Phase
Recall
p
GS
T
=
m mdG SdT Vdp d S dT V dp= =
( ) ( ) ( )m m mS s S l S g
mm
p
GS
T
=
m
p
ST
=
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5
Phase Diagram
Phase Diagram: Shows regions of pressure and temperature where
phases are thermodynamically stable
Phase Boundaries: Separate regions and showp and T where two
phases exist in equilibrium
Sublimation
Fusion
Liquefaction
Triple point,
3 phases in equilibrium
Critical Point (liquid and
vapor have same density)
6
How to find a Phase Boundary?
The thermodynamic criterion of equilibrium is:
At equilibrium, the chemical potential of a substance is the same
throughout a sample, regardless of how many phases are present
If amount dn is transferred from 1 to 2, Gibbs
energy changes by1dn in location 1 and by
+2dn in location 2
2 1dG dn =
If 1 > 2, G is negative, process is spontaneous
If 1 = 2, no change in G, system at equilibrium
1 (p,T)= 2(p,T) p = p(trans)
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7
How to find a Phase Boundary?
Since
How to predict the
phase boundary?
d= dd
- d
= 0
So
exact expression for the slope of a phase boundary -
applies to phase equilibrium of any pure substance
dp
dT=
trs
Sm
trsV
m
=
trs
Hm
TtrsV
m
, ,m md S dT V dp =
, ,m md S dT V dp =
So
We have
, , , ,m m m md d S S dT V V dp =
0trs m trs mS dT V dp= =
8
Liquid-Vapor Boundary
Va p
Va p
Hdp
dT T V
=
Clapeyron equation
1 1ln ln
Va p
f i
f i
Hp p
R T T
=
2
ln Va pHd p
dT RT
=
Vapor pressure
Clausius-Clapeyron Equation.
Integrate:
)ln(since xdx
dx
=
Large and positive
Dominated by the gas
V=RT/p
Approximate, because
weve assumed pV=nRT
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Clapeyron Equation Example
Calculate the slope of the solid-liquid phase boundary curve for
benzene. At the melting point (278.7 K) at one atmosphere pressure:
1-3
1
.mol10.3cm(benzene)
kJ.mol95.9(benzene)
=
=
mfus
mfus
V
H
From:
mtrs
mtrs
VT
H
dT
dp
=
1-
1-1-
-1-13
33
1
atm.K2.34
.K8.314J.mol
.K.atm.moldm08206.0
dm103.10K7.278
J.mol9950
=
=
dT
dp
p
T
solid
liquid
very steep!
Vapor Pressure Example
Find the vapor pressure of a liquid at some temperature
1 1ln ln
Va p
f i
f i
Hp p
R T T
=
At normal boiling point, the vapor pressure is one atmosphere.
For benzene, T* (b.p.) is 80 C, 353 K. vapH = 30.8 kJ mol-1. What is
the vapor pressure at 20 C, 293 K?
= 353
1
293
1
314.8
30800
)0.1ln(ln fp
atm117.0=fp or, about 90 torr
UNITS!
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Clausius-Clapeyron Example
For water vapH = 40.65 kJmol-1 and the normal boilingpoint (at one atm) is 373.15 K. At what temperature does
the vapor pressure of liquid water equal 380.0 Torr?
1 1ln ln
vap
f i
f i
Hp p
R T T
=
1atm = 760 Torr, 380 T = 0.5 atm
Find Tf= 354.3 K
Vapor pressure at different T?
Whats the vapor pressure of water at 25 C and 1 atm?
=
if
vap
ifTTR
Hpp
11)/ln(
Clausius-Clapeyron Eqn. Lets find
the vapor pressure at 298.15K. We
have vapH (100 C) = 40.7 kJmol-1 for
H2O(l).
3001.3
15.373
1
15.298
1
314.8
40700)1/ln( 298
=
=p So p298.15 = 0.03688 atm or 28.03 torr
How does this change when the pressure changes from 1 to 10 atm?
Need to think about how the chemical potential changes with pressure.
For a gas (ideal): For a liquid:
p
RTdpgd =)( dPVld m=)(
Where dP is the applied P
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Vapor pressure change with T (contd)
Liquid and vapor in equilibrium, we can equate these
eqns. Integrate:
Here, we have pi= 28.03 torr, Vm= 18.069 cm3 and we increase pressure
from 1 to 10 atm.
dPlVldp
RTdpgd m )()()( === =
f
i
f
i
p
p
m
p
p
dPlVp
RTdp)(
PlVpRT mp
pf
i= )(ln
Quite a small change!
(Increase)
torr22.28006669.103.28
006647.015.298314.8
910325.10110069.18
03.28ln
36
==
=
=
f
f
p
p
Example
trs
trs m
H
T V
=
trs
trs m
dp S
dT V
=
Clapeyron Equation:VT
H
dT
dp
sub
sub
sub =
)(
etc.
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Example, Contd
V
V
H
H
dT
dpdTdp
sub
vap
vap
sub
vap
sub
=
)(
)( subV =vapV=
volume of gas
78.7kJ/mol6.0
kJ/mol)0.67.40(
)(
)( =
=
=
H
H
dT
dpdT
dp
vap
sub
vap
sub
Mixtures
16
Energy of simple mixture: non-reactive, binary
* ?A A
= Change of due to mixing
Liquids, solutions, Raoults and Henrys laws.
Component A conventionally the solvent, B the solute. Ideal,
Ideal Dilute solutions.
Approach this material from several angles...
Changes in chemical potential in the mixture
drive all processes.
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17
Generalized fundamental equation
...A A B BdG dn dn = dG SdT Vdp
= At constant n
At constant p,T
However at constant T, p, dG also equal to maximum non-expansion work
,max ...add A A B Bdw dn dn =
This work can be achieved by chemical rxn, e.g. via a battery.
Electrical work results from changing chemical composition.
...A A B BdU TdS pdV dn dn = Similarly:
A
TnVSA Bn
U=
...,,,
Then:
...A A B BdG SdT Vdp dn dn =
Combining the two
A
VSpTAn
G=
,,,
Gibbs-Duhem Equation
18
A A B BG n n = Since and theJdepend on composition, for abinary mixture G may change by
However, we know that
A B A A B B
A B
G GdG dn dn dn dn
n n
= =
A A B BdG d n d n = A A B B A A B Bn dn ddn nd =
So 0A A B Bn d n d =
Gibbs-Duhem Equation(for a binary mixture at constant p, T)
BA B
A
nd d
n = .....BA B
A
ndV dV etc
n=
It is true for any molar quantities, for binary mixture, we have
More general form of G-D Eq. is
0j j
j
SdT Vdp n d =http://en.wikipedia.org/wiki/Gibbs-Duhem_relation
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Gibbs-Duhem Equation
19
0A A B Bn d n d =
Gibbs-Duhem Equation
(for binary mixture at constant p, T)
Significance is that the chemical potential of one component of
a mixture cannot change without affecting the chemical
potentials of the other components.
http://en.wikipedia.org/wiki/Gibbs-Duhem_relation
Review: Partial Molar Quantities
Concept of partial molar quantity and the relation can be applied to
any extensive state function: G, A, H, U, V, S, Cp etc at constant T,p
http://www.chem.neu.edu/Courses/1382Budil/PartialMolarQuantities.htm
Integral
V=nAVA+nBVB+..
G=nAA+nBB+..
S=nASA+nBSB+..
M=nAMA+nBMB+..Mass
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Change of due to mixing General expression of Gibbs energy for mixture
21
f A A B BG n n = 0A A B Bn d n d =Gibbs-Duhem
At const T, p
What is the difference between in its pure state (J* )and in
mix (J) ?
* *
i A A B BG n n =
* lnJ J JRT x = Ideal solution,
* ln lnoA A A A ART a RT a = =
*
*ln lnoBB B B B B
B
KRT a RT a
p = =
Real solution
What is the standard Gibbs energy change for mixing?
mix ln lnideal A A B BG nRT x x x x =
mix mix
E idealG G G= Use excess function
Chemical Potential and Mixing
*
*ln JJ J
J
pRT
p
=
Chemical potential drops for mixture
22
Real solutions
(chi) is the mole fraction here.
Summary, more details later
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G-D equation, Rauolt, Henrys Law
*
pa x
p= =
pa x
K= =
Treat as solvent
Treat as solute
*
*ln AA A
A
pRT
p =
*
*ln BB B
B
pRT
p =
p p x=
How to find?
0A A B Bn d n d =
0A A B Bx d x d =
B BA B
B B
d dx x
dx dx
=
Using Gibbs-
Duhem relation
Change of due to mixing General expression of Gibbs energy for mixture
24
f A A B BG n n = 0A A B Bn d n d =Gibbs-Duhem
At const T, p
What is the difference between in pure state ( )and in mixture () ?
* ?A A =
* *
i A A B BG n n =
*
*ln AA A
A
pRT
p
=
How to express this in terms of mole fractions: xA, xB,as we did for ideal gas mixtures?
Goal-I
component
A in solution
component
A pure liq.
component A above
soln. and abovepure liq. (ideal gas
assumed)At equilibrium, the chemical potential of each
component has the same value in each phase
where it is present
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25
Ideal solution: Raoults Law
Ideal solutions: both solute and
solvent obey Raoults LawMixture of benzene and toluene
*
*
AA A A A
A
pp x p or x
p= =
Ideal solutions obey Raoults Law:
i.e. the partial vapor pressure relative to
p* equals the molar fraction in the liquid.
pA*
*
*ln AA A
A
pRT
p
=
* lnA ART x=
mole fraction in liq.
Raoults Law Example
The vapor pressure of benzene is 53.3 kPa at 60.6 C but it fell
to 49.2 kPa when 51.2 g of an involatile organic compound was
dissolved in 500 g benzene. Calculate the molar mass of the
compound, assume the solution is ideal.
* and BB B B B
B A
np x p x
n n= =
where B = benzene and A
= solute
* *( ) orB B B B BB A
A B B
n p n p pp n
n n p
= =
Hence:A
A
A
mn
M=
* *( ) ( )
A B B A BA
B B B B B B
m p M m pM
n p p m p p= =
Subs. In:
MA=96.0 g mol-1
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Vapor Pressure Example
Consider a mixed solution of 1-propanol (1) and 2-propanol (2). Itsideal at all concentrations. Given p1
*=20.9 torr, p2*=45.2 torr at 25 C,
calculate the total vapor pressure and composition of the vapor for
x2=0.75 at this temperature.
Ideal, so p(total) = p1 + p2 = x1p1* + x2p2
*
= (0.25)(20.9) + (0.75)(45.2) = 39.1 torr
Daltons law of partial pressures:
y1(mole fraction of 1 in vapor) = p1/p(total) = x1p1*/p(total)
= (0.25)(20.9)/(39.1) = 0.13Similarly y2 = (0.75)(45.2)/(39.1) = 0.87
Note: y1+y2=1 and the vapor has more of the more volatile component
Another example
The Henrys law constant for a solute B in a solvent A is
8.21x103 kPa. What is the vapor pressure of B in a
solution of molality = 0.25 mol/kg? The molar mass of
the solvent A is 74.1 g/mol.
BBB Kxp = Weve been given K in mole fraction (pressure)units and we need to convert it to solve the
problem.BBB Kbp =
0.25 mol/kg is 0.25 moles of B in 1000g of A. 1000g of A
contains 1000/74.1 or 13.5 moles. Mole fraction of B is hence
.0182.0)5.1325.0(
25.0 ==Bx BBB Kxp =
Therefore kPa1491021.80182.0 3 ==Bp
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29
Departure from Raoults Law & Ideal Dilute Solutions
When mixture components are structurally different, see strong
deviation from Raoults law. But when1) The major component is nearly pure (its x1), it still obeys Raoults law
2) The minor component is nearly absent (its x0), its pressure obeys Henrys law
then this solution is called an ideal dilute solution.
Note: even when solute x 0, some solutions do not follow Henrys law
KB is some empirical
constant
30
Henrys Law
For an ideal-dilute solution: the solvent obeys
Raoults law and the solute obeys Henrys law
For real solutions at low concentrations,
although the vapor pressure of the solute is
proportional to the mole fraction, the
constant of proportionality is not the vapor
pressure of the pure substance*when 0,B B B B B Bp x K x K p=
KB is a measured slope as xB0 with
dimensions of pressure.
The solute behaves entirely differently from when its pure state,
since its molecules are surrounded by solvent molecules.
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Units
Henrys law constants are often expressed in terms of
molality:
Molality is the molar concentration per mass of
solvent, has units of moles/kg.
Compare Molarity, the molar concentration per
volume and has units of moles/dm3, for example.
Mole fraction is dimensionless, KB has
units of pressure.B B B
p x K=B (solute)
B (solute) BBB Kbp = Molality bB has dimensions of mol kg-1
,KB has units of Pa.kg.mol-1
Henrys Law Example
The mole fraction of CO2 in the atmosphere has risen from
0.000314 in 1960 to 0.000397 in 2013. Use Henrys law to
calculate the change in concentration in the oceans (in molality)
at 25 C, assuming K(CO2) = 3.01x103 kPa.kg.mol-1 in water at
25 C.
Henrys Law: pB = bBKB (concentration in molality)
The partial pressure of solute in vapor is its concentration in solution times
its Henrys law constant. So:
p(CO2)= 0.000314 x 101.3 kPa = 0.0318 kPa.
b(CO2) = 0.0318 kPa/ (3.01x103
kPa.kg.mol-1
) = 0.01057x10-3
mol.kg-1
For the higher concentration:
p(CO2)= 0.000397 x 101.3 kPa = 0.0402 kPa.
b(CO2) = 0.0402 kPa/ (3.01x103 kPa.kg.mol-1) = 0.01336x10-3 mol.kg-1
Increase of about 26%.
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33
More detailed summary for real solution
B B Bp x K=*AA Ap x p=
Ideal dilute solution: xA1 and xB0
The activity is an effective mole fraction
*
AA A A
A
pa x
p= =
* *ln lnA A A A A ART a RT x = =
Activity:
Activity coefficient:Aa
A
Solvent
As 1, , 1A A A Ax a x *o
A A =Standard state Limiting behavor
* *
*ln lnAA A A A
A
pRT RT x
p = =
* *
* *ln lnB B BB B B
B B
p K xRT RT
p p = =
lno
B BRT x= *
*lno BB B
B
KRT
p = wherelno
A ART x= *o
A A =where
B
B B BB
p
a xK = =
Activity:
Activity coefficient:Ba
B
ln lno o
B B B B B BRT a RT x = =
*
*lno BB B
B
KRT
p =
Standard state Limiting behavor
As 0, , 1B B B Bx a x
Solute
34
The Validity of Raoults & Henrys Laws
Can you read the figure?
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35
Colligative Properties
There is no reference in these properties to the identities of the
solutes and/or solvents - strictly dependent upon composition
The simplest example is 1+2 system:
(1) a pure solvent phase, which may be vapor, liquid, or solid
(2) a solution phase.(3) an interface between the two phases that is not crossed by the solute
Increase of boiling point
Osmosis(water purification)
Decrease of freezing point
Solubility(dissolve salt in water)
Eq. of solvent in: pure s & mix l
Eq. of solvent: in pure g & mix lEq. of solute in: pure s & mix l
Eq. of solvent in: pure l & mix l
, ,( ) ( ) ln AA s A lT T RT x =
, ,( ) ( ) ln AA g A lT T RT x = , ,( ) ( ) ln BB s B lT T RT x =
* *
, ,( ) ( ) ln
AA Ap p RT x
=
Chemical Potential of Solvent
Presence of solute decreases
(solvent), resulting in reduced
freezing point and increased boiling
point.
It is an entropy effect, because it
occurs even for ideal solutions. When
solute is present, there is an
additional contribution to the entropy
of the liquid. Hence a smaller changein entropy on vaporization.
For b.p. change, look for the temperature at which chemical potentials
of vapor and solvent molecules are the same at 1 atm pressure. Same
for melting
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37
Elevation of Boiling point
Boiling at T* Boiling at T
*
* *
( ) ( ) 1 1vap vapvap m
T TH
T T T T
=
*1 1
ln vap m
A
Hx
R T T
=
* * *
, ,( ) ( )
A g A lT T =
*( ) 0vap T =Pure liquid:
*
( ) ( ) ln AA l A lT T RT x = ( )
ln vap
A
Tx
RT
=In solution:
2
p
H
T T T
=
2
vap vap m
p
H
T T T
=
Gibbs-Helmholtz equation:
2
ln vapA Hd x
dT RT
=
Integral:
*
( )A g
T=
*2vap m
BH Tx
RT
Further simplification gives:
*2
;B
vap m
RTT Kx K
H = =
See
justification
5.1
39
Depression of freezing point
T depends on the solvent, change is biggest for solvents with high boiling
points. For practical applications, mole fraction of B is proportional to molality, b
(mol kg-1), in dilute solutions, and Kb is the empirical boiling-point constant
T =Kbb;K
b =
RT *2
fusH
mA
A =mol/kg for
solvent
*
, ,( ) ( ) ln
AA g A lT T RT x =
boiling Same argument applies also for freezing
*
, ,( ) ( ) ln AA s A lT T RT x =
(same ideas)
*2
;B
fus m
RTT Kx K
H = =
constant, K
*2
;f ffus m
RTT K b K
H A = =
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40
Osmosis
A+B AWhat is in balance?
pp+
gh =
*
,( ) ,( )( ) ( )A l A lp p = *
,( )( ) lnA l Ap RT x=
* *
,( ) ,( )( ) ( )
p
A l A l m m
p
p p V dp V
=
lnm A BV RT x RTx = /BB RT B n V = =
Vant Hoff Eq.
= molar concn.xB=nB/nA, approx.
solventparticle
Liquid(solid)/vapor
Solution/vapor:boiling point
Solution/solid:freezing point
Solution/solid: solubility
Solution/solvent: osmosis
, ,( , ) ( , ) lno o
A A o
pT p T p RT
p
=
2
ln vapHd p
dT RT =lnvap
o
pR
T p
=
Clausius-Clapeyron equation for
sublimation or vaporization
Vant Hoff equation
Applicability: dilute solution
Applicability: dilute solution
Applicability: dilute solution
Applicability: dilute solution
*
, ,( , ) ( , ) ln AA AT p T p RT x =
( )ln
vap
A
TR x
T
= 2
ln vapA Hd x
dT RT
=
*
, ,( , ) ( , ) ln AA AT p T p RT x =
( )( )ln
fusconA
TTR x
T T
= = 2
ln fusA Hd x
dT RT
=
*
, ,( , ) ( , ) ln BB BT p T p RT x =
( )( )ln
fusconB
TTR x
T T
= =
2
ln fusB Hd x
dT RT
=
* *
, ,( ) ( ) ln AA Ap p RT x =
*
, ( ) lnm AA p V RT x=
lnm A BV RT x R Tx =
Similarities
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Quiz (9th November, 2012)
What is the freezing point of a 0.2 molal solution of
sucrose in water? [fusH (H2O)=6.02 kJmol-1,
Tfus(H2O)=273.15K]
What is the value of Kffor an organic solvent, molar mass
= 84g, whose Tfus= 279.6 K and fusH = 2.68 kJ mol-1? If a
0.2 molal solution of hexane in this solvent was prepared,
calculate the freezing point depression.
Solutions
Approximate for small T-T* and small xB, then:
BfB
mfus
xKxH
RTT =
=
2*
This is our working equation. But we see the amount of solute is given in
mole fraction here, and we need to convert to molal concentration, i.e. mBmoles/kg of solvent. Remember:
1-kgg10001000AB
BA
BB
Mm
mM
mx
= For dilute solutions, where MA is the molar
mass of the solvent in g and mB
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Solutions, Contd
Now we can just substitute in the information we have:.
We find the freezing point depression in a 0.2 molal solution of sucrose is -0.37 K,
so the freezing point is 273.15 - 0.37 = 272.78 K.
For the organic solvent, we find
Kf= 8.314 x 279.62/2680 = 242.5 K
For 0.2 molal solution, T = 0.2x242.5x84/1000 = 0.2x20.3 K = 4.1 K, so
Tfus = 275.5K
Example
The addition of 100g of a compound (A) to 750g of CCl4 (B)lowered the freezing point of the solvent by 10.5K.Calculate the molar mass of the compound.
Note:Kf (CCl4) = 30 K kg mol-1
B
BB
n
mM = Depression of freezing point :
AH
RTK
bKT
mfus
f
f
=
=
2*
b is molality (mol/kg) of solute,
A=mol/kg of solvent
BAB bmn =
Hence:
Tm
KmM
A
fB
B
= and g/mol108.3K5.10kg75.0
)molkgK30(g100 21
=
=
BM
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Solubility example
The enthalpy of fusion of napthalene is 18.80 kJ mol
1
and its meltingpoint is 81C. Calculate its ideal solubility in benzene at 25C.
301.0
20.115.298
1
15.354
1
314.8
108.1811ln
20.1
3
*
==
=
=
=
ex
TTR
Hx
B
fus
B
B here is the solute,
naphthalene
Naphthalene dissolves until its chemical potential equalizes in the solution
and the vapor. Hence we can use the Gibbs-Helmholtz/modified Clausius-
Clapeyron equation:
Assume x(Nap) n(Nap)/n(Bz), so in 1000 g of solvent:
n(Nap) (0.301)(1000/78) = 3.85 mol. Hence molality is 3.85.
M(Nap) = 128 g, so 3.85 mol = 493 g in 1000g benzene.
Actually, xB=0.3 corresponds to closer to 5.5 mol/1000g
47
Constituents and Components
Constituent:A chemical species that is present in a system. For example,
a mixture of water and ethanol has 2 constituents
Component:A chemically independent component of the system. The
number of components in a system, C, is the minimum
number of independent species needed to define the
composition of all of the phases present in the system
When no reaction takes place, Constituents = Components When a reaction can occur, the number of components is the
minimum number of species which specifies the composition
of all of the phases (or: smallest number of independently variable
chemical species to describe the composition of each phase)
Number of Phases,P = 2
Number of Constituents = 3
Number of Components, C = 2*
*CaCO3
can be expressed in terms of 2 components in two different
phases from the stoichiometry of the reaction
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Examples
N.B., if additional HCl or NH3 were added to the system, decomposition
of the NH4Cl would not give the correct gas phase compositions, and
either HCl or NH3 would be invoked as a 2nd component
*NH3 and HCl are fixed in stoichoimetric proportions by the reaction, and
compositions of both phases can be specified by NH4Cl
Number of Phases, P = 2
Number of Constituents = 3
Number of Components, C = 1*
Consider the thermal decomposition of ammonium chloride:
Number of Phases, P = 2 Number of Constituents = 3
Number of Components, C = 3*
*At room temperature, O2 (g) and H2 (g) do not react to form water, so
they are not in equilibrium: regarded as independent constituents
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Variance and Degrees of Freedom
The variance, F, is the number of intensive variables in a system that
can be changed independently without disturbing the number of phases
in equilibrium.
In a single-component, single-phase system (C=1, P=1) the pressure
and temperature may be changed independently without disturbing the
number of phases in equilibrium:
F = 2, system is bivariant, or has two degrees of freedom
If two phases are in equilibrium in a single-component system (C=1,
P=2) (e.g., a liquid and its vapor), the temperature (or pressure) can be
changed, but there must be an accompanying dependent change inpressure (or temperature) to preserve the phases in equilibrium
F = 1, system has one degree of freedom
If three phases are in equilibrium (C=1, P=3) (e.g.: s, l and g), neither the
T nor p can be changed, otherwise the equilibrium is not preserved.
F = 0, system has no degrees of freedom
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Phase Rule
2F C P=
By J. W. Gibbs
1. Count the number of intensive variables (p and T count as 2)
2. Specify composition of the phase by mole fractions of C-1 components
x1, + x2, ++ xC,=
Since there are P phases, total number of composition variables is P(C - 1),
and total number of intensive variables is P(C - 1) + 2
3. At equilibrium, chemical potential of J must be same in every phase
J, = J, = .... for P phases
There are P - 1 equations of this kind for each component J. If there are C
components, total number of equations is C(P - 1)
4. Each equation reduces our freedom to vary any of the P(C - 1) + 2 intensive
variables, so the number of degree of freedom is
F = P(C - 1) + 2 - C(P - 1) = C - P + 2
Degrees of freedom for several phases in equilibrium
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Understand the changes given phase diagram
a In gas phase, cooled at constant p, F = 2
(for fixed p, F=1,T can be varied at will)
b Liquid appears at the phase transition
(at Tb), and F = 1. (for fixed p, F=0, T = Tb)
c Lowering T takes liquid to single phase
liquid region, F = 2, (for fixed p, F=1,T
can be varied at will)
d Liquid-solid phase transition at Tf, with
F =1, and at constant p, T is not under ourcontrol
e Lowering T further results in single solid
phase with F = 2, (for fixed p, F=1,
T can be varied at will)
Consider pure waterWhat happens going from a to e?
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Two Component SystemFor two component system, C = 2, and F = 4 P.
3 free variables for single phase, use T, P, x (mole fraction).
If either T or p is held constant, remaining variance is F = 3 P, 2 variables Phase diagrams of P & x for stable phases at constant T
Phase diagrams of T & x for stable phases at constant p
pB*
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Composition of the Vapor
Compositions of liquid and vapor in equilibrium are not necessarily the
same, vapor should be richer in the more volatile component
A Ap y p= B Bp y p=
For an ideal mixture, partial pressures may be expressed in terms of
mole fractions in the liquid (see equations on previous page)
*
* * * , 1
( )
A AA B A
B A B A
x py y y
p p p x= =
If pA*>pB*, yA>xA (vapor richer in
more volatile component).
The total vapor pressure as
composition of the vapor:
* *
* * *( )
A B
A B A A
p pp
p p p y=
A(l)+B(l)
A(g)+B(g)
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Combining the Vapor & Liquid diagrams
Better understand the phase transition between vapor and liquid by combining
the two previous diagrams into one. The horizontal axis shows the overall
composition, zA, of the system:
Liquid & vapor
coexist
p
A(l)+B(l)
A(g)+B(g)
* * *AB B A
p p p p x=
* *
* * *( )
A B
A B A A
p pp
p p p y=
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What happens as p is varied
Line of constant overall composition: isopleth
pA*
pB*
All points above the solid diagonal correspond to a system under high pressure where
there is only a liquid phase (zA = xA, composition of liquid)All points on lower curve correspond to a system with low pressure where only the
vapor phase exists (applied pressure less than the vapor pressure, so zA = yA)
Points between lines are systems where two phases exist: 1 liquid & 1 vapor
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Interpretations of the Diagrams
Lowering the pressure on a liquid by drawing out a piston:
start at point a: Here, F = 2 and P = 1; only the liquid phase exists; changes to system
do not affect overall composition, so system moves down vertical line that passes
through a, until point a1 is reached (pressure reduced top1)
The horizontal axis shows the overall composition, zA, of the system:
p3
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The Lever Rule
Can get n/n
T-x and p-x diagrams summary
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2 component system: F=2+2-Phase
Variables: p, T, x
p,T diagram
TB*
TA*
T,x diagram
p,T,x
A(l)+B(l)
A(g)+B(g)
xA,yA,zA
p,x diagram
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Critical Solution Temperatures
Tuc
61Weak compound
Critical Solution Temperatures, contd
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Phases
a. Ideal solution behavior.
b. A low (T)-boiling azeotrope
c. A eutectic point
d. Congruent melting
e. Incongruent melting
f. Partial miscibility in 1 or more phases
g. Compound formation
h. Complete miscibility in all phasesA+AB L+AB
AB+BL+ B
Solid-Liquid Phase Diagrams
A2BAB2
Eutectic Point
for A+A2B
system
Compound
formation
L+A
L+A2B AB2+B
AB2 melts
L+B
peritectic point
Peaks represent
compound formation.
Increasing temperature
(up) along dotted line
includes incongruent
melting as it crosses thehorizontal line. Here the
AB2 melts but there is
excess B, so the melt
contains L and B (solid).
A2B+AB2
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Completely Miscible Solids
TB*
TA*
T,x diagram
Looks like a gas/vapor diagram, but now solid/liquid
liquid
solid
Liquid and solid in equilibrium with
Proportions given by lever rule