Chapter 9
Solutions and Multicomponent systems
9.1
Ideal Solution
2. On the other hand, for a pure•  = 1 system
•1 () = •1 () = 1 () +  ln 1• 
An ideal solution (IS) is one that obeys Raoult’s Law.
3. Solving for 1 () and inserting this into 1),


 =  •  =  • 
1 () = 1 () = •1 () +  ln 1•
(9.1)
1

= •1 () +  ln
In either version, the statement of  is that the presence of the other component(s) does not influence the
escaping tendency accept by the factor one would expect based on fractional occupancy of the liquid surface
regions, and thus escape points.
1 1•
1•
or
1 () = •1 () +  ln 1 
(9.2)
4. The pure• liquid is the liquid reference state ,
1 () ≡ •1 () = •1 ().
∴
1 () = •1 () +  ln 1 
(9.3)
5. The concept behind these equations is simple. The
chemical potential is just equal to that of the chemical potential of a pure liquid scaled DOWN in proportion to the log of the mole fraction. Why do I
say scaled down? What is the sign of ln  ? Also
appreciate that the ln term is unitless and the 
gives the correction term the units of energy/mole.
O=
O= X
O= m
Remark 1 As you pollute a liquid  with  the chemical potential of  drops via the  ln   There is nothing but statistics at play here. This is the result you
would get if the hetero-interactions ( − ) were the
same as the homo-interactions ( −   − )
9.1.1
The IS is like IG
1. The thermodynamic quantities upon mixing:
Definition 1  is the vapor pressure of i above a mixed
liquid+ and • the vapor pressure of i above the pure
liquid .
Consider a vapor in equilibrium with an IS.


∆
 = ∆ = ∆ = 0
P
 ln   0
∆
 = −
P

 ln   0
∆ = +
2. Volumes are additive
1. Each component is in phase equilibrium. For the
solvent1
1 () = 1 () = 1 () +  ln 1
61
(a) Consider the mixing of two components  and
 Before mixing the total volume is,
•
•
 =  
+  
(b) As the volume is a homogeneous function of
degree  = 1 the volume after mixing is,
  =   +  
(c) So what are the  ?
2
(d) Boltzmann said
ˆ
  =  ln  =  ln (+)!
!! 
(e) Now I need to introduce an approximation for
lnN! called Stirling’s approximation, a result that we prove later.

( 
 )
 

) = ( 
 ≡ ( 
 ) =

(d) If we look at the (ideal) expressions for the
chemical potential, we find that the mixing
term ( ln ) has no pressure dependence.
 •
•

∴ ( 
 ) =  =  = (  ) ⇒   =  
ln  ! ≈  ln  − 
(f) Returning to
ˆ
  =  {[( + ) ln( + ) − ( + )]
−[ ln  −  +  ln  − ]}
=  {[ ln  +  ln  ]
−[ ln  +  ln ]}


= − {[ ln 
+  ln 
]}
3. Enthalpy of mixing ∆
 = 0
(a) We derived the G-H eq..
[
(  )
] =  
(1 )
(GH)
(b) The expression for  when we divide by 
has NO temperature dependence in the mixing term. Thus the values of   one for each
component, are unchanged by mixing and thus
there is no heat evolved upon the mixing of
ideal solutions1 .
(c) If the enthalpy of pure  and  in the mix
are the same (at all temperatures), the differentials must be the same. This tells us that the
contribution to the heat capacity is additive,
•
•
 () =  
+  

P
 ln   0
4. The entropy of mixing ∆
 = −
(g) Using number fractions,
ˆ
  = −  { ln  +  ln  }
ˆ
 = −
X
 ln  
(9.4)
5. As ∆
 = 0 the free energy of mixing, for an

ideal system is ∆
= − ∆ 
Remark 2 Benzene and Toluene mix in all proportions
because they form an  and ∆ of the mixed phase
is always less than any partition, including the unmixed!

The plots of ∆
 and ∆ are valid for  because
all interactions are equivalent.
This is the statistical (as well as IG) result. You are
given the task of picking up a child’s mess of  red
and  blue balls strewn all over the floor. For now 9.2
Equilibrium between a S-IS
assume the balls are numbered and that there are
separate boxes for the Blue and Red balls with num- Ideal solutions form between similar chemical species.
bered positions for each. Consider the “options” for Either both are polar or both are not. The solutions that
a proper pick-up job.
are closest to ideal are non-polar–non-polar mixtures,
e.g. toluene and benzene or napthalene and benzene.
(a) First you select one of the  =  +  balls The latter can also be used as to help imagine an equiliband put it in its proper place. You continue rium between a solid, formed from the component with
until they are all put away. There are ( + )! the higher   and an ideal solution with the minority
possible sequences you could have done this.
species (solute) the same as the solid. (e.g. Nap.(s) in eq.
(b) Now imagine that the numbers on the balls with a solution of Ben. with Nap. dissolved in it. (For


were erased (i.e. that they were all equivalent reference: 
() = 5  and 
( ) = 80 )
(like atoms or molecules are of a given type.)
We can ask two questions about the equilibrium sitThe number of distinguishable sequences is far uation. How does the solubility (amount of Nap. that
less than ( + )! It is less by the number of will dissolve in Benzene) depend on: 1) the pressure
permutations of balls in each box.
and 2) the temperature ? By making this an equilib(c) Therefore the number of distinguishable pick- rium problem, we are requiring some solid to be present.
up sequences (for indistinguishable balls of a
2
1. ( 
given color) is
 ) =?
(+)!
At eq. 2 () = 2 (soln)
 = !! 
1 This is related to a result used later for a(T), that the only
enthalpy for an ideal solution is that for the ∞-dilute solution.
62
If the pressure is changed, the extent of dissolution
(2 ) of the saturated solution will change so that
where ∆2 and ∆2 are the entropy and enthalpy
change upon dissolution. However if the solution
is ideal, there is no enthalpy change in going from
a pure liquid to the mixed solution so ∆2 =
•2 () − 2 () =  . (For reasons I have never
been able to understand the “melting” direction is
usually called “fusion”. Go figure!)
the chemical potentials are, once again, equal. If the
chemical potentials of component 2 were equal in
the two phases before and after the pressure change,
the change must be the same for the solid and solution.
2 ()
→
=
2 (soln)
The solid can only experience a pressure change
while the solution can also change its composition.

( 2 )  = (
ln

2
 ) 

+(
2 () = 2 (soln) +

2 () = 2• () +

2

2
Remark 4 The above can also be interpreted in the
inverse fashion. How does the freezing point of the
solvent vary with change in its mole fraction? Just
ln

2
2 ) 2
2

2 
The results for  have been used in evaluating a)
the mole fraction dependence as well as b) in the
realization that the partial molar volume is simply
that of the pure• solute.
 2


invert the relationship: (  ln 
) = ∆1 ()
 As
1
this enthalpy change is always  0, T decreases as
the mole fraction of the solvent decreases.
9.3
Real Solutions
Rearranging one can solve for the  of the desired Now we want to deal with the type of deviations
partial.
from ideality commonly found when not dealing with
2 ()−2• ()
2
electrolytes2 . While we do not have simple laws that

2 =  ln 2 =

are valid throughout the concentration range, we do find
•
simple limiting laws.

∆2
 ln 2  2 () − 2 ()
. 0
) =
=
(


 
(9.5)
1. For the solvent1 in the limit 1 → 1  (the longdashed lines) is always valid,
Remark 3 The logarithm of the equilibrium mole

fraction changes in proportion to ∆2 , the change in
(9.7)
lim 1 = 1 1• 
1 →1
volume upon freezing, a small number. As the later
is usually  0 (an exception being H2 O) increases
2. For the solute2 in the limit 2 → 0 a proportionin pressure slightly DECREASE the solubility, an
ality with  exists but with a solvent dependent
effect that decreases with increasing temperature.
constant (short-dashed lines),
2
2. ( 
 ) =?

(9.8)
lim 2 = 2 2 
2 →0
You may have a gut feeling for how the dissolution
changes with temperature, but to be quantitative
we need an equation. Again the logic starts with a This proportionality is called Henry’s Law.

statement of eq. (2 = soln
Deviations from ideality are classified by whether
2 ) both before and after
the  change,
the deviation is + or − relative to RL. A − de→
viation (see figure), the type common when the hetero.
2 = soln
2 
interactions are stronger than the homo. interactions,
The solid can only change  but the solution can leads to a reduced fugacity (pressure) for both species.
also change in composition.
type
solvent
solute
Azeo
•
•

soln
soln
2
2
Ideal

=



=


No
1
1
2
2
( 2 )  = ( 
)  + ( 
)

1
2

2
2
+ dev 1 ⇒ 1 1• 2 ⇒ 2 2 2  2• Low T



soln
−2  = −2  + 2 2 
− dev 1 ⇒ 1 1• 2 ⇒ 2 2 2  2• High T
Rearranging to solve for,
soln
−2
2
2
9.4 Solvent/Solute ref. states
 or
2 =  ln 2 =

 ln 2
∆2
∆2  ∆2 ()
(
=
À 0
) =
=


 2
 2
(9.6)
Equilibrium between the phases requires

 =   expanding the RHS and adding 0,
2 Electrolytes
63
are far from ideal even at very low concentration.
•
 =  () +  ln  + [ ln • ]


Switching the numerators of the ln terms, a new reference state with the equilibrium fugacity of the pure
material (rather than 1 bar) is generated.
3. − deviations, can have high-boiling Azeotropes.
Example: Acetone - CHCl3
polar-polar
•
( = •),
1. With this reference state, 12
Hetero.  homo.
•
 = [ () +  ln  ] +  ln • 


 = •
 () +  ln


•
(9.9)
This reference is a real state and is always chosen for
the solvent1 and is a good choice for liquid solutes.
On the other hand, this reference is it a poor choice
for the solute when interested in dilute solutions3 .
For dilute solute studies, it is wiser to select a reference closer to the states of interest.
2. Two more possible solute2 reference states can be
identified using extrapolations of Henry’s law. By

inserting 2 = 2 2 into the expression above,
2
2 = [•
2 ()+ ln  • ]+ ln 2 +   
2
(a) An exceedingly unreal hypothetical state results when a state is imagined with a fugacity
equal to the Henry’s law extrapolation all the
way to 2 = 1 While this reference state, (the
[] in the equation above and labelled by “2 ”
and  =  ) is remote from the dilute states
of interest, at least it scales correctly for dilute
solutions.
2 = [
2 ] +  ln 2 +    
(9.10)
(b) A second hypothetical state, one which both
scales correctly and is close to dilute states
of common interest, is the HL extrapolation
of the fugacity only to unit molality 2 = 1.
With this ( = ) reference state,
+  ln
2 = 
2
9.4.1
2
+    
2
(9.11)
Non-idealities & Azeotropes Survey
1. More or less ideal, partial pressures linear on X.
2. + deviations, can have low-boiling Azeotropes.
Example: Acetone - CS2
polar-nonpolar
3 Because
homo.  Hetero.
reference state is so remote from the states of interest.
64
4. Azeotropes have vapor and liquid phases with the
SAME composition and as a consequence, distillation cannot pass through this composition.

(a) If IS:  =  • ⇒  =  • .
(b) Lets calculate the vapor pressure as a function
of both the liquid (X) and vapor (Y) mole fractions.
i.  (1 ) =?
 ≡   (by definition)

=  •
 •
 •
 ≡  =   or  =    
Therefore,
 (1 ) = 1 + 2 = 1 1• + 2 2•
= 1 1• + (1 − 1 )2•
= [1 (1• − 2• ) + 2• ]
= linear in 1 (or 2 ) the straight
line in the LHS figure.
ii.  (1 ) =?
1 ≡ 1 ; 2 ≡ 2 ;  = 1 + 2
(1 1• )
1 = [1 ( • −
• )+ • ] 
1
1. The liquid composition () is the top axis for the
LHS (total) Pressure plots and the bottom axis for
the RHS Temperature plots.
2. The vapor composition ( ) is the bottom axis for
the LHS (total) Pressure plots and the top axis for
the RHS Temperature plots.
3.  (vapor) approaches:
(a) Ideal or almost IS, vapor enriched in more
volatile.
(b) the azeotrope for +ve deviations (II) and
2
2
iii. We need to invert to get  ( ) but before
doing so lets consider the case where,
•
1 ( = 12 ) = ( • − •1)+2 •
1
•
2
2
1
•
•
1
=  • +
•  2 if 1  2 
1
2
That is, the vapor is enriched in the more
volatile component. Brandy anyone!
iv. We have  () = 1 (1• − 2• ) + 2• 
To get need  ( ), we can use 1 (1 )
from 
1 1 1• − 1 1 2• + 1 2• − 1 1• = 0
1 [1• 1 − 2• 1 − 1• ] = −1 2•
 2•
1 = [ • +(1• −
which can be in•
1
2
1 )1 ]
serted into the expression for  ( )
1 1• 2•
 2• 2•
}1 − { 1 []
}2
[]
1• 2• +2• 2• 1 −1• 2• 1
=
+ {}1 − {}2
[]
1• 2•
 (1 ) = [] = a curved function, the
v.  = 2• + {
(c) the pure material in -ve deviations (III).
4. Tie-line or Lever rule proof
lower curve on LHS figure.
(c) Phase separation (ideal). The system variables that lie between the lines, have coexisting
phases. Consider an isothermal expansion with
total composition  see figure.
i. All liquid with  = 
ii. Some vapor, enriched in volatile component, begins to form.

iii. ()() = 
= opposite side tie line.
iv. Mostly vapor, liquid depleted in volatile
component
v. All ,  = 
(d) Tie-line proof.
65
(b)
i. Material conservation (its got to be somewhere.)
() =   +  
ii. ( +  ) =   +   =   +  
 [ −  ] =  [ − ]
iii. This gives the tie-line or lever rule,
[ − ]

[ − ]
=
=


[ −  ]
[ − ]
9.5
(c)

P

 ( 
) =


  (  )
=
 ≡ (
(9.12)
Consider our two component system and the fact that
we should have added only one variable but we seemed
to have added two. At equiliubrium  , variations in
P

  = 1 1 + 2 2 = 0
Therefore the extra relation between the new variables
is
P

  = 0

  = 0
P
6. The equations given above define Partial Molar
Quantities.

)

(9.16)
   represent the differential increment of a function  attributable to one component, 
Gibbs-Duhem Equation
X
P
Example 2 The molar volume of pure water at 15
 is 180 ml/mole. If you add one mole of water
to a HUGE vat of a 50:50 mix of water:glycerol, the
volume only increases by 17.67 ml. The PM-Volume
of water is then 1767ml for  = 12.
Exercise  → 
9.5.1
The Gibbs-Duhem relation is a powerful tool and we will
(9.13) now exercise to show how the solute and solvent chemical
potentials are related. For example, if one component
A more useful form of this relation is derived below. (See obeys RL so must the other. The proof (that it takes
pg 18 of notes for an alternate derivation.)
two to tango) is below.

  = 0
1.  = (  )

 = ( 
 )   + (  )   +
P
=   −  +   
P
  =   
1. If component 1 follows RL (pressure version),
P

 (  )  6= 
1 = 1 1• ⇒ 1 = •1 +  ln 1 
2. The  equation for a two component system,
2. The consequence of  being a first order Homo. fxn
P
 P

  =   ( 
) =    thus,

P
P
  =   +   
3. Equating the two differentials, yields the GibbsDuhem, “sum rule”, relation:
X
X
  =
  = 0
(9.14)
For a C=2 system,
=2
=2
1
1 1 + 2 2 = 0 ⇒ 2 = − 
2 1 .
4. We (or you) will show that the GD can be written,
X
  ln  = 0
2
1 = − 
1 2 
1
3. But as 1 = 0 +   ln 1 =  ( 
1 )
1
2
−
1 2 =  ( 1 )
1
4. Rearranging, 2 = − 
2
However as 1 + 2 = 1
Thus, 2 = 
2
2 
1 = −2 
5. This can be integrated from a pure liquid 2 (2 = 1
1 = 0 2 = •2 ) to a solution with some arbitrary
amount of component 1 with 2 
R
R 2
2
2 =  1 2 
2 , which gives
•
2
2 = •2 +  ln 2 
6. From which it follows that,
(9.15)
2 = 2 2•
(IDEAL behavior).
And as they used to say in ancient Rome, “quad erat
demonstrandum”. (Of course the Romans were usually just proving something the Greek’s had already
proved. But neither knew what we just proved.)
5. Additional GD relations can be derived for other
degree 1 homo. fxns.
P
P

(a)
  (  ) =
   = 0
66
Remark 5 We have proven that given P1 =X1 P•1 it the chemical potential for ideal solutions to be used for
follows that P2 =X2 P•2  There are few solutions that non-ideal solutions.
are ideal. On the other hand, there are many that
are strictly regular (SR). SR ≡ the logarithm of the
 =  +  ln 
activity coefficient is quadratic in the mole fraction
 2
 =  +  ln 
(9.17)
of the other, i.e. ln 1 = 
2  with  a constant.
You should be able to prove that if component1 is
 2
This equation is the definition of activity and the activity
SR so is the other2 (i.e. ln 2 = 
1 )
coefficient  is introduced to relate  to some concentration variable,
9.6
Colligative properties - I
 =  ∗ []
There are the four colligative properties.
(9.18)
 → 1 in the ideal solution limit. For now, lets consider the mole fraction concentration scale ( =X) and
the reference state as the pure material.
2. Boiling-point elevation,  %
 =  +  ln 
3. Vapor-pressure lowering,  &
=  +  ln  +  ln  
=(Ref. energy)
4. Osmotic pressure, Π %
+(ideal reduction due to pollution)
+(change due to non-idealities).
These properties have one basic explanation. As
The terms in the expression are:
a solute is added to a solvent, the chemical potential of
the solvent is reduced. The first three of these properties
1. A reference energy.
are a simple reflection of the extended   region over
which the liquid phase is stable when polluted. Osmotic
2. A [con] term is an adjustment for the presence of
pressure is the extra pressure that must be applied to
other components. This term assumes the form of
return the chemical potential to the “unpolluted” value.
an  and is always  0 As you “turn on” or add
These properties are the response of the solvent (and the
other components, the chemical potential drops.
( − ) equilibrium to the addition of the solute. This
3. A non-ideal correction (knob), adds an energy which
response can either be changing the ( − ) equilibrium
deals with all molecular interactions.  = [0 ∞] ,
 (at fixed  ) or  (at fixed  )
with  = 1 being the ideal case. The numerical
value of this term is linked to the reference state
choice!
1. Freezing-point depression,  &
9.8
9.8.1
Review - Reference states
Gases
The reference state is the pure gas at 1 bar. If we are
dealing with gases that are close to ideal we can replace
 by 
9.8.2
9.7
Activities
G.N. Lewis introduced fugacity to provide a “stand in”
for pressure to allow for calculation of  with the functional form appropriate for ideal gases. He did not stop
there. He introduced the activity to provide a “stand
in” for concentration to allow the functional form for
Solvent
The reference state of the solvent is always taken as the
pure liquid ( = 1) solvent under its own vapor pressure
 • rather than 1 bar. Therefore,
R •
1. 
= (pure l) +   • 
1
R
The correction is −ve (as  •    ) and usually
very small, as the molar volume of condensed phases
is small. Therefore to high accuracy we can write,
67
2. 
=  (of pure liquid at T and P=1 bar)
1
9.8.3
3.  = 
+  ln  = 
+  (ln  + ln  )


The reader must understand which of the common reference states is being used.
4. •1 () = 1 = 1 and the activity would
decrease linearly, with mole fraction, if the solution
were ideal.
Solute
1. A real reference state based on Raoult’s law. This
is useful for liquid solutes for which the pure solute
() is real at the temperature of interest.
5. In real solutions,  can increase more or less quickly
than linearly, or vary in a complex fashion due to
molecular interactions.
(a) 
=  (of pure liquid at T and P=1bar)
2
= a physical state
6.  =  see the figure section a.
(b) •2 () = 2 = 1
(c)  =  =
a.
real
 
see the figure section
2. A (factitious - hypothetical) reference state based on
an extrapolation of the Henry’s law slope to 2 =1.
Note this is an extrapolation into a region for which
Henry’s law is extremely ill suited.
(a) 
= extrapolated unphysical state.
2
(b)  =  see the figure section b.
3. A reference state based on a Henry’s law extrapolation to 1 molality (or 1 Molar.) This reference state
is factitious but often very close to a real state.
(a) 
= extrapolated state - almost real.
2
+  ln 2 
(b) 2 = 
2

+
 ln 
= 
 +  ln  
2
(c)  =  see the figure section c
9.9
The Eq. Constant 
We have already shown how  is related to the standard

change in the Gibbs function ∆
 Now we will do
the same (borrrrring) for   This equilibrium constant
reverts to [] for ideal solutions.
1. Consider our favorite reaction,
( ) + ( ) ® ( ) + ( )

2.  =   +   +   +   
P
3. ∆ =  +  −  −  =   
4. Inserting the forms for  ,
∆ = ( +  ln  ) + ( +  ln  )
−( +  ln  ) − ( +  ln  )
P
 
=   +  ln  
 
 
68
9.11.1
5. When equilibrium is attained, ∆ = 0
P
 

≡   = − ln[  
∆
 ]
1. For the solvent, neglecting the pressure dependence
(as the liquid is almost incompressible),
 
 
1 ≈ •1 
6. Defining  ≡ [  
 ]
 

∆
= − ln 

2. For the solute, lets work with a molality reference
AND with the idealization (IS) that  = 1 This
means we are working along the Henry’s law line.
(9.19)
or
(a)
 ( ) = exp{−
h

∆
}


2
2
 =  +  ln 

2 
2 
(  ) = ( 
)
ln 2
+ (  
) 
(b) Using the  −  equation,
(9.20)
 
(a) If the mole fraction is the concentration scale
being used,
 =   =







 =   =


 

 

∗[
 


 



2 
2
) = − 2 , ( 
) = −  22 .
( 
(c) Furthermore,
ln 2
) = 0 if on the  line ( = 1)
(  
7. Equilibrium constants can be decomposed into ideal
and activity coefficient parts.


3. Therefore, − 22 = −  22 ⇒ 2 = 2 = ∞
2
] 
The only enthalpy along Henry’s Law line is that
Note I am using “” for two different quantities of the infinitely dilute solution. (That is, the reference
enthalpy2 is when 2 never interacts with another 2 in the
in this (and the next two) equation. Sorry!
solution.)
(b) If the molality scale is being used,
 
∗ [  
 ] 
9.11.2
 
(c) For Molarity (a sloppy concentration unit),
 = []  =
9.10


[] []
[] []
∗



 =  +  ln 

ln 
 
 
( 
) = ( 
) + (  
) 

−

ln 
)
− 2 = −  2 +(  
1. As
 
[  
 ] 
 
*a(P)
Activities, and thus equilibrium constants, depend on 
1. As  =  +  ln  


 −

 −

 2
=−
 −∞

 2 
3. The numerator,  − ∞
  is the enthalpy of dilution.
ln 
(  
) = −
so
ln 
ln 
= (  
) = (  
)
3. The terms in the difference on the  are the
condensed-phase molar volumes. The difference itself is between the actual and reference state molar
volumes, a very small number. The point here is,
as always, the pressure dependence is encoded in
molar volumes and for condensed phases, the molar
volumes are small and changes even less.
9.11
ln 
2. (  
) = −
4. As the concentrations are held fixed, the dependence
of  on  is completely carried by the nonidealities,
 ln 


( 
 ) = (  ) +  (  ) 



2. But ( 
 ) =  and (  ) = 
a( )
*a(T)
You should expect the  dependence to be encoded in
the reaction enthalpies. We have to start by proving
something about the reference enthalpies.
69
 −∞

 2
or
1 −∞
1
 2
∞
 −
− 2 22 
ln 1
) = −
(a) for the solvent: (  
ln 2
(b) for the solute: (  
) =
5. Therefore, measurement of the heat evolved by dissolution provide the logarithmic (or relative) change
in the activity.
Fig. Enthalpy of dilution: h(X).
9.12
More on Binary Systems
(a) Fluid of comp. X
1. GPR
3
2
C=2
 = 2 + C − P = 4 − P =
1
0
(b) Some of () starts to freeze out.
(P=1)
(P=2)

(P=3)
(P=4)
Definition 3 A single phase has well defined molar
volume  and can be classified as   or The distinction between the phases is done on the basis of
the extent of long-range order. (Gas has none, liquid some and solid substantial.) This can be made
formal (more formal anyways) with the help of radial distribution functions. These functions tell you
the probability of finding another unit at a distance
 from a unit fixed at the origin.
2. A prototypical − phase diagram with no solid
solutions or compounds, at fixed  is shown for
Ca silicates ( (3 )2 + 2 Si2 8 ) below.
The curves separating the liquid region from the
regions with a solid, are determined by the melting enthalpy. This is a C= 2 system. Therefore,
 = 4 − P Thus if P = 2 there are 2  0  If
YOU specify the pressure, you have only one 
remaining. If you also specify the composition, the
temperature is fixed or if you specify the temperature you have lost control over the composition
and equilibrium will dictate the composition of the
species in the “pot”. The nature dictated compositions are the curved lines in the phase diagram.
Note that the regions “captured” behind these lines
are P = 2 and therefore you can specify only  (in
addition to  )
Imagine cooling at fixed composition (X in the figure). This is an isopleth. Consider the diagram
above.
70
(c) () and fluid of comp  Note YOU do not
determine the composition of the fluid.  =
4 − 2 = 2 = [  ] THERE IS NO THERMODYNAMIC SUBSYSTEM WITH COMPOSITION  The amounts of the two phases [()
and ] are determined by a “lever rule”.
(d) “lever rule”⇒

()
=
1−
−
(e) Eutectic point. (Eutectic - Greek for “easily melting” - is for solids, what a low-boiling
azeotrope is for liquids.)
(f) Total freezing of separate solids. The temperature drop will halt as the solids freeze. This
halt in the isopleth is called a eutectic halt.
(g) At the eutectic point the  and  coexist in
equilibrium. This composition has the lowest
melting point of any mixture. The most common eutectic is solder (˜40%  + 60%)
3. A prototypical  −  phase diagram with solid
solutions, at fixed  is shown for Ag:Cu below.
Solid solutions form if variable composition is thermodynamically favorable.   is mostly    is
mostly  Note there is never a pure  or  (Ag
and Cu in the example.)
4.  −  WITH solid compound.
A compound has a fixed stoichiometry.
9.12.1
Why some liquids do NOT mix.
If the free energy of mixing ( + ) has the form
of the heavy solid and (if mixed) dashed line in the figure above, the liquid will phase separate into two liquid
phases of compositions  and  The line with a common
tanget is given by  = (  −  ) +  with the



(common) intercepts 
 =  on the LHS and  = 
on the RHS. That is, both components are in chemical
equilibrium. Therefore, IF you try to prepare a solution inside the “blister” region (with a value of  inside
the blister) - you will fail to create any material with
this composition. (With one more phase, you have lost
composition control!)
Compare this figure to that for  ( ). The
Maxwell construction is determined by a similar cotangent construction in ( ) This construction is the one
that makes the free energies the same in the two phases.
The problem we have just dealt with is the C = 2 version of the same problem. As the new problem is two
component, we had to use  ’s rather than  and thus
there are two Gibbsian conditions rather than one.


The figure above illustrates how miscibility gaps
usually close with increasing  The figure below examines the cosolution of simple alcohols in water. These
systems have an upper “consolution” temperature () ,
i.e. above a certain temperature, the materials are miscible in all proportions. Below () the liquid separates
into two phases ( and ). The first figure in the set
a) (vapor pressure vs mole fraction) illustrates that the
miscibility gap (in mole fraction) at fixed  increases
with complexity of the alcohol.

=  and 
 =  
C=1
 =  
71
9.12.2
Survey of results
*Solute
Upper consolution temperatures are (by far) the most If the solute is volatile and its vapor pressure can be
common. A few mixtures have lower consolution tem- measured (X2 = [0,1]), its activity(X2 ) can be directly
peratures and a very few have both (nicotine and water.) measured. If limited volatility prevents this from being
done throughout the full X2 range, a limiting procedure
can be employed.
1. As always, (in general),
2 () = 2 ()
2 () +  ln 2 = 2 () +  ln 2 
This presumes that the vapor2 is ideal and  → 
2. Now let us choose the reference state to be the 2 =
1 (close to real) Henry’s law extrapolation.

2 () = 2 () +  ln 2 
where 2 is the Henry’s Law extrapolation of vapor
pressure of 2 to the reference state at 2 = 1 The
denominator is just the reference pressure of 1 bar.

3. Using 2 in 1, ln 2 + ln 2 = ln 2

ln 2 = ln 2 − ln 2 = ln 2 ⇒ 2 =
2
2
2 
This is not a simple expression as the denominator
can not be measured as it is a fictitious state.
9.13
Determination of Activities
9.13.1
From Vapor measurements
4. However HL provides an extrapolation logic,
2
2 = lim2 →0 ( 
)
2
The above discussion should not be taken to imply
that this limiting slope procedure is the most common
If the component is volatile (i.e. there is phase equilib- means of determination of 2  By far the most common
method employs either vapor pressure measurements of
rium) 1 can be readily determined.
the solvent1  or one of the colligative properties, and use
of
the GD relation. (A procedure for electrolytes will be
1. From the phase equilibria ( − )
presented in a subsequent chapter.)
1 () = 1 ()
Solvent
•1 () +  ln 1 = 1 () +  ln 1
9.13.2
1
2. For the pure liquid and its vapor,
The colligative properties should be thought of as correlated motions in ( ) or ( ) to preserve equilibrium.
•
•1 () = •1 () = 1 () +  ln 1 
1
So
1 ()
=
•1 ()
−  ln
1•
1 
3. Inserting this into 1), cancelling and / yields,
•
•
ln 1 + ln 1 = ln 1 ⇒ ln 1 = ln 1 − ln 1 
1
1
1
1
Thus the activity is directly measurable from the
fugacity or pressure if the vapor is ideal.
1 =
From Colligative Properties
1  1


= • =
= 1
•
1
1
1
72
(b) Solving for the temperature drop,
Freezing-Point depression
 • 
Imagine an initial phase ( − ) equilibrium with pure 1
with  =[1 = 1 • ] which we pollute to get the final
state phase equilibrium with  = [1 , ]

1. Initially 1 = 
1 ⇒

1

=

1
 
 
Θ = − ln 1 ( ) ∗ [ ∆
]
(c) If the solution is relatively dilute,
ln 1 ≈ ln 1 = ln(1 − 2 ) = −2 − 12 22 −
and  ∼ • . ∴
2. Add solute and maintain equilibrium .

( 1 )
=
Θ ≈ 2

( 1 )
3. Assume the solid remains pure and the solute (pollution) is totally in the liquid phase. This means
that to maintain eq. the d1 ( ) must match the
d1 ( ). This calls for the chain rule.

[ 
(
•
1
 ) ]
+
(9.22)
(d) We can change concentration variables. As
2 = + ≈  where  is the number of moles in 1 of solvent.
=

 1
[ 
(  )] 
(• )2
∆


1
[ 
(
 )] 1 
1
Θ ≈
 (• )2
 ∆
(9.23)
4. We now employ the Gibbs-Helmholtz equation, (see
Osmotic Pressure
pg 46),
•

In this case the reduction in  from the addition of the
“pollution”, is compensated by increasing the pressure
The last term was simplified by noting,
on the (polluted) solution.
•
1
1



1 = 1 +  ln 1 = 1 +  ln 1 +  ln 1 
1 [  ] = 1 [  +  ln 1 ] = 1 [ ln 1 ]
As 1 & 1 would go down if additional pressure were
R   =
5. Now integrate 1=1 =
• along a convenient path.
not
applied.

ln 1
[− 12 ] = [− 12 ] + [ 
] 1 
1
First a) we go from  →  (constant ) and
then b)  →  (constant  )
R  •
R 1 
R  
−  • 12  = − 1 • 12  +  11 


 ln 1
1 1 
(a)  →  : 1 = 1  & [• →  ]
•

1 = 1
(b)  →  :  =   1 & [1 → 1 ]
6. −
R 
•
1

∗  2
−
=
R 1
•
1

1•  2
+
R (1 )
(1)
 ln 1 ( )
1 
1
7. Collecting the enthalpy terms on the LHS and just
integrating the RHS (appreciating that the activity
of the pure reference state is 1 - thus the ln is 0)
Z

∗
• − •
( 1 2 1 ) =  ln 1 ( )

(a)   =  
(9.21)
8. Now we start making approximations.
(a) If ∆ 6=  ( )

ln ( ) = ∆
( 1• −


(
= − ∆
• −
• 
1

1. Start with thermal and chemical eq. with pure solvent on both sides of the membrane.
(b) 
= 
≡ 1
1
1
(c)   =  
2. Add solute to RHS and maintain thermal and chemical equilibrium   ( =  + Π) increases.
)
(a)   =  
(b) 
= 
1
1
Θ

) = − ∆
(  •   )

(c)  =      =  + Π
where Θ ≡ • −   0
73
4. Dropping these terms, the GD equation becomes,
3. As nothing is done on the LHS,

1

1
= 0 ∴ in order to maintain equilibrium,
=0=
1
{( 
) 1
1
+
2
 ln 1 = − 
1  ln 2 or
1
( 
 )1  }
X
4. The above expression can be conveniently integrated along the orthogonal (dotted lines) path.
R 1  +Π
1 = 0
1
R  +Π
R 1 
= 1 1 (1 +  ln 1 )1 +  1 
where 1 =
1

( 
 )
(a)
1 ∼1
1
≈ ( 
 )1
R
5. As the first term in the first has no dependences,
R
R  +Π
1 ( )
0 =  1 1  ln
1 +  1 
1
=
 ln 1 ( ) = −
Z



P
=  ln 
= [1 ln 1 ] + (1 − 1 ) ln 2
(b) Taking the derivative wrt 1 


 ln 1

1 (  ) = [ln 1 + 1 ( 1 )]+
ln 2
ln 2
) − ln 2 − 1 ( 
)
( 
1
1
1
 ln 1
ln 2
= ln( 2 ) +1 ( 1 ) + (1 − 1 )( 
)
1
ln 1
ln 2
) + 2 ( 
)]
= ln( 12 ) +[1 ( 
1
1
 +Π
1 

(9.26)
5. Defining an “excess” FE of mixing,
P


≡ ∆ − ∆
=    ln  




1 [(  ) ]

 (ln  ) = 0
(9.24)

(c) The [] = 0 ∴
6. To get more useful and familiar forms:



1 (  )
= ln( 12 )
6. A simple test that the activity coefficients are therR  2
modynamically consistent comes from  1 
R1
R1 
 =  0 ln( 12 ) = 0
0
(a) Assume that the coefficient of isothermal com
pressibility [ ≡ − 1 ( 
) ] is so small that
1 =  −  ∆ ≈  
Thus
 ln 1 ( ) = − [ + Π −  ]
Or
 Π = − ln 1

as 
= 0 in either extreme. This suggests that a
plot of the ln of the ratio of  0  with mole fraction
must have + and − contributions which cancel
in area. The following plot of ln(1 2 ) vs X1 , shows
just such a check.
(b) If the solution is sufficiently dilute, Raoult’s
Law is valid and 1 → 1 
 Π = − ln 1
(c) The ln can be expanded using a Taylor series,
 Π = − ln(1 − 2 )
2
≈
Π ≈  [
9.14
− (−2 ) =  2 ≈  21 ∴
2
2
] =  [ ] = []
1 

(9.25)
*Thermo. consistency
Changes in 1 are mated to changes in 2 via 
1. 0 = 1 1 + 2 2 = 1  ln 1 + 2  ln 2 
2. Dividing by  ⇒ 1  ln 1 + 2  ln 2 = 0,
9.15
1 ( ln 1 +  ln 1 ) + 2 ( ln 2 +  ln 2 ) = 0
*Partial-Molar Quantities
We will consider a specific example of a partial-molar
3. We can get rid of the  ln  terms if we appreciate, volume in a moment. Before we launch into an example
1
2
2 2
I need to lay some background. Consider the volume of
1 = − 1 = − 1 2 
a mixture (1 and 2) with a total of one mole. What an
2
 ln 1 = − 
1  ln 2 
experiment can provide is the total volume (for the gross
mole) as a function of the mix, say (2 )
1 ( ln 1 ) = −2 ( ln 2 )
74
2. The tangent lines at  are  =  0  + 
1.   =  = 1 1 + 2 2 

recall  = ( 
)

(a) We can get the slope by differentiation.




 0 = 
=  − + ( 
− 
)+ 





(2 ) = (1 − 2 )1 + 2 2 
= 1 + (2 − 1 )2


+  

=  −  + (1 −  ) 


Note this is of the form of the experimental data.
But how does one separate the two  0 ?
2. 1 =

( 
)2
1
=


+  
]
=  −  + [ 


[]
=   − 
1 +2 )
( (
)2
1
(b) The  = 0 intercept is deduced by requiring,
( ) =  ( )
 ( −  ) +  = ( −  ) + 
∴  = 

1 +2 )
= ( (
)2 + (1 + 2 )( 
)2
1
1

2
=  ∗ 1 + (1 + 2 )( 
)2 ( 
1 )2
2
3. The last derivative is,
(c) The equation for the tangent line is therefore,
 = ( −  ) + 
With intercepts of (see section a):
[2 (1 +2 )
2
( 
)2
1 )2 = (
1
∴
= −2 (1 + 2 )−2 

)2 2 (1 + 2 )−2
4. 1 =  − (1 + 2 )( 
2
=−

( 
)2 2 (1
2
+ 2 )
 at  = 0
9.15.3

=  − 2 ( 
)2 
2
To measure v1 all you need to do is measure (2 ) and
use the linear equation above which requires evaluating
the derivative as a function of 2 
9.15.1
and
 at  = 1
−1
Using the difference (b)
1. We can change the function of interest to a difference quantity such as ∆
•
•
 ≡ ∆ =  −  
−  
•
•
= [] − (1 −  )
−  
•
•
= [  +   ] − {(1 −  )
−  
}
Method of intercepts.
= [(1 −  ) +   ] − {}
I want to prove that a common tangent in the   ()
plot defines the the joint equilibrium condition that:




 =  and  =  
2. Again consider the tangent line,  =  0  + 
and extract both the slope and intercept.
(a) The slope, by differentiation of 1 w.r.t.  ,
•
•
 0 = ( −  ) + 
− 
•
•
= [( − 
) − ( − 
)]
(b) The intercept can be deduced by requiring,
( ) =  ( )
•
•
 − (1 −  )
−  
•
•
=  [( − 
) − ( − 
)] + 
•
 −  =   −   + 
=   − (1 −  ) + 
=  −  + 
•

(c) Therefore the LHS intercept is,  =  − 
9.15.2
Using the PMQ (a)
1. The volume per mole can be written as a linear function of  
(d) Plugging into  we get the RHS intercept,
thus the two intercepts are,
•
at  = 1  = 0
 − 
.
•
 − 
at  = 0  = 1
3. IF a tangent line is tangent at two points (cotangent) they must share these intercepts and thus
values of the partial molar quantities,  and  
 =  = (  +   ) =   +   
( ) = (1 −  ) +  
=  ( −  ) +  
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