1
Mixtures and Solutions
Partial Molar Quantities
Partial molar volume
The total volume of a mixture of substances is a function of the amounts of both
substances (among other variables as well). V  V  n1 , n 2 
Thus the change in volume of a mixture can be written as
 V 
 V 
dV  
 dn1  
 dn 2
 n1  n 2
 n 2  n1
The change in volume due to a change in the amount of a substance is known as its
partial molar volume.
 V 
V1  

 n1  n 2
 dV  V1dn1  V2 dn 2
The molar volume for a one-component system is the inverse of the density and is
independent of the amount of substance (it is an intensive quantity).
V
V 1

n 
However, when considering a mixture, the partial molar density can be greater than,
equal to or less than the molar density. The partial molar density can even be
negative as occurs when magnesium or aluminum salts are added to water.
58
18
56
16
54
14
0
0.2
0.4 0.6
XE tOH
0.8
1
Partial Molar Volume of EtOH (cm 3/mol)
Partial Molar Volume of EtOH (cm 3/mol)
The partial molar volume is not the same as the molar volume: V  V
2
Example: Calculate the total volume of solution when 0.050 mol of MgSO4 is added
to 1.00 kg of water. The partial molar volume of MgSO4 depends on its
concentration with the following expression: VH2O  18.063 cm3/mol
VMgSO4  69.38  m m   0.070  cm3 mol
V  VMgSO4 n MgSO4  VH2O n H2O
 69.38  0.050 m m  0.070  cm3 mol  0.050 mol  18.063cm3 mol 1000 g  18.02 g mol
  0.069 cm3   1002.38cm3  1002.31cm3
Other partial molar quantities
Other quantities such as partial molar heat capacities are useful for thermodynamic
work with solutions.
 Cp 
Cp1  

 n1  n 2 ,n3 ,n3
Partial molar Gibbs free energy
As seen in the last chapter, the partial molar Gibbs free energy at constant
temperature and pressure is the chemical potential
 G 
1  

 n1 T,p,n 2 ,n3 ,
So the total Gibbs free energy of a mixture can be found in terms of the chemical
potential.
G  1n1  2 n 2  3n 3  4 n 4 
3
Gibbs – Duhem equation
The chemical potentials of the components of a mixture are not independent of each
other. Change in the chemical potential depends on the changes of the other chemical
potentials.
Let us examine the Gibbs free energy of a two-component mixture.
G  1n1  2 n 2
The change in the Gibbs free energy (at constant temperature and pressure) can be
written in terms of changes in the chemical potentials and changes in the amounts.
dG  d  1n1  2 n 2   n1d1  1dn1  n 2d2  2dn 2
However recall that the Gibbs free energy is a function of temperature, pressure and
amount.
G  G  p, T, n1 , n 2   dG  Vdp  SdT  1dn1   2dn 2
 dG T,p  1dn1   2dn 2
Comparing the two expressions for dG leads to the Gibbs – Duhem equation.
n1d1  1dn1  n 2d 2   2dn 2  1dn1   2dn 2
n 2 d 2
n1
Similar relationships hold for the partial molar volumes and partial molar entropies of
a mixture.
 n1d1  n 2 d 2  0  d1  
dV1  
n 2 dV2
n1
dS1  
n 2 dS2
n1
Changes in one partial molar quantity depend on the changes of the other partial
molar quantities.
4
Activities and ideal solutions
Consider a solution of a volatile solvent with a nonvolatile solute. It is an
experimental fact that the vapor of the solvent above the solution will be reduced as
more solute is added to the solution.
For the solvent at equilibrium, the chemical potential of the vapor is equal to the
chemical potential of the liquid. (“A” is a customary label for the solvent when
examining solutions.)
A  g   A  l 
The chemical potential of the vapor is generally lower than its standard state.
p 
A  g   A  g   RT ln  A 
p 
Thus the chemical potential of the solvent in solution can be written in terms of the
chemical potential of its vapor.
p 
A  l   A  g   RT ln  A 
p 
For a pure solvent, a similar expression can be written for the solvent vapor in
equilibrium with the pure liquid. (The “*” indicates pure solvent.)
 p* 
*A  l   A  g   RT ln  A 
p 
Examining the difference between the chemical potential of the liquid in solution and
the chemical potential of the pure liquid yields
 p*A
  


g

RT
ln
 
  A 
 
p
p 
 p* 
p 
 RT ln  A   RT ln  A   RT ln  *A 
p 
p 
 pA 
p
 A  l   *A  l    A  g   RT ln  A
p



The difference in the chemical potential of the solvent can be expressed in terms of
activity.
p 
A  l   *A  l   RT ln  *A   RT ln a A
 pA 
5
Thus activity is
A  l   *A  l 
p 
ln a A 
 a A   *A 
RT
 pA 
If the activity of the solvent in the solution is equal to the mole fraction, the solution
is called an ideal solution as it follows Raoult’s law.
aA  xA
PA
 pA  x p
*
A A
PA0
Properties of ideal solution
Volume
 G 
Recall that 
V
 p 
Then
0
XA
1
1
  G    G  1


V  V1 
 V1


 
n1  p  p  n1  p n1
p
 A  l   *A  l   RT ln x A

 A  l   *A  l   RT ln x A 
p 
  A  l   *A  l  
p
 V1  V1* 
 RT ln x A RT  ln x A

p
p
 ln x A
 0  V1  V1*  0  V1  V1*
p
The equality of the partial molar volumes implies that in an ideal solution, all the
molecules are the same size.
6
Enthalpy
Recall the Gibbs-Helmholtz equation:
    G
  H
   G  H
  2  

 
 
 
n1  T p T n1  T 2 
 T p T  T 
H
   G
   1   21
T
 T p T
H
   
   1   21
T
 T p T
*
 H A  H*A   RT ln x A RT  ln x A
    A  l    A  l  





 
2
T
T
T
 T 
 T

 ln x A
 0  H1  H1*  0  H1  H1*
T
The equality of the partial molar enthalpies implies that in an ideal solution, all the
interactions between molecules are the same. That is,
Hsolvent solvent  Hsolutesolvent  Hsolutesolute
Aside: Very often the solute – solute interactions are assumed to be zero.
Hsolutesolute  0
Compare the microscopic conditions of an ideal solution to that of an ideal gas.
volume
intermolecular forces
ideal solution
nonzero, all the same
nonzero, all the same
ideal gas
none
none
7
Thermodynamics of Mixing
Gibbs free energy of mixing
The chemical potential of a substance changes when it is part of a solution and the
difference between the two chemical potentials is what is used to define the activity.
Let G be the Gibbs free energy of a mixture
G  n11  n 22  n 33 
G* be the Gibbs free energy of all of the components before mixing.
G*  n11*  n 2*2  n 3*3 
Then the Gibbs free energy of mixing can be defined as the difference between two
quantities.
G mix  G  G*   n i  i  *i    n i RT ln a i  nRT  x i ln a i
i
i
i
The Gibbs free energy can be rewritten in terms of the mole fraction by recalling that
n i  n x i , where n is the total number of moles of all substances in the solution.
G mix  RT n i ln a i  nRT x i ln a i
i
If the solution is ideal, then
i
G mix  nRT x i ln x i
i
Entropy of mixing
 G 
Recall that 
  S
 T  p
Let us taking the derivative of the Gibbs free energy of mixing to find the entropy of
mixing.
 

G mix  nRT  x i ln x i 

T 
i

G mix


nRT  x i ln x i
T
T
i
Smix  nR  x i ln x i
i
8
Enthalpy of mixing
Recall the Gibbs – Helmholtz equation
   G  H
  2 
 
 T p T  T 
Apply the equation to the Gibbs free energy of mixing.
G mix

T
nRT  x i ln x i
i
T
 nR  x i ln x i  Smix
i
 G mix H mix



nR  x i ln x i  0
2
T T
T
T
i
H mix
 0  H mix  0
T2
Therefore an ideal solution has no enthalpy of mixing. The mixing of an ideal
solution is strictly an entropic process.
Colligative Properties
Colligative properties are properties of solutions that depend on the amount of solute
and not the identity of the solute (at first approximation).
Adding solute decreases the chemical potential of the solvent.
A  l   *A  l   RT ln a A  RT ln x A
Lower the chemical potential of a liquid component in a solution, lowers the freezing
point and raises the boiling point.

solid
liquid
solution
gas
Tm
Tm
TbTb T
9
Freezing point depression
At equilibrium (at the melting point) 1  l   1  s  [Subscript “1” labels the solvent.]
The chemical potential of the liquid is related to its mole fraction in solution.
A  l   *A  l   RT ln x A
 1  s   1*  l   RT ln x1
1*  l   1  s 

  ln x1
RT
However, consider that G fus  1*  l   1  s 
G fus
  ln x1
RT
Take the temperature derivative, considering the Gibbs – Helmholtz equation.
 G fus


ln x1 
T RT
T
 ln x1
1  G fus

R T T
T
 ln x1 H fus

T
RT 2

 ln x1
1 H fus

2
R T
T
Integrate the equation assuming Hfus is constant. (Excellent assumption considering
the temperature range involved.)
Tm*
1
Hfus
Hfus  1
1 
dT  ln1  ln x1  

2
*

RT
R  Tm Tm 
Tm
 d ln x  
1
x1
ln x1 
H fus  1
1 

*

R  Tm Tm 
Change the emphasize from the amount of solvent to the amount of solute (labeled “2”),
remembering that x1  x 2  1
ln 1  x 2  
Hfus  Tm  Tm* 


R  Tm* Tm 
Assume that Tm*  Tm so that Tm* Tm  Tm*2 Also, from a Taylor series, ln 1  x 2    x 2
Hfus  Tm  Tm* 
x 2 


R  Tm*2 
10
Rearranging yields
Tm*  Tm 
x 2 RTm*2
H fus
Consider a relatively dilute solution such that n2 << n1. Then x 2  n 2 n1
The moles of solute can be calculated in terms of molality (c2) and mass of the
solvent, n 2  c2  m1 .
The moles of solvent can be calculated in terms of mass of the solvent and its molar
mass, n1  m1 M1 .
Thus
Tm 
x 2 RTm*2 n 2 RTm*2
c2  m1RTm*2
c M RT*2


 2 1 m
H fus
n1H fus  m1 M1  H fus
H fus
M1RTm*2
Tm 
c2  k f c2
H fus
The kf is called the cryoscopic constant and it is unique for each solvent.
kf 
M1RTm*2
H fus
At equilibrium (at the boiling point) 1  l   1  v 
From here, a similar analysis can be done to find the boiling point elevation as
Tb  k bc2
where kb, the ebullioscopic constant, is
kb 
M1RTb*2
H vap
11
Osmotic pressure
Osmosis – solutions in contact with a semipermeable membrane that allows flow of
only solvent through the membrane.
Osmosis

At equilibrium, the chemical potential of the solution “on the left” of the
semipermeable membrane is equal to the chemical potential of the solution “on the
right” of the membrane. Assume the solution “on the right” is actually the pure
solvent.
1  T, patm  , x1   1*  T, patm 
The chemical potential on the left has a concentration dependence.
1  T, patm  , x1   1*  T, patm    RTln x1
Thus,
1*  T, patm   1*  T, patm    RT ln x1
1*  T, patm    1*  T, patm   RTln x1  0
The difference in chemical potentials can be cleverly rewritten as an integral.
1*  T, patm     1*  T, patm  
patm 

d1*  T 
patm
 G 
  
However, since 
 V    V 
 p T
 p T
patm 

patm
d  T   RT ln x1  0 
*
1
patm 

 d   Vdp
V1*dp  RT ln x1  0  V1*p
patm
V1*  patm    patm   RTln x1  V1*  RTln x1  0
Recall our trick with the freezing point depression,
 ln x1   ln 1  x 2     x 2   x 2
patm 
patm
 RT ln x1  0
12
V1*  RT ln x1  RTx 2  RT
V1*  RT
n2
n1
n2
n1
 V1*n1  n 2 RT  V  n 2 RT   
n2
RT
V
  c RT
Another one of the van’t Hoff equations.
The special case of ionic solutions
Because colligative properties are independent of the specific nature of the solute, no
distinction is made between a molecule in solution and an ion in solution.
Thus, when taking into account the colligative properties of ionic solutions, the
dissociation of the ionic compound must be considered.
0.12 M NaCl  0.12 M Na   0.12 M Cl  0.24 M particles
0.45M K2SO4  0.90 M K  0.45M SO2  1.35M particles
4
The proper way to write the van’t Hoff equation for osmotic pressure is
  i c RT
where “i” is the called the van’t Hoff factor which describes the amount of
dissociation of electrolytes (strong and weak).
Example: 4.02 mg of human insulin is dissolved in 10.7 mL of water. The osmotic
pressure of the solution is 1.25 torr at 37 C. Calculate the molar mass
of the protein.
Calculate concentration of solute
1atm

760 torr
c

  6.47 x105 M
0.08206
L

atm
RT
 310 K
mol  K
1.25 torr 
Calculate moles of solute
n  c  V  6.47 x105
mol
 0.0107 L  6.92 107 mol
L
Calculate molar mass
4.02x10 3 g
m
M 
  4160g mol
n 6.92x10 7 mol