Chapter 07: Simple Mixtures
The thermodynamic description of mixtures:
Partial molar quantities
The thermodynamics of mixing
The chemical potentials of liquids
The properties of solutions:
Liquid mixtures
Colligative properties
Activities:
The solvent activity
The solute activity
The activities of regular solutions
Assignment for Chapter 07
• 7.2(b),7.6(a),7.10(b),7.15(a),7.20(b),7.22(b)
• 7.2,7.7,7.13,7.16,7.19,7.20
Simple Mixtures Studied in This
Chapter
Non-reactive: No chemical reaction would occur.
Binary:
x A  xB  1
Non-electrolyte: the solute is not present as ions.
Concentration Units
• There are three major concentration units in use
in thermodynamic descriptions of solutions.
• These are
– molarity
– molality
– mole fraction
• Letting J stand for one component in a solution
(the solute), these are represented by
• [J] = nJ/V (V typically in liters)
• bJ = nJ/msolvent (msolvent typically in kg)
• xJ = nJ/n (n = total number of moles of all species
present in sample)
Mass
m = n (MM)
Solute
Moles
Volume
n = m/MM
(MM = molar mass)
m(g) = n (MM) n = m/MM
Solvent m(kg) = m(g)
1000
Related to
mass sol'n =
nsol'n =
Solution msolute + msolv nsolute + nsolv msol'n through
density
Exercise
What mass of glycine should be used to make 250 mL of a solution of
molar concentration 0.15M NH2CH2COOH(aq)?
Mass
Moles
Volume
n = (0.15 mol/L)
m = (0.0375
(0.250L) =
(MM = 107 mol) (107
g/mol)
g/mol) = 4.01 g 0.0375 mol
Solute
Solvent
Solution
250 mL =
0.250 L
Exercise Calculate the mole fraction of sucrose
in an aqueous sample of molality 1.22 mol kg-1.
Mass
Solute
Water
(MM =
18.015 g/mol)
Solution
..
Moles
Volume
n = 1.22 mol
m = 1 kg = n = (1000 g) x (1
mol/18.015 g) =
1000 g
55.51 mol
nsol'n = 1.22 mol +
55.51 mol =
56.73 mol
Therefore, x sucrose = 1.22/56.73 = 0.0215
Partial Molar Volume
 V
V J  
 n J



 P,T ,n'
Unit: L/mol or mL/mol
 V 
 V


dV  
dn

A

 n

n
 A  P,T ,nB
 B


dnB

 P,T ,nB
 V A dnA  VB dnB .
V  n AV A  n BV B
Exercise
• Use the figure in slide 8 to calculate the density of
a mixture of 20 g of water and 100 g of ethanol.
• Solution: First calculate the mole fractions.
– 20 g H2O = 1.11 mol; 100 g EtOH = 2.17 mol
– xH2O = 0.34; xEtOH = 0.66
• Then interpolate from the mixing curve (next
slide):
– VH2O = 17.1 cm3 mol-1; VEtOH = 57.4 cm3 mol-1
• Then plug the moles and partial molar volume
– (1.11 mol)(17.1 cm3/mol) + (2.17 mol)(57.4 cm3/mol)
=19.0 cm3 + 125 cm3 = 144 cm3
• Finally, the total mass is divided by the total
volume: 120 g/144 cm3 = 0.83 g/ cm3
Exercise: the Interpolation
•
.
.
.
read
water
here
read
EtOH
over
here
Illustration: Ethanol and water ( 1 kg at 25C)
2
3
V / mL  1002.93  54.6664b  0.36394b  0.028256b
b: molality
The partial molar volume of ethanol:
-1
VE / mL mol

V
/
mL
(
) p,T , nW
b
 54.6664 0.72788b
 0.084768b 2
Partial Molar Gibbs Energy
 G
 J  
 n J



 P,T ,n'
G  n A A  nB  B
Fundamental Equation of
Chemical Thermodynamics
dG  Vdp  SdT 
dG  Vdp  SdT   AdnA   B dnb  ...
At constant pressure and temperature:
dG   Adn A   B dn B  .........
dWadd, max  dG
Therefore,
dWadd,max   A dnA   B dnB  ....
(non-expansion ) Work can be done by changing the composition.
The Wider Significance of the Chemical
Potential
G  U  pV  TS
dU   pdV  Vdp  SdT  TdS  dG
dU   Adn A   B dn B  .............
 U
 J  
 n J



 S ,V , n '
 H
(a )  J  
 n
J




 S , P ,n '
 A
(b)  J  
 n
J



.

V ,T ,n '
(Classroom exercise)
The Gibbs-Duhem Equation
dG   A dn A   B dn B  n A d A  n B d B
n A d A  n B d B  0
 n J d J
0
nA
B  
d A
nB
J
Gibbs-Duhem equation applies to ALL partial molar quantities:
 n dX
J
J
J
0
X=V, μ,H,A,U,S, etc
Using The Gibbs-Duhem Equation
Aqueous solution of K2SO4:
VK2 SO4 / cm3mol  32.380 18.216b1/ 2
Given the molar volume of water at 298K is 18.079 mL/mol,
find the partial molar volume of water.
n A dV A  n B dV B  0
n
dVB   A dVA
nB
nA
VB  VB  
dVA
nB
*
dVA
 9.108b1/ 2
db
n dV
dVA  VB*   A ( A )db
nB db
A  K 2SO 4 , B  water
n
VB  VB*   A
nB
nA
 VB  
(9.108b 1/ 2 )db
nB
*
b
nA
, M B : molar mass of water
nB M B
VB  VB*  9.108M B  b1/ 2db  VB*  6.075M Bb3 / 2  18.079 0.1094b3 / 2
VB  18.079  0.1094b3 / 2
The Thermodynamics of Mixing
Is mixing ( composition change) spontaneous?
p
  0  RT ln
p0

 
Gi  n A A  nB  B


P
P
 n A   A0  RT ln   nB   B 0  RT ln 
P0 
P0 


Gi  n A  A0  RT ln p  nB  B 0  RT ln p

 

G f  nA  A0  RT ln pA  nB B0  RT ln pB
p
p
mixG  nART ln A  nB RT ln B
P
P
pJ
 xJ
p
mixG  nRTxA ln xA  xB ln xB 
mixG  0

Mixing of two perfect gases or two liquids that form an ideal solution:
mixG  0
Exercise
• Suppose the partial pressure of a perfect gas
falls from 1.00 bar to 0.50 bar as it is consumed
in a reaction at 25 oC. What is the change in
chemical potential of the substance?
• Solution: We want J,f - J,i But J,i = Jo so we
want J ! Jo = RT ln (pJ/po)
=
(2.479 J/mol) (ln 0.50) = -1.7 kJ/mol
Exercise: at 25C, calculate the Gibbs energy change when the
partition is removed
Gi  n A  A0  RT ln p A  nB  B 0  RT ln pB .
 


 


G f  nA  A0  RT ln p f , A  nB B0  RT ln p f , B .
n A  3 mol, nB  1 mol
p A  3 p, p B  1 p
pf  2 p
p f , A  3 p, p f , B  1 p
2
2
p f ,A 
p f ,B 


Gmix  G f  Gi  n A  RT ln
  nB  RT ln

3
p
p




n A RT ln 1  nB RT ln 1
2
2
Gmix  6.9kJ
Other Thermodynamic Mixing
Functions
Entropy of mixing
  mixG 
 mixS  

 T  P,, n A ,, nB
mixG  nRTxA ln xA  xB ln xB 
mixS  nRx A ln x A  xB ln xB 
Entropy of mixing of perfect gases
mixS  nRx A ln x A  xB ln xB 
x A  xB  1
xA
1 x A
mixS  nRln x A (1  x A )
Other Thermodynamic Mixing
Functions
Enthalpy of mixing
G  H  TS
H  G  TS
For perfect gases,
mixG  nRTxA ln xA  xB ln xB 
mixS  nRx A ln x A  xB ln xB 
mixH  0
Understandable?
Ideal Solutions
Pure substance:
 A*   A0  RT ln p A*
Solute:
Vapor pressure
 A   A0  RT ln p A.
 A   A*  RT ln
pA
p A*
.
Raoult’s Law and Ideal Solutions
• Raoult found that the partial vapor pressure of a
substance in a mixture is proportional to its mole
fraction in the solution and its vapor pressure
when pure: pJ = xJ pJ*
• Any solution which obeys Raoult’s law
throughout its whole range of composition (from
xJ = 0 to xJ = 1) is an ideal solution.
p A  x A p A*  A   A*  RT ln x A
Raoult’s Law
p A  x A p A*
Raoult’s Law: Examples
p A  x A p A*
 A   A*  RT ln x A
Exercise
• A solution is prepared by dissolving 1.5 mol C10H8
in 1.00 kg benzene. The v.p. of pure benzene is
94.6 torr at this temperature (25oC). What is the
partial v.p. of benzene in the solution?
• Solution: We can use Raoult’s law, but first we
need to compute the mole fraction of benzene.
MM benzene = 78.1 g/mol, so 1.00 kg = 12.8 mol.
• xbenz = 12.8 mol / (12.8 mol + 1.5 mol) = 0.895
• pbenz = xbenz p*benz = (0.895)(94.6 torr) = 84.7 torr
Exercise
• By how much is the chemical potential of
benzene reduced at 25oC by a solute that is
present at a mole fraction of 0.10?
• Solution: We want benz = benz ! *benz
• But benz = *benz + RT ln xbenz so benz !
*benz = RT ln xbenz
• And if xsolute = 0.10, then xbenz = 0.90
• Thus benz = (2.479 kJ/mol) (ln 0.90)
.
= ! 0.26 kJ/mol
Raoult’s Law: Molecular Interpretation
rateof vaporazation :
v A  kx A
rateof condesation :
cA  k' pA
At equilibrium:
cA  vA, pA  k xA
k'
x A  1  p A*  k
k'
*
pA  xA pA
Raoult’s Law:
Molecular Interpretation
p A  x A p*A
Ideally Dilute Solutions
• Solutions of dissimilar liquids can show strong deviations from Raoult’s law
(green line at left) unless a
substance has x > 0.90.
• However, the v.p. usually
starts off as a straight line.
• This is embodied in Henry’s
law,
– pB = KB xB
– and the slope of the line is
the Henry’s-law constant.
• A solution which obeys
Henry’s law is called an
ideal-dilute solution.
Henry’s Law:
Molecular Interpretation
Solute
Solvent
In ideally dilute solution,
the solvent is almost like
a pure liquid whereas the
solute behaves very differently
from a pure liquid.
Exercise
• The v.p. of chloromethane at various mol fracs
in a mixture at 25oC was found to be as follows.
• x
0.005 0.009
0.019
0.024
• p/torr
205
363
756
946
• Estimate the Henry’s law constant for chloromethane at
25oC in this particular solvent.
• Solution: The v.p.’s are plotted vs. mol fraction.
The data are fitted to a polynomial curve (using
a computer program that has a curve-fitting
function) and the tangent (slope) is calculated
by evaluating the first derivative of the polynomial at xCHCl3 = 0.
1000 800
600
400
200
0.005
0.01
0.015
How to estimate the Henry’s law constant.
0.02
x
0.025
Exercise
• What partial pressure of methane is needed to
achieve 21 mg of methane in 100 g benzene at
25oC?
• Solution: From Table 7.1, KB = 4.27 × 105 torr.
mol CH4 = 0.021 g / 16.04 g mol!1 = 0.0013
mol. mol C6H6 = 100 g / 78.1 g mol!1 = 1.28
mol
• xCH4 = 0.0013 mol / (1.28 mol + ~0 mol)
.
= 0.0010
• pCH4 = KCH4 xCH4 = (4.27 × 105 torr)(0.0010) =
436 torr = 4.3 × 102 torr
Validity of Raoult’s and Henry’s Laws
• The v.p. of propanone (acetone,A) and
trichloromethane (chloroform, CHCl3, C) at
various mol fracs in a mixture at 35oC was
found to be as follows.
•
•
•
•
x
0.0 0.2 0.4 0.6 0.8 1.0
pC/torr
0.0 35 82 142 200 273
pA/torr
347 250 175 92
37
0
Confirm that the mixture conforms to Raoult’s law for
the component in large excess and to the Henry’s law
for the minor component. Find the Henry’s law
constants.
Validity of Raoult’s and Henry’s Laws: Result
Liquid Mixtures
For two liquids (A+B) forming an ideal solution:
Gi  n A  A  nB  B
0

0

G f  n A  A  RT ln x A  nB  B 0  RT ln xB 
0
 mixG  nRT x A ln x A  xB ln xB 
 mix S  nR x A ln x A  xB ln xB 
[The ideality of a solution holds well if interactions A-A, B-B are
the same as A-B]
For real solutions, that’s not true.
Excess Functions and Regular Solutions
S E   mix S   mix S ideal
V E   mixV   mixV ideal
H E   mix H   mix H ideal
G E   mixG   mixG ideal
Excess Enthalpy
H E   mix H   mix H ideal
Excess Volume
V E   mixV   mixV ideal
Excess Functions and Regular Solutions
S E   mix S   mix S ideal
A regular solution is the one which
is not ideal solution but has zero
excess entropy:
S E  0, H E  0
H E  nRTx A xB
w

RT
Excess Enthalpy
H E  nRTx A xB
w

RT
Excess Gibbs Energy
 mixG  nRT x A ln x A  xB ln xB  x A xB 
Colligative Properties
• Vapor pressure lowering is one of the four
colligative properties of the solvent. These are
properties that depend only on the number concentration
of particles of solute and not at all on the nature of the
solute. (They do depend on the nature of the solvent.)
• The other three are:
– Boiling-point elevation Tb = KbB
– Freezing-point depression Tf = KfbB
• where bB is the molality of the solute B in the solution
– Osmotic pressure
P . [B] RT
• where [B] is the molarity of the solute B in the solution
Boiling-point elevation
Tb = KbbB
Freezing-point depression Tf = KfbB
where bB is the molality of the solute B in the solution
Elevation of Vapor Pressure
Exercise
• Estimate the lowering of the freezing point of the
solution made by dissolving 3.0 g sucrose in 100 g
of water.
• Solution: From Table 4.3, Kb for water = 1.86 K kg
mol-1. 3.0 g sucrose = 3.0 g / 342 g mol!1 = 0.0088
mol sucrose
• bsuc = 0.0088 mol / 0.100 kg = 0.088 mol kg!1
• Tf = KfbB = (1.86 K kg mol-1)(0.088 mol kg!1) .
= 0.16 K
• Note: New f.p. = 0.00oC ! 0.16 oC = !0.16oC
• Note also: It is assumed that pure water freezes.
Exercise
• The heights of the solution in an osmometry experi-ment on
a solution of an enzyme in water at 25oC were as follows.
• c/g dm!3 0.50
1.00
1.50
2.00
2.50
• h/cm
0.18
0.35
0.53
0.71
0.90
• The density of the solution is 0.9998 g cm!3. What is the
molar mass of the enzyme?
• Solution: At each concentration, P is found from P =
rgh. [B] is found from P / RT and MM from c(g/L) / [B]
(mol/L)
• The molar masses are plotted against concentration and
extrapolated to zero conc’n.