Key points For today’s lecture; EDR Ch 8.1-8.3, 8.10-8.11
• Set up and Read Phase Diagrams
•Ideal solutions
*
P

x
P
•Raoult’s Law:
i
i i
• Raoult’s Law for a Solvent of a Dilute Solution:
i (l )  i* (l , P* )  RT ln  xi 
•Henry’s Law for a solute.
Pi  xi K H ,i
Chemical Equilibrium for reactions involving ideal dilute solutions
(For solvents)
Raoult’s Law:
Pi  xi Pi*
Vapor pressure
above solution of
component i
(For solutes)
Henry’s Law: Pi
Vapor pressure
above solution of
component i
Vapor pressure of
pure component i
Mole fraction of
component i in
solution
 K H ,i xi
Mole fraction of
component i in
solution
Henry’s Law
constant

*
H 2O
 EtOH ( g ) PEtOH ( g )
*
*
ethanol
( g ), Pethanol
*
H 2O
(g) P
 H O ( g ) PH O ( g )
2
+
=
 H* O
*
ethanol
2
At equilibrium:

*
H 2O
(l )  
*
H 2O
(g)

(l )  
ethanolxethanol
 H O xH O
2
At equilibrium:
*
ethanol
2
*
ethanol
2
At equilibrium:
(g)
soln
soln
 EtOH
(l )   EtOH
(g)
 HsolnO (l )   HsolnO ( g )
2
2
Focus on one species as a solvent. The chemical potentials of the
gas and liquid phases must be equal at equilibrium.
Consider a solvent, initially pure, at a given T there will be a vapor
pressure above the liquid, or a partial pressure (if there is air above the
liquid). The partial pressure is determined by the balance (equality) of
the chemical potential of the liquid and the vapor (or gas).
*


P
solvent
0
*
0
0
i
(l , pure)  i (l )  i ( g , P )  i ( g , P )  RT ln  0 
P 
This defines the partial pressure of the gas above the liquid P*.
If we add a solute to the solvent, this lowers the chemical potential
of the liquid. Therefore the gas chemical potential must lower as
well to keep the two chemical potentials balanced.
 Pi 
0 
P 
isolution (l , i )  i0 (l )  RT ln  i   i ( g , P)  i0 ( g , P 0 )  RT ln 
Subtracting the top equation from the bottom one shows the
differences due to the presence of the solute.
 Pi 
 Pi * 
 Pi 
RT ln  i   RT ln  0   RT ln  0   RT ln  * 
P 
P 
 Pi 
i 
Pi
P*
or
Pi  i Pi *
This is Raoult’s Law
Applying the Ideal Solution Model to Binary Solutions
Assume both solvent and solute obey Raoult’s Law
*
*
PTotal   benzene Pbenzene
 1   benzene  Ptoluene
*
Pbenzene   benzene Pbenzene
*
Ptoluene  1   benzene  Ptoluene
This is a coexistence line for vapor and liquid. At fixed T
but X is varying.
Dalton’s Laws and the components
N
PTotal   Pi and Pi  yi PTotal (Dalton's Laws)
i 1
Pi   i Pi *
 Rauolt's Law 
Pi is the vapor pressure of i.
 i Pi
*
i
PTotal
N
PTotal    i Pi *
i 1
 yi PTotal
N
yi
1
yi
 * or, after summing
 *
Pi
PTotal i 1 Pi
The partial pressure of species i is the same from both
Dalton’s and Rauolt’s Laws. “y” are the mole fractions in the
vapor phase, and “x” are the mole fractions in the solution.
Using both of these laws simultaneously implies that the mole
fraction of i in the gas phase is in general not the same as the
mole fraction of the same species in the liquid phase.
What is the mole fraction of each component in the gas phase?
P1
x1 P1*
y1 
 *
Ptotal P2   P1*  P2*  x1
P1  x1 P1*
P2  x2 P2*  1  x1  P2*
y1 P2*
x1  *
P1   P2*  P1*  y1
Ptotal
P1* P2*
 P   P  P  x1  *
P1   P2*  P1*  y1
*
2
*
1
*
2
P1* Ptotal  P1* P2*
y1 
Ptotal  P1*  P2* 
We can find the composition in the gas phase based on the composition in
the liquid phase. If the pressure of the pure substance is very larger
(compared to the other pure substance) the mole fraction of that species will
dominate in the gas phase.
PTotal  
*
benzene benzene
P
 1   benzene  P
*
toluene
x1 P1*
y1  *
P2   P1*  P2*  x1
The mole fraction of benzene in the vapor, and the total
Pressure along the coexistence line are both functions of the
mole fraction of benzene in the solution in this benzenetoluene solution.
P – x phase diagram
xbenzene
ybenzene
xbenzene  or ybenzene 
Example Problem
An ideal solution is made from 5.00 moles of benzene and 3.25
moles of toluene. At 300 K, the vapor pressures of the pure
*
*
substances are Pbenzene
 103Torr and Ptoluene
 32.1Torr .
a) The pressure above this solution is reduced from 760 Torr. At
what pressure does the vapor phase first appear?
b) What is the composition of the vapor under these conditions?
Solution
a) The mole fractions of the components in the solution are
xbenzene = 0.606 and
xtoluene = 0.394. The vapor pressure above this solution is
*
*
Ptotal  xbenzene Pbenzene
 xtoluene Ptoluene
 0.606 103Torr  0.394  32.1Torr  75.1Torr
No vapor will be formed until the pressure has been reduced to
this value.
b) The composition of the vapor at a total pressure of 75.1 Torr is
given by
P1
x1P1* 0.606 103
y1 


 0.83
Ptotal Ptotal
75.1
ytoluene  1  ybenzene  0.17
Note that the vapor is enriched in the more volatile component.
Expressing Raoult’s Law for dilute solutes in terms of concentration
ni
concentration of solute i, ci 
Vtotal
where, Vtotal = Vsolvent +  Vi  nsolvent V solvent +  ni V i
solutes
for a dilute solution:
solutes
 n  n
i
solvent
solute
ni
ni
cV
ci nsolvent V solvent
i total
xi =



 ci V solvent
ntotal nsolvent nsolvent
nsolvent
Neglecting these extra terms is not as important as it might seem:
The approximations underestimate both the numerator and denominator,
The effects tend to cancel.
Chemical potential for a solute in a dilute solution:
i (l )  i0 (l , P0 )  RT ln  xi 

i (l )  i0 (l , P 0 )  RT ln ci V solvent

 V solvent 
ci


  (l , P )  RT ln 

RT
ln



1
L/mole
1
mole/L




0
i
0
ci


i (l )   (l , P )  RT ln 

1
mole/L


0
i ,c
0
Chemical potential of component i in its
standard state at 1 M and P0
Ideal dilute solution: A recap
System
Gas
Chemical Potential
 Pi 
i ( g , Pi )   ( g , P )  RT ln  
 P0 
0
i
0
Solvent
i (l )  i0 (l )  RT ln  xi 
Solute
i (l )  i0 (l , P0 )  RT ln  xi 
ci



1
mole/L


i (l )  i0,c (l , P 0 )  RT ln 
Raoult’s Law:
Pi  xi Pi*
Henry’s Law: Pi
 xi K H ,i
What happens with real solutions?
We define an activity of component I, ai, which acts like an
effective mole fraction. It is related to the real mole fraction
by an activity coefficient, gi:
ai
activity of component i
gi  
xi mole fraction of component i
For solvents, asolvent =g solvent xsolvent  xsolvent
ci
ci
ci
For solutes, ai  g i


-1
1 mole L
1 M c0
Pi
Pi
For gases, ai  g i 0  0
P
P
For solids, ai  1 (and pure liquids )
This results in:
System Chemical Potential
Gas
Solvent
i ( g , Pi )  i0 ( g , P0 )  RT ln  ai 
i (l )  i* (l , P* )  RT ln  ai 
As the solute becomes
more dilute:
Pi  0, then
ai  Pi / P 0  0
xi  1, then
ai  xi  1
Solute
i (l )   (l , P )  RT ln  ai 
0
i
0
xi  0, then
ai  ci /1 mole L-1  0
Real Solutions Exhibit Deviations from Raoult’s Law
Why we need Henry’s Law (for “ideal dilute solution”)
Gmixing  0
Smixing  0
Vmixing  0
H mixing  0
Chemical equilibrium of solution phase chemical reactions
vi

 products  ai 
0
G  G  RT ln 
vi

 reactants  ai 
At equilibrium,
vi

 products  ai 
0
G   RT ln 
vi

 reactants  ai 
Example:

   RT ln K a

eq
10 A  7 B  8C  9D
8
9

aC   aD  
0

G  G  RT ln 

  a 10  a 7 
B
 A

  aC 8  aD 9 
Ka  

  a 10  a 7 
B
 A
eq




Example: Consider a reaction among dilute solutes at
constant temperature and pressure
A B C  D
  aC  aD  
K a  

a
a
  A  B  eq
 [C ] / moleL1[ D] / moleL1 

1
1 
 [ A] / moleL [ B] / moleL eq
Example: What is the G per mole for the reaction below at
25 °C?
Zn(solid )  2H  (aq)  Zn2 (aq)  H 2 ( g )
PH 2  105 atm
 Zn 2   103 M
 H    5.0 M
K eq  5.3 x1025
This problem concerns dilute solutions of two volatile liquids A and B, At
50 °C, the vapor pressure of pure liquid A is 0.67 atm and the vapor
pressure of pure liquid B is 1.1 atm. You may assume Raoult’s law is
valid for the solvent species in this solution and Henry’s Law is valid for
the solute species in this solution. The gas phase is ideal.
A) At 50 °C, a solution with mole fraction Xa=0.9 and Xb=0.1 has a total
vapor pressure of 0.75 atm. Calculate the mole fraction Yb of B in the
vapor phase that is in equilibrium with the solution.
B) Calculate the total vapor pressure at 50 °C for a solution with mole
fractions Xa=0.95 and Xb=0.05.
Use Henry’s Law for Oxygen to determine the concentration of oxygen
in water and compare that with the concentration (M) of oxygen in the
Air.
kH  O2 / water   50kBar
Example: Using Raoult’s Law to determine Molecular Weight of
large molecules with no vapor phase (proteins)
Adding a small amount of solute (x2) decreases the vapor
pressure of the solvent (x1). This can be used to measure
the molecular weight of the solute.
P1  x1P1*
P1  (1  x2 ) P1*
P1
x2  1  *
P1
n2
n2
w2
x2 
 
n1  n2 n1 M 2 n1
P1
w2
1 * 
P1 M 2 n1
 w2 
 V
w2


M2 


P1   n1  
P1 
n1 1  *    1  * 
 P1   V   P1 