1. Lecture 3 30-09-2008 Lecture: Last lecture problems
simple mixtures (cont)
colligative properties
membrane potential
Debye-Hückel limiting law
two-component phase diagrams
new problems
Last lecture problems

2. Chemical potential of liquid
Ideal solutions
Let’s consider vapour (treated as perfect gas) above the solution. At equilibrium the chemical potential of a substance in vapour phase must be equal to its potential in the liquid phase
For pure substance:
In solution:
Francouis Raoult experimentally found that:
Raoult’s law:
Mixtures obeying Raoult’s law called ideal solutions

3. Chemical potential of liquid
Molecular interpretation of Raoult’s law
rate of condensation
rate of evaporation

4. Chemical potential of liquid
Dissimilar liquid often show strong deviation
Similar liquid

5. Chemical potential of liquid
Ideal-dilute solutions: Henry’s law
In a dilute solution the molecule of solvent are in an environment similar to a pure liquid while molecules of solute are not!
empirical constant

6. Chemical potential of liquid
Using Henry’s law
Example: Estimate molar solubility of oxygen in water at 25 0C at a partial pressure of 21 kPa.
molality

7. Liquid mixtures Ideal solutions If Raoult’s law applied to we have:
From molecular prospective it means that interactions of A-A, A-B, and B-B are the same.

8. Liquid mixtures In real solutions we can define excess functions, e.g.
excess entropy:
Benzene/Cyclohexane
Model: regular solution
Molecules are randomly distributed but A-B interaction is different from A-A and B-B
suppose:
Then:

9. Colligative properties
Elevation of boiling point
Depression of freezing point
Osmotic pressure phenomenon
All stem from lowering of the chemical potential of the solvent due to presence of solute (even in ideal solution!)
Larger

10. Colligative properties
Elevation of boiling point
(Here we neglect temperature dependence)
For pure liquid:

11. Colligative properties
Depression of freezing point
Cryoscopic constant
Can be used to measure molar mass of a solute

12. Colligative properties
Solubility

13. Colligative properties: Osmosis
Osmosis – spontaneous passage of pure solvent into solution separated by semipermeable membrane
Van’t Hoff equation:

14. Osmosis For dilute solution: Van’t Hoff equation: More generally:
Osmotic virial coefficients

15. Osmosis: Examples
Calculate osmotic pressure exhibited by 0.1M solutions of mannitol and NaCl.
Mannitol (C6H8(OH)6)

16. Osmosis: Examples Hypotonic conditions:
cells burst and dye haemolysis (for blood)
Decreasing salt concentration
Increasing salt concentration
Hypertonic conditions:
cells dry and dye
Isotonic conditions
Internal osmotic pressure keeps
the cell “inflated”

17. Application of Osmosis
Using osmometry to determine molar mass of a macromolecule
Osmotic pressure is measured at a series of mass concentrations c and a plot of vs. c is used to determine molar mass.

18. Membrane potential Electrochemical potential
Fext
Fcyt
Example: membrane potential
Na+
Na+ P- P- P- P-
Na+
Na+
Na+
Na+ P- P-
Na salt of a protein

19. Activities
the aim: to modify the equations to make them applicable to real solutions
Generally:
vapour pressure of A above solution
vapour pressure of A above pure A
For ideal solution
(Raoult’s law)
For real solution
activity of A
activity coefficient of A

20. Activities
Ideal-dilute solution: Henry’s law
Real solutes

21. Example: Biological standard state
Biological standard state: let’s define chemical potential of hydrogen at pH=7

22. Activities
Margules equation
Raoult’s law
Henry’s law

23. Ion Activities standard state: ideal solution at molality b0=1mol/kg
Alternatively:
ideal solution of the same molality b
In ionic solution there is no experimental way to separate contribution of cations and anions
In case of compound MpXq:

24. Debye-Hückel limiting law
Coulomb interaction is the main reason for departing from ideality
Oppositely charged ions attract each other and will form shells (ionic atmosphere) screening each other charge
The energy of the screened ion is lowered as a result of interaction with its atmosphere

25. Debye-Hückel limiting law
In a limit of low concentration the activity coefficient can be calculated as:
Ionic strength of the solution
Example: calculate mean activity coefficient of 5 mM solution of KCL at 25C.

26. Debye-Hückel limiting law
Extended D-H law:

27. Phase Diagrams

28. Phase diagrams
what is the composition (number of phases and their amount and composition) at equilibrium at a given temperature;
what happens to the system when is cools down/heats up
we can predict the structure and the properties of the system at low temperature.
iron-carbon diagram

29. Phase diagrams That’s the base of all modern
water-surfactant-oil
That’s the base of all modern
engineering from swiss knife to food and cosmetics!
iron-carbon diagram

30. Phase diagrams Constituent – a chemical species that is present
Component – a chemically independent constituent of the system (i.e. not connected by a chemical reaction)
Phase1
Phase2
Phase3
Variance – the number of intensive variables that can be changed independently without disturbing the number of phases at equilibrium.
Phase rule (J.W. Gibbs): F=C-P+2
number of phases
variance
number of components
Indeed: number of variables would be: P*(C-1)+2
number of equations: C*(P-1)

31. One component diagrams
C=1 therefore F=C-P+2=3-P

32. One component diagrams
Detection of phase transitions and building a phase diagram is based on calorimetry measurements

33. Two-components diagrams
C=2 therefore F=4-P.
We have to reduce degree of freedom e.g. by fixing T=const
Vapour pressure diagrams
Raoult’s Law

34. Two-components diagrams
The composition of vapour
From Dalton’s law:
From Raoult’s law:

35. Two components diagrams

36. Two components diagrams

37. Two components diagrams
The lever rule

38. Two-components diagrams
Temperature-composition diagrams
Distillation of mixtures

39. Two-components diagrams
Temperature-composition diagrams
Azeotropes
Azeotrope, evaporation w/o change in composition
A-B interacation stabilize the mixture
A-B interacation destabilize the mixture

40. Two components diagrams
Immiscible liquids
Will boil at lower temperature!

41. Two components diagrams
Liquid-liquid phase diagrams

42. Two components diagrams
Upper critical solution T

43. Two components diagrams
Lower critical temperature is usually caused by breaking a weak complex of two components

44. Two components diagrams
Upper critical temperature is less than the boiling point
Boiling occur before liquids are fully miscible

45. Liquid-solid phase diagrams
Eutectic
composition

46. Liquid-solid phase diagrams
Eutectic halt

47. Liquid-solid phase diagrams
Reacting systems
Incongruent melting: compounds melts into components

48. Liquid crystals
Mesophase – an intermedediate phase between solid and liquid. Example: liquid crystal
Liquid crystal – substance having a liquid-like imperfect order in at least one direction and long-range positional or orientational order in at least one another direction
Nematic
Smectic
Cholesteric

49. Nematic crystals in LCD

50. Problems I
5.6a The addition of 100 g of a compound to 750 g of CCl4 lowered the freezing point of the solvent by 10.5 K. Calculate the molar mass of the compound.
5.14a The osmotic pressure of solution of polystyrene in toluene were measured at 25 C and the pressure was expressed in terms of the height of the solvent of density 1.004g/cm3. Calculate the molar mass of polystyrene: c [g/dm3] h [cm]
5.20(a) Estimate the mean ionic activity coefficient and activity of a solution that is mol kg–1 CaCl2(aq) and mol kg–1 NaF(aq).

51. Problems II
Atkins 6.9b: sketch the phase diagram of the system NH3/N2H4 given that the two substances do not form a compound and NH3 freezes at -78C, N2H4 freezes at +2C, eutectic formed with mole fraction of N2H and melts at -80C.
Atkins 6.10b Describe the diagram and what is observed when a and b are cooled down

52. Home problem analysis
4.7a: An open vessel containing (a) water, (b) benzene, (c) mercury stands in a laboratory measuring 5.0 m  5.0 m  3.0 m at 25C. What mass of each substance will be found in the air if there is no ventilation? (The vapour pressures are (a) 3.2 kPa, (b) 13.1 kPa, (c) 0.23 Pa.)
4.9a Calculate the melting point of ice under a pressure of 50 bar. Assume that the density of ice under these conditions is approximately 0.92 g cm–3 and that of liquid water is 1.00 g cm–3.
5.2a At 25C, the density of a 50 per cent by mass ethanol–water solution is g cm–3. Given that the partial molar volume of water in the solution is 17.4 cm3 mol–1, calculate the partial molar volume of the ethanol.
4.7a, , 4.11