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

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

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

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

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

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

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.

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:

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

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

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

Colligative properties Solubility

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

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

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

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”

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.

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

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

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

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

Activities Margules equation Raoult’s law Henry’s law

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:

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

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.

Debye-Hückel limiting law Extended D-H law:

Phase Diagrams

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

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

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)

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

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

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

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

Two components diagrams

Two components diagrams

Two components diagrams The lever rule

Two-components diagrams Temperature-composition diagrams Distillation of mixtures

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

Two components diagrams Immiscible liquids Will boil at lower temperature!

Two components diagrams Liquid-liquid phase diagrams

Two components diagrams Upper critical solution T

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

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

Liquid-solid phase diagrams Eutectic composition

Liquid-solid phase diagrams Eutectic halt

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

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

Nematic crystals in LCD

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] 2.042 6.613 9.521 12.602 h [cm] 0.592 1.910 2.750 3.600 5.20(a) Estimate the mean ionic activity coefficient and activity of a solution that is 0.010 mol kg–1 CaCl2(aq) and 0.030 mol kg–1 NaF(aq).

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 N2H4 0.07 and melts at -80C. Atkins 6.10b Describe the diagram and what is observed when a and b are cooled down

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 0.914 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