Brief reminder of the previous lecture

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Charles Glenn
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1 Brief reminder of the previous lecture partial molar quantities: contribution of each component to the properties of mixtures V j V = G µ = j n j n j pt,, n pt,, n dg = Vdp SdT + µ dn + µ dn +... A A B B ndµ J J = J Gibbs-Duhem equation: 0 Mixing energy (entropic) G = nrt( x lnx + x ln x ) mix A A B B Raoult s and Henry s laws: p p = x p A A A = xk A A A

2 Brief reminder of the previous lecture Non-ideal mixtures: excess of enthalpy G = nrt( κ lnκ + κ ln κ + βκ κ ) mix A A B B A B Elevation of boiling point Depression of freezing point Osmotic pressure phenomenon

3 Lecture 4

4 Colligative properties: Osmosis Osmosis spontaneous passage of pure solvent into solution separated by semipermeable membrane Van t Hoff equation: Π = [ B] RT, [ B] = n / V B

5 Osmosis µ µ κ dg = SdT + Vdp A ( p) = A ( p+π ) + RTln A p+π +Π = + µ ( p ) µ ( p) V dp A A m p For dilute solution: RTκ B n = ΠV B / n A m V / na Van t Hoff equation: Π = [ B] RT, [ B] = n / V B More generally: Π = [ BRT ] (1 + bb [ ] +...) Osmotic virial coefficients

6 Osmosis: Examples Calculate osmotic pressure exhibited by 0.1M solutions of mannitol and NaCl. Π= [ B] RT, [ B] = n / V B Mannitol (C 6 H 8 (OH) 6 )

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

8 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. Π / c Π= [ BRT ] (1 + bb [ ] +...) ρ gh c/ M h RT brt c... 2 c = ρgm + ρgm +

9 Membrane potential Electrochemical potential µ = µ + zneφ = µ + RTln[ j] + zfφ 0 j j j A j j F ext Example: membrane potential F cyt µ + RT ln[ Na ] + z FΦ = µ + RT ln[ Na ] + z FΦ Na in + Na in + Na out + Na out + RT [ Na out ] Φ = ln F [ + Na in ] Na + P - Na + P - P - Na salt of a protein Na + Na + P - P - P - Na + Na +

10 Activities Let s consider solvent in equilibrium with its vapour: P A P A µ = µ sol gas Pure solvent Solution Generally: A µ = µ + A RT ln p p A A vapour pressure of A above solution vapour pressure of A above pure A For ideal solution µ = µ + RT lnκ A A A (Raoult s law)

11 Activities the aim: to modify the equations to make them applicable to real solutions For ideal solution µ = µ + RT ln x A A A (Raoult s law) For real solution: we introduce effective concentration related to the partial vapour pressure µ = µ + RT ln a activity of A A A A a κ as κ 1 A A A µ = µ + RT ln x + RT ln γ A A A A activity coefficient of A

12 Solute activity Ideal-dilute solution: Henry s law B B B p = K κ p K µ = µ + RT ln = µ + RT ln + RT ln x B B B B B B p B p B µ = µ + RT ln x 0 B B B 0 µ B Real solutes µ = µ + RT ln a 0 B B B µ = µ + RT ln x + RT ln γ 0 B B B B a B = p K B B Other definitions of standard state Activity in terms of molality Biological standard state µ = µ + RT ln a 0 H H H bb µ B = µ B + RT ln ab; ab = γb 0 b µ = µ 7RT ln(10) = µ 40 kj / mol H H H

13 Activities Margules equation ln γ = βκ ln γ = βκ 2 2 A B B A G = nrt( κ lna + κ ln a ) mix A A B B G = nrt( κ lnκ + κ lnκ + κ lnγ + κ ln γ ) mix A A B B A A B B E G = nrt( κ lnκ + κ ln κ + βκ κ ) H mix A A B B A B = nβ RTκκ A B Raoult s law Henry s law p = κ e p β A A A

14 Ion Activities µ µ 0 = + RT ln a a b = γ b 0 standard state: ideal solution at molality b 0 =1mol/kg Alternatively: 0 µ = µ + RT ln b + RT ln γ = µ ideal + RT ln γ ideal solution of the same molality b In ionic solution there is no experimental way to separate contribution of cations and anions Gm ideal ideal = µ + µ = µ + µ + RTln γ γ ideal ideal µ = µ + RT ln γ ; µ = µ + RT ln γ + + ± ± 2 γ ± In case of compound M p X q : G = pµ + qµ = G + RTln γ γ m ideal p q + m +

Debye-Hückel limiting law Coulomb interaction is the main reason for departing from ideality Oppositely charged ions attract each other and will

15 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

16 Debye-Hückel limiting law In a limit of low concentration the activity coefficient can be calculated as: 1 2 log = z z AI, A = for water γ ± + 1 = Ionic strength of the solution (activity depends on all ions present! e.g. salting-in effect) 2 0 where: I zi ( bi / b ) 2 i Example: calculate mean activity coefficient of 5 mm solution of KCL at 25C. 1 I = ( b+ + b )/ b = b / b = 510 i /2 logγ ± = zz + AI = 0.509(5i10 ) = γ ± = 0.92

17 Debye-Hückel limiting law logγ ± = zz + AI 1 2 Extended D-H law: zz AI logγ = + ± 1+ BI

18 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

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

20 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) CaCO3() s CaO() s + CO2( g) Phase1 Phase2 Phase3 C = 2 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 variance number of phases number of components Indeed: number of variables would be: number of equations: P(C-1)+2 C(P-1)

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

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

23 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 p = κ p p = κ p A A A B B B p = p + p = p + κ ( p p ) A B B A A B

24 Two-components diagrams The composition of vapour pa From Dalton s law: ya = ; yb = p From Raoult s law: p p p = x p ; p = x p A A A B B B B pa y = ; y = 1 y p + ( p p ) x A B A B A B A p p A B

25 Two components diagrams

26 Two components diagrams

27 Two components diagrams The lever rule nl α α = nl β β Proof: nz = n x + n y A α A β A nz = n z + n z A α A β A nx + ny = nz + nz α A β A α A β A n ( x z ) = n ( z y ) α A A α A A

28 Two-components diagrams Temperature-composition diagrams Distillation of mixtures Simple distillation: liquid boiled, vapour collected and withdrawn Fractional distillation: boiling/condensation cycles repeated successively

29 Two-components diagrams Temperature-composition diagrams Azeotropes Azeotrope, evaporation w/o change in composition nl α α A-B interaction stabilizes the mixture A-B interaction destabilizes the mixture

30 Two components diagrams Immiscible liquids Will boil at lower temperature! p = p + p A B

31 Two components diagrams Liquid-liquid phase diagrams

32 Two components diagrams G = nrt( x lnx + x ln x +β x x ) mix A A B B A B x mixg = 0 ln + β (1 2 x) = 0 κ 1 x Upper critical solution T

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

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

35 Two components (liquid-liquid) Example: what happens when solution of composition a 1 is boiled and vapour condensed?

36 Eutectic composition Liquid-solid phase diagrams

37 Liquid-solid phase diagrams Eutectic halt

38 Liquid-solid phase diagrams Reacting systems Solid solution of Na in K (reach in K) Solid Na 2 K + solid solution of Na in K Solid solution of K in Na (reach in Na) Solid Na 2 K + solid solution of K in Na Incongruent melting: compounds is not stable up to melting temperature and melts into components

39 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 longrange positional or orientational order in at least one another direction Nematic Smectic Cholesteric

40 Nematic crystals in LCD

41 Zone refining (or how to make ultrapure materials) Equipment for Float Zone (FZ) Crystal Growth Carrier lifetime: 1-20 ms O and C impurity: <10 16 atoms/cm 3

42 Problems 5.2(a) 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 cm 3 mol 1, calculate the partial molar volume of the ethanol. 5.5(a) The vapour pressure of benzene is 53.3 kpa at 60.6 C, but it fell to 51.5 kpa when 19.0 g of an involatile organic compound was dissolved in 500 g of benzene. Calculate the molar mass of the compound. 5.20(a) Estimate the mean ionic activity coefficient and activity of a solution that is mol kg 1 CaCl 2 (aq) and mol kg 1 NaF(aq).

43 Problems Atkins 6.9b: sketch the phase diagram of the system NH 3 /N 2 H 4 given that the two substances do not form a compound and NH 3 freezes at -78C, N 2 H 4 freezes at +2C, eutectic formed with mole fraction of N 2 H and melts at -80C. Atkins 6.10b Describe the diagram and what is observed when a and b are cooled down

Lecture 6. NONELECTROLYTE SOLUTONS

Lecture 6. NONELECTROLYTE SOLUTONS NONELECTROLYTE SOLUTIONS SOLUTIONS single phase homogeneous mixture of two or more components NONELECTROLYTES do not contain ionic species. CONCENTRATION UNITS percent