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1 Electrochemistry & Solutions 1. Solutions and Mixtures Year 1 – Module 3 8 Lectures Dr Adam Lee Department of Chemistry.

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Presentation on theme: "1 Electrochemistry & Solutions 1. Solutions and Mixtures Year 1 – Module 3 8 Lectures Dr Adam Lee Department of Chemistry."— Presentation transcript:

1 1 Electrochemistry & Solutions 1. Solutions and Mixtures Year 1 – Module 3 8 Lectures Dr Adam Lee Department of Chemistry

2 2 Aims To: Understand physical chemistry of solutions and their thermodynamic properties  predict/control physical behaviour  improve chemical reactions Link electrochemical properties to chemical thermodynamics  rationalise reactivity.

3 3 Synopsis Phase rule Clapeyron & Clausius-Clapeyron Equations Chemical potential Phase diagrams Raoults law (Henry’s law) Lever rule Distillation and Azeotropes Osmosis Structure of liquids Interactions in ionic solutions Ion-ion interactions Debye-Huckel theory Electrodes Electrochemical cells Electrode potentials Nernst Equation Electrode types Recommended Reading R.G. Compton and G.H.W. Sanders, Electrode Potentials Oxford Chemistry Primers No 41. P. W. Atkins, The Elements of Physical Chemistry, OUP, 3 rd Edition, Chapters 5, 6 & 9. P. W. Atkins, Physical Chemistry, OUP, 7 th Edition, Chapters 7, 8 & 10 OR 8 th Edition, Chapters 4, 5, 6 & 7.

4 4 Phase Diagrams

5 5

6 6 Vacuum Pump

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8 8

9 9 No. of degrees of freedom No. of components No. of phases

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12 Josiah Willard Gibbs Gibbs Free Energy American mathematical physicist developed theory of chemical thermodynamics. First US engineering PhD…later Professor at Yale.

13 Benoit Paul Emile Clapeyron Parisian engineer and mathematician. Derived differential equation for determining heat of melting of a solid

14 14

15 15

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17 17 Modified eqn. forLiquid/Gas and Solid/Gas Lines: Clausius-Clapeyron Eqn. Maths dx/x = dlnx dp/p = dlnp = = = = =  dxx n 1n x 1n    2 x 12 x 12   x 1   2 T dT T 1  dp VT H   Clapeyron Equation Rudolf Julius Emmanuel Clausius

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19 19 Clausius-Clapeyron Equation

20 20 99

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24 24

25 25 Consider ideal gas at constant temperature: dG = Vdp – SdT = Vdp Since pV = nRT, If initial state 1 = STP (1 atm) In general for a mixture AB: G A = G o A + n A RT ln p A, G B = G o B + n B RT ln p B since G A = n A  A  A =  A o + RT ln p A

26 26 p = p A + p B

27 27 yAyA yAyA yByB yByB nAnA nBnB n A + n B G initial = n A  A + n B  B = n A [   + RTlnp] + n B [  B  + RTlnp] G final = n A [   + RTlny A p] + n B [  B  + RTlny B p]  G = G final - G initial = RT[n A lny A p – n A lnp + n B lny B p - + n B lnp] yaya ybyb yaya yaya ybyb ybyb InitialFinal A B  G mixing

28 28 yAyA yAyA yByB yByB yaya ybyb yAyA yAyA yByB yByB  G mixing  S mixing yaya ybyb

29 29 Chemical Potential (in English!) Gibbs Free Energy Pressure Molecules acquire more spare energy Greater “chemical potential” Low Pressure Constant Temperature High Pressure G  ln(pressure) G molar = G  molar + RT lnp Energy free for molecules to “do stuff”at STP Effect of environment on this free energy 

30 30 For single component e.g. pure H 2 O Why do we use Chemical Potential? No real need to use  Gibbs Free Energy (G) is total energy in entire system available to “do stuff” - includes all molecules, of all substances, in all phases G = n A  A + n B  B For mixtures e.g. H 2 O/C 2 H 5 OH G = n H2O  H2O Free energy only comes from H 2 O Free energy from 2 sources G = n H2O  H2O +n EtOH  EtOH  tells us how much from H 2 O versus C 2 H 5 OH

31 31 Chemical Potential  : 1. A measure of "escaping tendency" of components in a solution 2. A measure of the reactivity of a component in a solution Why do we different molecules have different Chemical Potentials? Ethanol can soak up much more energy in extra vibrational modes and chemical bonds - will respond differently to pressure/temperature increases Free Energy (G) &  Gas-phase molecule Liquid-phase Solid-state Pressure Volatile Involatile

32 32 Raoult's law eqns. of straight lines passing thru origin Total pressure above boiling liquid Volatility

33 33 x A runs between 0 (none present) to 1 (pure solution) Mixing (diluting a substance) always lowers x A  this means mixing ALWAYS gives a negative lnx A Pure A 0 lnx A Pure B xAxA 10 Dilution Mixing always lowers 

34 34

35 35 VolatileInvolatile

36 36 VolatileInvolatile

37 37

38 38

39 39 A B A B Case 2: -ve deviation A more attracted by B (e.g. CHCl 3 + acetone)  mix H = < 0 A B A B Case 3: +ve deviation A less attracted by B (e.g. EtOH + water)  mix H = > 0 b.pt. > ideal b.pt. < ideal

40 40 Summary: Raoult’s Law for Solvents p A = x A. p A Θ Liquid p o A 1 0 x o A   p Partial pressure of A Total pressure 0 1 x o B Partial pressure of B p B = x B. p B Θ pBΘpBΘ Proportionality constant Involatile low vapour pressure Volatile high vapour pressure A B

41 41 p 0 ө = vapour pressure = tendancy of system to increase S High order: low SLess order: higher S Strong desire to  S Boiling of A favoured Less need to  S A happier in liquid A A A A High p 0 ө (A)Low p 0 ө (A)

42 42 Proportionality constant Amount in solution

43 43 For dissolution of oxygen in water, O 2 (g) O 2 (aq), enthalpy change under standard conditions is kJ/mole. Dissolution is EXOTHERMIC

44 44 Consider O 2 dissolution in water: Important in Green Chemistry for selective oxidation Cinnamyl Alcohol Cinnamic Acid Cinnamaldehyde Solvent: H 2 OSolute: O 2 p H2O p O2 Aspects of Allylic Alcohol Oxidation Adam F. Lee et al, Green Chemistry 2000, 6, 279 Henry's law accurate for gases dissolving in liquids when concentrations and partial pressures are low. As conc. and partial pressures increase, deviations from Henry's law become noticeable Similar to behavior of gases - deviate from the ideal gas law at high P and low T. Solutions obeying Henry's law are therefore often called ideal dilute solutions. H2OH2O O2O2

45 45

46 46  A xAxA yAyA

47 47

48 48 AA A B A A B A BB B A B B A B Tie-line LIQUID GAS LIQUID GAS Lever Rule A-B CompositionLiquid-Gas Distribution BA VolatileInvolatile

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50 50 Example Problem The following temperature/composition data were obtained for a mixture of octane (O) and toluene (T) at 760 Torr, where x is the mol fraction in the liquid and y the mol fraction in the vapour at equilibrium The boiling points are  C for toluene and  C for octane. Plot the temperature/composition diagram of the mixture. What is the composition of vapour in equilibrium with the liquid of composition: 1. x(T) = x(O) = 0.25 P.W. Atkins, Elements of Phys.Chem. page 141 Boiling/ Condensation Temperature LiquidVapour x(T) = 1, T =  C x(O) = 1, x(T) = 0, T =  C

51 51  x1x1 y2y2 y1y1 boiling x2x2 cool

52 52

53 53 A B A B Case 2: -ve deviation A more attracted by B (e.g. CHCl 3 + acetone)  mix H = < 0 A B A B Case 3: +ve deviation A less attracted by B (e.g. EtOH + water)  mix H = > 0 b.pt. > ideal b.pt. < ideal

54 54 azeotropic composition

55 55 A B A B Residue Distlllate

56 56 Topics Covered (lectures 2-4)  Chemical Potential -  A (l) =   A (l) + RTlnx A -  A (g) =   A (g) + RTlnp A  Mol fractions - A = n A / n A +n B  Raoult’s Law - p A = p o A x A p B = p o B x B - ideal solutions - +ve/-ve deviations  Vapour-pressure diagrams - Tie-lines - Lever Rule

57 57 Solutions Equations Phase Rule F = c - p + 2 c = components degrees of freedomp = no. of phases Clapeyron Equation  H = enthalpy of phase change  V = volume change associated with phase change Clausius-Clapeyron Equation or Mol fractions x A = n A / n A +n B n i = mols of i y A = p A / p A + p B p i = partial pressure of i Raoults Law p A = p o A x A and p B = p o B x B Lever Rule (for tie-line joining phases via point a) n l =no. moles in liquid phase n v =no. moles in liquid phase

58 58 Contains solute (e.g. NaCl, glucose) WHY?!!! Solvent A  A (l) =   A (l) + RTlnx A A  A (solution) <  A (solvent)

59 59


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