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The thermodynamic description of mixtures: Partial molar quantities The thermodynamics of mixing The chemical potentials of liquids Chapter 07: Simple.

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Presentation on theme: "The thermodynamic description of mixtures: Partial molar quantities The thermodynamics of mixing The chemical potentials of liquids Chapter 07: Simple."— Presentation transcript:

1 The thermodynamic description of mixtures: Partial molar quantities The thermodynamics of mixing The chemical potentials of liquids Chapter 07: Simple Mixtures The properties of solutions: Liquid mixtures Colligative properties Activities: The solvent activity The solute activity The activities of regular solutions

2 Assignment for Chapter (b),7.6(a),7.10(b),7.15(a),7.20(b),7.22(b) 7.2,7.7,7.13,7.16,7.19,7.20

3 Simple Mixtures Studied in This Chapter Non-reactive: No chemical reaction would occur. Binary: Non-electrolyte: the solute is not present as ions.

4 Concentration Units There are three major concentration units in use in thermodynamic descriptions of solutions. These are –molarity –molality –mole fraction Letting J stand for one component in a solution (the solute), these are represented by [J] = n J /V (V typically in liters) b J = n J /m solvent (m solvent typically in kg) x J = n J /n (n = total number of moles of all species present in sample)


6 Exercise Exercise What mass of glycine should be used to make 250 mL of a solution of molar concentration 0.15M NH 2 CH 2 COOH(aq)?

7 Exercise Exercise Calculate the mole fraction of sucrose in an aqueous sample of molality 1.22 mol kg -1.

8 Partial Molar Volume Unit: L/mol or mL/mol


10 Exercise Use the figure in slide 8 to calculate the density of a mixture of 20 g of water and 100 g of ethanol. Solution: First calculate the mole fractions. –20 g H 2 O = 1.11 mol; 100 g EtOH = 2.17 mol –x H2O = 0.34; x EtOH = 0.66 Then interpolate from the mixing curve (next slide): –V H2O = 17.1 cm 3 mol -1 ; V EtOH = 57.4 cm 3 mol -1 Then plug the moles and partial molar volume –(1.11 mol)(17.1 cm 3 /mol) + (2.17 mol)(57.4 cm 3 /mol) =19.0 cm cm 3 = 144 cm 3 Finally, the total mass is divided by the total volume: 120 g/144 cm 3 = 0.83 g/ cm 3

11 Exercise: the Interpolation read. EtOH. over. here read water here

12 Illustration: Ethanol and water ( 1 kg at 25C) b : molality The partial molar volume of ethanol:

13 Partial Molar Gibbs Energy

14 Fundamental Equation of Chemical Thermodynamics At constant pressure and temperature: Therefore, (non-expansion ) Work can be done by changing the composition.

15 The Wider Significance of the Chemical Potential (Classroom exercise)

16 The Gibbs-Duhem Equation Gibbs-Duhem equation applies to ALL partial molar quantities: X=V, μ,H,A,U,S, etc

17 Using The Gibbs-Duhem Equation Given the molar volume of water at 298K is mL/mol, find the partial molar volume of water. Aqueous solution of K 2 SO 4 :


19 The Thermodynamics of Mixing Is mixing (  composition change) spontaneous?

20 Mixing of two perfect gases or two liquids that form an ideal solution:

21 Exercise Suppose the partial pressure of a perfect gas falls from 1.00 bar to 0.50 bar as it is consumed in a reaction at  o C. What is the change in chemical potential of the substance? Solution: We want  J,f -  J,i  But  J,i =  J o so we want  J !  J o = RT ln (p J /p o ) = (2.479 J/mol) (ln 0.50) = -1.7 kJ/mol

22 Exercise: at 25C, calculate the Gibbs energy change when the partition is removed

23 Other Thermodynamic Mixing Functions Entropy of mixing

24 Entropy of mixing of perfect gases

25 Other Thermodynamic Mixing Functions Enthalpy of mixing For perfect gases, Understandable?

26 Ideal Solutions Pure substance: Solute: Vapor pressure

27 Raoult’s Law and Ideal Solutions p J = x J p J *Raoult found that the partial vapor pressure of a substance in a mixture is proportional to its mole fraction in the solution and its vapor pressure when pure: p J = x J p J * Any solution which obeys Raoult’s law throughout its whole range of composition (from x J = 0 to x J = 1) is an ideal solution.

28 Raoult’s Law

29 Raoult’s Law: Examples

30 Exercise A solution is prepared by dissolving 1.5 mol C 10 H 8 in 1.00 kg benzene. The v.p. of pure benzene is 94.6 torr at this temperature (25 o C). What is the partial v.p. of benzene in the solution? Solution: We can use Raoult’s law, but first we need to compute the mole fraction of benzene. MM benzene = 78.1 g/mol, so 1.00 kg = 12.8 mol. x benz = 12.8 mol / (12.8 mol mol) = p benz = x benz p* benz = (0.895)(94.6 torr) = 84.7 torr

31 Exercise By how much is the chemical potential of benzene reduced at 25 o C by a solute that is present at a mole fraction of 0.10? Solution: We want  benz =  benz !  benz But  benz =  benz + RT ln x benz so  benz !  benz = RT ln x benz And if x solute = 0.10, then x benz = 0.90 Thus  benz = (2.479 kJ/mol) (ln 0.90). = ! 0.26 kJ/mol

32 Raoult’s Law: Molecular Interpretation At equilibrium:

33 Raoult’s Law: Molecular Interpretation

34 Ideally Dilute Solutions Solutions of dissimilar li- quids can show strong devi- ations from Raoult’s law (green line at left) unless a substance has x > However, the v.p. usually starts off as a straight line. This is embodied in Henry’s law, –p B = K B x B –and the slope of the line is the Henry’s-law constant. A solution which obeys Henry’s law is called an ideal-dilute solution.

35 Henry’s Law: Molecular Interpretation Solvent Solute In ideally dilute solution, the solvent is almost like a pure liquid whereas the solute behaves very differently from a pure liquid.


37 Exercise The v.p. of chloromethane at various mol fracs in a mixture at 25 o C was found to be as follows. x p/torr Estimate the Henry’s law constant for chloromethane at 25 o C in this particular solvent. Solution: The v.p.’s are plotted vs. mol fraction. The data are fitted to a polynomial curve (using a computer program that has a curve-fitting function) and the tangent (slope) is calculated by evaluating the first derivative of the poly- nomial at x CHCl3 = 0.

38 x p How to estimate the Henry ’ s law constant.

39 Exercise What partial pressure of methane is needed to achieve 21 mg of methane in 100 g benzene at 25 o C? Solution: From Table 7.1, K B = 4.27 × 10 5 torr. mol CH 4 = g / g mol ! 1 = mol. mol C 6 H 6 = 100 g / 78.1 g mol ! 1 = 1.28 mol x CH4 = mol / (1.28 mol + ~0 mol). = p CH4 = K CH4 x CH4 = (4.27 × 10 5 torr)(0.0010) = 436 torr = 4.3 × 10 2 torr

40 Validity of Raoult ’ s and Henry ’ s Laws The v.p. of propanone (acetone,A) and trichloromethane (chloroform, CHCl 3, C) at various mol fracs in a mixture at 35 o C was found to be as follows. x p C /torr p A /torr Confirm that the mixture conforms to Raoult’s law for the component in large excess and to the Henry’s law for the minor component. Find the Henry’s law constants.

41 Validity of Raoult ’ s and Henry ’ s Laws: Result

42 Liquid Mixtures For two liquids (A+B) forming an ideal solution: For real solutions, that ’ s not true. [The ideality of a solution holds well if interactions A-A, B-B are the same as A-B]

43 Excess Functions and Regular Solutions

44 Excess Enthalpy

45 Excess Volume

46 Excess Functions and Regular Solutions A regular solution is the one which is not ideal solution but has zero excess entropy:

47 Excess Enthalpy

48 Excess Gibbs Energy

49 Colligative Properties Vapor pressure lowering is one of the four colligative properties of the solvent. These are properties that depend only on the number concentration of particles of solute and not at all on the nature of the solute. (They do depend on the nature of the solvent.) The other three are: –Boiling-point elevation  T b = Kb B –Freezing-point depression  T f = K f b B where b B is the molality of the solute B in the solution –Osmotic pressure . [B] RT where [B] is the molarity of the solute B in the solution

50 Boiling-point elevation  T b = K b b B Freezing-point depression  T f = K f b B where b B is the molality of the solute B in the solution

51 Elevation of Vapor Pressure

52 Exercise Estimate the lowering of the freezing point of the solution made by dissolving 3.0 g sucrose in 100 g of water. Solution: From Table 4.3, K b for water = 1.86 K kg mol g sucrose = 3.0 g / 342 g mol ! 1 = mol sucrose b suc = mol / kg = mol kg ! 1  T f = K f b B = (1.86 K kg mol -1 )(0.088 mol kg ! 1 ). = 0.16 K Note: New f.p. = 0.00 o C ! 0.16 o C = ! 0.16 o C Note also: It is assumed that pure water freezes.

53 Exercise The heights of the solution in an osmometry experi-ment on a solution of an enzyme in water at 25 o C were as follows. c/g dm ! h/cm The density of the solution is g cm ! 3. What is the molar mass of the enzyme? Solution: At each concentration,  is found from  =  gh. [B] is found from  / RT and MM from c(g/L) / [B] (mol/L) The molar masses are plotted against concentration and extrapolated to zero conc’n.



















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