The thermodynamic description of mixtures: Partial molar quantities The thermodynamics of mixing The chemical potential

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

2. Assignment for Chapter 07 • 7.2(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] = nJ/V (V typically in liters) • bJ = nJ/msolvent (msolvent typically in kg) • xJ = nJ/n (n = total number of moles of all species present in sample)

6. Exercise What mass of glycine should be used to make 250 mL of a solution of molar concentration 0.15M NH2CH2COOH(aq)?

7. 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 H2O = 1.11 mol; 100 g EtOH = 2.17 mol • xH2O = 0.34; xEtOH = 0.66 • Then interpolate from the mixing curve (next slide): • VH2O = 17.1 cm3 mol-1; VEtOH = 57.4 cm3 mol-1 • Then plug the moles and partial molar volume • (1.11 mol)(17.1 cm3/mol) + (2.17 mol)(57.4 cm3/mol) =19.0 cm3 + 125 cm3 = 144 cm3 • Finally, the total mass is divided by the total volume: 120 g/144 cm3 = 0.83 g/ cm3

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 Aqueous solution of K2SO4: Given the molar volume of water at 298K is 18.079 mL/mol, find the partial molar volume of water.

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 25 oC. What is the change in chemical potential of the substance? • Solution: We want mJ,f - mJ,i But mJ,i =mJo so we want mJ !mJo = RT ln (pJ/po) = (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: Vapor pressure Solute:

27. Raoult’s Law and Ideal Solutions • 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: pJ = xJpJ* • Any solution which obeys Raoult’s law throughout its whole range of composition (from xJ = 0 to xJ = 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 C10H8 in 1.00 kg benzene. The v.p. of pure benzene is 94.6 torr at this temperature (25oC). 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. • xbenz = 12.8 mol / (12.8 mol + 1.5 mol) = 0.895 • pbenz = xbenz p*benz = (0.895)(94.6 torr) = 84.7 torr

31. Exercise • By how much is the chemical potential of benzene reduced at 25oC by a solute that is present at a mole fraction of 0.10? • Solution: We want Dmbenz = mbenz !m*benz • But mbenz =m*benz + RT ln xbenz so mbenz !m*benz = RT ln xbenz • And if xsolute = 0.10, then xbenz = 0.90 • Thus Dmbenz = (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 > 0.90. • However, the v.p. usually starts off as a straight line. • This is embodied in Henry’s law, • pB = KB xB • 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 In ideally dilute solution, the solvent is almost like a pure liquid whereas the solute behaves very differently from a pure liquid. Solute Solvent

37. Exercise • The v.p. of chloromethane at various mol fracs in a mixture at 25oC was found to be as follows. • x 0.005 0.009 0.019 0.024 • p/torr 205 363 756 946 • Estimate the Henry’s law constant for chloromethane at 25oC 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 xCHCl3 = 0.

38. 1000 800 600 400 200 p 0.005 0.01 0.015 0.02 0.025 x 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 25oC? • Solution: From Table 7.1, KB = 4.27 × 105 torr. mol CH4 = 0.021 g / 16.04 g mol!1 = 0.0013 mol. mol C6H6 = 100 g / 78.1 g mol!1 = 1.28 mol • xCH4 = 0.0013 mol / (1.28 mol + ~0 mol) . = 0.0010 • pCH4 = KCH4 xCH4 = (4.27 × 105 torr)(0.0010) = 436 torr = 4.3 × 102 torr

40. Validity of Raoult’s and Henry’s Laws • The v.p. of propanone (acetone,A) and trichloromethane (chloroform, CHCl3, C) at various mol fracs in a mixture at 35oC was found to be as follows. • x 0.0 0.2 0.4 0.6 0.8 1.0 • pC/torr 0.0 35 82 142 200 273 • pA/torr 347 250 175 92 37 0 • 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: [The ideality of a solution holds well if interactions A-A, B-B are the same as A-B] For real solutions, that’s not true.

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 DTb = KbB • Freezing-point depression DTf = KfbB • where bB is the molality of the solute B in the solution • Osmotic pressure P. [B] RT • where [B]is the molarity of the solute B in the solution

50. Boiling-point elevation DTb = KbbBFreezing-point depression DTf = KfbBwhere bB is the molality of the solute B in the solution