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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 Slide 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 Slide 3 Simple Mixtures Studied in This Chapter Non-reactive: No chemical reaction would occur. Binary: Non-electrolyte: the solute is not present as ions. Slide 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) Slide 5 Slide 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)? Slide 7 Exercise Exercise Calculate the mole fraction of sucrose in an aqueous sample of molality 1.22 mol kg -1. Slide 8 Partial Molar Volume Unit: L/mol or mL/mol Slide 9 Slide 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 3 + 125 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 Slide 11 Exercise: the Interpolation read. EtOH. over. here read water here Slide 12 Illustration: Ethanol and water ( 1 kg at 25C) b : molality The partial molar volume of ethanol: Slide 13 Partial Molar Gibbs Energy Slide 14 Fundamental Equation of Chemical Thermodynamics At constant pressure and temperature: Therefore, (non-expansion ) Work can be done by changing the composition. Slide 15 The Wider Significance of the Chemical Potential (Classroom exercise) Slide 16 The Gibbs-Duhem Equation Gibbs-Duhem equation applies to ALL partial molar quantities: X=V, ,H,A,U,S, etc Slide 17 Using The Gibbs-Duhem Equation Given the molar volume of water at 298K is 18.079 mL/mol, find the partial molar volume of water. Aqueous solution of K 2 SO 4 : Slide 18 Slide 19 The Thermodynamics of Mixing Is mixing ( composition change) spontaneous? Slide 20 Mixing of two perfect gases or two liquids that form an ideal solution: Slide 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 Slide 22 Exercise: at 25C, calculate the Gibbs energy change when the partition is removed Slide 23 Other Thermodynamic Mixing Functions Entropy of mixing Slide 24 Entropy of mixing of perfect gases Slide 25 Other Thermodynamic Mixing Functions Enthalpy of mixing For perfect gases, Understandable? Slide 26 Ideal Solutions Pure substance: Solute: Vapor pressure Slide 27 Raoults 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 Raoults law throughout its whole range of composition (from x J = 0 to x J = 1) is an ideal solution. Slide 28 Raoults Law Slide 29 Raoults Law: Examples Slide 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 Raoults 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 + 1.5 mol) = 0.895 p benz = x benz p* benz = (0.895)(94.6 torr) = 84.7 torr Slide 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 Slide 32 Raoults Law: Molecular Interpretation At equilibrium: Slide 33 Raoults Law: Molecular Interpretation Slide 34 Ideally Dilute Solutions Solutions of dissimilar li- quids can show strong devi- ations from Raoults 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 Henrys law, p B = K B x B and the slope of the line is the Henrys-law constant. A solution which obeys Henrys law is called an ideal-dilute solution. Slide 35 Henrys 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. Slide 36 Slide 37 Exercise The v.p. of chloromethane at various mol fracs in a mixture at 25 o C was found to be as follows. x0.005 0.0090.019 0.024 p/torr 205 363 756 946 Estimate the Henrys 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. Slide 38 0.005 0.01 0.015 0.02 0.025 1000 800 600 400 200 x p How to estimate the Henry s law constant. Slide 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 = 0.021 g / 16.04 g mol ! 1 = 0.0013 mol. mol C 6 H 6 = 100 g / 78.1 g mol ! 1 = 1.28 mol x CH4 = 0.0013 mol / (1.28 mol + ~0 mol). = 0.0010 p CH4 = K CH4 x CH4 = (4.27 10 5 torr)(0.0010) = 436 torr = 4.3 10 2 torr Slide 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. x0.0 0.2 0.4 0.6 0.8 1.0 p C /torr0.0 35 82 142 200 273 p A /torr347 250 175 92 37 0 Confirm that the mixture conforms to Raoults law for the component in large excess and to the Henrys law for the minor component. Find the Henrys law constants. Slide 41 Validity of Raoult s and Henry s Laws: Result Slide 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] Slide 43 Excess Functions and Regular Solutions Slide 44 Excess Enthalpy Slide 45 Excess Volume Slide 46 Excess Functions and Regular Solutions A regular solution is the one which is not ideal solution but has zero excess entropy: Slide 47 Excess Enthalpy Slide 48 Excess Gibbs Energy Slide 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 Slide 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 Slide 51 Elevation of Vapor Pressure Slide 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 -1. 3.0 g sucrose = 3.0 g / 342 g mol ! 1 = 0.0088 mol sucrose b suc = 0.0088 mol / 0.100 kg = 0.088 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. Slide 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 ! 3 0.50 1.001.50 2.002.50 h/cm0.18 0.35 0.53 0.710.90 The density of the solution is 0.9998 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 concn. Slide 54 Slide 55 Slide 56 Slide 57 Slide 58 Slide 59 Slide 60 Slide 61 Slide 62 Slide 63 Slide 64 Slide 65 Slide 66 Slide 67 Slide 68 Slide 69 Slide 70 Slide 71