Presentation on theme: "Chapter 07: Simple Mixtures"— Presentation transcript:
1 Chapter 07: Simple Mixtures The thermodynamic description of mixtures:Partial molar quantitiesThe thermodynamics of mixingThe chemical potentials of liquidsThe properties of solutions:Liquid mixturesColligative propertiesActivities:The solvent activityThe solute activityThe 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 UnitsThere are three major concentration units in use in thermodynamic descriptions of solutions.These aremolaritymolalitymole fractionLetting 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)
10 ExerciseUse 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 = molxH2O = 0.34; xEtOH = 0.66Then interpolate from the mixing curve (next slide):VH2O = 17.1 cm3 mol-1; VEtOH = 57.4 cm3 mol-1Then plug the moles and partial molar volume(1.11 mol)(17.1 cm3/mol) + (2.17 mol)(57.4 cm3/mol) =19.0 cm cm3 = 144 cm3Finally, 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 b: molality Illustration: Ethanol and water ( 1 kg at 25C) The partial molar volume of ethanol:
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 ExerciseSuppose 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
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 = xJ pJ*Any solution which obeys Raoult’s law throughout its whole range of composition (from xJ = 0 to xJ = 1) is an ideal solution.
30 ExerciseA 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 mol) = 0.895pbenz = xbenz p*benz = (0.895)(94.6 torr) = 84.7 torr
31 ExerciseBy 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*benzBut mbenz = m*benz + RT ln xbenz so mbenz ! m*benz = RT ln xbenzAnd if xsolute = 0.10, then xbenz = 0.90Thus Dmbenz = (2.479 kJ/mol) (ln 0.90) = ! 0.26 kJ/mol
32 Raoult’s Law: Molecular Interpretation At equilibrium:
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 xBand 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 likea pure liquid whereas thesolute behaves very differentlyfrom a pure liquid.SoluteSolvent
37 ExerciseThe v.p. of chloromethane at various mol fracs in a mixture at 25oC was found to be as follows.xp/torrEstimate 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.
39 ExerciseWhat 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 = g / g mol!1 = mol. mol C6H6 = 100 g / 78.1 g mol!1 = 1.28 molxCH4 = mol / (1.28 mol + ~0 mol) =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.xpC/torrpA/torrConfirm 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.
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 = KbBFreezing-point depression DTf = KfbBwhere bB is the molality of the solute B in the solutionOsmotic pressure P . [B] RTwhere [B] is the molarity of the solute B in the solution
50 Boiling-point elevation. DTb = KbbB Freezing-point depression Boiling-point elevation DTb = KbbB Freezing-point depression DTf = KfbB where bB is the molality of the solute B in the solution
52 ExerciseEstimate 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, Kb for water = K kg mol g sucrose = 3.0 g / 342 g mol!1 = mol sucrosebsuc = mol / kg = mol kg!1DTf = KfbB = (1.86 K kg mol-1)(0.088 mol kg!1) = KNote: New f.p. = 0.00oC ! 0.16 oC = !0.16oCNote also: It is assumed that pure water freezes.
53 ExerciseThe heights of the solution in an osmometry experi-ment on a solution of an enzyme in water at 25oC were as follows.c/g dm!h/cmThe density of the solution is g cm!3. What is the molar mass of the enzyme?Solution: At each concentration, P is found from P = rgh. [B] is found from P / RT and MM from c(g/L) / [B] (mol/L)The molar masses are plotted against concentration and extrapolated to zero conc’n.