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Chapter 07: Simple Mixtures

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1 Chapter 07: Simple Mixtures
The thermodynamic description of mixtures: Partial molar quantities The thermodynamics of mixing The chemical potentials of liquids The properties of solutions: Liquid mixtures Colligative properties 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 = 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 cm 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 b: molality Illustration: Ethanol and water ( 1 kg at 25C)
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 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 = 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.

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 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 p/torr 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 p 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 = g / g mol!1 = mol. mol C6H6 = 100 g / 78.1 g mol!1 = 1.28 mol xCH4 = 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. x pC/torr pA/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:
[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 = 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

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, Kb for water = K kg mol g sucrose = 3.0 g / 342 g mol!1 = mol sucrose bsuc = mol / kg = mol kg!1 DTf = KfbB = (1.86 K kg mol-1)(0.088 mol kg!1) = K Note: New f.p. = 0.00oC ! 0.16 oC = !0.16oC 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 25oC 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, 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.



















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