Chapter 7 : Simple Mixtures
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1 Chapter 7 : Simple Mixtures Using the concept of chemical potential to describe the physical properties of a mixture. Outline 1)Partial Molar Quantities 2)Thermodynamics of Mixing 3)Chemical Potentials of Solutions 4)Colligative Properties
2 Partial Molar Quantities a) Partial Molar Volume dd 1 mol of water to 100 moles of water V dd 1 mol of water to 100 moles of EtOH n V n j Partial Molar Quantities
3 The definition of partial molar volume is V ~ V j V = n j T,P,n' n j V n j t 298 K, the density of a 50%, by mass, EtOH/water mixture is g/cm 3. The partial molar volume of water at this composition is known to be 17.4 cm 3 /mol. What is the partial molar volume of EtOH for this mixture? Partial Molar Quantities
4 b) Partial Molar Gibbs Free Energy G Pure substance n G Mixture n j For a binary mixture G = μ n + μ n a a b b Partial Molar Quantities
5 Partial Molar Quantities Parameter Molar quantity Partial Molar quantity Volume Gibbs Free Energy Entropy Partial Molar Quantities
6 Gibbs-Duhem Equation For a pure substance G = G(T,P) For a mixture G = G(T,P,n a,n b ) nj μ j = j d 0 n equivalent analysis can be done for all partial molar quantities. For a binary system then. a = EtOH b = water Partial Molar Quantities
7 Thermodynamics of Mixing Two ideal gases ( and ) are introduced to a common container. We ve looked at the entropy of mixing, can we determine the Gibbs free energy of mixing? Recall o G G nrtln P o μ P = + P Tln P o = μ + R o Where P o is 1 bar. ( ) ( ) ΔmixG = RT naln χa + nbln χb Thermodynamics of Mixing
8 Earlier we derived the change in entropy for the mixing of an ideal gas. Δ mix 1 1 S = R n a ln + n b ln χa χb ΔG = ΔH-TΔS ΔH = ΔG+ TΔS 1 1 ΔH = RT n aln( χa) + nbln( χb) + RT naln + nbrtln χa χb Thermodynamics of Mixing
9 Chemical Potentials of Ideal Solutions efore we develop expressions for the chemical potential of real solutions we first need to develop expressions for the chemical potentials of solutes and solvents. Consider a pure solvent (). Gas phase () Liquid phase () Chemical Potentials of Solutions
10 t Equilibrium Mixture of Two Liquids ( and ) Pure Liquid () Rate of vaporization μ P P o = μ + RT ln o surface, = = Rate of condensation = k P k n n tot, surface kχ Mixture of Two Liquids ( and ) Rate of vaporization n surface, = = k n tot, surface kχ Rate of condensation = k P Chemical Potentials of Solutions
11 Can now determine the chemical potential of a solvent in the presence of a solute for an ideal solution. μ = μ + RT ln χ In an ideal solution both the solute and the solvent should obey Raoult s Law. (solvent) (solute) Chemical Potentials of Solutions
12 Henry s Law Real solutes don t obey Raoult s Law. In the limit of low solute concentration though, the solute s partial pressure is directly proportional to its mole fraction. = solute P Raoult s Law P = χ P P 0 1 Mole Fraction of Henry s Law P = χ K Chemical Potentials of Solutions
13 μ μ RT ln P = + P Henry s Law P = χ K In an ideal-dilute solution the solute should obey Henry s Law and the solvent should obey Raoult s Law. (solvent) (solute) Chemical Potentials of Solutions
14 Chemical Potentials of Real Solutions What about non-ideal solutions? Solvents in Real Solutions μ μ = μ + RT ln = μ + RT ln P P a The activity is defined as an effective mole fraction. a = γ χ μ = μ + RT ln γ χ Chemical Potentials of Solutions
15 a = γ χ P P = a μ = μ + RT ln γ χ We can now define the standard state to be the pure solvent. i.e. χ 1 and γ 1. μ o = μ + RT ln a Chemical potential of the solvent in a real solution Chemical Potentials of Solutions
16 Solutes in Real Solutions μ P = μ + RT ln ut let s define the solute P activity to be Solving for P nd if we also introduce the solute activity coefficient We get an eqn. that accounts for deviations to ideal-dilute, Henry s Law behaviour. K μ = μ + RT ln γ χ P Chemical Potentials of Solutions
17 ll that s left to do is define a standard state. K μ = μ + RT ln γ χ P a = γ χ P s χ 0, Henry s Law applies and γ must 1. P 0 1 χ K μ = μ = μ + RT ln + ln γ + ln χ ( ) ( ) o P
18 μ = μ + RT γ χ μ = μ + RT ln a o ln ( ) o Solvent () Chemical potential of the solute in a real solution Solute () Ideal solutions Ideal-dilute solution Real solutions a a = γ χ P = P Chemical Potentials of Solutions a a = γ χ P = K
19 ctivities in Terms of Molaties Sometimes it is preferable to use concentrations rather than mole fractions. Molarity is not the best unit of concentration for thermodynamics. Why not? M = moles of solute volume of solution Chemical Potentials of Solutions
20 The choice of standard state is entirely at our discretion. To express the chemical potential of a solute in a solution in terms of molality, μ = μ + RT ln a o only requires a slightly different definition of the activity and a new definition of the standard state. Chemical Potentials of Solutions
21 Colligative Properties μ What happens if the chemical potential of the liquid phase is lowered while maintaining the chemical potentials of the solid and gas phases? Solid Liquid Temperature Gas Colligative properties stem from the lowering if the pure solvent s chemical potential. They depend only on the number of solute particles and not their chemical identity. Colligative Properties
22 The starting point for an analysis of all colligative properties is the fundamental equation of thermodynamics. dg = SdT + VdP + μ dn + μ dn + μ dn... C C For a two component system ( = liquid ; = solute); dg = SdT + VdP+ μ dn + μ dn Now take the partial derivative with respect to n Colligative Properties
23 Colligative Property #1 : oiling Point Elevation ΔT = ( T ) R ΔH vap 2 χ Equation found from experiment. Can we derive it? Colligative Properties
24 Colligative Property #2 : Freezing Point Depression ΔT = ( T ) R ΔH fus 2 χ Colligative Properties
25 Colligative Property #3 : Osmosis Pure Solvent () Solvent () + Solute () Membrane is permeable to solvent (), but not to solute (). Solvent will flow from the pure solvent side to the mixture s side. s a result there will be a pressure difference, π. Π = RT n V Colligative Properties
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