Boiling point elevation and freezing point depression

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1 Class 4.2 More Colligative Properties, Solutions & Distillation CHEM 102H T. Hughbanks Boiling point elevation and freezing point depression When nonvolatile solutes are present in a solvent, the solution s boiling point rises and freezing point declines. For very dilute solutions, these effects depend only on what the solvent is and the the solute concentration. The origin of this behavior is basically the same as the origin of osmotic pressure. Entropy of the solute gas decreases when volume decreases in vaporization V V V m n moles of solute molecules are forced into the decreased solvent volume when a mole of solvent leaves (V m = volume of a mole of solvent). S solute = nrln[(v-v m )/V] = nrln[1 - (V m /V)] Since V >> V m, ln[1 - (V m /V)] - V m /V S solute = - nr(v m /V) = - (n/v)rv m

2 Total Entropy of vaporization is less for the solution than for the pure solvent S vap,soln = S vap,pure + S solute ( S solute < 0) T b,soln = H vap,pure / S vap,soln - for dilute solution = H vap,pure /[ S vap,pure + S solute ] [ H vap,pure / S vap,pure ][1 - ( S solute / S vap,pure )] = T b,pure - T b,pure ( S solute / S vap,pure ) Therefore, T b,soln = T b,pure + T b,pure [(n/v)rv m / S vap,pure ] Boiling point elevation T b,pure [(n/v)rv m / S vap,pure ] V m = L solvent ; mol solvent n V V mol solute m! mol solution n mol solute = V L solution mol solute kg solvent = " kg solvent mol solvent = m M solvent ; where m is the molality 1000 (M solvent /1000)[RT b,pure / S vap,pure ] m (M solvent /1000)[RT 2 b,pure/ H vap,pure ] m Calculated and Observed k b s Solvent Boiling point at 1 atm, K!H vap kj/mol Calc. Obs. Water (CH 3 ) 2 CO CCl CH 3 Cl CH 3 CH 2 OH CH 3 OH (CH 3 CH 2 ) 2 O C 6 H

3 Freezing point depression T f,pure [(n/v)rv m / S fus,pure ] (M solvent /1000)[RT 2 f,pure/ H fus,pure ] m Note: On p. 307, it states that the freezing point constant of the solvent is different for each solvent and must be determined experimentally This is true, but it doesn t mean that it isn t theoretically predictable from other properties of the solvent, as this equation s derivation shows. Boiling point elevation and freezing point depression Raoult s Law The vapor pressure, P, of a solvent over a solution (with a nonvolatile solute) is proportional to the solvent mole fraction, x solvent : P = x solvent P pure Like the colligative properties already discussed, this can be shown by considering the change in entropy associated with the presence of the solute. Applies well to dilute solutions and to certain concentrated ideal solutions.

4 Derivation of Raoult s Law liq(soln ) gas - RT ln K = - RT ln P = ΔG vap = ΔH vap - TΔS vap For the solution, ΔS vap = ΔS vap,pure + ΔS solute For a dilute solution, ΔH vap = ΔH vap,pure Therefore, -RT ln P = ΔH vap,pure - TΔS vap,pure - TΔS solute -RT ln P = -RT ln P pure - TΔS solute -RT ln (P/P pure ) = - nrtln[1 - (V m /V)] ln (P/P pure ) = ln[1 - (V m /V)] n Derivation of Raoult s Law, cont. liq(soln ) gas (P/P pure ) = [1 - (V m /V)] n (P/P pure ) [1 - n(v m /V)] = [1 - (n/v)v m ] n/v = (mol solute)/(l solution) V m = (L solvent)/(mol solvent) 1 - (n/v)v m (mol solute)/mol(solute + solvent) (P/P pure ) = x solvent P = x solvent P pure Example (a) Calculate the vapor pressure of water over a solution prepared by dissolving 10.0 g of glucose (C 6 H 12 O 6 ) in 100 g of water. (b) Sucrose is a disaccharide formed by condensation of one molecule of glucose and one molecule of fructose. Will the vapor pressure of water of a solution made from 10.0 g of sucrose and 100 g H 2 O be greater or less than the in the case of glucose?

5 Binary Liquid Mixtures When a mixture of two liquids, A and B, both obey Raoult s Law (ideal mixture), the vapor pressures of each gas over the solution are: P A = x A,liquid P A,pure P B = x B,liquid P B,pure Binary Liquid Mixtures P total = P A + P B = x A,liquid P A,pure + x B,liquid P B,pure The gas phase mole fractions are: x A,gas = P A /(P A + P B ); x B,gas = P B /(P A + P B ) x A,gas = x A,liquid P A,pure /(x A,liquid P A,pure + x B,liquid P B,pure ) x B,gas = x B,liquid P B,pure /(x A,liquid P A,pure + x B,liquid P B,pure ) A is more volatile, B is less volatile (i.e., P A,pure > P B,pure ). Do some basic algebra: x A,gas = x A,liquid /[x A,liquid + x B,liquid (P B,pure /P A,pure )] x B,gas = x B,liquid /[x A,liquid (P A,pure /P B,pure ) + x B,liquid ] Binary Liquid Mixtures, cont. Remember, x A,liquid + x B,liquid = 1 x A,liquid + x B,liquid (P B,pure /P A,pure ) < 1 x A,gas > x A,liquid x A,liquid (P A,pure /P B,pure ) + x B,liquid > 1 x B,gas < x B,liquid The vapor is richer in the more volatile A - as we would expect!

6 Ideal Binary Liquid Behavior A is more volatile than B (P A,pure > P B,pure ) x A,liquid + x B,liquid = 1 x A,gas > x A,liquid x B,gas < x B,liquid Distillation This figure (from text) shows how the vapor phase is richer in the volatile benzene (b.p. 80.1) in a mixture with toluene (b.p ). Note that if the vapor mixture is condensed it would boil at lower T. Fractional Distillation Successive distillations give vapor fractions that are increasingly rich in the volatile component, benzene (b.p. 80.1) in a mixture with toluene (b.p ).

7 ΔH mix > 0 Non-ideal Behavior; Azeotropes When the interactions between solute and solvent molecules are dissimilar, Raoult s Law doesn t apply (ΔH mix 0). Negative deviations from Raoult s Law are seen in this low-boiling azeotrope.

Colligative Properties. Vapour pressure Boiling point Freezing point Osmotic pressure

Colligative Properties. Vapour pressure Boiling point Freezing point Osmotic pressure Colligative Properties Vapour pressure Boiling point Freezing point Osmotic pressure Learning objectives Describe meaning of colligative property Use Raoult s law to determine vapor pressure of solutions