Phase Equilibrium II - Two Component System

Mixture of ethanol & water glucose Mixture of ethanol & water fermentation filtration How many components and phases in this system? 2 components and 1 liquid phase Method to separate ethanol from water?? Fractional distillation? How? obtained only about 95% of ethanol Why not 100%???

21.1 Two Component System - Mixture of 2 Miscible Liquids vapour pressure of a liquid (Pv) - is a measure of the escape tendency of the liquid molecules - influenced by its composition and the intermolecular interaction between the 2 components composition of a solution - shown by the mole fraction ( ) Ideal solutions - mixtures of liquids of similar chemical structures - linear relationship of boiling point and composition

non-ideal solutions?? - mixtures of liquids having dissimilar structures - have marked deviations from the linear boiling point - composition relationship

21.2 Raoult's Law for Ideal Solutions states that the vapour pressure of a component in a mixture at a given temperature is directly proportional to its mole fraction, and is equal to the product of its mole fraction and the vapour pressure of the pure component at that temp. partial pressure of A: PA = XAPAo partial pressure of B: PB = XBPBo (Po is vapour pressure; X is mole fraction) total pressure of the mixture Pt = PA + PB (Pt = XAPAo + XBPBo)

Example 21.1 nA = 0.2mol ; nB = 0.5 mol ; Pt = 40mmHg ; PAo = 20mmHg ; PBo = ? XA = 0.2 / (0.2+0.5) = 0.286 XB = 0.5 / (0.2+0.5) = 0.714 Pt = XAPAo + XBPBo 40 = (0.286)(20)+(0.714) PBo PBo= 48mmHg XA+ XB=1

Ex 21.1 Pmethanolo = 90mmHg : Pethanolo = 45mmHg ; Pt = ? nmethanol = 64/32 = 2mol nethanol = 92/46 = 2mol Xmethanol = 2 / (2+2) = 0.5 Xethanol = 2 / (2+2) = 0.5 Pt = XmethanolPmethanolo + XethanolPethanolo Pt = (0.5)(90)+(0.5)(45) Pt= 67.5mmHg

Ex 21.1 PPo = 20kPa : PQo = 8kPa ; Pt = ? nP = 6mol nQ = 2mol XP = 6 / (6+2) = 0.75 XQ = 2 / (6+2) = 0.25 Pt = XPPPo + XQPQo Pt = (0.75)(20)+(0.25)(8) Pt= 17 kPa

Mixtures of liquid that obey Raoult's Law are known as ideal solutions + Strength of Intermolecular attraction between particle A and particle B in mixture  Strength of Intermolecular attraction in pure A Strength of Intermolecular attraction in pure B 

===> no enthalpy change upon mixing the escape tendency of molecule A or B in mixture equals to their respective escape tendency in pure A and pure B ===> no enthalpy change upon mixing ===> no volume change upon mixing Reasons: interaction doesn’t change upon mixing more nearly alike the two substances are chemically, the more nearly they exhibit ideal behaviour Actually, ideal solutions are seldom found and their existence only holds for very dilute solutions.

Ex. 21.3 Which of the following solution mixtures are ideal solutions? a) benzene and methylbenzene b) cyclohexane and ethanol c) bromomethane and iodomethane d) propanone and trichloromethane e) propan-1-ol and propan-2-ol f) n-hexane and n-heptane it’s a rough prediction only!!! Which of the following solution mixtures are ideal solutions? a) benzene and methylbenzene b) cyclohexane and ethanol c) bromomethane and iodomethane d) propanone and trichloromethane e) propan-1-ol and propan-2-ol f) n-hexane and n-heptane

Vapour pressure-composition curve for ideal solution 1. PA a XA & PB a XB 2. Pt = PA + PB ===> gives a straight line in the vapour pressure-composition curve Information from graph: pure A is more volatile than pure B

boiling point-composition curve For ideal solution: a linear variation of the b.p. with the composition of the solution slope of the graph is opposite to the vapour pressure-composition curve more volatile liquid, the vapour pressure is higher at a fixed temperature ==> a lower temperature is required to set up the Pv ==> the liquid has a lower b.p.

Ex 21.4 a) b) P(octane) = (0.3)(19.0) =5.7mmHg P(2-methylheptane) =(1-0.3)(27.0)=18.9mmHg Pt = 5.7 + 18.9 = 24.6mmHg c) It is assumed that the intermolecular interaction between octane and 2-methylheptane molecules in the mixture is the same as the that among octane and 2-methylheptane molecules in their pure solutions. (it is assumed that the solution mixture is ideal solution.) d) b.p. = (126)(0.3) +(112)(0.7) = 116.2oC

Ex 21.5 a) nmethanol = 20/32 = 0.625mol nethanol = 100/46 = 2.174mol Xmethanol = 0.625 / (0.625+2.174) = 0.223 Xethanol = 2.174 / (0.625+2.174) = 0.777 Pmethanol=(12.530)(0.223) = 2.79kPa Pethanol = (5.866)(0.777) = 4.56kPa b) Pt = 2.79 + 4.56 = 7.35kPa c) compositon of methanol in vapour = (2.79)/(7.35) = 0.380 (38%) composition of ethanol in vapour = (4.56)/(7.35) = 0.620 (62%)

Liquid contains 1 mol A ( )and 1 mol B ( ) PAo =10kPa ;PBo = 30kPa Composition of liquid: XA = 0.5 ; XB = 0.5 Composition of vapour? Partial pressure of A = 10 x 0.5 = 5kPa Partial pressure of B = 30 x 0.5 = 15kPa Total vapour pressure = 5 + 15 = 20 kPa mole fraction of A in vapour = 5/20 = 0.25 mole fraction of B in vapour = 15/20 =0.75

21.3 Deviations from Raoult's Law - Non-ideal Solution How about those liquid mixtures do not obey Raoult's law?? ===> Non-ideal Solution 1) Solution with positive deviation from Raoult’s Law 2) Solution with negative deviation from Raoult’s Law

Positive Deviation from Raoult's Law Occurs when Pt > that predicted by Raoult's law i.e. PA > XAPAo & PB > XBPBo ===> Pt > XAPAo + XBPBo Molecules A & B in the mixture escape from the liquid surface more than that expected for an ideal mixture. Strength of Intermolecular attraction in pure A Strength of Intermolecular attraction in pure B + Strength of Intermolecular attraction between particle A and particle B in mixture > Average

The vapour pressure would be higher and the boiling point would be lower compared with ideal behaviour

Example: Binary mixture of cyclohexane and ethanol Weak Van der Waal’s force in pure cyclohexane Strong hydrogen bonding in pure ethanol Weak dipole-induced dipole interaction in the mixture

the interaction is now weaker after mixing the solution ===> escape tendency of molecule is higher ===> Pt is greater than expected ===> enthalpy change is endothermic Energy(breaking interaction between molecules in pure solution) > Energy(formation of new interaction between 2 kinds of molecules in mixture) ===>Volume expansion weaker attraction between molecules in the mixture => intermolecular distance increase

Negative Deviation from Raoult's Law Occurs when Pt < that predicted by Raoult's law i.e. PA < XAPAo & PB < XBPBo ===> Pt < XAPAo + XBPBo Molecules A & B in the mixture has a lower escape tendency from the liquid surface compared with that of ideal mixture. Strength of Intermolecular attraction in pure A Strength of Intermolecular attraction in pure B + Strength of Intermolecular attraction between particle A and particle B in mixture < Average

The vapour pressure would be lower and the boiling point would be higher compared with ideal behaviour

Example: mixture of ethyl ethanoate & trichloromethane Weak Van der Waal’s force in pure ethyl ethanoate in pure trichloromethane Strong hydrogen bonding in the mixture

the interaction is now stronger after mixing the solution ===> escape tendency of molecule is lower ===> Pt is lower than expected ===> enthalpy change is exothermic Energy(breaking interaction between molecules in pure solution) < Energy(formation of new interaction between 2 kinds of molecules in mixture) ===>Volume contraction stronger attraction between molecules in the mixture => intermolecular distance decrease

Ex. 21.6 a) Bromopropane and 2-bromobutane obey b) Tetrachloromethane and cyclohexane c) Benzene and methylbenzene d) Water and benzene positive deviation from Raoult’s Law e) Propanone and trichloromethane negative deviation from Raoult’s Law

Explanation The intermolecular force in pure water is strong hydrogen bond. The attraction between benzene molecules is weak Van der Waals’ force in the pure solution. However, the intermolecular force between water and benzene is weak Van der Waals’ force in the mixture. As the average strength of intermolecular force in the two pure solutions is greater than that in the mixture,

5. - the escape tendency of molecules in the mixture is higher than expected from the Raoult’s Law. Therefore, this mixture has a positive deviation compared with ideal solution. - the interaction between molecules in the mixture is weaker and the space between molecules is larger. Thus, there will be an expansion in volume after mixing. - the energy absorbed for breaking the intermolecular force between molecules in pure solutions is greater than the energy released for forming new intermolecular force between water and benzene molecules in the mixture. Thus, the process is endothermic and heat absorbed from the surroundings.

Richer with B comparing with than in liquid Ex. 21.7 Richer with B comparing with than in liquid

PAo B.p.(B) B.p.(A) PBo What is the mole fraction of A in liquid and in vapour when the vapour pressure is 1atm at ToC? Mole fraction of A in liquid is about 0.45 Mole fraction of A in vapour is about 0.55

(3) vapour y condense to liquid y (1) X boils at Tx (2) vapour y is obtained (3) vapour y condense to liquid y (4) liquid y boils at Ty and vapour z is obtained (5) vapour z condense to liquid z (6) liquid z boils at Tz (7) ...... Pure A is obtained as final distillate XB

Repeating boiling and condensation x(l) y(l) Tx y(g) z(l) Ty z(g) y(l) Repeating boiling and condensation Tedious!!! Tz z(l)

Fractional distillation Decreasing b.p.

fractionating tower for petroleum refining

21.5 Azeotropic Mixtures Liquid mixtures which deviate negatively from Raoult's law show a maximum in the boiling point-composition curve Start from x ==> distillate = pure B Start from w ==> distillate = pure A Start from M ==> distillate = remain M M is azeotropic mixture/ constant boiling mixture (vapour composition = liquid composition upon boiling)

Start from x, residue? Azeotropic mixture M Start from w, residue? start from M, residue? What to do for M? Discard?? Think about it !!!

Positive Deviation from Raoult’s Law Liquid mixtures which deviate positively from Raoult's law show a minimum in the boiling point-composition curve Start from X, distillate and residue? Distillate = Azeotropic mixture M Residue = pure B Start from Y, distillate and residue? Distillate = Azeotropic mixture M Residue = pure A start from M, distillate and residue? Azeotropic mixture M X Y

impossible to separate a non-ideal solution into pure components by fractional distillation! What should we do with azeotropic mixture?? Discard??? Solve the problem??? Add a little amount of one of the component and fractional distillation can be continued!!!

Ex 21.8 a) A constant boiling mixture (azeotropic mixture) is a mixture of liquids with a fixed composition and it cannot be separated by fractional distillation since the vapour composition is the same as the liquid composition. b) b.p. - composition diagram for HNO3 - water mixture

c) According to the diagram, it is known that the mixture deviates negatively from Raoult’s Law. There will be a stronger attraction between molecules in the mixture. When nitric acid is added to water, there is an evolution of heat. That means the reaction is exothermic. Moreover, there is a reduction is volume when nitric acid and water are mixed. d) Raoult’s Law states that the partial vapour pressure of a component in a mixture is directly proportional to its mole fraction, and is equal to the product of its mole fraction and the vapour pressure of the pure component at that temperature.

e) The total vapour pressure above the nitric acid - water mixture is less than the predicted value based on the ideal behaviour. It shows a negative deviation from Raoult’s law meaning that there is less tendency for the molecules to escape from the solution than from the pure liquids. f) The temperature of the mixture would rise gradually until it reaches 122oC. The residual mixture is more concentrated in nitric acid as water is distilled off until its composition is 65% by mass of nitric acid.