phase equilibrium ii - two component system

Phase Equilibrium II- Two Component System

• 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%???

glucoseMixture of

ethanol & waterfermentation filtration

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

BA

AA nn

n

• 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 + XBPB

o)

Example 21.1

• nA = 0.2mol ; nB = 0.5 mol ; Pt = 40mmHg ; PA

o = 20mmHg ; PBo = ?

• XA = 0.2 / (0.2+0.5) = 0.286XB = 0.5 / (0.2+0.5) = 0.714

• Pt = XAPAo + XBPB

o

40 = (0.286)(20)+(0.714) PBo

• PBo= 48mmHg

XA+ XB=1

Ex 21.1

• Pmethanolo = 90mmHg : Pethanol

o = 45mmHg ; Pt = ?

• nmethanol = 64/32 = 2mol nethanol = 92/46 = 2mol

• Xmethanol = 2 / (2+2) = 0.5Xethanol = 2 / (2+2) = 0.5

• Pt = XmethanolPmethanolo + XethanolPethanol

o

Pt = (0.5)(90)+(0.5)(45)

• Pt= 67.5mmHg

Ex 21.1

• PPo = 20kPa : PQ

o = 8kPa ; Pt = ? nP = 6mol nQ = 2mol

• XP = 6 / (6+2) = 0.75XQ = 2 / (6+2) = 0.25

• Pt = XPPPo + XQPQ

o

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 in pure A

Strength of Intermolecular attraction in pure B

Strength of Intermolecular attraction between particle A and particle B in mixture

+

• 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 ar

e 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

• 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!!!

Vapour pressure-composition curve

• for ideal solution1. PA XA & PB 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%)

Composition of liquid:XA = 0.5 ; XB = 0.5

Liquid contains 1 mol A ( )and1 mol B ( )PA

o =10kPa ;PBo = 30kPa

Composition of vapour?

Partial pressure of A = 10 x 0.5 = 5kPaPartial pressure of B = 30 x 0.5 = 15kPaTotal vapour pressure = 5 + 15 = 20 kPamole fraction of A in vapour = 5/20 = 0.25mole 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 > XBPB

o ===> Pt > XAPA

o + XBPBo

• Molecules A & B in the mixture escape from the liquid surface more than that expected for an ideal 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

CH3CH2 O

H

CH3CH2 O

H

CH3CH2

OH

Weak Van der Waal’s forcein pure cyclohexane Strong hydrogen bonding

in pure ethanol

CH3CH2 O

H

CH3CH2 O

H

Weak dipole-induced dipoleinteraction 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 endothermicEnergy(breaking interaction between molecules in pure solution) > Energy(formation of new interaction between 2 kinds of molecules in mixture)

• ===>Volume expansionweaker 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 < XBPB

o ===> Pt < XAPA

o + XBPBo

• Molecules A & B in the mixture has a lower escape tendency from the liquid surface compared with that of ideal 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 forcein pure ethyl ethanoate Weak Van der Waal’s force

in pure trichloromethane

Strong hydrogen bonding in the mixture

O CCH3

OC2H5

O CCH3

OC2H5CHCl

Cl

Cl CHCl

Cl

Cl

HC

Cl

ClCl

O CCH3

OC2H5

• the interaction is now stronger after mixing the solution

• ===> escape tendency of molecule is lower

• ===> Pt is lower than expected

• ===> enthalpy change is exothermicEnergy(breaking interaction between molecules in pure solution) < Energy(formation of new interaction between 2 kinds of molecules in mixture)

• ===>Volume contractionstronger attraction between molecules in the mixture=> intermolecular distance decrease

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

Explanation

1. The intermolecular force in pure water is strong hydrogen bond.

2. The attraction between benzene molecules is weak Van der Waals’ force in the pure solution.

3. However, the intermolecular force between water and benzene is weak Van der Waals’ force in the mixture.

4. 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.

Ex. 21.7Set 1 2 3 4 5 6 7 8

Composition of liquid

mixture (XA)1 0.9 0.7 0.6 0.5 0.3 0.1 0

Composition of liquidmixture (XB)

Partial pressure of A(PA)

Partial pressure of B (PB)

Vapour pressure of themixture (Pt)

Composition of vapour(XA')

Composition of vapour(XB')

Set 1 2 3 4 5 6 7 8

Composition of liquid

mixture (XA)1 0.9 0.7 0.6 0.5 0.3 0.1 0

Composition of liquidmixture (XB)

0 0.1 0.3 0.4 0.5 0.7 0.9 1

Partial pressure of A(PA)

30 27 21 18 15 9 3 0

Partial pressure of B (PB) 0 5 15 20 25 35 45 50

Vapour pressure of themixture (Pt)

30 32 36 38 40 44 48 50

Composition of vapour(XA')

1 0.844 0.583 0.474 0.375 0.205 0.0625 0

Composition of vapour(XB')

0 0.156 0.417 0.526 0.625 0.795 0.9375 1

Richer with B comparing with than in liquid

• 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 0.45Mole fraction of A in vapour is about 0.55

PBo

PAo

B.p.(A)

B.p.(B)

• (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

x(l)

y(l)

Tx

y(g)

z(l)

Ty

z(g)

y(l)

Tz

z(l)

Repeating boiling and condensationTedious!!!

Dec

reas

ing

b.p

.Fractional distillation

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 BStart 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 MResidue = pure B

• Start from Y, distillate and residue?

• Distillate =Azeotropic mixture MResidue = pure A

• start from M, distillate and residue?


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 21.8a) A constant boiling mixture (azeotropic mixture) is a mixt

ure 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.

phase equilibrium ii - two component system

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