CHEMISTRY FORM 6 CHAP 08 NEW.pdf

CHEMISTRY FORM 6

PHYSICAL CHEMISTRY

CHAPTER 8

PHASE EQUILIBRIA

12.1 Introduction to Ideal solution

Miscible liquids Immiscible liquids Partially miscible liquid

- Formed when 2 solution

can dissolve completely

with each other and

formed a homogeneous

solution

- Example : ethanol and

water

- Formed 2 separate layers

of solution when 2

immiscible solutions are

added together.

- Example : Benzene and

water

- When 2 solutions are

mixed, only a little

amount of the solution

can mixed with each

other. There are still 2

visible layers observed

- Example : ether and

water

• An ideal solution formed when 2 solutions are mixed together,

▫ Intermolecular forces formed between A---A = B---B =A---B.

▫ There’s no change of volume on mixing

▫ No enthalpy change occur on mixing the solutions

• As a result, the partial pressure exerted by each individual solution of an ideal solution will be lesser than the pressure exerted by its pure liquid, as the volume of the mixing solution increased.

• Since it is a mixing solution, the vapour pressure exerted by each individual solution depends on the composition of A and B. The relationship between the vapour pressure and the composition of liquid mixture is given by Raoult’s Law which state that the partial pressure of A in a solution at a given temperature is equal to the vapour pressure of pure A multiplied with the mole fraction of A in the solution (or equation, PA = xA . P

0A )

• From the pressure exerted through vaporisation, using Dalton’s Law Partial pressure of gas.Law Partial pressure of gas.

• Total vapour pressure, Ptot = PA + PB• Partial pressure of A = mol fraction of A x Vapour pressure of

PA = xA • PoA• Partial pressure of B = mol fraction of B x Vapour pressure of B

PB = xB • PoB• From the pressure exerted through vaporisation, using Dalton’s Law Partial pressure of gas.

Total vapour pressure, Ptot = PA + PB

Vapour Pressure / kPa

P0B

PoAA

0 1

mol fraction of B

• Example : Given the vapour pressure of hexane and heptane are 10 kPa and 3 kPa respectively. If the mole fraction of hexane in liquid mixture of hexane and heptane is 0.4, calculate the mole fraction of heptane in the vapour that is in equilibrium with liquid mixture

Solution : mol fraction of heptane in liquid mixture = 1 – 0.4 = 0.6

Phex = xhex • Phex = 0.4 (10 kPa) = 4.0 kPa

Phep = xhep • Phep = 0.6 (3 kPa) = 1.8 kPa

P = P + P = ( 4.0 + 1.8 ) kPa = 5.8 kPaPtotal = Phex + Phep = ( 4.0 + 1.8 ) kPa = 5.8 kPa

Mol fraction of heptane in vapour = PB / Ptot= 1.8 kPa / 5.8 kPa = 0.31

• Example : Mixing benzene with methylbenzene gives an ideal solution. If the vapour pressure of benzene and methylbenzene are 90 kPa and 78 kPa respectively. Calculate the mole fraction of benzene in the vapour at a mixture composition of

a) mole fraction of benzene = 0.20

PB = xB . P0B PMB = xMB . P0

MB

PB = 0.20 x 90 kPa PMB = 0.80 x 78 kPa

PB = 18 kPa PMB = 62.4 kPa

Ptot = 18 + 62.4 = 80.4 kPa

In vapour

b) mole fraction of methylbenzene = 0.35

PB = xB . P0B PMB = xMB . P0

MB

PB = 0.65 x 90 kPa PMB = 0.35 x 78 kPa

PB = 58.5 kPa PMB = 27.3 kPa

Ptot = 58.5 + 27.3 = 85.8 kPa

In vapourIn vapour

Mol fraction, xB = PB / Ptot

xB = 18 / 80.4

= 0.224

In vapour

Mol fraction, xB = PB / Ptot

xB = 58.5 / 85.8

= 0.682

• For an ideal solution, the boiling point – composition graph is the inverse of vapour pressure – composition graph

▫ If solution A is more volatile than solution B, the vapour pressure of A is ……………….. than in B.

▫ In the vapour pressure – composition graph, the straight line Ptot also represent the composition of the liquid phase, whereas the curve represent the composition of the vapour phase

kPa oC

HIGHER

0 1 0 1

Vapour pressure – composition graph Boiling point – composition graph

12.2 Fractional Distillation

• The most efficient method for separating an ideal solution is by using fractional distillation. The typical apparatus for carry out fractional distillation is shown in figure below

i. The distillation column is a long

tube filled with glass beads or short

pieces of glass tubing. The

distillation column provides a large

surface area where condensed liquid

and ascending vapour can have

maximum contact.maximum contact.

ii. The length of the distillation

column is largely determined by the

difference in boiling point of the

components to be separated, the

smaller the difference ; the longer

the distillation column needed to

achieve good separation.

12.3 Non – ideal solution

• In a non-ideal solution :

▫ The intermolecular forces of attraction between A---B could be stronger or weaker than A---A and B---B

▫ Thus, the partial vapour pressures and the total pressure measured can be lower or higher than expected.

▫ When this occur, we said that the mixture solution formed deviates from Ideal Solution.

• In the deviation from ideal solution, it can be categorise into 2 • In the deviation from ideal solution, it can be categorise into 2 type of deviations

• a) Positive deviation b) Negative deviation

Negative deviation Positive deviation

Example

HCl – H2O ; HNO3 – H2O ; H2SO4 –

HCOOH ; phenol – aniline ; acetone

(CH3COCH3) – CHCl3

Propanol – H2O ; ethanol – H2O ;

ethanol – toluene ; ethanol and

benzene

Happen

when

Intermolecular forces between A---B

molecules are stronger than in

between A---A and B---B

Intermolecular forces between A---B

molecules are weaker than in

between A---A and B---B

Partial

pressure

of

mixture

Since the intermolecular forces are

stronger in A---B ; the partial vapour

pressure are lower than expected in

Raoult’s Law


weaker in A---B ; the partial vapour

pressure are higher than expected in

Raoult’s Law

Heat

changes


stronger, meaning that A --- B are

more stable than their individual

forces, thus the process is exothermic


weaker, meaning that A --- B are less

stable than their individual force thus

the process is endothermic


Reason

HCl and H2O

The forces hold between HCl – HCl

are pure Van DerWaals forces

whereas the forces between H2O –

H2O are pure hydrogen bonding

However, when HCl and H2O are

mixed together, it will ionised to form

HCl + H O � H- O+ + Cl-

Ethanol, C2H5OH and H2O

The intermolecular forces between

ethanol – ethanol are hydrogen

bonding and same goes to between

water – water.

However, when this 2 solutions are

mixed, the hydrogen between water HCl + H2O � H-3O+ + Cl-

The ionic interactions between

hydrated ions are stronger than both

Van Der Waals forces and hydrogen

bond.

mixed, the hydrogen between water

are broken by ethanol and reduce the

intermolecular forces formed between

ethanol – water.

Volume

of

mixture

Since mixing the solution causes a

stronger attraction forces formed

between A---B ; so when 50 cm3 A +

50 cm3 B < 100 cm3.

Since mixing the solution causes a

weaker attraction forces formed

between A---B ; so when 50 cm3 A +

50 cm3 B > 100 cm3.


Vapour

pressure –

compositi

on curve

Boiling

point –

compositi

on curve


Mixture b.p. (oC)Azeotrope

b.p. (oC) composition

Water 100120.5

31.8

HNO3 86 68.2

CHCl3 61.2

64.5

80

CH3COCH3 56.2 20

Mixture b.p. (oC)Azeotrope

b.p. (oC) composition

C2H5OH 78.578.0

95.6

Water 100 4.4

CHCl3 61.259.4

93

C2H5OH 78.5 7

Negative deviation

vapour

Boiling

First distillate :Pure B2nd Distillate :Azeotropic mixtureResidue

liquid

Boiling point of pure A

Boiling point of pure B

First distillate :Pure A2nd Distillate :Azeotropic mixtureResiduePure B

ResiduePure A

Positive deviation

vapour

Boiling point of

Boiling point of pure B

First distillate :Azeotropic mixture2nd Distillate :Pure AResiduePure B

liquid

point of pure A

First distillate :Azeotropic mixture2nd Distillate :Pure BResiduePure A

12.4 Fractional distillation under reduced pressure

• Distillation under reduced pressure is usually used to distil out substances that have high boiling point.

• A substance will only boiled when its vapour pressure reaches the external pressure

• Some organic solution decomposed at temperature below its boiling point at atmospheric pressure. It is thereforeundesirable to distil such liquid at normal boiling point.

• In such case, there’s a necessity to distil at reduced pressure• In such case, there’s a necessity to distil at reduced pressure

• To do this a vacuum pump is connected to side arm to reduce the pressure.

• So, if the external pressure is reduced, the boiling point is also reduced (less heat is required), there’s no need to distillate at its normal boiling point and therefore, may prevent the decomposition of the organic substance

• The disadvantage of reduced pressure distillation• The apparatus has to be strong to withstand reduced pressure• The liquid boils irregularly• Other impurities with low boiling point may also be distilled.

P/atm

1.0

T/0C

1.0

0.8

0.5

a < b < c

1. The vapour pressure of water at 26°C is 3 350 Pa. What is the vapour pressure of a solution containing 72 g glucose, C6H12O6, in 100 cm3 water? (Given density of water = 1 g / cm3)

2. The vapour pressure of pure benzene at 30°C is 0.1600 atm. When 20 g of a non-volatile solvent, X, is dissolved in 180 g benzene, the vapour pressure of the resulting solution is 0.1572 atm. Determine the relative molecular mass of the solvent X if the relative molecular mass of benzene is 78.

Mol of glucose = 72 g / 180

= 0.40 mol

Mol of water = 100 g / 18

= 5.56 mol

Xwater = 5.56 / (5.56 + 0.40)

= 0.93

[email protected] B

B

0

BBB ==water

= 0.93

Pwater = Xwater x P0w

= 0.93 x 3350

= 3120 Pa

6.486M

1600.0

1572.0

[email protected]

R

M20

78180

78180

0

B

BBBB

R

=

=+

==

3. Calculate the mass of potassium chloride, KCl, that must be added to 200 cm3 of water to lower its vapour pressure by 350 Pa at 37°C? (The vapour pressure of water at 37°C is 7 310 Pa).

4. At temperature T, the vapour pressure of pure benzene and ethanol are 120 mm Hg and 50 mm Hg respectively. Assuming that these liquids form an ideal solution when mixed, plot a vapour pressure-composition graph for the mixture. If mole fraction of ethanol in the solution mixture is 0.5, determine from the graph(a) the mole fraction of benzene in the vapour phase(b) the partial vapour pressure of ethanol

w

w0

w

ww

n

952.0x

7310

6960x

P

Px

=

=⇒=

(b) the partial vapour pressure of ethanol(c) the total vapour pressure

Mass of KCl = 41.7 g

5.74m

18200

18200

w

KClw

ww

x

nn

nx

+=

+=

a) 0.71

b) 25 mm Hg

c) 85 mm Hg

5. Methylbenzene and benzene forms an ideal solution. The vapour pressure of pure benzene and methylbenzene are 122 mm Hg and 37 mm Hg respectively at room temperature. A solution mixture contains 0.40 mole fraction of benzene.(a) Determine the total vapour pressure of the solution mixture.(b) What is the composition of the vapourwhich is in equilibrium at room temperature with the above

6. The vapour pressure of pure methanol, CH3OH, at 298 K is 12.8 kPa. How many moles of a non-volatile solute X per mole of methanol is required to prepare a solution of methanol having a vapourpressure of 10.6 kPa at 323 K?

xM = PM / P0M

= 10.6 / 12.8

= 0.828

nx = 1.00 – 0.828

= 0.172 moltemperature with the above methylbenzene-benzene mixture?

a) PB = 0.40 x 122 mm Hg

= 48.8 mm Hg

PMB = 0.60 x 37 mm Hg

= 22.2 mm Hg

Ptot = 48.8 + 22.2 = 71.0 mm Hg

b) XB = 48.8 / 71.0

= 0.687

XMB = 1.0 – 0.687 = 0.313

= 0.172 mol

There’s 0.172 mol of solute / 0.828 mol of

benzene. Mole of solute / benzene

0.172 / 0.828 = 0.208 mol solute / benzene

7. The vapour pressure of water at 75°C is 0.39 atm. What are the partial vapourpressures of water in mixtures of(a) 100.0 g water and 20.0 g glucose(b) 100.0 g water and 20.0 g NaCl(c) Comparing the same mass of glucose and sodium chloride dissolved in water which solution will boil at a lower temperature?

a) nw = 100.0 g / 18 = 5.56 mol

ng = 20.0 g / 180 = 0.11 molg

Pw = xw x P0w

= 5.56 / (5.56+0.11) x 0.390

= 0.382 atm

b) nNaCl = 20.0 g/ 58.5 = 0.342 mol

Pw = xw x P0w

= 5.56 / (5.56+0.342) x 0.390

= 0.368 atm

c) Glucose solution – the water vapour

pressure is only slightly lowered.

CHEMISTRY FORM 6 CHAP 08 NEW.pdf

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