physical chemistry lecture note_02

8/12/2019 Physical Chemistry Lecture Note_02

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Physical Chemistry II(TKK-1237)

13/14 Semester 3

Instructor: Rama Oktavian

Email: [email protected]

Office Hr.: M.13-15, W. 13-15 Th. 13-15, F. 13-15

Outlines

1. Review

2. Liquid-liquid equilibria (2-components)

3. Liquid-liquid equil ibria (3-components)

4. Ternary diagrams

Review Review

Ch. 12Equilibrium condition

the chemical potential of each substance must have the samevalue in every phase in which that substance appears

a state in which there are no observable changes as time goes by.

Review

Ch. 12Phase diagram

Review

Ch. 12Phase rule

the phase rule for a one-component system

Gibbs Phase Rule


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Review

Ch. 13

SolutionSolution - homogeneous mixture of chemical species

One phase

Review

Ch. 13

Raoults Law and Ideal Solution (only one volatile componet)

Raoults law

Review

Ch. 14Raoults Law and Binary Ideal Solution

Review

Ch. 14

Gaseous phase

Partial pressure of component 1

Review

Ch. 14

Review

Ch. 14

P-x,ydiagram


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Review

Ch. 14

T-x,ydiagram

Review

Ch. 14

Azeotropes

Liquid-liquid equilibria

Basic concept of miscibility

1. Miscible e.g: Toluene-benzene

2. Partially miscible e.g: water-phenol

3. Immiscible e.g: water-nitrobenzene


Basic concept

A + B

Liquid (bottom layer)

A + B

Liquid (upper layer)

1

Ax

In equilibrium condition

2

A

1A

21

AA

Partially miscible solution

2

Ax


Partially miscible liquid

P= 2, F= 1 the selection of temperaturemakes the compositions of theimmiscible phases fixed

P= 1, F= 2 (two liquids are fullymixed) both temperature andcomposition can be changed


Partially miscible liquid

1. Add small amount of nitrobenzene tohexane at 290 K, it still dissolvescompletely, P= 1

2. Add more nitrobenzene to hexaneand mixture of nitrobenzene-hexanebecomes saturated, add morenitrobenzene, the mixture willbecome two phases (line 2-3).

3. In point 3, the mixture will becomesaturated (more nitrobenzene)

4. In point 4, the mixture will becomeone phase (hexane will dissolve innitrobenzene)


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Representation of liquid liquid phase diagram

Point A - Mixture of 50 g hexane (0.59mol C6H14) and 50 g nitrobenzene (0.41mol C6H5NO2) was prepared at 290 K

A

There will be two phases solution withthe composition at point 2 and point 3

xN= 0.35 and xN= 0.83 (these arethe compositions of the two phases



Use Lever-Rule to determine the ratio ofamount of each phase:

A

735.041.0

41.083.0

l

l

n

n

There is 7 times as much hexane-richphase as there nitrobenzene-richphase

If the mixture is heated to 292 K, wego into a single phase region




Critical solution temperature

1. The upper critical solution temperature, Tuc

2. The lower critical solution temperature, Tlc



1. The upper critical solution temperature, Tuc

The upper critical solution temperature, Tuc, is the highesttemperature at which phase separation occur



2. The lower critical solution temperature, Tuc

The lower critical solution temperature, Tlc, is the lowesttemperature at which phase separation occur

For triethylamine and water, thesystem is partially miscible above Tlc,and single phase below


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Some systems have both Tuc and Tlc, with a famous example being

nicotine in water, where Tuc= 210oC and Tlc= 61

oC


0 1Xnicotine

Temperature(oC)

X2X1 X3

T1210

o

C

61oC

T2

T3

nicotine /

water solution

nicotine

saturated

water rich

phase in

equilibrium

with a water

saturated

nicotine rich

phaseT4

lower

consulate

temperature

we cool a nicotine water solution of

composition X2 from sometemperature above the upper

consulate temperature of 210oC.

At temperatures greater than T1 the

nicotine and water are miscible

When T1 is reached water saturated

nicotine rich phase just begins to form and

is in equilibrium with the predominant

nicotine saturated water rich phase

As the system is further cooled there will be

two phase region. In the two phase region the

relative amounts of the phases present are again

given by the lever law, e.g. at T2 we have:

nX1 (X2 - X1) = nX3 (X3 - X2)


Distillation of partially miscible liquids

First case - the Tuc is lower than the azeotrope temperature



a1 initial composition and temperature

one phase

a2 the point where boil ing begins and the

vapor will have composition at b1

When the distillate is cooled enough

to cause condensation, a single phasefirst forms, represent by point b2

point b3 represents the overallcomposition once the temperature i slowered back to the startingtemperature



Another case - the Tuc is higher than the azeotrope temperature



a1 initial composition and temperature

one phase

It will start boiling at point a2 withvapor having composition given bypoint b1

This distillate will condense into a twophase liquid directly (b3).


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A system at e1 forms two phases up to the

boiling point at e2

condensing a vapor of composition e3gives a two-phase liquid of the sameoverall composition

At e2, F = 0, their compositions andthe temperature are fixed


Distillation of immiscible liquids

Immiscible liquids



Immiscible liquids

The total vapor pressures of liquids is





Example: Aniline(1)-water(2) system, we want to distill 100 g of

water from this mixture at 98.4C under atmospheric condition

mmHgp 4201

mmHgp 71802

The mass of aniline that distills for each 100 g of water


System of three components

Call Gibbs Phase Rule

P= 1, F= 4 T, P, x1, x2

P= 2, F= 3 T, P, x1


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Ternary phase diagram

How to read it

100% C

100% A

100% B


Ternary phase diagram

Ternary phase diagram for methyl isobutyl ketone + acetone + water

Liquid-liquid pha se

separation occurs

Binodal / cloud point curve

Plait point

physical chemistry lecture note_02

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