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PHASE EQUILIBRIA PHASE EQUILIBRIA

PHASE EQUILIBRIA - PowerPoint Presentation

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PHASE EQUILIBRIA - PPT Presentation

INTRODUCTION At sea level At top of a mountain Water boils at 100 C Water boils at lt 100 C bP fP P 10133 bar bp 100 C P 026 bar bp 69 C elevation 8848 m ID: 276829

solid diagram phase solution diagram solid solution phase liquid system composition water solvent solute crystal temperature component enthalpy point eutectic binary btu

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Slide1

PHASE EQUILIBRIASlide2

INTRODUCTION

At sea level

At top of a mountain

Water boils at 100

C

Water boils at < 100C

b.P

= f(P)Slide3

P = 1.0133 bar

bp

= 100CSlide4

P = 0.26 bar

bp

= 69C

elevation = 8848 mSlide5

P = 2

atm , b.p = 124C+ EG 50/50 (v/v), b.p =129.5  C =>

b.p = f(P, c)Slide6

stability map

stability of the state of aggregation

Depending on the circumstances, molecules aggregate to form solid, liquid, or gas

If I specify P, T, and c, what are the stable phases?

Phase diagramSlide7

Examples of single-phase system (

 = 1):Pure water (l)White gold (alloy of Au-Ag-Ni) => Ag and Ni substitute Au in its FCC lattice => homogeneous composition throughout => solution

Air (N

2, O

2,

Ar, CO2

)

Phase = region of a substance that is:

Uniform in chemical composition

physically distinct

mechanically separable

Examples of dual-phase system (

 = 2)

:

Ice cubes in liquid water

Milk (fat globules in aqueous solution) Slide8

Equilibrium =

the condition which represents the lowest energy level the properties are invariant with time

Component = the measure of chemical complexity

N

1

2

1

> 1

Water (l)

diamond

Ice cube in liquid water (slush)

White gold

(Au-Ag-Ni)

CCl

4

– H

2

OSlide9

ONE-COMPONENT SYSTEM (N = 1)

Phase diagram of water

P (

atm

)

T (

C

)

1

100

Solid

 = 1

Gas

 = 1

Liquid

 = 1

l = v coexistence curve

 = 2

s = v coexistence curve

 = 2

s = l coexistence curve

 = 2

0

0.01

Triple point

 = 3

4.58 mm Hg

Critical point

(374

C, 218

atm

)Slide10

POLYMORPHS

Phase diagram of sulfur

B : 95,5

C and 0,51 PaC :

115C and 2,4 PaE :

151C and 1,31108

Pa

Different atomic arrangement at constant composition Slide11

The transformation from one polymorph to another can be :

Reversible : two crystalline forms are said to be enantiotropicIrreversibel : two crystalline forms are said to be

monotropicSlide12

Pressure-temperature diagram for

dimorphous substances: (a) enantiotropy, (b) monotropySlide13

TWO-COMPONENT SYSTEM (N = 2)

There are tree variables that can affect the phase

equilibria

of a binary system: T, P, and C

T

T

P

P

cSlide14

Two-component phase diagram

3 dimensionT-P-c

2 dimension

T-P

P-c

T-c

Crystallization: liquid and solid

The effect of P can be ignored

T-c Diagram Slide15

BINARY SYSTEM– TYPE 1

Complete solubility in solid and liquid states

Change of state (s = l)

Metal (

Hume-

Rothery

Rule)

Similar crystal structure

Similar atomic volumes

Small values of

electronegativity

For

interstitial

solid solutions, the Hume-

Rothery

rules are:

Solute

atoms must be smaller than the pores in the solvent

lattice.

The

solute and solvent should have similar

electronegativity

.Slide16

The

atomic radii of the solute and solvent atoms must differ by no more than 15

%:

For

substitutional solid solutions, the Hume-

Rothery rules are:

The

crystal structures

of solute and solvent must match.

Maximum

solubility

occurs when the solvent and solute have the same

valency

. Metals with lower

valency

will tend to dissolve in metals with higher

valency

.

The

solute and solvent should have similar

electronegativity

. If the

electronegativity

difference is too great, the metals will tend to form

intermetallic

compounds

instead of solid solutions.Slide17

T

c

A

B

Liquid

 = 1

Solid

 = 1

Liquid + solid

 = 2

Liquidus

SolidusSlide18

Liquidus

: l  s + l : the lowest T for only liquid (at any c)Solidus : s  s + l : the highest T for only solid (at any c)

The diagram represents the phase behavior of two components having many similarities (type of bonding, atomic size, crystal structure, etc), we call it

ISOMORPHOUS DIAGRAM

.

The shape of the diagram looks like a lens  lens-shape diagram / lenticular

diagram.

Example: naphthalene - -

naphtholSlide19

Solid solution of naphthalene -

-

naphthol

Similar

Slide20

Solid solution of naphthalene -

-naphthylamine

 >Slide21

BINARY SYSTEM– TYPE 2

Partial or limited solubility

 miscibility gap

No change of state

 always liquid or always solid

T

c

A

B

 = 2

l

= 1

s

l

1

+

l

2

 + 

Coexistence curveSlide22

Hexane - nitrobenzene

upper critical solution temperature (upper consolute temperature)Slide23

lower critical solution temperature (lower consolute temperature)Slide24
Slide25

BINARY SYSTEM– TYPE 3

Partial or limited solubility

Change of state

T

c

A

B

l

l

+

l

+

Solid solution (A rich)

Solid solution (B rich)

Liquid solution

= 1

= 1

= 1

= 2

= 2

= 2

Eutectic

l

EUTECTIC DIAGRAMSlide26

Freezing point depression of both components (A and B)

Eutectic point is equilibrium of , , and l

Eutectic point is unique; it only happens at one T, P, and cSlide27

Phase diagram for the simple eutectic system naphthalene - benzeneSlide28

Phase diagram for system

NaCl – H2O Slide29

ENTHALPY-COMPOSITION DIAGRAMSlide30

Solution A having composition of

xA and enthalpy of HA

is mixed adiabatically with solution A having composition of x

B

and enthalpy of HB

.The composition and the enthalpy of the mixture is calculated using material balance:

Total balance:

Component balance:

Combining both equations yields:Slide31

Similarly, if mixture A were to be removed adiabatically from mixture C, the enthalpy and composition of residue B can be located on the straight line through points A and C by means of the equation:Slide32

EXAMPLE

Calculate (a) the quantity of heat to be removed and (b) the theoretical crystal yield when 5000 lb of a 30 per cent solution of MgSO4

by mass at 110F is cooled to

70F. Evaporation and radiation losses may be neglected.

SOLUTION

(a)

Initial solution (A)

x

A

= 0.3 H

A

= - 31 Btu/lb

Cooled system (B)

x

B

= 0.3 H

B

= - 75 Btu/lb

Enthalpy change

H = - 44 Btu/lb

Heat to be removed = (- 44) (5000) = - 220000 Btu

(b)

The cooled system B is located in the region where MgSO

4

.7H

2

O is in equilibrium with solution.Slide33
Slide34

The crystal MgSO

4.7H2O contains:

x

C

= 0.49

From the graph:

x

A

= 0.26

The MgSO

4

.7H

2

O crystal yield is 869.6 lb