PHASE EQUILIBRIA

INTRODUCTION

At sea level At top of a mountain

Water boils at 100C Water boils at < 100C

b.P = f(P)

P = 1.0133 barbp = 100C

P = 0.26 barbp = 69C

elevation = 8848 m

P = 2 atm , b.p = 124C+ EG 50/50 (v/v), b.p =129.5 C => b.p = f(P, c)

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 diagram

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 (N2, O2, 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)

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)

CCl4 – H2O

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(374C, 218 atm)

POLYMORPHS

Phase diagram of sulfur

B : 95,5C and 0,51 PaC : 115C and 2,4 PaE : 151C and 1,31108 Pa

Different atomic arrangement at constant composition

The transformation from one polymorph to another can be :

1. Reversible : two crystalline forms are said to be enantiotropic

2. Irreversibel : two crystalline forms are said to be monotropic

Pressure-temperature diagram for dimorphous substances: (a) enantiotropy, (b) monotropy

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

c

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

BINARY SYSTEM– TYPE 1

Complete solubility in solid and liquid states

Change of state (s = l)

Metal (Hume-Rothery Rule)Similar crystal structureSimilar atomic volumesSmall values of electronegativity

For interstitial solid solutions, the Hume-Rothery rules are:

1. Solute atoms must be smaller than the pores in the solvent lattice.

2. The solute and solvent should have similar electronegativity.

1. 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:

2. The crystal structures of solute and solvent must match.

3. 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.

4. 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.

T

cA B

AmpT

BmpT

Liquid = 1

Solid = 1

Liquid + solid = 2

Liquidus

Solidus

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 - -naphthol

Solid solution of naphthalene - -naphthol

Similar

Solid solution of naphthalene - -naphthylamine

>

BINARY SYSTEM– TYPE 2

Partial or limited solubility miscibility gap No change of state always liquid or always solid

T

cA B

= 2

lp= 1

s

l1 + l2

+

Coexistence curve

Hexane - nitrobenzene

upper critical solution temperature (upper consolute temperature)

lower critical solution temperature (lower consolute temperature)

BINARY SYSTEM– TYPE 3 Partial or limited solubility Change of state

T

cA B

l

l +

l +

Solid solution (A rich)

Solid solution (B rich)

Liquid solution

= 1

= 1 = 1

= 2

= 2

= 2

AmpT

BmpT

Eutectic l

EUTECTIC DIAGRAM

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 c

Phase diagram for the simple eutectic system naphthalene - benzene

Phase diagram for system NaCl – H2O

ENTHALPY-COMPOSITION DIAGRAM

Solution A having composition of xA and enthalpy of HA is mixed adiabatically with solution A having composition of xB and enthalpy of HB.

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

BBAACC xmxmxm

Total balance: BAC mmm

Component balance:

Combining both equations yields:

BA

BBAAC mm

xmxmx

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:

AC

AACCB mm

xmxmx

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 110F is cooled to 70F. Evaporation and radiation losses may be neglected.

SOLUTION

(a) Initial solution (A) xA = 0.3 HA = - 31 Btu/lbCooled system (B) xB = 0.3 HB = - 75 Btu/lbEnthalpy change H = - 44 Btu/lbHeat to be removed = (- 44) (5000) = - 220000 Btu

(b) The cooled system B is located in the region where MgSO4.7H2O is in equilibrium with solution.

The crystal MgSO4.7H2O contains:

1263.1203.120

MW

MWMgSOoffractionWeight

O.7HMgSO

MgSO4

24

4

xC = 0.49

From the graph: xA = 0.26

6.869500026.049.026.03.0

BAC

ABC m

xxxx

m

The MgSO4.7H2O crystal yield is 869.6 lb