Phases

and

Solution

Learning outcomes

Able to determine the degree of freedom of a system

Able to distinguish the types of phases and components in a system

Able to perform calculations involved in phase transformations and equilibria.

Contents

3.1 Phases

3.2 Vapour pressure

3.3 Ideal solutions

3.4 Equilibrium between phases

3.5 Partial molar quantities

3.6 Clausius-Clapeyron equation

3.7 Interphase region

Phase

- a form of matter that is uniform (homogeneous portion) throughout, not only in chemical composition but also in physical state (system)

3.1 Phases

Number of phases is denoted by P :

P = 1 for gas, gaseous mixture, crystal, two miscible liquids, ice

P = 2 for slurry of ice and water, immiscible metal alloys

A phase can consist of several disconnected pieces (i.e. AgBr crystals in equilibrium with an aqueous solution)

Note that the defination of a phase does not mention solids, liquids or gases.

A system can be entirely liquid (or entirely solid) and still have more than one phase.

A system composed of two or more phases is heterogeneous.

Phase recognition

System Phase

Water (H2O) One water phase

Ice + water Ice and water phases

Ice + water + oil Ice, water and oil phases

Water + CCl4 Two liquid phases of water and CCl4

Solid diamond and graphite

Two solid phases of diamond and graphite

Chemical species, is denoted by c :

chemicals that present in a system. For example, a mixture of water and ethanol has 2 chemical species, c =2.

Independent Components, is denoted by cind:

chemically independent component of the system. The number of components in a system, Cind, is the minimum number of independent species needed to define the composition of all of the phases present in the system

Phase Rule

J.W. Gibbs, deduce the phase rule, which isgeneral relation between degree of freedom, f,the number of chemical species, c , and thenumber of phases at equilibrium, p, for a systemof any composition:

f = c - p + 2(phase rule)the no. of phases at

equilibrium

no. of independent intensive variables needed to specify the intensive state of the system.

the no. of chemical species

(a) No reaction

f = c - p + 2

f = degree of freedom = no. of independent intensive variables needed to specify the intensive state of the system.

p = number of phase presentc = number of chemical species

Described by specifying P, T, xi

in each phase.

(b) With reaction

f = c - p + 2 - r - ar = number of independent chemical reaction. a = additional restrictionscind = c - r - a number of independent component

f = cind - p + 2

Arising from

stoichiometric and

electroneutrality

conditions

example

Pure H2O, no reaction

c = 1,

r = 0,

p = 1

cind = c - r - a = 1

f = cind - p + 2 = 1 - 1 + 2 = 2

example

Pure H2O, dissociation of H2OH2O H+ + OH-

r = 1, c = 3x (H+) = x (OH-) electroneutrality condition, a = 1

So,cind = 3 - 1 - 1 = 1f = 1 -1 + 2 = 2

Pure H2O only has 1 independent component and 2 degree of freedom (T, P).

Evaporation is the process of vaporisation in which liquids areconverted into gas (vapour).

In an open system, evaporation continues until all liquid isgone

In a closed system, evaporation and condensation occursimultaneously until the system achieved equilibrium.

3.2 Vapour Pressure

Vapour

Liquid

Equilibrium is achieved when the evaporation rate equals condensation rate

Vapour Pressure is the pressure of vapour in equilibrium with the condensed phase

The vapour can be assumed to obey the ideal gas law. For a given liquid, the value of its vapour pressure depends on the strength of its intermolecular forces. (the higher the intermolecular forces, the lower the vapour pressure).The vapour pressure is independent of the volume of the container but depends on the temperature ( n/V is constant at T).The vapour pressure of a liquid always increases as temperature rises.

A solution is a homogeneous mixture of asolute dissolved in a solvent.

A solution may be solid, liquid or gas(physical state).

It can be consisting of one phase or two ormore phases

The components of a solution can be anynumber.

3.3 Ideal Solutions

Phase rule (Degrees of Freedom, f)

The Degrees of Freedom, f, is the number of intensive variables in a system that can be changed independently without disturbing the number of phases in equilibrium.

f = c - p + 2(w reaction)f = cind - p + 2(wo reaction)

Where, cind = c-r-a

In a single-component, single-phase system (C=1, P=1) the pressure and temperature may be changed independently without disturbing the number of phases in equilibrium:

Raoult’s Law

Ideal solutions obey Raoult’s law

For an ideal solution formed betweentwo volatile liquids A and B, thepartial vapour pressure of component Ais proportional to the mole fraction of Ain the solution:

oAAA PxP o

BBB PxP

xA = mole fraction of component A in the solution

xB = mole fraction of component B in the solution

xA + xB =1

PA=partial vapour pressure of component A

PB=partial vapour pressure of component B

PoA, Po

B =vapour pressure of pure A and B

PA + PB = PTotTotal Pressure

oAAA PxP o

BBB PxP

0 1xB=1

oBP

oAP

Mole fraction, xA

Vapour pressure of a mixture of two

volatile liquids A and B (fixed T)

A is more volatile than B

The vapour pressure of component B is lowered by the presence of component A (and vice versa)

As xA increases, the total pressure of the solution increases.

xA=1

PTot

Dalton’s Law

For an ideal solution of a two componentsystem, the composition of the vapour inequilibrium with the solution is not the sameas that of the solution.

The vapour phase is richer in the morevolatile component.

TAA PyP TBB PyP

TAA PyP TBB PyP

yA = mole fraction of component A in the vapour yB = mole fraction of component B in the vapouryA + yB =1PA=partial vapour pressure of component APB=partial vapour pressure of component BPT= PA + PB= Total pressure of the system

Henry’s Law

In an ideally dilute solution, the solvent obeys Raoult’slaw and the solute obeys Henry’s law.

Henry’s law states that the vapour pressure of solute iabove an ideally dilute solution is proportional to themole fraction of i in the solution.

iii xKP

xi = mole fraction of i in the solution Pi= partial pressure of i in the vapour in equilibrium with an

ideally dilute solutionKi= Henry’s law constant (unit in pressure)

exercise

At 20oC, Pobenzene=74.7 torr and Po

toluene=22.3 torr

A certain solution of benzene and toluene at 20oC has a vapour pressure 46.0 torr. Find the benzene mole fraction in this solution and in the vapour above this solution.

ans

Let, benzene = b and toluene =t

Total pressure PTot =Pb + Pt

From Raoult’s law,

Pb = xbPob

Pt = xtPot

PTot =Pb + Pt =xbPob + xtP

ot

=xbPob + (1-xb)P

ot ,where(xA + xB =1)

Cont,

PTot= xbPob + (1-xb)P

ot ,where(xA + xB =1)

46.0torr= xb(74.7torr) + (1-xb)(22.3torr)

xb=0.452 (mole fraction of benzene in the sol.)

partial vapour pressure of benzene in solution, Pb

Pb = xbPob

=(0.452)(74.7torr)

=33.8torr

Mole fraction of benzene in vapour phase is

Dalton’s Law

Pb=ybPTot

Yb= Pb/PTot

= 33.8torr/46.0torr

=0.735

Phase Diagram

-Shows regions of pressure and temperature where phases are thermodynamically stable.

3.4 Equilibrium between phases

Phase diagrams are extremely useful for systemswith multiple components, and serve to describephysical and chemical equilibria over a range ofdifferent compositions, as well as points wheresubstances are mutually miscible, or even when asystem has to be brought to a specific set ofconditions for equilbrium to exist (e.g., pressure,temperature and composition)

Phase Diagrams

All phase diagrams are developed around onerelationship, the phase rule, which wasderived by J.W. Gibbs - it can be applied to awide variety of systems

Phase diagrams: pictorial way ofunderstanding the properties of a system

Phase diagram for CO2, Fire

Extinguishers & Coffee

example

Phase diagrams

Phase diagrams are very important in the development of:

# alloys # ceramic materials

# steels # semiconductors

# plastics # superconductors

# cosmetics # petroleum product separations

# food products # glasses

Equilibrium between phases

One-component system

solid

gas

liquid

0

D

A

C

(f = 2)

(f = 0)

(f = 2)

(f = 2)

218 atm

4.6 Torr

P

T (°C)0.01 374

A: triple point (3 phase in eqm)

C: critical point (surface between two phases disappears)

•0 A : Sublimation curve

•A D : Melting point line

•A C : Boiling point curve

f = 1

One-phase regions, p =1, f = 2 – both

P and T must be specified todescribed the intensive state.

Along the lines except point A, 2

phases are present in equib. Thus, f

=1. E.g. Can vary T anywhere along

AC, but once T is fixed, then P is also

fixed.

Line AC gives b.p of water as a

function of P.

Line AD gives m.p. of ice as a function

of P.

At any temperature above Tc, liquid

and vapour phases cannot coexist in

equilibrium.

solid

gas

liquid

0

D

A

C

(f = 2)

(f = 0)

(f = 2)

(f = 2)

218 atm

4.6 Torr

P

T (°C)0.01 374

critical point

-point at which a liquid surface

disappears and above which a liquid

does not exist at whatever the

pressure. Critical pressure is vapour

pressure at critical temp.

triple point

-three phase

boundaries meet

and all three phases

are in mutual

equilibrium

Definition

Critical Points

The container need to be strong. Tc of water is

374oC and vapour pressure is 218atm

Boiling Points

In an open vessel, at the temperature where thevapour pressure is equal to the external pressure,free vapourization occurs throughout the liquid, andvapour can expand freely into the surroundings.Free vapourization is known as boiling

The normal boiling point, Tb, is the temperatureof free vapourization at a pressure of 1.0 atm

The standard boiling point is the temperature offree vapourization at a pressure of 1.0 bar (0.987atm) (e.g., water: 99.6oC for Tb = 100oC)

Melting Temperature

The temperature at a specified pressure where the liquid and solid phases coexist in equilbrium

The normal freezing point, Tf, is the temperature of freezing at a pressure of 1.0 atm (for water, Tf = 0 oC)

The standard freezing point is the temperature of freezing at a pressure of 1.0 bar (0.987 atm) (negligibly different from Tf)

Two-component system

A two-component system is called a binary system

The phase rule, f = c - p + 2 (without reaction) becomes f = 4 - p

For a one-phase, two-component system, f = 3 (P, T, mole fraction).

Multicomponent systems have important applications in chemistry, geology and material science.

Metal, semiconductor, polymers, ceramics and composites e.g SiC, Si3N4, Al2O3 etc

TB

*

BBBPyPxP TC

*

CCCPyPxP CBT PPP

B

*

C

*

B

*

CTxPPPP

*

B

TB

BP

Pyx

Eqn 1 - bubble point line

*

B

*

B

*

CB

*

C

*

B

T

*

B

TB*

C

*

B

*

CT

PPPy

PPP

P

PyPPPP

Eqn 2 – dew point line

where

(a)Two-component liquid-vapour equilibrium

Two liquids B and C form an ideal solution.

(i) Ideal solution at fixed temperature

B

*

C

*

B

*

CTxPPPP

Points on the dew point line and bubble point line at the same pressure, P represent the compositions of vapour and liquid phases in equilibrium and they are connected by a horizontal line called tie line.

Bubble point line –

above this line, the

system is in liquid state

Dew point line – below

this line, the system is in

vapour state

PC*

PB*

0 mole fraction B 1

P

liquid

vapour

*

B

*

B

*

CB

*

C

*

B

TPPPy

PPP

(ii) Ideal solution at fixed pressure

The upper curve gives the T as a function xB while the lower curve gives the T as a function of yB.

The tie linegives the composition of the vapour and liquid phases in equilibrium at T

At fixed pressure, the phase diagram is a boiling point diagram.

TB* and TC

* are normal boiling point of pure B and C if the fixed pressure is 1 atm.

Bubble point line –below this line, the system is in liquid state

Dew point line –above this line, the system is in vapour state

TC*

TB*

0 mole fraction B 1

T

liquid

vapour

T vs. yB

T vs. xB

(b) Two-component liquid-liquid equilibrium

When ethanol and water are shaken and mixed together at room temperature, a single phase is always obtained - complete miscibility

When 1-butanol and water are shaken together at room temperature, two liquid phases are obtained: one phase is water containing some dissolved 1-butanol and the other 1-butanol containing small amount of dissolve water – partial miscibility

Two phases

One phaseTC

T

0 xB 1

As the temperature is raised, the region of liquid-liquid immiscibility decreases, until at TC (critical solution temperature) it shrinks to zero. Above Tc, the liquids are completely miscible.The two-phase region is called miscibility gap.

(c) Two-component liquid-solid equilibrium

The effect of pressure on solids and liquids is small.

Normally, pressure is fixed at 1 atm and the T-xB solid-liquid phase diagram is examined.

(i) Solid solutions

A pair of solids are miscible in all proportions in the solid/liquid form.

(ii) Liquid phase miscibility and solid-phase immiscibility

Substances B and C are miscible in all proportions in the liquid phase and completely immiscible in the solid phase. Cooling a liquid solution of B and C will cause either pure B or pure C to freeze out of the solution.

(ii)TC

* and TB* are freezing

points of pure C and pure

B.

E is the eutectic point.

AE is the freezing point

depression curve of C due

to solute B

DE is the freezing point

depression curve of B due

to solute C

l.s.+solid Cl.s.+

solid B

Liquid solution (l.s.) of B + C

Solid B + solid C

TB*

TC*

xB0 1

A

E

D

TTHxlnR

*

B

B,fus,mB

11

TTHxlnR

*

C

C,fus,mC

11

(iii) Liquid-phase miscibility and solid-phase miscibility

l.s. + s.s.

Liquid solution (l.s.)

solid solution (s.s.)

0 xNi 1

Cu Ni

Pair of substance thatcompletely miscible inthe solid state. E.g. Cu-Ni, Sb-Bi, Pd-Ni etc.

With completemiscibility in both theliquid and the solidphases, the T-XB binaryphase diagram maylook like for Cu-Niphase diagram.

1000oC

1200oC

1400oC

T

3.5 PARTIAL MOLAR QUANTITIES

The solvent’s partial molar properties are the same as for an ideal solution-component.

For the solute partial molar properties:

iV is the change in V when 1 mol of i is

added to an infinitely large amount of solution at constant T, P

= infinite dilutionO = standard state

oA,mA VV o

A,mA HH Ao

A,mA xlnRSS

i

oii VVV

is the molar volume of the pure solvent Ao

A,mV

i

oii HHH i

oii xlnRSS

Relation between solution volume

and partial molar volume

V=niVii

3.6 CLAUSIUS-CLAPEYRON EQUATION

A phase change at constant T and P is accompaniedby an enthalpy change, H - latent heat of transition(heat of fusion: solidliquid, heat of sublimation:solidvapour, heat of vaporisation: liquidvapour,heat of transition: solid1 solid2).

The Clausius-Clapeyron equation is used to calculatethe T or P or H for vaporisation, sublimationand melting processes.

Derivation of Clapeyron Equation

Consider two infinitesimally close points 1 and 2 on such a line:

+ Phase

Phase

21

T

P

dP

dT

One component2 phase equilibrium

2

For equilibrium phase change:

Clapeyron equation

3.6.1 Liquid/vapour and solid/vapour equilibria.

If ideal gas:

since

VT

H

VT

H

dT

dP

m

m

gas,mliqmgas,mm V,VVV

P

RTV gas,m

2RT

HP

P

RTT

H

dT

dP mm

PlndP

dP

or

Clausius-Clapeyron equation.

since

lnP

1/T

2RT

H

dT

Plnd m

dTTT

d

2

11

R

H

Td

Plnd m

1

R

Hslope m

If H is constant:

gives vapour pressure vs T for solid & liquid.

2T

dT

R

HPlnd m

121

2 11

TTR

H

P

Pln m

If Hfus, Vfus are constant:

• Determine melting point at different pressure.

• Normal melting point of a solid is the melting point at 1 atm pressure.

3.6.2 Solid-liquid equilibrium

For melting:

fus

fus

VT

H

dT

dP

dT

VT

HdP

fus

fus

1

212

T

Tln

V

HPP

fus

fus

3.7 INTERPHASE REGION

3.7.1 Planar interface

BULKPHASE

BULKPHASE

A

C D

B

INTERFACIAL LAYER, SURFACE LAYER OR INTERPHASE

REGION

• Transition region between bulk phases and .

• Properties vary from those of and .

• Because of differences in intermolecular interactions, molecules in the interphase region have different average interaction energy.

C

BD

A

Liquid

Vapour

z

Ci

Boundaries of

interfacial layer

Intermolecular interactions in a liquid lower the internal energy.

Molecules at the surface have fewer attractions from other liquid-phase molecules have a higher average energy.

The stronger the intermolecular attractions in a liquid, the greater the work needed to bring molecules from the bulk to the surface.

vapour

liquid

It requires work to increase the area of the interface.

The work done on the closed system of phases and is:

dwrev = -PdV + dA (plane interface)

where = surface tension (N/m)

dA = the increase in interfacial area

P = pressure in each bulk phase

dV = the change in volume

At the critical temperature (Tc), the liquid-vapour interface disappears, only one phase is present.

At Tc, = 0

3.7.2 Curved Interface Surface tension causes equilibrium pressure in

phases and to differ

If lower piston is pushed, more phase goes into the conical region, area of interface increases. P > P

The relation between Pand P was derived using the modified setup (Fig b)

P

(Fig a) (Fig b)

RP

P

For spherical interface

P = surface tension

R•Is the basis for the capillary-rise method of measuring surface tension for liquid-vapour and liquid-liquid interfaces.

•As R -->, pressure difference -->0•For H2O, if R= 1 cm, P- P = 0.1 Torr•if R = 0.01 cm, P- P = 10 Torr

RPP

2

3.7.3 Capillary Rise

r = R cos

where r = radius of capillary tube

= contact angle

and = densities of phases and , respectively

g = gravitational acceleration = 981 cm/s2,

= surface tension or interfacial tension

h = capillary rise

ghR 2

1

Capillary riseContact angles between a liquid and and a glass tube

(a) 0 oC < 90oC (b) 90 oC < 180oC

cos

ghr

2

The surface tension has unit of work (or energy) divided by area.

cgs unit for : ergs cm-2

SI unit for : J m-2 = N m-1

1 erg = 1 dyn cm

1 erg cm-2 = 1 dyn cm-1 = 10 -3 J m-2 = 10-3 N m-1

material (dyn cm-1) at 20oC

organic/inorganic liq 15 - 50

water 73

mercury 490

decreases as T increases

At 20 oC, the interfacial tension for n-hexane and water is 52.2 ergs cm-2. The densities of n-hexane and water at 20 oC are 0.6599 and 0.9982 g cm3,

respectively. Assuming a zero contact angle, calculate the capillary rise at 20 oC in a 0.350 mm inside diameter tube inserted into a two-phase n-hexane-water system.

= 52.2 ergs cm-2

= 0.6599 = 0.9982

g = 980.7 cm/s2

r = 0.0175 cm