Partial Molar Properties of solutions

The relationships for pure component are not applicable to solutions.

Which needs modification because of the change in thermodynamic

properties of solution.

The properties of a solution are not additive properties, it means

volume of solution is not the sum of pure components volume.

When a substance becomes a part of a solution it looses its identity

but it still contributes to the property of the solution.

A solution is a homogeneous mixture; that is, a solution is a one-

phase system with more than one component.

A homogeneous mixtures of two or more components in the gas,

liquid or soild phase

1

A mole of component “i” in a particular solution at specified temperature and pressure has got a set of properties associated with it like.

These properties are partially responsible for the

properties of solution and it is known as partial

molar property

The PMP of a particular component in a mixture

measures the contribution of that component to the

mixture property.

, , , ....i i i iV H U S etc

The partial molar value expresses how that property

(volume, pressure, enthalpy, entropy) depends on changes

in amount of one component

2

It is defined as

iM = Partial molar property of component i.

Mt = Total value of any extensive thermodynamic property of the

solution

n = Total number moles in a solution

ni=Number of moles of component i in the solution

ijnPTi

t

ijnPTi

in

M

n

nMM

,,,,

Intensive property,

value depends only on the composition at the given Temp and Pressure

3

Physical Significance of Partial Molar

Properties To understand the physical

meaning of molar properties,

consider a open beaker

containing huge volume of

water, if one mole of water is

added to it, the volume increase

is 18x 10-6m3 .

If the same amount of water is

added to pure ethanol the

volume increased was

approximately 14 x 10-6m3 .

This is the partial molar volume

of H2O in pure ethanol.

0 25 50 75 100 V (ethanol) cm3

100

99

98

97

96

75 50 25 0

V (H2O) cm3

Vt cm3

Addition of 50.0 cm3 of water to 50.0 cm3 of ethanol at

20 oC and 1 atm gives a solution of 96.5 cm3 .

4

The Partial molar property changes with composition. The intermolecular forces also changes Results in change in thermodynamic property.

wV = Partial molar volume of the water in

ethanol water solution = Molar volume of pure water at same

temperature and pressure wV

tV =Total volume of solution when water added to

ethanol water mixture and allowed for sufficient

time so that the temperature remains constant

5

The partial molar volumes of the components of a

mixture vary with the composition of the mixture,

because the environment of the molecules in the

mixture changes with the composition.

It is the changing molecular environment (and the

consequent alteration of the interactions between

molecules) that results in the thermodynamic

properties of a mixture changing as its composition

is altered

6

Partial molar properties and properties of the solution

the total property of the solution nMMt

321 nnnn

1,2,3 represents number of constituents

Thermodynamic property is a j321 nn,n,n,P,Tf

Let M be the molar property of a solution, ( may be Volume,

free energy, heat capacity…

7

For small change in the pressure and temperature and amount of various

constituents can be written as

i

inTPi

t

nnpT

t

np

t

nT

tt

dnn

M

dnn

MdT

T

MdP

P

MdM

j,,

1

,,,1,, 32

1 2 3, , , , .......T

jM f T P n n n n

8

At constant temperature and pressure dP and

dT are equal to zero.

The above equation reduces to

dni

n

MdM

ijn,T.P

ni

1i i

tt

i

n

1i

i

tdnMdM

9

iM is an intensive property depends on composition

and relative amount of constituents.

All constituent properties at constant temperature

and pressure are added to give the property of

the solution

332211

tdnMdnMdnMdM

dnxMxMxMdM 332211

t

dnxdn

dn

dn

n

nx

ii

iii

xi – mole fraction of component i in the solution

10

i

n

1i

i

tdnMdM

1 1 2 2 3 3

1 1 2 2 3 3

tM M x M x M x n

M n M n M n

t

i iM n M

For binary system

2211

,2

VxVxV

propertytheisVolumei

MxM

t

ii

t

Integrating yields

at constant temp and press, the total property is equal to the sum of partial

molar property of the species and its mole fraction; its not equivalent to

mole fraction and pure component property

11

dnxMxMxMdM 332211

t

12

Problem

A 30% mole by methanol –water solution is to be prepared.

How many m3 of pure methanol (molar volume = 40.727 x 10-6m3/mol) and pure water (molar volume = 18.068 x 10-6m3/mol) are to be mixed to prepare 2m3 ( 2000 L)of desired solution.

The partial molar volume of methanol and water in 30% solution are 38.362 x 10-6 m3/mol and 17.765 x 10-6 m3/mol respectively.

13

Methanol = 0.3 mole fraction

Water = 0.7 mole fraction

Vt = 0.3 x 38.632 x10-6 + 0.7 x 17.765 x 10-6

=24.0251 x 10-6 m3/mol

Total moles for 2 m3 solution

3

6

283.2463 10

24.0251 10mol

2211 VxVxV t

14

Number of moles of methanol in 2m3solution

=83.2463 x 103 x 0.3 = 24.9739 x 103 mol

Number of moles of water in 2m3solution

= 83.246 x 103 x 0.7 = 58.2724 x 103 mol

Volume of pure methanol to be taken

= 24.97 x 103 x 40.7 x10-6 =1.0717 m3

Volume of pure water to be taken

= 58.272x103 x 18.068x10-6 =1.0529 m3

To prepare 2m3 ( 2000L), 30% mol methanol-water solution,

one should add 1.0717 m3 ( 1071.7 L) of pure methanol and 1.0529 m3

(1052.9 L)of pure water 15

Estimation of Partial molar properties for a binary

mixture

Analytical Method : if the volume of a solution is known as

function of its composition, partial differentiation with respect to the amount

of that constituent.

jnPTi

t

in

VV

,,

partial molar volume

partial molar enthalpy

partial molar entropy

16

Analytical method 2

1 2( )tV nV n n V

Consider binary solution n1 moles of component 1 and

n2 moles of component 2

Let Vt = the total volume

V = the molar volume

1... 2, ,Diff wrt to n Keeping n T P

1 1 221 , , 2

( )1

t

nT P n

V VV V n n

n n

2

22

1 2

but mole fraction x by definition

nx

n n

2

2 2 21 2

1 1 21 2

. .

n

x n xdiff w r t n

n n nn n

1 2 2

1 2

n n xrearranging

dn dx

1 2

2

VV V x

x

17

2 1

1

VV V x

x

1 2

2

VV V x

x

2 2

2

(1 )V

or V V xx

18

Tangent -Intercept method

Widely used method to estimate PMP of both components in a binary system

The molar volume V is plotted against mole fraction of one of the components,

(Let x2, the mole fraction component 2)

Draw the tangent to the curve at the desired mole fraction

A

B

C D

E F

P

0 1 X2 mole fraction of comp 2

V

Mo

lar

Volu

me,

V m

3/k

mol

The intercept with vertical axis

gives pure component volume,

At x2 =1(x1=0) , but V1=0

X2 =0(x1=1) , but V2=0

;

; 12

EDBEBDlength

VACVBD

1V2V

BE is the slope of the tangent at P*PE

2)1( 2 x

VxBE

ED = V, the molar volume at the mole fraction x2, 19

Tangent -Intercept method

12

2

2

2

2)1(

VxVFAFCAC

VBD

xVBD

xV

xV

A

B

C D

E F

P

0 1 X2 mole fraction of comp 2

V

;

; 12

EDBEBDlength

VACVBD

20

Limiting cases: For infinite dilution of component

when a tangent is drawn at x1=0, will give the

partial molar property of component 1 at infinite

dilution )M( 1

Tangent is drawn at x2=0 or x1=1 will give infinite

dilution the partial molar property of component 2 )M( 2

21

Problem

The Gibbs free energy of a binary solution is given by

mol

cal)xx10(xxx150x100G 212121

(a) Find the partial molar free energies of the components at x2=0.8 and

also at infinite dilution.

(b) Find the pure component properties

mol

cal)xx10(xxx150x100G 212121

Substitute

150x49x8x9G 1

2

1

3

1

1

11dx

dG)x1(GG 2

1 1

1

27 16 49G

x xx

12 x1x

1 1

1

2 2 2 1

2 1

(1 )

(1 ) ;

VV V x

x

V VV V x V V x

x x

22

101x16x35x18G 1

2

1

3

11

1

12dx

dGxGG 150x8x18G

2

1

3

12

To find the partial molar properties of components 1 and 2 x2=0.8,

x1= 1 - 0.8 = 0.2

101x16x35x18G 1

2

1

3

11

mol

cal944.102G1

150x8x18G2

1

3

12

mol

cal824.149G2

1

11dx

dG)x1(GG

23

At infinite dilution

0111

atxGGmol

cal101G1

1122

atxGG or x2=0 mol

cal160G2

To find the pure component property

1atxGG 111 mol

cal100G1

0atxGG 122 mol

cal150G2

24

Problem

The enthalpy at 300K and 1 bar of a binary liquid mixture is

]x20x40[xxx600x400H 212121

Where H is in J/mol. For the stated temperature determine

1.Expression in terms of and in terms of x1.

2.Numerical Values of pure component enthalpies.

3. Numerical values for partial molar properties at infinite

dilution.

1H 2H

25

Solution:

]x20x40[xxx600x400H 212121

Substitute 12 x1x

600x180x20H 1

3

1

1

11dx

dH)x1(HH

1

12dx

dHxHH

420x60x40H2

1

311 600x40H

3

12

26

At infinite dilution

0atxHH 111

mol

J420H1

1atxHH 122

or x2=0

mol

J640H2

To find the pure component property

1atxHH 111

mol

J400H1

0atxHH 122 mol

J600H2

27

Chemical Potential of species i

It is widely used as a thermodynamic property. It is used

as a index in chemical equilibrium, same as pressure

and temperature.

The chemical potential of component i

jnPTi

t

iin

GG

,,

The chemical potential of a substance is an intensive property.

Intensive properties are spatially uniform under equilibrium conditions.

Temperature gradients lead to heat conduction to achieve thermal

equilibrium.

Pressure gradients lead to fluid flow to achieve mechanical

equilibrium.

Differences in i between phases leads to the diffusion of component

i between phases (or chemical reactions) to achieve chemical

equilibrium.

28

1 2, , ,t

kG f P T n n n

1, , , , j

t t ti kt

i

i iT n P n P T n

G G GdG dP dT dn

P T n

ii

nP

t

nT

tt dndT

T

GdP

P

GdG

,,

jnPTi

t

in

Gce

,,

,sin

Total Gibbs free energy , Gt

29

For closed system there will be no exchange of constituents

(n is constant)

dTSdPVdG ttt

at constant temperature t

t

T

GV

P

at constant pressure tt

P

GS

T

30

ii

ttt dndTSdPVdG

At constant temperature and pressure

iiPT

t dndG ,

For binary solution the molar free energy of the solution

2211 xxGt Leads to GIBBS

DUHEM relation

…fundamental relationship for changes in the free energy of a solution

the change in free energy is entirely because of the changes in the number of moles

t

i iG n

31

Effect of temperature and pressure on chemical potential

Effect of temperature:

jnPTi

t

iin

GG

,,

differentiating equation (1) with respect to T at constant P

----------(1)

----------(2)

2

,

i

P n i

G

T T n

32

SdTVdPdG ---------(3)

G with respect to T at constant P

ST

G

P

differentiating again w r t ni 2

, , j

t

ii

P n i i P n

G SS

T T n n

is partial molar entropy of component i

2

,

ii

i

P n

TT T

T T

In a useful form

2( )U

V

VdU UdVd

V

33

2T

ST ii TSHG

In terms of partial molar properties

i i iG H T S

i i i

i i i

H T S

H T S

2

,

( )i i

P n

T H

T T

This equation represents the

effect of temperature on

chemical potential.

34

Effect of Pressure:

jnPTi

t

in

GGi

,,

-------------------(4)

differentiating equation (4) with respect to P at constant T

inT

i

Pdn

G

P

2

,

---------------------(5)

SdTVdPdG ---------(3)

G with respect to P at constant T

VP

G

T

35

differentiating again w r t ni

i

nTii

Vn

V

nP

G

j

,

2

i

nT

i VP

,

This equation represents the effect of pressure on chemical potential.

The rate of change of chemical potential with pressure is equal to the

partial molar volume of the constituent.

36