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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