3. APLICATION OF GIBBS-DUHEM EQUATION Gibbs-Duhem …users.metu.edu.tr/olguny/files/2043.pdf ·...
Transcript of 3. APLICATION OF GIBBS-DUHEM EQUATION Gibbs-Duhem …users.metu.edu.tr/olguny/files/2043.pdf ·...
3. APLICATION OF GIBBS-DUHEM EQUATION
Gibbs-Duhem Equation
When extensive property of a solution is given by;
Q' = Q'(T,P,n1,n2 ...)
The change in extensive property with composition was;
dQQ
ndn
Q
ndn
T P n n j T P n n j
12
1
21
2, , ,.. , , ,.. (1)
Partial molar value of an extensive property of component “i” was
defined as;
ni
iT P n j i
, , (2)
Combination of equations (1) and (2) yields;
dQ' = Q1 dn1 + Q2dn2 + ... (3)
But, Q' = n1Q1 + n2Q2 + … (4)
Differentiation of equation (4) gives
dQ' = Q1 dn1+ Q2dn2 + ... + n1dQ1 + n2dQ2 + ... (5)
Subtraction of equation (3) from equation (5) yields
n1dQ1 + n2dQ2 + ... = 0
In general,
ni dQi = 0 (6)
Dividing by n, total number of moles of all the components, gives
Xi dQi = 0 (7)
Equations (6) and (7) are equivalent expressions of the Gibbs-
Duhem equation.
3.1. Integration of Gibbs-Duhem Equation
It is performed to calculate partial properties from each other.
Consider a 2 component system, say Q1 is given. Q2 is asked.
Gibbs-Duhem equation: X1 dQ1 + X2 dQ2 = 0
equivalents can be written for mixing (relative molar) and excess
properties. Rearrangement of Gibbs-Duhem equation yields:
dQX
XdQ2
1
2
1
Integration of both sides
Q at X
Q at X
dQ2 2
2 2
2( )
( )
-Q at X
Q at XX
XdQ
1 2
1 21
2
1
( )
( )
Integration yields:
Q at X Q at X2 2 2 2( ) ( )
-
Q at X
Q at XX
XdQ
1 2
1 21
2
1
( )
( )
The value of Q2 must be known as boundary condition to integrate
this function. Usually properties of pure substances are easier to
find. Say property at X2 = 1 is known. Then let X2' as X2=1 and
X2" to any X2;
Q at X Qo2 2 2( ) -
Q X
Q at XX
XdQ
1 2 1
1 21
2
1
( )
( )
The value of integral may be determined graphically or analytically
(if analytical dependence of Q1 to composition is given).
Analytical determination involves the following steps:
i. Organize Q1 as a function of only one composition variable (X1
or X2).
ii. Take the derrivative of Q1 and replace it into the integral.
iii. Organize the function inside the integral in such a way to leave
only one variable. At the same time, change, if and when necessary,
the limits of the integral to make it in accord with the derrivative of
the integral. Use the relationship dX1 = -dX2 when necessary.
iv. Integrate the function, replace the limits to determine the value of
the integral.
Analytical integration with relative partial and partial excess
properties are also possible. The same steps can be used, by
replacing Q1 with Q M1 and/or Q
xs1 in the case of relative partial
molar properties and partial excess properties respectively.
Graphical determination requires plot of X1/X2 vs. Q1. The value of
the integral is the area under the curve between the specified limits.
However there are some problems with this integration:
i. the value of Q1 becomes plus or -, if Q1 has logarithmic
composition terms.
ii. X1/X2 becomes , when X2 becomes zero.
Therefore, the area under the curve may not be bounded well with
the given limits. To determine the area under the curve properly;
either alternative limits may be given or ways to resolve the
problems of tails to infinity should be found. Furthermore, Gibbs-
Duhem integration with partial molar properties can only be used for
entropy and volume. For other properties Gibbs-Duhem integrations
involving relative partial and/or excess properties have to be used.
Then with relative partial molar properties, Gibbs-Duhem
integration is (in general form):
Q at X Q at XM M2 2 2 2( ) ( )
-
Q M at X
Q M at X
MX
Xd Q
1 2
1 21
2
1
( )
( )
By setting the lower limit X2' as X2=1 and X2" to any X2, Q M2 (at
X2=1) is zero, then
Q M2 (at X2) = -
Q M X
Q M at X
MX
Xd Q
1 2 1
1 21
2
1
( )
( )
Therefore, area under X1/X2 vs. Q M1 curve gives the value of
integral. Two of the above problems are still valid in this case (for
some properties), but this form of the integral may be used with any
thermodynamic property. To illustrate the problem, consider the
following data for Fe-Ni alloys at 1600oC.
XNi 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
aNi 1 0.89 0.766 0.62 0.485 0.374 0.283 0.207 0.136 0.067 0
GNiM
0 -432 -989 -1773 -2684 -3647 -4681 -5841 -7399 -10024 -
cal/mol
XNi/XFe 9 4 2.33 1.5 1 0.67 0.43 0.25 0.11 0
XNi/XFe
-GNiM
The value of GNiM at lower limit (i.e. XFe=1) is -, results in an
unbounded area under the curve. Therefore, to get precise value for
GFeM
(at XFe); either GFeM
at a composition other than XFe=1
should be given (known) or other alternatives should be considered.
Use of excess properties: (in general form) Gibbs-Duhem integration
with excess properties are:
Q at X Q at Xxs xs2 2 2 2( ) ( )
-
Q xs at X
Q xs at X
xsX
XdQ
1 2
1 21
2
1
( )
( )
By setting the lower limit X2' as X2=1 and X2" to any X2, Qxs
2 (at
X2=1) is zero, then
Q at Xxs2 2( ) -
Q xs at X
Q xs at X
xsX
XdQ
1 2 1
1 21
2
1
( )
( )
Area under X1/X2 vs. Qxs
1 curve gives the value of integral. One of
the above problems is solved; by using excess properties, a finite
value is assigned to the value of Qxs
1 (at X2=1). Therefore the area is
bounded from the lower end of the integral. The problem of tail to
infinity for X1/X2 at X2=0 is still valid in this case.
Then, previous problem for Fe-Ni alloys at 1600oC requires
integration of;
XNi 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
aNi 1 0.89 0.766 0.62 0.485 0.374 0.283 0.207 0.136 0.067
GNixs 0 -41.4 -161 -450 -789 -1077 -1283 -1376 -1430 -1485 RT ln
Ni cal/mol
XNi/XFe 9 4 2.33 1.5 1 0.67 0.43 0.25 0.11 0
log Ni 0 -.005 -.019 -.053 -.092 -.126 -.15 -.161 -.167 -.174 log Ni
G at XFexs
Fe( ) -G
Nixs at X Fe
GNixs at X Fe
Ni
Fe
NixsX
XdG
( )
( )
1
XNi/XFe
RT ln Nio
-1530
-GNixs
This is good for values XFe=1 to XFe > 0. As XFe 0; XNi/XFe
. Therefore, G at XFexs
Fe( ) 0 is mathematically indeterminate with
excess properties.
The use of -Function:
Problems arising from X1/X2 as X20 may be resolved using
this function. For any component i, i is defined as:
iixs
i
Q
X
( )1 2
For 1-2 binary solution from previous relationships
Q at Xxs2 2( ) -
Q xs at X
Q xs at X
xsX
XdQ
1 2 1
1 21
2
1
( )
( )
From above relationship Q Xxs1 1 2
2
Then dQ d X X dXxs1 1 2
21 2 22
Replacing into the integral yields
Q at Xxs2 2( ) -
1 2 1
1 21
2
22
1
( )
( )
at X
at XX
XX d
- X
XX
XX dX
2 1
21
2
1 2 22
Q at Xxs2 2( ) -
1 2 1
1 2
1 2 1( )
( )
at X
at X
X X d
- X
X
X dX2 1
2
1 1 22
By virtue of identity d(xy) = y dx + x dy, the first integral
1 2 1
1 2
1 2 1( )
( )
at X
at X
X X d
= d X X( )1 2 1 - 1 1 2d X X( )
Then,
Q at Xxs2 2( ) - d X X( )1 2 1 + 1 1 2d X X( ) - 2 1 1 2 X dX
= - X X1 2 1 + 1 1 2X dX + 1 2 1X dX - 2 1 1 2 X dX
= - X X1 2 1 - 1 1 1 2 22( )X X X dX
= - X X1 2 1 - X
X
dX2 1
2
1 2
Numerical value for -X1X21 can be determined, then the value of
the integral can be determined graphically or analytically (if an
analytical expression for the composition dependence of 1 or other
properties that leads to determine 1 given). Analytical integration
with -function can be done by replacing 1 into the equation and
integrating it between the given limits. Graphical determination can
be done from the area under X2 vs. 1 graph.
1
X2
X
Then, previous problem for Fe-Ni alloys at 1600oC requires
determination of;
G at XFexs
Fe( ) = - X XNi Fe Ni - X Fe
X Fe
Ni FedX1
XNi 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
Ni 0/0 -1 -1.07 -1.3 -1.33 -1.16 -.96 -.76 -.62 -.49 ln Ni
GNixs
0 -41.4 -161 -450 -789 -1077 -1283 -1376 -1430 -1485 RT ln Ni cal/mol
XNi/XFe 9 4 2.33 1.5 1 0.67 0.43 0.25 0.11 0
log Ni 0 -.005 -.019 -.053 -.092 -.126 -.15 -.161 -.167 -.174 log Ni
Ni XFe
.5
1
1.5
For XFe = 0.7
ln Fe = -0.3 x 0.7 (-0.76) - (0.3 x 0.4 + 0.36 x 0.3 x 0.5) = 0.1596 - 0.174
ln Fe = - 0.0144; Fe = 0.985; aFe = 0.6895
Direct Calculation of the Integral Property
Consider a 2 component system, say Q1 is given. Q is asked. Indirect
method involves determination of Q2 by Gibbs-Duhem integration,
then computation of the integral property from
Q = X1 Q1 + X2 Q2
but
Q1 = Q + (1 - X1) (dQ/dX1)
multiply both sides by dX1 and divide by X22, then
Q dX
X
QdX X dQ
X
1 1
22
1 1
22
1
( )
Replacing (1 - X1) = X2 and dX1 = - dX2
Q dX
X
QdX X dQ
Xd
Q
X
1 1
22
2 2
22
2
( )
Integration of both sides
Q
Xat X
Q
X
dQ
X2
2 1
2
2( )
( )
= X
X Q dX
X1 0
11 1
22
Q
XQo
2
2 = X
X Q dX
X1 0
11 1
22
Using relative partial molar properties and setting QM at X2=1 to
zero;
Q
X
M
2
= X
X MQ dX
X1 0
11 1
22
or QM = X2 X
X MQ dX
X1 0
11 1
22
Equivalent relationship from partial excess properties and setting
Qxs at X2=1 to zero;
Qxs = X2 X
X xsQ dX
X1 0
11 1
22
Values of integrals can be obtained either analytically (if analytical
dependence of Q1 or Q M1 or Q xs
1 to composition is given) or
graphically. Analytical integration involves the following steps:
i. The replacement of the property (Q1 or Q M1 or Q
xs1 ) into the
integral.
ii. Organization of the function inside the integral in such a way to
leave only one variable.
iii. Integration and replacement of the limits.
Graphical integration involves the determination of the area under
the curve (Q X1 22
or Q XM1 2
2 or Q Xxs
1 22
vs. X1 between the limits.
X1 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1
Gxs
1 RT ln 1 .. .. .. .. .. .. .. .. .. ..
G Xxs1 2
2RT ln 1 .. .. .. .. .. .. .. .. .. ..
G Xxs1 2
2
X X1