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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;
dQ
Q
n
dn
Q
n
T P n n j T P n n jdn
1 , ,^2 ,..^1 2 , ,^1 ,..^2 (1)
Partial molar value of an extensive property of component “i” was defined as;
Q
Q
i ni T P n j i
^ , ,^ (2)
Combination of equations (1) and (2) yields;
dQ' = Q 1 dn1 + Q 2 dn2 + ... (3)
But, Q' = n 1 Q 1 + n 2 Q 2 + … (4)
Differentiation of equation (4) gives
dQ' = Q 1 dn1+ Q 2 dn2 + ... + n1d Q 1 + n2d Q 2 + ... (5)
Subtraction of equation (3) from equation (5) yields
n1d Q 1 + n2d Q 2 + ... = 0
In general,
ni d Qi = 0 (6)
Dividing by n , total number of moles of all the components, gives
Xi d Q (^) i = 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 Q 1 is given. Q 2 is asked.
Gibbs-Duhem equation: X1 d Q 1 + X2 d Q 2 = 0
equivalents can be written for mixing (relative molar) and excess properties. Rearrangement of Gibbs-Duhem equation yields:
dQ
X X 2 1 dQ 2
1
Integration of both sides
Q at X
Q at X
dQ
2 2
2 2 ( )^2
( )
- Q at X
Q at X X
X
dQ
1 2
1 2 1 2
1 ( )
( )
Integration yields:
Q at X 2 ( (^) 2 ) Q at X 2 ( 2 ) - Q at X
Q at X X
X
dQ
1 2
1 2 1 2
1 ( )
( )
The value of Q 2 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 2 ( (^) 2 ) Q 2 o - Q X
Q at X X
X
dQ
1 2 1
1 2 1 2
1 ( )
( )
The value of integral may be determined graphically or analytically
(if analytical dependence of Q 1 to composition is given). Analytical determination involves the following steps:
i. Organize Q 1 as a function of only one composition variable (X or X2).
ii. Take the derrivative of Q 1 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,
Therefore, area under X1/X2 vs. Q 1^ M 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
X Ni /X Fe
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=
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 2 xs^ ( at X (^) 2 ) Q 2 xs ( at X 2 ) - Q xs at X
Q xs at X
X xs
X
dQ
1 2
(^1 2 )
2
1 ( )
( )
By setting the lower limit X2' as X2=1 and X2" to any X2, Q 2^ xs (at
X2=1) is zero, then
Q 2 xs ( at X 2 ) - Q xs at X
Q xs at X
X xs
X
dQ
1 2 1
(^1 2 )
2
1 ( )
( )
Area under X1/X2 vs. Q^
xs 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 Q 1^ xs (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. 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 (^) Fexs ( at X (^) Fe ) - GNixs at X Fe
GNixs at X Fe Ni Fe
Ni
X xs
X
dG
( )
( )
1
X Ni /X Fe
RT ln (^) Ni^ o
-
This is good for values XFe=1 to XFe > 0. As XFe 0; XNi/XFe . Therefore, G^ Fexs ( at X^ Fe ^0 ) is mathematically indeterminate with excess properties.
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
X (^2)
X
Then, previous problem for Fe-Ni alloys at 1600oC requires determination of;
G Fexs ( at X Fe ) = - X Ni XFe Ni -
X Fe
X Fe
1 Ni^ dXFe
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
.
1
For XFe = 0. 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. ln Fe = - 0.0144; Fe = 0.985; aFe = 0.
Direct Calculation of the Integral Property
Consider a 2 component system, say Q 1 is given. Q is asked. Indirect method involves determination of Q 2 by Gibbs-Duhem integration, then computation of the integral property from Q = X1 Q 1 + X2 Q 2 but
Q 1 = Q + (1 - X1) (dQ/dX1)
multiply both sides by dX1 and divide by X 2
2 , then
Q dX
X
QdX X dQ
X
1 1 2
2
1 1 2
2
Replacing (1 - X1) = X2 and dX1 = - dX
Q dX
X
QdX X dQ
X
d
Q
X
1 1 2
2
2 2 2
2 2
Integration of both sides
X
- G 1 xs RT ln
- G 1 xs X 22 RT ln
- G 1 xs X - X
