Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Gibbs-Duhem Equation, Lecture notes of Chemistry

Application and Integration of Gibbs-Duhem Equation

Typology: Lecture notes

2020/2021

Uploaded on 05/24/2021

torley
torley 🇺🇸

4.6

(41)

258 documents

1 / 13

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
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
ndn Q
ndn
T P n n jT P n n j

121212
, , ,.. , , ,..
(1)
Partial molar value of an extensive property of component “i” was
defined as;
QQ
n
i
iT P n j i
, ,
(2)
Combination of equations (1) and (2) yields;
dQ' =
Q1
dn1 +
Q2
dn2 + ... (3)
But, Q' = n1
Q1
+ n2
Q2
+ … (4)
Differentiation of equation (4) gives
dQ' =
Q1
dn1+
Q2
dn2 + ... + n1d
Q1
+ n2d
+ ... (5)
Subtraction of equation (3) from equation (5) yields
n1d
Q1
+ n2d
Q2
+ ... = 0
In general,
ni d
Qi
= 0 (6)
Dividing by n, total number of moles of all the components, gives
Xi d
Qi
= 0 (7)
Equations (6) and (7) are equivalent expressions of the Gibbs-
Duhem equation.
pf3
pf4
pf5
pf8
pf9
pfa
pfd

Partial preview of the text

Download Gibbs-Duhem Equation and more Lecture notes Chemistry in PDF only on Docsity!

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

  • 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=

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

-

  • GNixs

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 -X1X21 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