3.012 Fundamentals of Materials Science
Fall 2003
The Gibbs-Duhem equation for a binary solution

Free energy-composition diagrams give us a lot more information than just the decrease in free energy on mixing.
We can also use them to determine the partial molar free energy/chemical potential of each component as a
function of composition. To see how we obtain this information, let’s calculate the Gibbs-Duhem equation for a
binary system.
o
Recall the Gibbs Duhem equation is:
C
0  SdT  VdP   n i di
(Eqn 1)
i1
o
For a single-phase binary system, this becomes:
(Eqn 
2)
0  SdT VdP  nA dA  nB dB
(Eqn 3)
0  S dT VdP  X A dA  X B dB

o

(Eqn 4)

Suppose we have the common case that the system is closed- a fixed number of total A and B molecules
is present which cannot change: NA + NB = constant. We can simplify the Gibbs-Duhem expression by
expanding the chemical potentials into their components. Similar to our approach in several previous
instances, let’s write the algebraic form of the differential for µ A and µB. The chemical potential is a
function of 3 variables: µA = µA(P,T,XB) and µB = µB(P,T,XB):
 
 
 
dA   A  dP   A  dT   A  dX B
 P T ,X B
 T P ,X B
X B P,T
 
 
 
dB   B  dP   B  dT   B  dX B
 P T ,X B
 T P ,X B
X B P,T

Note that we pick XB to represent composition arbitrarily, we could also have picked XA. We can
substitute some of the identities we’ve previously identified:

(Eqn 5)
G 
A 
    A   SA
 T P,X B  T P,X B
(Eqn 6)
G 
A 
    A   VA
P T ,X B  P T ,X B


(Eqn 7)


…similarly for the B component terms. Substituting into the Gibbs-Duhem equation:


A  
B  


0  S dT  V dP  X A 
V
dP

S
dT


X
V
dP

S
dT




 
A
B  B
B
 A
X B P,T 
X B P,T 




Lecture 17 – Solutions
1 of 2
2/12/16
3.012 Fundamentals of Materials Science

(Eqn 
8)


Fall 2003
 
 
0  S dT  X A SA  X B SB dT  V dP  X AVA  X BVB dP   A  X A   B  X B
X B P,T
X B P,T
 
 
0  S dT  S dT  V dP  V dP   A  X A   B  X B
X B P ,T
X B P,T
 
 
0   A  1 X B    B  X B
X B P,T
X B P ,T
GIBBS-DUHEM FOR BINARY SOLUTIONS
Now, let’s see how this equation can help us determine the chemical potential from a free energy-composition
diagram.
Lecture 17 – Solutions
2 of 2
2/12/16