A fully disordered state

Order-disorder transformations

Describing ordering using a two-sublattice model









Major constituents are bolded. In plain English, A wants to go to the 1st sublattice, and B prefers to reside on the 2nd sublattice.

There is a reason why it is advantageous to use the normalizing condition1:

1

p q









This is how the long-range order (LRO) parameter  is typically defined:

A

1 1

y x p   

 (2)

It is clear from (1) that:



 





A 1 1 1

x  p y  p z  p py  1 z p  pz py  1 z pz (3)

Let us use (3) in (2):

A

1 1

1 y x

p    



1 y py

  



 



1 1

y p z p

z pz

z y

p p

  

  _{  }

 

The order parameter is a difference between the site fractions of B in the second and the first

sublattice. It is worth remembering that A is the major constituent on the first sublattice, and B is

the major constituent on the second sublattice. In view of this, it is apparent that:

0  1

The order parameter can be defined using the site fractions of A as well:



 



z y y z

      (4)

1

A fully disordered state

Site fractions of A and B are equal to their molar fractions. For A, this means that

A

1    y 1 z 1 x . It immediately follows from (4) that for a fully disordered (random)

solutions the order parameter is equal to zero:

0

 

Stoichiometric composition

The stoichiometric composition corresponds to the case when the first sublattice is completely

occupied by A, and the second one is fully filled by B:

   

It is clear that 1 y 1 and that 1 z 0. In immediately follows from one that the order

parameter for the fully ordered solution having the stoichiometric composition is equal to unity:

1



A partial ordering is characterized by 

 

Non-stoichiometric compositions

In this case, x_A  p and x_B  1 p. What can be said about the maximum value of the order parameter? Obviously, two situations must be analyzed.

Excess of A (deficit of B)

If x_A  p, then the first sublattice is completely filled with A:

  



This means, of course, that:

1 y 1 (5)

In order to find the site fraction of A on the second sublattice, let us consider the mass balance

with respect to A:







A 1 1

x   p p z

A

1

1

x p z

p   

Let us make use of (5) and (6) in (4):

A max

1 1

1

p x p

p

     



A x p

  1 _A

1 1

x

p p

 

 

Since xA p, max 1. Let us summarize our finding by stating that in the case of

non-stoichiometric composition and an excess of A, the following is true:



max

A

1

0

1

x

p







 



Deficit of A (excess of B)

If x_A p, then the second sublattice is completely filled with B:





 

What is means is that:

1 z 0 (7)

Let us use mass balance with respect to A to find 1y;





A 1

x  p y

A

1 y x

p

  (8)

By inserting (7) and (8) in (4), we arrive at:

A A

max 0

x x

p p

   

Since x_A p, _max 1. Let us summarize our finding by stating that in the case of non-stoichiometric composition and an excess of B, the following is true:



max

A

0

x

p





Renaming the sublattices









nd st

1 1

2 sublattice 1 1 sublattice

B A

y y z z p p

 



A



 B (9)

Mother Nature does not care how we name sublattices. In other words, the properties of the

solution do not depend on what sublattice is called the first, and what sublattice is name the

second.









ST nd

1 1

1 sublattice 1 2 sublattice

A _{z z y y}

p p

 



B A

 B (10)

For (9), the order parameter is defined as:

 9



 



    

For (10), the order parameter is defined as:

 10



 



      

If the lattices are renamed, then the order parameter changes its sign2. Is it important? No, it is

not important, it is very important! Why? Well, let us consider the Gibbs energy of ordering,

which, of course, should depend on the order parameter . A corresponding expression cannot

change if the lattices are renamed, which imposes limitations on our choice of G

 

for instance, that it is decided to use

 

0

n i i i

G  c 





If the lattices are renamed, then the following must hold:

 

0 0

n n

i i

i i

i i

c c 

 

 





If is even, then i i  

 

, but if is odd, then i i  

 

. Apparently, there should be no

terms with odd powers in (11), which means that it should be written as:

 

0 2 4

G   c c c 

You understand that c0 must be equal to 0, don’t you?!

2

First- and second-order phase transitions









Phase transformation

Disordered state  0 Ordered state 0 

If  suddenly changes from 0 to a finite value, then we have a phase transformation of the first order.

If  gradually (continuously) changes from 0 to _max, then we have a phase transformation of the second order.

Landau

theory of the second-order phase transformations

Let us consider a binary A–B solution, which is fully disordered above , and which starts to

order below this critical temperature. Apparently, we are dealing with the order-disorder

transformation, which is a second order transition. Let us postulate that the Gibbs energy of this

solution depends on temperature, pressure, composition and the LRO parameter:

0 T





GG T P x 

Let us also postulate that is continuously differentiable in the vicinity of the phase transition.

This postulate, which is not self-evident, indeed, limits the generality and applicability of the

Landau’s theory.

G

Since in the vicinity of T₀, 1, we can use our favourite ultimate weapon:





A 0 1

, , ,

2 3 4

A A

G T P x  G A     A44 ₍₁₂₎

where G A0, 1,,A4 are functions of T, P and xA.

First of all, let us make use of the equilibrium condition:

2 3

1 2 3 4 1

0 0 G

A A A A A



  

 _

 

      

_ 

  0 (13)

In view of (13), one can simplify (12):





A 0

, , ,

2 3 4

A A

G T P x  G      A44

(14)

3

At this moment, you should revisit the section Renaming the sublattices on page 4 and realize

that all terms with odd powers in (14) must be identical to 0, which, in turn, means that A₃ 0.





A 0

, , ,

2 4

A A

G T P x  G    4 (15)

Since the solution is in equilibrium, the following two conditions must be fulfilled:   G  0

and 2G 2 0. With respect to (15), these conditions yield:





2 4 3 2

2 4

0 2 4 2 0

2 4

A A

G   A A  A 



  _{ } _{    }

 

   A4 (16)

2

2 4 2

2 4

0 2

2 3

2 4

A A

G   A 



  

4 0

A

   

 

    (17)

The equation (16) has two solutions. The first solution

0

  (18)

corresponds to the disordered state. By substituting (18) in (17) we conclude that for the

disordered state:

0 2

IfT T , then A 0 (19)

The second solution

2

2 4 A A

   (20)

corresponds to the ordered state. Substitution of (20) in (17) gives:

2

2 4 2

4

3 A 2

A A A

A 0

   

It is clear that for the ordered state:

0 2

IfT T A, 0 (21)

From (21) and (20) one instantly concludes that for the ordered state:

4 0 A 

It is apparent from (19) and (21) that depends on temperature and that . Let

be a function of

2

A A T₂





2

A TT₀. Since in the vicinity of the phase transition the difference T T₀ is

small, let us decompose A₂ into the Taylor’s series and restrict our appetite by the first term:





2

The constant in a (22) is always positive, because if T T₀, then A₂ 0, and if T T₀, then

. Let us use

2 0

A  (20) and assume that in the vicinity of T₀ is constant:

0 4

A A₄ (23)

From (20), (22) and (23) one gets:









2 2

0

4 4 4

a T T a T T A

A A A

       ₀ (24)

Let us substitute (22) and (24) in (15):







 









 







_

_

2 0

0 0 4 0

A 0 0 0

4 4

2

0 2

2

0 0 4 0

0 0 0 0 0

4 4 4

, , ,

2 4

2 4 4

a T T a T T A a T T G T P x G

A A

a T T a T T A a T T a

G G

A A A

      _  _

 

    

   _{ }   

  T0 T

Theory vs. experiment

Figure 1 hardly deserves extensive comments. It is clear that when 1 the theory works fine.

T

2



0

T

experiment

theory

is not

small here!

