A9.1 A Statistical Mechanical Interpretation of the Entropy of Mixing in an Ideal Mixture   CDA9.1-1

Appendix A9.1 A Statistical Mechanical Interpretation of the Entropy of Mixingin an Ideal Mixture

The entropy change of mixing for an ideal binary gaseous or liquid mixture,

mix S IM = − R

 x i   ln x i

= − R( x 1   ln x 1 +  x 2   ln x 2)   (A9.1-1)

(cf. Eq. 9.1-8 and the appropriate entry in Table 9.3-1) has a very simple statistical

mechanical interpretation.

To demonstrate this, we consider a collection of  N 1  molecules of species 1 and   N 2molecules of species 2 maintained at constant total internal energy  U  and constant

volume V . Furthermore, it is assumed either that the molecules do not interact (this is

the case for an ideal gas mixture) or that the molecules do interact, but the interaction

energy function is the same for all species; that is, the species 1–species 1, species 2–

species 2, and species 1–species 2 interactions are all alike (this is the assumption that

underlies the ideal mixture model).

In courses on advanced physical chemistry and statistical mechanics, it is shownthat the entropy function for a binary mixture at constant internal energy, volume, and

number of particles is

S  =  N 1 S 1,intra +  N 2 S 2,intra + S conf    (A9.1-2)

Here  S i,intra is the entropy of one molecule of species i due to its intramolecular struc-

ture (which can be calculated from detailed structural and spectroscopic information

on the species), and   S conf  is the entropy contribution due to the configuration of the

system, that is, the way species 1 and 2 are distributed in the mixture. The statisticalmechanical expression for the configurational entropy for a specific macroscopic state

of the mixtures being considered here is

S conf = k   ln   (A9.1-3)

where    is the number of different arrangements of the molecules that result in the

desired macroscopic state and   k   is the Boltzmann constant. Our interest here is in

computing the value of   S conf  for different distributions of the two species within the

system.

To compute  S conf  we will imagine that within the volume  V  there is a three-dimen-

sional lattice with N  =   N 1 +  N 2  equally spaced lattice points. Different macroscopicstates for this model system correspond to different arrangements of the  N 1  molecules

of species 1 and the  N 2  molecules of species 2 among the  N  lattice points. By the ideal

mixture assumption, each distribution of molecules among the lattice points has the

same energy as any other.

The number of ways of distributing the  N 1  identical molecules of species 1 and the

 N 2  identical molecules of species 2 among the   N 1  +  N 2  lattice sites in a completely

random fashion is, from simple probability theory, equal to

( N 1 +  N 2)!

 N 1! N 2!

Therefore, the configurational entropy of the completely random mixture is

S conf 

random

mixture

= k   ln( N 1 +  N 2)!

 N 1! N 2!(A9.1-4)

 

CDA9.1-2   Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture

This equation can be simplified by using Stirling’s approximation for the logarithm of a factorial number,

ln M ! =  M   ln M −  M    (A9.1-5)

which is valid for large  M . Using Eq. A9.1-5 in Eq. A9.1-4 yields

S conf 

random

mixture

 = k [ln( N 1 +  N 2)! − ln N 1! − ln N 2!]

= k [( N 1 +  N 2)  ln( N 1 +  N 2)− ( N 1 +  N 2)

−  N 1   ln N 1 +  N 1 −  N 2   ln N 2 +  N 2]

= −k 

 N 1   ln N 1

 N 1 +  N 2+  N 2   ln

 N 2

 N 1 +  N 2

= −k ( N 1 +  N 2)[ x 1   ln x 1 + x 2   ln x 2]

(A9.1-6)

Now dividing by the sum of   N 1  and   N 2  and multiplying by Avogadro’s number   ˜ N 

(  =  6.022 × 1023), we obtain an expression for the configurational entropy per mole

of mixture

S conf 

random

mixture

 = − ˜ N k [ x 1   ln x 1 + x 2   ln x 2]

= − R[ x 1   ln x 1 +  x 2   ln x 2]

(A9.1-7)

where  R  =   ˜ N k  is the gas constant of Table 1.4-1.

Now consider the completely ordered configuration in which the molecules of species

1 are restricted to the first N 1  lattice sites and the molecules of species 2 to the remain-

ing N 2 lattice sites; that is, the two species are not mixed. The number of different ways

in which this can be accomplished is

 N 1! N 2!

 N 1! N 2!= 1   (A9.1-8)

so that

S conf 

completelyordered

=   R   ln 1 = 0   (A9.1-9)

Therefore, the entropy change on going from the completely ordered (unmixed) state

to the randomly ordered (completely mixed) state is

mix S  =  S conf 

random

mixture

− S conf 

completely

ordered

= − R[ x 1   ln x 1 +  x 2   ln x 2]

(A9.1-10)

which is in agreement with Eq. A9.1-1.

One can easily establish, though we will not do so here, that any partially ordered

state will have a molar configurational entropy intermediate to the randomly mixed and

completely ordered states. Therefore, the randomly mixed state is the state of highest

entropy. Since the criterion for equilibrium at constant internal energy and volume is

that the entropy of the system achieve a maximum, the randomly mixed or completely

disordered state is the equilibrium state in an ideal mixture.

It is tempting to try to generalize this result by suggesting that the completely disor-

dered state is always the equilibrium state. However, this is not correct!  In a mixture

 

A9.1 A Statistical Mechanical Interpretation of the Entropy of Mixing in an Ideal Mixture   CDA9.1-3

in which not all interactions are alike, different distributions of molecules among thelattice sites will result in different total energies of the system. In this case, energetic,

as well as entropic, effects are important in determining the equilibrium state. Thus,

in a system at constant internal energy and volume, the equilibrium state will be the

state of maximum entropy (or maximum randomness) among  only those states   that

have the required internal energy. For systems at constant temperature and volume, the

equilibrium state is a state of minimum Helmholtz energy. Since   A  =   U  − T S , it is

evident that increasing the entropy (or disorder) in the system decreases the Helmholtz

energy only if it does not simultaneously increase the internal energy of the system.For example, consider the lattice model used here, but now let  u ii  be the interaction

energy of a species i–species i interaction, where  u 11  =   u12  =   u22, so the mixture is

not ideal. Clearly, if  u12  is greater than the arithmetic average of  u11  and  u22, that is, if 

u12  >  1

2(u11 + u22)

increasing the randomness of the mixture (and the number of 1–2 interactions at the

expense of 1–1 and 2–2 interactions) increases both the entropy and the internal energy

of the system. Therefore, the equilibrium state for this system will be that compromisebetween energetic and entropic effects for which the Helmholtz energy is a minimum.

In real mixtures this balance between energetic and entropic effects is illustrated, forexample, in liquid-liquid phase equilibrium (to be discussed in Chapter 11), in which

the equilibrium state of some liquid mixtures is as two coexisting phases of different

composition, a distinctly ordered rather than random state. The equilibrium state is not

a single phase of uniform composition, since the increase in randomness (and hence

decrease in −T S ) of producing such a state from the two-phase mixture would be less

than the increase in the internal energy of the system. Consequently, the random single-phase mixture would have a higher Helmholtz energy than the more ordered two-phase

mixture, and therefore would not be a stable equilibrium state.