Physical Chemistry , 1st ed.

Algebraically, we can bring the reactants to the other side of the “equation” and

rewrite equation 18.55 as

(^) CC (^) DD AA BB  0
(It is the convention to subtract the reactants from the products.) In terms of
classical thermodynamics, chemical equilibrium is given by equation 5.4, which
requires that

no. of components
i 1
(^) i
i 0
Using the general equation above, we have for this equilibrium
(^) C (^) C (^) D (^) DA (^) AB (^) B 0 (18.56)
We would like to substitute for (^) iin the equation above. For a mixture of gases,
we will assume that the overall partition function of the system Qsyscan be writ-
ten as the product of the molecular partition functions of each component:
18.9 Equilibria 641

Qsys

components

(Q

N

i)

i!

Ni





QA(N

N

A,

A

V

!

,T)NA



QB(N

N

B,

B

V

!

,T)NB



QC(N

N

C,

C

V

!

,T)NC



QD(N

N

D,

D

V

!

,T)ND

 (18.57)

In this equation, we are labeling each molecular partition function with the la-

bel of the relevant component. Also, we are reminded that each component is

occupying the same volume and has the same temperature (otherwise the sys-

tem is not at equilibrium), but that each component has its own characteris-

tic amount Niat equilibrium. Statistical thermodynamics gives an expression

for the chemical potential of a component:

(^) ikTln 




N

Q

i



Nji,V,T

(18.58)

where in the case of a multicomponent mixture, the partial derivative is taken

with respect to only one component,Ni, and the other components remain as

constants. This has the effect of eliminating all other species’ partition func-

tions from the evaluation of each particular (^) i. After substituting equation
18.57 into equation 18.58 for each of the four components and applying
Stirling’s approximation, we have
(^) ikTln 
Qi(
N

V

i

,T)

 (18.59)

where we have dropped the NA.. ., labels from each Q. Substituting for each

in equation 18.56, we have

(^) CkTln QC
N

(V

C

,T)

^ DkTln 

QD

N

(V

D

,T)

AkTln 

QA

N

(V

A

,T)

BkTln 

QB(

N

V

B

,T)

^0

The kTterms cancel to yield

(^) Cln 

QC

N

(V

C

,T)

 (^) Dln 

QD

N

(V

D

,T)

Aln 

QA

N

(V

A

,T)

Bln 

QB(

N

V

B

,T)

 0

We can use the properties of logarithms to take each coefficient and make it

an exponent inside the logarithm term. Removing the remaining labels from

Qfor clarity gives