Physical Chemistry Third Edition

1102 26 Equilibrium Statistical Mechanics. II. Statistical Thermodynamics

version of Eq. (26.3-5) for each substance is substituted into Eq. (26.3-2) we obtain the

following equation for a reacting dilute gas mixture at equilibrium

0 

∑s

a 1

νa

[

ε 0 a−kBTln

(

z′a

Na

)]

(26.3-6)

where all of theε 0 values must be taken with respect to the same zero of energy. We

define∆ε 0 as the difference in the energies of the ground states of the product and

reactant molecules:

∆ε 0 

∑s

a 1

νaε 0 a (26.3-7)

This quantity is equal to the energy change whenνamolecules of substanceaappear if

ais a product, and|νa|molecules of substanceaare consumed ifais a reactant, with

all molecules in their ground states.

The sum of logarithms is the logarithm of a product. Use of the identity

yln(x)ln(xy) (26.3-8)

and division bykBTgives

0 

∆ε 0

kBT

−ln

[s

∏

a 1

(

z′a

Na

)νa]

(26.3-9)

This equation can be manipulated into an equation containing the equilibrium constant.

The molecular partition functions are proportional toV, since the translational factor is

ztr

(

2 πmkBT

h^2

) 3 / 2

V (26.3-10)

and the other factors are independent ofV. We define

z′′a

z′a

V

(26.3-11)

so thatz′′ais independent ofV. Since dilute gases obey the ideal gas law,

V

Na



kBT

Pa



kBT

P◦

P◦

Pa



Vm◦

NAv

P◦

Pa

(26.3-12)

wherePais the partial pressure of substancea,NAvis Avogadro’s constant, andVm◦is

the volume occupied by 1.000 mol of ideal gas at the standard pressureP◦(exactly 1

bar) and temperatureT. Equation (26.3-9) can now be written as

0 

∆ε 0

kBT

−ln

[s

∏

a 1

(

z′′aVm◦

NAv

)νa]

+ln

[ s

∏

a 1

(

Pa

P◦

)νa]

(26.3-13)

The last term in this expression is the logarithm of the equilibrium constant in

Eq. (7.2-2). We can now write

ln(K)−

∆ε 0

kBT

+ln

[s

∏

a 1

(

z′′aVm◦

NAv

)νa]

(26.3-14)