Physical Chemistry , 1st ed.

Equation 18.62 shows that statistical thermodynamics can calculate temperature-

dependent equilibrium constants from partition functions. Since the partition

functions themselves are ultimately determined from the energy levels of

the chemical species, we see once again how a knowledge of energy levels—

obtained from spectroscopy—helps us make thermodynamic predictions about

chemical reactions.

A word about units is necessary. It is important to keep track of all units in

equation 18.62, and when comparing calculated results to experimental ones,

units must be consistent. Equation 18.62 is the concentration-based equilib-

rium constant,Kc. Equilibrium constants can also be expressed in terms of the

partial pressures of the gas-phase reactants and products. The pressure-based

equilibrium constant,Kp, is related to Kcby the expression

KpKc(kT)i (18.63)

where irepresents the proper combination of the stoichiometric coefficients

of gas-phase substances in the balanced chemical reaction. Remember that (^) i
values are positive for products and negative for reactants.
Example 18.10
Of necessity, examples of equation 18.62 must be relatively straightforward.
Calculate the equilibrium at 298 K for the reaction
H 2 D 2 2HD
Assume that the electronic and nuclear partition functions cancel. Compare
the calculated value with the experimental value of 3.26 (unitless).
Solution
Since i 2  1  1 0,KpKcin this case. The remaining three par-
tition functions for each species are calculated as follows:
q H 2 D 2 HD
trans/V 2.737
1030 7.741
1030 5.028
1030
vib 2.959
10 ^5 6.283
10 ^4 1.197
10 ^4
rot 1.746 3.489 4.656
where qtranshas been divided by volume. For H 2 and D 2 , the rotational par-
tition functions are scaled by a symmetry number of 2. HD has a symmetry
number of 1. We have for the equilibrium constant

JQPJ

18.9 Equilibria 643

K(T) 



Notice that the volume terms cancel out of the expression for K(T). We get

(5.028

1030 1.197
10 ^4 4.656)^2



(2.737 

1030 2.959 

10 ^5 1.746) (7.741 

1030 6.283 

10 ^4 3.489)



qtransq

V

vibqrot



2

HD





qtransq

V

vibqrot

H^2 

qtransq

V

vibqrot

D^2

K(T) 3.273

This is very close to the experimentally determined value.

Statistical thermodynamics thus provides tools for us to predict equilibrium