Physical Chemistry Third Edition

26.3 Chemical Equilibrium in Dilute Gases 1101

26.13Estimate the difference in the rotational contribution to

the entropy of 1.000 mol of^14 N^15 N and 1.000 mol of

(^14) N 2 at 298.15 K. Ignore the difference in the
masses.
26.14Calculate the standard-state energy, entropy, and Gibbs
energy of 1.000 mol of xenon gas at 298.15 K.
26.15Calculate the molar entropy of water vapor at 25. 0 ◦C and
1.000 bar. The bond distances are equal to 95.8 pm and
the bond angle is equal to 104.45◦. The vibrational
frequencies are 4.7817× 1013 s−^1 , 1.0947× 1014 s−^1 ,
and 1.1260× 1014 s−^1.
26.16Calculate the translational, rotational, and vibrational
contributions toCV,mof SO 2 gas at 298.15 K and
1.000 bar. The vibrational frequencies are in Figure 22.6.
26.17Calculate the standard-state enthalpy (relative to the
value at 0 K) of argon gas at 298.15 K and at 500.0 K.
26.18Using data in Example 25.10, make a graph of the
electronic contribution to the molar energy of NO and a
graph of the electronic contribution to the molar heat
capacity of NO between 0 K and 500 K. The behavior of
this electronic heat capacity as a function of temperature
is known as theSchottky effect.

26.3 Chemical Equilibrium in Dilute Gases

In previous chapters we have written a chemical reaction equation in the form

0 

∑s

a 1

νaFa (26.3-1)

wheresrepresents the number of substances involved in the reaction andFarepresents

the chemical formula of substancea. In Eq. (7.1-10), we established that at equilibrium

(

∂G

∂ξ

)

T,P



∑s

a 1

νaμa(eq) 0 (26.3-2)

Equation (26.1-27) for the chemical potential corresponds to an unspecified choice

of the zero of energy and gives a different value for a different choice of the zero

of energy. We must use a consistent zero of energy for all substances. We write the

molecular partition function as in Eq. (25.3-28):

z

∑

i

e−εi/kBT

∑

i

e−(εi−ε^0 +ε^0 )kBTe−ε^0 /kBT

∑

i

e−(εi−ε^0 )kBT (26.3-3)

whereε 0 is the energy eigenvalue of the molecule’s ground state relative to a fixed zero

of energy. We denote the sum in the second version of Eq. (26.3-3) byz′, so that

ze−ε^0 /kBTz′ (26.3-4)

The partition functionz′is the partition function relative to the ground-state energy

of the molecule. It is independent of the choice of the zero of energy since it contains

only energy differences. Equation (26.1-27) for the chemical potential now becomes

μ−kBTln

(z

N

)

−kBT

[

−ε 0

kBT

+ln

(

z′

N

)]

ε 0 −kBTln

(

z′

N

)

(26.3-5)

In a dilute gas mixture, each substance behaves as though the other substances were

absent, so we can use our formulas for each one of the gases in the mixture. When a