25.4 The Calculation of Molecular Partition Functions 1071
partition function. If additional electronic terms must be included, the same rotational
and vibrational partition functions do not necessarily apply to different electronic states
since all electronic states do not usually have the same force constant and internuclear
distance. If different values ofBeandνeapply to different electronic states we must
compute a different rotational and vibrational partition function for each electronic
state. The partition function becomeszztr[g 0 e−ε^0 /kBTzrot,0zvib,0+g 1 e−ε^1 /kBTzrot,1zvib,1+...] (25.4-21)wherezrot,0andzvib,0are the rotational and vibrational partition functions correspond-
ing to electronic state number 0,zrot,1andzvib,1correspond to electronic state num-
ber 1, and so on. If the excited electronic states make a fairly small contribution, it
might be a usable approximation to assume that all electronic states have roughly the
same values ofBeandνe. If so, we can write the partition function in the factored
form of Eq. (25.4-6) even if more than one term is included inzel. The values of
Beandνeof the ground electronic level are used since this term makes the largest
contribution.EXAMPLE25.10
The ground electronic level of NO (a relatively stable molecule with an unpaired electron) is a(^2) Π 1 / 2 term. The first excited level is a (^2) Π 3 / 2 term with an energy 2. 380 × 10 − (^21) J above the
ground-level. Both levels have a degeneracy equal to 2. All other electronic states are more
than 7× 10 −^19 J above the ground-level. For the ground-levelνe 5. 7086 × 1013 s−^1 and
Be 5. 0123 × 1010 s−^1.
a.Find the molecular partition function of NO at 298.15 K and at a volume equal to the
molar volume at 101325 Pa. Assume that the ground-level values ofBeandνeapply to
both electronic states.
b.Find the molecular partition function for NO for the conditions of part a, using
the values for the first excited state,νe 5. 7081 × 1013 s−^1 andBe 5. 1569 ×
1010 s−^1.
Solution
a.
m(NO)
0 .014003074 kg mol−^1 + 0 .015994915 kg mol−^1
6. 02214 × 1023 mol−^1
4. 98127 × 10 −^26 kg
V
nRT
P
(1.000 mol)(8.3145 J K−^1 mol−^1 )(298.15 K)
101325 Pa
0 .024466 m^3
ztr
(
2 πmkBT
h^2
) 3 / 2
V
(
2 π(4. 98127 × 10 −^26 kg)(1. 3807 × 10 −^23 JK−^1 )(298.15 K)
(
6. 6261 × 10 −^34 Js
) 2
) 3 / 2
×(0.024466 m^3 )
3. 8893 × 1030