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478 13 STATISTICAL THERMODYNAMICS
account.
E0(H2O)vib = 1
2NAhc(3656.7 + 1594.8 + 3755.8) = NAhc(4503.65 cm
−1)
E0(HDO)vib = 1
2NAhc(2726.7 + 1402.2 + 3707.5) = NAhc(3918.2 cm
−1)
E0(HCl)vib = NAhc(1495.5 cm−1) E0(DCl)vib = NAhc(1072.5 cm−1)
∆rE0 = NAhc(3918.2 + 1495.5 − 4503.65 − 1072.5)
= NAhc(−162.5 cm−1)
�us the term −∆rE0/RT evaluates as
−∆rE0
RT
= −NAhc(−162.5 cm−1)
RT
= −(6.0221 × 1023mol−1) × (6.6261 × 10−34 J s) × (2.9979 × 1010 cms−1)
× (−162.5 cm−1)
(8.3145 JK−1mol−1) × T
= (233.7 K)/T
With these expressions the equilibrium constant is evaluated using mathemat-
ical so�ware to give K = 3.90 at 298 K and K = 2.42 at 800 K.�e value of the
equilibrium constant is dominated by the symmetry factors and the e−∆rE0/RT
term.
P13F.4 �e standard molar Gibbs energy is computed from the partition function us-
ing [13F.9b–569], G−○m(T) = G−○m(0) − RT ln q−○m/NA. As usual, the partition
function is factored into separate contributions from translation, rotation and
so on.�e factor of 1/NA is usually taken with the translational contribution.
�e standard molar translational partition function is given by q−○m = V−○m/Λ3 =
RT/p−○Λ3. Taking the mass of BSi as 10.81 + 28.09 = 38.90 mu, Λ is given by
[13B.7–541]
Λ = h/(2πmkT)1/2
= 6.6261 × 10−34 J s
[2π(38.90×1.6605 × 10−27 kg)×(1.3806 × 10−23 JK−1)×(2000 K)]1/2
= 6.25... × 10−12 m
q−○m/NA =
RT
p−○Λ3NA
= (8.3145 JK−1mol−1) × (2000 K)
(105 Nm−2) × (6.25... × 10−12 m)3 × (6.0221 × 1023mol−1)
= 1.12... × 109
For a heteronuclear diatomic the rotational partition function in the high tem-
perature limit is given by [13B.13b–544] qR = kT/hcB̃.�e rotational constant