1112 26 Equilibrium Statistical Mechanics. II. Statistical Thermodynamics
(
2 π(1. 674 × 10 −^27 kg)(1. 3807 × 10 −^23 JK−^1 )(500.0K)
(6. 6261 × 10 −^34 Js)^2
) 3 / 2
(2)
2. 127 × 1030 m−^3 (2) 4. 254 × 1030 m−^3
The partition function of HBr has a translational and a rotational factor:
z′′tr,Br
(
2 πmHBrkBT
h^2
) 3 / 2
(
2 π(1. 327 × 10 −^25 kg)(1. 3807 × 10 −^23 JK−^1 )(500.0K)
(6. 6261 × 10 −^34 Js)^2
) 3 / 2
1. 501 × 1033 m−^3
zrot,HBr
kBT
hcB ̃e
(1. 3807 × 10 −^23 JK−^1 )(500.0K)
(6. 6261 × 10 −^34 J s)(2. 9979 × 1010 cm s−^1 )(8.4649 cm−^1 )
41. 05
z′′HBr(1. 501 × 1033 m−^3 )(41.05) 6. 161 × 1034 m−^3
The partition function of the activated complex has a translational, a rotational, and a vibra-
tional factor:
z‡HHBr′′ z‡HHBr,tr′ z‡HHBr,rot′ z‡HHBr,vib′
z‡tr,HHBr′′
(
2 πmHHBrkBT
h^2
) 3 / 2
(
2 π(1. 360 × 10 −^25 kg)(1. 3807 × 10 −^23 JK−^1 )(500.0K)
(6. 6261 × 10 −^34 Js)^2
) 3 / 2
1. 558 × 1033 m−^3
To obtain the rotational partition function, we require the moment of inertia of the activated
complex. The center of mass of three objects is found as in Eq. (D-25) of Appendix D.
xc
m 1 x 1 +m 2 x 2 +m 3 x 3
m 1 +m 2 +m 3
If the origin of coordinates is taken at the hydrogen nucleus at the end of the activated complex,
we have
xc
0 +mH(150 pm)+mBr(292 pm)
mH+mH+mBr
(1.008 amu)(150 pm)+(79.904 amu)(292 pm)
2 .016 amu+ 79 .904 amu
286 .7pm
The moment of inertia is
IemH(286.7pm)^2 +mH(136.7pm)^2 +mBr(5.3pm)^2
(1.008 amu)(286.7pm)^2 +(1.008 amu)(136.7pm)^2 +(79.904 amu)(5.3pm)^2
(1. 039 × 104 amu pm^2 )
(
0 .001kg
6. 02214 × 1023 amu
)(
1m
1012 pm
) 2
1. 73 × 10 −^46 kg m^2