Physical Chemistry Third Edition

22.5 The Equilibrium Populations of Molecular States 943

Solution

Ratio

e−E^1 /kBT

e−E^2 /kBT

e−(E^1 −E^2 )/kBT

exp

(

( 2 .128 eV)

(

1. 602 × 10 −^19 J(eV)−^1

)

(

1. 3807 × 10 −^23 JK−^1

)

(298 K)

)

e−^82.^87  1. 03 × 10 −^36

The result of this example is typical for most atoms and molecules except for a few molecules

such as NO that have one low-lying excited level.

Exercise 22.9

Calculate the ratio of the population of the 2shydrogen state to that of the 1sstate at 298 K.

The Population of Rotational States of Diatomic Molecules

To the rigid-rotor approximation, the rotational energy of a diatomic molecule is

EvhBeJ(J+1)hc ̃BeJ(J+1) (22.5-5)

The rotational levels have a degeneracy

gJ 2 J+ 1 (22.5-6)

Therefore,

(

population of rotational

energy levelJ

)

∝( 2 J+ 1 )e−EJ/kBT (22.5-7)

EXAMPLE22.11

a.Using the value of ̃Befor CO in Table A.22, find the ratio of the population of one of the

J2 states to that of theJ0 state at 298 K.

b.Find the ratio of the population of theJ2 level to that of theJ0 state at 298 K.

Solution

a.LetN(J,M) be the number of molecules with quantum numbersJandM.

N(2, 0)

N(0, 0)

e−(E^2 −E^0 )/kBTe−^6 hc ̃Be/kBT

exp

(

− 6

(

6. 6261 × 10 −^34 Js

)(

2. 9979 × 1010 cm s−^1

)(

1 .931 cm−^1

)

(

1. 3807 × 10 −^23 JK−^1

)

(298 K)

)

e−^0.^05594  0. 9456

b.LetN(J) be the population of levelJ:

N(J)(2J+1)N(J,M)

N( 2 )

N( 0 )



5 N(2, 0)

N(0, 0)

 5 ( 0. 9456 ) 4. 728