Electronic Partition Function
qe

g e
  j ,e
j
j ,electronic energies
 g 0  g1e
  1,e
 g 2e
  2 ,e

In nearly all cases electronic energies are very
large and all terms except the first are ~ 0
 qe  g 0
the degeneracy of the ground state.
A rare exception is NO:
2
140 cm-1
(…π ) - Π
2
2
2
 qe  2  2e
140 / kT
Π3/2
Π1/2
 2  2(0.509)  3.018
Overall expression for the molecular partion
function of a diatomic molecule:
 2 mkT 

q  
2
h


3/ 2
1
 kT 
V

h / kT

B
1

e


and the System partition function:
Q  q or Q  q / N!
N
N
is defined in terms of:
m, B, ν, g0 - properties of the molecule
N, T, V – properties of the System

g0

Magnitude of q.
Consider H-35Cl at 300 K
m = (36 x 10-3)/NA = 5.978 x 10-26 kg
B = 10.54 cm-1
ν ~ 3000 cm-1
g0 = 1
q t = 2.109 x 10 32 V
q r = 19.78
qv =1
qe =1
Compare 127 I 2
B = 0.0374 cm-1
ν = 214.5 cm-1
q t = 3.952 x 10 33 V
q v = 1.556
q r = 5.575 x 103
qe =1
Units of q t : m (kg), k (J K-1), h (J s)
( kg J K-1 K / J2 s2 )3/2 V  (kg J-1 s-2)3/2 V
but J – work – force x distance –
mass x acceleration x distance – kg m2 s-2
(kg J-1 s-2)3/2 V  (kg kg-1 m-2 s2 s-2)3/2 V
 m-3 V
If system contains 1 mole at 1 atmos at 298K,
then V ~ 24 l but has to be in m3
( 1 l = 10-3 m3 ).
Justifying the value of β
Consider a very simple system and calculate
its Internal Energy.
System with equally spaced energy levels:
E  nE ,
( n  0,1,2,)
Qe
e
n
  En
n
n
1 e  e
 (1  e )
 E
 E
but e
 E
  nE
2  E
e
3  E

1
 1  E  
 Q ~ ( E )
1
which is a very simple result for a very simple
system.
Now calculate the Internal Energy:
 1 Q
 1  (  E ) 1
U 

Q  (  E ) 1

1
2
1

(

E

)


(  E ) 1
another simple result.
U is a state function of the system; (N, T, V )
fixed, U fixed.
To change U have to change (N, T, V ) to new,
fixed values.
But: d U = T dS + p dV and dV = 0 as V is
fixed (not variable).
Change U of the system by changing T .
Therefore:
U~T,
β -1 ~ T ; β ~ T -1
β = ( kT ) -1
The Boltzmann distribution for individual
molecules.
Have been using the Boltzmann distribution
of Systems in an Ensemble.
  Ei
n
e

N
e
i
  Ei
i
T
to define the average energy per System < E>.
Now consider the N (identical) molecules in a
System with average energy < ε >
 
E

N
n
j
j
j
N


  e j
 N  q   N  j
E 
 
q   
q  





E
N
n

 e
j j
q

j
N


e


j
n 

j
j


 N  e j
j j


q



j
N
j
q
The fraction of molecules in the nth energy
level.
An application of the Internal Energy
expression.
The Equipartition Theorem and Heat
Capacities.
The Equipartition Theorem states: That
when quantum effects can be ignored, the
average energy of each ‘squared term’ of
molecular energy is ½ kT .
translational energy:
½ mvx2 + ½ mvy2 +½ mvz2  3/2 kT
rotational energy:
½ IA ωA2 + ½ IB ωB2 + ½ IC ωC2  3/2 kT
(kT for a linear as only 2 rotational axes)
vibrational energy:
½ μ v2 + ½ kx2  kT per vibrational mode
This means that at a temperature T there is
(on average) as much energy in translation as
there is in rotation (can seldom ignore the
quantum effects in vibrational motion).
 1 Q  N q
U 

Q 
q 
q  qt qr qvt qe
 N  (qt qr qvt qe )
U 
qt qr qv qe

N  (qt ) N  (qr )


qt 
qr 
N

N  ( qv ) N  ( qe )


qv 
qe 
t
 r  v  e

average translational energy per molecule,
etc.
Now consider each ‘bit’ of the molecular
energy.