1
Boltzmann Distribution
Equipartition of Energy
At a given temperature, the energy of a collection of molecule is distribution among the
various modes of motion (that is, translational, rotational, vibrational, electronic, …). All
distributions are equally likely; however, the particles in the distributions are
indistinguishable. Therefore, some distributions are more probable than others. This
principle is known as the equipartition of energy.
Illustration 1: Consider the following illustration putting two balls into three different boxes.
If each ball is equally likely to be in any box, what is the most likely arrangement if we
examine the box after shaking it?
a) Two balls in one box.
b) Two balls in separate boxes.
Each ball is equally likely to be in any box so let’s look at all the possibilities.
By looking at all the possibilities, we see that finding the two balls in separate boxes is twice
as likely finding two ball in one box.
Illustration 2: Consider the following illustration putting three balls into three different
boxes.
If each ball is equally likely to be in any box, what is the most likely arrangement if we
examine the box after shaking it?
a) Three balls in one box.
b) Two balls in one box and one ball is a separate box.
c) Three balls in separate boxes.
2
Three balls in one box.
Two balls in one box and one ball is a separate box.
Three balls in separate boxes.
Probability
3/27 = 11%
18/27 = 67%
6/27 = 22%
Boltzmann Distribution
As the numbers get larger, that is 1023 , the most probable distribution for a given amount of
macroscopic energy among microscopic energy states is given by the Boltzmann distribution.
−
ni
e
=
N
ni – number of molecules in the ith energy state.
N – total number of molecules.
Ei – energy of the ith energy state.
k – Boltzmann’s constant: 1.38 × 10-23 J/K
Ei
kT
∑e
−
Ej
kT
j
Note: R = kNA
Let us examine how the energy of 106 HCl molecules is distributed among the different
vibrational states at T = 298 K.
1
1


E ν = hcν% e  ν +  = ( 6.626 ×10 −34 J ⋅ s )( 2.997 × 1010 c m / s )( 2991cm −1 )  ν + 
2
2


1

= ( 5.617 ×10 −20 J )  ν + 
2

kT = (1.381× 10−23 J / K ) ( 298K ) = 4.118 × 10−21 J
ν
0
1
2
3
4
Eν
2.802 × 10-20 J
8.425 × 10-20 J
1.404 × 10-19 J
1.966 × 10-19 J
2.528 × 10-19 J
Eν/kT
6.82
20.46
34.10
47.74
61.38
exp(-Eν/kT)
1.09 × 10-3
1.31 × 10-9
1.55 × 10-15
1.85 × 10-21
2.20 × 10-27
n0 n0
e −6.82
1.09 ×10−3
= 6 = − 6.82
=
= 1 ⇒ n 0 = 106 ⋅ 1 = 106
−20.46
−34.10
−47.74
−61.38
−3
N 10
e
+e
+e
+e
+e
1.09 ×10
n1 n1
e −20.46
1.31×10 −9
= 6 = −6.82 −20.46 − 34.10 −47.74 −61.38 =
= 1.21× 10−6
−3
N 10
e
+ e
+ e
+e
+e
1.09 ×10
6
−6
⇒ n1 = 10 ⋅1.21×10 = 1
The vibrational energy is large compared to the thermal energy; therefore, approximately
999,999 molecules out of a million are in the ground vibrational state at 298 K and 1
molecule is in the first excited state.
3
Let us consider what happens to the energy distribution at a higher temperature such as
2000 K?
kT = (1.381×10 −23 J / K ) ( 2000K ) = 2.762 × 10−20 J
−
ν
0
1
2
3
4
5
6
Eν
kT
1.02
3.04
5.08
7.11
9.15
11.17
13.21
n ν = 106
e
Eν
kT
∑e
−
Eν
kT
ν
868998
113843
14911
1950
260
34
4
Role of Degeneracy in the Boltzmann Distribution
The Boltzmann distribution includes all degenerate states as equally probable as well.
Therefore the more precise formulation of the Boltzmann distribution is
−
Ei
ni
g e kT
= i
Ej
N
−
∑ g je kT
gi – degeneracy of the ith energy state.
j
Degeneracy is important when partitioning the rotational energy of sample of molecules.
Remember that each rotational state J can have an M value that ranges from J, …, -J. Thus
each rotational state J has 2J +1 M values; thus the degeneracy for rotational energy levels is
2J +1.
4
Example:
What is the distribution of rotational energy in a sample of HCl at 298 K? The
rotational constant of HCl is 10.44 cm-1
Recall that
E J = hcBJ ( J + 1) = ( 6.626 ×10−34 J ⋅ s )( 2.997 ×1010 c m / s )(10.44cm −1 ) J ( J + 1)
= ( 2.073 ×10 −22 J ) J( J+ 1)
J
g
E
0
1
2
3
4
5
6
7
8
9
10
11
1
3
5
7
9
11
13
15
17
19
21
23
0J
4.146 × 10-22 J
1.244 × 10-21 J
2.488 × 10-21 J
4.146 × 10-21 J
6.219 × 10-21 J
8.707 × 10-21 J
1.161 × 10-20 J
1.493 × 10-20 J
1.866 × 10-20 J
2.283 × 10-20 J
2.736 × 10-20 J
(2J + 1) ×
exp(EJ/kT)
1
2.713
3.696
3.826
3.288
2.430
1.569
0.895
0.493
0.205
0.083
0.030
cumulative
sum
1
3.713
7.409
11.235
14.523
16.953
18.522
19.417
19.910
20.115
20.198
20.228
% of
population
4.94
13.41
18.27
18.91
16.25
12.01
7.76
4.42
2.44
1.01
0.48
0.15
Most of the molecules are in the J = 3 state, with the J = 2 and J = 4 states also well
populated.
5
The fact that most of our molecules are not in the rotational ground state accounts for the
shape of the rotovibrational spectrum. The intensities of the lines are not the highest for the
transitions involving the J = 0 state. Below is the acetylene rotovibrational spectrum. The
highest intensities involve the J = 10 states. Therefore we surmise that the J = 10 states are
the most populated. (The most populated states will yield the most transitions; therefore, the
highest intensities.)
(Taken from L. Willard Richards, J. of Chemical Education, Vol 43, pp. 644-647 (1966))