Foundations of Chemical Kinetics Lecture 5: The Boltzmann - - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 5: The Boltzmann distribution

Marc R. Roussel Department of Chemistry and Biochemistry

Ensembles

◮ Even without quantum mechanics, matter has some inherently

probabilistic qualities.

◮ In every collision between molecules, energy and momentum

are exchanged.

◮ Even if we knew the positions and momenta of every particle

in a container at some time t, our ability to predict the energies, positions and momenta of particles after a few collisions would be essentially zero, except in a statistical sense.

◮ We treat such systems using ensembles, which are large,

imaginary sets (approaching infinite size) of copies of a system

• f interest which have some variables in common and which

have been similarly prepared.

The Boltzmann distribution

◮ Imagine an ensemble of systems all of which are held at the

same temperature or an ensemble of non-interacting molecules in a container at a fixed temperature T. = ⇒ canonical ensemble

◮ According to Boltzmann, the probability that one of these

systems (molecules) has energy ǫi is given by P(ǫi) = gi exp

• − ǫi

kBT

• /Q

where gi is the degeneracy of energy ǫi, i.e. the number

• f different microscopic states that give the

same energy, kB is Boltzmann’s constant, Q is a constant chosen such that

i P(ǫi) = 1.

The partition function

◮ The normalization constant Q is called the partition function. ◮ We must have

i P(ǫi) = 1, so

• i

P(ǫi) =

• i

gi exp

• − ǫi

kBT

• /Q = 1

∴ 1 Q

• i

gi exp

• − ǫi

kBT

• = 1

∴ Q =

• i

gi exp

• − ǫi

kBT

Meaning of the partition function

Q =

• i

gi exp

• − ǫi

kBT

• ◮ Suppose that, at some temperature of interest, kBT ≫ ǫi for

i ≤ n and kBT ≪ ǫi for i > n.

◮ For states well below kBT, exp(−ǫi/kBT) ≈ 1. ◮ For states well above kBT, exp(−ǫi/kBT) ≈ 0. ◮ Thus,

Q ≈

n

• i=1

gi.

◮ Roughly speaking, the partition function counts the number of

states accessible at temperature T.

Example: The partition function for a two-level system

◮ Suppose that ǫ1 = 0, g1 = 1 and ǫ2 = 1 × 10−19 J, g2 = 3.

Q = 1 + 3 exp

• − ǫ2

kBT

• = 1 + 3 exp(−7243 K/T)

1 1.5 2 2.5 3 3.5 4 100 1000 10000 100000 1e+06 Q T/K

A minor rewrite of the partition function

We can drop the degeneracy if we sum over quantum states (distinct sets of quantum numbers) rather than over energy levels: Q =

• i

exp

• − ǫi

kBT

Separability of partition functions

◮ Suppose that the molecular energy can be decomposed into a

sum of independent contributions, i.e. ǫ = ǫ(1) + ǫ(2) + . . .

◮ Then, using the second form of the partition function, we have

Q =

• i
• j
• k

· · · exp

• −ǫi + ǫj + ǫk + . . .

kBT

• =
• i
• j
• k

· · · exp

• − ǫi

kBT

• exp
• − ǫj

kBT

• exp
• − ǫk

kBT

• · · ·

=

• i

exp

• − ǫi

kBT

j

exp

• − ǫj

kBT

k

exp

• − ǫk

kBT

• · · ·

= Q(1)Q(2)Q(3) . . .

The translational partition function

◮ Consider some molecules in a rectangular container in the gas

phase. We can treat their translational degrees of freedom using a particle-in-a-box treatment. In three dimensions, we just include three particle-in-a-box terms, one for each dimension and with its own quantum number.

◮ For an oxygen molecule with an average kinetic energy at

room temperature in a 1 cm box of 2 × 10−21 J, the translational (particle-in-a-box) quantum number is nx ≈ 4 × 108.

◮ Enx+1 − Enx = (2n + 1)h2/8mL2

x ≈ 9 × 10−30 J

The translational partition function (continued)

◮ Since the energy levels are very close together,

Qx =

nx exp(−ǫnx/kBT) is a very accurate Riemann sum

for the corresponding integral (with ∆nx = 1).

◮ Thus,

Qx ≈ ∞ exp

• −

n2

xh2

8mL2

xkBT

• dnx

◮ This is a known integral:

Qx = Lx h

• 2πmkBT

◮ The translational partition function is therefore

Qtr = QxQyQz = LxLyLz h3 (2πmkBT)3/2 = V h3 (2πmkBT)3/2

The vibrational partition function

◮ For a harmonic oscillator, Evi = ω(i)

• vi + 1

2

• .

We are allowed to set our zero of energy wherever we want. In particular, we could set it so that E0 = 0, i.e. use Evi = ω(i)

0 vi:

Qi =

∞

• vi=0

exp

• −ω(i)

0 vi

kBT

• =

∞

• vi=0
• exp
• −ω(i)

kBT vi =

• 1 − exp
• −ω(i)

kBT −1

Rotational partition function

◮ The treatment of rotation in quantum mechanics is complex,

with several cases to consider.

◮ The most general case (a nonlinear molecule with no special

symmetries) leads to the following partition function: Qrot = π1/2 8π2IakBT h2 1/2 8π2IbkBT h2 1/2 8π2IckBT h2 1/2

◮ Ia, Ib and Ic are the moments of inertia of the molecule

around its three principal rotational axes.