CHE-20028: PHYSICAL & INORGANIC CHEMISTRY
STATISTICAL THERMODYNAMICS: LECTURE 1
Dr Rob Jackson
Office: LJ 1.16
r.a.jackson@keele.ac.uk
http://www.facebook.com/robjteaching
Statistical Thermodynamics:
topics for lecture 1
• A link between quantum mechanics and
thermodynamics and kinetics
• The Boltzmann Distribution
• Partition Functions
– Vibrational
– Translational
– Rotational
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Thermodynamics Lecture 1
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Why is it useful?
• There are 2 strands of Physical Chemistry:
– Macroscopic: thermodynamics and kinetics (do
reactions occur, and if so, at what rate?)
– Microscopic: atomic structure-spectroscopy and
their explanation by quantum mechanics.
• These (apparently) unrelated strands can be
linked by statistical thermodynamics.
• E.g. we can calculate the enthalpy of a
reaction instead of measuring it !
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Thermodynamics Lecture 1
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Additional useful reference
http://pubs.rsc.org/en/content/ebook/978-0-85404-632-4
‘Thermodynamics & Statistical
Mechanics’ by J M Seddon & J D
Gale. (An RSC Primer)
• Some page references will be
given to this book, but
alternatives will be given on the
Teaching Pages.
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Thermodynamics Lecture 1
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The Boltzmann Distribution - 1
• This is the formula which tells us how to
calculate the number of molecules in each
state of a system at a temperature T:
N exp( Ei / kT )
Ni 
q
Ni= no. of molecules in a state with energy Ei,
N= total number of molecules, k=
Boltzmann’s constant, q= partition function.
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Thermodynamics Lecture 1
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The Boltzmann Distribution - 2
• We have come across all these terms before
except for q, which is defined as:
q   exp( Ei / kT )  exp( E0 / kT )  exp( E1 / kT )  ...
i
 The sum of the exponential of the negative of each
energy level divided by kT
• (This is a general expression for q, but we will
look at how to calculate it and what it means
later.
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Thermodynamics Lecture 1
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The Boltzmann Distribution - 3
• Molecules in a substance are arranged over a
range of molecular energy levels, and there
are many possible arrangements of
molecules that will give a particular energy.
• The Boltzmann distribution gives the
statistically most probable distribution of
molecules in each level for a given overall
energy E and temperature T.
• The total number of molecules, N = Ni, and
the total energy, E = Ei, is constant.
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Thermodynamics Lecture 1
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Boltzmann Distribution example
(see Seddon & Gale, p 72-3, ex. 9.2 & 9.3)
• A diagram will be drawn of a system
containing 5 energy levels, spacing 1 x 10-20 J
(n=0 to n=4), and we will see how many ways
6 particles can be arranged such that the total
energy is 4 x 10-20 J.
• Once all the combinations are drawn, we
have to decide if they are all equally
probable.
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Thermodynamics Lecture 1
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How do we decide which distribution
has the highest probability?
•
•
•
•
All distributions have equal energy
What about Free Energy, G?
G = H – TS, where S is entropy
Entropy can be thought of as a measure of
disorder
• Which state has the highest entropy?
– Hint: consider distribution over energy levels.
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Thermodynamics Lecture 1
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The Boltzmann Distribution:
Occupation of energy levels as a function
of temperature
• As the temperature increases, more
energy levels are occupied.
• q is the partition function, which we
will meet later. This is a measure of
how many levels are occupied.
• You can see that q goes from 1.05 to
3.86 as the temperature increases.
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Thermodynamics Lecture 1
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An example of the use of the
Boltzmann Distribution
• The simplest application of the Boltzmann
distribution is to calculate the relative number
of molecules in two states separated by an
energy difference E = E2 – E1
• From the equation on slide 4, we consider 2
states E1 and E2, and obtain:
N2
 ( E 2  E1 ) 
  E 
 exp  
  exp 

N1
kT
 kT 


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Thermodynamics Lecture 1
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The boat and chair conformations
of cyclohexane
• The boat conformation of cyclohexane lies 22
kJ mol-1 above the chair conformation (why?)
• The relative population of the two
conformations at a given temperature (e.g.
293 K) can be found from the expression on
slide 11:
 Nboat/Nchair = exp(-22x103/8.314x293) = 1.2 x 10-4
• This shows the very low population of the
boat conformation at 293K – try a higher
temperature!
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Thermodynamics Lecture 1
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Note on use of k and R
• In the previous calculation we used R instead
of k.
• Which to use depends on if we are dealing
with single atoms/molecules (k) or moles (R).
• Make sure you use the right one! Examples
will be given.
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Thermodynamics Lecture 1
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Notes on energy levels and
degeneracies
• An important convention is that all energies
are expressed relative to the ground state,
even if there is a zero point energy.
• For the harmonic oscillator, energies are:
½ h, 3/2 h, … but written as 0, h, …
• Energy levels that are degenerate have
several states. The number of molecules in a
particular state is multiplied by the
degeneracy, g, to get the overall population of
the energy level (illustrated in examples later).
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Thermodynamics Lecture 1
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Where do partition functions come
in? - 1
• As long as we only want to calculate the
relative populations of levels and states, we
do not need the partition function, q because
it cancels out in the expression on slide 11.
• However, if we need to know the actual
population of a state, we need to know q.
• Going back to the expression on slide 6, and
remembering that the ground state energy is
zero, we can write a new expression for q:
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Thermodynamics Lecture 1
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Where do partition functions come
in? - 2
• The new expression is:
q  1  exp( E1 / kT )  exp( E2 / kT )  ...
• The first term is 1 since E0=0, and exp(0)=1
• If T=0 K it follows that q=1 because all terms
apart from the first are equal to 0.
• At 0 K only the ground state would be
occupied, so q tells us the number of
occupied states at a particular value of T.
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Thermodynamics Lecture 1
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Where do partition functions come
in? - 3
• If T is very high, however, all the values of
Ei/kT ~ 0, so in this case:
q= 1+1+1+1+1 … (= the number of states of
the molecule).
• In this case all available states are occupied.
• Consider an intermediate case, where kT is
large compared to E1 and E2 but small
compared to E3:
q = 1+1+1+ 0 = 3 (so 3 states are occupied).
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Thermodynamics Lecture 1
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An example of the calculation of q
• In the case of cyclohexane, there are 2
energy levels, at 0 and 22 kJ mol-1
• How are the levels populated at different
temperatures?
q = 1 + exp(-22 x 103/8.314 x T)
• When T=293 K, q=1.0001, which suggests
that only one energy level will be occupied at
this temperature, but what about at higher
temperatures?
 note use of R since E is in kJ mol-1
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Thermodynamics Lecture 1
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Wavelength, wavenumber and
frequency units
• You need to be able to convert between
wavelength, wavenumber and frequency, as
all these are used!
• Basic formulas
– Wavelength and frequency:  x  = c
– Wavelength and wavenumber:
= 1/
– Wavenumber and frequency:  = c x


• Examples for you to try will be given on the
problem sheet.
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Thermodynamics Lecture 1
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Examples of partition functions - 1
• We can obtain expressions for the partition
function which describe particular types of
energy level – e.g. vibrational, translational or
rotational:
1. The vibrational partition function
From quantum mechanics we know that the
permitted energy levels, relative to a ground
state energy of zero, are:
E0=0, E1=h, E2=2h, E3=3h, etc.
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Thermodynamics Lecture 1
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Examples of partition functions - 2
• So the expression for q is:
q= 1 + exp(-h/kT) + exp(-2h/kT) + exp(-3h/kT) + ...
• This can be simplified to give:
q
1
1  exp( h / kT )
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Thermodynamics Lecture 1
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Calculation of a vibrational
partition function
The vibrational partition function of 35Cl2 at
298K
• For 35Cl2,
 = 560 cm
-1
(wavenumber units*)
So  = 3 x 1010 x 560 Hz = 1680 x 1010 Hz
 q = 1 / [1 - exp (-6.626 x 10-34 x 1.68 x 1013 / 1.381 x
10-23 x 298)]
 q = 1.07, which tells us that at 298K, one vibrational
energy level is occupied.
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Thermodynamics Lecture 1
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Examples of partition functions - 3
2. The translational partition function
The partition function for a molecule of mass
m in a flask with volume V is given by:
( 2mkT )3 / 2V
q
h3
An example calculation will be included on
the problem sheet.
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Thermodynamics Lecture 1
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Examples of partition functions - 4
3. The rotational partition function
For a linear molecule at high temperature, the
rotational partition function is written as:
kT
q 
hB
• Where B is the rotational constant, and  is
the symmetry number (1 for an
unsymmetrical linear molecule, e.g. HCl, and
2 for a symmetrical linear molecule, e.g. H2).
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Thermodynamics Lecture 1
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Calculation of a rotational partition
function
• For H35Cl, B= 318 GHz*
(unsymmetrical linear molecule)
• T= 298 K
and
=1
q = (1.381 x 10-23 x 298)/(1 x 6.626 x 10-34 x 318 x 109)
 q = 19.5, so about 20 rotational states are
occupied at 298 K.
• *Note that B is sometimes given in
wavenumber units, so be prepared to
convert!
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Thermodynamics Lecture 1
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Summary
• Statistical
Thermodynamics
has
been
introduced.
• The Boltzmann distribution has been
discussed and used in calculation of the
separation of energy levels.
• Partition functions have been introduced, and
expressions given for vibrational, translational
and rotational forms, and some* example
calculations carried out.
• * But see also the problem sheet.
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Thermodynamics Lecture 1
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