chem2503_oct05.ppt

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G. H. CHEN
Department of Chemistry
University of Hong Kong
Intermediate Physical Chemistry

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Intermediate Physical Chemistry
Contents:
Distribution of Molecular States
Perfect Gas
Fundamental Relations
Diatomic Molecular Gas

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Distribution of Molecular States
Configurations and Weights
Boltzmann Distribution, and Physical
Meanings of 
Molecular Partition Function and its
Interpretation
The Internal Energy and the Entropy
Independent Molecular and their
Partition Function

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Perfect Gas
Partition Function
Energy
Heat Capacity
Pressure and the gas law

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Diatomic Molecular Gas
Factorization of Partition Function
Rotational Partition Function
Vibrational Partition Function
Electronic Partition Function
Mean Energy and Heat Capacity

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Fundamental Relations
Helmholtz Energy
Pressure
Enthalpy
Gibbs Energy

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Statistical Mechanics provides the link between the
microscopic properties of matter and its bulk
properties.
Statistical Mechanics:

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THE DISTRIBUTION OF MOLECULAR STATES
Consider a system composed of N molecules,
and its total energy E is a constant. These
molecules are independent, i.e. no interactions
exist among the molecules. Countless collisions
occur. It is hopeless to keep track positions,
momenta, and internal energies of all molecules.

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Principle of equal a priori probabilities:
All possibilities for the distribution of energy
are equally probable provided the number of
molecules and the total energy are kept the
same.
That is, we assume that vibrational states of a
certain energy, for instance, are as likely to be
populated as rotational states of the same energy.

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For instance, four molecules in a three-level
system: the following two conformations have
the same probability.
---------l-l-------- 2 ---------l--------- 2
---------l----------  ---------1-1-1---- 
---------l---------- 0 ------------------- 0

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Configurations and Weights
Imagine that there are total N molecules among
which n0 molecules with energy 0, n1 with
energy 1, n2 with energy 2, and so on, where 0
< 1 < 2 < .... are the energies of different states.
The specific distribution of molecules is called
configuration of the system, denoted as { n0, n1,
n2, ......}

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For instance, a system with 17 molecules,
and each molecule has four states.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16 17
1
2
3
4
STATE
The above configuration is thus, { 4, 6, 4, 3 }

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{N, 0, 0, ......} corresponds that every molecule
is in the ground state, there is only one way to
achieve this configuration; {N-2, 2, 0, ......}
corresponds that two molecule is in the first
excited state, and the rest in the ground state,
and can be achieved in N(N-1)/2 ways.
A configuration { n0, n1, n2, ......} can be achieved in W
different ways, where W is called the weight of the
configuration. And W can be evaluated as follows,
W = N! / (n0! n1! n2! ...)

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Justification
1. N! different ways to arrange N molecules;
2. ni! arrangements of ni molecules with energy i
correspond to the same configuration;

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1. Calculate the number of ways of distributing
3 objects a, b and c into two boxes with the
arrangement {1, 2}.
Answer:
| a | b c |, | b | c a |, | c | a b |.
Therefore, there are three ways 3! / 1! 2!
Example:
| a | c b |, | b | a c |, | c | b a |
To eliminate overcounting of these configurations

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2. Calculate the number of ways of distributing
20 objects into six boxes with the
arrangement {1, 0, 3, 5, 10, 1}.
Answer:
20! / 1! 0! 3! 5! 10! 1! = 931170240
note: 0! = 1
Example:

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Stirlings Approximation:
When x is large, ln x!  x ln x - x
x ln x! x ln x - x ln A
1 0.000 -1.000 0.081
2 0.693 -0.614 0.652
4 3.178 1.545 3.157
6 6.579 4.751 6.566
8 10.605 8.636 10.595
10 15.104 13.026 15.096
Note: A = (2)1/2 (x+1/2)x e-x

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Therefore,
ln W  ( N ln N - N ) -  ( ni ln ni - ni )
= N ln N -  ni ln ni
x ln x! x ln x - x ln A
12 19.987 17.819 19.980
16 30.672 28.361 30.666
20 42.336 39.915 42.332
30 74.658 72.036 74.656
Stirlings approximation(cont’d):

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The Dominating Configuration
Imagine that N molecules distribute among
two states. {N, 0}, {N-1, 1}, ..., {N-k, k}, ... ,
{1, N-1}, {0, N} are possible configurations,
and their weights are 1, N, ... , N! / (N-k)! k!,
... , N, 1, respectively. For instance, N=8, the
weight distribution is then

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0 1 2 3 4 5 6 7 8
0
20
40
60 N = 8
W
k

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0 1 2 3 4 5 6 7 8 9 10111213141516
0
2000
4000
6000
8000
10000
12000
N =16
W
k

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0 4 8 12 16 20 24 28 32
0.00E+000
1.00E+008
2.00E+008
3.00E+008
4.00E+008
5.00E+008
6.00E+008
N = 32
k

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When N is even, the weight is maximum at k = N/2,
i.e.
Wk=N/2 = N! / [N/2)!]2.
When N is odd, the maximum is at k = N/2  1
As N increases, the maximum becomes sharper!
The weight for k = N/4 is
Wk=N/4 = N! / [(N/4)! (3N/4)!]

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| N | 4 | 8 | 16 | 32 | 256 | 6.0 x 1023
|R(N) | 1.5 | 2.5 | 7.1 | 57.1 | 3.5 x 1014 | 2.6 x 103e+22
Therefore, for a macroscopic molecular system
( N ~ 1023 ), there are dominating configurations
so that the system is almost always found in or
near the dominating configurations, i.e. Equilibrium
The ratio of the two weights is equal to
R(N) Wk=N/2 / Wk=N/4
=(N/4)! (3N/4)! / [(N/2)!]2
If the system has more states, would we reach the same
conclusion as above? Why?

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1. The total energy is a constant, i.e.
 ni i = E = constant
2. The total number of molecules is conserved,
i.e.
 ni = N = constant
How to maximize W or lnW under
these constraints?
Two constraints for the system

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The Boltzmann Distribution
ni / N = Pi = exp ( - i )
Interpretation of Boltzmann Distribution
Meaning of : ensure total probability is ONE
1 = i ni / N = i exp( -i)
exp( ) = 1 / i exp(-i)
 = - ln [i exp(-i)]
1 /  = kT

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Starting from the principle of equal a priori
probabilities, we evaluate the probabilities of
different configurations by simple counting,
and find that the dominating configuration
whose population obeys the Boltzmann
distribution which relates the macroscopic
observables to the microscopic molecular
properties, and is capable of explaining the
equilibrium properties of all materials.
Summary

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To find the most important configuration, we vary { ni } to
seek the maximum value of W. But how?
E.g. One-Dimensional Function: F(x) = x2
dF/dx = 0
Enough?

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Two-Dimensional Case: for instance, finding the
minimum point of the surface of a half water
melon F(x,y).
F/x = 0,
F/y = 0.
Multi-Dimensional Function: F(x1, x2, …, xn)
F/xi = 0, i = 1,2,…,n
To find the maximum value of W or lnW,
 lnW / ni = 0, i=1,2,3,...

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Let’s investigate the water melon’s surface:
cutting the watermelon
how to find the minimum or maximum
of F(x, y) under a constraint x = a ?
L = F(x, y) -  x
 L /  x = 0
 L /  y = 0
x = a
The method of Lagrange Multipliers

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Generally, to minimize or maximize a function
F(x1, x2, …, xn) under constraints,
C1(x1, x2, …, xn) = Constant1
C2(x1, x2, …, xn) = Constant2
.
.
.
Cm(x1, x2, …, xn) = Constantm
L = F(x1, x2, …, xn) - i iCi(x1, x2, …, xn)
L/xi = 0, i=1,2, ..., n

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JUSTIFICATION
dL = dF - i i dCi
under the constraints, dCi = 0, thus
dF = 0
i.e., F is at its maximum or minimum.

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The method of Lagrangian Multiplier
Procedure
Construct a new function L,
L = lnW +  i ni -  i ni i
Finding the maximum of L by varying { ni }, 
and  is equivalent to finding the maximum of
W under the two constraints, i.e.,
L/ni = lnW/ni +  -i = 0

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Since
ln W  ( N ln N - N ) - i ( ni ln ni - ni )
= N ln N - i ni ln ni
lnW/ni = (N ln N)/ni - (ni ln ni)/ ni
= - ln (ni/N)
Therefore,
ln (ni / N) +  -i = 0
ni / N = exp( -i)
The method of Lagrangian Multiplier

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Therefore, E = N <  > = 3N/2, where <  > is
the average kinetic energy of a molecule.
Therefore,
<  > = <mv2/2> = 3/2.
On the other hand, according to the Maxwell
distribution of speed, the average kinetic energy
of a molecule at an equilibrium,

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This is the physical meaning of , the reciprocal
temperature.
1 /  = kT
where k is the Boltzmann constant. Thus,
<mv2>/2 = 3kT/2
(This is actually the definition of the temperature)

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Two friends A and B are drinking beer in a pub one
night. Out of boredom, the two start to play
“fifteen-twenty”. A mutual friend C steps in, and the
three play together. Of course, this time one hand is
used by each. A is quite smart, and studies chemistry
in HKU. He figures out a winning strategy. Before
long, B and C are quite drunk while A is still pretty
much sober. Now, what is A’s winning strategy?
When A plays with B and try the same strategy, he
finds that the strategy is not as successful. Why?
Example 1:

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Example 2:
Consider a molecular whose ground state energy is
-10.0 eV, the first excited state energy -9.5 eV, the
second excited state energy -1.0 eV, and etc.
Calculate the probability of finding the molecule in
its first excited state T = 300, 1000, and 5000 K.

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The summation is over all possible states
(not the energy levels).
If the energy level is gi-fold degenerate, then the
molecular partition function can be rewritten as
The Molecular Partition Function
The Boltzmann distribution can be written as
pi = exp(-i) / q
where pi is the probability of a molecule being found
in a state i with energy i. q is called the molecular
partition function,
q = i exp(-i)
q = i gi exp(-i)

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As T  0, q  g0,
i.e. at T = 0, the partition function is equal to the
degeneracy of the ground state.
As T  , q  the total number of states.
Therefore, the molecular partition function gives an
indication of the average number of states that are
thermally accessible to a molecule at the temperature of
the system. The larger the value of the partition function
is, the more the number of thermally accessible states is.
The relationship between q and :
exp() = q-1
Interpretation of the partition function

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Consider a proton in a magnetic field B. The proton’s
spin (S=1/2) has two states: spin parallel to B and spin
anti-parallel to B. The energy difference between the
two states is  = pB where p is proton’s magneton.
Calculate the partition function q of the proton.
Example 3
Example 4
Calculate the partition function for a uniform ladder
of energy levels
Calculate the proportion of I2 molecules in their ground,
first excited, and second excited vibrational states at
25oC. The vibrational wavenumber is 214.6 cm-1.
Example 5

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Partition function contains all the thermodynamic
information!
Statistical Thermodynamics
The Internal Energy, the Heat Capacity & the Entropy
The relation between U and q
If we set the ground state energy 0 to zero, E
should be interpreted as the relative energy to
the internal energy of the system at T = 0,
E = i ni i = i Nexp(-i) i / q = - Ndlnq/d.

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Therefore, the internal energy U may be expressed as
The Internal Energy, the Heat Capacity & the Entropy
U = U(0) + E = U(0) - N (lnq/)V
Where, U(0) is the internal energy of the system at
T = 0. The above equation provides the energy as
a function of various properties of the molecular
system (for instance, temperature, volume), and
may be used to evaluate the internal energy.

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The relation between CV and q
Cv is the constant-volume heat capacity which measures
the ability of a system to store energy. It is defined as the
rate of internal energy change as the temperature T
varies while the volume is kept constant:
Cv  (U/T)V = N(1/kT2) (2lnq/2)V
Note that, d/dT = (d/dT) d/d = -(1/kT2) d/d

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Example 6:
Calculate the constant-volume heat capacity of a
monatomic gas assuming that the gas is an ideal gas.
U = U(0) + 3N / 2 = U(0) + 3NkT / 2
where N is the number of atoms, and k is the Boltzmann constant.
Cv  (U/T)V = 3Nk / 2 =3nNAk/2= 3nR/2
where, n is the number of moles, R  NAk is the gas constant, and
NA = 6.02 x 1023 mol-1 is the Avogadro constant

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The relation between S and the partition function q
The Statistical Entropy
According to thermodynamics, entropy S is
some measurement of heat q. The change of
entropy S is proportional to the heat absorbed
by the system:
dS = dq / T
The above expression is the definition of thermodynamic entropy.

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Boltzmann Formula for the entropy
S = k lnW
where, W is the weight of the most probable configuration of the system.
Boltzmann Formula
(1) indicates that the entropy is a measurement of the weight
(i.e. the number of ways to achieve the equilibrium
conformation), and thus a measurement of randomness,
(2) relates the macroscopic thermodynamic entropy of a system
to its distribution of molecules among its microscopic states,
(3) can be used to evaluate the entropy from the microscopic
properties of a system; and
(4) is the definition of the Statistical Entropy.

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JUSTIFICATION
The energy of a molecular system U can be expressed as,
U = U(0) + i nii
where, U(0) is the internal energy of the system at T=0,
ni is the number of molecules which are in the state with
its energy equal to i
Now let’s imagine that the system is being heated while
the volume V is kept the same. Then the change of U
may be written as,
dU = dU(0) + i nidi + i idni = i idni
[dU(0) = 0 because U(0) is a constant; di = 0 because i
does not change as the temperature of the system arises.]

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According to the First Law of thermodynamics, the
change of internal energy U is equal to the heat
absorbed (q) and work received (w), i.e.,
dU = dq + dw
dq = TdS
(thermodynamic definition of entropy; or more the heat
absorbed, the more random the system)
dw = -PdV = -Force * distance
(as the system shrinks, it receives work from the environment)

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dU = TdS - PdV = TdS (dV = 0)
dS = dU/T = k i idni
= k i (lnW/ni)dni + k i dni
= k i (lnW/ni)dni
= k dlnW
( lnW/ni = - ln (ni/N) = - +  i )
d(S - lnW) = 0
S = k lnW + constant
What is the constant?

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According to the Third Law of thermodynamics,
as T  0, S  0;
as T  0, W  1 since usually there is only one ground state,
and therefore,
constant = 0.

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Relation between S and the Boltzmann distribution pi
S = k lnW = k ( N lnN -i ni lnni )
= k i ( ni lnN - ni lnni )
= - k i ni ln(ni /N)
= - Nk i (ni /N)ln(ni /N)
= - Nk i pi ln pi
since the probability pi = ni /N.
The above relation is often used to calculate the
entropy of a system from its distribution function.

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The relation between S and the partition function q
According to the Boltzmann distribution,
ln pi = - i - ln q
Therefore,
S = - Nk i pi (- i - ln q)
= k i ni i + Nk ln qi pi
= E / T + Nk ln q
= [U-U(0)] / T + Nk ln q
This relation may be used to calculate S from the known entropy q

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Independent Molecules
Consider a system which is composed of N identical molecules.
We may generalize the molecular partition function q to the
partition function of the system Q
Q = i exp(-Ei)
where Ei is the energy of a state i of the system, and summation
is over all the states. Ei can be expressed as assuming there is no
interaction among molecules,
Ei = i(1) + i(2) +i(3) + … + i(N)
where i(j) is the energy of molecule j in a molecular state i

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The partition function Q
Q = i exp[-i(1) - i(2) - i(3) - … -i(N)]
= {i exp[-i(1)]}{i exp[-i(2)]} … {i exp[-i(N)]}
= {i exp(-i)}N
= qN
where q  i exp(-i) is the molecular partition function. The
second equality is satisfied because the molecules are independent
of each other.

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The relation between U and the partition function Q
U = U(0) - (lnQ/)V
The relation between S and the partition function Q
S = [U-U(0)] / T + k ln Q
The above two equations are general because they not only apply to
independent molecules but also general interacting systems.

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Perfect Gas
Perfect gas is an idealized gas where an individual molecule is
treated as a point mass and no interaction exists among molecules.
Real gases may be approximated as perfect gases when the
temperature is very high or the pressure is very low.
The energy of a molecule i in a perfect gas includes only
its kinetic energy, i.e.,
i = i
T
q = qT
i.e., there are only translational contribution to the energy
and the partition function.

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Translational Partition Function of a molecule qT
Although usually a molecule moves in a three-dimensional space,
we consider first one-dimensional case. Imagine a molecule of mass
m. It is free to move along the x direction between x = 0 and x = X,
but confined in the y- and z-direction. We are to calculate its
partition function qx.
The energy levels are given by the following expression,
En = n2h2 / (8mX2) n = 1, 2, …

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Setting the lowest energy to zero, the relative energies
can then be expressed as,
qx = 1 dn exp [ -(n2-1) ] = 1 dn exp [ -(n2-1) ]
= 0 dn exp [ -n2 ] = (2m/h22)1/2 X
n = (n2-1) with  = h2 / (8mX2)
qx = n exp [ -(n2-1) ]
 is very small, then

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Now consider a molecule of mass m free to move in a
container of volume V=XYZ. Its partition function qT
may be expressed as
qT = qx qy qz
= (2m/h22)1/2 X (2m/h22)1/2 Y (2m/h22)1/2 Z
= (2m/h22)3/2 XYZ = (2m/h22)3/2 V
= V/3
where,  = h(/2m)1/2, the thermal wavelength. The thermal
wavelength is small compared with the linear dimension of the
container. Noted that qT  as T . qT  2 x 1028 for an O2 in
a vessel of volume 100 cm3,  = 71 x 10-12 m @ T=300 K

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Partition function of a perfect gas,
Q = (qT) N = VN / 3N
Energy
E = - (lnQ/)V = 3/2 nRT
where n is the number of moles, and R is the gas constant
Heat Capacity
Cv = (E/T)V = 3/2 nR

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Fundamental Thermodynamic Relationships
Consider an equilibrium system which is consistent of N
interacting molecules. These molecules may or may not
be the same.
Relation between energy and partition function
U = U(0) + E = U(0) - (lnQ/)V
Relation between the entropy S and the partition function Q
S = [U-U(0)] / T + k lnQ

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Helmholtz energy
The Helmholtz free energy A  U - TS. At constant temperature
and volume, a chemical system changes spontaneously to the
states of lower Helmholtz free energy, i.e., dA  0, if possible.
Therefore, the Helmholtz free energy can be employed to assess
whether a chemical reaction may occur spontaneously. A system
at constant temperature and volume reaches its equilibrium when
A is minimum, i.e., dA=0. The relation between the Helmholtz
energy and the partition function may be expressed as,
A - A(0) = -kT ln Q

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Pressure
dA = dU - d(TS) = dU - TdS - SdT
dU = dq + dw
dq = TdS
dw = -pdV
dA = - pdV - SdT
Therefore, pressure may be evaluated by the following expression,
p = -(A/V)T
= kT( lnQ/V)T
This expression may be used to derive the equation of state
for a chemical system.

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Consider a perfect gas with N molecules. Its partition
function Q is evaluate as
Q = (1/N!) (V / 3)N
the pressure p is then
p = kT( lnQ/V)T
= kT N ( lnV/V)T
= NkT / V
pV = NkT = nNAkT = nRT
which is the equation of the state for the perfect gas.

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The enthalpy
During a chemical reaction, the change in internal energy is
not only equal to the heat absorbed or released. Usually, there
is a volume change when the reaction occurs, which leads
work performed on or by the surroundings. To quantify the
heat involved in the reaction, a thermodynamic function, the
enthalpy H, is introduced as follows,
H  U + pV
Therefore,
H - H(0) = -( lnQ/)V + kTV( lnQ/V)T

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The Gibbs energy
Usually chemical reactions occur under constant temperature. A
new thermodynamic function, the Gibbs energy, is introduced.
G  A + pV
At constant temperature and pressure, a chemical system changes
spontaneously to the states of lower Gibbs energy, i.e., dG  0, if
possible. Therefore, the Gibbs free energy can be employed to
access whether a chemical reaction may occur spontaneously. A
system at constant temperature and pressure reaches its
equilibrium when G is minimum, i.e., dG = 0. The relation
between the Helmholtz energy and the partition function may be
expressed as,
G - G(0) = - kT ln Q + kTV( lnQ/V)T

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Example 7
Calculate the translational partition function of an H2 molecule
confined to a 100-cm3 container at 25o
C
Example 8
Calculate the entropy of a collection of N independent harmonic
oscillators, and evaluate the molar vibraitional partition function
of I2 at 25o
C. The vibrational wavenumber of I2 is 214.6 cm-1
Example 9
What are the relative populations of the states of a two-level system
when the temperature is infinite?
Example 10
Evaluate the entropy of N two-level systems. What is the entropy
when the two states are equally thermally accessible?

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Example 11
Calculate the ratio of the translational partition functions of D2 and
H2 at the same temperature and volume.
Example 12
A sample consisting of five molecules has a total energy 5. Each
molecule is able to occupy states of energy j with j = 0, 1, 2, ….
(a) Calculate the weight of the configuration in which the molecules
share the energy equally. (b) Draw up a table with columns headed
by the energy of the states and write beneath then all configurations
that are consistent with the total energy. Calculate the weight of
each configuration and identify the most probable configuration.
Example 14
Given that a typical value of the vibrational partition function of one
normal mode is about 1.1, estimate the overall vibrational partition
function of a NH3.

GHC@copyright
Example 15
Consider Stirling’s approximation for lnN! In the derivation of
the Boltzmann distribution. What difference would it make if
the following improved approximation is used ?
x! = (2)1/2 (x+1/2)x e-x
Example 16
Consider N molecules in a cube of size a.
(a) Assuming molecules are moving with the same speed v
along one of three axises (x-, y-, or z-axis). The motion may be
along the positive or negative direction of an axis. On the
average, how many molecules move parallel to the positive x
direction?

GHC@copyright
(b)When a molecule collides with one of six sides of the
cube, it is reflected. The reflected molecule has the same
speed v but moves in the opposite direction. How many
molecules are reflected from the side within a time interval
t ? What is the corresponding momentum change of these
molecules?
(c) Force equals the rate of momentum change. Calculate the
force that one side of the cube experiences.
Pressure is simply the force acting on a unit area. What is
the pressure of the gas? Assuming the average kinetic energy
of a molecule is 3kT/2, derive the equation of state for the
system.

GHC@copyright
!
/ N
q
Q N

Diatomic Gas
Consider a diatomic gas with N identical molecules. A molecule is
made of two atoms A and B. A and B may be the same or different.
When A and B are he same, the molecule is a homonuclear diatomic
molecule; when A and B are different, the molecule is a heteronuclear
diatomic molecule. The mass of a diatomic molecule is M. These
molecules are indistinguishable. Thus, the partition function of the gas
Q may be expressed in terms of the molecular partition function q,
The molecular partition q
where, i is the energy of a molecular state i, β=1/kT, and ì is the
summation over all the molecular states.
 
 i i
q )
exp( 

GHC@copyright

GHC@copyright
where T denotes translation, R rotation, V vibration, and E the
electronic contribution. Translation is decoupled from other
modes. The separation of the electronic and vibrational motions
is justified by different time scales of electronic and atomic
dynamics. The separation of the vibrational and rotational
modes is valid to the extent that the molecule can be treated as a
rigid rotor.
)
(
)
(
)
(
)
(
)
( j
j
j
j
j E
V
R
T




 



Factorization of Molecular Partition Function
The energy of a molecule j is the sum of contributions from its
different modes of motion:

GHC@copyright

 





 i
E
i
V
i
R
i
T
i
i i
q )]
(
exp[
)
exp( 





    



 i i i i
E
i
V
i
R
i
T
i )]
exp(
)][
exp(
)][
exp(
)][
exp(
[ 



E
V
R
T
q
q
q
q


GHC@copyright
R
V
T
q
q
q
q 
1
/ 
E
q
w
3
/ 
 V
qT
2
/
1
)
2
/
( M
h 



kT
/
1


where
The translational partition function of a molecule
ì sums over all the translational states of a molecule.
The electronic partition function of a molecule
 ì sums over all the electronic states of a molecule.
 
 i
T
i
T
q )
exp( 
The rotational partition function of a molecule
 ì sums over all the rotational states of a molecule.
The vibrational partition function of a molecule
 ì sums over all the vibrational states of a molecule.
 
 i
R
i
R
q )
exp( 
 
 i
V
i
V
q )
exp( 
 
 i
E
i
E
q )
exp( 

GHC@copyright
5--------------5hv
4--------------4hv
3--------------3hv
2--------------2hv
1--------------hv
0--------------0
hv
n
V
n )
2
/
1
( 


nhv
V
n 

n= 0, 1, 2, …….
kT hv


Vibrational Partition Function
Two atoms vibrate along an axis connecting the two atoms.
The vibrational energy levels:
If we set the ground state energy to zero or measure energy
from the ground state energy level, the relative energy levels
can be expressed as

GHC@copyright
Then the molecular partition function can be evaluated

 





 n
n n
v
hv
hv
n
q )]
exp(
1
/[
1
)
exp(
)
exp( 


...
1 3
2 

 





 e
e
e
qv
1
....
3
2





 


 v
v
q
e
e
e
q
e 



hv
v
e
e
q 







1
1
)
1
/(
1
Consider the high temperature situation where kT >>hv, i.e.,
hv
kT
hv
hv v
/
/
1
q
,
1 

 

Vibrational temperature v
High temperature means that T>>v
hv
k v 




h
e hv



1

I2 F2 HCl H2
v/K 309 1280 4300 6330
v/cm-1 215 892 2990 4400
m
k
v 
where
Therefore,
Vibrational Partition Function

GHC@copyright
where B is the rotational constant. J =0, 1, 2, 3,…
)
1
( 
 J
hcBJ
R
J

 
 states
rotational
all
]
exp[ R
J
R
q 
 
 levels
energy
rotational
all
]
exp[ R
J
J
g 
 


 J
)]
1
(
exp[
)
1
2
( J
hcBJ
J 






0
)]
1
(
exp[
)
1
2
( dJ
J
hcBJ
J
qR

dJ
/
)]}
1
(
{exp[
)
/
1
(
0
dJ
J
hcBJ
d
hcB 


 







 0
]}
1
(
){exp[
/
1
( l
J
hcBJ
hcB 

Bh/8cI2
c: speed of light
I: moment of Inertia

i
i
r m I
i
2
hcB<<1
where gJ is the degeneracy of rotational energy level εJ
R
Usually hcB is much less than kT,
Note: kT>>hcB
Rotational Partition Function
If we may treat a heteronulcear diatomic molecule as a rigid rod,
besides its vibration the two atoms rotates. The rotational energy
=kT/hcB

GHC@copyright
For a homonuclear diatomic molecule
Generally, the rotational contribution to the molecular partition
function,
Where  is the symmetry number.
Rotational temperature R
hcB
kT
qR
2
/

hcB
kT
qR

/

12
CH
3
NH
2
4
3
2O
H
hcB
k R 


Rotational Partition Function

GHC@copyright
where, gE = g0 is the degeneracy of the electronic ground state,
and the ground state energy 0
E is set to zero.
If there is only one electronic ground state qE = 1, the partition
function of a diatomic gas,
]
exp[
]
exp[
states
electronic
all energies
electronic
all
E
j
j
E
j
E
g
q 

  



]
exp[ 0
0
E
g 


N
hv
N
N
e
hcB
kT
V
N
Q 



 )
1
(
)
/
(
)
/
)(
!
/
1
( 3 

At room temperature, the molecule is always in its ground state
Electronic Partition Function
=g0
=gE

GHC@copyright
The internal energy of a diatomic gas (with N molecules)
v
e
n
N
N
n
N
U
U hv
v
v ]
/
)
1
(
1
[
)
/
ln
(
)
/
1
(
3
)
0
( 


 











 
)
1
/(
/
1
/
1
)
2
/
3
( 


 hv
e
Nhv
N
N 


)
1
/(
)
2
/
5
( 

 hv
e
Nhv
NkT 
kT
N
)
2
/
7
(

qV = kT/hv
qR = kT/hcB
The rule: at high temperature, the contribution of one
degree of freedom to the kinetic energy of a molecule
(1/2)kT
Mean Energy and Heat Capacity
(T>>1)

GHC@copyright
the constant-volume heat capacity
(T>>1)
Contribution of a molecular to the heat capacity
Translational contribution
(1/2) k x 3 = (3/2) k
Rotational contribution
(1/2) k x 3 = k
Vibrational contribution
(1/2) k + (1/2) k = k
kinetic potential
Thus, the total contribution of a molecule to the heat capacity is (7/2) k
v
v T
U
C )
/
( 


2
2
)
1
/(
hv)
(
K
N
k
)
2
/
5
( 

 hv
hv
e
e
N 


k
)
2
/
7
( N


GHC@copyright
Translational energy
2
2
2
2
2
1
8
mVx
mx
h
n
T
n 


mx
h
n
Vx
2


where n = 1, 2, …
n is a measurement of the speed of the molecule
mx
50h
mx
h
2mx
h
Vx
....
100
....
2
1
n
sec
/
10
9
.
1
2
5
m
mx
h 


Quantum Classical
for a H in a one-dimonsional box x= 1cm

GHC@copyright
Vibrational energy
2
2
1
)
2
1
( kA
hV
n
v
n




Quantum Classical
A is the amplitude of the vibration
k
hv
n
k
hv
n
A )
1
2
(
)
2
1
(
2 



Here the vibrational quantum number n is a measurement
of the vibrational amplitude.

GHC@copyright
 





mj
J J
R
e
e
q
R
J
R
J
J
,
)
1
2
( 
 

mj=-J,-J+1…….
Rotational energy
2
2
1
)
1
( 
 I
J
hcBJ
R
J



Quantum Classical
 is the rotational angulor velocity
)
1
(
2

 J
J
I
h


J is a measurement of angular velocity
mJ is a measurement of the projection of the angular velocity of the z-
axis. i.e. a measurement of the rotation’s orientation.

GHC@copyright
Molecule with N atoms
Degree of freedom:
Translation: 3
Rotation: 3 nonlinear
2 linear
vibration: 3N – 6 nonlinear
3N – 5 linear
Diatomic Molecule A-B
Symmetry  =
Supplementary note
0 if A  B
1 if A = B

GHC@copyright
Summary
Principle of equal a priori probabilities:
All possibilities for the distribution of energy are equally
probable provided the number of molecules and the total
energy are kept the same.
A configuration { n0, n1, n2, ......} can be achieved in W
different ways or the weight of the configuration
Dominating Configuration vs Equilibrium
Pi = exp (-i ) / q
W = N! / (n0! n1! n2! ...)

GHC@copyright
q = i exp(-i) = j gjexp(-j) Q = i exp(-Ei)
Partition Function
Energy
E= N i pi i = U - U(0) = - (lnQ/)V
Heat Capacity
CV = (E/T)V = k2 (2lnQ/2)V
Entropy
S = k lnW = - Nk i pi ln pi = k lnQ + E / T
A= A(0) - kT lnQ
Helmholtz energy
Summary

GHC@copyright
H = H(0) - (lnQ/)V + kTV (lnQ/V)T
Q = qN or (1/N!)qN
q = qTqRqVqE
Enthalpy
Molecular partition function
Summary

chem2503_oct05.ppt

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chem2503_oct05.ppt