CHAPTER 15 1

CHAPTER 15 Statistical Thermodynamics 1: The Concepts

I. Introduction.

A. Statistical mechanics is the bridge between microscopic and macroscopic world descriptions of nature.

B. Microstates.

1. Statistical mechanics predicts how a system will be distributed over all of its available microstates.

2. Microstate = particular spatial or spin or energy configuration of a system. Many are possible.

3. Simple example: system = a collection of 3 fixed, non-interacting particles, each of which can have either a spin-up (1/2) or spin-down (-1/2) configuration of equal energy.

4. Specifying the microstate of the system involves (in this simple example) assigning a spin value to each particle.

macroscopic world Thermodynamics example macroscopic variables: P = pressure V = volume T = temperature E = internal energy S = entropy G = Gibbs free energy A = Helmholtz free energy ρ = density η = viscosity ε = dielectric constant

microscopic world Quantum & classical mechanics example microscopic variables: xi = position of particle i pi= momentum of particle i n, l, ml , ms =H-atom quantum #s v = vibrational quantum # J= rotational quantum #

Statistical mechanics

CHAPTER 15 2

5. Here are the possible microstates of the system. There are 8 possible:

(1/2,1/2,1/2) (1/2,1/2,-1/2) (1/2,-1/2,1/2) (-1/2,1/2,1/2) (1/2,-1/2,-1/2) (-1/2,1/2,-1/2) (-1/2,-1/2,1/2) (-1/2,-1/2,-1/2)

or we could use the following notation for simplicity:

(+++) (++-) (+-+) (-++) (+--) (-+-) (--+) (---)

6. Assigning the quantum numbers to particles is the microscopic way of specifying the state of the system, that is, the microstate.

7. Now suppose we are interested in a “macroscopic” property of the system, such as the total spin of the system.

Total spin of system = S = S1 + S2 + S3

The total spin S of the system could then have the following values:

S = 3/2, 1/2, 1/2, 1/2, -1/2 -1/2, -1/2, -3/2

8. Now the statistical “weight” function of the system W(S) is the total number of possible ways (or microstates) that the system can have a total spin value of S.

W(S) = number of ways of having total spin value S

W(3/2) = 1 way W(1/2) = 3 ways W(-1/2) = 3 ways W(-3/2) = 1 way

9. Notice that the function W(S) “peaks” in the middle, that is, there are more ways of having lower total spin values 1/2 and -1/2 than there are to have higher spin values such as 3/2 or -3/2. This is a typical property of distribution functions, which becomes even more pronounced when the system has a large number of particles rather than just 3.

10. A similar statistical function we could define is the probability P(S) of finding the system having a total spin of S.

P(S) = probability of system being found in state having total spin = S

P(S) = ways of having total spin S total microstates of system

e.g., P(3/2) = W(3/2)/8 = 1/8

P(1/2) = W(1/2)/8 = 3/8

CHAPTER 15 3 CHAPTER 15 3

P(S)

S-3/2 3/2

C. The Binomial Theorem.

1. If instead of having only 3 particles that can each be in two states, we had a total of n particles, the function W could be determined using the binomial theorem.

W(n+,n-) = number of ways of arranging a system of n particles in

which n+ particles are in (+) state and n- particles are in (-) state.

!

W n+,n"( ) =n!

n+!n"!

n! = n – factorial $ 1 x 2 x 3…x n 2. Let’s try the binomial theorem out on our simple 3-particle system first:

!

W 3 2( ) = W 3,0( ) =3!

3!0!=

3"2"13"2"1( ) " 1( )

!

0!"1

!

W 3 2( ) =1

W 1 2( ) =3!

2!1!= 3

W "1 2( ) =3!

1!2!= 3

W "3 2( ) =3!

0!3!=1

C. The Binomial Theorem.

1. If instead of having only 3 particles that can each be in two states, we had a total of n particles, the function W could be determined using the binomial theorem.

W(n+,n-) = number of ways of arranging a system of n particles in

which n+ particles are in (+) state and n- particles are in (-) state.

W n+,n

−( ) = n!n+!n

−!

n! = n – factorial ≡ 1 x 2 x 3…x n 2. Let’s try the binomial theorem out on our simple 3-particle system first:

W 3 2( ) =W 3,0( ) = 3!3!0!

=3 ×2 ×1

3 ×2 ×1( ) × 1( )

0! ≡ 1

W 3 2( ) = 1

W 1 2( ) = 3!2!1!

= 3

W −1 2( ) = 3!1!2!

= 3

CHAPTER 15 4

3. Now lets demonstrate the use of the binomial theorem on a much larger system. Consider a system of n=100 coins, each of which can be either in a heads or tails state. Let n+ be the number of coins that are heads, and n- be the tails. Question: How many different ways can 100 coins be arranged so that 50 are heads and 50 are tails?

n = 100; n+ = 50; n- = 50

€

W 50,50( ) =100!

50!50!

Use Stirling’s approximation to evaluate the factorial of a large number:

ln x! ≈ x ln x - x Now solve for W:

€

W =n!

n!+n!−lnW = lnn!−lnn+!−lnn−!

= nlnn−n− n+ lnn+ −n+( ) − n− lnn− −n−( )

Now for n=100; n+ = n- = 50; ln W = 69.3; so W = e69.3 ≈ 1030 ways 4. So we see an enormous number of microstates that are evenly

distributed between heads and tails. There are 1030 arrangements of the coins that all give a 50/50 distribution.

5. Now lets compare this very random distribution with a more orderly

distribution in which there is one coin which is “heads” and the other 99 are “tails.”

€

W 1,99( ) =100!1!99!

=100 × 99!

1!99!= 100 ways

so there are a relatively small number of microstates with this orderly

distribution, even though a coin is just as happy being a “heads” as a “tails.”

6. Now we can see demonstrated for this simple system of 100 coins the

statistical basis of entropy. That is, even though coins can be heads or tails with equal preference, a large system of coins will more likely be found with an even distribution near 50/50 than with a 1/99 distribution. A box of coins, shaken and allowed to sample its microstates at random, will almost always be found near a 50/50 distribution.

CHAPTER 15 5

7. Entropy maximization is the consequence of a very important theorem

of statistical mechanics. The statistical law of equal a priori probabilities states that:

A system at equilibrium is evenly distributed over all

microstates accessible to it at fixed total energy.

D. Statistical Definition of Entropy.

1. The entropy S of a system in a given configuration is defined as follows:

S = kB lnW

where kB = Boltzmann’s constant = 1.38 x 10-23 joule/K-mol ln is the natural logarithm W = number of microstates when system is in a given configuration 2. Example: For a spin system (or coin system), S is a function of n+ and

n-, so that: S(n+, n-) = kB ln W(n+, n-) e.g. S(0,100)=kB ln W(0,100) = kB ln (1)=0 (the zero point of entropy) S(1,99) kB ln W(1,99) = kB x 4.6 S(50,50) = kB ln e69.3 = kB x 69.3 3. The shaking of the box will cause the system of coins to tend toward a

50/50 distribution. The system entropy S is spontaneously increasing towards its maximum value, Smax = 69.3kB.

4. This is a manifestation of the Second Law of Thermodynamics = a

closed isolated system will tend to increase its entropy in any spontaneous process.

CHAPTER 15 6

E. Energy Considerations.

1. Let’s look at a more complicated example system. This example is complicated by energy considerations.

2. Consider a system of 3 independent harmonic oscillators. For example, this could be three atoms in a crystal.

3. The allowed quantum energy states of a harmonic oscillator are evenly spaced by increments of hν such that the energy of one oscillator is given by

E = (v+1/2)hν

Here h = Planck’s constant = 6.626 x 10-34 J-s ν = the classical oscillation frequency of the oscillator v = the vibrational quantum number of the oscillator = 0,1,2,3,...

4. Energy diagram of three oscillator system might be depicted as follows:

CHAPTER 15 6

E. Energy Considerations.

1. Let’s look at a more complicated example system. This example is complicated by energy considerations.

2. Consider a system of 3 independent harmonic oscillators. For example, this could be three atoms in a crystal.

3. The allowed quantum energy states of a harmonic oscillator are evenly spaced by increments of hn such that the energy of one oscillator is given by

E = (v+1/2)h&

Here h = Planck’s constant = 6.626 x 10-34 joule-sec & = the classical oscillation frequency of the oscillator v = the vibrational quantum number of the oscillator = 0,1,2,3,...

4. Energy diagram of three oscillator system might be depicted as follows:

5. In the above example, the total system has 3 increments of energy

above the absolute minimum of energy. We would say the system has 3 quanta of energy above the ground state, or total energy = 3h& above the zero-point energy.

6. Let’s figure out how many microstates are available to a system of three oscillators given that it has a fixed amount of energy 3h& above the zero-point energy.

Here are the 10 possible microstates of the system, all having a total amount of energy = 3h&.

5. In the above example, the total system has 3 increments of energy

above the absolute minimum of energy. We would say the system has 3 quanta of energy above the ground state, or total energy = 3hν above the zero-point energy.

6. Let’s figure out how many microstates are available to a system of three oscillators given that it has a fixed amount of energy 3hν above the zero-point energy.

CHAPTER 15 6

E. Energy Considerations.

1. Let’s look at a more complicated example system. This example is complicated by energy considerations.

2. Consider a system of 3 independent harmonic oscillators. For example, this could be three atoms in a crystal.

3. The allowed quantum energy states of a harmonic oscillator are evenly spaced by increments of hn such that the energy of one oscillator is given by

E = (v+1/2)h&

Here h = Planck’s constant = 6.626 x 10-34 joule-sec & = the classical oscillation frequency of the oscillator v = the vibrational quantum number of the oscillator = 0,1,2,3,...

4. Energy diagram of three oscillator system might be depicted as follows:

5. In the above example, the total system has 3 increments of energy

above the absolute minimum of energy. We would say the system has 3 quanta of energy above the ground state, or total energy = 3h& above the zero-point energy.

6. Let’s figure out how many microstates are available to a system of three oscillators given that it has a fixed amount of energy 3h& above the zero-point energy.

Here are the 10 possible microstates of the system, all having a total amount of energy = 3h&.

Here are the 10 possible microstates of the system, all having a total amount of energy = 3hν.

CHAPTER 15 7

The statistical law of equal a priori probabilities states that at equilibrium a system is to be found with equal probability in any one of its microstates having the appropriate energy.

7. In our system then, over a period of time the system will randomly

range over the 10 accessible microstates depicted above, found with equal likelihood in any one of these at any given moment of time.

8. Let NV = average number of atoms found occupying vibrational level v

(after taking many observations).

N0 = 12/10 N1 = 9/10 N2 = 6/10 N3 = 3/10

Total N = Σ NV = 0.3 + 0.6 + 0.9 + 1.2 = 3.0 = tot number of particles 9. Then the fraction of particles in level v is NV/N:

N0/N = 0.4 N1/N = 0.3 N2/N = 0.2 N3/N = 0.1

which when plotted becomes our distribution function of vibrational

energy state occupations:

CHAPTER 15 7

The statistical law of equal a priori probabilities states that at equilibrium a system is to be found with equal probability in any one of its microstates having the appropriate energy.

7. In our system then, over a period of time the system will randomly

range over the 10 accessible microstates depicted above, found with equal likelihood in any one of these at any given moment of time.

8. Let NV = average number of atoms found occupying vibrational level v

(after taking many observations).

N0 = 12/10 N1 = 9/10 N2 = 6/10 N3 = 3/10

Total N = ' NV = 0.3 + 0.6 + 0.9 + 1.2 = 3.0 = total number of atoms 9. Then the fraction of atoms in level v is NV/N:

N0/N = 0.4 N1/N = 0.3 N2/N = 0.2 N3/N = 0.1

which when plotted becomes our distribution function of vibrational

energy state occupations:

10. Note the very important and typical property of distributions over

energy states - the lower energy states have a higher occupation probability. As the state energy increases, the population decreases. Why?

10. Note the very important and typical property of distributions over

energy states - the lower energy states have a higher occupation probability. As the state energy increases, the population decreases. Why?

CHAPTER 15 8

Simply because there are many more ways to spread the energy equitably between all the molecules that to give a few molecules all the energy.

This is just like the coin example, where we found that there were many more ways to arrange the coins with the same number of heads and tails than any other arrangements.

II. The Boltzmann Distribution Law for large numbers of molecules.

A. Macroscopic-Sized Systems.

1. For realistic large systems having N molecules, where N may be as large as a mole of particles (1023 particles), the quantum states i of the molecule are occupied according to the Boltzmann Distribution Law.

2. The fraction of molecules in state i is defined as Ni/N and is given by the following equation:

Ni

N= Ce−εi /kBT

where C is a normalization constant, εi is the energy of quantum state i,

kB is the Boltzmann constant = 1.38 x 10-23 J/K, and T is the system temperature in K.

Thus the population of each quantum state i is dictated by the ratio of the energy of state i to the energy factor kBT. This ratio appears in the exponent.

For convenience it is customary to define:

β ≡1kBT

and to refer to the exponential term as the “Boltzmann factor” for state I

e−βεi

CHAPTER 15 9

3. Here is a sketch of Ni/N versus i:

CHAPTER 15 9

3. Here is a sketch of Ni/N versus i:

5432100.0

0.2

0.4

0.6

0.8

1.0

1.2

Boltzmann Distribution

Energy of level i

Ni/N

4. Note that this shows the same trend as our simpler three-atom system.

The lower levels are more highly occupied because there are many more ways of spreading the energy among all the molecules than by giving more to one than another.

5. Now we need to determine C the normalization constant: We do this by enforcing the conservation condition that

'Ni = N

or in other words, the number of molecules occupying each of the states, summed over all the states, should equal the total number of molecules N. Another way of saying this is by enforcing the condition:

!

Ni

Ni

molec_ states

" = 1

Ni

Ni

molec_ states

" = C e#$%ii

molec_ states

"

Then

C =1

e#$%ii

molec_ states

"

Then the Boltzmann Law may be written as

Ni

N=

e#$%i

e#$%ii

molec_ states

"

4. Note that this shows the same trend as our simpler three-atom system.

The lower levels are more highly occupied because there are many more ways of spreading the energy among all the molecules than by giving more to one than another.

5. Now we need to determine C the normalization constant: We do this by enforcing the conservation condition that

ΣNi = N

or in other words, the number of molecules occupying each of the states, summed over all the states, should equal the total number of molecules N. Another way of saying this is by enforcing the condition:

Ni

Ni

molec _states

∑ = 1

Ni

Ni

molec _states

∑ = C e−βεi

i

molec _states

∑

Then

C =1

e−βεi

i

molec _states

∑

Then the Boltzmann Law may be written asN

i

N=

e−βεi

e−βεi

i

molec _states

∑

CHAPTER 15 10

6. Finally the denominator sum is defined as q, the molecular partition function.

Ni

N=e−βεi

qwhere

q ≡ e−βεii

molec _states

∑

Interpretation of q ~ effective number of states that are thermally accessible to a molecule at the temperature T.

CHAPTER 15 11

7. Now for systems with degenerate energy levels, the Boltzmann Law can be rewritten in terms of levels rather than states. For example here is the first few energy levels for the rotational energy of a diatomic molecule:

j=2 _ _ _ _ _ j=1 _ _ _

j=0 _ 8. Each j value is a level. Each line is a quantum state. Now the

Boltzmann Law is:

€

Nj

N=

Ωje−εj kBT

Ωje−εj kBT

j∑

Boltzmann Law in terms of levels

Q = Ωje−εj kBT

j∑

Nj/N = fraction of molecules in level j εj = energy of level j q = partition function rewritten in terms of summation over levels rather

than states Ωj = degeneracy of level j = number of quantum states of identical energy in level j = 2j+1 for a diatomic molecule, for example

9. Now for example, the Boltzmann distribution over rotational energy

levels of a diatomic molecule can be written as:

Nj

N=2j +1( )e− j j+1( )2 2IkBT

2j +1( )e− j j+1( )2 2IkBT

j

∑

Q = 2j +1( )e− j j+1( )2 2IkBT

j

∑

Since:

εj= j j +1( ) 2 2I; I = rotational moment of inertia

Ωj= 2j +1 = degeneracy factor of level j

CHAPTER 15 12

B. Sample calculation using the Boltzmann distribution function for rotation of a diatomic molecule. 1. Here is a sample FORTRAN program for implementing the previous

equation. Something similar could be done in EXCEL or other software.

c rotation.for c Predicts rotational level population fractions Nj/N for a diatomic c molecule without making the classical rotor approximation. write(6,*) &' Input the mass in amu of each of the two atoms in the diatomic' read(5,*)am1,am2 write(6,*)' Input the bond distance of the molecule in Angstroms' read(5,*)r theta=228.0/(r**2)/(am1*am2/(am1+am2)) write(6,*)' Rotational temperature of the molecule = ',theta write(6,*)' This is in units of K' write(6,*)' input highest level for which you want information' read(5,*)maxlevel write(6,*)' input the temperature in K' read (6,*)T c First let's evaluate the partition function Q by summing over the

first c 100 terms in the series: Q=0.0 do j=0,100 arg=j*(j+1)*(theta/T) if(arg.gt.20)expo=0.0 if(arg.le.20)expo=exp(-arg) term=(2*j+1)*expo Q=Q+term end do write(6,*)'Partition function = ',Q c Now lets loop through levels, one at a time and calculate fraction write(6,*)' level population fraction' do j=0,maxlevel arg=j*(j+1)*(theta/T) if(arg.gt.20)expo=0.0 if(arg.le.20)expo=exp(-arg) term=(2*j+1)*expo fraction=term/Q write(6,*)j,fraction end do stop end

2. In the previous program the quantity theta, called the “rotation

temperature,” is set equal to a collection of constants:

θR≡h2

8π2IkB

units Kelvin( ) = 2

2IkB

CHAPTER 15 13

I = µr2;µ =m1m2

m1+m

2

; r = bond length

3. Here is a sample execution of the FORTRAN program, showing the

input of parameters by the user, followed by the output of results:

$ exe rotation.for $ Fortran/Check/NoOptimize/NoShow ROTATION.FOR $ Link ROTATION $ Run ROTATION Input the mass in amu of each of the two atoms in the diatomic 14,14 Input the bond distance of the molecule in Angstroms 1.0 Rotational temperature of the molecule = 32.57143 This is in units of K input highest level for which you want information 10 input the temperature in K 298.15 Partition function = 9.494500 level population fraction 0 0.1053241 1 0.2539569 2 0.2734202 3 0.1987428 4 0.1066300 5 4.3710325E-02 6 1.3925157E-02 7 3.4811632E-03 8 6.8701972E-04 9 1.0746627E-04 10 1.3361231E-05 FORTRAN STOP

CHAPTER 15 13

3. Here is a sample execution of the FORTRAN program, showing the

input of parameters by the user, followed by the output of results:

$ exe rotation.for $ Fortran/Check/NoOptimize/NoShow ROTATION.FOR $ Link ROTATION $ Run ROTATION Input the mass in amu of each of the two atoms in the diatomic 14,14 Input the bond distance of the molecule in Angstroms 1.0 Rotational temperature of the molecule = 32.57143 This is in units of K input highest level for which you want information 10 input the temperature in K 298.15 Partition function = 9.494500 level population fraction 0 0.1053241 1 0.2539569 2 0.2734202 3 0.1987428 4 0.1066300 5 4.3710325E-02 6 1.3925157E-02 7 3.4811632E-03 8 6.8701972E-04 9 1.0746627E-04 10 1.3361231E-05 FORTRAN STOP

1210864200.0

0.1

0.2

0.3

0.4

0.5 Rotational level distribution for a diatomic molecule with theta = 9.49 K at T=298 K

j

N j / N

This Boltzmann formula for the distribution of a diatomic over its levels J explains the appearance of the peak heights in a gas phase rotational spectrum of diatomics (microwave spectrum):

CHAPTER 15 14

This Boltzmann formula for the distribution of a diatomic over its levels J explains the appearance of the peak heights in a gas phase rotational spectrum of diatomics (microwave spectrum):

CHAPTER 15 14

frequency

increase dueto degeneracy

decrease due toexponential factor

Absorption

C. Detailed Derivation of Boltzmann Distribution Law. 1. For systems with large number of particles N in which each can be in an

arbitrary number of molecular states s(which are countable), the number of ways of achieving a system configuration:

{n1, n2, n3…ns} )# of molecules in molecular state 1...etc. is given by:

!

W n1,n2,n3...ns( ) =N!

n1!n2!n3!...ns!

2. There are constraints, however, on what configurations are allowed, or

accessible. System may have a fixed total energy such that:

!

E = ni"ii=1

s

# is a constraint

E=total energy ni=number in molecular state i ei=energy of a molecule in state i 3. Further, there may be fixed number of particles. For example:

!

ni = Ni=1

s

" is another constraint

#2 and #3 will serve as restrictions, or constraints

C. Detailed Derivation of Boltzmann Distribution Law. 1. For systems with large number of particles N in which each can be in an

arbitrary number of molecular states s (which are countable), the number of ways of achieving a system configuration:

{n1, n2, n3…ns} ↑# of molecules in molecular state 1,2,3,...etc. is given by:

W n1,n2,n3...n

s( ) = N!n1!n2!n3!...n

s!

2. There are constraints, however, on what configurations are allowed, or

accessible. System may have a fixed total energy such that:

€

E = niεii=1

s

∑ is a constraint

E=total energy ni=number in molecular state i ei=energy of a molecule in state i 3. Further, there may be fixed number of particles. For example:

€

ni = Ni=1

s

∑ is another constraint

CHAPTER 15 15

#2 and #3 will serve as restrictions, or constraints 4. The number of accessible microstates is generally strongly peaked.

Remember the heads/tails analogy:

This peak represents the statistically most typical configuration of the

system, so much so that most other configurations are inconsequential. 5. Let’s find, for general system, where W reaches a peak: i.e. find the configuration {n1, n2, n3…ns}* for which W reaches a

maximum. i.e. find the derivative d(ln W) and set = 0. 6. We get the following variation equation:

d ln W( ) = ∂ln W∂n1

"

#$$

%

&''dn1 +

∂ln W∂n2

"

#$$

%

&''dn2 + ...

=∂ln W∂ni

"

#$$

%

&''

i=1

s

∑ dni

7. We also need to subject Eq(6) to constraints #2 and #3. Do this by Lagrange’s method of undetermined multipliers = A

constraint should be multiplied by some constant and added to the main variation equation.

particle # energy constr constraint

0 = ∂ln W∂ni

"

#$$

%

&''

i=1

s

∑ dni

+ α dni

− β εidn

ii=1

s

∑i=1

s

∑

CHAPTER 15 16

or we now have a series of equations, one for each i:

€

0 =∂ ln W∂ni

#

$ %

&

' ( + α − βεi

ln W = lnN!

n1!n2!...ns!

= ln N!− ln njj=1

s

∑ !

Use Stirling’s formula for factorials. 8. Final Result:

€

ni = Constant×e-βε i

at equilibrium, where we have defined

€

β ≡1

kBT

Verbal meaning: occupancy of state i at equilibrium diminishes exponentially with energy εi. of state i.

9. Constant = C determined by Normalization, that is, by applying the

constraint on # of particles:

€

nii=1

s

∑ = N

C e−βεii=1

s

∑ = N

C =N

e−βεii=1

s

∑

10. And so we get the final form of Boltzmann’s Law:

fraction of molec in molec state

€

ni

N=

e−βεi

e−βεii=1

s

∑

CHAPTER 15 17

2.0

3.0

1.0

11. Molecular Partition Function q.

q = e−βεii=1

s

∑

-verbal meaning: an indication of the number of states thermally

accessible to each molecule in the system.

D. Example: simple two level system. A collection of molecules that have only two levels available to them. The lower one is nondegenerate and has energy =0, by definition. The upper one is doubly degenerate and has energy E.

Partition function:

€

q = Ωje−βεj

j=1

2

∑

q = 1⋅ e−0 + 2⋅ e−βE

q = 1 + 2e−βE

Fraction of molecules in level j is:

€

Nj

N=Ωje

−βεj

q

Nj

N=

Ωje−βεj

1 + 2e−βE

N1

N=

1

1 + 2e−βE

N2

N=

2e−βE

1 + 2e−βE

CHAPTER 15 18

Fraction of molecules in level 1 and level 2:

E. Example: A large collection of pure harmonic

oscillators.

q = e−βεii=1

∞

∑

q = e−0 + e−βε + e−2βε + ...

q = 1+ x + x2 + x3 + ...where

x ≡ e−βε

CHAPTER 15 19

This summation has a simple solution:

q = 1+ x + x2 + x3 + ...

q = 11− x

so:

q = 11− e−βε

which is plotted below as function of T:

CHAPTER 15 20

III. Simple Illustration: Application to Chemical Equilibrium.

Allowed Energies in Chemical Species A and B E

A(v) = vε

a v=0,1,2,3,4...

E

B(v) = ΔE + vε

b v=0,1,2,3,4...

Partition function for species A and B and for whole system of states:

qA= e−βEA(v)

v

A−states

∑ = e−βvεav

A−states

∑ = (1− e−βεa)−1

qB= e−βEB(v)

v

B−states

∑ = e−β(ΔE+vεb)

v

B−states

∑ = e−βΔE(1− e−βεb )−1

q = qA+ q

B

Population in any given state v in Species A or B:

Nv

N=e−βEA(v)

qA+ q

B

Nv

N=e−βEB(v)

qA+ q

B

CHAPTER 15 21

Now, what is the equilibrium constant for a chemical reaction AB?

Keq=

NB

NA

Keq=

Nv

v

B−states

∑

Nv

v

A−states

∑

Keq=

N e−βEB(v)

v

B−states

∑ / (qA+ q

B)

N e−βEA(v) / (qA+ q

B)

v

A−states

∑

and finally, the important equation:

Keq=

qB

qA

The equilibrium constant for a reaction is simply the ratio of the partition functions!

Substituting in the expressions for the species A and B partition functions:

Keq=e−βΔE(1− e−βεb )−1

(1− e−βεa)−1

CHAPTER 15 22

Let’s look at a specific case: Suppose zero-point energy of B is kBT above A as in drawing. So ΔE = kBT. Energy spacings of species A are also εa = kBT. Observe the equilibrium constant & how it’s affected by the spacing of levels εb in species B.

Notice that, even though the species B manifold of states is ABOVE those

of A, the Keq can be > 1, favoring the product side, IF the spacings in B are small enough. This is an ENTROPY effect. The states on the B side are denser allowing more “ways” of distributing the system, even though the B side starts at a higher energy.

A states

B states

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

1 2

3 4

5 6

7 state #

Population of states when B spacing is 0.1 kT

A states

B states

CHAPTER 15 23

IV. Thermodynamic energy and entropy.

The partition function contains all the information needed to calculate the thermodynamic properties of a system. Partition function is to stat mech as the wavefunction is to quantum mechanics. Let’s now see the relationship of the molecular partition function q to thermodynamic internal energy U and entropy S.

A. The Internal Energy U.

1. This is the energy found in the first law of thermodynamics, (ΔU= q-w).

We first obtain an expression for energy E = total energy of the system above its zero-point energy (lowest energy state).

E = n

ii

∑ εi

ni= number of molecules in state i

εi= energy of molecule in state i

Substituting in the Boltzmann Law for the number of molecules in state

i, and then performing some mathematical operations, we have:

E = Ne−βεi

qi

∑ εi=Nq i

∑ εie−βεi

But now we use math trick to rewrite this:

εie−βεi = −

∂

∂βe−βεi

so

E = −Nq

∂

∂βe−βεi

i

∑ = −Nq∂

∂βe−βεi

i

∑

E = −Nq∂q∂β

and finally

E = −N∂lnq∂β

CHAPTER 15 24

And so now adding the zero-point energy of the system we get our final result for the thermodynamic energy U:

U = U(0) −N ∂lnq∂β

$

%&&

'

())N,V fixed

where we have indicated the partial derivative is taken with volume V

and number of molecules N fixed. So to find the thermodynamic internal energy we must take a

derivative of the partition function q with respect to β. 2. It is convenient to rewrite this derivative in terms of temperature T

rather than β.

Remembering the definition β ≡1kBT

we can rewrite the partial:

∂

∂β=∂T∂β

∂

∂T=∂ βk

B( )−1

∂β

∂

∂T= −k

BT2 ∂

∂T

And so our final internal energy result is written in the more convenient

form:

U = U(0)+NkBT2 ∂lnq

∂T

"

#$$

%

&''N,V _ fixed

B. The Entropy S. 1. The Boltzmann formula for the entropy was given earlier as: S = k

BlnW

where W is the number of ways of distributing the total energy among a system of molecules in the most probable configuration.

CHAPTER 15 25

2. The W term can be rewritten in terms of the molecular partition

function q. See Atkins/dePaula 10th ed. P. 639.

S = U −U(0)T

+NkBlnq

C. The Canonical Partition Function Q.

1. Up to now we have assumed that our individual molecules are

statistically independent, and all our equations have been based on the molecular partition function q.

2. There is a more general partition function Q that is more appropriate

for interacting systems of molecules (such as in condensed phases).

Q = e-βEi

i

system states

∑

3. This looks deceptively the same as q, but Q differs from q in that the

summation is not over the quantum states possessed by an individual molecule. It is over configurations of the collective system of interacting molecules, and Ei is the energy of the entire system in one of those configurations i, not the energy of one molecule in quantum state i.

4. Therefore Q can be thought of as a TOTAL partition function of N

molecules, not the partition function of an individual molecule. 5. The thermodynamic internal energy and entropy in terms of Q are:

U = U(0) − ∂lnQ∂β

$

%&&

'

())V fixed

or

U = U(0)+kBT2 ∂lnQ

∂T

$

%&&

'

())V.fixed

and

S =U −U(0)

T+k

BlnQ

CHAPTER 15 26

6. But wait a minute, there is a relationship between Q and q. When the

N molecules are independent and indistinguishable.

Q =qN

N!

Independent means that the total energy can be obtained by a sum of

independent energy terms of independent molecules, giving rise to the

power of N.

Indistinguishable means that the molecules are identical and free to

move and exchange positions. This would cause a statistical overcount

of unique states of the collection of molecules, so we divide by N!

7. When the molecules are independent but distinguishable, such as

each molecule different OR molecules are fixed on a lattice:

Q = qN

8. Note that when this last equation is inserted for Q in the equations for

U and S, those equations revert to what we had originally.