Chem%350:%Statistical%Mechanics…scienide2.uwaterloo.ca/~nooijen/website_new_20_10_2011/Chem350... · Statistical’Mechanics’of’diatomic’molecules

Winter 2015 Chem 350: Statistical Mechanics and Chemical Kinetics

Chapter 5:Partition Functions and Properties of Real Molecules

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Chapter 5: Partition Function and Properties for Real Molecules ............................................................. 41 Thermodynamics of a system of independent particles ........................................................................ 41 The Partition function ............................................................................................................................ 42

Origin of factor 1!N for identical particles ............................................................................................ 42

Reworking the Helmholtz free energy ................................................................................................... 43 Translational Partition Function: Particle in the box wave function (atoms) ........................................ 44 Relating to thermodynamic properties (translational motion, particle in the box) .............................. 46 Statistical Mechanics of diatomic molecules ......................................................................................... 48 Vibrational Partition function in harmonic approximation ................................................................... 48 Rotational Partition Function for a diatomic ......................................................................................... 52 The origin of the symmetry factor in rotational partition function ....................................................... 55 Polyatomic Systems ............................................................................................................................... 59 Chemical Reactions and Equilibrium (recall from thermodynamics). .................................................... 63 Statistical Mechanics of Gibbs free energy ............................................................................................ 64 Connect to thermodynamics of chemical equilibrium .......................................................................... 65

Chapter 5: Partition Function and Properties for Real Molecules Thermodynamics of a system of independent particles -‐ Neglect internal degrees of, in particular rotations, vibrations. Easiest atoms, e.g. rare gases -‐ Will look at molecules in the gas phase, which are dilute and at high temperature (ideal gases)

Quantum Hamiltonian ( )ˆi

H h i=∑ no inter-‐atomic interactions

→ Solve ( ) ( ) ( )1 1 1a a ah ϕ ε ϕ=

→ ( ) ( ) ( )( )ˆ 1 2 .....a a zH Nϕ ϕ ϕ

( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( )1 1 2 .... 1 2 2 ....a a z a b zh N h Nϕ ϕ ϕ ϕ ϕ ϕ= +

( ) ( ) ( ) ( )( ).... 1 1 ....a b zh N Nϕ ϕ ϕ+ +

( ) ( ) ( ) ( )... 1 2 ...a b z a b z Nε ε ε ϕ ϕ ϕ= + +



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-‐ product of single-‐particle wavefunctions is eigenfunction -‐ sum of one-‐particle eigenvalues → total energies The Partition function

/

allE kT

all statesQ e−= ∑

( ). ... /

, , , ....

a b c d kT

a b c dQ e ε ε ε ε− + + += ∑

/ / / ..a b ckT kT kT

a b ce e eε ε ε− − −⎛ ⎞⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠∑ ∑ ∑

....a b cQ q q q= ⋅ ⋅ Since every atom/molecule is the same ....a b Nq q q q= =

( )NQ q=

Where N is the number of particles, q is the partition function for a single component (molecule). The overall partition function Q of the system is just the product of the individual partition functions. General feature: If Hamiltonian is a sum of terms without cross terms (interactions) → partition function will be a product of terms corresponding to terms in H → form of the total wavefunction is also a product function -‐ This is a big simplification. However this is only partially correct as it only applies if the particles were distinguishable from one another. In many cases the particles in multi molecule systems are indistinguishable. This is a very important feature of quantum mechanics.

Origin of factor 1!N for identical particles

All particles in nature should be viewed as either bosons or fermions.

→ Quantum mechanical wavefunctions are either symmetric or antisymmetric under interchange of particle coordinates

Fermions → antisymmetric, Bosons → symmetric If ( ) ( ) ( )( )1 2 3a b cAψ ϕ ϕ ϕ= a b c≠ ≠

In our sum we counted all permutations of , ,a b c as distinct states 3!→ contributions

But there is only 1 fully antisymmetryic wavefunction abc (Slater determinant) (fermions, e.g. electrons) and also only 1 symmetric function (bosons, e.g. certain atoms). → For 3 particles divide by 3!



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For N particles divide by !N If the number of available states >> the number of particles then it is a very good approximation to simply divide by !N

!

NqQN

=

This was done even for classical partition functions, but the reasons were not clear (although [erroneous] arguments were made) This statistics is known as Boltzmann Statistics. The procedure is not rigourously correct for either bosons or fermions. It does the counting wrong if single-‐particle states are identical. eg. a bϕ ϕ= a b= , same product wavefunction → For fermions: wavefunction = 0 For bosons: different factor to count symmetric wavefunctions (not 2 but 1)

In general many more states than # of particles ( >>> 2310 ). In that case simple Boltzmann counting is almost exact.

However, Boltzmann approximation is not always valid Eg. -‐ Electrons in metals -‐ photons in a light source -‐ very light particles (more elaborate discussion later on) In general if we have species ,A B etc. the Boltzmann partition function is given by

( ) ( )

! !

A BN NA BAB

A B

q qQ

N N=

Reworking the Helmholtz free energy

Using Stirling’s approximation for !N A = −kBT lnQ ( )ln ! ln ln 1N N N N N N= − = −

A = −kBT ln qN

N !⎛⎝⎜

⎞⎠⎟ ( )ln ln ln NN N e N

e⎛ ⎞= − = ⎜ ⎟⎝ ⎠

= −kBT lnqN − ln N !( ) ln

NNe

⎛ ⎞= ⎜ ⎟⎝ ⎠

= −kBT lnqN − ln N

e⎛⎝⎜

⎞⎠⎟

N⎛

⎝⎜

⎞

⎠⎟ !

NNNe

⎛ ⎞≈ ⎜ ⎟⎝ ⎠ (another Stirling approximation)

= −kBT ln qe

N⎛⎝⎜

⎞⎠⎟

N⎛

⎝⎜

⎞

⎠⎟



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= −NkBT ln qe

N⎛⎝⎜

⎞⎠⎟

NkBT = nN AkBT = nRT 8.314 /AN k R J molK= =

ln qeA nRTN

⎛ ⎞= − ⎜ ⎟⎝ ⎠

The Helmholtz free energy A seems to contain terms that do not scale linearly with N , i.e. non interacting particles seem to interact! This is a consequence of (anti) – symmetry requirement of many-‐particle wavefunction. Later we will see ~q V , and q/N is independent of the size of the system.

A = −NkBT ln qe

N⎛⎝⎜

⎞⎠⎟ ~

−NkBT ln V

N⎛⎝⎜

⎞⎠⎟

A will be proportional to N , in the end, as should be the case. Simple system: Non-‐interacting atoms

n e ti α β γε ε ε ε= + +

n : nuclear wave function (only nuclear spin is important. Degeneracy factor) e : electronic (important for open shell atoms, e.g. O or C atoms) t : translational (kinetic energy. Most important)

Translational Partition Function: Particle in the box wave function (atoms) Look at the quantum problem 1-‐D system, then extrapolate to 3D

−

2

2md 2

dxψ = Eψ

Eigenfunctions sin n xLπψ = ,

22

sind n n xdx L Lψ π π⎛ ⎞= −⎜ ⎟⎝ ⎠

2

2mnπL

⎛⎝⎜

⎞⎠⎟

2

sin nπ xL

= En sin nπ xL

En =

2π 2

2mL2

⎛⎝⎜

⎞⎠⎟

n22

228

h nmL

= = h

2π

qT ,1D = e

− h2

8mL2kBTn2

n∑ 1,2,3...n =

Define 2

28hmL kT

Δ = qt ,1D = e−Δ2n2

n∑

Since the energy spacings are small relative to kBT it is possible to use an integral in the place of summation



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qt ,1D = e−Δ2n2

n∑ ≈ e−Δ2x2

dx0

∞

∫ = 1Δ

12

π1 (look up the integral in chapter 3)

qt ,1D = 1

2Δπ = 1

Δπ4=

8mkBTL2

h2

π4= LΛ

Λ = Δ 4

π= h2

2mπkBT where Λ is called the thermal deBroglie wavelength

,1x

t DLq =Λ

Moving into 3D ,3 ,1 ,1 ,1t D x D y D z Dq q q q= ⋅ ⋅

,3 3 3x y z

t D

L L L Vq⋅ ⋅

= =Λ Λ

the molecular partition function for translational motion

Approximation is best if particle is heavy, box is large → classical limit (many energy levels, high density of states, sum is integral, also we neglected interactions.)

1Λ3 =

2πmkBTh2

⎛

⎝⎜

⎞

⎠⎟

3

=2πmkB

h2

⎛⎝⎜

⎞⎠⎟

3/2

T 3/2 =αT 3/2

define α =

2πmkB

h2

⎛⎝⎜

⎞⎠⎟

3/2

(constant, m is mass of particles)

3/2,3 3t D

Vq VTα= =Λ

Other contributions to partition function (less important usually) o

n nq g= (nuclear degeneracy, think NMR nuclear spins)

qe = e−Ei /kBT

i∑ (sum over states) (electronic, only for open-‐shell

atoms/molecules, e.g. O2 molecule)

/E kTg e αα

α

−=∑ (sum over energy levels)

/iE Eα : molecular energies, do not depend on ,N V !



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Relating to thermodynamic properties (translational motion, particle in the box)

ln qeA nRTN

⎛ ⎞= − ⎜ ⎟⎝ ⎠ 3/2

,3T Dq VTα=

3/2

ln VT eA nRTN

α⎛ ⎞= − ⎜ ⎟

⎝ ⎠

3ln ln ln2

VnRT e TN

α⎛ ⎞= − + +⎜ ⎟⎝ ⎠

1A NP nRTV V N∂⎛ ⎞ ⎛ ⎞= − = ⋅⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠

PV nRT= (ideal gas law!)

3/2

,

lnN V

A d VT eS nRTT dT N

α⎡ ⎤∂⎛ ⎞= − = ⎢ ⎥⎜ ⎟∂⎝ ⎠ ⎣ ⎦

3ln ln ln ln2

d nRT e V T NdT

α⎡ ⎤⎛ ⎞= + + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

3/2 3ln2

VT e nRTnRN T

α= +

3/23 ln2

VT eS nR nRN

α= +

2 AU TT T

⎛ ⎞∂ ⎛ ⎞= − ⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

3/22 ln VTT nRT N

α⎛ ⎞⎛ ⎞∂= − ⎜ ⎟⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

232nRTT

=

32

U nRT= 32V

UC nRT

∂⎛ ⎞= =⎜ ⎟∂⎝ ⎠

H U PV= +

3 52 2

H nRT nRT nRT= + =

52P

HC nRT

∂⎛ ⎞= =⎜ ⎟∂⎝ ⎠



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A = −nRT ln qe

N⎛⎝⎜

⎞⎠⎟= −NkBT ln qe

N⎛⎝⎜

⎞⎠⎟

µ = ∂A

∂N⎛⎝⎜

⎞⎠⎟ T ,V

= ∂∂N

−NkBT ln αVT 3/2eN

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟

= −kBT lnαVT 3/2e

N+ −NkBT ⋅ − 1

N⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = − kBT lne − kBT αVT 3/2

N+ kBT

µ = −kBT lnαVT 3/2

N

µ N , P,T( ) = −kBT lnαT 3/2NkT

NP= −kBT lnαT 3/2kT

P= −kBT lnαkT 5/2

P

µ = N Aµ = −N AkBT lnαT 5/2k

P= −RT ln

αT 5/2kB

P

G = nµ = −nRT ln

αT 5/2kB

P0

⋅P0

P⎛

⎝⎜⎞

⎠⎟

G = −nRT ln αT 5/2k

P0

⎛

⎝⎜⎞

⎠⎟− nRT ln

P0

P⎛⎝⎜

⎞⎠⎟

G = G0 − nRT ln

P0

P⎛⎝⎜

⎞⎠⎟= Go + nRT ln P

Po

⎛

⎝⎜⎞

⎠⎟

Check formula for S compared to thermodynamics: The formulas below check with thermodynamics as taught in chem254.

3/23 ln

2VTS nR nRNα⎛ ⎞

= + ⎜ ⎟⎝ ⎠

5/23 ln2

kTnR nRPα⎛ ⎞

= + ⎜ ⎟⎝ ⎠

1 2 V V→ , T constant 2

1

ln VS nRV

⎛ ⎞Δ = ⎜ ⎟

⎝ ⎠

1 2 T T→ , V constant 2

1

3 ln2

TS nRT

⎛ ⎞Δ = ⎜ ⎟

⎝ ⎠ =

CV ln

T2

T1

⎛

⎝⎜⎞

⎠⎟

1 2 T T→ , P constant 2

1

5 ln2

TS nRT

⎛ ⎞Δ = ⎜ ⎟

⎝ ⎠=

CP ln

T2

T1

⎛

⎝⎜⎞

⎠⎟

1 2 P P→ , T constant 2

1

ln PS nRP

⎛ ⎞Δ = − ⎜ ⎟

⎝ ⎠

All of ideal gas thermodynamics follows from A = −NkBT ln

qteN

⎛⎝⎜

⎞⎠⎟, ( )

!

NtqQN

= translational

partition function. This is a very satisfying result: particle in the box quantum mechanics!



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Statistical Mechanics of diatomic molecules

Molecules have: translational, rotational, vibrational, electronic and nuclear spin degrees of freedom To good approximation: t r v e nE E E E E E= + + + + This is not quite true, in particular there could be a coupling between rotational and vibrational motions (certainly for higher levels).

t r v e nq q q q q q= ⋅ ⋅ ⋅ ⋅ (not quite true as well, complications do arise)

tq is the same as for atoms.. 3/2tq VTα= ,

α =

2π MkB

h2

⎛⎝⎜

⎞⎠⎟

3/2

. Same ideal gas

assumptions, works best at low density, light molecules, and works only for gases. At low temperatures, gases condense or solidify due to (weak) interactions.

eq : typically only one electronic level contributes. This would be different for radicals or triplet states. Even then: simply account for degeneracies. Unexpected complication: strong coupling between nuclear and rotational wavefunction (another manifestation of (anti)symmetry).

nrq rotation + nuclear should be treated together “Pauli principle for nuclei”.

Note: the factor 1!N was also result of requiring (anti)symmetric wavefunctions. The

nuclear+rotation aspect will be discussed later. Simple results are obtained when vibrations and rotations can be treated separately (no coupling) and harmonic oscillator is used for vibrations.

Vibrational Partition function in harmonic approximation

Eharm = 1

2gx2

where ( )0x R R= − (Harmonic approximation)

H = −

2

2µd 2

dx2 +12

gx2⎛⎝⎜

⎞⎠⎟

where 1 2

1 2

m mm m

µ =+

(reduced mass)

12nE n ω⎛ ⎞= +⎜ ⎟⎝ ⎠h 0,1,2...n = and

ω = g

µ



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( g is the force constant and depends on the molecule) (only need g and µ and you will find all the energies) This would be discussed in a quantum mechanics class Finding the partition function

Energy levels harmonic oscillator: En =

12ω + nω 0,1,2...n =

e−En /kBT = e

−12!ω /kBT

⋅e−n!ω /kBT

qV = e

−12!ω /kBT( )

e−n !ω /kBT( )n=0,1,2..∑

Set y = e− !ω /kBT( )

e−2 !ω /kBT( ) = e− !ω /kBT( )e− !ω /kBT( ) = y2 so e

−n !ω /kBT( ) = yn

qV = e

−12!ω /kBT( )

1+ y + y2 + y3 + ....( ) = e−1

2!ω /kBT( )

yn

n=0,1,2∑

( )2 3

0,1,2...

11 ....1

n

ny y y y

y=

= + + + + =−∑ (known math relation)

qV = e

−12!ω /kBT( )

⋅ 11− y

⎛⎝⎜

⎞⎠⎟

qV = e

−12!ω /kBT( )

⋅ 11− e− !ω /kBT( )

⎛⎝⎜

⎞⎠⎟ partition function with zero point energy included

Define vibrational temperature, TV = !ω

kB

→ 1/J JK− = K (kelvins) so VTT

is dimensionless

qV = e

−12

TV /T( )⋅ 1

1− e− Tv /T( )⎛⎝⎜

⎞⎠⎟

In p-‐chem vibrational and rotational ‘energies’ are often expressed in cm-‐1. There is a conversion factor between energy (in J) and cm-‐1. If we use the same conversion for kB everything makes

sense. Hence in practice we use the formula:

TV =

!ω!kB

(units : cm−1

cm−1K −1 = K ), !kB = 0.6950348 cm−1K −1

More careful discussion is found in the section on rotational energies. For some molecules, the internal rotation can range from relatively small !ω ≈ 200cm−1 , to

something like C-‐H stretching !ω ≈ 3100cm−1 , hence TV =

!ω!kB

can range from about 300 – 6000K.



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e−1

2!ω /kBT

⋅e−n!ω /kBT = e−1

2TV /T( )

⋅e−n TV /T( ) , since VT ranges from 300 – 6000K, typically a large T value is needed for a decent population of excited states (n=1 or higher).

Energy scale is not convenient if we want to consider mixtures of molecules, as we have chosen the zero of energy as the bottom of the well. From quantum chemical calculations one can obtain good estimates for the total ‘electronic’ energy at the equilibrium distance. We can denote this energy as

E0el = Eel (R0 ) . The energy of the ground vibrational level is then denoted as

Eg = E0

el + 12!ω = E0 + Ezp . Here

Ezp denotes the so-‐called zeropoint energy. It is the energy of the

lowest vibrational level. The same kind of formula will be applicable for general poyatomic molecules. In practice it is very useful to treat the total electronic energy and the vibrational energies together. This will lead to formulas we will use for polyatomic molecules, and in particular when we discuss chemical reactions and the associated change in the Gibbs free energy. Then the electronic energy will play a primary role.

The electronic/vibrational partition function becomes qe/V = e−Eg /kT ⋅ 1

1− e− TV /T( )⎛⎝⎜

⎞⎠⎟

Three cases are of special interest: a.) The low temperature limit:

If ( )/VT T is large (low T , or high TV ) e−(TV /T ) → 0 ; qe/V = e−Eg /kT

This means that all molecules are in the lowest vibrational state in the electronic ground state. This is the usual case. In general very few molecules are vibrationally excited. If we neglect the excited molecules we get the so-‐called low temperature limit. It follows:

A = −NkBT ln(qe/V ) = −NkBT (−Eg

kBT) = NEg

S = − ∂A∂T

⎛⎝⎜

⎞⎠⎟= 0; U = A+TS = A = NEg ; Cv = 0

This is all perfectly consistent with the fact that all molecules are in the ground electronic and vibrational state. b) The high T limit. This is historically interesting.

Consider the large T limit of better / VT T is large

eEg /kBT 1

1− e−Tv /T

⎛⎝⎜

⎞⎠⎟ /1 , 1 /VT Tx

Ve x e T T−− ≈ − − ≈



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qe/V = e−Eg /kBT T

TV

⎛⎝⎜

⎞⎠⎟

!AV = −NkBT lnqV = NEg − NkBT ln T

TV

⎛⎝⎜

⎞⎠⎟

!SV = +NkB ln T

TV

⎛⎝⎜

⎞⎠⎟+ NkB

!UV = !AV +T !SV = NEg + NkBT = NEg + nRT

CV = NkB = nR

Famous classical Equipartition theorem: For every ( )2xP and 2x in Hamiltonian, the contribution to

internal energy is 12RT per mole → contribution to VC is

12R

“Every vibrational mode contributes nRT to U , and nR to VC at high temperature”. This is exactly the prediction from classical mechanics (for any temperature), and this result is typically very wrong. You can think of the classical limit when the energy level spacing or ω is very small (0 in the limit). This was a much disputed discrepancy during the time of Maxwell and Boltzmann when they derived the kinetic theory of gases. Classical mechanics assumes a vibrating particle can have any energy. In quantum mechanics the energy is quantized, and the first excited vibrational state is often so high in energy that only the ground state is populated at room temperature or below. This discrepancy greatly hindered the acceptance of statistical mechanics (and the molecular hypothesis): The theory made some totally wrong predictions. How can it be correct? Of course what was missing was quantum mechanics, to describe the internal motion of molecules. c) Exact thermodynamic values for the electronic / vibrational partition function for a non-‐degenerate electronic state.

qV = e−Eg /kBT ⋅ 1

1− e− TV /T( )⎛⎝⎜

⎞⎠⎟

AV = −NkBT lne−Eg /kBT − ln 1− e−TV /T( )( )

= NEg + NkT ln 1− e−TV /T( )

( ) ( ) ( )/ /2/

ln 11

V V

V

T T T TV VV T T

A TNkTS kN e eT Te

− −−

∂⎛ ⎞ ⎛ ⎞= − = − − −⎜ ⎟ ⎜ ⎟∂ −⎝ ⎠ ⎝ ⎠

If we multiply this by /

/

V

V

T T

T Tee

then we get

SV = −NkB ln 1− e−Tv /T( ) + NkBTV / TeTV /T −1( )



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U = A+TS = NEg +NkBTV

eTV /T −1( )

µ =

∂AV

∂N⎛⎝⎜

⎞⎠⎟ T ,V

= Eg + kBT ln 1− e−TV /T( ) This is all exact, for harmonic oscillator. On a computer these formulas are easily evaluated exactly. However, in questions on paper (midterm and final) we will usually invoke the low temperature approximation. All that remains then from the vibrational contribution is the zero-‐point energy contribution to the electronic energy. This is always an important contribution. Rotational Partition Function for a diatomic

Use the so called rigid rotor approximation, neglect coupling between rotations and vibrations (small error)

The Quantum Mechanical Hamiltionian for rotations

H = L2

2µR2 µ =

m1m2

m1 + m2

2L : angular momentum operator depending on ,θ ϕ , R : internuclear distance Note: 2L is the same operator that shows up for the H-‐atom orbitals: s,p,d,f functions

L2Yl

m θ ,ϕ( ) = !2l l +1( )Ylm θ ,ϕ( )

.....lm l l= − + 0l = s 0 1l = p -‐1, 0, +1

2l = d -‐2, -‐1, 0, +1, +2 3l = f -‐3, -‐2, -‐1, 0 , +1, +2, +3

Known solutions for energy eigenvalues

EJ =

!2

2µR2 J J +1( ) ( )1BJ J= + 0,1,2...J =

Has degeneracy gJ = 2J +1( ), from M = −J ...J

States: YJ

M θ ,ϕ( ) M = −J ,−J +1.....J −1, J ( ) 2 1J→ +

Energy levels are often expressed in cm−1 hυ = hc

λ= hc !k

EJ = BJ J +1( )

= h

8π 2cµR2 (in 1cm− ) 2R Iµ = (moment of inertia)

A convenient conversion: 11 8065.5eV cm−≈



53

Also the Boltzmann constant can be converted to cm-‐1 K-‐1 units: !kB =

kB

hc≈ 0.695cm−1K −1

These are very convenient units for rotational energies (also for vibrational energies). Rotational partition function

qR = 2J +1( )e−BJ J+1( )/kBT

J=0,1,2..∑ = 2J +1( )e− BJ J+1( )/ kBT

J=0,1,2..∑

Note: we sum over energy levels J , and need to explicitly include degeneracies Using a math program, one can carry out the sum explicitly (eg. Run until maxJ = 100)

In practice, in the “high temperature” limit one replaces the sum by an integral

qR ≈ 2x +1( )e− Bx x+1( )/ kBT dx

0

∞

∫

Substitute ( ) 21y x x x x= + = +

( )2 1dy x dx= +

qR = e− By/ kBT dy = −

kBTB

e− By/ kT⎡⎣ ⎤⎦0

∞

∫0

∞

=kBTB

qR =

kBTB

= TTR

where , B , kB are in units of cm−1

( B = !2

2µR2 for diatomic, !B = B / (hc) )

TR = B

kB

=BkB

Rotational temperature (in Kelvins)

The high temperature limit is quite accurate for most molecules (we will check in Matlab). Hower there is a complication. This formula is “correct” for heteronuclear diatomics like CO , but for homonuclear case, like 2H it is off by a factor of 2. One can correct for this

qR =

kBTσ B

= TσTR

where σ is the (mysterious) symmetry factor ; 1σ = for heteronuclear, 2σ = for homonuclear

To understand the symmetry factor one has to take nuclear spin into consideration. It is really a

consequence of the Pauli principle for nuclei. It is comparable to the Boltzmann factor 1!N in

!

NqQN

= .



54

In the next lecture I will discuss the rotational partition function for 2 2,H D and HD isotopes. This will give us a better idea of the origin of the mysterious σ .

Thermodynamics in high temperature limit: Contributions due to rotational degree of freedom

AR = −nRT ln T

σTR

= −NkBT ln TσTR

(R is the gas constant again, not the interatomic distance)

,

lnRR

V N R

A TS nR nRT Tσ

∂⎛ ⎞= − = +⎜ ⎟∂⎝ ⎠

R R RU A TS nRT= + = ,V RC nR=

µR = −N A

∂AR

∂N⎛⎝⎜

⎞⎠⎟ T ,V

= −N AkBT ln TσTR

= −RT ln TσTR

Let me note that the high temperature limit for rotations is excellent in practice. It agrees with the

classical treatment of rotations. According to the equipartition theorem we have have 12

RT of energy

for each kinetic energy degree of freedom. Rotation of a rod (or diatomic) corresponds to 2 degrees of freedom, so the result would be U R = nRT as above. The rotational energy levels are so finely spaced, that the classical treatment works well. Except: it cannot account for the symmetry factor. This is a quantum mechanical effect.

Probability to find molecules in energy level ( )JP E

( ) ( ) ( )1 /2 1RJ J T T

JR

JP E e

q− ++

=

( ) ( )1 /2 1 RJ J T TRTJ eTσ − +≈ +

We can also plot the probability to find a particular state (one from the 2 1J + ) YJ

M (θ ,ϕ )



55

This peaks at the ground state, which always has the highest probability The origin of the symmetry factor in rotational partition function

Nuclei can be bosons (consisting of even number of fermions) or fermions (consisting of odd number of fermions). This character is reflected by nuclear spin: Bosons will have integer spin, Fermions have half integer spin. Nuclei are described by Quantum Mechanical wave functions and they obey fundamental symmetries of nature -‐ ψ is symmetric under interchange of identical bosons -‐ ψ is anti symmetric under interchange of identical fermions Consider a system consisting of 2 nuclei diatomics → nuclear wavefunction has both a spatial and a spin part. Focus first on H -‐ atom, spin ½, ,α β functions. Nuclear spin functions for H2 molecule:

( ) ( )1 2α α

orthohydrogen → ( ) ( ) ( ) ( )1 2 1 2 2α β β α⎡ + ⎤⎣ ⎦ symmetric

= triplet ( ) ( )1 2β β Parahydrogen

α 1( )β 2( )− β 1( )α 2( )⎡⎣ ⎤⎦ 2 antisymmetric

= singlet Nuclear spin functions are virtually degenerate (even in presence of a magnetic field)

For us the symmetry of the spin-‐eigenfunctions are most important. Consider 2 nuclei of general spin I



56

Symmetric ( )1 2 2 1 2mm m m+ ( m1 ≤ m2 m1 = m2 is included)

( )( ) ( )( )1 2 1 2 2 1 2 12

I I I I+ + = + + (symmetric functions) , e.g. 1 3 2 32 2

I = → ⋅ =

Antisymmetric ( )1 2 2 1 2mm m m− ( m1 < m2 , m1 = m2 is excluded)

( )( ) ( )1 2 1 2 2 12

I I I I+ = + (antisymmetricfunctions), e.g. 1 1 2 12 2

I = → ⋅ =

(see above)

(compare square n x n matrix which has ( )1 12n n + entries in upper block including the diagonal

(i>=j), while there are ( )1 12n n − entries in the lower block (i<j), excluding diagonal)

Symmetry of nuclear spin function under permutation is now understood. What about spatial part of nuclear wave function? Consider the nuclear coordinates

R1,R2

Center of mass coordinate

R1 + R2

2=Rcm (for identical nuclei)

R1 −R2 = sin cosR θ ϕ

sin sinR θ ϕ cosR θ

R =R2 −

R1

Nuclear wavefunction:

ψ t

Rcm( ) ⋅ψ V R( ) ⋅ψ R θ ,ϕ( ) (translational, vibrational, rotational parts)

If

R1 ↔

R2 (interchange) then

Rcm and R are unaffected

However: P12

R2 −

R1( ) = −

R2 −

R1( )

= sin cosR θ ϕ− sin sinR θ ϕ− cosR θ− Interchanging nuclei

R1 ↔

R2 is equivalent to



57

θ π θ→ − ( ) ( )cos cosπ θ θ− = −

( ) ( )sin sinπ θ θ− =

ϕ ϕ π→ + ( ) ( )cos cosϕ π ϕ+ = −

( ) ( )sin sinϕ π ϕ+ = −

Hence interchanging R1 and

R2 is equivalent to changing

θ π θ→ − ϕ ϕ π→ +

( ),mlY π θ π ϕ− + = ( ),m

lY θ ϕ+ l even ( ),m

lY θ ϕ− l odd Transformations are equivalent to

x xy yz z

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟→ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

(inversion)

where the inversion is expressed using the angular variables. s, d, g functions are even under inversion (like J=0,2,4,…) p, f functions are odd under inversion (like J=1,3,5,…) 0,2,4..J = even under 1 2R R↔ 1,3,5..J = odd under 1 2R R↔

Antisymmetric nuclear wavefunctions (fermions): ( ),triplet

spin Jφ ψ θ ϕ⋅ Even Odd

Or ( )sin glet ,spin Jφ ψ θ ϕ⋅ The only allowed combinations for 2H Odd Even

(H is fermion) Symmetric nuclear wavefunctions (bosons):

( ),tripletspin Jφ ψ θ ϕ⋅

Even Even Or ( )sin glet ,spin Jφ ψ θ ϕ⋅ Overall symmetric for 2D

Odd Odd

(D is boson) The restriction to either overall symmetric wavefunctions or overall antisymmetric wavefunctions amounts to a “coupling” between rotational and nuclear partition function. Hence they should be treated together. We refer to the combined nuclear spin and rotational partition function as qnR .

For 2H (fermions, antisymmetric) 12

I =



58

qnR

H2 = 3 2J +1( )e−BJ J+1( )/kT

J=1,3,5..∑ +1 2J +1( )e−BJ J+1( )/kT

J=0,2,4..∑

For 2D (bosons, symmetric) 1I =

qnR

D2 = 6 2J +1( )e−BJ J+1( )/kT

J=0,2,4..∑ + 3 2J +1( )e−BJ J+1( )/kT

J=1,3,5..∑

Remember number of spin states: ( )1/ 2H I = ( )1D I =

Odd ( )2 1I I + 1 3

Even ( )( )1 2 1I I+ + 3 6

Total ( )22 1I + For HD no symmetry requirement, 2 3 6ng = ⋅ = (degeneracies)

qnR

HD = 6 2J +1( )e−BJ J+1( )/kT

J=0,1,2..∑

Summary: In general for homonuclear diatomics with nuclear spin I

Fermion: ( )( ) ( )( )1 2 1 2 1Odd EvenR RI I q I I q+ + ⋅ + + ⋅

Bosons: ( )( ) ( )( )2 1 1 2 1Odd EvenR RI I q I I q+ ⋅ + + + ⋅

For heteronuclear diatomics: ( )( )2 1 2 1 total

nR A B Rq I I q= + + ⋅ For spin AI and spin BI How does this reduce to the symmetry factor σ for rotational wavefunction?

a) ( )22 1 nucleareven oddn n I q+ = + =

b) 12

even odd totalR R Rq q q≈ ≈

Proof of b):

2J +1( )e−BJ J+1( )/kBT

J=0,2,4∑ → substitute J = 2k

= 4k +1( )e−B2k 2k+1( )/kBT

k=0,1,2..∑ 24 2y k k= + , 8 2dy k= +

→ 1

2e− y B/kBT( ) dy = 1

2kBTB0

∞

∫

Similarly:

2J +1( )e−BJ J+1( )/kBT

J=1,3,5∑ 2 1J k= + , 0,1,2..k =

2 2k +1( ) +1( )e−B 2k+1( ) 2k+2( )/kBT

J=1,3,5∑

( )( )2 1 2 2y k k= + + , dy = 2 2k + 2( ) + 2 2k +1( ) = 2*[2(2k +1)+1)]



59

→ 1

2e− y B/kBT( ) dy = 1

2kBTB0

∞

∫

→ Same high temperature limit, multiplied by a factor ½. This analysis is quite involved. We will do a simulation in Matlab to clarify. Importantly for systems like H2, D2 and HD this has all been beautifully demonstrated by both theory and experiment. Polyatomic Systems

Consider system with N atoms 3N→ coordinates. 3 collective coordinates describe the overall translation of center of mass. If we choose to optimize the equilibrium geometry we can identify 3 collective coordinates that describe rigid rotation (2 for linear molecules). 3 6N − collective coordinates remain that describe internal vibrations (3 5N − for linear molecule) Solve electronic Schrodinger equation for fixed nuclear position iR , 1i = , 3N

H R{ }( )ψ r ,

R{ }( ) = E

R{ }( )ψ r ,

R{ }( )

3N→ dimensional potential energy surface (PES)

∂E∂Rj

= 0 ∀j → extrema on PES

Different isomers: different minima on PES Transition State: Saddle points on PES (Max in one direction, min in all others (like a mountain pass in 3d)

a) Vibrations

Taylor series of potential energy surface around minimum Re (3 N coordinates)

ER( ) = E

Re( ) + ∂E

∂Ri R=Re

R −Re( )i

+i∑ 1

2∂2 E

∂Ri ∂Rj R=Re

R −Re( )i

i∑

R −Re( ) j

Extremum →

∂E∂Ri

= 0 ∀i

Mass-‐weighted Hessian (This is a detail, to be complete)



60

Hij = Mi

∂2 E∂Ri ∂Rj R=

Re

M j

(to account for nuclear masses in nuclear kinetic energy term) Diagonalize “mass-‐weighted Hessian” ijH

6 (5 for linear molecule) eigenvalues are 0: correspond to overall translation, overall rotation. 3 6N − (or 3 5N − ) eigenvalues iε of Hessian correspond to normal modes i .

The eigenvalues ε i are analogous to the force constant for a diatomic molecule (denoted g) By diagonalizing the Hessian, the vibrational problem is reduced to 3N − 6 (or5) independent harmonic oscillator problems

− !

2

2d 2

dqi2 +

12ε iqi

2⎡

⎣⎢

⎤

⎦⎥ χn qi( ) = En

i( )χ qi( )

Define !ω i =

ε i

1 (analog of

!ω i =

gµ, with µ = 1)

En

i( ) = ni +12

⎛⎝⎜

⎞⎠⎟ω i

If all iε , iω > 0 , then stationary points is minimum. If precisely one of the iε is negative, or, iω is imaginary, then structure is transition state Vibrational frequencies and normal modes are obtained from Hessian. Can be calculated using quantum chemistry program (e.g. Gaussian). As usual the vibrational “energies” are usually expressed in wave numbers. We have seen that before. b) Rotations:

Position of minimum: , jRα α = x, y,z , 1,2,...j N= (number of nuclei)

Rcm

α = 1

mj∑mk Rα ,k

k∑

Moment of inertia tensor:

( )( ) ( )2, , , , ,1

N

j j cm j cm j j cmj j

I m R R R R m R Rα β γα β α β α β γ

γδ

=

= − − − + −∑ ∑ ∑

,Iα β → 3 x 3 matrix

Diagonalizing matrix I yields 3 eigenvalues , ,A B CI I I -‐ Spherical Top A B CI I I= = -‐ Symmetric Top A B CI I I= > (prolate cigar) or A B CI I I> = (oblate disk) -‐ Asymmetric Top A B CI I I> >



61

Rotational eigenvalues spectrum can be calculated purely from , ,A B CI I I . The various cases are somewhat complicated. ‘High’ Temperature partition function always has a simple form (used in practice. This is usually accurate)

qR T( ) = π

σTTA

⎛

⎝⎜⎞

⎠⎟

1/2TTB

⎛

⎝⎜⎞

⎠⎟

1/2TTC

⎛

⎝⎜⎞

⎠⎟

1/2

TA =

!2

2kB I A

This formula always works, except for linear molecules, where one uses RR

TqTσ

= , TR = !

2

2kB I

(only 2 rotational degrees of freedom. Rotation around molecular axis has no meaning)

Symmetry Factorσ : # of pure rotations in the point group of the molecule, known from group theory. (# of rotations that map molecule onto itself). This is again due to the interplay of rotational and nuclear spin patterns. The details are complicated (I myself don’t know how this works precisely). Examples:

2H O σ = 2 4CH 3 4 12σ = × = 4 3-‐fold rotation axes

3NH σ = 3 6 6C H 12σ = 1 6-‐fold axis + two times 3 2-‐fold axis! Summary: Overall partition function for polyatomic molecule (non-‐linear):

qmol = qtqRqnqvqe

Translational: 3/2tVq TNα=

α =

2π MkB

h2

⎛⎝⎜

⎞⎠⎟

3/2

jj

M m=∑

Vibrational: qv =

e−1

2hω i /kBT

1− e−hω i /kBTi=1

3N−6 (or5)

∏

One factor for each vibrational mode, including zeropoint frequency If we define the vibrational zeropoint energy

Ezp =

12i=1

3N−6

∑ !ω i

Then:

qv = e−Ezp /kBT (1− e−hω i /kBT )−1

i=1

3N−6

∏ = e−Ezp /kBT (1− e−TV( i ) /T )−1

i=1

3N−6

∏

By far the most important contribution is the zeropoint energy contribution. The other factors are typically close to unity. It is helpful to write

qV = qV ,zpqV ,rest

qV ,zp = e−Ezp /kBT ; qV ,rest = (1− e−TV( i ) /T )−1

i=1

3N−6 (or5)

∏ ≈1



62

Rotational: qnR = π

σTTA

⎛⎝⎜

⎞⎠⎟

12 T

TB

⎛⎝⎜

⎞⎠⎟

12 T

TC

⎛⎝⎜

⎞⎠⎟

12

TA =

!2

2kB I A

etc.

Nuclear Spin ( )2 1nq Iα

α= +∏ Iα : nuclear magnetic moment for nucleus α

Electronic: e−E0

el /kBT or, if we have very low-‐lying excited states (e.g. for magnetic systems)

e−E0

el /kBT e−( Eiel−E0

el )/kBT

i∑

E0el : The electronic energy at the minimum of the potential energy curve.

Often it is convenient to treat the vibrational zeropoint energy and the electronic energy together, and (as for diatomics) we define

Eg = E0

el + Ezp; qe/Vzp = e−Eg /kBT

This energy is the energy of the molecule in its ground electronic and ground vibrational state. The ground state energy

Eg plays its most important role when we compare the relative stability of two

different isomers, and we can calculate the difference in energy (see later). From electronic structure (quantum chemistry) calculations we can calculate the absolute energies. They are very large (negative) numbers however. Often we forget about the ground state energy, and just add excited electronic states in the sum (if necessary) if we are purely interested in the temperature dependence of U, Cv and so on. The total vibrational/electronic ground state energy plays a vital role when discussing chemical equilibria. Many quantities can be calculated accurately from contemporary electronic structure calculations. Geometries and vibrational frequencies are fairly accurate (but within harmonic approximations). Atomization energies/reaction energies would be the hardest to obtain accurately. The harmonic approximation is poor for floppy molecules. This is difficult to correct. Other very low frequencies of vibrations, e.g. internal rotation for example in ethane 3 3CH CH− can be treated in advanced calculations, e.g. Potential along torsional mode

( ) ( )cos 3V Aϕ ϕ= A : barrier height



63

This can be included (neglect mode-‐coupling). There are remaining problems for floppy molecules though (think molecules with many conformers, many systems in biology). There are also other problematic cases, where electronic structure calculations are difficult (e.g. when the molecular orbital picture breaks down). The bottom line however is that this type of approach works very well for many molecules. The current discussion is focused on molecules in the gas phase, neglecting interactions. In this case the expensive part of the calculation is the electronic structure calculation: optimizing geometry, calculating vibrational frequencies and obtaining an accurate electronic energy. The calculation of the thermal corrections due to statistical mechanics, do not take much time at all after this (seconds). The statmech procedure can also be readily developed for solid state materials, where the material is pefectly crystalline and periodic. Of course calculations will be more costly than for gas phase molecules. The most complicated case concerns liquids where molecules are strongly interacting. This situation is not so different from the case of molecules with a large number of conformers. In such cases one has to use extensive sampling procedures and simulations to obtain suitable averaged results. One mostly uses classical mechanics in this sampling, although quantum mechanical procedure using path integrals are also used. In all cases the calculations become far more expensive and only dedicated users typically performed such calculations. In contrast, the use of electronic structure calculations using for example the Gaussian program has become a routine tool for any chemist. Chemical Reactions and Equilibrium (recall from thermodynamics).

Consider reactions in gas phase a) Thermodynamics:

Prototype reaction: aA+ bB cC + dD

, , ,a b c d : stoichiometric coefficients , , ,A B C D : chemical species Reactants → Products

Write it in the form 0cC dD aA bB+ − − =

0AA

Aυ =∑ 0Aυ > for products 0Aυ < for reactants

Since we will consider equilibrium, reactants vs products is an arbitrary choice At chemical equilibrium

0reactionGΔ = and 0A AAυ µ =∑

( ) ( )ln /oA A A oT RT P Pµ µ= +

oP = standard pressure, AP = partial pressure of species A For ideal gases: A AP x P= Ax : mole fraction of species A

( ) ( )ln / lnoA A o AT RT P P RT xµ µ= + +

( ), lnoA AT P RT xµ= + (alternative expression in mole fractions. See

thermo)



64

0reaction A AA

G υ µΔ = =∑

( ) ( )ln / AoA A A o

A AT RT P P υυ µ= +∑ ∑

( ) ( )ln / AoA A A o

A AT RT P P υυ µ= +∑ ∏

Define equilibrium constant ( )/ A

p A oA

K P P υ=∏

( ) lnoA A p

AT RT Kυ µ = −∑ ( )o

A AA

Tυ µ∑ from stat mech

In practice we can calculate ( )o

A Tµ from QM and Stat-‐Mech → first principles theory of chemical equilibrium For previous example: aA+ bB cC + dD

( ) ( )( ) ( )/ /

/ /

c dC o D o

p a bA o B o

P P P PK

P P P P=

Often it is easier to work with molefractions A AP x P=

( ) ( )/ /A A

P A o A oA A

K P P x P Pυ υ= =∏ ∏

( ) / AAA o

A A

x P P υυ= ⋅∏ ∏

( ) /AA o

A

x P P υυ Δ= ⋅∏ AA

υ υΔ =∑

( )/P x oK K P P υΔ=

( ) ( )/A

x A p oA

K x K P Pυ υ−Δ= =∏

This is all a repeat of what was done in thermodynamics. Let us take the next step. Statistical Mechanics of Gibbs free energy Chemical potential = Gibbs Free energy G = A+ PV = A+ NkBT (ideal gas)

= −kBT ln qN

N !⎛⎝⎜

⎞⎠⎟+ NkBT

= −NkBT lnq + kBT N ln N − N( ) + NkBT

= −NkBT lnq + kBTN ln N

= −NkBT ln q / N( )G = −nRT ln q / N( )



65

tv R n e

qq q q q qN N

= ⋅

3/2tM

q V TN N

α= α M =

2π MkB

h2

⎛⎝⎜

⎞⎠⎟

3/2

PV = NkBT → VN

=kBTP

qt

N=α M kBT 5/2

P Gibbs free energy use P as variable, not V. The other, very important contribution is electronic + zeropoint

qe/Vzp = e−Eg /kBT ; Eg = E0

el + Ezp = E0el + 1

2i=1

3N−6 (or 5)

∑ !ω i

qnR ,qV ,rest : can both be calculated for each species in chemical reactions, exactly as we

did for polyatomic molecules in general.

tqN

determines the pressure dependence

−nRT ln

qt

N⎛⎝⎜

⎞⎠⎟= −nRT ln

α M kBT 5/2

Po

+ nRT ln PPo

⎛⎝⎜

⎞⎠⎟

qt ,0 = −nRT ln

α M kBT 5/2

P0

→ enters reactionGΔ (see previous thermo

discussion). Connect to thermodynamics of chemical equilibrium

( )ln op A A

ART K Tυ µ− =∑

ln AA

A

qRTN

υ ⎛ ⎞= − ⎜ ⎟⎝ ⎠∑

ln lnA A

A A

A A

q qRT RTN N

υ υ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∏

A

Ap

A

qKN

υ⎛ ⎞= ⎜ ⎟⎝ ⎠

∏

=

qt ,oA

N

⎛

⎝⎜

⎞

⎠⎟

υA

qe,VzpA( )υA qRn

A( )υA qV ,restA( )υA

A∏ qn

A( )υA



66

= Kt ,0Ke,Vzp KRn(KV ,rest Kn )

Each factor is a ratio of corresponding q ’s. This is a very convenient and insightful result! Let us analyze different factors:

-‐ nuclear spin factor: easiest, since the number of nuclei does not change between reactants and products and neither does nuclear spin

reactants 1productsn

nn

q Kq

= = (always)

-‐ rotational factors: Just has to be calculated for each molecule, (for atoms → qRA = 1 )

For linear molecules use qR,n =

TσTR

(Good enough. Include symmetry factor)

Polyatomics:

11 122 2

RA B C

T T TqT T T

πσ

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

-‐ temperature dependence of 3/2T υ⋅Δ (if no atoms/linear molecules. But always simple)

-‐ translational factor:

kBT 5/2

Po

⎛

⎝⎜⎞

⎠⎟

Δυ

α AυA

A∏ A

Aυ υΔ =∑

For both rotational and translation factors xT υΔ dependence reflects entropic contributions

0υΔ > → more species on product side → increase in entropy (# of states) upon reaction

-‐ vibrational + electronic factor:

These factors are numerically (by far) most important. Let us try to understand how the terms originate. Let us for definiteness consider a concrete reaction HCN CNH→ PES along reaction coordinate



67

We would make a harmonic oscillator model for reactant and products (3 normal modes each). The change in electronic energy is determined from the difference in energy at the bottoms of the well.

The zeropoint vibrations energy for each species A would be ( )12

AAzp i

iE ω=∑ h . Sum over

normal modes (here 4 for a linear molecules like HCN and HNC) In general energy difference for reaction can be written as sum over electronic energies at the respective minima and sum over zero point frequencies e zpE E EΔ = Δ + Δ

12

A AA e i A

A A iEυ ω υ⎛ ⎞= + ⎜ ⎟

⎝ ⎠∑ ∑ ∑ h

Kzp,e = e−ΔE /kBT

In this description of chemical equilibrium we limit ourselves to only considering non-‐degenerate ground states. Things will get slighty more complicated when degeneracies occur. Principles are the same.

For each species there would be in addition the factor qV ,rest = (1− e−!ω i /kBT )−1

i∏ . This

factor is usually very close to unity (1). In KV ,rest one would take the ratio of products

and reactants. I think it is clearest to write the electronic-‐vibrational contribution as

Ke ⋅Kzp ⋅KV ,rest

Ke = e−ΔEe /kBT Ae A o

AE EυΔ =∑



68

Kzp = e−ΔEzp /kBT

ΔEzp =

12ω i

i∈A∑⎛⎝⎜

⎞⎠⎟υA

A∑

Kv =

11− e−!ω i /kBT

i∈A∏⎛⎝⎜

⎞⎠⎟

υA

A∏

A : label for species i : normal mode for species A This clearly indicates the origin of the various terms K p = KeKzp Kt KR,nKV ,rest

This indicates the importance of various factors ~e zp t R vK K K K K>> > >

,e zpK K contribute to an exponential factor e−ΔE /kBT and this absolutely dominates the

equilibrium constant. Other factors depend on powers of T . Pressure dependence derives from

translational partition function lno

PRTP

. We have reached the essential usage of statistical

mechanics in chemistry. It is worthwhile to look at examples. These final results suggest that nowadays it is well possible to calculate gas phase molecular partition functions and equilibrium constants. We will go in the computer lab and run Gaussian calculations. We can also obtain the relevant data from the Gaussian program and do the statmech ourselves using a simple Matlab program. This way you can convince yourself of the possibilities of computational chemistry. In the Gaussian program all the statmech is done exactly as described here. Let me provide a somewhat more detailed discussion of the temperature dependence of the equilibrium constant. We know (neglecting the usually unimportant “rest of vibrational term) that

KP = Kel ,zp Kt ,P0KRn

Each of the individual K’s is a ratio of partition functions. This yields the following temperature dependences: a) Electronic + zero-‐point

Kel ,zp = e−ΔEel ,zp

rxn / RT, here

ΔEel ,zp

rxn is the reaction energy for 1 mole of reaction. This can

be calculated using a quantum chemistry program like Gaussian. This energy difference itself is rigorously independent of temperature. b) Translational motion

qt ,P0 ~ T 5/2 ,

Kt ,P0 ~ T Δν 5/2 , Δν = ν i

i∈reactants,products∑

c) Rotational

qR,n ~ T for linear molecules

qR,n ~ T 3/2 for non-linear molecules

It then follows that the equilibrium constant can be written generically as



69

KP = aT xe−ΔE / RT

Here a and x are constants that depend on the particular molecules involved. Hence:

−RT ln K p = −RT (ln a + x lnT −ΔEel ,zp

rxn

RT)

= ΔEel ,zprxn −T (R ln a + Rx lnT )

Form Thermodynamics we know −RT ln KP = ΔrG

0(T ) = Δr H −TΔrS

Often this expression is interpreted as ΔH is independent of temperature, while the other term depends linearly on Temperature using TΔrS . The Stat-‐mech expression is more pure.

ΔEel ,zprxn ≈ ΔH is rigorously independent of T

(R ln a + Rx lnT ) ≈ ΔS , has a weak temperature dependence

This is a non-‐standard treatment, but it provides insight in what is measured experimentally.

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