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Spring 2012 Chem 350: Statistical Mechanics and Chemical Kinetics

Chapter 5:Partition Functions and Properties of Real Molecules 1

Thermodynamics of a system of independent particles .......................................................................... 1

The Partition function .............................................................................................................................. 2

Origin of factor 1!N for identical particles .............................................................................................. 2

Reworking the Helmholtz free energy ..................................................................................................... 3

Translational Partition Function: Particle in the box wave function (atoms) .......................................... 4

Relating to thermodynamic properties ................................................................................................... 5

Statistical Mechanics of diatomic molecules ........................................................................................... 7

Vibrational Partition function in harmonic approximation ..................................................................... 8

Rotational Partition Function for a diatomic ......................................................................................... 11

The origin of the symmetry factor in rotational partition function ....................................................... 13

Polyatomic Systems ............................................................................................................................... 17

Chemical Reactions and Equilibrium ..................................................................................................... 20

Statistical Mechanics ............................................................................................................................. 21

Connect to thermodynamics ................................................................................................................. 22

Chapter 5: Partition Function and Properties for Real Molecules Thermodynamics of a system of independent particles -‐ neglect internal degrees of, in particular rotations, vibrations eg. Atoms, 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= + +

-‐ product of single particle wavefunction is eigenfunction -‐ sum of 1-‐particle eigen values → 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 wavefunction is product function too -‐ 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.

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 → anti symmetric, 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) and also only 1 symmetric function. → For 3 particles divide by 3! 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 eg. a bϕ ϕ= a b= , same wavefunction → For fermions: wavefunctions = 0 For bosons: different factor to count symmetric wavefunctions

Many more states than # of particles ( >> 2310 ) 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 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 lnA kT Q= − ( )ln ! ln ln 1N N N N N N= − = −

ln!

NqA kTN

⎛ ⎞= − ⎜ ⎟

⎝ ⎠ ( )ln ln ln NN N e N

e⎛ ⎞= − = ⎜ ⎟⎝ ⎠

( )ln ln !NkT q N= − − lnNN

e⎛ ⎞= ⎜ ⎟⎝ ⎠

ln lnN

N NkT qe

⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ !

NNNe

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

ln qeNkTN

⎛ ⎞= − ⎜ ⎟⎝ ⎠

ANkT nN kT nRT= = 8.314 /AN k R J molK= =

ln qeA nRTN

⎛ ⎞= − ⎜ ⎟⎝ ⎠



A contain terms that do not scale linearly with N , non interacting particles seem to interact! This is a consequence of (anti) – symmetry requirement of many-‐particle wavefunction. Later we will see ~q V .

ln qeA NkTN

⎛ ⎞= − ⎜ ⎟⎝ ⎠ ~ ln VNkT

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) e : electronic (important for open shell atoms) t : translational (kinetic energy)

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

2 2

2d E

m dxψ ψ− =h

sin n xLπψ =

22

sind n n xdx L Lψ π π⎛ ⎞= −⎜ ⎟⎝ ⎠

22

sin sin2

n n x n xEm L L L

π π π⎛ ⎞ =⎜ ⎟⎝ ⎠h

2 22

22nE nmLπ⎛ ⎞

= ⎜ ⎟⎝ ⎠

h 22

28h nmL

= 2hπ

=h

2

228

,1

h nmL kT

T Dn

q e−

=∑ 1,2,3...n =

Define 2

28hmL kT

Δ = and nx n= Δ 22

,1nxn

t Dn n

q e e−−Δ= =∑ ∑

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

( )nn

q f x=∑ ( )( )1( )n n n nn

f x dx f x x x+= = −∑∫

1 ( 1 1)n nx x n+ − = Δ + − = Δ

( ) ( )n nnf x f x dxΔ =∑ ∫

( ) 1 ( )n nnf x f x dx≈

Δ∑ ∫



This approx is correct if ( )nf x is smoothly varying Back to the partition function

2 2

,1 0

1 1 12 1

nx xt D

nq e e dx π∞− −= ≈ =

Δ Δ∑ ∫ (look up the integral)

2

,1 2

1 1 82 4 4t D

mkTL Lqh

π ππ= = = =Δ Δ Λ

242hm kTπ π

Λ = Δ = 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)

3 3/2

3/2 3/23 2 2

1 2 2mkT mk T Th h

π π α⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟Λ ⎝ ⎠⎝ ⎠

define 3/2

2

2 mkhπα ⎛ ⎞= ⎜ ⎟⎝ ⎠

(constant)

3/2,3 3t D

Vq VTα= =Λ

on nq g= (nuclear)

/iE kTe

iq e−=∑ (sum over states) (electronic, only for open-‐shell atoms)

/E kTg e αα

α

−=∑ (sum over energy levels)

/iE Eα : molecular energies, do not depend on ,N V ! Relating to thermodynamic properties

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

∂⎛ ⎞= =⎜ ⎟∂⎝ ⎠

ln lnqe qeA nRT NkTN N

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3/2

,

lnT V

A VT eNkTN N N

αµ⎛ ⎞⎛ ⎞∂ ∂⎛ ⎞= = −⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

3/2 1ln VT ekT NkTN N

α ⎛ ⎞⎛ ⎞= − + − ⋅ −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ lnkT e= −

3/2VTkT kTN

α− +

3/2

ln VTkTN

αµ = −

( )3/2 3/2 5/2

, , ln ln lnT NkT T kT kTN P T kT kT kTNP P P

α α αµ = − = − = −

5/2 5/2

ln lnA AT k T kN N kT RTP P

α αµ µ= = − = −

5/2

0

0

ln PT kG n nRTP P

αµ⎛ ⎞

= = − ⋅⎜ ⎟⎝ ⎠



5/2

0

0

ln ln PT kG nRT nRTP P

α⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

00 ln lno

o

P PG G nRT G nRTP P

⎛ ⎞⎛ ⎞= + = − ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

Check formula for S compound to thermodynamics:

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 = ⎜ ⎟

⎝ ⎠

1 2 T T→ , P constant 2

1

5 ln2

TS nRT

⎛ ⎞Δ = ⎜ ⎟

⎝ ⎠

1 2 P P→ , T constant 2

1

ln PS nRP

⎛ ⎞Δ = − ⎜ ⎟

⎝ ⎠

All of ideal gas thermodynamics follows from ln tq eA NkTN

⎛ ⎞= − ⎜ ⎟⎝ ⎠, ( )

!

NtqQN

= translational

partition function 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= ,

3/2

2

2 Mkhπα ⎛ ⎞= ⎜ ⎟⎝ ⎠

. 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. This

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

212harmE kx=

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

2 2

22

1ˆ2 2

dH kxdxµ

⎛ ⎞= − + ⎜ ⎟⎝ ⎠h

where 1 2

1 2

m mm m

µ =+

(reduced mass)

12nE n ω⎛ ⎞= +⎜ ⎟⎝ ⎠h 0,1,2...n = and

kωµ

=

( k is the force constant and depends on the molecule) (only need k 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: 12nE nω ω= +h h 0,1,2...n =

1 // /2n

kTE kT n kTe e eω ω−− −= ⋅h h

( ) ( )1 / /2

0,1,2..

kT n kTV

nq e e

ω ω− −

=

= ∑h h

Set ( )/kTy e ω−= h ( ) ( ) ( )2 / / / 2kT kT kTe e e yω ω ω− − −= =h h h so ( )/n kT ne yω− =h

( ) ( ) ( )1 1/ /2 32 2

0,1,21 ....

kT kT nV

nq e y y y e y

ω ω− −

=

= + + + + = ∑h h

( )2 3

0,1,2...

11 ....1

n

ny y y y

y=

= + + + + =−∑ (known math relation)



( )1 /2 1

1kT

Vq ey

ω− ⎛ ⎞= ⋅ ⎜ ⎟−⎝ ⎠

h

( )

( )

1 /2

/

11

kT

V kTq ee

ω

ω

−

−⎛ ⎞= ⋅ ⎜ ⎟−⎝ ⎠

h

h partition function with zero point energy included

Define vibrational temperature, VT kω= h → 1/J JK− = K (kelvins) so VT

T is dimensionless

( )

( )

1 /2

/

11

V

v

T T

V T Tq ee

−

−⎛ ⎞= ⋅ ⎜ ⎟−⎝ ⎠

For some molecules, the internal rotation can range from relatively small 1200cmω −≈h , to something like C-‐H stretching 13100cmω −≈h

Knowing 1 10.695k cm K− −= , VT kω= h can range from 400 – 6000K

( ) ( )1 1/ / //2 2 V V

kT T T n T Tn kTe e e eω ω− − −−⋅ = ⋅h h , since VT ranges from 400 – 6000K, a large T value is

needed for a decent population

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. More convenient: choose zero of energy as the energy of dissociated atoms → same energy scale for all molecules

12v eE D nω ω= − + +h h

0D n ω= − + h Results we obtain from partition function are independent of the overall shift in energy

11 // 22

eD kTkTe e

ωω⎛ ⎞−− ⎜ ⎟⎝ ⎠→

hh / oD kTe=



The vibrational partition function becomes ( )0 /

/

11 V

D kTV T Tq e

e−⎛ ⎞= ⋅ ⎜ ⎟−⎝ ⎠

If ( )/VT T is large (low T ) → 0 /D kTVq e=

If ( )/VT T is small (high T ) → 0 /D kTV

V

Tq eT

⎛ ⎞= ⎜ ⎟

⎝ ⎠ (high T limit)

Thermodynamic values

( )0 /

/

11 V

D kTV T Tq e

e−⎛ ⎞= ⋅ ⎜ ⎟−⎝ ⎠

( )( )0 / /ln ln 1 VD kT T TVA NkT e e−= − − −

( )/0 ln 1 VT TD N NkT e−= − + −

( ) ( ) ( )/ /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

( ) ( )/

/

/ln 11

v

V

T T VV T T

NkT TS Nk ee

−= − − +−

( )0 / 1V

VT T

NkTU A TS D Ne

= + = − +−

( )/0

,

ln 1 VT TV

T V

A D kT eN

µ −∂⎛ ⎞= = − + −⎜ ⎟∂⎝ ⎠

This is all exact, for harmonic oscillator Consider the large T limit of better / VT T is large

0 //

11 v

D kTT Te

e−⎛ ⎞⎜ ⎟−⎝ ⎠

/1 , 1 /VT TxVe x e T T−− ≈ − − ≈

0 /D kTV

V

Tq eT

⎛ ⎞= ⎜ ⎟

⎝ ⎠

0ln lnV VV

TA NkT q ND NkTT

⎛ ⎞= − = − − ⎜ ⎟

⎝ ⎠%

lnVV

TS Nk NkT

⎛ ⎞= + +⎜ ⎟

⎝ ⎠%

0 0V V VU A TS ND NkT ND nRT= + = − + = − +% %% VC Nk nR= =



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 mole contributes nRT to U , and nR to VC at high termperature” 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

2

2

ˆˆ2LHRµ

= 1 2

1 2

m mm m

µ =+

2L̂ : an angular momentum operator depending on ,θ ϕ , R : internuclear distance

Note: 2L̂ is the same operator that shows up in the H-‐atom orbital’s s,p,d,f functions ( ) ( ) ( )2 2ˆ , 1 ,m m

l lL y l l yθ ϕ θ ϕ= +h

.....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

( )2

2 12JE J JRµ

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

Has degeneracy ( )2 1Jg J= +

States: ( ),mJY θ ϕ , 1..... 1,m J J J J= − − + − ( ) 2 1J→ +

Energy levels are often expressed in 1cm− hch hckυλ

= = %

( )1JE BJ J= +%2 28hc Rπ µ

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

A convenient conversion: 11 8065.5eV cm−≈ Rotational partition function

( ) ( )1 /

0,1,2..2 1 BJ J kT

RJ

q J e− +

=

= +∑

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

( ) ( )1 /02 1 Bx x kT

Rq x e dx∞ − +≈ +∫

%%

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

( )2 1dy x dx= +

/ /

0 0

By kT By kTR

kT kTq e dy eB B

∞∞ − −⎡ ⎤= = − =⎣ ⎦∫ % %% %% %

% %

RR

kT TqTB

= =%% where B%, k% are in units of 1cm−

RBTk

= Rotational temperature (in Kelvins)

This formula is “correct” for heteronuclear diatomics with heavy masses like CO , but for homonuclear case, like 2H it is off by a factor of 2. Correcting for this

RR

kT TqB Tσ σ

= =

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

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

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

!

NqQN

= .

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

lnRR

TA nRTTσ

= −

,

lnRR

V N R

A TS nR nRT Tσ

∂⎛ ⎞= − = +⎜ ⎟∂⎝ ⎠

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

,

lnRR

T V R

A TRTN T

µσ

∂⎛ ⎞= − = −⎜ ⎟∂⎝ ⎠



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 2 1J + ) ,J Jmψ

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:



( ) ( )1 2α α

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

= triplet ( ) ( )1 2β β Parahydrogen ( ) ( ) ( ) ( )1 2 1 2 2α β β α⎡ + ⎤⎣ ⎦ 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

Symmetric ( )1 2 2 1 2mm m m+

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

I I I I+ + = + + (symmetric functions) 1 3 2 32 2

I = → ⋅ =

Antisymmetric ( )1 2 2 1 2mm m m−

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

I I I I+ = + (functions) 1 1 2 12 2

I = → ⋅ = (see above)

(compare symmetric ( )1 12n n + and antisymmetric ( )1 1

2n n − , including diagonal)

Symmetry of spin function under permutation is understood. What about spatial part of nuclear wave function? Consider the nuclear coordinates 1 2,R R

r r

1 2

2 cmR R R+ =

r (for identical nuclei)

1 2R R−r r

= sin cosR θ ϕ sin sinR θ ϕ cosR θ

2 1R R R= −r r

Nuclear wavefunction:

( ) ( ) ( ),t cm V RR Rψ ψ ψ θ ϕ⋅ ⋅r



If 1 2R R↔r r

then cmRr

and R are unaffected

However: ( ) ( )12 2 1 2 1P R R R R− = − −r r r r

= sin cosR θ ϕ− sin sinR θ ϕ− cosR θ− Interchanging nuclei 1 2R R↔

r r is equivalent to

θ π θ→ − ( ) ( )cos cosπ θ θ− = −

( ) ( )sin sinπ θ θ− =

ϕ ϕ π→ + ( ) ( )cos cosϕ π ϕ+ = −

( ) ( )sin sinϕ π ϕ+ = −

Hence interchanging 1Rr and 2R

r 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)

s, d, g functions are even under inversion p, f functions are odd under inversion 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

For 2H (fermions, antisymmetric) 12

I =

( ) ( ) ( ) ( )2 1 / 1 /

1,3,5.. 0,2,4..3 2 1 1 2 1BJ J kT BJ J kTH

nRJ J

q J e J e− + − +

= =

= + + +∑ ∑

For 2D (bosons, symmetric) 1I =

( ) ( ) ( ) ( )2 1 / 1 /

0,2,4.. 1,3,5..6 2 1 3 2 1BJ J kT BJ J kTD

nRJ J

q J e J e− + − +

= =

= + + +∑ ∑

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)

( ) ( )1 /

0,1,2..6 2 1 BJ J kTHD

nRJ

q J e− +

=

= +∑

In general for homonuclear diatomics with spin I Fermion: ( )( ) ( )( )1 2 1 2 1Odd Even

R 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): ( ) ( )1 /

0,2,42 1 BJ J kT

JJ e− +

=

+∑ 2J k→ =

( ) ( )2 2 1 /

0,1,2..4 1 B k k kT

kk e− +

=

= +∑ 24 2y k k= + , 8 2dy k= +

( )/

0

1 1 2 2

y B kT kTe dyB

∞ −→ =∫

Similarly:

( ) ( )1 /

1,3,52 1 BJ J kT

JJ e− +

=

+∑ 2 1J k= + , 0,1,2..k =



( )( ) ( )( )2 1 2 2 /

1,3,52 2 1 1 B k k kT

Jk e− + +

=

+ +∑

( )( )2 1 2 2y k k= + + , ( ) ( )2 2 2 2 2 1dy k k= + + +

( )/

0

1 1 2 2

y B kT kTe dyB

∞ −→ =∫

→ Same high temperature limit This analysis is quite involved, We will do a simulation in Matlab to clarify

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 r R E R r Rψ ψ=r r rr r

3N→ dimensional potential energy surface (PES)

0j

ER∂ =∂

→ extreme on PES

Different isomers: different minima on PES Transition State: Saddle points on PES (Max in one direction, min in all others) Taylor series of potential energy surface around minimum eR

r

( ) ( ) ( ) ( ) ( )21

2e e

e e e ei i ji ii i jR R R R

E EE R E R R R R R R RR R R

= =

∂ ∂= + − + − −∂ ∂ ∂∑ ∑

r r r r

r r r r r r r r

Extremum → 0i

ER∂ =∂

Mass-‐weighted Hessian



2

e

ij i ji j R R

EH M MR R

=

∂=∂ ∂ r r

(to account for nuclear masses in nuclear kinetic energy term) Diagonalize ijH

6 (5) eigenvalues are 0: correspond to overall translation, overall rotation. 3 6N − (3 5N − ) eigenvalues iε of Hessian correspond to normal mode i .

By diagonalizing the Hessian the vibrational problem is reduced to 3 6N − independent harmonic oscillator problems

( ) ( ) ( )2

22

1 12 2

ii i n i n i

i

d q q E qdq

ε χ χ⎡ ⎤− + =⎢ ⎥⎣ ⎦

Define 1i

iεω = (analog of i

km

ω = , with 1m = )

( ) 12

in i iE n ω⎛ ⎞= +⎜ ⎟⎝ ⎠

h

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

,1

cm k kk

j

R m Rm

αα= ∑

∑r

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> >

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)



( )1/21/2 1/2

RA B C

T T Tq TT T T

πσ

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2

2AA

TI

= h 2I Rµ=

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

TqTσ

= , 2

2RT I= h (2

rotational degrees of freedom)

Symmetry Factorσ : # of pure rotations in the point group of the molecule, known from group theory. (# of rotations that map molecule onto itself).

2H O σ = 2 4CH 3 4 12σ = × =

3NH σ = 3 6 6C H 12σ = Overall partition function for polyatomic molecule: ind

t R v n eq q q q q q=

Translational: 3/2tVq TNα=

3/2

2

2 Mkπα ⎛ ⎞= ⎜ ⎟⎝ ⎠h j

jM m=∑

Vibrational:

1 /3 6 2

1 /1 21

i

i

h kTN

vh kTi

eqe

ω

ω

−−

−=

=−

∏

One factor for each vibrational mode, including zeropoint frequency

Rotational:

11 122 2

RA B C

T T TqT T T

πσ

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2

2AA

TI

= h etc.

Nuclear Spin ( )2 1nq Iα

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

Electronic: /eD kTe or /iE kT

ie−Δ∑

eD : atomization energy, bottom of well → separated atoms 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, internal rotation for example in ethane 3 3CH CH− Potential along torsional mode



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

This can be included (neglect mode-‐coupling). There are problems for floppy molecules though. Chemical Reactions and Equilibrium

Consider reactions in gas phase a) Thermodynamics:

Prototype reaction: aA bB cC dD+ +Ä

, , ,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)

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 principle theory of chemical equilibrium For previous example: aA bB cC dD+ +Ä

( ) ( )( ) ( )/ /

/ /

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 Ax P P υυ= ⋅∏ ∏

( ) /AA o

Ax P P υυ Δ= ⋅∏ A

Aυ υΔ =∑

( )/P x oK K P P υΔ=

( ) ( )/A

x A p oA

K x K P Pυ υ−Δ= =∏

Statistical Mechanics Chemical potential = Gibbs Free energy G A PV A NkT= + = + (ideal gas)

ln!

NqkT NkTN

⎛ ⎞= − +⎜ ⎟

⎝ ⎠

( )ln lnNkT q kT N N N NkT= − + − + ln lnNkT q kTN N= − +

( ) ( )ln / ln /NkT q N nRT q N= − = −

tv R n e

qq q q q qN N

= ⋅

3/2tM

q V TN N

α= 3/22

2MhMk

απ

−⎛ ⎞

= ⎜ ⎟⎝ ⎠

PV NkT= → V kTN P

=

5/2

t Mq kTN P

α=



//

11

oD kTv h kTq e

e ω−=−

(diatomic)

, ,R n eq q q : can all be calculated for each species in chemical reactions

tqN

determines the pressure dependence

5/2

ln ln lnt M

o o

q kT PnRT nRT nRTN P P

α ⎛ ⎞⎛ ⎞− = − + ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

5/2

, ln Mt o

o

kTq nRTP

α= − → enters reactionGΔ

Connect to thermodynamics

( )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⎛ ⎞= ⎜ ⎟⎝ ⎠

∏

( ) ( ) ( ) ( ),A

A A A AAt o A A A A

v R n eA

qq q q q

N

υυ υ υ υ⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠∏

,t o v R n eK K K K K= each factor is a ratio of corresponding q ’s Let us analyze different factors:

-‐ nuclear 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 → 1A

Rq =

For diatomics use RR

TqTσ

= (good enough)

Polyatomics:

11 122 2

RA B C

T T TqT T T

πσ

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

-‐ temperature dependence of 3/2T υ⋅Δ (if no atoms/diatomics)



-‐ translational factor: 5/2

AA

Ao

kTP

υυα

Δ⎛ ⎞⎜ ⎟⎝ ⎠

∏ AA

υ 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 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

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 would be ( )12

AAzp i

iE ω=∑ h . Here

1,...3i = . Sum over normal modes 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

( ) /,o E kTv eK e−Δ=

( )o : indicates the contribution due to the ground states of various species



For each species there would be in addition the factor , /

11 i

dv x kT

iq

e ω−=−∏ h . This factor is

very close to unity (1).

I think it would be clearest to write the contribution as e zp vK K K⋅ ⋅

/eE kTeK e−Δ= A

e A oA

E EυΔ =∑

/zpE kTzpK e−Δ=

,

12zp i A

A i AE ω υ

⎛ ⎞Δ = ⎜ ⎟

⎝ ⎠∑ ∑ h

/,

11

A

iv kTA i A

Ke

υ

ω−

⎛ ⎞= ⎜ ⎟−⎝ ⎠∏ ∏ h

A : label for species i : level for normal mode This clearly indicates the origin of the various terms p e zp t R vK K K K K K=

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 kTe−Δ and this absolutely dominates the

equilibrium constant. Other factors depends on power of T . Pressure dependence derives from

translational partition function lno

PRTP

. We have reached the essential usage of statistical

mechanics in chemistry. Worthwhile to look at examples

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