slide 1 chapter 4 rigid-rotor models and angular momentum eigenstates

Chapter 4

Rigid-Rotor Models andAngular Momentum Eigenstates

Outline

• Math Preliminary: Products of Vectors

• Rotational Motion in Classical Physics

• The 3D Quantum Mechanical Rigid Rotor

• Angular Momentum in Quantum Mechanics

• Angular Momentum and the Rigid Rotor

• The 3D Schrödinger Equation: Spherical Polar Coordinates

• Rotational Spectroscopy of Linear Molecules

Not Last Topic

Outline (Cont’d.)

• Application of QM to Molecular Structure: Pyridine

• Statistical Thermodynamics: Rotational contributions to the thermodynamic properties of gases

Mathematical Preliminary: Products of Vectors

x y zA A i A j A k

x y zB B i B j B k

Scalar Product (aka Dot Product)

x x y y z zA B A B A B A B

Note that the productis a scalar quantity(i.e. a number)

c o s ( )A B A B

Magnitude:

Parallel Vectors: 0c o s ( 0 )A B A B

x y zA A i A j A k

x y zB B i B j B k

Cross Product

The cross product of two vectors is also a vector.

Its direction is perpendicular to both A and B and is given by the “right-hand rule”.

s i n ( )A x B A B

Magnitude:

Parallel Vectors: 0s i n ( 0 )A x B A B

0s i n ( 9 0 )A x B A B

Perpendicular Vectors:

x y zA A i A j A k

x y zB B i B j B k

AxB A A A

y z x yx z

A A A AA AA xB i j k

B B B BB B

Expansion byCofactors

( ) ( ) ( )y z z y z x x z x y y xA x B A B A B i A B A B j A B A B k

Outline

Rotational Motion in Classical Physics

s i n ( )L r p Magnitude:

Angular Momentum (L)

Circular Motion: 0s i n ( 9 0 )L r p r p

or: 2 2( )v

L rp rm v m r m rr

L I ω where 2 vI m r

Energy

2 2 2( )

m r m r

2 2( )

Momentof Inertia

AngularFrequency

Comparison of Equations for Linear and Circular Motion

Linear Motion Circular Motion

Mass 2I m r Moment of inertiam

Velocityv

r Angular velocityv

Momentum L I Angular momentump=mv

Energy2

I Energy

Energy

Modification: Rotation of two masses about Center of Mass

m1 m1r

2I m r 2I r

where 1 2

Outline

Angular Momentum in Quantum Mechanics

L r x p

Classical Angular Momentum

( ) ( ) ( )z y x z y xL y p z p i z p x p j x p y p k

r x i y j z k

x y zp p i p j p k

x z yL y p z p

y x zL z p x p

z y xL x p y p

Angular Momentum in Quantum Mechanics

QM Angular Momentum Operators

x z yL y p z p

y x zL z p x p

z y xL x p y p

ˆ xp ii x x

ˆ yp ii y y

ˆ zp ii z z

Classical QM Operator

ˆxL i y z

ˆyL i z x

ˆzL i x y

2 ˆ ˆ ˆ ˆ ˆ ˆx x y y z zL L L L L L L

Operator Commutation and Simultaneous Eigenfunctions

It can be shown that: ˆ ˆ, 0x yL L Do not commute

ˆ ˆ, 0y zL L Do not commute

ˆ ˆ, 0x zL L Do not commute

Because the operators for the individual components do not commute,one cannot determine two separate components simultaneously.

i.e. they cannot have simultaneous eigenfunctions.

In contrast, it can be shown that:2ˆ , 0zL L

^ Do commute

Because these operators commute, one can determine Lz and L2

simultaneously; i.e. they can have simultaneous eigenfunctions.

Outline

The 2D Quantum Mechanical Rigid Rotor

Assume that two masses are attached by a rigid rod (i.e. ignorevibrations) at a fixed distance, r,and are free to rotate about the Center of Mass in their x-y plane.

The angle represents the angle of rotation relative to the x-axis.

The 2D Schrödinger equation for therelative motion of two masses is:

2 2x y

Two Dimensional Laplacianin Cartesian Coordinates

If one (a) converts the Laplacian to polar coordinates

(b) assumes that the potential energy is constant (arbitrarily 0)

(c) holds r fixed (i.e. neglects derivatives with respect to r)

It can be shown that the Schrödinger Equation for a 2D RigidRotor becomes: 2 2

where 2I r is the moment of inertia

The Solution2 2

C o n s t a n t

2IEm or

Note: So far, m can have any value;

i.e. there is no energy quantization

Assume

( ) imdim Ae

0, 1, 2, 3,2

Application of the Boundary Conditions:Quantization of Energy

To be a physically realistic

solution, one must have: ( 2 ) ( )

Therefore: 2i m m i i mA e e A e

or 2 1m ie

c o s ( 2 ) s i n ( 2 ) 1m i m

This is valid only for: 0 , 1 , 2 , 3 ,m

Therefore, only certain values for the energy are allowed;i.e. the energy is quantized:

Zero Point Energy

0, 1, 2, 3,2

There is no minimum Zero Point Energy.

One encounters a ZPE only when the particle is bound (e.g. PIB,Harmonic Oscillator, Hydrogen Atom), but not in freely movingsystems (e.g. 2D and 3D Rigid Rotor, free particle)

Application of the 2D Rigid Rotor

We have solved the 2D Rigid Rotor primarily as a learning exercise,in order to demonstrate the application of angular Boundary Conditions.

However, the model has a real world application, in that it can be used to characterize the rotation of molecules adsorbed on surfaces.

Example

When an H2 molecule is chemisorbed on a crystalline surface, itsrotation can be approximated as that of a 2D rigid rotor.

The H2 bond length is 0.74 Å Calculate the frequency (in cm-1) of the lowest energy rotational transition of chemisorbed H2.

0, 1, 2, 3,2

Ir = 0.74 Å = 0.74x10-10 m1 amu = 1.66x10-27 kgħ = 1.05x10-34 J•sh = 6.63x10-34 J•sc = 3.00x1010 cm/s

2(1 )0 .50

amuamu

amu amu

2 72 81 .6 6 1 0

8 .3 0 1 0x kg

x kgam u

22 2 8 1 08 .3 0 1 0 0 .7 4 1 0I r x k g x m 4 8 24 . 5 5 1 0x k g m

2 2 2 22 1

m1 = 0

m2 = 12

1.05 10

2 4.55 10

x kg m

2 11 . 2 1 1 0x J

1.21 10

6.63 10 3.00 10 /

x J s x cm s

16 0 . 9 c m

Outline

The Three Dimensional Schrödinger Equation

In Cartesian Coordinates, the 3D Schrödinger Equation is:

2 2 2 2 2 2

2 2 2( , , )

2 2 2V x y z E

m x m y m z

( , , )x y z

The Laplacian in Cartesian Coordinates is: 22 2 2x y z

T(x) T(y) T(z) V(x,y,z)

Therefore:2

2 ( , , )2

V x y z Em

It is sometimes not possible to solve the Schrödinger exactly inCartesian Coordinates (e.g. the Hydrogen Atom), whereas itcan be solved in another coordinate system.

The “Rigid Rotor” and the Hydrogen Atom can be solved exactlyin Spherical Polar Coordinates.

Spherical Polar Coordinates

To specify a point in space requires three coordinates.In the spherical polar coordinate system, they are:

r 0 r < Distance of point from origin (OP)

0 < Angle of vector (OP) from z-axis

0 < 2 Angle of x-y projection (OQ) from x-axis

Relation of Cartesian to Spherical Polar Coordinates

cos( )z

c o s ( )z r

x-axis

y-axis

cos( )x

c o s ( )x O Q

s i n ( ) c o s ( )x r

sin( )y

s i n ( )y O Q

s i n ( ) s i n ( )y r

OQ=rsin()

The Volume Element in Spherical Polar Coordinates

In Cartesian Coordinates,

the volume element is: d V d x d y d z

In spherical polar coordinates,

the volume element is: d V d r r d O Q d

s i n ( )d V d r r d r d

2 s i n ( )d V r d r d d

The Laplacian in Spherical Polar Coordinates

22 2 2x y z

Cartesian Coordinates:

One example of a chain rule formula connecting a derivative withrespect to x, y, z to derivatives with respect to r, , is:

x r x x x

It may be shown that by repeated application of chain rule formulaeof this type (with 2-3 hours of tedious algebra), the Laplacian inspherical polar coordinates is given by:

2 2 2 2 2

1 1 1sin ( )

sin ( ) sin ( )r

r r r r r

Angular Momentum Operators in Spherical Polar Coordinates

ˆzL i x y

It may beshown that

ˆzL i

2 ˆ ˆ ˆ ˆ ˆ ˆx x y y z zL L L L L L L

It may beshown that

1 1sin( )

sin( ) sin ( )L

Outline

The 3D Quantum Mechanical Rigid Rotor

3D Schrödinger Equation for a particle (Sph. Pol. Coords.)

2 2 2 2 2

1 1 1sin( )

2 sin( ) sin ( )r V E

m r r r r r

22 ( , , ) ( , , )

2V r E r

Modification: Two masses moving relative to their CM

m1 m1r

2 2 2 2 2

1 1 1sin( )

r r r r r

2 2 2 2 2

1 1 1sin( )

r r r r r

The Schrödinger Equation in terms of the L2 operator^

1 1sin( )

sin( ) sin ( )L

The L2 operator is:^

2 2 22

2 2 2 2

1 1 1sin( )

2 2 sin( ) sin ( )r V E

r r r r

2 2r L V E

r r r I

2I rwhere^

Radial KE Rotational KE

The Quantum Mechanical Rigid Rotor

2 2 22

2 2 2 2

1 1 1sin( )

2 2 sin( ) sin ( )r V E

r r r r

The Rigid Rotor model is used to characterize the rotation ofdiatomic molecules (and is easily extended to linear polyatomic molecules)

It is assumed that: (1) The distance between atoms (r) does not change.

(2) The potential energy is independent of angle [i.e. V(,) = Const. = 0]

2 2r L V E

r r r I

Therefore:

1 1sin( )

2 sin( ) sin ( )E

1 1sin( )

2 sin( ) sin ( )E

This equation can be separated into two equations, one containing only and the second containing only .

Assume: ( , ) ( ) ( )

Solution of the Rigid Rotor Schrödinger Equation

Algebra + Separation of Variables

sin( ) sin( ) sin ( )2

We will only outline the method of solution.

Solution of the equation is rather simple.

However, solution of the equation most definitely is NOT.

Therefore, we will just present the results for the quantum numbers,energies and wavefunctions that result when the two equations are solved and boundary conditions are applied.

sin( ) sin( ) sin ( )2

The Rigid Rotor Quantum Numbers and Energies

The Quantum Numbers:

Note that because this is a two dimensional problem, thereare two quantum numbers.

0 , 1 , 2 , 3 , 0 , 1 , 2 , ,m

The Energy:2

Note that the energy is a function of l only. However, there are2 l + 1 values of m for each value of l . Therefore, the degeneracyof the energy level is 2 l + 1

Remember that for a classical Rigid Rotor:2

Comparing the expressions, one finds for the

angular momentum, that: ( 1)L

An Alternate Notation

The Quantum Numbers: 0 , 1 , 2 , 3 ,J

0 , 1 , 2 , ,M J

The Energy:2

( 1)2JE J JI

2 1Jg J

When using the Rigid Rotor molecule to describe the rotationalspectra of linear molecules, it is common to denote the twoquantum numbers as J and M, rather than l and m.

With this notation, one has:

The Wavefunctions

When both the and differential equations have been solved,the resulting wavefunctions are of the form:

,( , ) ( ) ( ) ( )mimmN e P

The are known as the associated Legendre polynomials.( )mP

The first few of these functions are given by:

01 c o s ( )P

11 s in ( )P

13 cos ( ) 1

1sin( ) cos( )

2 22 s in ( )P

We will defer any visualization of these wavefunctions untilwe get to Chapter 6: The Hydrogen Atom

Spherical Harmonics

( , ) ( ) ( ) ( )mimlm mY N e P

The product functions of and are called “Spherical Harmonics”,Ylm(, ):

They are the angular solutions to the Schrödinger Equation for anyspherically symmetric potential; i.e. one in which V(r) is independentof the angles and .Some examples are:

0 01 0 1 0 1 1 0( , ) ( ) c o s ( )iY N e P N

11 1 1 1 1 1 1( , ) ( ) s i n ( )i iY N e P N e

0 0 22 0 2 0 2 2 0( , ) ( 0 ) 3 c o s ( ) 1iY N e P N

12 1 2 1 2 2 1( , ) ( ) s i n ( ) c o s ( )i iY N e P N e

2 2 2 22 2 2 2 2 2 2( , ) ( ) s i n ( )i iY N e P N e

11 1 1 1 1 1 1( , ) ( ) s i n ( )i iY N e P N e

12 1 2 1 2 2 1( , ) ( ) s i n ( ) c o s ( )i iY N e P N e

2 2 2 22 2 2 2 2 2 2( , ) ( ) s i n ( )i iY N e P N e

( , ) sin( )iY Ne One of the Spherical Harmonics is:

Show that this function is an eigenfunction of the Rigid RotorHamiltonian and determine the eigenvalue (i.e. the energy).

cos( )iYNe

sin( ) sin( ) cos( )iYNe

2 2s in ( ) co s ( )iN e

21 2 s in ( )iN e

1sin( ) 2 sin( )

sin( ) sin( )

iiY Ne

1 1sin( )

2 sin( ) sin ( )

2HY L Y EY

( , ) s i n ( )iY N e

1 1sin( )

2 sin( ) sin ( )

1sin( ) 2 sin( )

sin( ) sin( )

iiY Ne

sin( )iYiNe

2sin( )iY

sin ( ) sin( )

1 1sin( )

2 sin( ) sin ( )

2 sin( )2 sin( ) sin( )

i iiNe Ne

2 sin( )2

Therefore:2

Note: Comparing to:2

1we see that:

Outline

Angular Momentum and the Rigid Rotor

The Spherical Harmonics, Ylm(, ), are eigenfunctions of theangular momentum operators:

1 1sin( )

sin( ) sin ( )L

ˆzL i

Note: It is straightforward to show that L2 and Lz commute;

i.e. [L2,Lz] = 0.

Because of this, it is possible to find simultaneouseigenfunctions of the two operators which are, as shown above,the Spherical Harmonics.

2 2( , ) ( 1 ) ( , )l m lmL Y Y ^

The eigenvalues are given by the equations:

ˆ ( , ) ( , )z lm lmL Y m Y

As discussed earlier, the restrictions on the quantum numbersare given by:

0 , 1 , 2 , 3 , 0 , 1 , 2 , ,m

Therefore, both the magnitude, |L|, and the z-component, Lz, of theangular momentum are quantized to the values:

( 1 0 , 1, 2 , 3 ,L

0 , 1 , 2 , ,zL m m z-axis

Lz |L|

If a magnetic field is applied, its directiondefines the z-axis.

If there is no magnetic field, the z-directionis arbitrary.

( , ) s i n ( )iY N e One of the Spherical Harmonics is:

1 1sin( )

sin( ) sin ( )L

Show that this function is an eigenfunction of L2 and Lz and determinethe eigenvalues.

We’ve actually done basically the first part a short while ago.

Remember: 21

2HY L Y EY

Therefore: 2 22L Y Y ^ 2( 1 ) Y 1

sin( )ii N ei

Preview: The Hydrogen Atom Schrödinger Equation

2 2 2 2 2

1 1 1sin( )

m r r r r r

3D Schrödinger Equation in Spherical Polar Coordinates

The Hydrogen atom is an example of a “centrosymmetric” system,which is one in which the potential energy is a function of only r, V(r).

In this case, the Schrödinger equation can be rearranged to:

2 2 22

2 2 2 2 2

1 1 1( ) sin( )

2 2 sin( ) sin ( )r V r E

m r r r m r r

Radial Part Angular Part

Note that the Angular part of the Hydrogen atom Schrödinger equationis the same as Rigid Rotor equation, for which the radial part vanishes.

Therefore, the angular parts of the Hydrogen atom wavefunctions arethe same as those of the Rigid Rotor

Outline

Rotational Spectroscopy of Linear Molecules

Energy Levels2

( 1)2JE J JI

2 1Jg J

Equivalent Form:2

hE J J

hE hc J J

( 1 )JE h c B J J

RotationalConstant (cm-1):

Note: You must use c in cm/s, even when using MKS units.

0 0 g0=1

1 g1=3B~

J EJ gJ

2 g2=5B~

3 g3=7B~

4 g4=9B~

Diatomic versus Linear Polyatomic Molecules

In general, for linear molecules, the moment of inertia is given by:

N is the number of atomsmi is the mass of the atom iri is the distance of atom i from the Center of Mass.

If N=2 (diatomic molecule) the moment of inertia reduces to:

2I r1 2

where r is the interatomic distance

Selection Rules

Absorption (Microwave) Spectroscopy

For a rotating molecule to absorb light, it must have a permanentdipole moment, which changes direction with respect to the electric vector of the light as the molecule rotates.

J = 1 (J = +1 for absorption)

e.g. HCl, OH (radical) and O=C=S will absorb microwave radiation.

O=C=O and H-CC-H will not absorb microwave radiation.

Rotational Raman SpectroscopyFor a rotating molecule to have a Rotation Raman spectrum, thepolarizability with respect to the electric field direction must change as the molecule rotates. All linear molecules have Rotational Raman spectra.

J = +2: Excitation (Stokes line)

J = -2: Deexcitation (Anti-Stokes line)

Intensity of Rotational Transitions

The intensity of a transition in the absorption (microwave) orRotational Raman spectrum is proportional to the number of moleculesin the initial state (J’’); i.e. Int. NJ’’

Boltzmann Distribution:''

kTJ JN g e

''( '' 1)

'' (2 '' 1)hcBJ J

kTJN J e

Rotational Spectra

0 0 g0=1

1 g1=3B~

J EJ gJ

2 g2=5B~

3 g3=7B~

4 g4=9B~

2 0Absorption (Microwave) Spectra

' ''J JE E E

J’’ J’

'( ' 1 ) ''( '' 1 )E h c B J J h c B J J

J’ = J’’+1

( '' 1 ) ( '' 2 ) ''( '' 1 )E h c B J J h c B J J

( 2 '' 2 ) '' 0 , 1 , 2 , 3 ,E h c B J J

(2 '' 2 ) '' 0 , 1, 2 , 3,E

B J Jhc

0 0 g0=1

1 g1=3B~

J EJ gJ

2 g2=5B~

3 g3=7B~

4 g4=9B~

2 0Rotational Raman Spectra

' ''J JE E E

J’’ J’

'( ' 1 ) ''( '' 1 )E h c B J J h c B J J

J’ = J’’+2

( '' 2 ) ( '' 3 ) ''( '' 1 )E h c B J J h c B J J

( 4 '' 6 ) '' 0 , 1 , 2 , 3 ,E h c B J J

(4 '' 6 ) '' 0 , 1, 2 , 3,E

B J Jhc

The HCl bond length is 0.127 nm.

Calculate the spacing between lines in the rotational microwave

absorption spectrum of H-35Cl, in cm-1.h = 6.63x10-34 J•sc = 3.00x108 m/sc = 3.00x1010 cm/sNA = 6.02x1023 mol-1

k = 1.38x10-23 J/K1 amu = 1.66x10-27

0.9721 35

amu amuamu

amu amu

2I r 22 7 91 .6 1 1 0 0 .1 2 7 1 0x k g x m 4 7 22 . 6 0 1 0x k g m

2 47 2 10

6.63 10

8 3.14 2.60 10 3.00 10 /

x kg m x cm s

As discussed above, microwave absorption lines occur at 2B, 4B, 6B, ...

Therefore, the spacing is 2B

0 . 9 7 2 a m u 27

271.66 101.61 10

x kgx kg

1 11 0 . 7 8 1 0 . 8c m c m

1S p a c i n g 2 2 1 0 . 8 2 1 . 6B x c m

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K

B = 10.8 cm-1~

Calculate the ratio of intensities (at 250C): 3 4

0 0 g0=11 g1=3B

J EJ gJ

2 g2=5B~

3 g3=7B~

4 g4=9B~

hcB kT

3hcB kTe

34 10 1

10 6.63 10 3.00 10 / 10.810

(1.38 10 / ) 298

x J s x cm s cmhcB

kT x J K K

= 0.52

0.523 4

Note: This is equivalent to asking for the ratio of intensites offourth line to the first line in the rotational microwave spectrum.

The first 3 Stokes lines in the rotational Raman spectrum of 12C16O2 are found at 2.34 cm-1, 3.90 cm-1 and 5.46 cm-1.

Calculate the C=O bond length in CO2, in nm.

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K1 amu = 1.66x10-27 kg

0 0 g0=11 g1=3B

J EJ gJ

2 g2=5B~

3 g3=7B~

4 g4=9B~

2012 . 3 4 6 0 6c m B B

13 . 9 0 1 2 2 1 0c m B B B

15 . 4 6 2 0 6 1 4c m B B B

10 .3 9B c m

2 1 10

6.63 10 /

8 3.14 0.39 3.00 10 /

x kg m sI

cm x cm s

4 6 27 . 1 8 1 0x k g m

rCO rCO

The first 3 Stokes lines in the rotational Raman spectrum of 12C16O2 are found at 2.34 cm-1, 3.90 cm-1 and 5.46 cm-1.

Calculate the C=O bond length in CO2, in nm.

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K1 amu = 1.66x10-27 kg

4 6 27 . 1 8 1 0I x k g m

I m r C OO 22 20O C O C O C Om r m m r 22 O C Om r

16 1.66 10 /

2.66 10

Om amu x kg amu

7.18 10

2 2.66 10CO

x kg mr

1 01 .1 6 1 0x m 0 . 1 1 6 1 . 1 6n m A n g s t r o m s

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K

Calculate the initial state (i.e. J’’) corresponding to the mostintense line in the rotational Raman spectrum of 12C16O2 at 25oC.

Hint: Rather than calculating the intensity of individualtransitions, assume that the intensity is a continuousfunction of J’’ and use basic calculus.

B = 0.39 cm-1~

'' '' 2 ''J J JI N '' '' 1

2 '' 1J J hcB

2'' ''2 '' 1

J J hcBJ e

NJ'' is at a maximum for dNJ''/dJ''=0.

2 2'' '' '' '''' 0 2 '' 1 2 '' 1

'' '' ''

J J J JJdN d dJ e e J

dJ dJ dJ

2 2'' '' '' ''0 2 '' 1 2 '' 1 2

J J J JJ e J e

2'' '' 2

0 2 '' 1 2J J

Therefore: 22 '' 1 2 0J

22 '' 1J

2'' '' 2

0 2 '' 1 2J J

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K

B = 0.39 cm-1~

34 10 1

2 1.38 10 / 2982 '' 1

6.63 10 3.00 10 / 0.39

x J K KJ

x J s x cm s cm

1 0 6 0 3 2 .6

32.6 1''

1 5 . 8 1 6

Consider the linear molecule, H-CC-Cl.

There are two major isotopes of chlorine, 35Cl (~75%) and37Cl (~25%). Therefore, one will observe two series of linesin the rotational spectrum, resulting from transitions ofH-CC-35Cl and H-CC-37Cl.

Can the structure of H-CC-Cl be determined from these two series?

No. There are 3 bond distances to be determined, but only 2 moments of inertia.

What additional information could be used to determine all threebond distances?

The spectrum of D-CC35Cl and D-CC37Cl

Non-Linear Molecules

Non-linear molecules will generally have up to3 independent moments of inertia, Ix, Iy, Iz.

The Hamiltonian will depend upon the angularmomentum about each of the 3 axes.

2 2 2yx z

The Schrödinger Equation for non-linear rotors is more difficult to solve,but can be done using somewhat more advanced methods, and therotational spectra can be analyzed to determine the structure(sometimes requiring isotopic species).

For small to moderate sized molecules (I would guess 10-15 atoms),rotational microwave spectroscopy is the most accurate method fordetermining molecular structure.

Outline (Cont’d.)

N 1.35 Å1.34

1.40 Å1.38

1.39 Å1.38

The Structure of Pyridine

Calculated: MP2/6-31G(d) – 4 minutes

Experimental: Crystal Structure

#MP2/6-31G(d) opt freq Pyridine 0 1 C -1.236603 1.240189 0.000458 C -1.236603 -0.179794 0.000458 C -0.006866 -0.889786 0.000458 C 1.222870 -0.179794 0.000458 C 1.053696 1.280197 0.000458 N -0.104187 1.989731 0.000458 H 1.980804 1.872116 -0.008194 H 2.205566 -0.673935 -0.009628 H -0.006866 -1.989731 -0.009064 H -2.189194 -0.729767 -0.009064 H -2.205551 1.760818 0.009628

The Command File for the Structure of Pyridine

The Frontier Orbitals of Pyridine

HOMO LUMO

Calculated: HF/6-31G(d) – <3 minutes

Experimental: Huh???

N 1.35 Å1.40

1.40 Å1.33

1.39 Å1.44

The Structure of Excited State Pyridine

S0: Calculated: MP2/6-31G(d) – 4 minutes

T1: Calculated: MP2/6-31G(d) – 11 minutes

Experimental: None

Outline (Cont’d.)

Statistical Thermodynamics: Rotational Contributions to Thermodynamic Properties of Gases

A Blast from the Past

lnV N T N

Q QH kT kT

S k QT

l n ( )A U T S k T Q

ln T N

QG H TS kT Q kT

The Rotational Partition Function: Linear Molecules

The Energy:2

( 1)2J J JI

2 1Jg J

Jrot kT

The Partition Function: 2( 1) / 2

2 1J J I

2 1RJ Jrot T

It can be shown that for most molecules at medium to hightemperatures:

1J J k T

Thus, the exponent (and hence successive terms in thesummation) change very slowly.

Therefore, the summation in qrot can be replaced by an integral.

2 1RJ Jrot T

2 1RJ Jrot Tq J e dJ

This integral can be solved analytically by a simple substitution.

( 1) Ru J JT

(2 1) Rdu J dJ

Therefore:( 1)

(2 1)RJ Jrot RT

Tq e J dJ

1rot u

A Correction

For homonuclear diatomic molecules, one must account for the factthat rotation by 180o interchanges two equivalent nuclei.

Since the new orientation is indistinguishable from the original one, onemust divide by 2 so that indistinguishable orientations are counted once.

For heteronuclear diatomic molecules, rotation by 180o produces adistinguishable orientation. No correction is necessary.

is the "symmetry number"

Homonuclear Diatomic Molecule: = 2

Heteronuclear Diatomic Molecule: = 1

How good is the approximate formula?

RJ Jrot T

Exact:rot

Approx:

H2: I = 4.61x10-48 kg•m2 R = 87.5 K

O2: I = 1.92x10-46 kg•m2 R = 2.08 K

Compd. T qrot(ex) qrot(app) Error

O2 298 K 71.8 71.7 0.2%

H2 298 1.88 1.70 10%

H2 100 0.77 0.57 26%

H2 50 0.55 0.29 47%

For molecules like H2, HCl, …, with small moments of inertia, the integral approximation gives poor results, particularly at lowertemperatures.

The Total Partition Function for N Molecules (Qrot)

and Nro t ro tQ q

Internal Energy

lnln ln(

Q dN T

ln rotrot

Therefore: 2rot NU kT

T N k T An N k T

r o tU n R T or r o tU R T

This illustrates Equipartition of Rotational Internal Energy.

[(1/2)RT per rotation].

ln lnN

ln ln ( )RN T

rot rotrot

V N T N

Q QH kT kT

Enthalpy

rotrot

Qrot independent of V

r o t r o tH U R T

Heat Capacities

rot rotrot VV

C UC R

rot rotrot PP

C HC R

Remember that these results are for Diatomic and LinearPolyatomic Molecules.

2tranPC R

Chap. 3

Experimental Heat Capacities at 298.15 K

Compd. CP (exp)

H2 29.10 J/mol-K

O2 29.36

I2 36.88

r o tPC R

7 78 .3 1 4 / 2 9 .1 0 /

2 2tra n ro tP PC C R J m o l K J m o l K

We can see that vibrational (and/or electronic) contributionsto CP become more important in the heavier diatomic molecules.

We'll discuss this further in Chap. 5

Entropy

rot rot US k Q

lnN rot

T US k

T RTNk

O2: r = 1.202 Å = 1.202x10-10 m (from QM - QCISD/6-311G*)

m m mamu

2 7 2 61 . 6 6 1 0 / 1 . 3 2 8 1 0x x k g a m u x k g

2 2 6 1 0 2 4 6 2(1 . 3 2 8 1 0 ) (1 . 2 0 2 1 0 ) 1 . 9 2 1 0I r x k g x m x k g m

= 2 (Homonuclear Diatomic)

2 2 46 2 23

6.63 102.08

8 8 (3.1416) 1.92 10 1.38 10 /R

x J shK

Ik x kg m x J K

O2 (Cont'd)

TS nR R

2 .0 8R K

For one mole of O2 at 298 K:

Stran = 151.9 J/mol-K (from Chap. 3)

Stran + Srot = 195.7 J/mol-K

O2: Smol(exp) = 205.1 J/mol-K at 298.15 K

Thus, there is a small, but finite, vibrational (and/or) electronic contribution to the entropy at room temperature.

2981(8.31 / ) ln 8.31 /

(2)(2.08 )

43.8 /

rot KS J mol K J mol K

J mol K

E (Thermal) CV S KCAL/MOL CAL/MOL-K CAL/MOL-K TOTAL 3.750 5.023 48.972 ELECTRONIC 0.000 0.000 2.183 TRANSLATIONAL 0.889 2.981 36.321 ROTATIONAL 0.592 1.987 10.459 VIBRATIONAL 2.269 0.055 0.008 Q LOG10(Q) LN(Q) TOTAL BOT 0.330741D+08 7.519488 17.314260 TOTAL V=0 0.151654D+10 9.180853 21.139696 VIB (BOT) 0.218193D-01 -1.661159 -3.824960 VIB (V=0) 0.100048D+01 0.000207 0.000476 ELECTRONIC 0.300000D+01 0.477121 1.098612 TRANSLATIONAL 0.711178D+07 6.851978 15.777263 ROTATIONAL 0.710472D+02 1.851547 4.263345

Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)

QCISD/6-311G(d)

)2 9 8(/4 8.2/5 9 2.0 KR Tm o lk Jm o lk c a lUU r o tt h e r m

RKm o lJKm o lc a lC r o tV /3 1 4.8/9 8 7.1

Km o lJKm o lc a lS r o t /8.4 3/4 5 9.1 0 (same as our result)

Translational + Rotational Contributions to O2 Entropy

Note that the other (vibration and/or electronic) contributionsto S are even greater at higher temperature.

0 1000 2000 3000 4000 5000100

Expt T TR

Temperature [K]

Translational + Rotational Contributions to O2 Enthalpy

There are also significant additional contributionsto the Enthalpy.

0 1000 2000 3000 4000 5000

Expt T TR

Temperature [K]

Translational + Rotational Contributions to O2 Heat Capacity

0 1000 2000 3000 4000 500015

Expt T TR

Temperature [K]

Note that the additional (vibration and/or electronic) contributionsto CP are not important at room temperature, but very significantat elevated temperatures.

Helmholtz and Gibbs Energies

l nr o t r o tA k T Q ,

rotrot rot

QG kT Q kT

Qrot independent of V

r o tA

rot rot

TA G kT

For O2 at 298 K: R = 2.08 K

298ln (8.314 / ) ln 35.5 /

2 2.08rot rot

T KA G RT J mol K J mol K

Non-Linear Polyatomic Molecules

It can be shown that:

1/ 21/ 2 3rot

The symmetry number is defined as the"number of pure rotational elements(including the identity) in the molecule's point group:NO2: = 2NH3: = 3CH4: = 12

I will always give you the value of for non-linear polyatomic molecules.

One can simply use the expression for qrot above in the sameway as for linear molecules to determine the rotational contributionsto the thermodynamic properties of non-linear polyatomic molecules(as illustrated in one of the homework problems).

slide 1 chapter 4 rigid-rotor models and angular momentum eigenstates

topic slide

angular momentum eigenstates

rigidrotor models

mv energy orenergy slide

d schrdinger equation

rotational contributions

parallel vectors

perpendicular vectors

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