Slide 1 Chapter 3 Particle-in-Box (PIB) Models. Slide 2 Carrots are Orange. Tomatoes are Red. But...

Chapter 3

Particle-in-Box (PIB) Models

Carrots are Orange.

Tomatoes are Red.

But Why?

Outline

• The Classical Particle in a Box (PIB)

• The Quantum Mechanical PIB

• PIB and the Color of Vegetables

• PIB Properties

• Multidimensional Systems: The 2D PIB and 3D PIB

• Other PIB Models

• The Free Electron Molecular Orbital (FEMO) Model

• Statistical Thermodynamics: Translational contributions to the thermodynamic properties of gases.

The Classical Particle in a Box

x

P(x

)

0 a0

a

E = ½mv2 = p2/2m = any value, including 0

i.e. the energy is not quantized and there is no minimum.

P(x) = Const. = C 0 £ x £ a

P(x) = 0 x < 0, x > a

P(x) = Const. = C 0 £ x £ a

P(x) = 0 x < 0, x > a

Normalization

<x>

<x2>

0

00( ) 1 0 0

a a

aP x dx dx Cdx dx C x Ca

1( )P x C

a

2 2

00

1 1 1( )

2 2 2

aa x a a

x xP x dx x dxa a a

3 3 22 2 2

00

1 1 1( )

3 3 3

aa x a a

x x P x dx x dxa a a

Note:22 2

22 2 03 2 12x

a a ax x

Outline




• PIB Properties





Quantum Mechanical Particle-in-Box

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0 , x > a

Outside the box: x<0, x>a

P(x) = *(x)(x) = 0

(x) = 0

Inside the box: 0 x a

Schrödinger Equation

2 2

2( )

2

dV x E

m dx

2 2

22

dE

m dx

or

2

2 2

2d mE

dx

Solving the Equation

Assume

So far, there is no restriction on and, hence none onthe energy, E (other than that it cannot be negative)

2

2 2

2d mE

dx

cos( ) sin( )d

A x B xdx

sin( ) cos( )A x B x

22 2

2sin( ) cos( )

dA x B x

dx

22

2

d

dx

22

2mE

2 2

2E

m

Applying Boundary Conditions (BC’s)

Nature hates discontinuous behavior;

i.e. It’s not really possible to stop on a dime

Because the wavefunction, , is 0 outside the box, x<0 and x>a,it must also be 0 inside the box at x=0 and x=a

BC-1: (0) = 0 = Asin(0) + Bcos(0) = B

Therefore, (x) = Asin(x)

BC-2: (a) = 0 = Asin(a) sin(n) = 0

Therefore: a = n n = 1, 2, 3, ...

Why isn’t n = 0 acceptable?

or: n = 1, 2, 3, ...n

a

sin( ) cos( )A x B x

Wavefunctions and Energy Levels

Note that, unlike the classical particle in a box:

(a) the allowed energies are quantized

(b) E = 0 is NOT permissible (i.e. the particle can’t stand still)

n = 1, 2, 3, ...n

a

n = 1, 2, 3, ...sin( ) sinn

nA x A x

a

2 22 2 1

2 2 2

n hE

m a m

n = 1, 2, 3, ...2 2

28n

n hE

ma

2 2 2

2 2

1

2 4

n h

a m

Energy Quantization results from application of the Boundary Conditions

2 2

28n

n hE

ma

2

1 21

8

hE

ma

2

2 24

8

hE

ma

2

3 29

8

hE

ma

2

4 216

8

hE

ma

2

5 225

8

hE

ma

2

2

1 8ma

1hE

2

2

2 8ma

4hE

2

2

3 8ma

9hE

a

xsinAψ 1

a

x2sinAψ 2

0 a 20 a

Wavefunctions

a

x3sinAψ 3

Note that: (a) The Probability, 2 is very different from the classical result.

(b) The number of “nodes” increases with increasing quantum number.

The increasing number of nodes reflects the higher kineticenergy with higher quantum number.

The Correspondence Principle

The predictions of Quantum Mechanics cannot violate the resultsof classical mechanics on macroscopically sized systems.

Consider an electron in a 1 Angstrom box.Calculate (a) the Zero Point Energy (i.e. minimum energy) (b) the minimum speed of the electron me = 9.1x10-31 kg

a = 1x10-10 mh = 6.63x10-34 J-s

Thus, the minimum speed of an electron confined to an atomis quite high.

234218

1 2 31 10 2

6.63 106.04 10

8 8(9.1 10 )(1 10 )

x J shE x J

ma x kg x m

2 18 2 21 1

1 1 31

2 2(6.04 10 /

2 9.1 10

mv E x kg m sE v

m x kg

6 71 3.6 10 / ( 1.7 10 / )v x m s x mi hr

Consider a 1 gram particle in a 10 cm box.Calculate (a) the Zero Point Energy (i.e. minimum energy) (b) the minimum speed of the particle m = 1x10-3 kg

a = 0.10 mh = 6.63x10-34 J-s

Thus, the minimum energy and speed of a macroscopic particleare completely negligible.

Let's perform the same calculation on a macroscopic system.

234263

1 2 3 2

6.63 105.50 10

8 8(1 10 )(0.1 )

x J shE x J

ma x kg m

2 63 2 2301 1

1 1 3

2 2(5.50 10 /3.3 10 /

2 10

mv E x kg m sE v x m s

m x kg

Probability Distribution of a Macroscopic Particle

Consider a 1 gram particle in a 10 cm box moving at 1 cm/s.

Calculate the quantum number, n, which represents the numberof maxima in the probability, 2.

m = 1x10-3 kga = 0.10 mv = 0.01 m/sh = 6.63x10-34 J-s

Thus, the probability distribution is uniform throughout the box,as predicted by classical mechanics.

22 3 81 11 10 0.01 / 5.0 10

2 2E mv x kg m s x J

23 82 2 22

22 2 34

8 1 10 0.1 (5.0 10 )8

8 6.63 10

x kg m x Jn h ma EE n

ma h x J s

2 54 279.1 10 3 10n x n x

Outline




• PIB Properties





Some Useful Integrals

2 1 1sin sin 2

2 4x dx x x

sin ( ) sin ( )sin sin

2 2

x xx x dx

21sin cos sin

2x x dx x

22

2

sin 2 cos 2sin

4 4 8

x x xxx x dx

3 22 2

3 2

cos 21sin sin 2

6 4 8 4

x xx xx x dx x

PIB Properties

Normalization of the Wavefunctions

)s i n ()(s i n xxd xx 24

1

2

120 x asinn

nA x

a

2

2 2 2

0 0 01 sin sin

a a a

n

n ndx A x dx A x dx

a a

2 2

0

1 1 1 1 1sin 2 sin 2 0 sin 0

2 4 2 4 4

an

A x x A a aa

212

aA

1/22 2

Aa a

Probability of finding the particle in aparticular portion of the box

This is significantly lower than the classical probability, 0.25.


1

2

12

Calculate the probability of finding a particle with n=1 in the region of the box between 0 and a/4

1 sinx

Aa

/ 4

2 2

0

1 1 1 1 1sin 2 sin 2 0 sin 0

2 4 2 4 4 / 4 4( / )

aa a

A x x Aa a a

/ 42 2

00 / 4 sin /

aP x a A x dx a

2 1 10 / 4 0.091

8 4 4 2

a aP x a

a

Orthogonality of the Wavefunctions

)(

])s i n [(

)(

])s i n [()s i n ()s i n (

22

xxd xxx

aa

a

aa

a 33

0Thus, 1 and 2 are orthogonal

Two wavefunctions are orthogonal if: * 0i jd i j

We will show that the two lowest wavefunctions of the PIB

are orthogonal: 1 2

2sin sin

x xA A

a a

22 10 0 0

2 2sin sin sin sin

a a ax xdx A A dx A x x dx

a a a a

2 2

2 10

sin ( ) sin ( )0 0

2 2

a a adx A A

22 10

sin sin 30

2 2

adx A

Thus, the PIB wavefunctions are orthogonal.

If they have also been normalized, thenthe wavefunctions are orthonormal:

Using the same method as above, it can be shown that:

* 0i jd For all i j

*i j ijd

or

i j ij

Positional Averages

We will calculate the averages for 1 and present the generalresult for n

<x>

1

2sin sin

xA A x A

a a a

sin sin 0 sin 2 sin 2 0a a a aa a

cos cos 1 cos 2 cos 2 1a a a aa a

*1 1 1 10 0

sin sina a x x

x x x dx A xA dxa a

2 2

0sin

aA x x dx

a

This makes sense because 2 is symmetric about thecenter of the box.

2

22

8

)2c o s (

4

)2s in (

4)(s in

xxxx

d xxx 2 2

0sin

ax A x x dx

2

2 2

sin 2 cos 2 cos 020 0

4 4 8 8

a a aax

a

2

2 2

2 1 1

4 8 8

ax

a

2

ax

General Case:2

ax

<x2 >

23

2322

4

22

8

1

46

)c o s ()s in()(s in

xxx

xxd xxx

2 2 * 2 2

1 1 1 10 0sin sin

a a x xx x x dx A x A dx

a a

2 2 2

0sin

aA x x dx

a

3 22

3 2

cos 22 1sin 2 0 0 0

6 4 8 4

a aa ax a

a

3 2 22 2 2

2 2 2

2 1 10.28

6 3 2 3 24

a a a ax a a

a

a

Note that the general QM value reduces to the classical resultin the limit of large n, as required by the Correspondence Principle.

2 2 22

1 10.28

3 2x a a

compared to the classical result2

2 20.332

ax a

General Case: 2 22 2

1 1

3 2x a

n

Momentum Averages

Preliminary

ˆ dp

i dx 2

2 22

dp

dx ^

1

2sin sinA x A x A

a a a

1 sin cosd d

A x A xdx dx

2

212

cos sind d

A x A xdx dx

sin sin 0 sin 2 sin 2 0a a a aa a

cos cos 1 cos 2 cos 2 1a a a aa a

)(s i n)c o s ()s i n ( xd xxx 2

2

1

<p> = 0

On reflection, this is not surprising.The particle has equal probabilities of moving in the + or - x-direction,and the momenta cancel each other.

(General case is the same)

<p> ˆ dp

i dx

* 11 1 10 0

sinˆ sin

a a

xdA

d ap p dx A x dx

i dx i dx a

2

0 0sin cos sin cos

a aA x A x dx A x x dx

i i

2 2 2 21 1sin sin 0 0 0

2 2p A a A

i i


1

2

12

<p2>2

2 22

dp

dx ^

^ 2

22 2 * 2 21

1 1 1 2 20 0

sinsin

a a

xd A

d ap p dx A x dx

dx dx a

2 2 2 2 2 2

0 0sin sin sin

a aA x A x dx A x dx

2 2 2 2 2 21 1 2sin 2 0 sin 0)

2 4 4 2

a ap A a

a

2 22 2 2

2p

a

General Case:2 2 2

22

np

a

Standard Deviations and the Uncertainty Principle

Heisenberg Uncertainty Principle:2x p

0.502

ax a

2 20.28x a

0p

2 22

2p

a

2 22 20.28 0.50 0.17x x x a a a

22 2 0 3.14p p p pa a

0.17 3.14 0.53x p aa

Outline




• PIB Properties





Other PIB Models

PIB with one non-infinite wall

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0

Vo

V(x) = Vo x > a

We will first consider the casewhere E < V0.

Region I

I

Region II

II

Boundary Conditions

The wavefunction must satisfy two conditionsat the boundaries (x=0, x=a, x)

1) must be continuous at all boundaries.

2) d/dx must be continuous at the boundaries unless V at the boundary.

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0

Vo

V(x) = Vo x > a


Region I

I

Region II

II

Region I

Ic o n s t

)c o s ()s i n ( xBxAI

canassume

22

20

2I

I

dE

m dx

2

2 2

2II I

d mEconst

dx

sin cosI A x B x

cos sinIdA x B x

dx

2

2 22

sin cosIdA x B x

dx

22

2I

I

d

dx

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0

Vo

V(x) = Vo x > a


Region I

I

Region II

II

Region I

Relation between and E

22

2I I

mE

22

2mE

2

2 2

2II

d mE

dx

22

2I

I

d

dx

sin cosI A x B x

2

2mE

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0

Vo

V(x) = Vo x > a


Region I

I

Region II

II

Region II

I Ic o n s t

22

022II

II II

dV E

m dx

2

02 2

2IIII

d mE V

dx

2

02 2

2IIII II

d mV E const

dx

canassume

x xII Ce De

x xIIdC e D e

dx

22 2

2x xIId

C e D edx

22

2II

II

d

dx

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0

Vo

V(x) = Vo x > a


Region I

I

Region II

II

Region II

Relation between and E

2

02 2

2IIII

d mV E

dx

x xII Ce De

22

2II

II

d

dx

202

2II II

mV E

202

2mV E

02

2mV E

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0

Vo

V(x) = Vo x > a


Region I

I

Region II

II

BC: x = 0

Boundary Conditions

sin cosI A x B x 2

2mE

(0) 0 sin 0 cos 0I A B B 0B

)s i n ( xAI Therefore: sinI A x

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0

Vo

V(x) = Vo x > a


Region I

I

Region II

II

BC: x = Boundary Conditions

x xII Ce De 02

2mV E

0II Ce De Ce 0C

Therefore: xII De

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0

Vo

V(x) = Vo x > a


Region I

I

Region II

II

BC’s: x = a

Boundary Conditions

sinI A x 2

2mE

xII De 02

2mV E

cosIdA x

dx

xIIdD e

dx

or

( ) ( )I IIa a

sin aA a De

or

I II

a a

d d

dx dx

cos aA a D e

x

V(x

)

0 a0

V(x) = 0 0 x a

V(x) x < 0

Vo

V(x) = Vo x > a


Region I

I

Region II

II

These equations can be solved analytically to obtain the allowedvalues of the energy, E. However, it’s fairly messy.

I’ll just show you some results obtained by numerical solutionof the Schrödinger Equation.

BC’s: x = a

Boundary Conditions

sinI A x 2

2mE

xII De 02

2mV E

sin aA a De cos aA a D e

0.05 0.30 0.55 0.80 1.05 1.30 1.55 1.80

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

x V

x0.05 0.30 0.55 0.80 1.05 1.30 1.55 1.80

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

x 2 V

2

x

x0.05 0.30 0.55 0.80 1.05 1.30 1.55 1.80

vals.1

-1.0

-0.5

0.0

0.5

1.0

x V

2

x0.05 0.30 0.55 0.80 1.05 1.30 1.55 1.80

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

x 2 V

x0.05 0.30 0.55 0.80 1.05 1.30 1.55 1.80

vals.1

-1.0

-0.5

0.0

0.5

1.0

x V

2

x0.05 0.30 0.55 0.80 1.05 1.30 1.55 1.80

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

x 2 V

This is an arbitrary value of V0

2

0 230

8

hV

ma

2

1 20.9

8

hE

ma

2

2 23.4

8

hE

ma

2

3 27.6

8

hE

ma

x

0.05 0.30 0.55 0.80 1.05 1.30 1.55 1.80

vals.1

-1.1

-0.6

-0.1

0.4

0.9

x V

x

2

0.05 0.30 0.55 0.80 1.05 1.30 1.55 1.80

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

x 2 V

What happened??

Why is the wavefunction so different?

The condition for the earlier solution, E < V0,is no longer valid.

2

0 230

8

hV

ma

2

6 238.2

8

hE

ma

0 a0

Vo

Region I

I

Region II

II

V(x) = 0 0 x a

V(x) x < 0

V(x) = Vo x > a

Region I Region II

E > Vo

2

0 230

8

hV

ma

2

6 238.2

8

hE

ma

22

20

2I

I

dE

m dx

2

2 2

2II I

d mEconst

dx

sin cosI A x B x

22

022II

II II

dV E

m dx

2

02 2

2IIII II

d mE V const

dx

IIconst

sin cosII C x D x

PIB with central barrier: Tunneling

Preliminary: Potential Energy Barriers in Classical Mechanics

V0 = mgh = 1430 J

Bowling Ball

m = 16 lb = 7.3 kgv = 30 mph = 13.4 m/s

h = 20 mKE = ½mv2 = 650 J

Will the bowling ball make it over the hill? Of course not!!

In classical mechanics, a particle cannot get to a position inwhich the potential energy is greater than the particle’s total energy.

x

V(x

)

00

x1

Vo

x2

a

I II III

V(x) x < 0

V(x) = 0 0 x x1

V(x) x > a

V(x) = V0 x1 x x2

V(x) = 0 x2 x a

I’ll just show the Boundary Conditions and graphs of the results.

BC: x=0 BC’s: x=x1 BC’s: x=x2 BC: x=a

(0) 0I ( ) 0III a 1 1( ) ( )I IIx x

1 1

I II

x x

d d

dx dx

2 2( ) ( )II IIIx x

2 2

II III

x x

d d

dx dx

x-0.1 0.1 0.3 0.5 0.7 0.9 1.1

0.0

0.2

0.4

0.6

0.8

1.0

x V

2

x-0.1 0.1 0.3 0.5 0.7 0.9 1.1

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

x 2 V

Note that there is a significant probability of finding the particleinside the barrier, even though V0 > E.

This means that the particle can get from one side of the barrierto the other by tunneling through the barrier.

2

0 220

8

hV

ma

2

1 27.2

8

hE

ma

We have just demonstrated the quantum mechanicalphenomenon called tunneling, in which a particle canbe in a region of space where the potential energy ishigher than the total energy of the particle.

This is not simply an abstract phenomenon, but is known tooccur in many areas of Chemistry and Physics, including:

• Ammonia inversion (the ammonia clock)

• Kinetic rate constants

• Charge carriers in semiconductor devices

• Nuclear radioactive decay

• Scanning Tunneling Microscopy (STM)

Outline




• PIB Properties





Carrots are Orange.

Tomatoes are Red.

But why??

Free Electron Molecular Orbital (FEMO) Model

So, does the solution to the quantum mechanical “particle in a box”serve any role other than to torture P. Chem. students???

Yes!!

To a good approximation, the electrons in conjugatedpolyalkenes are free to move within the confines of the orbital system.

H2CHC

HC

HC

HC CH2

a

Notes: Estimation of the box length, a, will be discussed later.

Each carbon in this conjugated system contributes 1 electron. Thus, there are 6 electrons in the -systemof 1,3,5-hexatriene (above)

Bonding in Ethylene

C C

+ +

- -

C C

+

+-

-

C C

C C

Bonding () Orbital

Anti-bonding (*) Orbital

Maximum electrondensity between C’s

PIB (12)

C CMaximum electrondensity between C’s

Electron densitynode between C’s

PIB (22)

C C

Electron densitynode between C’s

* Transition in Ethylene using FEMO

2

2

1 8ma1h

E

2

2

2 8ma4h

E

h = 6.63x10-34 J•s

m = 9.11x10-31 kg

a = ???

C CR = 0.134 nm

R½R ½R

a = 2R

It is common (although not universal) to assumethat the electrons are free to move approximately½ bond length beyond each outermost carbon.

= 0.268 nm

2 2 2

2 1 2 2 2

4 1 3

8 8 8

h h hE E E

ma ma ma

Calculation of max

= 7.9x10-8 m 80 nm

(exp) = 190 nm

The difficulty with applying FEMO to ethylene is that the resultis extremely sensitive to the assumed box length, a.

234218

2 31 9 2

3 6.63 1032.52 10

8 8(9.11 10 )(0.268 10 )

x J shE x J

ma x kg x m

Units:

22

2 2 2

2 2

kg mJ sJ s s J

kg m kg m

34 8

18

(6.63 10 / )

2.52 10

)(3.00 10phot

hc hc x J m sE E

E x J

s x

Application of FEMO to 1,3-Butadiene

H2CHC

HC CH2

R½R ½RR R

It is common to use R = 0.14 nm (the C-C bond length inbenzene,where the bond order is 1.5) as the average bond length in conjugated polyenes.

2

2

1 8ma1h

E

2

2

2 8ma4h

E

2

2

3 8ma9h

E

L = 3•R + 2•(½R) = 4•R

L = 4•0.14 nm = 0.56 nm

2 2 2

3 2 2 2 2

9 4 5

8 8 8

h h hE E E

ma ma ma

Calculation of max

= 2.07x10-7 m 207 nm

(exp) = 217 nm

2 34 219

2 31 9 2

5 5(6.63 10 )9.62 10

8 8(9.11 10 )(0.56 10 )

h x J sE x J

ma x kg x m

34 8

19

(6.63 10 / )

9.62 10

)(3.00 10phot

hc hc x J m sE E

E x J

s x

Application of FEMO to 1,3,5-Hexatriene

a = 5•R + 2•(½R) = 6•R

a = 6•0.14 nm = 0.84 nm

H2CHC

HC

HC

HC CH2

R½R ½RR RR R

2

2

1 8ma1h

E

2

2

2 8ma4h

E

2

2

3 8ma9h

E

2

2

4 8ma16h

E

E = 5.98x10-19 J

= 332 nm

(exp) = 258 nm

Problem worked out in detail inChapter 3 HW

2 2 2

4 3 2 2 2

16 9 7

8 8 8

h h hE E E

ma ma ma

The Box Length

a = 5•R + 2•(½R) = 6•R

H2CHC

HC

HC

HC CH2

R½R ½RR RR R

Hexatriene

General: a = nb•R + 2•(½R) =(nb+1)•R R = 1.40 Å = 0.14 nm

Outline




• PIB Properties





PIB and the Color of Vegetables

Introduction

The FEMO model predicts that the * absorption wavelength, ,increases with the number of double bonds, #DB

Compound #DB (FEMO)

Ethylene 1 80 nm

Butadiene 2 207

Hexatriene 3 333

roughly

N = # of electron pairsThis is because:

2 2

22

1 1N N nbE

a nbnb

and:1

nbE

Blue Yellow Red

(nm)400 500 600 700

% T

ran

smis

sio

n

Blue Yellow Red

(nm)400 500 600 700

% T

ran

smis

sio

n

If a substance absorbslight in the blue region ofthe visible spectrum, thecolor of the transmitted(or reflected) light will be red.

If a substance absorbslight in the red region ofthe visible spectrum, thecolor of the transmitted(or reflected) light will be blue.

Light Absorption and Color

Conjugated Systems and the Color of Substances

Ethylene 1 190 nm UV

1,3-Butadiene 2 217 UV

1,3,5-Hexatriene 3 258 UV

-Carotene 11 450 Vis.

Lycopene 13 500 Vis.

White light contains all wavelengths of visibleradiation ( = 400 - 700 nm)

If a substance absorbs certain wavelengths,then the remaining light is reflected, givingthe appearance of the complementary color.

e.g. a substance absorbing violet light appears to be yellow in color.

Carrots

Tomatoes

No. of Conj.Molecule Doub. Bonds max Region

Structure of Ground State Butadiene

R(C-C) = 1.53 Å

H2C

CH

HC

CH21.32 Å

1.32 Å1.47 Å

Calculation: HF/6-31G(d) – 1 minute

As well known, there is a degree of electron delocalization betweenthe conjugated double bonds, giving some double bond characterto the central bond.

0 20 40 60 80 100 120 140 160 180

0

5

10

15

20

25

Re

lativ

e E

ner

gy [

kJ/m

ol]

Dihedral Angle [Degrees]

Calculation: HF/6-31G(d) – 37 optimizations – 25 minutes

Potential Energy Surface (PES) of Ground State Butadiene

Minimum at~40o

0 20 40 60 80 100 120 140 160 180

0

5

10

15

20

25R

ela

tive

En

ergy

[kJ

/mol

]

Dihedral Angle [Degrees]

Energy Difference Between Trans and “Cis”-Butadiene

E(cal) = Ecis – Etrans = 12.7 kJ/mol

Experiment

At room temperature (298 K), butadiene is a mixture with approximately97% of the molecules in the trans conformation (and 3% “cis”).

We could get very close to experiment by using a higherlevel of theory.

(exp)Ecis RT

trans

Ne

N

(exp) ln cis

trans

NE RT

N

3 0.03

(exp) (8.31 10 / )(298 )ln 8.6 /0.97

E x kJ mol K K kJ mol

Structure of Excited State Butadiene

H2C

CH

HC

CH21.32 Å

1.32 Å1.47 Å

Ground State: S0

H2C

CH

HC

CH21.41 Å

1.41 Å1.39 Å

Excited State: S1

Structure of Excited State Butadiene

H2C

CH

HC

CH21.32 Å

1.32 Å1.47 Å

Ground State: S0

H2C

CH

HC

CH21.41 Å

1.41 Å1.39 Å

Excited State: S1

HOMO

LUMO

* Transition

Comparison of FEMO and QM Wavefunctions

a

xsinAψ 1

a

x2sinAψ 2

0 L

a

x3sinAψ 3

HOMO

LUMO

Outline




• PIB Properties





Multidimensional Systems: The 2D PIB

The Potential Energy

0 a0

b

x

y

V(x,y) = 0 0 x a and 0 y b

V(x,y) x < 0 or x > a or y < 0 or y > b

Outside of the boxi.e. x < 0 or x > a or y < 0 or y > b

(x,y) = 0

The 2D Hamiltonian:2 2 2

2 2( , )

2H V x y

m x y

The 2D Schr. Eqn.:2 2 2

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

2x y V x y x y E x y

m x y

Solution: Separation of Variables

Inside the box:

2 2 2 2

2 22 2 x yH H Hm x m y

V(x,y) = 0 0 x a and 0 y b

The 2D Schr. Eqn.:2 2 2

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

2x y V x y x y E x y

m x y

Note: On following slides, I will show how this two dimensionaldifferential equation can be solved by the method of Separation of Variables.

You are NOT responsible for the details of the soution, but only the assumptions and results.


Hx Hy

Inside the box:2 2 2 2


( , ) ( , ) ( , )x yH x y H x y E x y

Assume: ( , ) ( ) ( )x y X x Y y

( ) ( ) ( ) ( ) ( ) ( )x yH X x Y y H X x Y y EX x Y y

( ) ( ) ( ) ( ) ( ) ( )x yY y H X x X x H Y y EX x Y y

1 1( ) ( )

( ) ( )x yH X x H Y y EX x Y y


If: f(x) + g(y) = C

then: f(x) = C1 and g(y) = C2

C1 + C2 = C

f(x) g(y) C

=

Ex

=Ey

E = Ex + Ey



1 1( ) ( )

( ) ( )x yH X x H Y y EX x Y y

2 2 2 2

2 2

1 1

2 2

d X d YE

m X dx m Y dy

=

Ex

=

Ey

2 2 2 2

2 2

1 1

2 2

d X d YE

m X dx m Y dy

2 2

2

1

2 x

d XE

m X dx

Range0 x a

2 2

22 x

d XE X

m dx

2 2

21,2,3,...

8x

x x

n hE n

ma

2( ) sin x

x x

n xX x A A

a a

2 2

2

1

2 y

d YE

m Y dy

Range0 y b

2 2

22 y

d YE Y

m dy

2 2

21,2,3,...

8y

y y

n hE n

mb

2( ) sin y

y y

n yY y A A

b b

E = Ex + Ey( , ) ( ) ( )x y X x Y y

Two Dimensional PIB: Summary



Assume: ( , ) ( ) ( )x y X x Y y

Range0 x a

2 2

22 x

d XE X

m dx

Range0 y b

2 2

22 y

d YE Y

m dy

2 2

21,2,3,...

8x

x x

n hE n

ma

2( ) sin x

x x

n xX x A A

a a

2 2

21,2,3,...

8y

y y

n hE n

mb

2( ) sin y

y y

n yY y A A

b b

E = Ex + Ey( , ) ( ) ( )x y X x Y y

nx = 1, 2, 3,...ny = 1, 2, 3,...

Range0 x a0 y b

2nx = 1 ny = 1

2

2nx = 2 ny = 1

2

2

nx = 2 ny = 2

2

The Wavefunctions

2 2( , ) ( ) ( ) sin sin yx

x y x y

n yn xx y X x Y y A A A A

a b a b

4( , ) sin sin yx

n yn xx y

ab a b

The Energies: Wavefunction Degeneracy

nx = 1, 2, 3,...ny = 1, 2, 3,...

Square Box

a = b

2

2

8m ah

E

nx ny

1 12

51 2 2 1

g = 1

82 2

g = 1

101 3 3 1

g = 2

132 3 3 2

g = 2

g = 2

22 2 2 2 22

2 2 2 28 8 8yx x x

x y

nn h n h nhE E E

ma mb m a b

2

2 228 x y

hE n n

ma

Application: * Absorption in Benzene

1 1

nx ny

2

5

8

10

13

2 1 1 2

2 2

1 3 3 1

2 3 3 2

E [h

2/8

ma2

]

The six electrons in benzene can be approximated asparticles in a square box.

One can estimate the wavelength of the lowest energy* from the 2D-PIB model (See HW problem)

nx ny

1 12

5

8

10

13

2 1 1 2

2 2

1 3 3 1

2 3 3 2

E [h

2/8

ma2

]

ab

c

V(x,y,z) = 0 0 x a0 y b0 z c

V(x,y,z) Outside thebox

Three Dimensional PIB

2 2 2 2

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

2x y z V x y z x y z E x y z

m x y z

2 2 2 2

2 2 2( , , ) ( , , )

2x y z E x y z

m x y z

2 2 2 2

2 2 22 x y zH H H Em x y z

Assume: ( , , ) ( ) ( ) ( )x y z X x Y y Z z

2 2

22 x

XE X

m x

2 2

21,2,3,...

8

( ) sin

2

xx x

xx

x

n hE n

man x

X x Aa

Aa

2 2

22 y

YE Y

m y

2 2

21,2,3,...

8

( ) sin

2

yy y

yy

y

n hE n

mbn y

Y y Ab

Ab

2 2

22 z

ZE Z

m z

2 2

21,2,3,...

8

( ) sin

2

zz z

zz

z

n hE n

mcn z

Z z Ac

Ac

nx = 1, 2, 3, ...ny = 1, 2, 3, ...nz = 1, 2, 3, ...

Cubical Box

a = b = c

Note: You should be able to determine the energy levels and degeneracies for a cubical box and for various ratios of a:b:c

8( , , ) ( ) ( ) ( ) sin sin sinyx z

n yn x n zx y z X x Y y Z z

abc a b c

22 2 2 2 2 2 2 22

2 2 2 2 2 28 8 8 8yx x z x z

x y z

nn h n h n h n nhE E E E

ma mb mc m a b c

2

2 2 228 x y z

hE n n n

ma

Outline




• PIB Properties





Introduction to Statistical Thermodynamics

But this is a Quantum Mechanics course.

Why discuss thermodynamics??

Statistical Thermodynamics is the application of Statistical Mechanicsto predict thermodynamic properties of molecules.

1. The Statistical Thermodynamic formulas used to predict the properties come from the Quantum Mechanical energies for systems such as the PIB, Rigid Rotor (Chap. 4) and Harmonic Oscillator (Chap. 5).

2. Various properties of the molecules, including moments of inertia, vibrational frequencies, and electronic energy levels are required to calculate the thermodynamic properties

Often, these properties are not available experimentally,and must be obtained from QM calculations.

Some Applications of QM/Stat. Thermo.

Molecular Heat Capacities [CV(T) and CP(T)]

Molecular Enthalpies of Formation [Hf]

Reaction Enthalpy Changes [Hrxn]

Reaction Entropy Changes [Srxn]

Reaction Gibbs Energy Changes [Grxn] and Equilibrium Constants [Keq]

Kinetics: Activation Enthalpies [H‡] and Entropies [S‡]

Kinetics: Rate Constants

Bond Dissociation Enthalpies [aka Bond Strengths]

Frequencies -- 1661.9057 .............. ------------------- - Thermochemistry - ------------------- Temperature 298.150 Kelvin. Pressure 1.00000 Atm. Thermochemistry will use frequencies scaled by 0.9540. Atom 1 has atomic number 8 and mass 15.99491 Atom 2 has atomic number 8 and mass 15.99491 Molecular mass: 31.98983 amu. Principal axes and moments of inertia in atomic units: 1 2 3 EIGENVALUES -- 0.00000 41.27885 41.27885 X 0.00000 0.90352 0.42854 Y 0.00000 -0.42854 0.90352 Z 1.00000 0.00000 0.00000 THIS MOLECULE IS A PROLATE SYMMETRIC TOP. ROTATIONAL SYMMETRY NUMBER 2. ROTATIONAL TEMPERATURE (KELVIN) 2.09825 ROTATIONAL CONSTANT (GHZ) 43.720719 Zero-point vibrational energy 9483.1 (Joules/Mol) 2.26653 (Kcal/Mol) VIBRATIONAL TEMPERATURES: 2281.11 (KELVIN)Zero-point correction= 0.003612 (Hartree/Particle) Thermal correction to Energy= 0.005976 Thermal correction to Enthalpy= 0.006920 Thermal correction to Gibbs Free Energy= -0.016348 Sum of electronic and zero-point Energies= -150.022558 Sum of electronic and thermal Energies= -150.020195 Sum of electronic and thermal Enthalpies= -150.019250 Sum of electronic and thermal Free Energies= -150.042518

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)

Mini-Outline for Stat. Thermo. Development

• The Boltzmann Distribution

• The Partition Function (Q)

• Thermodynamic properties in terms of Q

• Partition function for non-interacting molecules

• The molecular partition function (q)

• The total partition function for a system of molecules (Q)

• The translational partition function

• Translational contributions to thermodynamic properties

The Boltzmann Distribution

E0

E1

E2

Normalization

Q

e kT

Ei

( )iE

kTip E e

23

23

1.3806 10 /

1.38 10 /

A

Rk x J K

N

x J K

( )iE

kTip E Ae

( ) 1iE

kTi

i i

p E A e

1

iE

kT

i

A e

( )

i i

i

E E

kt kt

i E

kt

i

e ep E

Qe

Therefore:

Boltzmann Distribution

The Partition Function

Q is called the "Partition Function". Thermodynamic properties,including, U, Cv, H, S, A, G, can be calculated from Q.

Alternative Form

One can sum over energy "levels" rather than individual energystates.

gj is the degeneracy of the j’th energy level, Ej

( )

iE

kt

i

ep E

Q

iE

kT

i

Q e

where

( )

iE

ktj

j

g ep E

Q

( )

jE

kTj

j level

Q g e

where

Thermodynamic Properties from Q: Internal Energy (U)

Derivative at constantV,N because energydepends on volumeand No. of Molecules

iE

kT

i

Q e

( )

iE

kTi

ii i

i

E eU E E p E

Q

, ,

,

ln( ) 1 1

iE

kT

i

V N V N

V N

eQ Q

T Q T Q T

2

,

1 1i

i

EEkT

i kT

i i

V N

Eee

Q T Q kT

2

,

lniE

kT

i i ii iV N

Q eU E kT E E p E

T Q

NOTE: On an exam, you will be provided with any required SM formulafor a thermodynamic property (e.g. the above expression for U)

Thermodynamic Properties from Q: Other Properties

The math involved in relating other thermodynamic propertiesto Q is a bit more involved.

It can be found in standard P. Chem. texts; e.g. Physical Chemistry,by R. A. Alberty and R. J. Silbey, Chap. 17

I'll just give the results.

Constant Volume Heat Capacity:,

VV N

UC

T

Pressure:ln

T

QP kT

V

Entropy: lnU

S k QT

Thermodynamic Properties from Q: Other Properties

Note: If Q f(V), then H = U

Note: If Q f(V), then G = A

Enthalpy: 2

, ,

ln ln

lnV N T N

Q QH kT kT

T V

Constant Pressure Heat Capacity:,

PP N

HC

T

Helmholtz Energy: lnA U TS kT Q

Gibbs Energy:,

lnln

ln T N

QG H TS kT Q kT

V

NOTE: On an exam, you will be provided with any required SM formulafor a thermodynamic property.

Partition Function for Non-interacting Molecules (aka IDG)

Case A: Distinguishable Molecules

The microstate of molecule 'a' is given by j, of molecule 'b' by k…

, , , ...i a j b k c lE

, , , , , ,...

, , ,... , , ,...

...a j b k c l a j b k c liE

kT kT kT kT kT

i j k l j k l

Q e e e e e

, , ,

...a j b k c l

kT kT kT

j k l

Q e e e

,

... , .a j

kTa b c a

j

Q q q q where q e etc

If the molecules are the same, a b cq q q q

Therefore, for N molecules we have: NQ q

a, b, c, ... designates the moleculei, j, k, ... designates the molecule's set of energies (e.g. trans., rot., vib., ...)

Partition Function for Non-interacting Molecules (aka IDG)

Case B: Indistinguishable Molecules

In actuality, one cannot distinguish between different molecules in a gas.

Let's say that there is one state in which the quantum numbers formolecule 'a' are m1,n1… and for molecule 'b' are m2,n2…

This state is the same as one in which the quantum numbers formolecule 'a' are m2,n2… and for molecule 'b' are m1,n1…

However, the two sets of quantum numbers above would give twoterms in the above equation for Q. i.e. we've overcounted thenumber of terms.

It can be shown that the overcounting can be eliminated by dividingQ by N! (the number of ways of permuting N particles). Therefore,

Indistinguishable Molecules:!

NqQ

N

Distinguishable Molecules: NqQ NQ q

The Molecular Partition Function (q)

To a good approximation, the Hamiltonian for a molecule canbe written as the sum of translational, rotational, vibrational andelectronic Hamiltonians:

Therefore, the molecule's total energy (i) is the sum of individualenergies:

tran rot vib electH H H H H

tran rot vib electi j k l m

The partition function is:tran rot vib elect tran rot vib electj k l m ji k l m

kT kT kT kT kT kT

i j k l m j k l m

q e e e e e e

.

tran rot vib electj k l m

kt kt kt kt

j k l m

q e e e e

, .

tranj

tran rot vib elect tran kt

j

q q q q q where q e etc

The Total Partition Function (Q)

It can be shown that the term,1/N! belongs to Qtran

!

NqQ

N tran rot vib electq q q q q

!

Ntran rot vib electq q q qQ

N

!

NtranN N Nrot vib elect

qQ q q q

N tran rot vib electQ Q Q Q

!

Ntran

tran

Nrot rot

Nvib vib

Nelect elect

qQ

N

Q q

Q q

Q q

Thermodynamic Properties of Molecules

The expressions for the thermodynamic properties all involve lnQ; e.g.

tran rot vib electQ Q Q Q Q

2 2

,,

lnlntran rot vib elect

V NV N

Q Q Q QQU kT kT

T T

2 2 2 2

, , , ,

ln ln ln lntran rot vib elect

V N V N V N V N

Q Q Q QU kT kT kT kT

T T T T

tran rot vib electU U U U U

where 2

,

ln, .

trantran

V N

QU kT etc

T

Similarly

etc.

tran rot vib electV V V V VC C C C C

The Translational Partition Function for an Ideal Gas

Consider that the molecules of a gas are confined in a cubical boxof length 'a' on each side.*

*One gets the same result if the box is not assumed to be cubical, with a bit more algebra.

The energy levels are given by: 2

2 2 2, , 28x y zn n n x y z

hE n n n

ma

The molecular translational partition function is:

2

2 2 222 22

28

21 1 1 1 1 1

,8

trani

x y zyx z

x y z x y z

hE n n n nn ntran makT

i n n n n n n

hq e e e e e

ma kT

22 2 2

3

1 1 1 1

yx z

x y z

nn ntran n

n n n n

q e e e e

One can show that the summand (term in summation) is an extremelyslowly varying function of n.

For example, for O2 [M = 32x10-3 kg/mol] in a 10 cm (0.1 m) box at 298 K:

n Δ[Expon]

1 8x10-20

106 5x10-14

1012 5x10-8

This leads to an important simplification in the expression for qtran.

2

3 2

21 8

tran n

n

hq e

ma kT

220

22.5 10

8

hx

ma kT 2 21 2 1Expon n n n

The extremely slowly changing exponentresults from the fact that:

is the change in energy betweensuccessive quantum levels.

1 or kTkT

1 2 3 4 5 6 7 8

ni

f(ni)

Relation between sum and integral for slowly varying functions

If f(ni) is a slowly varying function of ni, then one may replacethe sum by the integral:

22

11

( ) ( )n

n

i ni n

f n f n dn

2/1

0 42

12

dxe x

V is the volume of the box

Note: The translation partition function is the only one which depends upon the system’s volume.

2 2

3 23

201 8

tran n n

n

hq e e dn

ma kT

3/23 3/21/2 3/22

32 22

2

2 2

4 48

tran ma kT mkTq a

h hhma kT

3/2

2

2tran mkTq V

h

This is the translational partitition functionfor a single molecule.

A note on the calculation of V (in SI Units)

One is normally given the temperature and pressure, from whichone can calculate V from the IDG law.

One should use the pressure in Pa [1 atm. = 1.013x105 Pa],in which case V is given in m3 [SI Units]

3/2

2

2tran mkTq V

h

R = 8.31 J/mol-K = 8.31 Pa-m3/mol-KnRT

VP

Calculate qtran for one mole of O2 at 298 K and 1 atm.NA = 6.02x1023 mol-1

h = 6.63x10-34 J-sk = 1.38x10-23 J/KR = 8.31 Pa-m3/mol-KM = 32 g/mol1 J = 1 kg-m2/s2

1 atm. = 1.013x105 Pa

3/2

2

2tran mkTq V

h

326

23 1

32 10 /5.32 10

6.02 10A

M x kg molm x kg

N x mol

3

5

(1 )(8.31 / )(298 )

1.013 10

nRT mol Pa m mol K KV

P x Pa

2 32.45 10V x m

3/226 23

2 3234

2(3.14)(5.32 10 )(1.38 10 / )(298 )2.45 10

6.63 10

tran x kg x J K Kq x m

x J s

3/221 2 2 3 303.13 10 2.45 10 4.28 10tranq x m x m x

------------------ - Thermochemistry - ------------------- Temperature 298.150 Kelvin. Pressure 1.00000 Atm. 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

G-98 Output

3/226 23

2 3 30234

2(3.14)(5.32 10 )(1.38 10 / )(298 )2.45 10 4.28 10

6.63 10

tran x kg x J K Kq x m x

x J s

Units:

3/23/2

3 3 32 2 2 2 2 2

( / )

/ ( )tran kg J K K kg kg

q m m mJ s J s kg m s s

3/22 3 1 ( )m m Unitless

Their output is actuallyqtran / NA

306

23

4.28 107.13 10

6.02 10

tran

A

q xx

N x

Partition Function for a system with N (=nNA) molecules

This is the translational partitition functionfor a single molecule.

3/2

2

2tran mkTq V

h

Stirling’s Approximation

NN eNN !

!

Ntran

tranq

QN

NN tran

NtranN

e q eq

N N

3/2

2

2N

Ntrantran

mkTV e

q e hQ

N N

This is an extremely large number.

However, as we’ll see shortly, ln(Q) is multiplied by the very smallnumber (k) in the thermodynamics formulas.

Calculate ln(Qtran) for one mole of O2 at 298 K and 1 atm.

qtran = 4.28x1030

3/2

2

2N

Ntrantran

mkTV e

q e hQ

N N

ln ln lnNtran tran

tranA

q e q eQ nN

N N

30

23 1 25

23 1

(4.28 10 )(2.72)1 6.02 10 ln 1.01 10

1 6.02 10

xmol x mol x

mol x mol

Translational Contributions to the Thermodynamic Properties of Ideal Gases

Preliminary

3/2

2

2N

Ntrantran

mkTV e

q e hQ

N N

23 3ln ln 2 / ln ) ln ln

2 2tranQ N mk h N T N V N N N

The Ideal Gas Equation

Notes: (1) Because none of the other partition functions (rotational, vibrational, or electronic) depend on V, they do not contribute to the pressure.

(2) We end up with the IDG equation because we assumed that the molecules don’t interact when we assumed that the total energy is the sum of individual molecules’ energies.



ln trantran

T

QP P kT

V

(because R = kNA)AnNN nRTP kT kT

V V V

PV nRT

Internal Energy

This demonstrates the Principle of Equipartition of TranslationalInternal Energy (1/2RT per translation)

It arose from the assumption that the change in the exponentis very small and the sum can be replaced by an integral.

This is always true for translational motions of gases.



2

,

ln trantran

V N

QU kT

T

2 3

2

NkT

T

3

2 AnN kT

3

2tranU nRT or

Molar Internal Energy

3

2

trantran U

U RTn

Enthalpy

Note: As noted earlier, if Q is independent of V, then H = U.

This is the case for all partition functions except Qtran



2

, ,

ln ln

ln

tran trantran

V N T N

Q QH kT kT

T V

,

ln

ln

trantran

T N

QU kT

V

3

2tranH nRT kTN

3

2 AnRT kTnN 3

2nRT nRT

5

2tranH nRT or

Molar Enthalpy

5

2

trantran H

H RTn

Heat Capacities

Experimental Heat Capacities at 298.15 K

Compd. CP (exp)

Ar 20.79 J/mol-K

O2 29.36

SO3 50.67

3

2tranU RT

5

2tranH RT

,

312.47 /

2

tran trantran VV

V N

C UC R J mol K

n T

,

520.79 /

2

tran trantran PP

P N

C HC R J mol K

n T


Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)QCISD/6-311G(d)

Utran(mol) = 3/2RT = 3.72 kJ/mol = 0.889 kcal/mol

CVtran(mol) = 3/2R = 12.47 J/mol-K = 2.98 cal/mol-K

ln(Qtran) = 1.01x1025 mol-1

Earlier, we showed that for one mole of O2 at 298 K and 1 atm.,

O2: Smol(exp) = 205.14 J/mol-K at 298.15 K

Entropy3/2

2

2N

Ntrantran

mkTV e

q e hQ

N N

lntran

tran tran US k Q

T

23 25 1

38.31 /

21.38 10 / 1.01 10tranJ mol K T

S x J K x molT

139.38 12.47 151.85 /tranS J mol K


Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)QCISD/6-311G(d)

G-98: Smoltran = 36.321 cal/mol-K = 151.97 J/mol-K

Difference is round-off error and less Sig. Figs. in our calculation.

Our Calculation (above): Smoltran = 151.85 J/mol-K

Helmholtz Energy ln(Qtran) = 1.01x1025 mol-1

For O2 at 298 K(one mol):

Gibbs Energy

For O2 at 298 K(one mol):

Note: As noted before, if Q is independent of V, then G = A.

This is the case for all partition functions except Qtran

lntran tran tran tranA U TS kT Q

23 25 11.38 10 / 298 1.01 10tranA x J K K x mol

44.154 10 / 41.54 /tranA x J mol kJ mol

,

lnln

ln

trantran tran tran tran

T N

QG H TS kT Q kT

V

ln trankT Q NkT A nRT

4

4

4.154 10 / (8.31 / )(298 )

3.906 10 / 39.06 /

tranG x J mol J mol K K

x J mol kJ mol

Frequencies -- 1661.9057 .............. ------------------- - Thermochemistry - ------------------- Temperature 298.150 Kelvin. Pressure 1.00000 Atm. Thermochemistry will use frequencies scaled by 0.9540. Atom 1 has atomic number 8 and mass 15.99491 Atom 2 has atomic number 8 and mass 15.99491 Molecular mass: 31.98983 amu. Principal axes and moments of inertia in atomic units: 1 2 3 EIGENVALUES -- 0.00000 41.27885 41.27885 X 0.00000 0.90352 0.42854 Y 0.00000 -0.42854 0.90352 Z 1.00000 0.00000 0.00000 THIS MOLECULE IS A PROLATE SYMMETRIC TOP. ROTATIONAL SYMMETRY NUMBER 2. ROTATIONAL TEMPERATURE (KELVIN) 2.09825 ROTATIONAL CONSTANT (GHZ) 43.720719 Zero-point vibrational energy 9483.1 (Joules/Mol) 2.26653 (Kcal/Mol) VIBRATIONAL TEMPERATURES: 2281.11 (KELVIN)Zero-point correction= 0.003612 (Hartree/Particle) Thermal correction to Energy= 0.005976 Thermal correction to Enthalpy= 0.006920 Thermal correction to Gibbs Free Energy= -0.016348 Sum of electronic and zero-point Energies= -150.022558 Sum of electronic and thermal Energies= -150.020195 Sum of electronic and thermal Enthalpies= -150.019250 Sum of electronic and thermal Free Energies= -150.042518

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

QCISD/6-311G(d)

These are thermal contributionsto U, H and G.

They represent the sum of thetranslational, rotational, vibrationaland (in some cases) electroniccontributions.

They are given in atomic units (au).1 au = 2625.5 kJ/mol

Translational Contributions to O2 Entropy

0 1000 2000 3000 4000 5000100

150

200

250

300

Expt T

S

[J/m

ol-K

]

Temperature [K]

Obviously, there are significant additional contributions(future chapters)

Translational Contributions to O2 Enthalpy


0 1000 2000 3000 4000 5000

0

50

100

150

200

Expt T

H

[kJ/

mol

]

Temperature [K]

Translational Contributions to O2 Heat Capacity


0 1000 2000 3000 4000 500015

20

25

30

35

40

45

Expt T

CP

[J/m

ol-K

]

Temperature [K]

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