Molecular Information Content Béla Viskolcz Department of Chemical Informatics University of Szeged

Molecular Information Content

Béla Viskolcz Department of Chemical Informatics

University of Szeged

Department of Chemical Informatics, University of Szeged Molecular statistic 2014

Degrees of freedom

Degrees of freedom: the total number of independent variables whose values have to be specified for a complete description of the system.

One atom: In the context of molecular motion these are the spatial coordinates of all the particles. Since we need three coordinates to describe the position of an atom, we say the atom has three degrees of freedom.


One atom


Diatomic molecules Diatomic molecule: If the atoms are not

bound to one another, there will be no relation among the coordinates of the two atoms. On the other hand, when the two atoms are bound, the displacement of each other is coupled to the other. The result is to give three translational, one vibrational, and two rotational degrees of freedom for the molecule. The vibrational and rotational degrees of freedom are also referred to as internal degrees of freedom.


Polyatomic molecules Polyatomic molecule with N atoms: Linear molecule: The 3N degrees of freedom

of the atoms become three translational degrees of freedom and (3N-3) internal degrees of freedom. By analogy with diatomic molecules, we expect two rotational degrees of freedom for any linear polyatomic molecule (e.g. CO2 and C2H2). The remaining (3N-5) internal coordinates must correspond to vibrations.


Polyatomic molecules Polyatomic molecule with N atoms: Bent molecule: A bent molecule loses a

vibrational degree of freedom while gaining a rotational degree of freedom. Therefore, a nonlinear polyatomic molecule has three rotational degrees of freedom, hence (3N-6) vibrational degrees of freedom.

Department of Chemical Informatics, University of Szeged Molecular statistic 2014e.g. vibrational motions for CO2 and

H2O:


Összefoglalás

Degrees of

freedomAtom

Linear

polyatomic

Bent

polyatomic

Translation

al3 3 3

Vibrational 0 3N-5 3N-6

Rotational 0 2 3


Molecular description in space1. Atom (origo)

2. Atom (two atoms in one line) DISTANCE (streching)

3. Atom (three atoms in one plane) ANGLE (bending)

4. Atom (four atoms in space – two plane angle DIHEDRAL ANGEL (torsion)


Introduction Statistical Mechanics

Properties of individual molecules

PositionMolecular geometry

Intermolecular forces

Properties of bulk fluid (macroscopic properties)

PressureInternal EnergyHeat Capacity

EntropyViscosity


t r e nq q q q q q v

, , , , ,i t i r i i e i n i v

, , , , ,i t i r i i e i n ig g g g g g vThe partition functions for 5 mode motions are expressed as

, , ,

, ,

, , ,

, ,

; ;

;

t i r i v i

e i e i

kT kT kTt t i r r i v v i

i i i

kT kTe e i e e i

i i

q g e q g e q g e

q g e q g e


, r,it,i r,i

v,i e,iv,i e,i

n,in.i

[ exp( )] [ exp( )]

[ exp( )] [ exp( )]

[ exp( )]

t i

i i

i i

i

q g gkT kT

g gkT kT

gkT

t r v e nq q q q q

1

1

n

e

q

q


Potential V(R) for nuclear motion in a diatomic molecule Harmonic oscillator potential


Energy level


Zero-point energy zero-point energy is the energy at ground state

or the energy as the temperature is lowered to absolute zero.

Suppose some energy level of ground state is ε0, and the value of energy at level i is εi, the energy value of level i relative to ground state is

Taking the energy value at ground state as zero, we can denote the partition function as q0.

00i i


The vibrational energy at ground state is

therefore

the number of distribution in any levels does not depend on the selection of zero-point energy.

1,0 2 h v

0 /2h kTq e qv v

0 00

0

/ ( )/ //0 0

i i ikT kT kTi i i ikT

N N Nn g e g e g e

q q e q

es

iielec hEU

mod

0 2

1 Born- Oppenheimer


0

0i

kTi

i

q g e

0

0 kTq e q

,0 ,0 ,0

,0 ,0

/ / /0 0 0

/ /0 0

; ;

;

t r v

e n

kT kT kTt t r r v v

kT kTe e n n

q e q q e q q e q

q e q q e q

Since εt,0≈0, εr,0=0, at ordinary temperatures.

0 0,t t r rq q q q


Translational partition function

222 2

,t 2 2 2( )

8yx z

i

nnh n

m a b c

,it ,i exp( )t

ti

q gkT

Energy level for translation

The partition function


22 22

2 2 21 1 1

2 2 22 2 2

2 2 21 1 1

, , ,

exp /8

exp exp exp8 8 8

x y z

x y z

yx zt

n n n

x y zn n n

t x t y t z

nn nhq kT

m a b c

h h hn n n

mkTa mkTb mkTc

q q q

22

, 21

exp8

x

t x xn

hq n

mkTa

22

, 21

exp8

y

t y yn

hq n

mkTb

22

, 21

exp8

z

t z zn

hq n

mkTc

32

t 2

2( )

mkTq a b c

h

32

2

2 ( )

mkTV

h


Rotational partition function The rotational energy of a linear molecule is given by

εr = J(J+1)h2/8π2I and each J level is 2J+1 degenerate. 2

r 2( 1) 0 1 2

8

hJ J J

I

，，，

, 2

, 20

(2 1)exp ( 1)8

r i

kTr r i

i J

hq g e J J J

IkT

define the characteristic rotational temperature

2

28r

h

Ik


rr

0

( 1)(2 1)exp( )

J

J Jq J

T

Θr<<T at ordinary temperature, The summation can be approximated by an integral

r0(2 1)exp ( 1) / drq J J J T J

Let J(J+1)=x, hence J(2J+1)dJ=dx, then

2 2r r0

exp( / )d 8rq x T x T IkT h


Vibrational partition function

v

1( ) 0,1,2,

2v h v

Vibrational energies for one dimensional oscillator are

Vibration is non-degenerate, g=1. The partition function is

,

,0

1exp /

2

i

kTi

i

q g e h kT

v

v vv

v


v,0 v0v /

1exp1 exp( )

kT qqhkT

take the ground energy level as zero,

For NO, the characteristic vibrational temperature is 2690K. At room temperature Θv/T is about 9; the , indicating that the vibration is almost in the ground state.

0 1q v


Thermodynamic energy and partition function

/

/

Independent particle system:

;

(8.48)

i

i

i ii

kTi i

kTi i

i

U n

Nn g e

q

NU g e

q


/

/2

/2

/2

1

1

i

i

i

i

kTi

iV V

kT ii

i

kTi i

i

kTi i

iV

qg e

T T

g ek T

g ekT

qkT g e

T

Substitute this equation into equation (8.48), we have

Thermodynamic energy and partition function


2 2 ln

V V

N q qU kT NkT

q T T

Substitute the factorization of partition function for q

2 ln t r v e n

V

q q q q qU NkT

T

Only qt is the function of volume, therefore


2 2 2

2 2

ln lnln

ln ln

t vr

V

e n

t r v e n

q d qd qU NkT NkT NkT

T dT dT

d q d qNkT NkT

dT dTU U U U U

If the ground energy is specified to be zero, then

00 2 ln

V

qU NkT

T

0 /0Substitute = into this equation, it follows thatkTq qe


00U U N

It tells us that the thermodynamic energy depends on the zero point energy. Nε0 　 is the total energy of system when all particles are localized in ground state. It (denoted as U0) can also be thought of as the energy of system at 0K. Then,

00U U U

es

iielec hEU

mod

0 2

1


U0 can be expressed as the sum of different energies

0 0 0 0 0 0

0 0 0

0 0

2

0 0

t r v e n

t t r r v v

e n

U U U U U U

NhvU U U U U U

U U


The calculation of (1) The calculation of

0 0 0, ,t r vU U U

0tU

0 2

3/2

22

ln

2ln

3

2

tt t

V

V

qU U NkT

T

mkTV

hNkT NkT

T

＝


The calculation of

0 2

2

ln

ln

rr r

V

r

qU U NkT

T

Td

NkT NkTdT

＝

0rU

The degree of freedom of rotation for diatomic or linear molecules is 2, the contribution to the energy of every degree is also ½ RT for a mole substance.


The calculation of 0vU

0 /0 2 2

/

1lnln 1

1

1

v

v

Tv

v

v T

dd q eU NkT NkTdT dT

Nke

Usually, Θv is far greater than T, the quantum effect of vibration is very obvious. When Θv/T>>1, Showing that the vibration does not have contribution to thermodynamic energy relative to ground state.

0 0vU


If the temperature is very high or the Θv is very small, then Θv/T<<1, the exponential function can be expressed as

/ 1TeT

v v

0v /

1 1

1 1 1T

U Nk Nk NkTe

T

vv vv


For monatomic gaseous molecules we do not need to consider the rotation and vibration, and the electronic and nuclear motions are supposed to be in their ground states. The molar thermodynamic energy is

0,

3

2m mU RT U

es

iielec hEU

mod

0 2

1


For diatomic gaseous molecules vibration and rotation must be considered. If only lowest vibrational levels are occupied, the molar thermodynamic energy is

00, v

5( 0)

2m mU RT U U

es

iielec hEU

mod

0 2

1


If all vibrational energies are equally accessible, the molar thermodynamic energy for vibration is

The molar thermodynamic energy for diatomic molecules is then

0vU RT

00, v

7( )

2m mU RT U U RT


Heat capacity and partition function

The molar heat capacity, CV,m, can be derived from the partition function.

,UmCV m T V

ln2,

qRT

T VV m T V

C

Replace q with 0 /0 kTq q e 0ln2

,q

RTT

VV m T

V

C

We can see from above equations that heat capacity does not depends on the selection of zero point of energy.


Electrons and nucleus are in ground state

002 2

, , ,

0ln2,

lnln vr

V V

V t V r V v

qtRTT

VV m T

V

qqC RT RT

T T T T

C C C


The calculation of CV,t, CV,r and CV,v

(1) The calculation of CV,t

32

2

2( )t

mkTq V

h

0 3ln22,

qtRT RT

VV m T

V

C


(2) The calculation of CV,r

2

2

8(linear molecules)r

r

IkT Tq

h

02

,

ln rV r

V

qC RT R

T T

If the temperature is very low, only the lowest rotation state is occupied and then rotation does not contribute to the heat capacity.


The calculation of CV,v

T

kT

eqeq

v

0,v

1

1v

0v

22

vv, 1

vv

TTV ee

TRC

02 v

,v

lnV

V V

qdC RT

dT T


v v v v

v

2 22 2v v

,v

2

v

1

0

T T T TV

T

C R e e R e eT T

R eT

It shows that under general conditions, the contribution to heat capacity of vibration is approximately zero.

Generally, Θv/T>>1, equation becomes


v

1 vTeT

When temperature is high enough,

v v2 2

v v,v

T TVC R e Re R

T T


3

2VC R

, ,

50

2V V t V vC C C R

In gases, all three translational modes are active and their contribution to molar heat capacity is

The number of active rotational modes for most linear molecules at normal temperature is 2

122VC R R

In most cases, vibration has no contribution to the heat capacity,

Entropy and partition function

Entropy and microstate

Boltzmann formula

k = 1.38062×10-23 J K-1

As the temperature is lowered, the Ω, and hence the S of the

system decreases. In the limit T→0, Ω=1, so lnΩ=0, because

only one configuration is compatible with E=0. It follows

that S→0 as T→0, which is compatible with the third law of

thermodynamics.

lnS k

• For example

maxln ln lnDD

W W

maxlnS k W

48

50

100.01

10

48

50

ln100.96 1

ln10

When N approaches infinity, maxln

1ln

W

Entropy and partition functionEntropy and partition function

• For a non-localized system, the most probable distribution number is

• Using Stirling equation ln N!=N ln N - N and Boltzmann distribution expression

•

!

ini

Di i

gW

n

ln ( ln ln !)D i i ii

W n g n

/i kTi i

Nn g e

q

• We have,

0 0

ln ( ln ln )

( ln ln ln )

ln

ln ln (non-localised system)

or

ln (non-localised system)

B i i i i ii

i ii i i i i i

i

B

W n g n n n

nNn g n n g n

q kT

q UN N

N kTq U

S k W Nk NkN T

q US Nk Nk

N T

• For localized system

• Entropy does not depend on the selection of zero point energy .

00

ln ln (localised system)

or

ln (localised system)

B

US k W Nk q

T

US Nk q

T

• Factorizing the partition function into different modes of motions and using

• We can give

0 0 0 0 0 0t r v e nU U U U U U

t r v e nS S S S S S

0 0

ln t tt

q US Nk Nk

N T

0 0

ln v vv

q US Nk

N T

0 0

ln e ee

q US Nk

N T

0 0

ln e et

q US Nk

N T

0 0

ln r rr

q US Nk

N T

For identical particle system, entropies for every mode of

motion can be expressed as

Calculation of statistical entropyCalculation of statistical entropy

• At normal condition electronic and nuclear motions are in ground state, and in general physical and chemical process the contribution to the entropy by two modes of motion keeps constant. Therefore only translational, rotational and vibrational entropies are involved in computation of statistical entropy.

vt rS S S S

Calculation of statistical entropyCalculation of statistical entropy

30 22

2( )t t

mkTq q V

h

0 0

ln t tt

q US Nk Nk

N T

3/2

3

2 5ln

2t

mkT VS Nk Nk

Nh

(1) Calculation of St

0 3

2tU NkT

1,

3 5ln / kg mol ln( / K) ln( / Pa) 20.723

2 2m tS R M T p

• For ideal gases, the Sackur–Tetrode equation is used to calculate the molar translational entropy.

(2) Calculation of (2) Calculation of SSrr

• For linear molecules

• When all rotational energy levels are accessible

• We obtain

0 0

ln r rr

q US Nk

N T

0 0/r r r rq q T U NkT

ln( / )r rS Nk T Nk

, lnm rr

TS R R

(3) Calculation of (3) Calculation of SSvv

• Substitute

• Into the following equation

1 1/ /0 01 and 1v vT Tv v rq e U Nk e

0 0v v

1 1/ /1v

ln /

ln 1 1v v

v

T T

S Nk q U T

Nk e Nk T e

v v1 1/ /1

m,v vln 1 1T TS R e R T e


Entropy contribution

Electronic: 1 Translational:

Rotational: Vibrational: Symmetry:

S = RT ln2


Rotational contribution„Linear” „Non linear“


Vibrational contribution K 0

B. Viskolcz, Sz. N. Fejer and I.G. Csizmadia: J. Phys. Chem.A.3808, 110, (2006).

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Molecular Information Content Béla Viskolcz Department of Chemical Informatics University of Szeged