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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.
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
One atom
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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.
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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.
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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:
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
Összefoglalás
Degrees of
freedomAtom
Linear
polyatomic
Bent
polyatomic
Translation
al3 3 3
Vibrational 0 3N-5 3N-6
Rotational 0 2 3
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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)
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
Introduction Statistical Mechanics
Properties of individual molecules
PositionMolecular geometry
Intermolecular forces
Properties of bulk fluid (macroscopic properties)
PressureInternal EnergyHeat Capacity
EntropyViscosity
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
, 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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
Potential V(R) for nuclear motion in a diatomic molecule Harmonic oscillator potential
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
Energy level
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
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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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
/
/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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
=
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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.
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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.
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
(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.
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
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
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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
Department of Chemical Informatics, University of Szeged Molecular statistic 2014
v
1 vTeT
When temperature is high enough,
v v2 2
v v,v
T TVC R e Re R
T T
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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
Department of Chemical Informatics, University of Szeged Molecular statistic 201457
Entropy contribution
Electronic: 1 Translational:
Rotational: Vibrational: Symmetry:
S = RT ln2
Department of Chemical Informatics, University of Szeged Molecular statistic 201458
Rotational contribution„Linear” „Non linear“
Department of Chemical Informatics, University of Szeged Molecular statistic 201459
Vibrational contribution K 0
B. Viskolcz, Sz. N. Fejer and I.G. Csizmadia: J. Phys. Chem.A.3808, 110, (2006).