Fundamental relations: The thermodynamic functions The molecular partition function

Fundamental relations: The thermodynamic functions The molecular partition functionUsing statistical thermodynamics Mean energies Heat capacities Equation of state Residual entropies Equilibrium constants

Chapter 20 Statistical thermodynamics: the machinery

Exercises for Chapter 20

• 20.1(b), 20.3(a), 20.6(a), 20.10(b), 20.12(a), 20.15(b),20.17(a)

• 20.3, 20.6, 20.10, 20.16, 20.19

Fundamental relations

• The thermodynamic functions

• The molecular partition function

The thermodynamic functions: A and p

)0( QkT

UUS ln

The Helmholtz energy

QkTAA ln)0(

Independent molecules:

(Distinguishable) (Indistinguishable)

A= U - TS A(0)=U(0)

The pressure

Thermodynamic Riddles (Ch.5)

dU=TdS-pdV, dH=TdS+VdpdG=Vdp-SdT, dA=-SdT-pdV

Enjoy!!

U=q+w, H=U+pV, G=H-TS, A=U-TS

Deriving an equation of state

q nRTkTnNNkT A

Derive the expression for the pressure of a gas of independent particles

VqNqQ NFor a gas of independent particles:

Where following relations have been used:

Classroom exercise• Deriving the equation of state of a sample for which

With 3/Vq where f depends on the volume

kTVnRT

NfqNkT

]0)()([

]!lnlnln[

ln)ln3(ln

The thermodynamic function: H

lnln)0(

nRTHH2

H=U+pV

For a gas of independent particles:

nRTUU2

The thermodynamic functions: G

QkTVQkTGG

nRTQkTGG ln)0(

!NqQ N !lnlnln NqNQ NNNN ln!ln

qnRTnRTNNNkTqnRT

nRTNkTqNkTGG

ln)ln(ln

!lnln)0(

qnRTGG ln)0(

G=H-TS=A+pVQkTAA ln)0(

For a gas of independent particles: nRTpV G=A+pV

)( AnNN

Define molar partition functionn

The molecular partition functionEi

qqqqeeee

Translational, Rotational, Vibrational, Electronic energies

The translational contribution

22 mkT

TTqAt room temperature, O2 in a vessel of 100 ml

28102Tq pm18m108.1

At room temperature, H2

Independent states (factorization of q)

Xnnnn 321321

ZYX qqq

321321

22 mkT

Three-dimensional box:

Thermal wavelength

(Translational partition function)

Typical Rotors

The Rotational Energy Levels2aaa I

1E Around a fixed-axis

1ccbbaa IIIE Around a fixed-point

aaa IJ c

(Spherical Rotors)

22 1JJJ ,...2,1,0J

j I21JJE

1JhcBJE j

BJJFJF 21

Linear Rotors

,...2,1,0J 1 JhcBJEJ

The rotational contribution

R eJegq J )1()12(

hcB/kT=0.05111

At room temperature, for H(1)Cl(35),B=10.591 1/cm

kT/hc=207.22 1/cm

(Linear rotors)

The rotational contribution: Approximation(Linear rotors)

hcB/kT<<1

)1()12( dJeJq JhcBJR

)1(1dJe

hcBq JhcBJR

R eJegq J )1()12(

Symmetric Rotors

JJE acb

2222J cba JJJ

ac JIIII

21, KBAJBJKJF

,...2,1,0J JK ,.....,1,0

//4 cIA

JJJMKJ

KBAhcJhcBJE

,...1,,...3,2,1,0

)()1( 2,,

The rotational contribution: Approximation(Symmetric rotors)

JJJMKJ

KBAhcJhcBJE

,...1,,...3,2,1,0

)()1( 2,,

JJKMeq

kTJhcBJKkTBAhc

kTKBAhcJBJhc

eJq JJKM

dJdKeJeqK

kTJhcBJKkTBAhcR

)1()12(2

hcB/kT<<1

KJfKJf||0

),(),(

The rotational contribution: Approximation(Symmetric rotors)

kTdJeJ kJJhcBJ

kTJhcBJ /)1(

/)1()12(

kJhcBK

kJKKhcBK

kJJhcBJ

/)1||(||||

dKeehcB

kTq kThcBKKkTBAhcR

dxehcA

kTq xKkThcAR

The rotational contribution: Approximation(Asymmetric rotors)

1ccbbaa IIIE

(If you have question here, just ignore it.The detailed derivation of this equation is beyond theScope of this course.)

Rotational temperature

khcBR /

The ``high temperature`` approximation means

From Table 20.1, it is clear that this approximation is indeed valid unless the temperature is not too low (< ~10K).

Symmetry number

• How to avoid overestimating the rotational partition function?

Symmetrical linear rotor

After deducting the indistinguishableStates,

In general,

molecules diatomicear heteronucl.....1

molecules diatomicr homonuclea......2

Symmetry number

Nonlinear molecules:

General cases:

the number of rotational symmetry elements.

Symmetry Group and Symmetry Numbers

C1 CI CS: 1 D2 D2d D2h: 4 C : 1

C2 C2v C2h: 2 D3 D3d D3h: 6 D : 2

C3 C3v C3h: 3 D4 D4d D4h: 8 T, Th Td: 12

C4 C4v C4h: 4 D5 D5d D5h: 10 O, Oh: 24

C5 C5v C5h: 5 D6 D6d D6h: 12 I, Ih: 60

C6 C6v C6h: 6 D7 D7d D7h: 14 S4: 2

C7 C7v C7h: 7 D8 D8d D8h: 16 S6: 3

C8 C8v C8h: 8 S8: 4

kT/1 226.207

A=4.828 1/cm, B=1.0012 1/cm, C=0.8282 1/cm, T=298 K

4ABC=4.0033 1/cm

Classroom exercise

kT/1 226.207

A=0.2014 1/cm, B=0.1936 1/cm, C=0.0987 1/cm, T=298 K

ABC=0.004 1/cm

4103.4 Rq

Quantum mechanical interpretation

)1()2()2()1()2/12/1()2,1( )2()1()2,1(

)1()2()2()1()2/12/1()2,1(

The wavefunction of fermions changes sign whenexchanged whereas the wavefunction of bosons does not change sign when exchanged.

)2()1()2,1(

JhcBJR eJeJq )1()1( )12(3)12(4

JhcBJR eJq )1()12(2

(for even J)

(for odd J)

Nuclear spin = 0

Only even J-states are admissible

CO2 Boson

)3()2()1(

There are 8 nuclear spin states:

,...11,8,5,2

with functionseigen wave rotational

with combiningby onswavefuncti

lsymmetrica-anti form that Those)3(

,...10,7,4,1

with combiningby on wavefuncti

lsymmetrica-anti form that Those)2(

.0,3,6,9,..J

with combiningby onswavefuncti

lsymmetrica-anti form that Those (1)

:rotation about thesymmetry

their toaccording unctionsspin wavefClassify

)......3,2,1(

:ion wavefunt totalsymmetric

-anti form toionseigenfunct rotational the

with combine that statesspin ofnumber The

33...,11,8,5,2

22...,10,7,4,1

01...,9,6,3,0

:rotation Cwith

invariant are that statesspin ofnumber The

:rotation Cwith

unctions)spin wavef ricantisymmet

ofnumber the( :rotation Cwith

n )3,2,1(),(, KJY

,...3,2,1,0

,...8,5,2 ,...9,6,3,0

,...7,4,1

])12([

...)12()8()12()12(8

JhcBJJhcBJ

JhcBJR

eJnneJneJnq

• Generally, for a molecule with NR rotational elements (including the identity operation), the symmetry number

Symmetry Group and Symmetry Numbers

C1 CI CS: 1 D2 D2d D2h: 4 C : 1

C2 C2v C2h: 2 D3 D3d D3h: 6 D : 2

C3 C3v C3h: 3 D4 D4d D4h: 8 T, Th Td: 12

C4 C4v C4h: 4 D5 D5d D5h: 10 O, Oh: 24

C5 C5v C5h: 5 D6 D6d D6h: 12 I, Ih: 60

C6 C6v C6h: 6 D7 D7d D7h: 14 S4: 2

C7 C7v C7h: 7 D8 D8d D8h: 16 S6: 3

C8 C8v C8h: 8 S8: 4

10 points!!

Derive from quantum mechanics thesymmetry number of benzene:

The vibrational contribution

vhcE ~)2

1( v＝ 0 , 1 , 2 ,

)(~~ vhcvhcV eeq

)limit re temperatu(11

1 )(0~ low

:limit ratureHigh tempe

Normal modes

3N-6 vibrational degrees of freedom

For a nonlinear molecule of N atoms, there are 3N degrees of freedom: 3 translatinal, 3 rotational and

For a linear molecule of N atoms, there are 3N degrees of freedom: 3 translatinal, 2 rotational and

3N-5 vibrational degrees of freedom

The total vibrational partition function is:

5321~~~

vhcvhcvhcvvvV

eeeqqqq

6321~~~

vhcvhcvhcvvvV

eeeqqqq

Exercise• The wave number of the three vibrational modes of H2O

3656.7 1/cm, 1594.8 1/cm, and 3755.8 1/cm. Calculate vibrational partition function at 1500 K.

1-cm 6.1042/~

)//(~/~~

hckTkThchc

The total vibrational partition function then is: 1.031x1.276x1.028=1.353

1.0281.2761.031

3.6021.5303.507

3755.81594.83056.7

321mode

1.0281.2761.031

3.6021.5303.507

3755.81594.83056.7

321mode1/~ cm

kThc /~Vq

At 1500 K, most molecules are at their vibrational ground state!

Classroom exercise

• The three vibrational normal modes of CO2 are 1388 1/cm, 667.4 1/cm (doubly degenerate), 2349 1/cm. Calculate the vibrational partition function at 1500K.

.)..........~1(1

Low temperature approximation

111~ VV qqhc i

kThc ~High temperature approximation

khcV /~

Only the zero-point level is occupied.

The electronic contribution

lsjenergylevej

For most cases, the excited energy is much larger than kTand the electronic energy level of the ground is not degenerate:

1,00 00 j

lsjenergyleve

jE eegq j 22

Degenerate case: NO

eegqlsjenergyleve

The overall partition function

kTVgq ~3 1

kT3elec vib rot

(Linear rotor, Single vib mode)

)53or(63

vhcABCE

(Nonlinear rotor, multiple vib mode)

M=NA*m

Exercise

qnRTGG ln)0(

7.486oRmq2 )exercis previous( 352.1 eq oV

22 mkT

)Table20.3 c.f.( mol 10706.1 1-8

kJmol -317.31.352}486.710ln{1.706

K1500molK3145.8

)ln()0()(

mom RTGTG

qRTGTG lnln)0()(

• Calculate the value of molar Gibbs energy for H2O(g) at 1500 K given that A=27.8788 1/cm, B=14.5092 1/cm, and C=9.2869 1/cm and the information of normal modes given in last exercise.

Classroom Exercise

• Calculate the value of molar Gibbs energy for CO2(g) at 1500 K given that B=0.3902 1/cm. The three vibrational normal modes of CO2 are 1388 1/cm, 667.4 1/cm (doubly degenerate), 2349 1/cm

Classroom Exercise

qnRTGG ln)0(

mq2 ??oVmq

22 mkT

)Table20.3 c.f.( mol ??? 1-

kJmol -366.6??}??ln{??

K1500molK3145.8

)ln()0()(

mom RTGTG

qRTGTG lnln)0()(

• Calculate the value of molar Gibbs energy for CO2(g) at 1500 K given that B=0.3902 1/cm. The three vibrational normal modes of CO2 are 1388 1/cm, 667.4 1/cm (doubly degenerate), 2349 1/cm

II. Using statistical thermodynamics

• Mean energies

• Heat capacities

• Equation of state

• Residual entropies

• Equilibrium constants

Mean energies

1 M=T,R,V,orE

The mean translational energy

1D case:

3D case:

The mean rotational energy

......531 62 hcBhcBR eeq

......531

...)306(62

hcBhcB

hcbhcBM

JhcBJR eJq )1()12(2

1 (Linear rotors)

......531

...)306(62

hcBhcB

hcbhcBM

kTBhcd

The mean rotational energy

At high temperature, the partition function of a linear rotor:

Classroom exercise: the mean rotational energy of an asymmetricrotor at high temperature:

2/32/32

CkTCABChc

31 2/3

The mean vibrational energy

1...)~1(

Heat capacities

EVRTMNkT

MV ,,,,2

kTdNC A

Translational contribution:V

kTBhcd

The rotational contribution:

......531

...)306(62

hcBhcB

hcbhcBM

At high temperature:

Common cases

At low temperature:

The vibrational contribution:

2, RfCVmV

RvvC VRmV )23(2

Common cases

Rare cases: Cv,mR

The overall heat capacity

RvvC VRmV )23(2

kTdNC A

, RfCVmV

moleculesnonlinear for

moleculeslinear for

moleculesnonlinear for 3

moleculeslinear for 2

RvC RmV )3(2

At fairly high temperature, the vibrational contribution Is cloase to zero: Common cases:

θR<<T<<θV

RvvC VRmV )23(2

The overall heat capacity(diatomic molecules)

Common cases

Exercise• Estimate the molar constant-volume heat

capacity of water vapor at 100C.

The wave number of the three vibrational modes of H2O:3656.7 1/cm, 1594.8 1/cm, and 3755.8 1/cm.

The rotational constants of H2O: A=27.9 1/cm, B=14.5 1/cmAnd C=9.3 1/cm.

RvvC VRmV )23(2

Common cases:θR<<T<<θV

J/K/mol 0.253)033(2

1, RRC mV

Experimental value: 26.1 J/mol/K

Classroom Exercise• Estimate the molar constant-volume heat capacity

of gaseous I2 at 25C.

The rotational constants of I2: B=0.037 1/cm.

A rare case: θV ~ T

Classroom Exercise• Estimate the molar constant-volume heat capacity

of gaseous I2 at 25C.

The rotational constants of I2: B=0.037 1/cm.

RvvC VRmV )23(2

J/K/mol 2948.3)96.123(2

1, RRC mV

2, RfCVmV

99.0)(1

298/309

596/309

1298309

A rare case: θV ~ T, Cv,mR

Experimental value: 29.06 J/mol/K

Equations of state

....12

Equation of state of perfect gas:

For real gases,

The purpose: to find expressions for B and C in terms of the intermolecular interactions.

NrrrEN drdrdreQ Npiii ...21

),...,(21 21

),...,(21 21 Npiiii rrrENE

N drdrdreZ Np ...21),...,(

) ...!

1(or 21 N

E drdrdreN

021)0( !.....

Vdrdrdr

For perfect gases, N! should be dropped forsystems of distinguishable particles.

Deriving an equation of state

q nRTkTnNNkT A

Derive the expression for the pressure of a gas of independent particles

VqNqQ NFor a gas of independent particles:

Where following relations have been used:

Pair interactions

jijipjipNp rrErrErrrE

1 1,21

21 ),(),(),...,(

),(),( all bapjip rrErrE

terms2

321),(...),(),(

21),...,,(

NrrErrErrE

drdrdrdre

drdrdrdreN

drdrdreZ

212353

4324232

),(),(...),(...),(

),(),(...),(),(

),(...),(),(

),...,(

NNpNNpNpp

rrErrErrErrE

rrErrErrE

Further approximation:

Mayer function

drdrdrrf

drdrdreZ jip

...)1)((

1),( jip rrEef

...])([

]......))()()(1[

12122!1

121212!1

drrfdrV

drdrrfVV

drdrdrrfrfrfZ

Nji lk

klijji

2112 )(2

1drdrrf

Second virial coefficient B

212drfdr

NbNB A

QkTAA ln)0( vV

...])()1([

...])([)(221

NNVNTV

VTbNNNV

...])(1[

...]1[

...)()1(

VTbNNV

VTbNNNV

....12

Second virial coefficient B(spherical potential)

|||||)(|2

4|||||)(|2

||sin|||)(|2

rdrrfN

VrdrrfV

drddrdrrfV

drfdrV

|)(|),(|)(|

ijp rE

ijpjip

The interaction potential dependson distance only.

Second virial coefficient B (Hard sphere potential)

1fwhen r σ, E≦ p= ∞:

1 pEe 0fwhen r σ ,E≧ p= 0:

drrrfNB

Application of virial coefficient B • For perfect gases, all virial coefficients except the

first one are zero.• The thermodynamic properties that depend on

intermolecular interactions are determined by second and higher order virial coefficients.

lnln)0(

...])([ 12!

1 TbVNVZ NNN

...][ )(12!

NZ VN ...])()1([ 221

TbVNNNV NN

(Classroom Exercise)

Answer

kTnNpVV

Z N )(2

)()1( 221 TbVNNNVVZZ

12)(22

221)(2

)(1112

)]()1([

)]()1([)0(

TNNTBVNNpV

NTTbVNpV

NTTbVNNkT

TbVNNNkTkTN

TbVNNNVkTVNHH

TTNBVN

TNNTBVNNV

)(1112

Residual entropy

• The entropy (disorder) at temperature zero.

2ln2ln2lnln nRNkkWkS Nresidual

AB AB AB ABAB AB

AB AB BA ABAB BA

There are 2^N microscopic states for N molecules with two equallypossible orientations at T=0

General cases

sRS mresidual ln,

ABC ABC ABC ABCABC ABC

ABC ABC ACB ABCABC BAC

There are s equally possible orientations at T=0:

snRsNkskS Nresidual lnlnln

The residual entropy of iceHOH HOH HOH HOHHOH HOH

H2O is nonlinear

Hydrogen bonds areoriented.

The acceptable arrangement:Out of the four hydrogen atoms around an oxygen atom, two are close and two are far.

There are sixteen (2^4) equally possible configurations for ice, but only six of them are allowed.

J/mol 4.3ln 23

, RS icemresidual

Equilibrium constants

KRTGomr ln

(gas-phase reactions)

qRTJGJG

0,ln)0,()( For species J,

aA+bB+…cC+dD+…

omC re

NqNqK /

BbGAaGDdGCcG

BbGAaGDdGCcGG

,,,, lnlnlnln

)0,()()0,()0,(

)()()()(

)0,()0,()0,()0,(0 BbUAaUDdUCcUE om

Standard reaction Gibbs energy

G(0)=U(0)

KRTGor ln

NqNqRT

NqNqRTE

qcRTEG

)()(ln

lnlnlnln

)()(lnln

X2(g)→2X(g) oX

omXRTE

mX rr eNq

, )()(

)()0,()0,(2 XX020 DXUXUE om

RTgVgq

qqRTgqq

XX eqqgp

kTgK 0

A dissociation equilibrium

(dissociation energy of bond X-X)

Equilibrium constant: an example• Evaluate the equilibrium constant for the dissociation

Na2(g)2Na(g) at 1000 K with data: B=0.1547 1/cm, v=159.2 1/cm, D0=70.4 kJ/mol. The Na atoms have doublet ground terms.

XX eqqgp

kTgK 0

1JPam 1 Pa,10,2 35 op

885.4,2246

pm 11.5Λpm, 8.14Λ

2 NaNa

)m1015.1(4.8852246Pa10

)m1014.8(4K1000JK1038.1 47.86115

312123

Classroom exercise• Evaluate the equilibrium constant for the

dissociation Na2(g)2K(g) at 1500 K with data: B=0.1547 1/cm, v=159.2 1/cm, D0=70.4 kJ/mol. The Na atoms have doublet ground terms.

1500100047.8

312123

)m1015.1(4.8852246Pa10

)m1014.8(4K1000JK1038.1

Contributions to the equilibrium constant:density of states

dEEdnE )()(

Gas-phase reaction:

The density of states==The number of states in agiven range of energy:

Similar densities of states for products and reactants

)()( RP

The equilibrium is dominatedby the species with lower zero-point energy.

Very different densities of states

)()( RP

The equilibrium is dominatedby the species with larger density of states.

Contributions to the equilibrium constant

PR The number ratio of product to reactant molecules:

which leads to the equilibrium constant:

00 ANE

Energy separation and state density on equilibria

RTErekT

)levelsenergy spacedevenly (

favoured.not is

product 1, large, is when 0r KE

1) (with dominant be

might product ratures,high tempeAt

favoured. is here) (smaller states

ofdensity larger with species the,

Gibss energy , rather than enthalpy, controls position of equilibrium.

Helix structure of polypeptides (proteins)

Helix-coil transition

The statistical physics of helix-coil transition:a tetrapeptide

040302010

qqqqqqqqq

qqqqqq

h h h h

h h c h

h c h h

c h h h

h h h c

c c h h

c h c h

h c h c

h c c h

c h h c

h h c c

c c h c

c h c c

h c c c

c c c h

c c c c

4321 4641 KKKKq

432 14641i

ii sNssssq

4,3,2,1,*

00 ieK RTGi

es (stability parameter)

Helix-coil transition of proteins with n amino acid residues

h h h h h c h hc h h h

h h h h c c h hc c h h

h h h h h c h cc h h h

Zipper model:

Zipper model

h h h h h c h hh c h h

h h h h h c h hc c h h

h h h c h c h hc c h h

h c c c h c h hc c h h

The coils are necessarily in a contiguous region.

Zipper modelh h h h h c h hh h h h

Nucleation with equilibrium constantσ<<1.

ii sNq

i issnq11

)1(1 1 inN i

The number of possibilities of placing a coil with i amino acidsIn a peptide of n amino acids: n-i+1

c c h h h c h hh h h h

h c c h h c h hh h h hh h c c h c h hh h h h

9.15 i

11)1(ln

Degree of conversion

1 (Classroom exercise)

Zimm-Bragg model

212 41

h h c c h c h hh h c c

Separate coil regions are considered

Fundamental relations: The thermodynamic functions The molecular partition function

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