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

kT/1

NqQ

v

QUU

ln

)0( QkT

UUS ln

)0(

!NqQ

N

The Helmholtz energy

QkTAA ln)0(

TV

Ap

TV

QkTp

ln

Independent molecules:

(Distinguishable) (Indistinguishable)

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

The pressure

Thermodynamic Riddles (Ch.5)

ST

A

V

TS

H

p

pV

U

S

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

Vp

G

T

Vp

H

p

G

ST

TS

H

S

U

pV

pV

A

V

U

TS

ST

A

T

G

Vp

Enjoy!!

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

Deriving an equation of state

V

nRT

V

NkT

V

NkT

V

q

q

NkT

V

Q

Q

kT

V

QkTp

TTT

3

3 1

ln

3

3 1/

TT V

V

V

q nRTkTnNNkT A

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

3 ,!/

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

!NfqQ

N

With 3/Vq where f depends on the volume

TVf

TVf

TVV

TVf

TVq

T

T

Nfq

T

kTVnRT

kTnRT

NkT

V

NfqNkT

VkT

V

QkTp

N

)(/

)()(

]0)()([

]!lnlnln[

)ln(

ln

ln

ln)ln3(ln

lnln

!

The thermodynamic function: H

TVV

QkTV

QHH

lnln)0(

nRTHH2

5)0(

H=U+pV

v

QUU

ln

)0(TV

QkTp

ln

For a gas of independent particles:

nRTUU2

3)0(

The thermodynamic functions: G

TV

QkTVQkTGG

ln

ln)0(

nRTQkTGG ln)0(

!NqQ N !lnlnln NqNQ NNNN ln!ln

N

qnRTnRTNNNkTqnRT

nRTNkTqNkTGG

ln)ln(ln

!lnln)0(

A

m

N

qnRTGG ln)0(

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

TV

QkTp

ln

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

Or

)( AnNN

Define molar partition functionn

qqm

The molecular partition functionEi

Vi

Ri

Tii

EVRT

iiii

ii

qqqqeeee

eeq

ETi

Vi

Ri

Ti

Ei

Vi

Ri

Tii

i

ieq

Translational, Rotational, Vibrational, Electronic energies

The translational contribution

3

VqT

2

1

2

1

22 mkT

h

mh

TTqAt room temperature, O2 in a vessel of 100 ml

28102Tq pm18m108.1

2

11

2

1

mkT

h

At room temperature, H2

pm71

222

2

2

2

2

1 Om

m

H

H H

O

kTm

h

?2H

Independent states (factorization of q)

Zn

Yn

Xnnnn 321321

ZYX qqq

q

Zn

Yn

Xn

Zn

Yn

Xn

Zn

Yn

Xn

3

-

2

-

1

-

--

n all

-

n all

---

321

321321

eee

eeee

XYZh

mq

23

2

2

3V

q

21

21

22 mkT

h

mh

Three-dimensional box:

Thermal wavelength

(Translational partition function)

Typical Rotors

The Rotational Energy Levels2aaa I

2

1E Around a fixed-axis

222

2

1

2

1

2

1ccbbaa IIIE Around a fixed-point

aaa IJ c

c

b

b

a

aI

J

I

J

I

JE

222

222

I2

J

I2

JJJE

22c

2b

2a

(Spherical Rotors)

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

2

j I21JJE

I2

hcB2

cI4

B

1JhcBJE j

BJJFJF 21

Linear Rotors

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

The rotational contribution

J

JhcBJ

JJ

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

0

)1()12( dJeJq JhcBJR

0

)1(1dJe

dJ

d

hcBq JhcBJR

hcBe

hcBq JhcBJR

1

|1

0)1(

J

JhcBJ

JJ

R eJegq J )1()12(

Symmetric Rotors

//

222

22 I

J

I

JJE acb

2222J cba JJJ

2

//

2

//

222

2

1

2

1

2

J

22

Ja

ac JIIII

J

I

JE

21, KBAJBJKJF

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

//4 cIA

cI

B4

JJJMKJ

KBAhcJhcBJE

J

MKJ J

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

)()1( 2,,

The rotational contribution: Approximation(Symmetric rotors)

JJJMKJ

KBAhcJhcBJE

J

MKJ J

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

)()1( 2,,

0J

J

JK

J

JM

kTER

J

JJKMeq

K KJ

kTJhcBJKkTBAhc

K KJ

kTKBAhcJBJhc

K KJ

kTER

eJe

eJ

eJq JJKM

||

)1()(

||

)()1(

||

)12(

)12(

)12(

2

2

dJdKeJeqK

kTJhcBJKkTBAhcR

||

)1()12(2

hcB/kT<<1

J K

0

1

2

3

4

J K

0

1

2

3

4

K KJJ

J

JK

KJfKJf||0

),(),(

=

The rotational contribution: Approximation(Symmetric rotors)

dJedJ

d

hcB

kTdJeJ kJJhcBJ

KK

kTJhcBJ /)1(

||||

/)1()12(

kJhcBK

kJKKhcBK

kJJhcBJ

ehcB

kT

ehcB

kTe

hcB

kT

/

/)1||(||||

/)1(

2

|

dKeehcB

kTq kThcBKKkTBAhcR

/22

21

2

23

21

21

22

ABhc

kT

dxehcA

kT

hcB

kTdKe

hcB

kTq xKkThcAR

21

23

ABChc

kTq R

The rotational contribution: Approximation(Asymmetric rotors)

222

2

1

2

1

2

1ccbbaa IIIE

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

Rotational temperature

cI

B4

khcBR /

The ``high temperature`` approximation means

RT

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

hcB

kTqR

hcB

kTq R

2

After deducting the indistinguishableStates,

In general,

hcB

kTq R

molecules diatomicear heteronucl.....1

molecules diatomicr homonuclea......2

Symmetry number

21

23

1

ABChc

kTq R

Nonlinear molecules:

General cases:

the number of rotational symmetry elements.

2

3

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

21

23

ABChc

kTq R

cmhc

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

661Rq

Classroom exercise

?

1 21

23

ABChc

kTqR

cmhc

kT/1 226.207

2

N

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(

oddJ

JhcBJ

evenJ

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

1

J

JhcBJR eJq )1()12(2

1

(for even J)

(for odd J)

2

CO2

Nuclear spin = 0

Only even J-states are admissible

2

CO2 Boson

?

)3()2()1(

)3()2()1(

)3()2()1(

)3()2()1(

)3()2()1(

)3()2()1(

)3()2()1(

)3()2()1(

8

7

6

5

4

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

J

J

8

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

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

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

:ion wavefunt totalsymmetric

-anti form toionseigenfunct rotational the

with combine that statesspin ofnumber The

210

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

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

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

nnn

nY

nY

nY

KJ

KJ

KJ

3240

2120

10

:rotation Cwith

invariant are that statesspin ofnumber The

:rotation Cwith


unctions)spin wavef ricantisymmet

ofnumber the( :rotation Cwith


o

o

o

n

n

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

,...3,2,1,0

)1(31

,...3,2,1,0

)1(38

81

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

)1(21

)1(2

,...7,4,1

)1(1

)12(

])12([

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

1

J

JhcBJ

J

JhcBJ

J J

JhcBJJhcBJ

J

JhcBJR

eJ

eJ

eJnneJneJnq

3


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


RN

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

vhcV

eq ~

1

1

~hch

vhcV

eq ~

1

1

)limit re temperatu(11

1 )(0~ low

eq T

vhcV

)0(~

1

1

:limit ratureHigh tempe

Tvhc

V

eq

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

5321~~~

1

1...

1

1

1

1...

N

N

vhcvhcvhcvvvV

eeeqqqq

6321

6321~~~

1

1...

1

1

1

1...

N

N

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.

vhcV

eq ~

1

1

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.

3221

~1

1

VVVVV

vhcV

qqqqq

eq

i

i

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

1

vhcqV

vhc

kT

vhcqV

~~1

Low temperature approximation

111~ VV qqhc i

kThc ~High temperature approximation

khcV /~

Only the zero-point level is occupied.

The electronic contribution

110

00

jj

lsjenergylevej

E

ege

egq j

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

00

kT1

kT2

kT3

lsjenergyleve

jE eegq j 22

Degenerate case: NO

Degenerate case: NO

T

T

eegqlsjenergyleve

jE j

4

02

22

The overall partition function

vhcE

ehcB

kTVgq ~3 1

1

00

kT1

kT2

kT3elec vib rot

(Linear rotor, Single vib mode)

)53or(63

1

~2/1

2/3

13 1

1)(

NN

ii

vhcABCE

ehc

kTVgq

(Nonlinear rotor, multiple vib mode)

m

M=NA*m

Exercise

A

m

N

qnRTGG ln)0(

21

23

1

ABChc

kTq R

7.486oRmq2 )exercis previous( 352.1 eq oV

m

3

VqT

2

1

2

1

22 mkT

h

mh

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

A

oTm

Nq

1-8

1-1-

kJmol -317.31.352}486.710ln{1.706

K1500molK3145.8

)ln()0()(

A

oRm

oVm

oTm

Nqqqo

mom RTGTG

A

oVm

oRm

oTm

A

omo

mom N

qqqRT

N

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

A

m

N

qnRTGG ln)0(

hcB

kTqR

??oR

mq2 ??oVmq

3

VqT

2

1

2

1

22 mkT

h

mh

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

A

oTm

Nq

1-

1-1-

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

K1500molK3145.8

)ln()0()(

A

oRm

oVm

oTm

Nqqqo

mom RTGTG

A

oVm

oRm

oTm

A

omo

mom N

qqqRT

N

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

V

M

MM q

q

1 M=T,R,V,orE

The mean translational energy

i

Mii

M n

kTdB

dX

XV

M

2

1

2

112

12

1

kTM

2

3

3

VqT

Vq T

D1

1D case:

3D case:

The mean rotational energy

......531 62 hcBhcBR eeq

......531

...)306(62

62

hcBhcB

hcbhcBM

ee

eehcB

J

JhcBJR eJq )1()12(2

1 (Linear rotors)

......531

...)306(62

62

hcBhcB

hcbhcBM

ee

eehcB

kTBhcd

dBhc

d

dq

q

R

RM

111

The mean rotational energy

hcB

kTq R

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

12

3

/)(1

CkTCABChc

kTqR

kTc

d

d

d

dq

q c

R

RM

23

2/3 2

31 2/3

The mean vibrational energy

2~

~

~

1

~

1

1vhc

vhc

vhc

V

e

evhc

edB

d

dB

dq

vhcV

eq ~

1

1

1

~~

vhc

V

e

vhc

kTvhc

vhcV

1

1...)~1(

~

1

~~

vhc

V

e

vhc

Heat capacities

d

dk

d

d

kTd

d

dT

d

dT

d 22

1

V

v

UkC

2

Vv T

UC

EVRTMNkT

NC

V

M

V

M

MV ,,,,2

RdT

kTdNC A

TmV 2

3)

3

2(

,

3

5

V

p

C

C

Translational contribution:V

T

V

T

TV Nk

TNC

2

kTBhcd

dBhc

d

dq

q

R

RM

111

The rotational contribution:

R

kT

NCC

V

R

NN

V

R

NN

AR

VR

mV AA

2, /

......531

...)306(62

62

hcBhcB

hcbhcBM

ee

eehcB

At high temperature:

V

R

V

R

NNR

mV RT

CA

2,

Common cases

At low temperature:

The vibrational contribution:

2, RfCVmV

T

TV

V

V

e

e

Tf

1

2

RvvC VRmV )23(2

1 **,

Common cases

Rare cases: Cv,mR

The overall heat capacity

RvvC VRmV )23(2

1 **,

RdT

kTdNC A

TmV 2

3)

3

2(

, 2

, RfCVmV

moleculesnonlinear for

moleculeslinear for

23, R

RCV

mV

0

moleculesnonlinear for 3

moleculeslinear for 2

*

*

V

R

v

v

RvC RmV )3(2

1 *,

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

θR<<T<<θV

RvvC VRmV )23(2

1 **,

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

1 **,

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

1 **,

J/K/mol 2948.3)96.123(2

1, RRC mV

2, RfCVmV

99.0)(1

298/309

596/309

1298309

2

ee

T

TV

V

V

e

e

Tf

A rare case: θV ~ T, Cv,mR

Experimental value: 29.06 J/mol/K

Equations of state

....12

mm

m

V

C

V

B

RT

pV

N

ZQ

3

Equation of state of perfect gas:

For real gases,

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

TV

QkTp

ln

i

NrrrEN drdrdreQ Npiii ...21

),...,(21 21

),...,(21 21 Npiiii rrrENE

N

ZQ

3

!0 N

VZ

N

NrrrE

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

!1 21

) ...!

1(or 21 N

E drdrdreN

Z p

021)0( !.....

!

1Z

N

Vdrdrdr

NZ

N

NE p

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

Deriving an equation of state

V

nRT

V

NkT

V

NkT

V

q

q

NkT

V

Q

Q

kT

V

QkTp

TTT

3

3 1

ln

3

3 1/

TT V

V

V

q nRTkTnNNkT A

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

3 ,!/

VqNqQ NFor a gas of independent particles:

Where following relations have been used:

Pair interactions

N

i ij

N

jijipjipNp rrErrErrrE

1 1,21

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

),(),( all bapjip rrErrE

terms2

),(

!1

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

21),...,,(

!1

...

...!

1

...

13221

21

N

Nkji

jirrE

N

NrrErrErrE

NrrrE

N

drdrdrdre

drdrdrdreN

drdrdreZ

jip

NNppp

Np

terms

1

212353

4324232

13121

21

2)1(

),(

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

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

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

),...,(

NN

NNp

NNpNNpNpp

pNppp

Nppp

Np

rrE

rrErrErrErrE

rrErrErrErrE

rrErrErrE

rrrE

Further approximation:

Mayer function

Nji

ijN

Nji

rrE

N

drdrdrrf

drdrdreZ jip

...)1)((

...

21!1

21),(

!1

1),( jip rrEef

...])([

...])([

...])([

...])([

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

12!

1

12122!1

121212!1

21122

2)1(

!1

21!1

12

22

TbVNV

drrfV

drrfdrV

drdrrfVV

drdrdrrfrfrfZ

NNN

VNNN

VNNN

NNNNN

Nji lk

klijji

ijN

N

N

2112 )(2

1drdrrf

Vb

Second virial coefficient B

212drfdr

V

NbNB A

A

QkTAA ln)0( vV

Ap

TV

QkTp

ln

TTT V

Z

Z

kT

V

Q

Q

kT

V

QkTp

ln

...])()1([

...])([)(221

!1

12!

1

NNN

NNVNTV

Z

VTbNNNV

VTbNV

...])(1[

...]1[

][

][

][

1

12

1

12

12

12

1

12

221

)(

...)(

...)(

...)(

...)(

...)(

...)()1(

...)(

...)()1(

TbNkT

NkT

NkT

NkTpV

kTp

VN

V

VTNb

VTbNV

VTNb

VTbNV

VTbNV

VTbNV

VTbNNV

VTbNV

VTbNNNV

N

N

NN

N

NN

NN

NN

NN

NN

NN

....12

mm

m

V

C

V

B

RT

pV

Second virial coefficient B(spherical potential)

|||||)(|2

4|||||)(|2

||sin|||)(|2

2

2

2

22

21

rdrrfN

VrdrrfV

N

drddrdrrfV

N

drfdrV

NB

A

A

A

A

1

|)(|),(|)(|

ijp rE

ijpjip

ef

rErrE

The interaction potential dependson distance only.

Second virial coefficient B (Hard sphere potential)

0 pEe

1fwhen r σ, E≦ p= ∞:

1 pEe 0fwhen r σ ,E≧ p= 0:

0

32

0

2

3

22

)(2

AA

A

NdrrN

drrrfNB

σ

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.

dT

dBTBT

p

0lim

TT p

Hμ

TV

V

QkTV

QHH

lnln)0(

...])([ 12!

1 TbVNVZ NNN

TVVZ

ZkTV

Z

ZHH

)0(

...][ )(12!

1

TbN

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

!1

TbVNNNV NN

NVZ

(Classroom Exercise)

Answer

)1(

)1(1

AVNNB

VnB

Am

m

NkTpV

kTnNpVV

B

RT

pV

V

TbZ

Z N )(2

)()1( 221 TbVNNNVVZZ

)(1

])(1[

)(1

])(1[

)(12

12)(22

221)(2

11

)(1112

11

)(12

])(1[

)]()1([

)]()1([)0(

TNBVN

TNNTBVNNpV

TNBVN

NTTbVNpV

TVTb

TVTb

VTb

A

TVTB

AA

A

TVTb

NTTbVNNkT

TbVNNNkTkTN

TbVNNNVkTVNHH

TTB

TTNBVN

TNNTBVNNV

T

T

TTB

p

H

p

Hμ

A

TVTB

AA

)(

)(1

])(1[

)(

)0(11

)(1112

Residual entropy

• The entropy (disorder) at temperature zero.

2ln2ln2lnln nRNkkWkS Nresidual

AB AB AB ABAB AB

AB AB BA ABAB BA

S=0

S>0

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

NsW

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

N

NNW

)(

)(*4

23

166

Equilibrium constants

KRTGomr ln

(gas-phase reactions)

A

mJom

om N

qRTJGJG

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

aA+bB+…cC+dD+…

RTEB

Ao

mBa

Ao

mA

dA

omD

CA

omC re

NqNq

NqNqK /

,,

,, 0

)()(

)()(

A

omB

A

omA

A

omD

A

omC

om

om

om

om

om

om

om

om

or

N

qb

N

qa

N

qd

N

qcRT

BbGAaGDdGCcG

BbGAaGDdGCcGG

,,,, lnlnlnln

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

)()()()(

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

om

om

omr

Standard reaction Gibbs energy

G(0)=U(0)

KRTGor ln

BA

omB

aA

omA

dA

omD

CA

omCr

BA

omB

aA

omA

dA

omD

CA

omC

r

A

omB

A

omA

A

omD

A

omC

ro

r

NqNq

NqNqRT

RT

ERT

NqNq

NqNqRTE

N

qb

N

qa

N

qd

N

qcRTEG

)()(

)()(ln

)()(

)()(ln

lnlnlnln

,,

,,0

,,

,,0

,,,,0

BA

omB

aA

omA

dA

omD

CA

omC

or

NqNq

NqNq

RT

EK

)()(

)()(lnln

,,

,,

RTeN

qK E

V

A

omj

J

r

j

0,

X2(g)→2X(g) oX

X

pp

pK

2

2

RTE

Ao

mX

omXRTE

Ao

mX

AO

mX rr eNq

qe

Nq

NqK 0

2

0

2 ,

2,

,

2

, )()(

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

omr

33,X

oX

X

om

Xo

mX p

RTgVgq

33,

2

222

22

2

22

Xo

VX

RXXV

XRX

X

om

xo

mX p

qqRTgqq

Vgq

RTD

XVX

RXX

o

XX eqqgp

kTgK 0

222

2

6

32

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.

RTD

XVX

RXX

o

XX eqqgp

kTgK 0

222

2

6

32

1JPam 1 Pa,10,2 35 op

2,1

885.4,2246

pm 11.5Λpm, 8.14Λ

2

22

2 NaNa

NaNa

VNa

RNa

gg

qq

42.2

)m1015.1(4.8852246Pa10

)m1014.8(4K1000JK1038.1 47.86115

312123

eK

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

6115

312123

)m1015.1(4.8852246Pa10

)m1014.8(4K1000JK1038.1

eK

Contributions to the equilibrium constant:density of states

PR

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

RTE

R

P

R

P req

q

N

N0

RTE

R

P req

qK 0

PR The number ratio of product to reactant molecules:

which leads to the equilibrium constant:

Proof

q

Nen

r

i

R R

RRRe

q

NnN

P PPP

Peq

NnN

'

q

NqN R

R

RTEP

P

PP

r

P

P

eq

Nq

eeq

N

eq

NN

0

0

0 )(

RTE

R

P

R

P req

q

N

NK 0

0' PP

00 ANE

Energy separation and state density on equilibria

RTErekT

K 0

1Rq

)levelsenergy spacedevenly (

/kTqP

favoured.not is

product 1, large, is when 0r KE

1) (with dominant be

might product ratures,high tempeAt

K

favoured. is here) (smaller states

ofdensity larger with species the,

Again

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

Helix structure of polypeptides (proteins)

Helix-coil transition

h c

The statistical physics of helix-coil transition:a tetrapeptide

)1(

464

040302010

43210

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

4

1

432 14641i

ii sNssssq

4,3,2,1,*

00 ieK RTGi

pp

qq

i

oii

RTGo

es (stability parameter)

Helix-coil transition of proteins with n amino acid residues

n

i

iisNq

1

1 !!

!

iin

nN i

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.

n

i

ii sNq

1

1

n

i

in

i

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

2

1

)1(

1)1(1

s

snssq

nn

qsin

qq

i

nn

iip

s

snssq

)1(

2

1

)1(

1)1(1

1.1 i

8.3 i

9.15 i

3105

nq

siin

n

q

n

in

i

in

i

11)1(ln

Degree of conversion

qsd

d

nln

)(ln

1 (Classroom exercise)

Zimm-Bragg model

212 41

2)1(1

2

1

ss

s

h h c c h c h hh h c c

Separate coil regions are considered

Documents

Fundamental relations: The thermodynamic functions The molecular partition function