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Chapter 20 Statistical thermodynamics: the machinery. Fundamental relations: The thermodynamic functions The molecular partition function Using statistical thermodynamics Mean energies Heat capacities Equation of state Residual entropies Equilibrium constants. - PowerPoint PPT Presentation
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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
Quantum mechanical interpretation
There are 8 nuclear spin states:
Quantum mechanical interpretation
,...11,8,5,2
with functionseigen wave rotational
with combiningby onswavefuncti
lsymmetrica-anti form that Those)3(
,...10,7,4,1
with functionseigen wave rotational
with combiningby on wavefuncti
lsymmetrica-anti form that Those)2(
.0,3,6,9,..J
with functionseigen wave rotational
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
invariant are that statesspin ofnumber The
unctions)spin wavef ricantisymmet
ofnumber the( :rotation Cwith
invariant are that statesspin ofnumber The
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
Quantum mechanical interpretation
• Generally, for a molecule with NR rotational elements (including the identity operation), the symmetry number
Quantum mechanical interpretation
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
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
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 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
h h h h h c h hc 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
h h h h h c h hc 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
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