slide 1 chapter 5 molecular vibrations and time-independent perturbation theory part b: the symmetry...
TRANSCRIPT
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Slide 1
Chapter 5
Molecular Vibrations andTime-Independent Perturbation Theory
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
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Slide 2
• Symmetry and Vibrational Selection Rules
• Spectroscopic Selection Rules
• The Vibrational Partition Function Function
• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene
• Internal Energy: ZPE and Thermal Contributions
• Time Independent Perturbation Theory
• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
• Applications to O2(g) and H2O(g)
• QM Calculations of Vibrational Frequencies and Dissoc. Energies
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Slide 3
Symmetry and Vibrational Selection Rules
2
??axxe dx
2a xx e
x
+
-
2
0axxe dx
By symmetry: f(-x) = -f(x)
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Slide 4
Consider the 1s and 2px orbitals in a hydrogen atom.
1 2 ? ?xs p d V
+-+
+- + x
z
1 2 0xs p d V
By symmetry: Values of the integrand for -x are the negative of values for +x; i.e. the portions of the integral to the left of the yz plane and right of the yz plane cancel.
If you understand this, then you know all (or almost all) there is toknow about Group Theory.
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Slide 5
The Direct Product
There is a theorem from Group Theory (which we won’t prove)that an integral is zero unless the integrand either:
(a) belongs to the totally symmetric (A, A1, A1g) representation
(b) contains the totally symmetric (A, A1, A1g) representation
0fd unless 1( )f A
The question is: How do we know the representation of an integrand when it is the product of 2 or more functions?
( ) ?a b a bf d d
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Slide 6
The product of two functions belongs to the representationcorresponding to the Direct Product of their representations
a bf
xf a b
How do we determine the Direct Product of two representations?
Simple!! We just multiply their characters (traces) together.
f a a b bf
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Slide 7
C2V E C2 V(xz) V’(yz)
A1 1 1 1 1
A2 1 1 -1 -1
B1 1 -1 1 -1
B2 1 -1 -1 1
B1xB2 1 1 -1 -1 A2
A2xB1 1 -1 -1 1 B2
A2xB2 1 -1 1 -1 B1
When will the Direct Product be A1?
A2xA2 1 1 1 1 A1
B1xB1 1 1 1 1 A1
Only when the two representations are the same.(This is another theorem that we won’t prove)
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Slide 8
x xf a b c
The representation of the integrand is the Direct Product of therepresentations of the three functions.
i.e.
What is f if one of the 3 functions belongs to the totallysymmetric representation (e.g. if (c) = A1) ?
1x x A xf a b a b
What if the integrand is the product of three functions?
e.g. ( ) ?a b c a b cf d d
Therefore unless1f A a b
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Slide 9
When will the Direct Product be A1?
Only when the two representations are the same.(This is another theorem that we won’t prove)
A minor addition.
When considering a point group with degenerate (E or T)representations, then it can be shown that the product of twofunctions will contain A1 only if the two representations are the same.
e.g. E x E = A1 + other
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Slide 10
( ) ?a b a bf d d
xf a b
Based upon what we’ve just seen: 1f A
unlessa b
Therefore, the integral of the product of two functions vanishesunless the two functions belong to the same representation.
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Slide 11
x xf a b c
The representation of the integrand is the Direct Product of therepresentations of the three functions.
i.e.
What is f if one of the 3 functions belongs to the totallysymmetric representation (e.g. if (c) = A1) ?
1x x A xf a b a b
What if the integrand is the product of three functions?
e.g. ( ) ?a b c a b cf d d
Therefore unless1f A a b
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Slide 12
• Symmetry and Vibrational Selection Rules
• Spectroscopic Selection Rules
• The Vibrational Partition Function Function
• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene
• Internal Energy: ZPE and Thermal Contributions
• Time Independent Perturbation Theory
• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
• Applications to O2(g) and H2O(g)
• QM Calculations of Vibrational Frequencies and Dissoc. Energies
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Slide 13
Spectroscopic Selection Rules
When light (of frequency ) is shined on a sample, the light’selectric vector interacts with the molecule’s dipole moment, whichadds a perturbation to the molecular Hamiltonian.
( 0 ) ( 1 ) ( 0 ) ( 0 ) c o s ( )H H H H E H E t
The perturbation “mixes” the ground state wavefunction (0 = init ) withvarious excited states (i = fin). A simpler way to say this is that thelight causes transitions between the ground state and the excitedstates.
Time dependent perturbation theory can be used to show thatthe intensity of the absorption is proportional to the squareof the “transition moment”, M0i.
Consider a transition between the ground state (0 = init ) and thei’th. excited state (i = fin).
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Slide 14
2
0 iI n t e n s i t y C M
The transition moment is: *0 ˆ ˆi f i n i n i t f i n i n i tM d
ˆ e r e x i y j z k
is the dipole moment operator:^
0 ˆi f i n i n i t f i n i n i tM e x i y j z k
0 i f i n i n i t f i n i n i t f i n i n i tM e x i e y j e z k
0 0 0 0x y z
i i i iM M i M j M k
0
0
0
xi fin init
yi fin init
zi fin init
M e x
M e y
M e z
Thus, the transition moment has x, y and z components.
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Slide 15
The above equation leads to the Selection Rules in various areasof spectroscopy.
The intensity of a transition is nonzero only if at least one componentof the transition moment is nonzero.
That’s where Group Theory comes in.
As we saw in the previous section, some integrals are zero due tosymmetry.
Stated again, an integral is zero unless the integrand belongs to(or contains) the totally symmetry representation, A, A1, A1g...
0 0 0 0x y z
i i i iM M i M j M k
0
0
0
xi fin init
yi fin init
zi fin init
M e x
M e y
M e z
2
0 iI n t e n s i t y C M
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Slide 16
Selection Rules for Vibrational Spectra
Infrared Absorption Spectra
0 0 0 0x y z
i i i iM M i M j M k
0
0
0
xi fin init
yi fin init
zi fin init
M e x
M e y
M e z
2
0 iI n t e n s i t y C M
The equation governing the intensity of a vibrational infrared absorption is allowed (i.e. the vibration is IR active) is:
init and fin represent vibrational wavefunctions of the initialstate (often the ground vibrational state, n=0) and final state (oftencorresponding to n=1).
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Slide 17
A vibration will be Infrared active if any of the three componentsof the transition moment are non-zero; i.e. if
1 1( ) x ( ) x ( ) ( , , . . . )f i n i n i t gx A A A
1 1( ) x ( ) x ( ) ( , , . . . )f i n i n i t gy A A A or
1 1( ) x ( ) x ( ) ( , , . . . )f i n i n i t gz A A A or
2
0 iI n t e n s i t y C M
0 0 0 0x y z
i i i iM M i M j M k
0
0
0
xi fin init
yi fin init
zi fin init
M e x
M e y
M e z
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Slide 18
Raman Scattering Spectra
When light passes through a sample, the electric vector createsan induced dipole moment, ind, whose magnitude dependsupon the polarizability, . The intensity of Raman scattering dependson the size of the induced dipole moment.
Strictly speaking, is a tensor (Yecch!!)
indx xx xy xz xindy xy yy yz yindz xz yz zz z
E
E
E
I have used the fact that the tensor is symmetric; i.e. ij = ji
There are 6 independentcomponents of : xx , yy , zz , xy , xz , yz
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Slide 19
indx xx xy xz xindy xy yy yz yindz xz yz zz z
E
E
E
Therefore, a vibration will be active if any of the followingsix integrals are not zero.
u vf i n i n i t u = x, y, z and v = x, y, z
uv has the same symmetry properties as the product uv;
i.e. xx belongs to the same representation as xx, yz belongs to the same representation as yz, etc.
Therefore, a vibration will be Raman active if the direct product:
1 1( ) x ( u v ) x ( ) ( , , . . . )f i n i n i t gA A A
for any of the six uv combinations; xx, yy, zz, xy, xz, yz
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Slide 20
• Symmetry and Vibrational Selection Rules
• Spectroscopic Selection Rules
• The Vibrational Partition Function Function
• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene
• Internal Energy: ZPE and Thermal Contributions
• Time Independent Perturbation Theory
• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
• Applications to O2(g) and H2O(g)
• QM Calculations of Vibrational Frequencies and Dissoc. Energies
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Slide 21
The Symmetry of Vibrational Normal Modes
We will concentrate on the four C-Hstretching vibrations in ethylene, C2H4.
C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
Ethylene has D2h symmetry. However,because all 4 C-H stretches are in theplane of the molecule, it is OK to usethe subgroup, C2h.
x
y
C C
H H
H H
1
(3030 cm-1)
C C
H H
H H
2
(3100 cm-1)
C C
H H
H H
3
(3110 cm-1)
C C
H H
H H
4
(2990 cm-1)
1 1 1 1 1 ag
ag
2 1 1 1 1 ag
ag
3 1 -1 -1 1 bu
bu
4 1 -1 -1 1 bu
bu
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Slide 22
Symmetry of Vibrational Wavefunctions
One Dimensional Harmonic Oscillator
n = 02 / 2
0 0xN e
n = 222 / 2
2 2 ( 4 2 ) xN x e
n = 12 / 2
1 1 ( 2 ) xN x e
Harmonic Oscillator Normal Mode: k
Normal Coordinate: Qk
n = 02 / 2
0 0k k kQN e
n = 22 / 22
2 2 ( 4 2 )k k kQk kN Q e
n = 12 / 2
1 1 ( 2 )k k kQk kN Q e
0 ( t o t . s y m m . r e p . )kgA
2 ( t o t . s y m m . r e p . )kgA
1k
kQ
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Slide 23
31 2
1 2 3 1 2 3, , , . . . 1 2 3, , ,v i bn n n n n nn n n
Total Vibrational Wavefunction of Polyatomic Molecules
31 2
1 2 3 1 2 3, , , . . . x x xv i bn n n n n n
GroundState: 31 2
0 , 0 , 0 , . . . 0 0 00 , 0 , 0 ,v i b
31 20 , 0 , 0 , . . . 0 0 0x x xv i b
x x x . . .g g g ga a a A
SinglyExcited State: 31 2
0 ,1 , 0 , . . . 0 1 00 , 1 , 0 ,v i b
31 20 ,1 , 0 , . . . 0 1 0x x xv i b
2 21 1x ( ) x x . . . ( )g ga a
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Slide 24
C-H Stretching Vibrations of Ethylene
C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
Chapter 13: Slide 44
C C
H H
H H1(3030 cm-1)
Ag
C C
H H
H H3
(3110 cm-1)
Bu
C C
H H
H H2(3100 cm-1)
Ag
C C
H H
H H4
(2990 cm-1)
Bu
IR Activity of Fundamental Modes
0 , 0 , 0 , 0 1 , 0 , 0 , 0Ag:
0 1 , 0 , 0 , 0 0 , 0 , 0 , 0x v i b v i biM e x
31 2 41 , 0 , 0 , 0 1 0 0 0x x xv i b
gA
31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b
gA
x xg u g uA B A B
0 1 , 0 , 0 , 0 0 , 0 , 0 , 0y v i b v i biM e y x xg u g uA B A B
0 1 , 0 , 0 , 0 0 , 0 , 0 , 0z v i b v i biM e z x xg u g uA A A A
Ag vibrations are IR Inactive
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Slide 25
C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
Chapter 13: Slide 44
C C
H H
H H1(3030 cm-1)
Ag
C C
H H
H H3
(3110 cm-1)
Bu
C C
H H
H H2(3100 cm-1)
Ag
C C
H H
H H4
(2990 cm-1)
Bu
IR Activity of Fundamental Modes
0 , 0 , 0 , 0 0 , 0 , 1 , 0Bu:
0 0 , 0 ,1 , 0 0 , 0 , 0 , 0x v i b v i biM e x
31 2 40 , 0 ,1 , 0 0 0 1 0x x xv i b
uB
31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b
gA
x xu u g gB B A A
0 0 , 0 ,1 , 0 0 , 0 , 0 , 0y v i b v i biM e y x xu u g gB B A A
0 0 , 0 ,1 , 0 0 , 0 , 0 , 0z v i b v i biM e z x xu u g gB A A B
Bu vibrations are IR Active and polarized perpendicular to the axis.
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Slide 26
C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
Raman Activity of Fundamental Modes
0 , 0 , 0 , 0 1 , 0 , 0 , 0Ag:
31 2 41 , 0 , 0 , 0 1 0 0 0x x xv i b
gA
31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b
gA
x xg g g gA A A A x xg g g gA B A B
Ag vibrations are Raman Active
u v 0i u = x, y, z and v = x, y, z0( ) x ( u v ) x ( )i
21 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i bx
21 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i by
21 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i bz
1 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i bx y
1 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i bx z
1 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i by z
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Slide 27
C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
Raman Activity of Fundamental Modes
x xu g g uB A A B x xu g g uB B A A
Bu vibrations are Raman Inactive
u v 0i u = x, y, z and v = x, y, z0( ) x ( u v ) x ( )i
20 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i bx
20 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i by
20 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i bz
v i bv i b x y 0,0,0,00,1,0,0
0 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i bx z
0 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i by z
31 2 40 , 0 ,1 , 0 0 0 1 0x x xv i b
uB
31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b
gA
0 , 0 , 0 , 0 0 , 0 , 1 , 0Bu:
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Slide 28
The Shortcut C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,yCorky: Hey, Cousin Mookie. Wotcha doing all that work for? There’s a neat little shortcut.
Mookie: Would you please stop looking for shortcuts? They can get you in trouble.
Corky: But look!! If there’s an x,y or z attached to the representation, then the vibration’s IR active; Bu vibrations are IR Active. Ag vibrations are IR Inactive.
If there’s an x2, y2, etc. attached to the representation,then the vibration’s Raman active; Ag vibrations are Raman Active.Buvibrations are Raman Inactive.
Mookie: How do you determine the activity of a combination mode?
Corky: What’s a combination mode?
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Slide 29
IR Activity of Combination Modes
0 0 ,1 , 0 ,1 0 , 0 , 0 , 0x v i b v i biM e x
31 2 40 ,1 , 0 ,1 0 1 0 1x x xv i b
31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b
gA
x xu u g gB B A A
0 0 ,1 , 0 ,1 0 , 0 , 0 , 0y v i b v i biM e y x xu u g gB B A A
0 0 ,1 , 0 ,1 0 , 0 , 0 , 0z v i b v i biM e z x xu u g gB A A B
2 + 4 is IR Active and polarized perpendicular to the axis.
C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
Chapter 13: Slide 44
C C
H H
H H1(3030 cm-1)
Ag
C C
H H
H H3
(3110 cm-1)
Bu
C C
H H
H H2(3100 cm-1)
Ag
C C
H H
H H4
(2990 cm-1)
Bu
0 , 0 , 0 , 0 0 , 1 , 0 , 1 2 + 4
x x xg g g u ua a a b B
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Slide 30
C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
IR Activity of Combination Modes
0 0 , 0 ,1 , 0 0 , 0 , 0 ,1x v i b v i biM e x
31 2 40 , 0 ,1 , 0 0 0 1 0x x xv i b
x xu u u uB B B B
0 0 , 0 ,1 , 0 0 , 0 , 0 ,1y v i b v i biM e y
0 0 , 0 ,1 , 0 0 , 0 , 0 ,1z v i b v i biM e z x xu u u uB A B A
3 - 4 is IR Inactive.
Chapter 13: Slide 44
C C
H H
H H1(3030 cm-1)
Ag
C C
H H
H H3
(3110 cm-1)
Bu
C C
H H
H H2(3100 cm-1)
Ag
C C
H H
H H4
(2990 cm-1)
Bu
0 , 0 , 0 , 1 0 , 0 , 1 , 0 3 - 4
x x xg g u g ua a b a B
31 2 40 , 0 , 0 ,1 0 0 0 1x x xv i b
x x xg g g u ua a a b B
x xu u u uB B B B
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Slide 31
C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
x xu g g uB A A B x xu g g uB B A A
u v 0i u = x, y, z and v = x, y, z0( ) x ( u v ) x ( )i
20 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i bx
20 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i by
20 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i bz
0 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i bx y
0 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i bx z
0 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i by z
Raman Activity of Combination Modes
2 + 4 is Raman Inactive.
31 2 40 ,1 , 0 ,1 0 1 0 1x x xv i b
31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b
gA
x x xg g g u ua a a b B
0 , 0 , 0 , 0 0 , 1 , 0 , 1 2 + 4
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Slide 32
C2h E C2 i h
Ag 1 1 1 1 x2,y2,z2,xy
Bg 1 -1 1 -1 xz,yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
x xu g u gB A B A x xu g u gB B B B 20 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i bx
20 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i by
20 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i bz
0 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i bx y
0 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i bx z
0 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i by z
Raman Activity of Combination Modes
u v 0i u = x, y, z and v = x, y, z0( ) x ( u v ) x ( )i
0 , 0 , 0 , 1 0 , 0 , 1 , 0
31 2 40 , 0 ,1 , 0 0 0 1 0x x xv i b
3 - 4
x x xg g u g ua a b a B
31 2 40 , 0 , 0 ,1 0 0 0 1x x xv i b
x x xg g g u ua a a b B
3 - 4 is Raman Active.
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Slide 33
• Symmetry and Vibrational Selection Rules
• Spectroscopic Selection Rules
• The Vibrational Partition Function Function
• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene
• Internal Energy: ZPE and Thermal Contributions
• Time Independent Perturbation Theory
• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
• Applications to O2(g) and H2O(g)
• QM Calculations of Vibrational Frequencies and Dissoc. Energies
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Slide 34
Time Independent Perturbation Theory
Introduction
One often finds in QM that the Hamiltonian for a particular problemcan be written as:
( 0 ) ( 1 )H H H
H(0) is an exactly solvable Hamiltonian; i.e. ( 0 ) ( 0 ) ( 0 ) ( 0 )H E
H(1) is a smaller term which keeps the Schrödinger Equation from being solvable exactly.
One example is the Anharmonic Oscillator:
2 22 3 4
2
1
2 2
dH kx x x
d x
H(0)
Exactly SolvableH(1)
Correction Term
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Slide 35
( 0 ) ( 1 )H H H ( 0 ) ( 0 ) ( 0 ) ( 0 )H E where
( 0 ) ( 1 ) ( 2 ) ( )nE E E E E
In this case, one may use a method called “Perturbation Theory” toperform one or more of a series of increasingly higher order correctionsto both the Energies and Wavefunctions.
( 0 ) ( 1 ) ( 2 ) ( )n
We will use the notation: ( 0 )E E E ( 0 ) and
Some textbooks** outline the method for higher order corrections.However, we will restrict the treatment here to first order perturbationcorrections
e.g. Quantum Chemistry (5th. Ed.), by I. N. Levine, Chap. 9
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Slide 36
First Order Perturbation Theory
( 0 ) ( 1 )H H H ( 0 ) ( 0 ) ( 0 ) ( 0 )H E where
( 0 )E E E ( 0 ) Assume: and
H E
( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 )H H E E
( 0 ) ( 0 ) (1) ( 0 ) ( 0 ) (1)
( 0 ) ( 0 ) ( 0 ) ( 0 )
H H H H
E E E E
One can eliminate the two terms involving the product of two small corrections.
( 0 ) ( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )H H H H E E E E
One can eliminate two additional terms because: ( 0 ) ( 0 ) ( 0 ) ( 0 )H E
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Slide 37
( 1 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )H H E E
( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )* * * *H d H d E d E d
Multiply all terms by (0)* and integrate:
H(0) is Hermitian. Therefore:
( 0 ) ( 0 ) ( 0 ) ( 0 )* *H d H d ( 0 ) ( 0 ) *E d ( 0 ) ( 0 ) *E d
( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )* * * *H d E d E d E d
Plug in to get:
Therefore: ( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 )* *H d E d E
( 0 ) ( 1 ) ( 0 )*E H d is the first order perturbation theorycorrection to the energy.
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Slide 38
Applications of First Order Perturbation Theory
PIB with slanted floor
Consider a particle in a box with the potential:
( ) 0 ,V x x x a
0( ) 0V
V x x x aa
V0For this problem:
2 2( 0 )
20
2
dH
m dx
2 2( 0 )
28n
n hE
ma
( 0 ) 2sinn
n x
a a
The perturbing potential is: (1) 0VH x
a
x
V(x
)
0 a0
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Slide 39
We will calculate the first order correction to the nth energy level.In this particular case, the correction to all energy levels is the same.
( 0 ) ( 1 ) ( 0 )*E H d 0
0
2 2sin sin
a Vn x n xx dx
a a a a a
20
0
2sin
aV n xx dx
a a a
sin 2 sin 2 0n
a aa
cos 2 cos 2 1n
a aa
22
2
sin(2 ) cos(2 )sin ( )
4 4 8
x x x xx x dx
Integral Info
202
0
2s in
aV nx x dx
a a
20
2 2 2
2 sin (2 ) co s(2 ) co s(0 )0 0
4 4 8 8
V a a a a
a
20
2 2 2
2 1 1
4 8 8
V a
a
0
2
VE Independent of n
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Slide 40
Anharmonic Oscillator
For this problem:2 2
( 0 ) 22
1
2 2
dH kx
m dx
( 0 )0
1
2E
21/ 4
( 0 ) / 20
xe
The perturbing potential is:( 1 ) 3 4H x x
2 3 41( )
2V x k x x x
Consider an anharmonic oscillator with the potential energy of the form:
We’ll calculate the first order perturbation theory correction to the groundstate energy.
kk
2 21/ 4 1/ 4
/ 2 3 4 / 2x xe x x e dx
( 0 ) ( 1 ) ( 0 )0 0*E H d
21/ 2
3 4 xx x e dx
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Slide 41
2 21/ 2 1/ 2
3 4x xx e dx x e dx
2420
3
8
xx e dx
23
4k
Note: There is no First order Perturbation Theory correction due to the cubic term in the Hamiltonian.
However, there IS a correction due to the cubic termwhen Second order Perturbation Theory is applied.
21/ 2
3 4 xE x x e dx
21/ 2
4
0
0 2 xx e dx
1/ 2 1/ 2
2
32
8
2
3
4E
2
3
4k
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Slide 42
Brief Introduction to Second Order Perturbation Theory
As noted above, one also can obtain additional corrections to the energyusing higher orders of Perturbation Theory; i.e.
The second order correction to the energy of the nth level is given by:
( 0 ) ( 1 ) ( 2 )n n n nE E E E
En(0) is the energy of the nth level for the unperturbed Hamiltonian
En(1) is the first order correction to the energy, which we have called E
En(2) is the second order correction to the energy, etc.
2( 0 ) (1) ( 0 )
( 2 )( 0 ) ( 0 )
1
k n
nk n k
HE for k n
E E
( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 1 ) ( 0 )*k n k nH H d where
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Slide 43
If the correction is to the ground state (for which we’ll assume n=1), then:
2( 0 ) (1) ( 0 )
1( 2 )1 ( 0 ) ( 0 )
2 1
k
k k
HE
E E
2 2 2( 0 ) (1) ( 0 ) ( 0 ) (1) ( 0 ) ( 0 ) (1) ( 0 )2 1 3 1 4 1( 2 )
1 ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )1 2 1 3 1 4
H H HE
E E E E E E
Note that the second order Perturbation Theory correction is actuallyan infinite sum of terms.
However, the successive terms contribute less and less to the overallcorrection as the energy, Ek
(0), increases.
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Slide 44
• Symmetry and Vibrational Selection Rules
• Spectroscopic Selection Rules
• The Vibrational Partition Function Function
• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene
• Internal Energy: ZPE and Thermal Contributions
• Time Independent Perturbation Theory
• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
• Applications to O2(g) and H2O(g)
• QM Calcs. of Vibrational Frequencies and Dissoc. Energies
![Page 45: Slide 1 Chapter 5 Molecular Vibrations and Time-Independent Perturbation Theory Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical](https://reader036.vdocuments.us/reader036/viewer/2022062223/551a03d25503462e378b4a0a/html5/thumbnails/45.jpg)
Slide 45
The Potential Energy Curve of H2
0.0 0.5 1.0 1.5 2.0 2.5
-1.2
-1.1
-1.0
-0.9E
nerg
y (h
artr
ees)
Bond Length (Angstroms)
Calculation at QCISD(T)/6-311++G(3df,3pd) level
Re(cal) = 0.742 Å
Re(exp) = 0.742 Å
Emin(cal) = -1.17253 hartrees (au)
E(cal) =2(-0.49982) hartrees = -0.99964 hartrees (au)
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Slide 46
0.0 0.5 1.0 1.5 2.0 2.5
-1.2
-1.1
-1.0
-0.9
Ene
rgy
(ha
rtre
es)
Bond Length (Angstroms)
De(cal) = E(cal) – Emin(cal) = 0.17289 au
The Dissociation Energy
•2625.5 kJ/mol / au
De(cal) = 453.9 kJ/mol
D0(exp) = 432 kJ/mol
De: Spectroscopic
Dissociation Energy
D0: Thermodynamic Dissociation Energy
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Slide 47
De = D0 + Evib = D0 + (1/2)h
0.0 0.5 1.0 1.5 2.0 2.5
-1.2
-1.1
-1.0
-0.9
Ene
rgy
(ha
rtre
es)
Bond Length (Angstroms)
Relating De to D0
De: Spectroscopic
Dissociation Energy
D0: Thermodynamic Dissociation Energy
versus (exp) = 4395 cm-1~
H2: (cal) = 4403 cm-1 [QCISD(T)/6-311++G(3df,3pd)]~
The experimental frequency is the “harmonic” value,corrected from the observed anharmonic frequency.
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Slide 48
De(cal) = 453.9 kJ/mol
D0(exp) = 432 kJ/mol
0.0 0.5 1.0 1.5 2.0 2.5
-1.2
-1.1
-1.0
-0.9
En
erg
y (h
art
ree
s)
Bond Length (Angstroms)
De: Spectroscopic
Dissociation Energy
D0: Thermodynamic Dissociation Energy
De = D0 + Evib = D0 + (1/2)h
1 1
2 2v ibE h h c
3 4 1 0 116 .6 3 1 0 3 .0 0 1 0 / ( 4 4 0 3 )
2x J s x cm s cm
2 0 2 3 13
14 .3 7 9 1 0 6 .0 2 1 0
1 0
k Jx J x m o l
J
2 6 . 4 /v i bE k J m o l
D0(cal) = De(cal) –Evib
= 453.9 – 26.4
= 427.5 kJ/mol
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Slide 49
Calculated Versus Experimental Vibrational Frequencies
H2O
(Harm)a
3943 cm-1
3833
1649
(a) Corrected Experimental Harmonic Frequencies
(cal)b
3956 cm-1
3845
1640
(b) QCISD(T)/6-311+G(3df,2p)
(cal)c
4188 cm-1
4070
1827
(c) HF/6-31G(d)
Lack of including electron correlation raises computed frequencies.
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Slide 50
Harmonic versus Anharmonic Vibrational Frequencies
H2O
(Harm)a
3943 cm-1
3833
1649
(a) Corrected Experimental Harmonic Frequencies
(cal)b
3956 cm-1
3845
1640
(b) QCISD(T)/6-311+G(3df,2p)
(cal)c
4188 cm-1
4070
1827
(c) HF/6-31G(d)
Lack of including electron correlation raises computed frequencies.
(exp)z
3756 cm-1
3657
1595
(z) Actual measured Experimental Frequencies
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Slide 51
Differences between Calculated and Measured Vibrational Frequencies
There are two sources of error in QM calculated vibrationalfrequencies.
(1) QM calculations of vibrational frequencies assume that the vibrations are “Harmonic”, whereas actual vibrations are anharmonic.
Typically, this causes calculated frequencies to be approximately5% higher than experiment.
(2) If one uses “Hartree-Fock (HF)” rather than “correlated electron” calculations, this produces an additional ~5% increase in the calculated frequencies.
To correct for these sources of error, multiplicative “scale factors”have been developed for various levels of calculation.
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Slide 52
An Example: CHBr3
(exp)a
3050 cm-1
1149 (E)
669 (E)
543
223
155 (E)
(a) Measured “anharmonic” vibrational frequencies.
(cal)b
3222 cm-1
1208 (E)
691 (E)
546
228
158 (E)
(b) Computed at QCISD/6-311G(d,p) level
(scaled)c
3061 cm-1
1148 (E)
656 (E)
519
217
150 (E)
(c) Computed frequencies scaled by 0.95
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Slide 53
Another Example: CBr3•
(cal)b
799 (E) cm-1
324
235
164 (E)
(b) Computed at QCISD/6-311G(d,p) level
(a) Measured “anharmonic” vibrational frequencies.
(exp)a
773 (E) cm-1
???
???
???
(scaled)c
759 (E) cm-1
308
223
156 (E)
(c) Computed frequencies scaled by 0.95
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Slide 54
Some Applications of QM Vibrational Frequencies
(1) Aid to assigning experimental vibrational spectra
One can visualize the motions involved in the calculated vibrations
(2) Vibrational spectra of transient species
It is usually difficult to impossible to experimentally measure the vibrational spectra in short-lived intermediates.
(3) Structure determination.
If you have synthesized a new compound and measured the vibrational spectra, you can simulate the spectra of possible proposed structures to determine which pattern best matches experiment.
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Slide 55
• Symmetry and Vibrational Selection Rules
• Spectroscopic Selection Rules
• The Vibrational Partition Function Function
• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene
• Internal Energy: ZPE and Thermal Contributions
• Time Independent Perturbation Theory
• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
• Applications to O2(g) and H2O(g)
• QM Calculations of Vibrational Frequencies and Dissoc. Energies
![Page 56: Slide 1 Chapter 5 Molecular Vibrations and Time-Independent Perturbation Theory Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical](https://reader036.vdocuments.us/reader036/viewer/2022062223/551a03d25503462e378b4a0a/html5/thumbnails/56.jpg)
Slide 56
Statistical Thermodynamics: Vibrational Contributions to Thermodynamic Properties of Gases
Remember these?
2
,
ln
V N
QU kT
T
2
, ,
ln ln
lnV N T N
Q QH kT kT
T V
,V
V N
UC
T
,P
P N
HC
T
lnU
S k QT
l nA U T S k T Q
,
lnln
ln T N
QG H TS kT Q kT
V
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Slide 57
• Symmetry and Vibrational Selection Rules
• Spectroscopic Selection Rules
• The Vibrational Partition Function Function
• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene
• Internal Energy: ZPE and Thermal Contributions
• Time Independent Perturbation Theory
• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
• Applications to O2(g) and H2O(g)
• QM Calculations of Vibrational Frequencies and Dissoc. Energies
![Page 58: Slide 1 Chapter 5 Molecular Vibrations and Time-Independent Perturbation Theory Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical](https://reader036.vdocuments.us/reader036/viewer/2022062223/551a03d25503462e378b4a0a/html5/thumbnails/58.jpg)
Slide 58
The Vibrational Partition Function
1
2 2
k
1, 0 ,1 , 2 ,
2nE n h n
HO Energy Levels:
0
nvib kT
n
q e
The Partition Function
1/ 2
0
n h
kT
n
e
(1/ 2 )
0
nh h
kT kT
n
e e
(1/ 2 )
0
h nh
kT kT
n
e e
2
0
v vnvib T T
n
q e e
v
h
k
where
hc
k
If v/T << 1 (i.e. if /kT << 1), we can convert the sum to an integral.
Let's try O2 ( = 1580 cm-1 ) at 298 K.~
2 2 7 77 .6 4
2 9 8v K
T K
3 4 1 0 1
2 3
(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7
1 .3 8 1 0 /v
h c x J s x cm s cmK
k x J K
Sorry Charlie!! No luck!!
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Slide 59
Simplification of qvib
v
h hc
k k
2
0
v vnvib T T
n
q e e
Let's consider:0 0
w h erev vn
nT T
n n
e x x e
( )
0
1( ) (0 )
!n n
n
f x f xn
We learned earlier in the chapter that anyfunction can be expanded in a Taylor series:
0
1
1n
n
xx
It can be shown that the Taylor series for 1/(1-x) is:
Therefore:0 0
vnnT
n n
e x
1 1
11
v
Tx
e
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Slide 60
v
h hc
k k
2
0
v vnvib T T
n
q e e
0
1
1
v
v
n
T
n T
e
e
and
2
1
v
v
Tvib
T
eq
e
For N molecules, the total vibrational partition function, Qvib, is:
2
1
v
v
N
TNvib vib
T
eQ q
e
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Slide 61
Some comments on the Vibrational Partition Function
v
h hc
k k
2
1
v
v
N
TNvib vib
T
eQ q
e
If we look back at the development leading to qvib and Qvib, we see that:
(a) the numerator arises from the vibrational Zero-Point Energy (1/2h)
(b) the denominator comes from the sum over exp(-nh/kT)The latter is the "thermal" contribution to qvib because all terms aboven=0 are zero at low temperatures
In some books, the ZPE is not included in qvib; i.e. they use n = nh.In that case, the numerator is 1 and they then must add in ZPEcontributions separately
2ln ln ln 1v v
vib T TQ N e N e
ZPE Term Thermal Term
In further developments, we'll keep track of the two terms individually.
ln 12
v
v TN N eT
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Slide 62
• Symmetry and Vibrational Selection Rules
• Spectroscopic Selection Rules
• The Vibrational Partition Function Function
• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene
• Internal Energy: ZPE and Thermal Contributions
• Time Independent Perturbation Theory
• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
• Applications to O2(g) and H2O(g)
• QM Calculations of Vibrational Frequencies and Dissoc. Energies
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Slide 63
Internal Energy
ln ln 12
vvib v TQ N N e
T
2
,
ln
vibvib
V N
QU kT
T
2 2
, ,
ln 12
vvib v T
V N V N
U kT N kT N eT T T
v i b v i bZ P E t h e r mU U
UZPEvib Utherm
vib
2
,2
vib vZPE
V N
U kT NT T
2 1
2v
dN kT TdT
ZPE Contribution
2
2
1
2vN kT
T
2vNk
2v ibZPE
N k hcU
k
2AN hc
n
This is just the vibrational Zero-Pointenergy for n moles of molecules
or2
v ib AZPE
N hcU
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Slide 64
2
,
ln 1v
vib Ttherm
V N
U kT N eT
Thermal Contribution
2 ln 1v
Td
NkT edT
2
1
1
v
v
T
T
NkT de
dTe
2
( 1)
1
v
v
T
T
NkT de
dTe
2
1
v
v
vT
T
NkT de
dT Te
22
1
v
v
Tv
T
eNkT
Te
1
v
v
T
v
T
eNk
e
1 1
v v
v v
T T
A v v
T T
e enN k nR
e e
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Slide 65
1
v
v
Tvibtherm v
T
eU nR
e
Although this expression is fine, it's commonly rearranged, as follows:
1
v v
v v
T Tvibtherm v
T T
e eU nR
e e
1v
vib vtherm
T
nRU
e
1v
vib vtherm
T
RU
e
or
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Slide 66
Limiting Cases
Low Temperature:(T 0)
1v
vib vtherm
T
RU
e
0
1v
vib vtherm
T
U R
e
Because the denominatorapproaches infinity
High Temperature:(v << T) 1 1
vib vtherm
v
RU
T
R T
The latter result demonstrates the principal of equipartition of vibrationalinternal energy: each vibration contributes RT of internal energy.
However, unlike translations or rotations, we almost never reach thehigh temperature vibrational limit.
At room temperature, the thermal contribution to the vibrational internalenergy is often close to 0.
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Slide 67
• Symmetry and Vibrational Selection Rules
• Spectroscopic Selection Rules
• The Vibrational Partition Function Function
• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene
• Internal Energy: ZPE and Thermal Contributions
• Time Independent Perturbation Theory
• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases
Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics
• Applications to O2(g) and H2O(g)
• QM Calculations of Vibrational Frequencies and Dissoc. Energies
![Page 68: Slide 1 Chapter 5 Molecular Vibrations and Time-Independent Perturbation Theory Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical](https://reader036.vdocuments.us/reader036/viewer/2022062223/551a03d25503462e378b4a0a/html5/thumbnails/68.jpg)
Slide 68
Numerical Example
1v
vib vtherm
T
RU
e
2
vib AZPE
N hcU
Let's try O2 ( = 1580 cm-1 ) at 298 K.~
3 4 1 0 1
2 3
(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7
1 .3 8 1 0 /v
h c x J s x cm s cmK
k x J K
2 3 1 3 4 1 0 1( 6 .0 2 1 0 )( 6 .6 3 1 0 )(3 .0 0 1 0 / ) (1 5 8 0 )
2v ibZ P E
x m o l x J s x cm s cmU
( 2 2 7 7 / 2 9 8 )
8 .314 / 2277
11
v
vib vtherm K K
T
R J m ol K KU
ee
9 4 5 9 / 9 . 4 6 /J m o l k J m o l Independent ofTemperature
9 . 1 / 0 . 0 0 9 /J m o l k J m o l
vs. RT = 2.48 kJ/mol
Thus, the thermal contribution to the vibrational internal energyof O2 at room temperature is negligible.
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Slide 69
A comparison between O2 and I2
I2: = 214 cm-1 v = 309 K~
T O2 I2 RT(K) (kJ/mol) (kJ/mol) (kJ/mol)
v ibth e rmU v ib
th e rmU
298 0.009 1.41 2.48
500 0.20 3.00 4.16
1000 2.16 7.10 8.31
2000 8.92 15.4 16.6
3000 16.7 23.7 24.9
Because of its lower frequency (and,therefore, more closely spaced energy levels), thermal contributions to the vibrational internal energyof I2 are much more significant than for O2 at all temperatures.
At higher temperatures, thermal contributions to the vibrational internalenergy of O2 become very significant.
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Slide 70
Enthalpy
2
, ,
ln ln
ln
vib vibv ib
V N T N
Q QH kT kT
T V
Qvib independent of V
Therefore: 2
,
ln vibvib vib
V N
QH kT U
T
and: vib vibZPE ZPE
vib vibtherm therm
H U
H U
Similarly:, ,
vib vibvib vibP V
P N V N
H UC C
T T
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Slide 71
Heat Capacity (CVvib and CP
vib)1
v
vib vtherm
T
nRU
e
2vib AZPE
nN hcU
, , ,
vibvibvibvib thermZPEV
V N V N V N
UUUC
T T T
Independent of T
1v
vib vV
T
nRdC
dTe
1
1v
Tv
dnR e
dT
2
( 1) 1 1v v
T Tv
dnR e e
dT
2 2
1
v
v
v vT
T
nRe
Te
2
2
1
v
v
Tv
T
enR
Te
It can be shown that in the limit, V << T, CVvib R
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Slide 72
2
2
1
v
v
Tvib vib vV P
T
eC C nR
Te
Numerical Example
7 .642
27 .648 .314 7 .64 0 .23 /
1
vib vibV P
eC C J mol K
e
Let's try O2 ( = 1580 cm-1 ) at 298 K (one mole).~
3 4 1 0 1
2 3
(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7
1 .3 8 1 0 /v
h c x J s x cm s cmK
k x J K
2 2 7 77 .6 4
2 9 8v
T
vs. R = 8.314 J/mol-K
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Slide 73
A comparison between O2 and I2
I2: = 214 cm-1 v = 309 K~
298 0.23 7.61
500 1.85 8.05
1000 5.49 8.25
2000 7.47 8.30
3000 7.93 8.31
T O2 I2 (K) (J/mol-K) (J/mol-K)
Vibrational Heat Capacities (CPvib = CV
vib)
Vibrational contribution to heat capacity of I2 is greater because ofit's lower frequency (i.e. closer energy spacing)
The heat capacities approach R (8.31 J/mol-K) at high temperature.
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Slide 74
Entropy
lnvib
v ib v ib US k Q
T
1v
vib vtherm
T
nRU
e
2 2v ib v vZ P E
N k nRU
2
1
v
v
N
TNvib vib
T
eQ q
e
2
ln ln
1v
Nv
Tvib
T
ek Q k
e
2ln ln 1vv
T TNk e N k e
/ 2 1ln 1
21
v
v
vib v v vT
T
Nk nR nRS Nk e
T T Te
ln 1
1
v
v
vvib T
T
nR TS nR e
e
ANk nN k nR
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Slide 75
Numerical Example
Let's try O2 ( = 1580 cm-1 ) at 298 K (one mole).~
3 4 1 0 1
2 3
(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7
1 .3 8 1 0 /v
h c x J s x cm s cmK
k x J K
2 2 7 77 .6 4
2 9 8v
T
ln 1
1
v
v
vvib T
T
nR TS nR e
e
7 .6 47 .6 4
8 .3 1 4 7 .6 48 .3 1 4 ln 1
1v ibS e
e
0 . 0 0 4 0 . 0 3 1 0 . 0 3 5 /J m o l K
1 5 1 . 9 4 3 . 8 0 . 0 3 5 1 9 5 . 7 /t r a n r o t v i bS S S J m o l K
"Total" Entropy
Chap. 3 Chap. 4
O2: Smol(exp) = 205.1 J/mol-K at 298.15 K
We're still about 5% lower than experiment, for now.
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Slide 76
The vibrational contribution to the entropy of O2 is very low at 298 K.It increases significantly at higher temperatures.
T Svib
298 K 0.035 J/mol-K
500 0.49
1000 3.07
2000 7.86
3000 10.82
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Slide 77
E (Thermal) CV S KCAL/MOL CAL/MOL-K CAL/MOL-K TOTAL 3.750 5.023 48.972 ELECTRONIC 0.000 0.000 2.183 TRANSLATIONAL 0.889 2.981 36.321 ROTATIONAL 0.592 1.987 10.459 VIBRATIONAL 2.269 0.055 0.008 Q LOG10(Q) LN(Q) TOTAL BOT 0.330741D+08 7.519488 17.314260 TOTAL V=0 0.151654D+10 9.180853 21.139696 VIB (BOT) 0.218193D-01 -1.661159 -3.824960 VIB (V=0) 0.100048D+01 0.000207 0.000476 ELECTRONIC 0.300000D+01 0.477121 1.098612 TRANSLATIONAL 0.711178D+07 6.851978 15.777263 ROTATIONAL 0.710472D+02 1.851547 4.263345
Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)
QCISD/6-311G(d)
0 . 0 5 5 / 0 . 2 3 /v i bVC c a l m o l K J m o l K (same as our result)
2 . 2 6 9 / 9 . 4 9 /v i b v i b v i bZ P E t h e r mU U U k c a l m o l k J m o l
We got 9.46 + 0.009 = 9.47 kJ/mol (sig. fig. difference)
0 . 0 0 8 / 0 . 0 3 3 /v i bS c a l m o l K J m o l K We got 0.035 J/mol-K (sig. fig. difference)
G-98 tabulates the sum ofUZPE
vib + Uthermvib even though
they call it E(Therm.)
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Slide 78
Helmholtz and Gibbs Energy
l nv i b v i bA k T Q ,
lnln
ln
vibvib vib
T N
QG kT Q kT
V
Qvib independent of V
v i bA
2
1
v
v
N
TNvib vib
T
eQ q
e
l nNv i b v i b v i bG A k T q l n v i bn R T q l n v i bN k T q
2
ln
1
v
v
T
T
enRT
e
O2 ( = 1580 cm-1 ) at 298 K (one mole).~
3 4 1 0 1
2 3
(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7
1 .3 8 1 0 /v
h c x J s x cm s cmK
k x J K
2 2 7 7
7 .6 42 9 8
v
T
7.64 / 22
7.640.0219
11
v
v
Tvib
T
e eq
ee
( 8 . 3 1 4 / ) ( 2 9 8 ) l n ( 0 . 0 2 1 9 ) 9 4 6 0 / 9 . 4 6 /v i b v i bG A J m o l K K J m o l k J m o l
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Slide 79
Polyatomic Molecules
It is straightforward to handle polyatomic molecules, which have3N-6 (non-linear) or 3N-5 (linear) vibrations.
1, 2 ,i iN
vib kTQ e
1 2 1 1 2 2( 1 / 2 ) ( 1 / 2 ) . . .v i b v i b v i b n h c n h c Energy:
Partition Function
1, 2 ,i iN
kT kTe e
1 , 1 ,i iN N
kT kTe e
Therefore: 1 2v i b v i b v i bQ Q Q
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Slide 80
1 2v i b v i b v i bQ Q Q
Thermodynamic Properties
1 2l n l n l nv i b v i b v i bQ Q Q
1 2
, , ,
ln lnln vib v ibv ib
V N V N V N
Q QQ
T T T
Therefore: 1 2
1 2
etc.
vib vib vib
vib vib vib
U U U
S S S
Thus, you simply calculate the property for each vibration separately,and add the contributions together.
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Slide 81
Comparison Between Theory and Experiment: H2O(g)
For molecules with a singlet electronic ground state and no low-lyingexcited electronic levels, translations, rotations and vibrations are theonly contributions to the thermodynamic properties.
We'll consider water, which has 3 translations, 3 rotations and 3 vibrations.
Using Equipartition of Energy, one predicts:
Htran = (3/2)RT + RT CPtran = (3/2)R + R = (5/2)R
Hrot = (3/2)RT CProt = (3/2)R
Hvib = 3RT CPvib = 3R
Htran + Hrot = 4RT CPtran + CP
rot = 4R
Htran + Hrot + Hvib = 7RT CPtran + CP
rot + CPvib= 7R
Low T Limit:
High T Limit:
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Slide 82
500 1000 1500 2000 2500 3000180
200
220
240
260
280
300
Exp Cal
S
[J/m
ol-K
]
Temperature [K]
Calculated and Experimental Entropy of H2O(g)
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Slide 83
Calculated and Experimental Enthalpy of H2O(g)
500 1000 1500 2000 2500 3000
0
20
40
60
80
100
120
140
160
180
High T: 7RT
Low T: 4RT
Cal Exp
H(T
) [
kJ/m
ol]
Temperature [K]
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Slide 84
Calculated and Experimental Heat Capacity (CP) of H2O(g)
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Low T: 4R
Cal Exp
CP
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Temperature [K]
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Slide 85
Translational + Rotational + Vibrational Contributions to O2 Entropy
Unlike in H2O, the remaining contribution (electronic) to theentropy of O2 is significant. We'll discuss this in Chapter 10.
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S
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Temperature [K]
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Slide 86
Translational + Rotational + Vibrational Contributions to O2 Enthalpy
There are also significant additional contributionsto the Enthalpy.
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H
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Temperature [K]
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Slide 87
Translational + Rotational + Vibrational Contributions to O2 Heat Capacity
Notes: (A) The calculated heat capacity levels off above ~2000 K. This represents the high temperature limit in which the vibration contributes R to CP.
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CP
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Temperature [K]
(B) There is a significant electronic contribution to CP at high temperature.