Transcript Components of the Atom - Department of Chemistry

Chapter 3
Particle-in-Box (PIB) Models
Slide 1
Carrots are Orange.
Tomatoes are Red.
But Why?
Slide 2
Outline
• The Classical Particle in a Box (PIB)
• The Quantum Mechanical PIB
• PIB Properties
• Other PIB Models
• The Free Electron Molecular Orbital (FEMO) Model
• PIB and the Color of Vegetables
• Multidimensional Systems: The 2D PIB and 3D PIB
• Statistical Thermodynamics: Translational contributions to
the thermodynamic properties
of gases.
Slide 3
P(x)
The Classical Particle in a Box
0
a
0
x
P(x) = Const. = C
0 x  a
P(x) = 0
x <0, x > a
a
E = ½mv2 = p2/2m = any value, including 0
i.e. the energy is not quantized
and there is no minimum.
Slide 4
P(x) = Const. = C
0 x  a
P(x) = 0
x <0, x > a
Normalization

0
a


0
a
 P( x )dx  1   0dx   Cdx   0dx  C  x 0  Ca

a
P( x )  C 
1
a
<x>
a

a

0
x   xP ( x )dx  
1
1  x2 
1 a2 a
x  dx    

a
a  2 0 a 2 2
<x2>
a

a

0
x 2   x 2P ( x )dx  
1  x3 
1 a3 a 2
2 1
x  dx    

a
a  3 0 a 3
3
2
Note:  x2  x 2  x
2
a2  a 
a2



0
3  2  12
Slide 5
Outline
• The Classical Particle in a Box (PIB)
• The Quantum Mechanical PIB
• PIB Properties
• Other PIB Models
• The Free Electron Molecular Orbital (FEMO) Model
• PIB and the Color of Vegetables
• Multidimensional Systems: The 2D PIB and 3D PIB
• Statistical Thermodynamics: Translational contributions to
the thermodynamic properties
of gases.
Slide 6
Quantum Mechanical Particle-in-Box
Schrödinger Equation
d 2

 V ( x )  E
2m dx 2

0
a
V(x)
2

Outside the box: x<0, x>a
P(x) = *(x)(x) = 0
(x) = 0
0
x
V(x) = 0
0xa
V(x)  
x<0,x>a
Inside the box: 0  x  a
d 2

 E
2
2m dx
2
or
d 2
2mE



2
dx 2
Slide 7
Solving the Equation
d 2
2mE



2
2
dx
Assume
  A sin( x )  B cos( x )
d
  A cos( x )   B sin( x )
dx
 2  
2mE
2

d 2
2
2



A
sin(

x
)


B cos( x )
2
dx
d 2
2




2
dx
E
2
2
2m
So far, there is no restriction on  and, hence none on
the energy, E (other than that it cannot be negative)
Slide 8
Applying Boundary Conditions (BC’s)
Nature hates discontinuous behavior;
i.e. It’s not really possible to stop on a dime
Because the wavefunction, , is 0 outside the box, x<0 and x>a,
it must also be 0 inside the box at x=0 and x=a
  A sin( x )  B cos( x )
BC-1:
(0) = 0 = Asin(0) + Bcos(0) = B
Therefore, (x) = Asin(x)
BC-2:
Therefore:
or:
(a) = 0 = Asin(a)
a = n

n
a
sin(n) = 0
n = 1, 2, 3, ...
n = 1, 2, 3, ...
Why isn’t n = 0 acceptable?
Slide 9
Wavefunctions and Energy Levels

n
a
n = 1, 2, 3, ...
 n 
x
 a 
 n  A sin( x )  A sin 
2
n = 1, 2, 3, ...
2 2
1 h2
 n  1  h  n 
 2 

E

a
2m 4 2
2m  a  2m  2 
2
2
n 2h 2
En 
8ma 2
2
n = 1, 2, 3, ...
Note that, unlike the classical particle in a box:
(a) the allowed energies are quantized
(b) E = 0 is NOT permissible (i.e. the particle can’t stand still)
Slide 10
Energy Quantization results from application
of the Boundary Conditions
n 2h 2
En 
8ma 2
h2
E1  1
8ma2
h2
E2  4
8ma 2
h2
E3  9
8ma 2
h2
E4  16
8ma2
h2
E5  25
8ma2
Slide 11
Wavefunctions
 3 x 
ψ 3  A  sin 

a


9h 2
E3 
8ma 2
 2 x 
ψ 2  A  sin 

a


4h 2
E2 
8ma 2
 x 
ψ 1  A  sin  
 a 
1h 2
E1 
8ma 2
0

a
0
2
a
Slide 12
Note that: (a) The Probability, 2 is very different from the
classical result.
(b) The number of “nodes” increases with increasing
quantum number.
The increasing number of nodes reflects the higher kinetic
energy with higher quantum number.
Slide 13
The Correspondence Principle
The predictions of Quantum Mechanics cannot violate the results
of classical mechanics on macroscopically sized systems.
Consider an electron in a 1 Angstrom box.
Calculate (a) the Zero Point Energy (i.e. minimum energy)
(b) the minimum speed of the electron
me = 9.1x10-31 kg
a = 1x10-10 m
2
 6.63x1034 J  s 
h = 6.63x10-34 J-s
h2
18
E1 
8ma2

8(9.1x1031 kg )(1x1010 m)2
mv12
E1 
 v1 
2
2E1

m
 6.04 x10
J
2(6.04 x10 18 kg  m 2 / s 2
9.1x10 31 kg
v1  3.6x106 m / s ( 1.7x107 mi / hr )
Thus, the minimum speed of an electron confined to an atom
is quite high.
Slide 14
Let's perform the same calculation on a macroscopic system.
Consider a 1 gram particle in a 10 cm box.
Calculate (a) the Zero Point Energy (i.e. minimum energy)
(b) the minimum speed of the particle
m = 1x10-3 kg
a = 0.10 m
h = 6.63x10-34 J-s
E1 
2

6.63 x1034 J  s

2
h
63


5.50
x
10
J
2
3
2
8ma
8(1x10 kg )(0.1m)
mv12
E1 
 v1 
2
2E1

m
2(5.50 x1063 kg  m 2 / s 2
30

3.3
x
10
m/s
3
x10 kg
Thus, the minimum energy and speed of a macroscopic particle
are completely negligible.
Slide 15
Probability Distribution of a Macroscopic Particle
Consider a 1 gram particle in a 10 cm box moving at 1 cm/s.
Calculate the quantum number, n, which represents the number
of maxima in the probability, 2.
m = 1x10-3 kg
a = 0.10 m
1
1
2
v = 0.01 m/s
E  mv 2  1x103 kg   0.01m / s   5.0 x108 J
2
2
h = 6.63x10-34 J-s
E
2
2
2
nh
8ma E
2

n


2
2
8ma
h


8 1x10 3 kg  0.1m  (5.0 x10 8 J )

2
6.63 x10 34 J  s

2
n2  9.1x1054  n  3x1027
Thus, the probability distribution is uniform throughout the box,
as predicted by classical mechanics.
Slide 16
Outline
• The Classical Particle in a Box (PIB)
• The Quantum Mechanical PIB
• PIB Properties
• Other PIB Models
• The Free Electron Molecular Orbital (FEMO) Model
• PIB and the Color of Vegetables
• Multidimensional Systems: The 2D PIB and 3D PIB
• Statistical Thermodynamics: Translational contributions to
the thermodynamic properties
of gases.
Slide 17
Some Useful Integrals
2
sin
  x  dx 
 sin  x  sin   x  dx 
1
1
x
sin  2 x 
2
4
sin (   )x 
2    
 sin  x  cos  x  dx 

sin (   )x 
2    
1
sin2  x 
2
x 2 x sin  2 x  cos  2 x 
 x sin  x  dx  4  4  8 2
2
x cos  2 x 
x3  x2
1 
x
sin

x
dx



sin
2

x

 


 

6  4 8 3 
4 2
2
2
Slide 18
PIB Properties
Normalization of the Wavefunctions
 n 
x
a


 n  A sin 
0xa
2
s
in
 (  x )d x 
1
1
x
s in ( 2  x )
2
4
2
1

a
0
 n2dx  
a
0
a

 n  
2
2
A
sin
x
dx

A
sin
 x  dx
 a 


0




n
a
 1
1
1
1
1

 n   

 A2  x 
sin  2 x    A2  a 
sin  2
a  0 
sin  0   
4
4
4
2
0
 a  

 2
a
a
1 A  
2
2
1/ 2
2
A 
a

2
a
Slide 19
Probability of finding the particle in a
particular portion of the box
Calculate the probability of finding a particle with   A sin   x 
1
 a 


n=1 in the region of the box between 0 and a/4
2
 s in (  x ) d x 
P  0  x  a / 4   A2 
a/4
0
sin2  x  dx
1
1
x
s in ( 2  x )
2
4
  /a
a/4
 1 a

1
1
1
1

  a  
2
A  x
sin  2 x    A   
sin  2
    0  4( / a ) sin  0   

2
4

2
4
4

/
a
a
4

 

0
 
 

2
P 0  x  a / 4 
2a a

a  8 4
 1 1
  4  2  0.091

This is significantly lower than the classical probability, 0.25.
Slide 20
Orthogonality of the Wavefunctions
Two wavefunctions are orthogonal if:
  d  0
*
i
j
i j
We will show that the two lowest wavefunctions of the PIB
are orthogonal:  1  A sin   x   2  A sin  2 x 
 a 
 a 
a
a
a
2

 2 x 
x 
2


dx

A
sin
A
sin
dx

A
sin

x
sin

x
dx








 a 
 a 
0 2 1 0
0
a
a




 s in(  x ) s in(  x ) d x 
s in[(    ) x ] s in[(    ) x ]

2(   )
2(   )
 sin (   )a  sin (   )a  
2
0  2 1dx  A  2      2       A  0  0 


a
2
 sin  
sin 3  
0  2 1dx  A  2      2       0

 0
a
2
   a  3
a  3
a
   a   a  
a
Thus, 1 and 2 are orthogonal
Slide 21
Using the same method as above, it can be shown that:
  d  0
*
i
j
For all i  j
Thus, the PIB wavefunctions are orthogonal.
If they have also been normalized, then
the wavefunctions are orthonormal:
  d  
*
i
j
ij
or
 i  j   ij
Slide 22
Positional Averages
We will calculate the averages for 1 and present the general
result for n

2
x 

A
sin

x


A




a
a
 a 
 1  A sin 
 
sin  a   sin  a   0
a 
 
cos  a   cos  a   1
a 
  
sin  2 a   sin  2 a   0
 a 
  
cos  2 a   cos  2 a   1
 a 
<x>
a
a
x 
x 
x   1 x  1    1* x 1dx   A sin 
xA
sin
dx



0
0
 a 
 a 
a

 A2  x sin2  x  dx  
0
a
Slide 23
x A
2

a
0
x sin  x  dx
2

x 2 x sin( 2  x ) c o s( 2  x )
x sin ( x ) d x 


4
4
8 2
2
cos  0   
2  a 2 a sin  2 a  cos  2 a   
x   

  0  0 

2
a  4
4
8 2
8

 
 
2  a 2
1   1 
x   

 
a  4 8 2   8 2  
x 
a
2
General Case: x 
a
2
This makes sense because 2 is symmetric about the
center of the box.
Slide 24
<x2 >
a
a
x  2
x 
x 2   1 x 2  1    1* x 2 1dx   A sin 
x
A
sin
dx



0
0
 a 
 a 
a

 A2  x 2 sin2  x  dx  
0
a
x3  x2
1

x
s
in
(

x
)
d
x




6  4  8  3
2
x
2
x2
2

x cos(2 x )
 s in( 2  x ) 
4 2


a cos  2 a  
2  a 3  a 2
1 
   

   0  0  0 
 sin  2 a  
a  6  4 8 3 
4 2






2  a3
a  a2 a2
1  2
1
2






a

0.28
a
 3 2 2 
2
2
a 6


    3 2
4  

a 

Slide 25
1 
1
x 2    2  a2  0.28a2
 3 2 
compared to the classical result
x
General Case:
2
a2

 0.33a2
2
1 
1
x 2    2 2  a2
 3 2n  
Note that the general QM value reduces to the classical result
in the limit of large n, as required by the Correspondence Principle.
Slide 26
Momentum Averages
Preliminary
d
pˆ 
i dx
p^2  
 
x   A sin  x 
a


 1  A sin 


a
2
d2
dx 2
A
2
a
d 1 d

 A sin  x    A cos  x 
dx
dx
d 2 1 d

A cos  x     A 2 sin  x 

2
dx
dx
 
sin  a   sin  a   0
a 
  
sin  2 a   sin  2 a   0
 a 
 
  
cos  a   cos  a   1 cos  2 a   cos  2 a   1
a 
 a 
Slide 27
<p>
p   1 pˆ  1
pˆ 
d
i dx
x 
dA sin 
a
a
d 1

a 

*
  1
dx   A sin  x 
dx  
0
0
i dx
i
dx
a
a

  A sin  x  A cos  x  dx  A2 
0
i
i
 a
  0 sin  x  cos  x  dx

 s in (  x ) c o s (  x ) d x 

p  A2 
i
<p> = 0
  1
  1

2
2
2

sin

a

sin
0

A




  2
  2

i
 
 


1
s in 2 (  x )
2

  0  0 

(General case is the same)
On reflection, this is not surprising.
The particle has equal probabilities of moving in the + or - x-direction,
and the momenta cancel each other.
Slide 28
p^2  
<p2>
p
2
^ 2   a * 
 1 p
1
0 1 

  A sin  x  
a
0
2

2

d 2 1
dx 
dx 2


a
0

A sin  x  
2

2
d2
dx 2
x 
d 2 A sin 

 a  dx   
dx 2
a
 A 2 sin  x  dx  A2 2 2  sin2  x  dx
a
0
2
s
in
 (  x )d x 
1
1
x
s in ( 2  x )
2
4
 a 1
1
 
  2 
a 
p2  A2 2 2  
sin  2 a     0 
sin  0)     2 2  
4
 
  a 
2
 2 4
p
2

 
2
2
2
2
a2
General Case: p
2
n 2 2

a2
2
Slide 29
Standard Deviations and the Uncertainty Principle
Heisenberg Uncertainty Principle:  x p 
x 
a
 0.50a
2
p 0
p
x 2  0.28a 2
x 
x2  x
p 
p2  p
2
2
2
2

2
2
a2
 0.28a2   0.50a   0.17a
2

p2  0 

a




 3.14
a
 x p   0.17a   3.14   0.53
a
Slide 30
Outline
• The Classical Particle in a Box (PIB)
• The Quantum Mechanical PIB
• PIB Properties
• Other PIB Models
• The Free Electron Molecular Orbital (FEMO) Model
• PIB and the Color of Vegetables
• Multidimensional Systems: The 2D PIB and 3D PIB
• Statistical Thermodynamics: Translational contributions to
the thermodynamic properties
of gases.
Slide 31
Other PIB Models
PIB with one non-infinite wall

V(x)
Vo
0
Region I
I
0
x
Region II
II
a
V(x)  
x<0
V(x) = 0
0xa
V(x) = Vo
x>a
We will first consider the case
where E < V0.
Boundary Conditions
The wavefunction must satisfy two conditions
at the boundaries (x=0, x=a, x)
1)  must be continuous at all boundaries.
2) d/dx must be continuous at the boundaries
unless V at the boundary.
Slide 32

V(x)
Vo
Region I
I
0
0
x
Region II
II
a
V(x)  
x<0
V(x) = 0
0xa
V(x) = Vo
x>a
We will first consider the case
where E < V0.
Region I
d 2 I

 0  E I
2
2m dx
2
d 2 I
2mE


 I  const  I
2
2
dx
  c o n s t 
can
assume
d I
 A cos  x   B sin  x 
dx
I
d 2 I
2
2


A

sin

x

B

cos  x 


2
dx
d 2 I
2



I
2
dx
 II A A
s i sin
n ( x)x 
B Bc ocos
s ( 
x )x 
Slide 33

V(x)
Vo
0
Region I
I
0
x
Region II
II
a
V(x)  
x<0
V(x) = 0
0xa
V(x) = Vo
x>a
We will first consider the case
where E < V0.
Region I
Relation between  and E
d 2 I
2mE


I
2
2
dx
 I  A sin  x   B cos  x 
d 2 I
2



I
2
dx
 2 I  
2 

2mE
2
I
2mE
2
2mE
2
Slide 34

V(x)
Vo
Region I
I
0
0
x
Region II
II
a
V(x)  
x<0
V(x) = 0
0xa
V(x) = Vo
x>a
We will first consider the case
where E < V0.
Region II
d 2 II

 V0 II  E II
2
2m dx
2
d 2 II
2m


E  V0  II
2 
dx 2
d 2 II
2m


V  E  II  const  II
2  0
dx 2
  c o n s t  II
can
assume
d II
 C  e x  D e  x
dx
d 2 II
2 x
2  x

C

e

D

e
2
dx
d 2 II
2


 II
2
dx
 II  Ce x  De  x
Slide 35

V(x)
Vo
0
Region I
I
0
x
Region II
II
a
V(x)  
x<0
V(x) = 0
0xa
V(x) = Vo
x>a
We will first consider the case
where E < V0.
Region II
Relation between  and E
d 2 II
2m


V  E  II
2
2  0
dx
 II  Ce  De
x
d 2 II
  2 II
2
dx
 x
 2 II  
2m
2
2  
2m

2m
2
2
V0  E  II
V0  E 
V0  E 
Slide 36

V(x)
Vo
0
Region I
I
0
x
Region II
II
a
V(x)  
x<0
V(x) = 0
0xa
V(x) = Vo
x>a
We will first consider the case
where E < V0.
Boundary Conditions
BC: x = 0
 I  A sin  x   B cos  x 

2mE
 I (0)  0  A sin 0  B cos 0  B
2
B0
i n (xx)
Therefore: I I  AAssin
Slide 37

V(x)
Vo
0
Region I
I
0
x
Region II
II
V(x)  
x<0
V(x) = 0
0xa
V(x) = Vo
x>a
We will first consider the case
where E < V0.
a
Boundary Conditions
BC: x = 
 II  Ce  De
x

 x
 II    0  Ce  De  Ce
Therefore:
2m
2
V0  E 
C 0
 II  De  x
Slide 38

V(x)
Vo
Region I
I
0
0
x
 I  A sin  x 
Region II
II
V(x)  
x<0
V(x) = 0
0xa
V(x) = Vo
x>a
We will first consider the case
where E < V0.
a
Boundary Conditions
BC’s: x = a

2mE
d I
 A cos  x 
dx
2
 II  De  x

2m
2
V0  E 
d II
 D  e   x
dx
 I (a)   II (a)
 d I   d II 
 dx    dx 

a 
a
or
or
A sin  a   De  a
A cos a  D e a
Slide 39

V(x)
Vo
0
Region I
I
0
x
 I  A sin  x 

Region II
II
V(x)  
x<0
V(x) = 0
0xa
V(x) = Vo
x>a
We will first consider the case
where E < V0.
a
Boundary Conditions
BC’s: x = a
2mE
2
A sin  a   De  a
 II  De  x

2m
2
V0  E 
A cos a  D e a
These equations can be solved analytically to obtain the allowed
values of the energy, E. However, it’s fairly messy.
I’ll just show you some results obtained by numerical solution
of the Schrödinger Equation.
Slide 40
h2
V0  30
8ma2
h2
E1  0.9
8ma 2

This is an arbitrary value of V0
1.0
1.0
0.8
0.8
2
0.6
0.4
0.6
0.4
0.2
0.2
0.0
0.0
0.05
0.30
0.55
0.80
1.05
1.30
1.55
1.80
0.05
0.30
0.55
vals.1
x
x

1.30
1.55
1.80
x
2 V 
1.0
h2
E2  3.4
8ma 2
1.05
vals.1
x
 V
0.80
1.0
0.8
0.5
2
0.0
0.6
0.4
-0.5
0.2
-1.0
0.0
0.05
0.30
0.55
0.80
1.05
1.30
1.55
1.80
x
0.05
0.30
0.55
vals.1
x
 V
0.80
1.05
1.30
1.55
1.80
1.05
1.30
1.55
1.80
x
vals.1
x
2 V 
1.0
1.0
2
h
E3  7.6
8ma 2

0.5
0.8
2
0.0
0.6
0.4
-0.5
0.2
-1.0
0.0
0.05
0.30
0.55
0.80
1.05
1.30
1.55
1.80
0.05
vals.1
x
 V
x
0.30
0.55
0.80
x
vals.1
x
2 V 
Slide 41
h2
V0  30
8ma2
1.0
0.9
0.8
E6  38.2
2
h
8ma 2

0.4
2
-0.1
0.6
0.4
-0.6
0.2
0.0
-1.1
0.05
0.30
0.55
0.80
1.05
1.30
1.55
1.80
0.05
vals.1
x
 V
x
0.30
0.55
0.80
1.05
1.30
1.55
1.80
vals.1
x
2 V 
x
What happened??
Why is the wavefunction so different?
The condition for the earlier solution, E < V0,
is no longer valid.
Slide 42

Region I
I
0
0
h2
V0  30
8ma2
V(x)  
x<0
Vo
V(x) = 0
0xa
V(x) = Vo
x>a
Region II
II
a
E6  38.2
E > Vo
2
h
8ma 2
Region I
Region II
d 2 I

 0  E I
2
2m dx
d 2 II

 V0 II  E II
2
2m dx
2
d 2 I
2mE


 I  const  I
2
2
dx
 I  A sin  x   B cos  x 
2
d 2 II
2m


E  V0  II  const  II
2
2 
dx
 const  II
 II  C sin   x   D cos   x 
Slide 43
PIB with central barrier: Tunneling
Preliminary: Potential Energy Barriers in Classical Mechanics
Bowling Ball
m = 16 lb = 7.3 kg
v = 30 mph = 13.4 m/s
KE = ½mv2 = 650 J
V0 = mgh = 1430 J
h = 20 m
Will the bowling ball make it over the hill? Of course not!!
In classical mechanics, a particle cannot get to a position in
which the potential energy is greater than the particle’s total energy.
Slide 44

Vo
V(x)
0

I
0
II
x1
x
III
x2
a
V(x)  
x<0
V(x) = 0
0  x  x1
V(x) = V0
x1  x  x2
V(x) = 0
x2  x  a
V(x)  
x>a
I’ll just show the Boundary Conditions and graphs of the results.
BC: x=0
BC’s: x=x1
BC’s: x=x2
BC: x=a
 I (0)  0
 I ( x1)   II ( x1)
 II ( x2 )   III ( x2 )
 III (a)  0
 d I 
 d II 

 dx 



 x1  dx  x1
 d II 
 d III 

 dx 



 x2  dx  x2
Slide 45
h2
V0  20
8ma2
h2
E1  7.2
8ma 2

1.0
1.0
0.8
0.8
2
0.6
0.6
0.4
0.4
0.2
0.2
0.0
-0.1
0.0
0.1
0.3
0.5
0.7
0.9
1.1
-0.1
x
 V
x
0.1
0.3
0.5
0.7
0.9
1.1
x
vals.1
x
2 V 
Note that there is a significant probability of finding the particle
inside the barrier, even though V0 > E.
This means that the particle can get from one side of the barrier
to the other by tunneling through the barrier.
Slide 46
We have just demonstrated the quantum mechanical
phenomenon called tunneling, in which a particle can
be in a region of space where the potential energy is
higher than the total energy of the particle.
This is not simply an abstract phenomenon, but is known to
occur in many areas of Chemistry and Physics, including:
•
•
Ammonia inversion (the ammonia clock)
•
Charge carriers in semiconductor devices
•
Nuclear radioactive decay
•
Scanning Tunneling Microscopy (STM)
Kinetic rate constants
Slide 47
Outline
• The Classical Particle in a Box (PIB)
• The Quantum Mechanical PIB
• PIB Properties
• Other PIB Models
• The Free Electron Molecular Orbital (FEMO) Model
• PIB and the Color of Vegetables
• Multidimensional Systems: The 2D PIB and 3D PIB
• Statistical Thermodynamics: Translational contributions to
the thermodynamic properties
of gases.
Slide 48
Carrots are Orange.
Tomatoes are Red.
But why??
Slide 49
Free Electron Molecular Orbital (FEMO) Model
So, does the solution to the quantum mechanical “particle in a box”
serve any role other than to torture P. Chem. students???
Yes!!
To a good approximation, the  electrons in conjugated
polyalkenes are free to move within the confines of the
 orbital system.
H2C
H
C
H
C
H
C
H
C
CH2
a
Notes: Estimation of the box length, a, will be discussed later.
Each carbon in this conjugated system contributes 1 
electron. Thus, there are 6 electrons in the -system
of 1,3,5-hexatriene (above)
Slide 50
 Bonding in Ethylene
Bonding () Orbital
PIB (12)
+
+
C
C
-
-
C
C
C
Maximum electron
density between C’s
C
Maximum electron
density between C’s
Anti-bonding (*) Orbital
PIB (22)
+
-
C
C
-
+
C
C
C
Electron density
node between C’s
C
Electron density
node between C’s
Slide 51
  * Transition in Ethylene using FEMO
4h 2
E2 
8ma 2
1h 2
E1 
8ma 2
4h2
1h2
3h2
E  E2  E1 


2
2
8ma
8ma
8ma2
R = 0.134 nm
h = 6.63x10-34 J•s
m = 9.11x10-31 kg
a = ??? = 0.268 nm
C
½R
C
R
½R
a = 2R
It is common (although not universal) to assume
that the  electrons are free to move approximately
½ bond length beyond each outermost carbon.
Slide 52

34

2
3 6.63 x10 J  s
3h
E 

 2.52x1018 J
2
31
9
2
8ma
8(9.11x10 kg )(0.268 x10 m )
2
Units:
kg  m 2 2
J
s
2
J 2  s2
s

J
kg  m 2
kg  m 2
Calculation of max
E  Ephot 
hc

hc (6.63x1034 J  s )(3.00x108 m / s )
 

E
2.52x1018 J
 = 7.9x10-8 m  80 nm
(exp) = 190 nm
The difficulty with applying FEMO to ethylene is that the result
is extremely sensitive to the assumed box length, a.
Slide 53
Application of FEMO to 1,3-Butadiene
H
C
H2C
½R
R
H
C
R
CH2
R
½R
It is common to use R = 0.14 nm (the C-C bond length in
benzene,where the bond order is 1.5) as the average bond length
in conjugated polyenes.
L = 3•R + 2•(½R) = 4•R
9h 2
E3 
8ma 2
4h 2
E2 
8ma 2
1h 2
E1 
8ma 2
L = 4•0.14 nm = 0.56 nm
9h2
4h2
5h2
E  E3  E2 


2
2
8ma
8ma
8ma2
Slide 54
5h2
5(6.63 x1034 J  s )2
19
E 


9.62
x
10
J
2
31
9
2
8ma
8(9.11x10 kg )(0.56x10 m)
Calculation of max
E  Ephot 
hc

hc (6.63x1034 J  s )(3.00x108 m / s )
 

E
9.62x1019 J
 = 2.07x10-7 m  207 nm
(exp) = 217 nm
Slide 55
Application of FEMO to 1,3,5-Hexatriene
H
C
H2C
½R
R
H
C
R
H
C
R
H
C
R
CH2
R
½R
a = 5•R + 2•(½R) = 6•R
16h 2
E4 
8ma 2
a = 6•0.14 nm = 0.84 nm
16h2
9h2
7h2
E  E4  E3 


2
2
8ma
8ma
8ma2
9h 2
E3 
8ma 2
4h 2
E2 
8ma 2
1h 2
E1 
8ma 2
E = 5.98x10-19 J
(exp) = 258 nm
 = 332 nm
Problem worked out in detail in
Chapter 3 HW
Slide 56
The Box Length
Hexatriene
H
C
H2C
½R
R
H
C
R
H
C
R
H
C
R
CH2
R
½R
a = 5•R + 2•(½R) = 6•R
General: a = nb•R + 2•(½R) =(nb+1)•R
R = 1.40 Å = 0.14 nm
Slide 57
Outline
• The Classical Particle in a Box (PIB)
• The Quantum Mechanical PIB
• PIB Properties
• Other PIB Models
• The Free Electron Molecular Orbital (FEMO) Model
• PIB and the Color of Vegetables
• Multidimensional Systems: The 2D PIB and 3D PIB
• Statistical Thermodynamics: Translational contributions to
the thermodynamic properties
of gases.
Slide 58
PIB and the Color of Vegetables
Introduction
The FEMO model predicts that the * absorption wavelength, ,
increases with the number of double bonds, #DB
Compound #DB
(FEMO)
Ethylene
1
80 nm
Butadiene
2
207
Hexatriene
3
333
roughly
This is because:
and:
 N  1
E 
2
a2

 N2

nb
 nb 
2

1
nb
N = # of electron pairs
1
 nb
E
Slide 59
Light Absorption and Color
Blue
Red
% Transmission
Yellow
If a substance absorbs
light in the blue region of
the visible spectrum, the
color of the transmitted
(or reflected) light will be red.
400
500
600
700
 (nm)
Blue
Red
% Transmission
Yellow
If a substance absorbs
light in the red region of
the visible spectrum, the
color of the transmitted
(or reflected) light will be blue.
400
500
600
 (nm)
700
Slide 60
Conjugated  Systems and the Color of Substances
White light contains all wavelengths of visible
radiation ( = 400 - 700 nm)
If a substance absorbs certain wavelengths,
then the remaining light is reflected, giving
the appearance of the complementary color.
e.g. a substance absorbing violet light appears
to be yellow in color.
Molecule
No. of Conj.
Doub. Bonds
Ethylene
1
1,3-Butadiene
2
1,3,5-Hexatriene 3
-Carotene
11
Lycopene
13
max
Region
190 nm
217
258
450
500
UV
UV
UV
Vis.
Vis.
Carrots
Tomatoes
Slide 61
Structure of Ground State Butadiene
R(C-C) = 1.53 Å
H2C
1.32 Å
1.47 Å
C
H
H
C
1.32 Å
CH2
As well known, there is a degree of electron delocalization between
the conjugated double bonds, giving some double bond character
to the central bond.
Calculation: HF/6-31G(d) – 1 minute
Slide 62
Potential Energy Surface (PES) of Ground State Butadiene
Minimum at
~40o
Relative Energy [kJ/mol]
25
20
15
10
5
0
0
20
40
60
80
100
120
140
160
180
Dihedral Angle [Degrees]
Calculation: HF/6-31G(d) – 37 optimizations – 25 minutes
Slide 63
Energy Difference Between Trans and “Cis”-Butadiene
Relative Energy [kJ/mol]
25
20
15
10
E(cal) = Ecis – Etrans = 12.7 kJ/mol
5
0
0
20
40
60
80
100
120
140
160
180
Dihedral Angle [Degrees]
Experiment
At room temperature (298 K), butadiene is a mixture with approximately
97% of the molecules in the trans conformation (and 3% “cis”).
 N 
E (exp)  RT ln  cis 
 Ntrans 
 0.03 
E (exp)  (8.31x103 kJ / mol  K )(298 K )ln 
 8.6 kJ / mol

 0.97 
E (exp)

Ncis
 e RT
Ntrans
We could get very close to experiment by using a higher
level of theory.
Slide 64
Structure of Excited State Butadiene
H2C
1.32 Å
1.47 Å
H
C
C
H
1.32 Å
CH2
Ground State: S0
H2C
1.41 Å
1.39 Å
H
C
C
H
Excited State: S1
1.41 Å
CH2
Slide 65
Structure of Excited State Butadiene
H2C
1.32 Å
H
C
1.47 Å
C
H
1.32 Å
CH2
Ground State: S0
HOMO
* Transition
H2C
1.41 Å
1.39 Å
H
C
C
H
Excited State: S1
1.41 Å
CH2
LUMO
Slide 66
Comparison of FEMO and QM Wavefunctions
 3 x 
ψ 3  A  sin 

a


LUMO
 2 x 
ψ 2  A  sin 

 a 
HOMO
 x 
ψ 1  A  sin  
 a 
0

L
Slide 67
Outline
• The Classical Particle in a Box (PIB)
• The Quantum Mechanical PIB
• PIB Properties
• Other PIB Models
• The Free Electron Molecular Orbital (FEMO) Model
• PIB and the Color of Vegetables
• Multidimensional Systems: The 2D PIB and 3D PIB
• Statistical Thermodynamics: Translational contributions to
the thermodynamic properties
of gases.
Slide 68
Multidimensional Systems: The 2D PIB
 2
2 
H
 2  2   V ( x, y )
2m  x
y 
2
The 2D Hamiltonian:
 2
2 



 ( x, y )  V ( x, y ) ( x, y )  E ( x, y )
2m  x 2 y 2 
2
The 2D Schr. Eqn.:
The Potential Energy
V(x,y) = 0
b
0  x  a and 0  y  b
V(x,y)  x < 0 or x > a or y < 0 or y > b
y
0
0
x
a
Outside of the box
i.e. x < 0 or x > a or y < 0 or y > b
(x,y) = 0
Slide 69
Solution: Separation of Variables
 2
2 

 2  2  ( x, y )  V ( x, y ) ( x, y )  E ( x, y )
2m  x
y 
2
The 2D Schr. Eqn.:
Inside the box:
V(x,y) = 0
0  x  a and 0  y  b
2
2
2
H

 H x  Hy
2m x 2 2m y 2
2
Note: On following slides, I will show how this two dimensional
differential equation can be solved by the method of
Separation of Variables.
You are NOT responsible for the details of the soution, but only
the assumptions and results.
Slide 70
Solution: Separation of Variables
2
2
2
H

 H x  Hy
2
2
2m x
2m y
2
Inside the box:
Hx
Hy
Hx ( x, y )  Hy ( x, y )  E ( x, y )
Assume:  ( x, y )  X ( x )Y ( y )
Hx X ( x )Y (y )  Hy X ( x )Y (y )  EX ( x )Y (y )
Y (y )Hx X ( x )  X ( x )HyY (y )  EX ( x )Y (y )
1
1
Hx X ( x ) 
HyY ( y )  E
X(x)
Y (y )
Slide 71
Solution: Separation of Variables
2
2
2
H

 H x  Hy
2m x 2 2m y 2
2
Inside the box:
1
1
Hx X ( x ) 
HyY ( y )  E
X(x)
Y (y )
f(x)
g(y)
C
If:
f(x) + g(y) = C
then: f(x) = C1 and g(y) = C2
2
1 d2X
1 d 2Y


E
2m X dx 2 2m Y dy 2
=
2
=
C1 + C2 = C
Ex
Ey
E = Ex + Ey
Slide 72
=
2
1 d2X
1 d 2Y


E
2
2
2m X dx
2m Y dy
=
2
Ex
Ey
2
1 d 2Y

 Ey
2
2m Y dy
2
1 d2X

 Ex
2
2m X dx
2
d2X

 Ex X
2m dx 2
Range
0xa
nx2h2
Ex 
nx  1,2,3,...
8ma 2
2
n x 
X ( x )  Ax sin  x  Ax 
a
 a 
 ( x, y )  X ( x )Y ( y )
2
d 2Y

 EyY
2m dy 2
Ey 
ny2h2
8mb
2
Range
0yb
ny  1,2,3,...
n y 
2
Y ( y )  Ay sin  y
A

 y
b
b


E = Ex + Ey
Slide 73
Two Dimensional PIB: Summary
2
2
2
H

 H x  Hy
2
2
2m x
2m y
2
Inside the box:
Assume:  ( x, y )  X ( x )Y ( y )

2
2
d X
 Ex X
2
2m dx
Range
0xa
nx2h2
Ex 
nx  1,2,3,...
8ma 2
2
n x 
X ( x )  Ax sin  x  Ax 
a
 a 
 ( x, y )  X ( x )Y ( y )
2
d 2Y

 EyY
2m dy 2
Ey 
ny2h2
8mb
2
Range
0yb
ny  1,2,3,...
 ny  y 
2
Y ( y )  Ay sin 
A

 y
b
b


E = Ex + Ey
Slide 74
The Wavefunctions
 ny  y  

 n x x   
 ( x, y )  X ( x )  Y ( y )   Ax sin 
    Ay sin  b  
a

 



 ( x, y ) 
4
n x  n y 
sin  x  sin  y

ab
 a   b 
2
nx = 1 ny = 1
2
a
Ay 
2
b
Range
0xa
0yb
nx = 1, 2, 3,...
ny = 1, 2, 3,...
2
2
Ax 
2
2
nx = 2 ny = 1
2
nx = 2 ny = 2
Slide 75
The Energies: Wavefunction Degeneracy
2
nx2h 2
nx2h 2
h 2  nx2 ny 
E  Ex  Ey 


 2  2 
2
2
8ma
8mb
8m  a
b 
nx = 1, 2, 3,...
ny = 1, 2, 3,...
Square Box
a=b
h2
2
2
E
n

n
x
y
8ma 2

13

10
 h 
 8m a 2  8


2
E
23
32
13
31
22
5
1 2
2
2 1
g=2
g=2
g=1
g=2
g=1
11
nx ny
Slide 76
Application: * Absorption in Benzene
13
13
10
8
32
13
31
22
5
21
E [h2/8ma2]
E [h2/8ma2]
23
10
8
32
13
31
22
5
21
12
2
23
2
12
11
11
nx ny
nx ny
The six  electrons in benzene can be approximated as
particles in a square box.
One can estimate the wavelength of the lowest energy
* from the 2D-PIB model (See HW problem)
Slide 77
Three Dimensional PIB
 2
2
2 

 2  2  2  ( x, y , z )  V ( x, y , z ) ( x, y , z )  E ( x, y , z )
2m  x
y
z 
2
V(x,y,z) = 0
0xa
0yb
0zc
V(x,y,z)  
Outside the
box
c
b
a
 2
2
2 

 2  2  2  ( x, y , z )  E ( x, y , z )
2m  x
y
z 
2
Slide 78
 2
2
2 




   H x  Hy  Hz   E
2m  x 2 y 2 z 2 
2
Assume:  ( x, y, z)  X ( x ) Y ( y )  Z(z)
2 X

 Ex X
2
2m x
2
nx2h 2
Ex 
nx  1,2,3,...
2
8ma
n x
X ( x )  Ax sin  x 
 a 
Ax 
2
a
 2Y

 EyY
2
2m y
2
Ey 
ny2h 2
8mb 2
ny  1,2,3,...
 ny  y 
Y ( y )  Ay sin 

 b 
Ay 
2
b
 2Z

 Ez Z
2
2m z
2
nz2h 2
Ez 
nz  1,2,3,...
8mc 2
n z 
Z ( z )  Az sin  z 
 c 
Az 
2
c
Slide 79
8
 nx x   ny  y   nz z 
 ( x, y , z )  X ( x )  Y ( y )  Z( z) 
sin 
 sin  b  sin  c 
abc
a

 

 
nx = 1, 2, 3, ...
ny = 1, 2, 3, ...
nz = 1, 2, 3, ...
2
nx2h 2
nx2h 2
nz2h 2
h 2  nx2 ny nz2 
E  E x  E y  Ez 



 
 
8ma 2 8mb 2 8mc 2 8m  a 2 b 2 c 2 
Cubical Box
a=b=c
Note: You should be able to determine the energy
levels and degeneracies for a cubical box
and for various ratios of a:b:c
h2
E
nx2  ny2  nz2
2
8ma


Slide 80
Outline
• The Classical Particle in a Box (PIB)
• The Quantum Mechanical PIB
• PIB Properties
• Other PIB Models
• The Free Electron Molecular Orbital (FEMO) Model
• PIB and the Color of Vegetables
• Multidimensional Systems: The 2D PIB and 3D PIB
• Statistical Thermodynamics: Translational contributions to
the thermodynamic properties
of gases.
Slide 81
Introduction to Statistical Thermodynamics
Statistical Thermodynamics is the application of Statistical Mechanics
to predict thermodynamic properties of molecules.
But this is a Quantum Mechanics course.
Why discuss thermodynamics??
1. The Statistical Thermodynamic formulas used to predict the
properties come from the Quantum Mechanical energies for
systems such as the PIB, Rigid Rotor (Chap. 4) and
Harmonic Oscillator (Chap. 5).
2. Various properties of the molecules, including moments of inertia,
vibrational frequencies, and electronic energy levels are required
to calculate the thermodynamic properties
Often, these properties are not available experimentally,
and must be obtained from QM calculations.
Slide 82
Some Applications of QM/Stat. Thermo.
 Molecular Heat Capacities [CV(T) and CP(T)]
 Molecular Enthalpies of Formation [Hf]
 Bond Dissociation Enthalpies [aka Bond Strengths]
 Reaction Enthalpy Changes [Hrxn]
 Reaction Entropy Changes [Srxn]
 Reaction Gibbs Energy Changes [Grxn]
and Equilibrium Constants [Keq]
 Kinetics: Activation Enthalpies [H‡] and Entropies [S‡]
 Kinetics: Rate Constants
Slide 83
Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)
Frequencies -- 1661.9057
QCISD/6-311G(d)
..............
------------------- Thermochemistry E (Thermal)
CV
S
------------------KCAL/MOL CAL/MOL-K CAL/MOL-K
Temperature 298.150 Kelvin. Pressure 1.00000 Atm.
TOTAL
3.750
5.023
48.972
Thermochemistry will use frequencies scaled by 0.9540.
ELECTRONIC
0.000
0.000
2.183
Atom 1 has atomic number 8 and mass 15.99491
TRANSLATIONAL 0.889
2.981
36.321
Atom 2 has atomic number 8 and mass 15.99491
ROTATIONAL
0.592
1.987
10.459
Molecular mass: 31.98983 amu.
VIBRATIONAL
2.269
0.055
0.008
Principal axes and moments of inertia in atomic units:
Q
LOG10(Q)
LN(Q)
1
2
3
TOTAL BOT
0.330741D+08
7.519488
17.314260
EIGENVALUES -- 0.00000 41.27885 41.27885
TOTAL V=0
0.151654D+10
9.180853
21.139696
X
0.00000 0.90352 0.42854
VIB (BOT)
0.218193D-01
-1.661159
-3.824960
Y
0.00000 -0.42854 0.90352
VIB (V=0)
0.100048D+01
0.000207
0.000476
Z
1.00000 0.00000 0.00000
ELECTRONIC
0.300000D+01
0.477121
1.098612
THIS MOLECULE IS A PROLATE SYMMETRIC TOP.
TRANSLATIONAL 0.711178D+07
6.851978
15.777263
ROTATIONAL SYMMETRY NUMBER 2.
ROTATIONAL
0.710472D+02
1.851547
4.263345
ROTATIONAL TEMPERATURE (KELVIN)
2.09825
ROTATIONAL CONSTANT (GHZ)
43.720719
Zero-point vibrational energy
9483.1 (Joules/Mol)
2.26653 (Kcal/Mol)
VIBRATIONAL TEMPERATURES:
2281.11 (KELVIN)
Zero-point correction=
0.003612 (Hartree/Particle)
Thermal correction to Energy=
0.005976
Thermal correction to Enthalpy=
0.006920
Thermal correction to Gibbs Free Energy=
-0.016348
Sum of electronic and zero-point Energies=
-150.022558
Sum of electronic and thermal Energies=
-150.020195
Sum of electronic and thermal Enthalpies=
-150.019250
Sum of electronic and thermal Free Energies= -150.042518
Slide 84
Mini-Outline for Stat. Thermo. Development
•
•
•
•
•
•
•
•
The Boltzmann Distribution
The Partition Function (Q)
Thermodynamic properties in terms of Q
Partition function for non-interacting molecules
The molecular partition function (q)
The total partition function for a system of molecules (Q)
The translational partition function
Translational contributions to thermodynamic properties
Slide 85
The Boltzmann Distribution
p(Ei )  e
E2
E1

p(Ei )  Ae
Ei
kT
k
R
 1.3806 x1023 J / K
NA
 1.38 x1023 J / K
E
 i
kT
E0
Normalization
 p(E )  1  A e
E

 i
kT
A    e
 i
E
 i
kT
i
i
i
Therefore:
p(Ei ) 
e

E
 i
kt
e

Ei
kt

e

Ei
kT

Ei
kt



1
eQ
Q
i
Boltzmann Distribution
Slide 86
The Partition Function

Ei
kt
e
p(Ei ) 
Q
where
Q  e

Ei
kT
i
Q is called the "Partition Function". Thermodynamic properties,
including, U, Cv, H, S, A, G, can be calculated from Q.
Alternative Form
One can sum over energy "levels" rather than individual energy
states.
p(E j ) 
g je

Q
Ei
kt
where
Q

g je

Ej
kT
j ( level )
gj is the degeneracy of the j’th energy level, Ej
Slide 87
Thermodynamic Properties from Q: Internal Energy (U)
U  E   Ei p(Ei ) 
i
E e
i
i
Q

Ei
kT
Q  e

Ei
kT
i
E

 i 
   e kT 
1  Q 
1 i
  ln(Q ) 



 T 




V ,N Q  T V ,N Q  T



V ,N
Ei
  kT

E
1  e
1  Ei   kTi

 
  2 e

Q i  T 
Q i  kT 

V ,N

Derivative at constant
V,N because energy
depends on volume
and No. of Molecules
Ei
kT
e
  ln Q 
U  E  kT 2 

E
  Ei p  Ei 

i


T
Q

V ,N
i
i
NOTE: On an exam, you will be provided with any required SM formula
for a thermodynamic property (e.g. the above expression for U)
Slide 88
Thermodynamic Properties from Q: Other Properties
 U 
Constant Volume Heat Capacity: CV  

 T V ,N
The math involved in relating other thermodynamic properties
to Q is a bit more involved.
It can be found in standard P. Chem. texts; e.g. Physical Chemistry,
by R. A. Alberty and R. J. Silbey, Chap. 17
I'll just give the results.
Pressure:
Entropy:
  ln Q 
P  kT 

 V T
S  k ln Q 
U
T
Slide 89
Thermodynamic Properties from Q: Other Properties
  ln Q 
  ln Q 

kT

  lnV 
 T V ,N

T ,N
Enthalpy: H  kT 2 
Note: If Q  f(V), then H = U
Constant Pressure Heat Capacity: CP   H 
 T P,N
Helmholtz Energy:
Gibbs Energy:
A  U  TS  kT ln Q
  ln Q 
G  H  TS  kT ln Q  kT 

  lnV T ,N
Note: If Q  f(V), then G = A
NOTE: On an exam, you will be provided with any required SM formula
for a thermodynamic property.
Slide 90
Partition Function for Non-interacting Molecules (aka IDG)
Case A: Distinguishable Molecules
The microstate of molecule 'a' is given by j, of molecule 'b' by k…
Ei  a, j   b,k  c,l  ...
a, b, c, ... designates the molecule
i, j, k, ... designates the molecule's set of energies (e.g. trans., rot., vib., ...)
Q  e

Ei
kT
i
Q  e
j

e

 a , j   b ,k   c ,l  ...
j ,k ,l ,...

a, j
kT
 e
k
Q  qa  qb  qc ...

 b ,k
kT
e

kT

a, j
kT
e

 b ,k
kT
e

 c ,l
kT
...
j ,k ,l ,...
 e

 c ,l
kT
...
l
where
qa   e

a, j
kT
, etc.
j
If the molecules are the same, qa  qb  qc    q
Therefore, for N molecules we have: Q  qN
Slide 91
Partition Function for Non-interacting Molecules (aka IDG)
Case B: Indistinguishable Molecules
Distinguishable Molecules: Q Q qN q N
In actuality, one cannot distinguish between different molecules in a gas.
Let's say that there is one state in which the quantum numbers for
molecule 'a' are m1,n1… and for molecule 'b' are m2,n2…
This state is the same as one in which the quantum numbers for
molecule 'a' are m2,n2… and for molecule 'b' are m1,n1…
However, the two sets of quantum numbers above would give two
terms in the above equation for Q. i.e. we've overcounted the
number of terms.
It can be shown that the overcounting can be eliminated by dividing
Q by N! (the number of ways of permuting N particles). Therefore,
Indistinguishable Molecules:
qN
Q
N!
Slide 92
The Molecular Partition Function (q)
To a good approximation, the Hamiltonian for a molecule can
be written as the sum of translational, rotational, vibrational and
electronic Hamiltonians:
H  H tran  H rot  H vib  H elect
Therefore, the molecule's total energy (i) is the sum of individual
energies:
 i   tran
  krot   lvib   melect
j
The partition function is:
q  e

i
kT
i
q  e
j
  e
j

 tran
j
kt
k
e
k

l
 krot
kt

elect
 tran
  krot   lvib   m
j
kT
m
. e
l
q  q tran  q rot  q vib  q elect

  e
j
 lvib
kt
e

k
l

 tran
j
kT
e

 krot
kT
e

 lvib
kT
e

elect
m
kT
m
elect
m
kt
m
where q tran   e

 tran
j
kt
, etc.
j
Slide 93
The Total Partition Function (Q)
qN
Q
N!
q

Q
Q

q tran
N!

tran
q  qtran  q rot  qvib  q elect
q
rot
q
vib
q
elect

N
N!
N
    
 q rot
N
Q
 q vib


tran
N
 q elect
q tran

 
 q 
 q 
Q
vib
Q
elect
vib
 Qtran  Qrot  Qvib  Qelect
N
It can be shown that the term,
1/N! belongs to Qtran
N!
Q rot  q rot

N
N
N
elect
N
Slide 94
Thermodynamic Properties of Molecules
Q  Qtran  Qrot  Qvib  Qelect
The expressions for the thermodynamic properties all involve lnQ; e.g.
  ln Qtran  Qrot  Qvib  Qelect  

ln
Q




U  kT 2 
 kT 2 



T
 T V ,N


V ,N
rot
vib
elect
  ln Q tran 



2   ln Q
2   ln Q
2   ln Q
U  kT 

kT

kT

kT








T

T

T

T

V ,N

V ,N

V ,N

V ,N
2
U  U tran  U rot  U vib  U elect
where
Similarly
U
tran
  ln Q tran 
 kT 
, etc.

 T V ,N
2
CV  CVtran  CVrot  CVvib  CVelect
etc.
Slide 95
The Translational Partition Function for an Ideal Gas
Consider that the molecules of a gas are confined in a cubical box
of length 'a' on each side.*
The energy levels are given by: En ,n
x
y
,nz
h2

nx2  ny2  nz2
2
8ma


The molecular translational partition function is:
q
tran
 e
i

Eitran
kT




  e

h2
8 ma2
n
2
x
 ny2  nz2


nx 1 ny 1 nz 1
q tran 

e
nx 1
 nx2



  e
 nx2
e
 ny2
nx 1 ny 1 nz 1

e
ny 1
 ny2

e
nz 1
 nz2
   n 2 
 e

 n 1

e
 nz2
h2
, 
8ma 2kT
3
*One gets the same result if the box is not assumed to be cubical,
with a bit more algebra.
Slide 96


   e  n 
 n 1


q
tran
2
3
h2

8ma 2kT
One can show that the summand (term in summation) is an extremely
slowly varying function of n.
For example, for O2 [M = 32x10-3 kg/mol] in a 10 cm (0.1 m) box at 298 K:
h2

 2.5 x1020
2
8ma kT
n
Δ[Expon]
1
8x10-20
106
5x10-14
1012
5x10-8
 Expon     n  1   n 2    2n  1
2
The extremely slowly changing exponent
results from the fact that:

<< 1
kT
or
 << kT
 is the change in energy between
successive quantum levels.
This leads to an important simplification in the expression for qtran.
Slide 97
Relation between sum and integral for slowly varying functions
If f(ni) is a slowly varying function of ni, then one may replace
the sum by the integral:
n2
 f (n )  
i  n1
i
n2
n1
f (n )dn
f(ni)
1
2
3
4
5
6
7
8
ni
Slide 98
3
q
   n 2 
 e
 
 n 1

tran
 e

 n 2
0
dn


0
1/ 2 3
q tran
  
 

 4 







2
4 h
8ma2kT








e

h2

8ma 2kT
3
 x 2
1    
dx 


2   4 
1/ 2
3/ 2
 2 ma2kT 


h2


 2 mkT 


2
 h

3/ 2
 2 mkT 


2
 h

3/ 2
a3
3/ 2
q tran
V
V is the volume of the box
Note: The translation partition function is the only one which
depends upon the system’s volume.
Slide 99
 2 mkT 


2
 h

3/ 2
q
tran
V
This is the translational partitition function
for a single molecule.
A note on the calculation of V (in SI Units)
One is normally given the temperature and pressure, from which
one can calculate V from the IDG law.
V 
nRT
P
R = 8.31 J/mol-K = 8.31 Pa-m3/mol-K
One should use the pressure in Pa [1 atm. = 1.013x105 Pa],
in which case V is given in m3 [SI Units]
Slide 100
 2 mkT 


2
 h

3/ 2
q tran
V
Calculate qtran for one mole of O2 at 298 K and 1 atm.
M 32x103 kg / mol
26
m


5.32
x
10
kg
23
1
NA 6.02x10 mol
nRT (1mol )(8.31Pa  m3 / mol  K )(298 K )
V

P
1.013x105 Pa
NA = 6.02x1023 mol-1
h = 6.63x10-34 J-s
k = 1.38x10-23 J/K
R = 8.31 Pa-m3/mol-K
M = 32 g/mol
1 J = 1 kg-m2/s2
1 atm. = 1.013x105 Pa
V  2.45x102 m3
q tran


26
23
2(3.14)(5.32
x
10
kg
)(1.38
x
10
J
/
K
)(298
K
)


2
34


6.63
x
10
J

s



q tran  3.13 x1021 m2 
3/ 2


3/ 2

 2.45 x102 m3


 2.45 x102 m3  4.28 x1030
Slide 101
q tran


26
23
2(3.14)(5.32
x
10
kg
)(1.38
x
10
J
/
K
)(298
K
)


2
34


6.63
x
10
J

s




 kg  (J / K )  K  3  kg 
m  
2
2
2

J s

J s 
Units: q tran  

 m2 
3/ 2
3/2
3/ 2


 2.45 x102 m3  4.28 x1030


kg
3

m  
2
2
2
 kg  m / s (s ) 


3/2
 m3
 m3  1 (Unitless )
G-98 Output
------------------- Thermochemistry ------------------Temperature 298.150 Kelvin. Pressure 1.00000 Atm.
Q
TOTAL BOT
0.330741D+08
TOTAL V=0
0.151654D+10
VIB (BOT)
0.218193D-01
VIB (V=0)
0.100048D+01
ELECTRONIC
0.300000D+01
TRANSLATIONAL 0.711178D+07
ROTATIONAL
0.710472D+02
LOG10(Q)
7.519488
9.180853
-1.661159
0.000207
0.477121
6.851978
1.851547
Their output is actually
qtran / NA
LN(Q)
17.314260
21.139696
-3.824960
0.000476
1.098612
15.777263
4.263345
q tran 4.28 x1030

 7.13 x106
23
NA
6.02x10
Slide 102
 2 mkT 


2
 h

3/ 2
q
tran
This is the translational partitition function
for a single molecule.
V
Partition Function for a system with N (=nNA) molecules
q 


tran
Q tran
N!
N

 q tran
N ! N
N

N
 q tran  e 
e


N
N
 N 
N
N
eN
Stirling’s Approximation
Qtran
  2 mkT 3/ 2

N
V  e 


2
 q tran  e 

 h


 

N
 N 




N
Slide 103
Calculate ln(Qtran) for one mole of
O2 at 298 K and 1 atm.
qtran = 4.28x1030

ln Q tran

Qtran
  2 mkT 3/ 2

N
V  e 


2
 q tran  e 

 h


 

N
 N 




N
N
 q tran  e 
 q tran  e 
 ln 
  nN A ln 

N
N





(4.28 x1030 )(2.72)
23
1
 1mol  6.02 x10 mol ln 
 1mol  6.02 x1023 mol 1





  1.01x1025

This is an extremely large number.
However, as we’ll see shortly, ln(Q) is multiplied by the very small
number (k) in the thermodynamics formulas.
Slide 104
Translational Contributions to the
Thermodynamic Properties of Ideal Gases
Preliminary
Qtran
ln Qtran  
  2 mkT 3/ 2

N
V  e 


2
 q tran  e 

 h


 

N
 N 




N
3
3
N  ln  2 mk / h2   N  ln T )   N ln V   N  N ln  N 
2
2
Slide 105
ln Qtran  
3
3
N  ln  2 mk / h2   N  ln T )   N ln V   N  N ln  N 
2
2
The Ideal Gas Equation
P
tran
P  kT 
  ln Q tran 
 P  kT 


V

T
nN A nRT
N
 kT 

V
V
V
(because R = kNA)
PV  nRT
Notes: (1) Because none of the other partition functions (rotational,
vibrational, or electronic) depend on V, they do not
contribute to the pressure.
(2) We end up with the IDG equation because we assumed
that the molecules don’t interact when we assumed that
the total energy is the sum of individual molecules’
energies.
Slide 106
ln Qtran  
3
3
N  ln  2 mk / h2   N  ln T )   N ln V   N  N ln  N 
2
2
Internal Energy
U
tran
  ln Q tran 
3N
3
 kT 
 nN AkT
  kT 2 
2T
2
 T V ,N
2
U
tran
3
 nRT
2
or
U
tran
U tran 3

 RT
n
2
Molar Internal Energy
This demonstrates the Principle of Equipartition of Translational
Internal Energy (1/2RT per translation)
It arose from the assumption that the change in the exponent
is very small and the sum can be replaced by an integral.
This is always true for translational motions of gases.
Slide 107
ln Qtran  
3
3
N  ln  2 mk / h2   N  ln T )   N ln V   N  N ln  N 
2
2
Enthalpy
H
tran
  ln Q tran 
  ln Q tran 
  ln Q tran 
tran
 kT 
  kT 
  U  kT 


T

ln
V

ln
V

V ,N

T ,N

T ,N
2
H tran 
3
3
3
nRT  kTN  nRT  kTnN A  nRT  nRT
2
2
2
H
tran
5
 nRT
2
or
H
tran
H tran 5

 RT
n
2
Molar Enthalpy
Note: As noted earlier, if Q is independent of V, then
H = U.
This is the case for all partition functions
except Qtran
Slide 108
U tran 
3
RT
2
H tran 
5
RT
2
Heat Capacities
tran
V
C
tran
P
C
CVtran  U tran 
3



R  12.47 J / mol  K

n

T
2

V ,N
CPtran  H tran 
5



R  20.79 J / mol  K

n
 T P ,N 2
Experimental Heat Capacities at 298.15 K
Compd.
CP (exp)
Ar
20.79 J/mol-K
O2
29.36
SO3
50.67
Slide 109
Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)
QCISD/6-311G(d)
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
Utran(mol) = 3/2RT = 3.72 kJ/mol = 0.889 kcal/mol
CVtran(mol) = 3/2R = 12.47 J/mol-K = 2.98 cal/mol-K
Slide 110
Entropy
S
tran
 k ln Q
tran
U tran

T
Qtran
  2 mkT 3/ 2

N

 V  e 
2
 q tran  e 
h




 

N
 N 




N
Earlier, we showed that for one mole of O2 at 298 K and 1 atm.,
ln(Qtran) = 1.01x1025 mol-1


S tran  1.38 x1023 J / K 1.01x1025 mol 1

3
  8.31J / mol  K   T
2

T
Stran  139.38  12.47  151.85 J / mol  K
O2: Smol(exp) = 205.14 J/mol-K at 298.15 K
Slide 111
Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)
QCISD/6-311G(d)
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
G-98: Smoltran = 36.321 cal/mol-K = 151.97 J/mol-K
Our Calculation (above): Smoltran = 151.85 J/mol-K
Difference is round-off error and less Sig. Figs. in our calculation.
Slide 112
ln(Qtran) = 1.01x1025 mol-1
Helmholtz Energy
For O2 at 298 K
(one mol):

Atran  U tran  TS tran  kT ln Q tran

Atran   1.38 x10 23 J / K

  298 K  1.01x10
25
mol 1

Atran  4.154x104 J / mol  41.54 kJ / mol
Gibbs Energy
G
For O2 at 298 K
(one mol):
tran
  ln Q tran 
 H  TS  kT ln Q  kT 


ln
V

T ,N
 kT lnQtran  NkT  A  nRT
tran
tran
tran
Gtran  4.154 x104 J / mol  (8.31J / mol  K )(298 K )
 3.906 x104 J / mol  39.06 kJ / mol
Note: As noted before, if Q is independent of V, then
G = A.
This is the case for all partition functions
except Qtran
Slide 113
Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)
Frequencies -- 1661.9057
..............
------------------- Thermochemistry ------------------Temperature 298.150 Kelvin. Pressure 1.00000 Atm.
Thermochemistry will use frequencies scaled by 0.9540.
Atom 1 has atomic number 8 and mass 15.99491
Atom 2 has atomic number 8 and mass 15.99491
Molecular mass: 31.98983 amu.
Principal axes and moments of inertia in atomic units:
1
2
3
EIGENVALUES -- 0.00000 41.27885 41.27885
X
0.00000 0.90352 0.42854
Y
0.00000 -0.42854 0.90352
Z
1.00000 0.00000 0.00000
THIS MOLECULE IS A PROLATE SYMMETRIC TOP.
ROTATIONAL SYMMETRY NUMBER 2.
ROTATIONAL TEMPERATURE (KELVIN)
2.09825
ROTATIONAL CONSTANT (GHZ)
43.720719
Zero-point vibrational energy
9483.1 (Joules/Mol)
2.26653 (Kcal/Mol)
VIBRATIONAL TEMPERATURES:
2281.11 (KELVIN)
Zero-point correction=
0.003612 (Hartree/Particle)
Thermal correction to Energy=
0.005976
Thermal correction to Enthalpy=
0.006920
Thermal correction to Gibbs Free Energy=
-0.016348
Sum of electronic and zero-point Energies=
-150.022558
Sum of electronic and thermal Energies=
-150.020195
Sum of electronic and thermal Enthalpies=
-150.019250
Sum of electronic and thermal Free Energies= -150.042518
QCISD/6-311G(d)
These are thermal contributions
to U, H and G.
They represent the sum of the
translational, rotational, vibrational
and (in some cases) electronic
contributions.
They are given in atomic units (au).
1 au = 2625.5 kJ/mol
Slide 114
Translational Contributions to O2 Entropy
300
S [J/mol-K]
250
200
Expt
T
150
100
0
1000
2000
3000
4000
5000
Temperature [K]
Obviously, there are significant additional contributions
(future chapters)
Slide 115
Translational Contributions to O2 Enthalpy
200
H [kJ/mol]
150
100
50
Expt
T
0
0
1000
2000
3000
4000
5000
Temperature [K]
Obviously, there are significant additional contributions
(future chapters)
Slide 116
Translational Contributions to O2 Heat Capacity
45
40
CP [J/mol-K]
35
30
Expt
T
25
20
15
0
1000
2000
3000
4000
5000
Temperature [K]
Obviously, there are significant additional contributions
(future chapters)
Slide 117