H. Scott Fogler
10/27/03
Chapter 3
Professional Reference Shelf
B. Transition State Theory
Overview
Transition state theory provides an approach to explain the temperature and
concentration dependence of the rate law. For example, for the elementary reaction
ABC  ABC
The rate law is
rA  kC A C BC  Ae E A

RT
C A C BC
For simple reactions transition state can predict E and A in concert with computational
chemistry. In transition state theory (TST) an activated molecule is formed during the
reaction at the transition
state between forming products from reactants.

 A  B C #  AB  C
A  BC 
The rate of reaction is equal to the product of the frequency, v I, of the activated complex
crossing the barrier and the concentration of the transition state complex

rA  vI C ABC #
The transition state molecule (A B C# ) and the reactants are in pseudo equilibrium at
the top of the energy barrier.


Combining
K C# 
C ABC #
C A C BC
rA  v I K C# C A C BC
We will now use statisticaland quantum mechanics to evaluate K C# to arrive at the
equation

k T  
rA   B e
 h 
E 0
kT
q ABC #
q A q BC
C A C BC

where q is overall the partition function per unit volume and is the product of
translational, vibration, rotational and electric partition functions, i.e.,

q  q T q V q R q E
The individual partition functions to be evaluated are:

1
CD/TransitionState/ProfRef.doc
Translation
2mk B T

32
q T
h3
9.84 10 29  m
3 2  T 3 2
AB
 


 
3
 m
1 amu  300K 
Vibration
qV 

1
 hv 
1 exp

 k B T 
 
˜
˜ 300K 
hv
hc

 4.8 103 


1 cm1  T 
k BT k BT

Rotation
 1 
 T  I AB
8 2 Ik B T
qR 
12.4



 

300K 1amu  Å 2 
Sy h 2
Sy 

I AB   m i ri2
I AB   ABd 2 for diatomic molecules

The Eyring Equation 
k T e S R e H
k   B 
 h 
K CT 0
#

Liquids
Gases

k B T RT e S
k  
 
 h  P 

2
CD/TransitionState/ProfRef.doc
#
#
RT
R H # RT
e
K
HOT BUTTONS
I.
II.
Introduction
A. The Transition State
B. Procedure to Calculate the Frequency Factor
Background
A. Molecular Partition Function
B. Relating S˜ , ni and N
C. Relate S˜ and q
D. Canonical Partition Function for Interacting Molecules
˜ , S˜ and q
E. Thermodynamic relationships to relate G

F.
 Relate G, the molar partition function qm
G. Relating the dimensionless equilibrium constant K and the molar partition
function qm
 
H. Relate the molecular partition function on a basis of per unit volume, q and the
equilibrium constant K
I. Recall the relationship between K and KC from Appendix C.
J. The Loose Vibration, vI
K. Evaluating the Partition Functions
III. The Eyring Equation
References for Collision Theory, Transition State Theory and Molecular Dynamics
P. Atkins, Physical Chemistry, 6th ed. (New York: Freeman, 1998)
P. Atkins, Physical Chemistry, 5th ed. (New York: Freeman, 1994).
G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical
Kinetics (New York: Wiley, 1996).
P.W. Atkins, The Elements of Physical Chemistry, 2nd ed. (Oxford: Oxford Press, 1996).
K. J. Laidler, Chemical Kinetics, 3rd ed. (New York: Harper Collins, 1987).
G. Odian, Principles of Polymerization, 3rd ed. (New York: Wiley 1991).
R. I. Masel, Chemical Kinetics and Catalysis, Wiley Interscience, New York, 2001.
References Nomenclature
A5p403 Means Atkins, P. W. Physical Chemistry, 5th ed. (1994) page 403.
A6p701 Means Atkins, P. W. Physical Chemistry, 6th ed. (1998) page 701.
L3p208 Means Laidler, K. J., Chemical Kinetics, 3rd ed. (1987) page 208.
M1p304 Means Masel, R.I., 1st Edition (2001) page 304.
I.
INTRODUCTION
While the idea of an activated complex has been postulated for years, the first real
definitive observation was made by Nobel Prize Laureate, Ahmed Zewail. He used
3
CD/TransitionState/ProfRef.doc
femtosecond spectroscopy to study the formation of ethylene from cyclobutane.†
The reaction is shown schematically in Figure PRS.3B-1.
H2 C CH 2
|
|
H2 C – CH 2
H2 C = CH 2
+
H2 C = CH 2
Energy
H2 C – CH 2
|
|
H2 C – CH 2
Reaction Coordinate
Figure PRS.3B-1 Evidence of Active Intemediate.
The active intermediate is shown in transition state at the top of the energy barrier.
A class of reactions that also goes through a transition state is the SN2 reaction.
A. The Transition State
We shall first consider SN2 reactions [Substitution, Nucleophilic, 2nd order]
because many of these reactions can be described by transition state theory. A
Nucleophile is a substance (species) with an unshared electron. It is a species
that seeks a positive center.
Nu :  R X 
 Nu : R  X

An example is the exchange of Cl for OH–, i.e.,



OH  CH3Cl 
 HOCH3  Cl
The nucleophile seeks the carbon atom that contains the halogen. The
nucleophile always approaches from the backside, directly opposite the leaving
group. As the nucleophile approaches the orbital that contains the nucleophile
electron pairs, it begins to overlap the empty antibonding orbital of the carbon
atom bearing the leaving group (Solomon, T.W.G, Organic Chemistry, 6/e Wiley
1996, p.233).
†
Science News, Vol 156, p.247.
4
CD/TransitionState/ProfRef.doc


–
Nu + C –




L 
 Nu – – C – – L 
 Nu – C + L –






–
OH + C – Cl 
 HO – – C – – Cl 
 HO – C + Cl
–
Transition State

rate  k CH3Cl OH
CH 3 Inverted
From CH 3 Cl


The Figure PRS.3B-1 shows the energy of the molecules along the reaction
coordinate which measures the progress of the reaction. [See PRS.A Collision
theory-D Polyani Equations]. One measure of this progress might be the
distance between the CH3 group and the Cl atom.
–
+

–
HO -- C -- Cl
|
A – B – C±
Eb
Energy
Energy
OH–
CH 3 Cl
CH 3 OH, Cl–
A + BC
Reaction Coordinate
Reaction Coordinate
Figure PRS.3B-2 Reaction coordinate for (a) SN2 reaction, and (b) generalized reaction
We now generalize
A BC  ABC #  AB  C
with the reaction coordinate given in terms of the distance between the B and C
molecules. The reaction coordinate for this reaction was discussed in
PRS.3-A Collision
Theory when discussing the Polyani Equation.

The energy barrier shown in Figure PRS.3B-2 is the shallowest barrier
along the reaction coordinate. The entire energy diagram for the A–B–C system
is shown in 3-D in Figure PRS.3B-3. To obtain Figure PRS.3B-2 from Figure
PRS.3B-3 we start from the initial state (A + BC) and move through the valley
up over the barrier, Eb, (which is also in a valley) over to the valley on the other
side of the barrier to the final state (A + BC). If we plot the energy along the
dashed line pathway through the valley of Figure PRS.3B-3 we arrive at Figure
PRS.3B-2.
5
CD/TransitionState/ProfRef.doc
AB + C
ABC#
Energy
BC
+
A
AB
+
C
Reactants
Products
Reaction Coordinate
Figure PRS.3B-3 3-D energy surface for generalized reaction.
The rate of reaction for the general reaction (Lp90) is the rate of crossing the
energy barrier


vI
#
A BC 

AB  C
 A B C 
We consider the dissociation of the activated complex A  B  C# as a loose
vibration of frequency vI, (s–1). The rate of crossing the energy barrier is just the
vibrational
 frequency, vI, times the concentration of the activated complex,
C ABC #

rA  vI C ABC #
(1)
We assume the activated complex ABC# is in virtual equilibrium with the
reactants A and BC so we can use the equilibrium concentration constant K C# to

relate these concentrations,
i.e.,
ABC  C



#
K C#
AB
ABC #
C AC BC

(2)
Combining Eqns. (A) and (B) we obtain

rA  v I K #c C A C BC
The procedure to evaluate vI and K C# is shown in the table T.S.1


6
CD/TransitionState/ProfRef.doc
(3)
B. Procedure to Calculate the Frequency Factor
Table PRS.3B-1 Transition State Procedure to Calculate vI and K C#
Step 1)
A. Molecular partition function. The number of ways, W, of arranging
N molecules in m energy states, with ni molecules
in the i energy

state is
W
Step 2)
N!
n 1!n 2 !
The distribution that gives a maximum in W is the Boltzmann
distribution from which we obtain the molecular partition function,
q.

ni ei
1


, q  e i ,  
N
q
kT
B. Relating S˜ , ni and N.
The entropy or the system is given by the fundamental postulate
S˜  k ln W  k ln
Step 3)
n m!
N!
n 1!n 2 ! ... n m !
Next we manipulate the Boltzmann equation for N molecules
distributed in m energy states using Stirling’s approximation to
arrive at


n 
S˜  k n i ln i 

N 
˜–
C. Relate S˜ and q. Starting with the total energy of the system E= U
˜ =nii, relative to the ground state, substitute for the number of
U
o
molecules, ni, in energy state I, using the Boltzmann distribution in
the last equation of Step 3


ln
ni
  i  ln q
N
and then sum to arrive at
˜ U
˜o
U
S˜ 
 kN ln q
T

˜ is the ground state energy.
for non-interacting molecules. U
Step 4)
o
D. Canonical partition function for interacting molecules. We need to
consider interacting molecules and to do this we have to use

Canonical partition function

k
Q   e  i E i
i1
with the probability of finding a system with energy Ei is

CD/TransitionState/ProfRef.doc
7
e  i E i
Pi 
Q
These relationships are developed with the same procedure as that
used for the molecular partition function. For indistinguishable
molecules, the canonical and molecular partition functions are

related by
N
q
Q
N!
using the above equation we can arrive at
 q 
˜ U
˜o
U
S˜ 
 kNln
 

T
N! 
Step 5)
˜, U
˜ and qi, the molecular
E. Thermodynamic relationship to relate G
partition function
We begin by combining the Maxwell relationship, i.e.,
˜ U
˜  TS˜  PV
G
 
˜ ) represents the
with the last equation in Step 4 where the tilde (e.g. G
symbols are in units of kcal or kJ without the tilde is in units per
mole, e.g., kJ/mol. We first use the last equation for S in Step 4 to
substitute in the Maxwell Eqn. We next use the relationship between
Q and q, i.e.,
Q
qN
N!
˜ to q, the molecular partition function. For N
To relate G
indistinguishable molecules of an ideal gas
˜ U
˜ o  nRT ln q
 G
N
Step 6)
Step 7)
F. Relate G to the molar partition function qm. We define qm as
q
q m  , where N  n NAvo
n

and then substitute in the last equation in Step 5.
qm
G  U o  RT ln
NAvo
Note: The tilde’s have been removed.
where n = Number of moles, and NAvo = Avogadro’s number and
where G and Uo are on a per mole basis, e.g. (kJ/mole).
G. Relate the dimensionless equilibrium constant K and the molar
partition function qmi
For the reaction
 cC  dD
aA  bB 
the change in the Gibb’s free energies is related to K by

CD/TransitionState/ProfRef.doc
8
 v i G i  G  RT ln K
G i  U i0  RT ln q mi N Avo
Combining the last equation in Step 6 and the above equations

K  e E o

q cmC q dmD
RT
q amA q bmB
where
N
Avo
  d c  b a
Step 8)
H. Relate
 the partition function on a per unit volume basis, q, and the
equilibrium constant K

V 
q
 q   q  Vm
n 
n
qm 
Where Vm is the molar volume (dm3/mol). Substituting for qmi in
the equation for K in Step 7 we obtain

Step 9)
K e

E o
RT
qC  qD  V 
a
b m
qA  qB 
c
d
N 
Avo
I. Recall the relationship between K and KC from Appendix C
RT 
K  K  K C    K  K CVm
 f 

Equate the equilibrium constant K given in the last equation of Step 8
to the thermodynamic K for an ideal gas, ( K  1) to obtain KC in
terms
the partition functions, i.e.,
For the transition state A – B – C#, with  = –1,
K C#
e
E o RT
#
q ABC
 N Avo
q A q BC
we also know
K C# 

C ABC #
C A C BC
Equating the two equations and solving for C ABC #
C ABC #  e

E o RT
#
q ABC
N AvoC A C BC
q A q BC
The prime, e.g., q, denotes thepartition functions are per unit
volume.
Step 10) J. The loose vibration.

The rate of reaction is the frequency, vI, of crossing the barrier times
the concentration of the activated complex C ABC #
rA BC  vI C ABC #
9
CD/TransitionState/ProfRef.doc


This frequency of crossing is referred to as a loose (imaginary)
vibration. Expand the vibrational partition function to factor out the
partition function for the crossing frequency.
q #v  q v # q vI 
k BT
q v#
hv I
Note that # has moved from a superscript to a subscript to denote the
imaginary frequency of crossing the barrier has been factored out of
both the vibrational, q #V and overall partition functions, q#, of the

activated complex.
#
q ABC
 q #E q #R q #V qT#

k BT
hv I
Combine with rate equation, rABC  vI C ABC # noting that vI cancels
#
q ABC
 q ABC#
out, we obtain
A  k B T N Avo q ABC# e E o
r
h
q A q BC

RT
C A C BC
A
where A is the frequency factor.
Step 11) K. Evaluate the partition functions (qT , q V , q R )
Evaluate the molecular partition functions.
 Using the Schrödenger Equation
d 2  2m
  E V(x)  0
h
dx 2
we can solve for the partition function for a particle in a box, a
harmonic oscillator and a rigid rotator to obtain the following
partition functions

Translation:
2mk B T

32
q T
h3
9.84 10 29  m
2 3  T 3 2
AB
 


 
3
 m
1 amu  300K 
Vibration:
1
 hv 
1 exp

 k B T 
 
˜
˜ 300K 
hv
hc

 4.8 103 


1 cm1  T 
k BT k BT
qV 

Rotation:

 1 
 T  I AB
8 2 Ik B T
qR 
12.4




S 

2
2 


300K


S
h
1amu

Å
 y 

y
I AB   m i ri2
10
CD/TransitionState/ProfRef.doc


The end result is to evaluate the rate constant and the activation energy in
the equation
rA 
k BT
q 
N Avo ABC# e E o
h
q A q BC
RT
C A C BC
A
We can use computational software packages such as Cerius2 or Spartan to
calculate the partition functions of the transition state and to get the vibrational
frequencies
of the reactant and product molecules. To calculate the activation

energy one can either use the barrier height as EA or use the Polyani Equation.
Example: Calculating the Frequency Factor Using Transition State Theory
Use transition state theory to calculate the frequency factor A at 300K for the reaction
H + HBr  H2 + Br
Additional Information Literature values.† (Note: Most of this information can be
obtained from computational chemistry software packages such as Cerius2, Spartan or
Cache.)
Reactants. H, HBr
H atom (mass)
1 amu
HBr (mass)
80.9 amu
HBr vibration wave number
2650 cm–1
H – Br separation distance = 142 pm
Transition State Complex. H–H–Br
Vibration wave numbers
2340 cm–1
460 cm–1 (degenerate
Separation distances
H
150 pm
142 pm
H
Br
Solution
The reaction is
H + HBr  H–H–Br  H2 + Br

The specific reaction rate is
†
Laidler (lit.cit)

11
CD/TransitionState/ProfRef.doc
k T q  # N avo   E o
k   B  HHBr
 e RT
 h  q H q HBr 
A
Reactants
Hydrogen
Translation 
2mk B T

32
q T
h3
9.88 10 29  m 2 3  T 3 2 9.88 10 29 1 amu 3 2 300K 
 
 H  
 

 

3
1 amu  300K 
m3
 m
1 amu  300K 
qT,H  9.88 1029 m3
Hydrogen Bromide

Translation
2mk B T

32
q T
h3
32
32
32

9.88 10 29  m HBr   T 
9.84 10 29 80.9 amu  300K 
 

 

 

 
3
 1 amu  300K 
m3
 m
1 amu  300K 
qT,HBr  71891029 m3
Vibration

qV 


1
 hv 
1 exp

 k B T 
1 

 
300K 
˜
˜ 300K 
hv
hc 
3 2650 cm

 4.8 103 

4.8
10






1
1
1 cm  T 
k BT k BT
 1 cm
300K 

hv
 4.8 103 2650 12.7
k BT
qV,HBr 
Rotation

1
1.0
1 exp12.7)
 1 
 T  I AB
8 2 Ik B T
q

12.4



 

 R
300K 1amu  Å 2 
Sy h 2
Sy 
I HBr   m i ri2   HBr d 2HBr

I   HBr d 2HBr 

79.91 1.42Å 2 1.99

79.9 1
12
CD/TransitionState/ProfRef.doc


amu Å 2
Sy 1
300K 1.99 amu Å 2 
q r 12.4


300K  1 amu Å 2 

q r  24.6
Total partition function

718910 29 

q HBr  q T q V q r  
124.6
m3


q HBr 1.76 10 34 m 3
Transition State Complex
Translation
q T 

32
2mk B T
h3
9.88 10 29 m
3 2  T 3 2 9.88 10 29 81.9 amu 3 2 300K 
HHBr
 

 

 

 
3
 1 amu  300K 
m3
 m
 1 amu  300K 
q #T,1  73221029 m3

Vibration
q #V  q V1q 2V 2

qV 

1)  = 2340 cm–1
1
 hv 
1 exp

 k B T 
 
˜
˜ 300K 
hv
hc

 4.8 103 


1 cm1  T 
k BT k BT

hv
11.2
k BT

q V1 

2)  = 460 cm2
hv
 2.2
k BT

q V2 

CD/TransitionState/ProfRef.doc

1
1.0
1 e 11.2
1
1.235
1 e 2.2
13
q #V q V1q 2V 2  11.1235 1.26
2
Rotation
 1 
 T  I AB
8 2 Ik B T
q

12.4



 
 R

300K 1amu  Å 2 
Sy h 2
Sy 
I AB   m i ri2
below.
Calculate
the rotational partition function, q #r , for the transition state shown

 150 pm
H
H
142 pm

Br
x
First we find the Center of mass at x
m Br  x  292  xm H  142  xm H
79.9x  292  x1 142  x1
x  5.34 pm  0.0534Å
The moment of inertia
I  m H1 r12  m H 2 r22  m Br r32

r1  292 5.34  286.6pm  2.866Å
r2 142 5.34 136.6pm 1.366Å

r3  0.0534Å

I  1 a.m.u.
 2.866Å  1 a.m.u.1.366Å  79.9 a.m.u.0.0534Å
2
2
2
 Å2 1.866 a.m.u. Å2  0.228 a.m.u. Å2
I  8.21 a.m.u.
I 10.3 a.m.u. Å2

The rotational partition function is


q #r  12.410.3127.8
The total partition function for the transition state is


#
q HHBr
 q T# q #v q #r  732210 29 1.26127.8

1.17 10 35 m 3
We now calculate the frequency factoring A.

CD/TransitionState/ProfRef.doc
14
k B T  q #HHBr
A  
N Avo

 h q H q HBr
1.38 1023 kg m 2 s 2 molecule K 300K 
1.17 10 35 m 3
 
 N Avo

h  6.626 1034 kg  m 2 s

 9.88 10 29 m 3 1.76 10 34 m 3



6.78 1030 m 3 
23 molecules 
A  62.5 1011 s 
6.02 10

mol 
 molecule 



 25310 5

3
m3
1000dm 3
10 dm


2.5310
mols
mols
m3
A  2.5310

10
dm 3
mol s
Data From Computational Chemistry
Now let’s calculate A and E using the parameters from cache.
For the reaction:

H + HBr  H2 + Br
The 3D potential energy surfaces of the reacting particles along the reaction coordinates
was calculated using the MOPAC PM3 method:
The transition state structure was found at the saddle point, refined by using the
DFT/B88-PW91 method as:
15
CD/TransitionState/ProfRef.doc
In the transition state, the three atoms are linear and the H-Br distance is 1.48 Å while
the H-H distance is 1.55 Å.
The transition state was further proved by vibrational analysis (PM3 FORCE)
showing one and only one negative vibration (imaginary frequency of crossing the
barrier). Moreover, the negative vibration corresponds to the movement of the atoms
on the two reaction coordinates.
16
CD/TransitionState/ProfRef.doc
Summary of Information from Cache Software
Reactants. H, HBr
H atom (mass)
1 amu
HBr (mass)
80.9 amu
HBr vibration wave number
2122 cm–1
H – Br separation distance = 147 pm
Transition State Complex. H–H–Br
Vibration wave numbers
1736 cm–1
289 cm–1
Separation distances
H
155 pm
H
148 pm
Br
17
CD/TransitionState/ProfRef.doc
The translational partition functions remain the same
q H  9.88 10 29 m 3
q HBr  718910 29 m 3
q HHBr  732210 29 m 3
HBr
Vibration
h
10.19
k BT

q v 1
Rotation
79.91 1.47Å 2  2.13 amu Å 2
I HBr 
 
 79.9 1
q r  12.4 2.13  26.46


q HBr  718910 29 126.46 1.90 10 34 m 3

HHBr
Vibration

(1) 

h
 1736 4.8 103  8.33
k BT
q v1 1.0002

(2)


h
 289 4.8 103 1.387
k BT

q v 2 1.33
Rotation
 rotational partition function, q #r , for the transition state shown
Calculate the

below.
H
155 pm
H

148 pm
Br
x
First we find the Center of mass at x
m Br  x  303 xm H  148  xm H
79.9x  303 x1 148  x1
x  5.51 pm  0.0551Å
18
CD/TransitionState/ProfRef.doc

r1  303 5.51 2.97pm  2.97Å
r2 148  5.51142.5pm 1.425Å
r3  0.0551Å
I  12.97  11.425  79.90.0551
2
2
I  8.82  2.03 0.24 11.1

qr = 137.5



q #HHBr  732210 29 1.00021.33137.5 m 3
q #HHBr 1.339 10 35 m 3


1.339 10 35 m 3
 1.9 10 9.88 10 6.02 10 



 62.5 10 s7.1310 m 6.02 10 
A  62.5 10
11
s
34
30
11
A  2.66 1010

Chemical
23
29
3
23
dm 3
mol s
Heat of formation at 298K (kcal/mol)
Energy of zero point level (a.u.)
MOPAC PM3
 method
Experiments
DFT/B88-PW91 method
HBr
5.3
-8.71
-2574.451933
H
52.1
52.1
-0.502437858
H-H-Br
Transition
state
59.6
N/A
-2574.953345
Therefore, the standard enthalpy of activation is:
H‡298  59.6  5.3 52.1 2.2 kcal/mol = 9.2 kJ/mol
The intrinsic Arrhenius activation energy is:

‡
Eaint rinsic  H 298
 298R  2.8 kcal/mol = 11.7 kJ/mol
Barrier height E0 (difference between zero-point levels of activated complexes and
reactants) (because the conversion between the a.u. and the kcal/mol units is very large,
we need to maintain a high number of decimal point):
19
CD/TransitionState/ProfRef.doc
E 0  627.5 * (( 2574.953345)  (2574.451933)  (0.502437858))  0.64 kcal/mol = 2.7
kJ/mol
20
CD/TransitionState/ProfRef.doc
II.
BACKGROUND
A. Molecular Partition Function
In this section we will develop and discuss the molecular partition function for
N molecules with a fixed total energy E in which molecules can occupy
different energy states, i.
Total Energy of System
Total number of molecules, N,
N  n i
(4)
n i = Number of molecules with energy i
The total Energy E is
E   n i i
(5)
The number of ways, W, arranging N molecules among m energy states
(1, 2 . . . m) is

N!
(6)
W
n 1!n 2 ! ... n m !
For example, if we have N=20 molecules shared in four energy levels (1, 2, 3,
4) as shown below
4
E
n4 = 2
3
n3 = 4
2
n2 = 8
1
n1 = 6
n
W
N!
20!

n 1!n 2 !n 3 !n 4 ! 6!8!4!2!
W 1.75 10 9
there are 1.75x109 ways to arrange the 20 molecules among the four energy
levels shown. There are better ways to put the 20 molecules in the four energy
 at a number of arrangements greater than 1.75x10 9. What are
states to arrive
they?
For a constant total energy, E, there will be a maximum in W, the
number of possible arrangements and this arrangement will overwhelm the
rest. Consequently, the system will most always be found in that arrangement.
Differentiating Eqn. (3) and setting dW = 0, we find the distribution that gives
this maximum [see A6p571]. The fraction of molecules in energy state i is
21
CD/TransitionState/ProfRef.doc
ni
e i
e i


,  1 k B T
N e i
q
Boltzmann
Distribution
q  e
(7)
 i
(8)
The molecular partition function q, measures how the molecules are distributed

(i.e. partitioned) over the available energy states.
Equation (7) is the Boltzmann Distribution. It is the most probable
distribution of N molecules among all energy states i from i=0 to i= subject to
the constraints that the total number of molecules N and the total energy, E are
Total Energy constant.
E  n i i
(5)
This energy E =  ni i is relative to the lowest energy, U0, (the ground state)
the value at T=0. To this internal energy, E, we must add the energy at zero
degrees Kelvin, U0 (A6p579) to obtain the total internal energy
˜ U
˜ 0   n i i
U
(9)
˜ , represent that this is the total energy not the energy per mole.
The tildes, U
Comments on the Partition
Function q

The molecular partition function gives an indication of the average number of
states that are thermally accessible to a molecule at the temperature of the
system. At low temperatures only the ground state is accessible. Consider what
happens as we go to the extremes of temperature.
(a) At high temperatures, (kT ≥≥ ), most all states are accessible.
q   e i   e  i
kT
 e  0
kT
 e  1
 1 e  1
Now as
T  , e i
i.e.
q 111 11 . . .
kT
1, q  
kT
kT
 e  2
 e  2
kT
kT
 e 3
kT
. . .
 . . .
because   
and we see the partition function goes to infinity as all energy states are

accessible.
(b) At
the other extreme, very very low temperatures (kT << i).
– kT
as
T  0, e i
0
q  g0
then
and we see that none of the states are accessible with one exception,
namely degeneracy in the ground state, i.e., q  go for o  0

 S˜ , ni and N
B. Relating
22
CD/TransitionState/ProfRef.doc
W is the number of ways of realizing a distribution for N particles distributed
on i levels for a total energy E
E = 1n1 + 2n2 + 3n3 + . . .
W
N!
n 1!n 2 ! . . . n i!
(6)
Boltzmann formula for entropy
Ludwig Boltzmann 1896
Recall n i = number of particles in energy level i
The Basic Postulate is
S˜  k lnW  kln
N!
n1!n 2! . . . n i!
(10)
Next we relate S˜ and q through W
lnW  lnN! ln n1 !n 2 !
 lnN!  ln n0 ! lnn 1!  . . .

lnW  lnN!  lnn i!
(11)
Stirling’s approximation for the ln of factorials is
12
X! 2
X
X1 2
X
e
or approximately
ln X! X ln X  X,
Stirling’s
Approximation
(12)
For our system this approximation becomes



ln W  N ln N  N  
 n i ln n i   n i 


N 
(13)
N  ni
(4)
Recall substituting Eqn. (4) in Eqn. (13) we find

ln W  N ln N   n i ln n i

  n i ln N   n i ln n i

ln W  ni ln ni  ni ln N
Further rearrangement gives

ln W    ni ln
combining Eqns. (10) and (14)
23
CD/TransitionState/ProfRef.doc
ni
N
(14)
n
S˜  k  ni ln i
N
(15)
C. Relate S˜ and q
Recall that the fraction of molecules in the ith energy state is
n i e i

N
q

(8)
Taking the natural log of Eqn. (5)
ln
ni
  i  ln q
N
n i 
in Eqn. (13)
N 

S˜  k  n i i  ln q 
Substituting for ln
Rearranging

 k n i  i  k  n i ln q


 n i i
S˜ 
 kN ln q
T
˜ – Uo, where Uo is the ground state
Recall from Eqn. (9) for  n ii = E = U
energy in kcal.
˜ U
˜

U
o
S˜ 
 kNlnq
(16)
T

This result is for non interacting molecules. We now must extend/generalize
our conclusion to include systems of interacting molecules. The molecular
partition function, q, is based on the assumption the molecules are independent
and don’t interact. To account for interacting molecules distributed in different
energy states we must consider the Canonical partition function, Q.
D. Canonical Partition Function for Interacting Molecules
Canonical ensemble (collection) (A6p583)
We now will consider interacting molecules and to do this we must use the
Canonical ensemble which is a collection of systems at the same temperature T,
volume V, and number of molecules N. These systems can exchange energy
with each other.
24
CD/TransitionState/ProfRef.doc
1
4
N
T
V
2
N
T
V
3
System of specified volume, temperature
and composition (Number of Molecules)
~ times.
which we will “replicate” N
N
T
V
Energy exchange between ensembles
15 Replications
15
Let
E i = Energy of ensemble i
Eˆ = Total energy of all the systems  nˆ i E i = a constant
nˆ i = Number of members of the ensemble with energy EI
˜ = Total number of ensembles
N

 Let Pi be the probability of occurrence

that a member of the ensemble has an
 energy E . The fraction of members of the ensemble with energy E can be
i
i

derived in a manner similar to the molecular partition function.
nˆ i e Ei
Pi 

˜
N
Q
Q e
E i
(17)
Q is the Canonical Partition Function.
We now rate the Canonical partition function the molecular partition
function (A6p858). The energy of ensemble i, Ei, is the sum of the energies of
each of the molecules in the ensembles
E i   i 1  N 2 . . .  N N

Q 
Energy of
Energy of molecule
molecule (1)
(N) when it is in
when it is in
state N
state i

E
 1   i 2. . . i N 
e i  e i 

i 1
Expanding the i=1 and i=2 terms
Qe
1 1  1 2
. . . e
2 1
e
2 2

. . .   e i
i3

CD/TransitionState/ProfRef.doc
25
Each molecule, e.g. molecule (1), is likely to occupy all the states available to it.
Consequently, instead of summing over the states i of the system we can sum
over the states i of molecule 1, molecule 2, etc.
  e i
Q  
molecule
  e i 


 . . . . .   e i
1 molecule 2 


N
Q  qN
This result (Eqn. (17)) is for distinguishable molecules.
However, for indistinguishable molecules it doesn’t matter which molecule is in
which state, i.e., whether
 molecule (1) is in state (a) or (b) or (c)

A5p605


A6p585
(1)  (a) (1)  (b) (1)  (c)
(2)  (b) (2)  (c) (2)  (a)
(3)  (c) (3)  (a) (3)  (b)
E = a + b + c
(etc)
Consequently we have to divide by N!

N
q
Q
N!
(18)
The molecular partition function is just the product of the partition functions
for translational (qT), vibrational (qV), rotational (qR) and electronic energy (qE)
partition functions.
q  qT q V qR qE
(19)
This molecular partition function, q, describes molecules that are not
interacting. For interacting particles we have to use the canonical ensemble. We
can do a similar analysis on the canonical ensemble [collection] to obtain
[A5p684]
˜ U
˜
U
o
S˜ 
 k lnQ
(20)
T
Combining Equations (17) and (20) thus
Which is a result we
have been looking
for.
 N 
˜ ˜
˜S  U  U o  k lnq 
 N! 
T
 
(22)
˜ , S˜ and q
E. Thermodynamic relationships to relate G
We now are going to use the various thermodynamic relationships to relate the
molecular partition function to change in free energy G. Then we can finally
relate the molecular partition function to the equilibrium constant K. From
 
thermodynamic relationships we know that the Gibbs Free Energy, G˜ , can be
written as
26
CD/TransitionState/ProfRef.doc
˜ U
˜  TS˜  PV
G
(23)
For an ideal gas with n total moles
˜ U
˜  TS˜  nRT
G
(24)
˜ are energy (e.g. kcal or kJ) and not
Again we note the dimensions of G
˜
energy/mol (e.g. kcal/mol). Combining Equations (20) and (24) for S˜ and G
we obtain
˜ U
˜  kTlnQ  nRT
(25)
G
o
Recalling the relationship of Q to the molecular partition function
Q  q N N!
(18)
˜ U
˜ o  NkTln q  kTln N!  nRT
G
(26)
We use Avogadro’s number to relate the number of molecules N and moles n,

i.e., N = nNavo, along with the Stirling approximation to obtain

˜ U
˜ o  N avo n kT ln q  kTN ln N  N nRT
G
(27)
Now : kTN  kTN avo n  RTn
˜ o  nRTln q  nRTln N  nRT nRT
U


˜ U
˜ o  nRTln q
G
N
(28)
N  nN , where N  6.032 10 23 molecules mole
avo
avo
F. Relate G, the molar partition function qm
We divide by the number of moles, n to get

qm 
q
n
(29)
q
q
q

 m
N nN avo Navo
Substituting for (q/N) in Eqn. (28)
˜ U
˜  nRTln q m
G
o
Navo
(30)
To put our thermodynamic variables on a per mole basis (i.e., the Gibbs free
energy and the internal energy) we divide by n, the number of moles.
˜
˜
G
U
 G and o  Uo
n
n
This is a result we have
been looking for
CD/TransitionState/ProfRef.doc
27
G  Uo  RTln
qm
Navo
(31)
where G and Uo are on a per mole basis and are in units such as (kJ/mol) or
(kcal/mol).
G. Relating the dimensionless equilibrium constant K and the molar partition
function qm
Applying Eqn. (31) to species i
 q 
G i  U i0  RT ln im 
N Avo 
For the reaction


aA  bB 
 cC  dD
the change in 
Gibbs free energy is
G  cGC  dG D  bG B  aG A
(32)
Combining Eqns. (31) and (32)


G  cUC0  dU D0  bU B0  aU A0 
E 0

 q 
 q 
 q

 q 
cRT ln cm  dRT ln dm  bRT ln bm  aRT ln am 
N Avo 
N Avo 
N Avo 
N Avo 


q c q d


Dm
G  E 0  RT ln Cm
N
avo 
a
b
q Am q Bm

(33)
where again
  c  d b  a

From thermodynamic
and Appendix C we know
G  RT lnK
Dividing by RT and taking the antilog
K e

E o
RT
qcCmqdDm
a
b
qAmq Bm
N
Avo
(34)
H. Relate the molecular partition function on a basis of per unit volume, q and
the equilibrium constant K
The molecular partition function q is just the product of the electronic, q E,
translational, qT, vibrational, qV, and rotational, qR partition functions
28
CD/TransitionState/ProfRef.doc
q  e i  e
 E i  T i  Vi  R i 
 e
E i
e
T i
e
 Vi
e
R i
q  q Eq T q Vq R

(19)
Equations for each of these partition functions q E , qT , ...will be given later.
We now want to put the molecular partition function on a per unit volume
basis. We will do 
this by putting the translational partition function on a per
unit volume basis. This result comes naturally when we write the equation for
qT
qT  q
(35)
TV
therefore
q  q T Vq E q V q R  q  V
(36)
and
qm 

q
V
 q   q  Vm
n
n
(37)
By putting q T on a per unit volume basis we put the product q  q T q E q R q V
on a per unit volume basis. The prime again denotes the fact that the
transitional partition
function, and hence the overall molecular partition

function, is on per unit volume.

The molar volume is
RT
Vm 
, f  1 atm
f
where f° is the fugacity of the standard state of a gas and is equal to 1 atm.
K e

E o
RT
c
d
q C qD   RT 
a
b
q A  q B  1 atm
(38)
(See Appendix TS2 page 29 of Transition State Theory Notes for derivation)
I. Recall the relationship between K and KC from Appendix C.
The equilibrium constant and free energy are related by
G  RT lnK
c
K
d
 fC   fD 
   
fC0  fD0 
acC adD

a aAa bB  f a  f b
 A   B 
fA0  fB0 
(39)
The standard state is fi0 = 1 atm. The fugacity is given by fA = APA [See
Appendix C of text]
29
CD/TransitionState/ProfRef.doc
K
acC adD
aaA a bB

d
cC dD PCcPD

1 fi0
a b
a b
A B PAPB
 
Ka
K

,   d c a  b
(40)
Kp
Pi  CiRT

c d 
RT 
 CC CD RT 
  K 1 atm
K  KKC 
a b   
1 f 
CACB  fi0 
(41)
For an ideal gas K   1
RT 
Equating Eqns. (40) and (41) and canceling   on both sides
 fi 
KC  e

E o
RT
qC c q D d
a
b
qA  qB 


Navo
(42)
Now back to our transition state reaction


#
A  BC 
 AB  C
 ABC 
111 1
#
rA  vICABC#  vIKc CACB
K#C

C
ABC #
 e

CA CBC 
E o
RT
qABC
#
N avo
q
A q
BC
(c)
(43)
#
where q ABC
is the molecular partition function per unit volume for the
activated complex.
#
# # # #
(44)
qABC  q Eq VqR q
T
 Rearranging Eqn. (43), we solve for the concentration of the activated complex
C ABC#
C ABC#  v I C A C B e


E o
RT
#
q ABC
N avo
q A q BC
(45)
J. The Loose Vibration, vI
We consider
 the dissociation of A–B–C# as a loose vibration with frequency vI
in that the transition state molecule A–B–C# dissociates when it crosses the
barrier. Therefore the rate of dissociation is just the vibrational frequency at
which it dissociates times the concentration of ABC# (Lp96)
30
CD/TransitionState/ProfRef.doc
rABC  vI C ABC #
(46)
Substituting Eqn. (45) into Eqn. (46)
E
rABC
  v I C A C B e o
RT
#
q ABC
N Avo
q A q BC
Where q is the partition function per unit volume. Where vI is the “imaginary”
dissociation frequency of crossing the barrier
 The vibrational partition, q #v , function is the product of the partition
function for all vibrations
#
q v  q vIq v1q v2
(47)
Factoring out qvI for the 
frequency of crossing the barrier
qv#
#
qv
 q vI q v1q v2
#
q v  q vIq v#
(48)
q ABC#  q #E q vI q v # q #R q T#
(49)
Note we have moved the # from a superscript to subscript to denote that q v# is
the vibrational partition function less the imaginary mode vI. q#v is the

vibrational partition
function for all modes of vibration, including the
imaginary dissociation frequency (Lp96).
q vI 
1

1 e
hv I
k BT

k T
 B
 hv  hv I
1 1 I 
 kT 
1
#
qABC
 qABC#

kT
h vI
(50)
(51)
Substituting for q#ABC in Eqn. (45) and canceling vI
This is the result
we have been
looking for!
rABC
E


k B T   RTo q ABC # N avo 

 rA  
C A C BC
e



h

q
q


A
BC


(52)
where q ABC# is the partition function per unit volume with the partition
function for the vibration frequency for crossing removed.

31
CD/TransitionState/ProfRef.doc

Nomenclature Notes:
q = molecular partition function  q E qT q V q R
q = molecular partition function per unit volume
q = q V
q
q m   q Vm molar partition function
n
#
q = Partition function (per unit volume) of activated complex that includes
partition function of the vibration frequency vI, the frequency of
crossing
q# = Partition function (per unit volume) of the activated complex but does not
include the partition function of the loose vibration for crossing the
barrier
What are the Equations for qT, qV, qE, and qR
K. Evaluating the Partition Functions
Schrödinger Wave Equation
We will use the Schrödinger wave equation to obtain the molecular partition
functions. The energy of the molecule can be obtained from solutions to the
Schrödinger wave equation (A5p370)
 h 2 2

 2   Vx,y,z  Ex,y,z
2 m

(51)
This equation describes the wave function, , for a particle (molecule) of mass
m and energy E traveling in a potential energy surface V(x,y,z). “h” is Planck’s
constant.The one dimensional form is
d 2  2m
 2  2 E Vx  0
dx
h
h  h 2  1.05 10
34
(52)
Js
The probability of finding a particle in a region between x and x+dx is

2
Probability =  dx
2
 is the probability density (A5p373)
Molecular Partition Function
We shall use this equation to obtain the translational, vibrational, and
rotational energies, (t, V, and R) used in the partition function q. The
equation is solved for three special cases
1. Translational energy, t. Particle in a Box.
32
CD/TransitionState/ProfRef.doc
2. Vibrational energy, V. Harmonic Oscillator.
3. Rotational Energy, R. Rigid Rotator.
4. Electronic Energy, E.
Recall q  q Eq VqR qT
The electronic partition functions qE, is most always close to one.
Table PRS.3B-2Overview of Q
Parameter Values
1 atomic mass unit  1 amu = 1.67x10–27 kg, h = 6.626x10–34 kg•m2/s,
kB = 1.38x10–23 kg•m/s2/K/molecule
1. Transitional Partition Function qT Derive
2mk B T

32
q T


1
 the order of 10 30 m 3
3
3
h

h
  Thermal wavelength 
12
2mk B T

(53)
(Derive)
32
2k B 3 2
300
3 2 1 amu 
32
q T   2  m  
 T 
1 amu 
 h 
300
2k B  3001 amu 3 2  m AB 3 2  T 3 2
 

  
 

 1 amu  300 
h2
Substituting for kB, 1 amu and Plank’s constant h
9.84 10 29  m
3 2  T 3 2
AB
qT  

  
2
 m
1 amu  300 

(54)
  7.12 1011 m
for H 2 at 25 C
2. Vibrational 
Partition Function qv Derive

qv 
1
1 e
hv k BT
Expanding in a Taylor series

33
CD/TransitionState/ProfRef.doc
,
(55)

 1
q v  
hv
  k T
1 e B

 k B T
 hv  the order of 1 to 10


 
˜
˜ 300 
h
hc

 4.8 103 
 
1 cm –1  T 
k BT k BT

(Derive)
(56)
3. Electronic Partition Function
q E  q E (q E 1)

(Derive)
4. Rotational Partition Function qR Derive

For linear molecules
8 2 Ik B T k B T
qR 

Sy hcB
Sy h 2
B  Rotational constant 


(57)
h
8 2cI
I   m i ri2
For non-linear molecules
1 kB T    
qR
 
 the order of 10 to 1000
 
Sy  hc  ABC 
32
(Derive)
where ABC are the rotational constants (Eqn. 57) for a nonlinear molecule about
the three axes at right angles to one another

I   i ri2
For a linear molecule
 1 
 T  I AB

q R 12.4 
 

300 1 amu  Å 2 
Sy 
(58)
Sy = symmetry number of different but equivalent arrangements that can be
made by rotating the molecule. (See Laidler p. 99)
 and hydrogen molecule Sy = 2
For the water
1
2
2
1
H
H
H
H
1
2
2
H , H H1
, H
O
O
For HCl Sy = 1
34
CD/TransitionState/ProfRef.doc
Estimate A from Transition State Theory
Let’s do an order of magnitude calculation to find the frequency factor A.
rA 
q  #
k BT
N Avo ABC e E RT C A C B
h
q A q BC


A m 3 mol•s
rA  Ae E RT C A C B
Let’s first calculate the quantity
k B T 
13

10
 h 

at 300K

k B T 1.38 1023 J K  molecule  300K

 6.25 1012 s1 molecule
34
h
6.62
 10 J s
13
k BT
10 s 1 molecule
h
rA  kC A C BC
E


  RTo q ABC # 
k
T
B

k  
e
N

 avo 




h

q
q



A BC 


  1  



  dm 3   molecules
1

Units k   
 
s  molecule   1  1 
mole

 3  3 
dm dm 
k  
N Avo  6.02 10 23
dm 3
mol  s
molecules
mole
At 300K


k BT
1013 s 1 molecule 1
h
q T 10 32 m 3


qv 6
q R 100


CD/TransitionState/ProfRef.doc
35
(51)
A
q  #
k BT
N Avo ABC
h
q A q BC
rA  Ae E
RT
C AC B

A
BC
ABC
q T 6 x 1032 m–3
6 x 1032
6 x 1032
qv
-5
5
qR
-20
200
q  6 x 1032 m–3 600 x 1032 m–3 6000 x 1032 m–3





q ABC #
q A q BC


600010 32
1.71032 m 3
32
32
6 10  600 10
q  # 
k T 
A   ABC  N Avo   B 
 h 
q A q BC 


molecule  13
1
A  1.7 1032 m 3 6 10 23
10


mole 
molecule s 

A 10 104 m 3 moles 108 dm 3 mole s


whichcompares with A predicted by collision theory
the equation for q , q , q , and q
Derivation of
T V E
R
CLICK BACKS
1. Translational Partition Function qT
(Click Back 1)
HOT BUTTON
To show
2k B Tm 3 2 3
qT  
 m
 h 2 
 
(T1)
Translational Energy
We solve the Schrödinger equation for the energy of a molecule trapped in an
infinite potential well. This situation is called “a particle in a box”. For a particle in a
box of length a 
36
CD/TransitionState/ProfRef.doc


V(X)
X=0
•
X=a
The potential energy is zero everywhere except at the walls where it is infinite so
that the particle cannot escape the box. Inside the box
h d 2
(T2)

 E
2m dx 2
The box is a square well potential where the potential is zero between x = 0 and x = a
but infinite at x = 0 and x = a (A5p392). The solution is to the above equation is
  A sinkx B coskx
k 2h 2
2m

2mE 

from k 

h 2 

We now use the boundaryconditions
At x = 0 :  = 0, sin 0 = 0, and cos 0=1
 B=0

the wave equation is now
 = A sin kx

At x = a  = 0
 will be zero provided
 ka = n
where n is an integer, 1, 2, 3
0
a
n
k
a
Substituting (T5) into (T4) we see that only certain energy states are allowed
n2 h 2
En 
8ma2
For particle in a a 3-D box of sides a, b, and c
h 2 n 21 n 22 n 23 
 2  2  2 
En 
8m 
a

b
c 
Back to one dimension
n2 h 2
En 
8m a 2
Therefore relative to the lowest energy level n = 1, the energy is
E
where

Then

h2
 n  E n  E 1   n 1 , where  
8ma 2

CD/TransitionState/ProfRef.doc
2
37
(T3)
(T4)
(T5)
(T6)
(T7)

q Tx 
e


 n 2 1 
(T8)
n1
2kTm 
q Tx   2 a
 h

Derive
(T9)
(Click Back within a Click Back) – Hot Button within a Hot Button
We will assume that the energy states are sufficiently close together, such that

there is a continuous distribution of energies. Replace  by 


  n 2 1 
q Tx   e
1
  n 2
dn
0
dn   e
12
Let: x 2  n 2, dn  dx 
qTx
 1 1 2  x2 dx  1 1 2 1 2
 
 
 
  0 e
 
 

 
2
2m 1 2

1 2
 a  2kTm a
 
 2 
 h 2 
h  
(T10)
For 3-D
The translation partition function for y and z directions are similar to that
for qTx
This is the
result we have
been looking
for!
2kTm 1 2 3
 
  abc
2


 
 h
 V
q Txyz  qTx qTy q Tz
 q 
TV
(T11)
2mkT 3 2
qT  

 h 2 
We define  as thermal wave length
12
 h2 
 (m)
  
2mkT 
qT 
1
3
Order of Magnitude and Representative Values
For O 2 @25C q T  2 10 28 m 3

at 25C for H 2 :   71 pm, for O 2 :  18 pm (pm  pico meter)
38
CD/TransitionState/ProfRef.doc

(T12)
2. Vibration Partition Function qv (Click Back 2)
To show
qv 
1
1 e hv
Derive
(A12)
(Click Back)
Again we solve the wave equation for two molecules undergoing oscillation about
an equilibrium position 
x = 0. The potential energy is shown below as a function of
the displacement from the equilibrium position x = 0 (A5p402)
The uncertainty principle says that we cannot know exactly where the particle is
located. Therefore zero frequency of vibration in the ground state, i.e.  = 0 is not an
option (A5p402 and pA22). When vo is the frequency of vibration, the ground state
energy is
1
(V1)
E o  hv
2
Harmonic oscillator (A5p402)
Spring Force F  kx , potential energy from equilibrium position x = 0
dx 2
m 2  kx
dt

the solution is of the form for t=0 then x=0

where
x  A sin t
1
k
m

Spring contant 
12
  
 k m  ,   2v


mass

Classical vibration
The potential energy is
1
V  kx 2
2
 to show
We now want
39
CD/TransitionState/ProfRef.doc

(V2)
 1 
E v    hv
 2 
(V3)
We now solve the wave equation:
h2 d2
1
 
 kx 2   E
2
2m dx
2
(V4)
to find the allowable energies, .
h 2 1 4
k
x
E
Let y  ,  
, where     ,   , and w  2

1
m

mk 
h
2
With these changes of variables Eqn. (A15) becomes




2

d 
dy
2

  y   0
2
(V5)
The solutions to this equation (A5 pA22, i.e., Appendix 8)will go to infinity unless
 = 2+1
 = 0, 1, 2, 3 . . .
 1 
E v     hv,
 2 
˜
Wave number  

v
 c

2 
A5p541
[c = speed of light]

s 1
1 
˜



 cm s cm 
v 1

c 
 = wave length
 1 
 1 
˜
E     hv    hc 
 2 
 2 

(V6)
Measuring energy relative to the zero point vibration frequency, i.e.,  = 0
 1  1
E   hv   hv
 2  2

E 
hv
kT
Substituting for E  in the partition function summation e

qv 


e
hv

0
e
0
40
CD/TransitionState/ProfRef.doc

hv



E 

1
 Xi  1 X
i 0
This is the result we
have been looking for!
˜
For hc 
1 then
qv 
1
1e
hv
(V7)
1  e  1  1  X X , we can make the approximation
X
qv 
kT kT

˜
hv hc 
(V8)
For m multiple frequencies of vibration
q v  qv1qv2 ... q vm

Order of Magnitude and Representative Values
For H2O we have three vibrational frequencies with corresponding wave numbers,
v˜ i , v˜ 2 and v˜ 3 .
˜ 1  3656 cm 1  q v1 1.03 ,


˜ 2 1594.8 cm 1  q v 2 1.27

and

˜ 3  3755.8 cm 1  q v3  1.028


q v  1.031.271.028  1.353
3. Electronic Partition Function (A5p701) (Click Back 3)
From the ground state electronic energy separation are very large.
qE = gE
where gE is the degeneracy of the ground state.
For most cases
qE = 1
4. Rotational Partition Function (Click Back 4)
qR 
k BT
hcB
Rigid Rotation (A5p409, 413, 557 and A24)
To show

41
CD/TransitionState/ProfRef.doc
(E1)
qR 
where
k BT
hcB
(R1)
h
B  2 = Rotational Constant
 8 cI
(R2)
Consider a particle of mass m rotating about the z axis a distance r from the origin.


2r
m
(R3)
Jz=pr
X
r
P=mr
m
4
2
0
This time we convert the wave equation to spherical coordinate to obtain (A5p410)
d 2
2IE
 2
2
d
h
(R4)
Classical Energy of a rigid rotator is
1 2
 I
2

where w is the angular velocity (rod/s) and I is the
Moment of inertia(A5p555)
2
I   mi ri
E
(R5)
(R6)
where mi is the mass located and distance ri from the center of mass.
Quantum mechanics solutions to the wave equation gives two quantum numbers,
and m.
Magnitude of angular momentum = 
 1 h
12
z component of angular momentum = mh
E

h2
 1
2I
42
CD/TransitionState/ProfRef.doc
(A5p408,p413)
(R7)
 clockwise rotation
with  counter clockwise rotation
m  0, 
Let J  
For a linear rigid rotator
JJ 1h 2
E=
2I
 hcB J (J + 1)
(R8)
Where B is the rotation constant:
B

h
h
 2
4Ic 8 cI
(A5p557)
(R2)
with
c = speed of light
I = moment of inertia about the center of mass   m i ri2
The rotational partition function is

q R   2J  1 e

Replacing the

 hcBJ J 1
(A5p414,563,671)

J 0
(R9)
by an integral from 0 to  and integrating, we obtain the
J 0
rotational partition function qR for a linear molecule (A5p694)
This is the result we
have been looking for!
qR 
k B T 8 2 Ik B T

Sy hcB
Sy h 2
(R10)
where Sy is the symmetry number which is the number of different but equivalent
arrangements that can
 be made by rotating the molecules.
For HCl
Sy 1
For H 2 O
Sy  2
1
2
H
2
H
O

1
H
H
O
For a non-linear molecule
 1 kT 3 2   1 2
q R  

S 
  
 y hc  ABC 
A
where

h
2
8 cI A
CD/TransitionState/ProfRef.doc
, B
h
2
8 cI B
43
, C
h
2
8 cI C
Sy = symmetry number. [For discussion of  see Laidler 3rd Ed. p.99.] For a
hetronuclear molecule  = 1 and for a homonuclear diatomic molecule or a
symmetrical linear molecule, e.g., H2, then  = 2.
Order of Magnitude and Representative Values
For HCl @25C (A6p695), B 10.591 cm1 then q R 19.9 and at 0C q R 18.26.
For Ethylene at 25 C then q R  661

III. THE EYRING EQUATION
For the reaction
ABC ABC
The rate law is
rA  kC A C BC




k  Ae E A
RT
(59)
(60)
ln k  A ln E A RT
(61)
d ln k E A

dT
RT 2
(62)
Now let’s compare this with transition state theory.
The rate of reaction is the rate at which the activated complex crosses the
barrier

A BC  ABC #  AB  C
rA  vI C ABC #

K C# 

C ABC #
C A C BC
rA  v I K C# C A C BC
(1)
(2)
(3)
Factoring out partition
 function for the loose vibration frequency, v I, from the
vibrational partition function, q #v

q #v 
k BT
q v#
hv I
(63)
 can obtain
then from Eqn. (43) we

K C#  K C #
44
CD/TransitionState/ProfRef.doc

k BT
hv I
(64)
k T 
rA   B K C # C A C B
 h 
(65)
which is referred to as the Eyring Equation.
From thermodynamics
G  H TS

K #  e G
#
S #
RT
e R
(66)

e
H #Rx
RT
(67)

The overall dimensionless terms of mole fraction x i and the activity coefficients i

K
 ABC x ABC #
 A x A  BC x BC
K  K CT

 K
C ABC #
C A C BC
x ABC #
x A x BC

C 2T
CT 2
(68)
 K CT K C#
(69)
K
e G RT e S R e H RT



K CT
K CT
K  C 2T
#
KC#

#
(70)
S# will be negative because we are going form 0 less ordered system of A, BC
moving independently as reactants to a more ordered system of A, B and C being
 in the transition state. The entropy can be thought of as the number of
connected
configurations/orientations available for reactions. That is
#
Number of Configurations Leading to Reaction
 e S
Total Number of Configurations

H #  H ABC #  H A  H BC
R

will
 be positive because the energy of the transition state is greater than that of the
reactant state.
Case I Liquid 
For liquid CT = a constant = CT0 (Recall for water that Cw = 55.5 mol/dm3
k T  e S R e H RT
k   B 
 h 
K CT 0
#

S#  SABC #  SA  SBC
Here we see the
temperature dependence as
Case II Gases
kT ~ T e H

45
CD/TransitionState/ProfRef.doc


RT

(71)
(72)
From gases
CT 
P
RT
k B R  T 2 e S R e H
k  

 hP 
K

#
#
RT
Here we see the temperature dependence as
kT ~ T2 e H

#
RT
As with liquids S# is negative and H #Rx is positive.
Relating EA and H
Rx
Now lets compare 
the temperature dependent terms. The heat of reaction
will be positive because the activated state is at a higher energy level than the
reactants. See figure PRS.3B-2.
#
#
k R 
k   B  T 2 e S R e H Rx
 h P 

RT
(73)
#
k BR 
H #Rx
2 S
ln k  ln

 ln T 
 hP 
R
RT
(74)
dln k 2 H #Rx E A
 

dT
T RT 2 RT 2
(75)

Comparing Eqns (1) and (16), the activation energy the Erying Equation is
E A  2RT  H #Rx

with the frequency factor

k R  #
A   B  e S
 h P 

46
CD/TransitionState/ProfRef.doc
R
(76)
(77)
Some Generalizations May Be Eliminated
For reactants where there are N atoms (Lp108)
Atoms
3
q  qT
Linear polyatomic molecules
3
2
3N5
3
3
3N6
q  qTq Rq v
Nonlinear polyatomic molecules
q  qTq Rq v
Let qT = 10–8 cm–1 for each degree of transitional freedom, qR = 10, qv = 1 and
kT 
= 1013 s–1 with NA = 6.02  1023 molecules/mol.
 h 

Case 1 Atom + Atom 

 Diatomic Activated Complex
A calc
kT q 3T q2R  E 
k  Navo   3 3 e 0
 h q T qT
RT
Assuming all translational partition functions are approximately the same
kR 
2
kT 
qR 


NA
 h q 3T 

e E 0
RT
Acalc 6 1014 cm 3 mol s


Case 2 Atom + Linear Molecule 
 Linear Complex

kT  qTq Rq v
E 0
k  NA
e
3
3
2
3N5
 h q  q q q
3
T
k
NA
3 N1  6
2
RT
T R v
2

kT 
q R

3
 h qT 

e E 0
RT
A calc  61012 m3 mols
Case 3 Atom + Nonlinear Molecule
Same results as for a linear molecule


Case 4 Linear Molecule + Linear Molecule 
 Linear Complex
47
CD/TransitionState/ProfRef.doc
k
NA
kT  q 4v
 h q 2Rq 3T
e E 0
RT
A calc  61010 cm3 mols

Case 5 Linear Molecule + Nonlinear Molecule 

 Nonlinear Complex
k  NA
kT  q5v
 h q3R q3T
A calc  6109 cm 3 mols
48
CD/TransitionState/ProfRef.doc

e E 0
RT
Appendix
For the general reaction
M
G   vi G i   U iovi  RT  vi ln
i1
 a i lnbi   lnbai
i 1
i
qim
NA
 ln b1a1  lnb a2 2  lnb a2 3 . . .
N
 lnb a1 1 b a2 2 b aNN  ln  b ai i
i 1
qim v i

G  E o  RT ln  

N
A 
i 1
M
 q v i
 RTlnK  Eo  RT ln  mi 
i1  Nav g 
m
v

q mi  i 
qmi
q mi
q mi
qm2


 v inRT ln
 nRT  vi ln
 v1 ln
 v 2 ln
 . . .  nRT 

NA
NA
NA
NA
N A 


q vi
mi 
˜
˜
 vi Go   Uo vi  nRT ln


NA 
˜ nG
˜ n  Uo
 by n
G
U
 q
vi
mi

G  RT lnK   Uio vi  RT 


NAvo 


A  BC
ABC



C
KC  ABC  e
CA C BC
E o
RT

q ABC 
NAvo
qA qBC
  1  1  1  1

mol
 3

dm
dm3 qABC NAvo 
K

 C  mol  q  q  
 A BC 
E
q ABC
 NAvo  RTo
C
KC 
e
 ABC
q Aq BC
C AC BC
C ABC  C AC BC e

Eo
RT
q#ABC  q ABC#
49
CD/TransitionState/ProfRef.doc
mol 2 

 
3
dm  


q ABCNAvo
q Aq BC
kT
h
kT 
rA  vCABC
e
h
E o
RT
q
ABC NAvo
OK
qA
 qBC

Molar partition function
q
 q mol
n
q
V  q dm 3 
q m   q    m3 
 q m 1 mol
h
n dm mol 
 cd ba
qm 
fCc fDd

a Cc a dD fC fC
K a b  a b  f
a A a B fA fB
fA fB


K K  K P f ,
fCc fDd  Cc  dD PCc PDd

 a b
f
a b
a b  
fA fB  A  B PA PB
PA  C A RT
cdab
RT cdab
PCc PDd
CCc CdD RT 

K K  K P  K  a b f  K  a b  
 K  K C  
 f 
PA PB
C A C B  f 
K  K  KC

c d a b
RT 
 f 
Eo   viU io
RT lnK  E o  RT ln
lnK  
 q vi
E
mi 
 ln


RT
NAvo 
K e

E o
RT
v
q mii
NAvo
q m  mol 1
 q v i
mi 



NAvo 
E
RT 
 o qc qd
K KC    e RT C D N
av g ,
 f 
qaA q bB
V
 qVm 
n
E
c dad
RT 
 o q c q d V 

K  K  KC  e RT Ca Db  
NAvo
 f 
n 
q
A q 
B
q m  q
m
However,
qV  q
50
CD/TransitionState/ProfRef.doc
V RT

n f
KC  e

Eo
RT
c d
q 
C q
D
a b
q 
Aq 
B
51
CD/TransitionState/ProfRef.doc
N
Avo