1. Reaction Rate Theory D E
reaction coordinate + k A B AB

2. The Arrhenius Equationk +
A B AB
d [AB]
r = = k [A] [B]
k = v e
d t
- Eact / RT E
Svante
Arrhenius
Nobel Prize 1903
Eact +
Empirical !
reaction parameter

3. Transition State Theory
To determine the rate we must know the concentration on top
of the barrier.
The relative concentration between a reactant and product in a
Chemical reaction is given by the Chemical Equilibrium
The Chemical Equilibrium is given by the chemical potential
of the reactant and the product. That we know how to calculate.

4. The Chemical Equilibrium
The chemical potential for the reactant and the product can be
determined if we know their Partition Functions Q.
Here Qi is the partition function for the gas i and qi the
partition function for the gas molecule i
Let us assume that we know qi then

5. The Chemical Equilibrium
If we assume an ideal gas
and normalize the pressure with p0=1 bar
We obtain the important result that the Equilibrium Constant K(T)
is given by the Partitions Functions of the reactants and products
Thus we can determine the concentration of a product on top of the
barrier if we know the relevant Partion Functions

6. Partition Functions
Obviously are Partition Functions relevant. We shall here deal
with the Canonical Partition Function in which N, V, and T are
fixed.
Remember, that although we talk of a partition function for an
individual molecule we always should keep in mind that this only
applicable for a large ensample of molecules, i.e. statistics
Consider a system with i energy levels with energy ei and
degeneration gi
Where Pi is the probability for finding the system in state i

7. å e = P Ludwig Boltzmann (1844-1906) Boltzmann Statistics:RT = - å /
Boltzmann Statistics:
The high temperature/diluted limit of
Real statistical thermodynamics
There is some really interesting Physics here!!
S = k ln (W)

8. Partition Functions Why does the partition function look like this?
Lets see if we can rationalize the expression:
Let us consider a system of N particles, which can be distributed
on i states with each the energy ei and Ni particles. It is assumed
the system is very dilute. I.e. many more available states than particles.
Constraint 1
Constraint 2
Requirement: The Entropy should be maximized (Ludwig Boltzmann)

9. Partition Functions Where we have utilized Stirling approximation:
Only valid for huge N
Problem: Optimize the entropy and fulfill the two constraints at the
same time. USE LAGRANGE UNDERTERMINED MULTIPLIERS

10. Partition Functions Result: The Entropy Maximized when
If we now utilize the first constraints:
Which reminds us of q the partition function

11. Partition Functions The second constraint:
We have to relate the average energy to some thermodynamical data
This can be found by considering
a particle confined in a box
Now if we wants to perform the sum above we need to have an
analytical expression for the energy in state i

12. Partition Functions By inserting this in the
result of constraint 1 and
assuming close lying states

13. Partition Functions Utilizing this in constraint 2 Thus
I.e. temperature is just a Lagrange multiplyer

14. Partition Functions Since constraint 1 gave Since constraint 2 gave
and the entropy is max for
Thus the form of the partition
function comes as a result of
maximizing the entropy with
2 constraints

15. Translational Partition Functions
As we have assumed the system to be a particle capable of moving
in one dimension we have determined the one-dimensional
partition function for translational motion in a box of length l
Now what happens when we have several degrees of freedom?
If the different degrees of freedom are independent the Hamiltonian
can be written as a sum of Hamiltonians for each degree of freedom
Htot=H1+H2+….
Discuss the validity of this: When does this not work? Give examples

16. Translational Partition Functions
If the hamiltonian can be written as a sum the different coordinates
are indrependant and
Thus for translational motion in 3. Dimensions.
qtrans3D = qtransx qtransy qtransz=

17. Partition Functions
It is now possible to understand we the Maxwell-Boltzman
distribution comes from

18. Average: Maxwell-Boltzmann distribution of velocities
500 – 1500 m/s at 300 K

19. Partition Functions
Similarly can we separate the internal motions of a molecule in
Part involving vibrations, rotation and nuclei motion, and
electronic motion i.e. for a molecule we have
Now we create a system of many molecules N that are in principle
independent and as they are indistinguishable we get an overall
partition function Q
What if they were distinguishable ???

20. Partition Functions
What was the advantage of having the Partition Function?

21. Partition Functions
Similarly can we separate the internal motions of a molecule in
Part involving vibrations, rotation and nuclei motion, and
electronic motion i.e. for a mulecule we have
Now we create a system of many molecules N that are in principle
independent and as they are indistinguishable we get an overall
partition function Q

22. The Vibrational Partition Function
Consider a harmonic potential
If there are several normal modes:

23. The Rotational (Nuclear) Partition Function
Notice: Is not valid for H2 WHY?
TRH2=85K, TRCO=3K

24. The Rotational (Nuclear) Partition Function
Molecular symmetry 
Types of molecules
C1, Ci, and Cs 1
CO, CHFClBr, meso-tartraric acid, and CH3OH
C2, C2v, and C2h 2
H2, H2O2, H2O, and trans-dichloroethylene
C3v and C3h 3
NH3, and planar B(OH)3
The Symmetri factor:
This has strong impact
on the rotational energy
levels. Results in fx
Ortho- and para-hydrogen
For a non-linear molecule:

25. Effect of bosons and fermions
If two fermions (half intergral spin) are interchanges the total wave
function must be anti symmetric i.e. change sign.
Consider Hydrogen each nuclei spin is I=1/2
From two spin particles we can form 2 nuclear wave function:
and which are (I+1)(2I+1)=3 and I(I+1)=1
degenerate respectively
Since the rotation wave function has the symmetry
is it easily seen that if the nuclear function is even must j be odd and
visa versa

26. Ortho and Para Hydrogen
This means that our hydrogen comes in two forms: Ortho Hydrogen
Which has odd J and Para Hydrogen which has even J incl. 0
Notice there is 3 times as much Ortho than Para, but Para has the
lowest energy a low temperature.
If liquid Hydrogen should ever be a fuel we shall see advertisements
Hydroprod Inc.
Hydroprod Inc.
Buy your liq. H2 here!!!
Absolute Ortho free Hydrogen for longer mileages

27. Liquid Hydrogen
This has severe consequences for manufacturing Liq H2 !!
The ortho-para exchange is slow but will eventually happen so if we
have made liq. hydrogen without this exchange being in equilibrium
we have build a heating source into our liq. H2 as ¾ of the H2 will
End in J=1 instead of 0.
i.e. 11% loss due to the internal conversion of Ortho into Para hydrogen

28. The Electronic Partition Function
Does usually not contribute
exceptions are NO and fx. H
atoms which will be twice
degenerate due to spin
What about He, Ne, Ar etc??

29. Partition Functions Summary

30. Partition Functions Example
Knowing the degrees of internal coordinates and their energy
distribution calculate the amount of molecules dissociated into
atoms a different temperatures.
T(K)
KH2(T)
pH/p0
KN2(T)
pN/p0
KO2(T)
pO/p0
298
5.81*10-72
2.41*10-36
6.35*10-160
2.52*10-80
6.13*10-81
7.83*10-41
1000
5.24*10-18
2.29 *10-9
2.55*10-43
5.05*10-22
4.12*10-19
6.42*10-10
2000
3.13*10-6
1.76*10-3
2.23*10-18
1.80*10-9
1.22*10-5
3.49*10-3
3000
1.77*10-3
1.72*10-1
1.01*10-9
3.18*10-5
5.04*10-1
5.01*10-1
We see why we cannot make ammonia in the gas phase but O radicals
may make NO at elevated temperatures