Reaction Rate Theory

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Reaction Rate Theory
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The Arrhenius Equation
k

A B AB
d AB
r k A B
k v e
d t
- Eact / RT
E
Svante Arrhenius 1859 - 1927 Nobel Prize 1903
Eact

Empirical !
reaction parameter
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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.
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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
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The Chemical Equilibrium
Thus we can determine the concentration of a
product on top of the barrier if we know the
relevant Partion Functions
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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
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Ludwig Boltzmann (1844-1906)
Boltzmann Statistics The high
temperature/diluted limit of Real statistical
thermodynamics There is some really interesting
Physics here!!
S k ln (W)
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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)
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Partition Functions
Where we have utilized
Problem Optimize the entropy and fulfill the two
constraints at the same time. USE LAGRANGE
UNDERTERMINED MULTIPLIERS
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Partition Functions
Result The Entropy Maximized when
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Partition Functions
Now if we wants to perform the sum above we need
to have an analytical expression for the energy
in state i
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Partition Functions


By inserting this in the result of constraint 1
and assuming close lying states
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Partition Functions
Utilizing this in constraint 2
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Partition Functions
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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 HtotH1H2.
Discuss the validity of this When does this not
work? Give examples
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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.
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Partition Functions
It is now possible to understand we the
Maxwell-Boltzman distribution comes from
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Maxwell-Boltzmann distribution of velocities
Average
500 1500 m/s at 300 K
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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 ???
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Partition Functions
What was the advantage of having the Partition
Function?
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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
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The Vibrational Partition Function
Consider a harmonic potential
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The Rotational (Nuclear) Partition Function
Notice Is not valid for H2 WHY?
TRH285K, TRCO3K
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The Rotational (Nuclear) Partition Function
The Symmetri factor This has strong impact on
the rotational energy levels. Results in
fx Ortho- and para-hydrogen
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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 I1/2
From two spin particles we can form 2 nuclear
wave function and
which are (I1)(2I1)3 and
I(I1)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
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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.
Absolute Ortho free Hydrogen for longer mileages
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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 J1
instead of 0.
i.e. 11 loss due to the internal conversion of
Ortho into Para hydrogen
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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??
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Partition Functions Summary
s
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Partition Functions Example
Knowing the degrees of internal coordinates and
their energy distribution calculate the amount of
molecules dissociated into atoms a different
temperatures.
 
We see why we cannot make ammonia in the gas
phase but O radicals may make NO at elevated
temperatures
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Surface Collisions
Consider a box with volume V
What are the numbers?
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Surface Collisions
How many are successful in reacting? Simple
Maxwell-Boltzman distribution
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Transition State Theory
Consider the following reaction
q
How?
We assume that R and R are in Equilibrium
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Transition State Theory
By splitting the partition function in the
transition state
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Transition State Theory
The partition function q can conveniently be
split further
Which basically is the Arrhenius form If q0 q
n 1x1013s-1
Relation to Thermodynamics
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Transition State Theory
Think of some examples
Temperature dependence of prefactor
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Transition State Theory on Surfaces
Indirect adsorption of atoms
An atom adsorbs into a 2-dim mobile state, we
have Ng gas atoms, M sites on the surface, and N
atoms in the transition state
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Transition State Theory on Surfaces
Now what is K ?
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Transition State Theory on Surfaces
This corresponds to the collision on a surface
since the atoms are still free to move in two
dimensions
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Transition State Theory on Surfaces
Direct adsorption of atoms
M is total number of sites M is number of free
sites
Why?
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Transition State Theory on Surfaces
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Transition State Theory on Surfaces
Notice adsorption always result in loss of entropy
There may also be steric hindrance leading to
reduced S
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Transition State Theory on Surfaces
What happens in the regime between direct and
indirect adsorption?
The atoms breaks free of the site and start to
diffuse around in
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Transition State Theory on Surfaces
Indirect adsorption of molecules
Notice that if the precursor is sufficiently
loose S0(T)1.
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Transition State Theory on Surfaces
Direct adsorption of molecules
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Transition State Theory on Surfaces
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Transition State Theory on Surfaces
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Transition State Theory on Surfaces
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Transition State Theory on Surfaces
Notice how the Keq is alone determined from
initial and final state partition functions.
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Transition State Theory on Surfaces
Desorption
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Transition State Theory on Surfaces
System Prefactor s-1 Ea
kJ/mol CO/Co(0001) 1015
118 CO/Ni(111) 1015
130 CO/Ni(111) 1017
155 CO/Ni(111) 1015
126 CO/Ni(100) 1014
130 CO/Cu(100) 1014
67 CO/Ru (001) 1016
160 CO/Rh(111) 1014 134
How?
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Transition State Theory on Surfaces
If the details of the transition state can be
determined can the rate over the barrier be
calculated.

• Details of the transition state are difficult to
  access
• Low concentration
• Short lifetime.

Often determined by First Principle
calculations, but are only accurate to say 0.1
eV or 10 kJ/mol.