Reaction Mechanism
I. Reaction Mechanism - is the step by step sequence of elementary reactions by which overall chemical change
occurs.
A. The reaction mechanism must satisfy two requirements to be considered valid.
i. The sum of the elementary steps must give the overall balanced equation for the reaction.
ii. The mechanism must agree with the experimentally determined rate law.
B. Elementary steps- pathway reactions that add up to the target reaction.
i. The rate law for each elementary step is written from its molecularity.
C. Molecularity – The number of species that must collide to produce the reaction indicated by that step.
i. unimolecular step – a reaction involving one molecule.
ii. bimolecular step- reaction involving the collision of two species.
iii. termolecular step- reaction involving the collision of three species.
Elementary Step
Molecularity
Rate Law
A  products
unimolecular
Rate = k[A]
A + A  products
bimolecular
Rate = k[A]2
A + B  products
bimolecular
Rate = k[A][B]
A + A + A  products
termolecular
Rate = k [A]3
A + A + B  products
Rate = k[A]2[B]
A + B + C  products
termolecular
Rate = k[A][B][C]
D. Intermediates- a species that is neither a reactant nor a product. It is formed and consumed during a
chemical reaction.
E. Rate determining step- The slowest step in a reaction determines the rate of the reaction. The reaction
cannot go faster than the slowest step.
F. Catalyst- A species which is present in the first elementary step as a reactant and the last elementary step as
a product. It speeds up the rate of reaction.
G. Inhibitor- A species which is present in the first elementary step as a reactant and the last elementary step as
a product. It slows down the rate of reaction.
H. Example :What is the rate law for this reaction
2A + B  C occurs in the following two steps
Step 1
Step 2
A + B  C Occurs
A + C  D Occurs
slow
fast
1. Rate determining step is the slow step the rate law stops at that step.
Disregard any subsequent steps.
Rate = k [A][B]
Practice
2NON2O2 (fast)
N2O2 + H2  N2O + H2O (slow)
N2O + H2  N2 + H2O (fast)
a. Write the overall reaction
b. What is the rate determining step
c. List the intermediates
d. Write the rate law
6. Practice
CH3CH2OH + H+  CH3CH2OH2+ (fast)
CH3CH2OH2+  CH3CH2+ + H2O (fast)
CH3CH2+  CH2CH2 + H+ (fast)
a. Write the overall reaction
b. What is the catalyst and how do you know?
Exercise
Nitrogen oxide is reduced by hydrogen to give water and nitrogen,
2 H2(g) + 2 NO(g)  N2(g) + 2 H2O(g)
and one possible mechanism to account for this reaction is
2 NO(g) N2O2(g)
N2O2(g) + H2(g)  N2O(g) + H2O(g)
N2O(g) + H2(g)  N2(g) + H2O(g)
What is the molecularity of each of the three steps? Show that the sum of these elementary steps is the net
reaction.
Exercise 12.6
The balanced equation for the reaction of the gases nitrogen dioxide and fluorine is
2 NO2 (g) + F2 (g)  2 NO2F (g)
The experimentally determined rate law is
Rate = k [NO2][F2]
A suggested mechanism for the reaction is
NO2 + F2  NO2F + F
Slow
F + NO2  NO2F
Fast
Is this an acceptable mechanism? That is, does it satisfy the two requirements? Justify.
II. Factors affecting reaction rate
The following conditions affect the speed of a chemical process:
1. Nature of the reactants-o Physical state- K2SO4(s) + Ba(NO3)2(s)  no rxn.; however when these solid are dissolved in water,
they react.
o Chemical identity - Ions of opposite charge react very rapidly. Usually, the more bonds between
reacting atoms in a molecule, the slower the reaction rate. Substances with strong bonds (larger
bond energies) will react much more slowly
2. Concentration of reactants--more molecules, more collisions.
3. Temperature--heat >em up & speed >em up; the faster they move, the more likely they are to
collide.
o An increase in temperature produces more collisions that have the needed activation energy (minimum
energy required to react), therefore, there is an increase in reaction rate with increasing temperature.
Maxwell-Boltzmann CurveShows the distribution of particles having the needed kinetic energy
to react (aka. activation energy).
The higher the temperature, which is a measure of average kinetic
energy, the greater number of particles with the energy required to
cause a reaction.
−Ea/RT
Mathematically stated: f = e
f= fraction of particles
Ea = activation energy
e= natural antilog of
R = 8.314 J/mol-K
T = temperature in K
o
o In fact, a general rule of thumb is that a 10°C increase in temperature will double the reaction rate.
4. Surface area of reactants--exposed surfaces affect speed.
o Except for substances in the gaseous state or solution, reactions occur at the boundary, or interface,
between two phases.
o The greater surface area exposed, the greater chance of collisions between particles, hence, the reaction
should proceed at a much faster rate. Ex. coal dust is very explosive as opposed to a piece of charcoal.
Solutions are ultimate exposure
III. Collision Theory
A. Molecules must collide in order to react
.
i. they must collide with sufficient energy – to climb the hill (activation energy)
ii. they must collide with the correct orientation
B. Activation Energy- The minimum energy required to cause a reaction. This energy comes from the
kinetic energies of the colliding molecules. Upon collision, the kinetic energy of the molecules can be
used to stretch, bend and ultimately break bonds leading to chemical reactions. The value of activation
energy, Ea, changes from reaction to reaction.
C. Kinetics and Energy
D. The Arrhenius Equation- ties the collision model with reaction rate. Mr. Arrhenius found that reaction rate
data obeyed an equation based on three factors:
1. the fraction of molecules possessing Ea or higher.
2. the number of collisions occurring per second.
3. the fraction of molecules with the correct orientation.
4. k = Ae –Ea/RT
k= rate constant
Ea= Activation energy
T = temperature in Kelvin
Ae= frequency of effective collisions
R = 8.31 x 10-3kJ/Kmol
As Ea (activation energy) goes up, k (rate) goes down
As T (temperature) goes up, k (rate) goes up
5. Do some tricky math and it’s easier to solve for Ea:
ln k = - Ea (1) + ln [A]
R (T)
Taking this equation, plot 1/T vs. ln [k] , and get a straight line. From the straight line, find the slope and
then find the activation energy.
slope = - Ea
so ... Ea = - ( R ) (slope)
6. The equation allows you to calculate Ea from two known temperatures and corresponding values of k.
You want to be able to do this. 
k 
E 1 1
ln  2  = a   
R  T1 T2 
 k1 
Example - Arrhenius
Calculate the activation energy for the following set of data:
T (C)
3
13
24
33
k (l/mol- s)
1.4 x 10-3
2.9 x 10-3
6.2 x 10-3
1.2 x 10-2
Exercise
The colorless gas dinitrogen tetroxide decomposes to the brown gas NO2 in a first order reaction with a value of
k of 4.5 x 103/s at 274K. If k is 1.00 x 104/s at 283K, what is the energy of activation?
Exercise 12.7
The reaction
2 N2O5 (g)  4 NO2 (g) + O2 (g)
was studied at several temperatures and the following values of k were obtained:
k (s-1)
2.0 x 10-5
7.3 x 10-5
2.7 x 10-4
9.1 x 10-4
2.9 x 10-3
T (C)
20
30
40
50
60
Calculate the value of Ea for this reaction. Sketch your graph.
Exercise 12.8
The gas-phase reaction between methane and diatomic sulfur is given by the equation
CH4 (g) + 2 S2 (g)  CS2 (g) + 2 H2S (g)
At 550 C the rate constant for this reaction is 1.1 M-1s-1 and at 625 C the rate constant is 6.4 M-1s-1. Using
these values, calculate Ea for this reaction.
IV. CATALYSIS
A. Catalyst- A substance that speeds up a reaction without being
consumed itself.
B. How it works
1. Lower activation energy but do not change ∆E for the reaction.
2. Provide alternate reaction pathways
3. Result in higher percentage of effective collisions.
C. HETEROGENEOUS CATALYST--different phase than reactants,
usually involves gaseous reactants adsorbed on the surface of a solid
catalyst
adsorption—refers to the collection of one substance on the surface of another
absorption—refers to the penetration of one substance into another; water is absorbed by a sponge
hydrogenation of unsaturated hydrocarbons—especially important in converting unsaturated fats [oils]
into saturated fats [solids like Crisco]
C=C bonds are converted into C-C bonds by adding a pair of hydrogens “across the double bond”
A simple example of hydrogenation involves ethylene:
D. HOMOGENEOUS CATALYST—exists in the same phase as the reacting molecules.
Freons or chlorofluorocarbons [CFC’s] were used until recently as refrigerants and as propellants in
aerosol cans. Freon-12 (CCl2F2) is relatively inert and thus remains in the environment for a long time.
Eventually they migrate upward into the upper atmosphere and are decomposed by high-energy light.
Among the decomposition products are chlorine atoms: