B. Activation Energy: Ea
a) Example reaction: the burning of charcoal in the BBQ
C(s) + O2(g)  CO2
(remember, burning is VERY exothermic)
Question: Will charcoal in
your BBQ spontaneously
catch fire? ________
You need to add a
flame (and usually some
lighter fluid).
This provides the
________________________
Question: Do you need to continue to add more energy to keep the reaction going?
________
The energy released from the initial portion of the exothermic reaction provides the Ea
for further reactants to become products.
Activation Energy:
This is the energy required to form the transition state complex from the reactants.
This is an energy “barrier” that must be climbed over for the reaction to proceed.
for an _____________
reaction
For an _____________
reaction
Kinetics - 25
C. Arrhenius Equation
What It Tells Us
The Arrhenius Equation gives the relationship between _____________
and ____________________.
Rxn Rate Constant = k = A e-Ea/RT Arrhenius Equation
 Ea = activation energy in J/mol
 R = universal gas constant = _________________
 A = Arrhenius pre-exponential (frequency) factor (very large #)
This accounts for the frequency of collisions that have the correct
_____________________ to react.
 e-Ea/RT accounts for the fraction of collisions that have with enough
_____________________ to react
The reaction rate increases exponentially as the temperature increases.
This equation can also be linearized:
  Ea  1 
ln k  ln A  
 
 R  T 
y = b + (m) x
Arrhenius Plots
We can obtain a line by plotting
___________ vs __________.
To obtain this graph:
 Do experiments at different
____________________.
 Determine k at each temp., take its
natural log, and plot vs 1/T.
These plots are used to determine the value of Ea
(multiply the slope of the graph by ________ to get Ea.)
Kinetics - 26
D. Alternate (2-point / Slope) Version of Arrhenius Equation
The activation energy can be related to the rate constants at just two temperatures using
another form of the equation:
k 2  Ea  1
1 


ln   
  
 k1   R T2 T1 
For lab: Don’t use 2 point slope form. Use best-fit line!!!
LP # 9.
The Ea for the reaction: 2 ClO2F(g)  2 ClOF(g) + O2(g) is 186 kJ/mol.
-4 -1
o
o
If the value of k is 6.76  10 s at 322 C, what is the value of k at 50 C?
SOLUTION:
T1 = ___________
T2 = ___________
Ea =
Kinetics - 27
Now It’s Your Turn (We’ve left out a different variable!)
A certain process was used to predict temperatures.
At 25.0 C the rate was 0.179 M/min. At 22 C it was 0.142 M/min.
What is the Ea of the process? (Assume a zero order process.)
a) 0.32 kJ
b) 51 kJ
c) 0.749 kJ
d) 6.16 kJ
Hint:_________________
Kinetics - 28
VIII. Reaction Mechanisms
Definition: The sequence of molecular events, or elementary reaction steps that defines
the pathway from reactants to products.
A. Elementary Reaction Steps
1. Definition: A “molecular event”; (most often times it is a collision)
2. It is a single step in a reaction mechanism.
Molecularity
Steps are classified on the basis of their molecularity, or the number of particles
involved in the collision.
 ____________________ reaction - elementary reaction step that
involves a single reactant molecule
 ____________________ reaction - elementary reaction step that
involve two atoms or molecules
____________________ reaction - elementary reaction step that
involve three atoms or molecules; rare
An ____________________________
 Is a reaction only having a single elementary reaction step.
B. Multiple Step Reactions
Multi-step rxns will have
more than one “hump” in the
reaction coordinate diagram.
___________________
 A species that is formed
in one step of a reaction
mechanism and
consumed in a
subsequent step.
a. do not appear in the
net equation for the
overall reaction
b. presence is only
noticed in the
elementary steps
Kinetics - 29
C. An Overall Reaction
It describes reaction stoichiometry. (The types of reactions we are used to writing.)
It is the sum of all of the Elementary Reaction Steps
 Experimentally observed rate law for an overall reaction depends on the reaction
mechanism.
 It will correspond to the rate determining elementary step.
(This is the step with the largest _______.)
(This is the __________________ step)
Since elementary reactions have only one step, we can write the rate law directly from
the reaction equation.
LP# 10.
Write the rate law for the following elementary reaction:
O3 + NO  O2 + NO2
__________________________
D. Rate Law from Reaction Mechanism
For elementary reaction steps, reaction law coefficients are equal to the rate law
exponents.
For multi-step reaction, the rate law is based on the slowest step.
Mechanism with a Slow Initial Step
If the first step is slow, then the overall reaction will proceed at the rate of this slow
initial step. The subsequent steps can only proceed as the first one provides products for
it.
Step Molecularity
Example:
2NO(g) + H2(g)  N2O(g) + H2O(g)
Slow
N2O(g) + H2(g)  N2(g) + H2O(g)
Fast
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
Overall
Rate Law: ___________________________
The rate law will only include species that are part of the overall reaction.
Kinetics - 30
Mechanism with a Fast Initial Step
 The rate law is based on a step other than the first step.
 The rate law will include species that are reaction ________________
that do not appear in the overall reaction.
 Through a series of substitutions, these reaction intermediate factors can be replaced
in the rate law with factors of species that do appear in the overall reaction.
Note: Although a proper rate law should not include any intermediates, for the purpose
of this course, we will not spend the time on the process of substitutions.
You may include intermediates in rate laws that you use for this course.
E. Procedure used for establishing a reaction mechanism.
1. Determine the overall rate law experimentally.
2. Devise a series of elementary steps.
3. Predict the rate law based on the reaction mechanism.
4. If observed and predicted rate laws agree, the proposed mechanism is a
probable pathway for the reaction.
5. Easy to disprove a mechanism; impossible to "prove" a mechanism.
IX. Catalysis
A. How Do Catalysts Work?
They work (generally) by
OR
Occasionally, they can make collisions more effective.
Remember
-Ea/RT
k=Ae
Lowering Ea:


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Example: 2 H2O2(aq)  2 H2O(l) + O2(g) + energy
Reaction coordinate diagrams w/ and w/out the catalyst
The two humps show that there are 2 steps in the reaction mechanism w/ catalyst.
 The # humps = # of steps in the reaction mechanism
 Hump sizes may be different.
 The larger hump is the slower, rate-determining step.
Catalysts decrease Activation Energy by allowing access to a lower
energy reaction mechanism.
Hypothesized catalyzed Reaction Mechanism
Step #1
Step #2
Overall
H2O2 + I-  H2O + IOH2O2 + IO-  H2O + O2 + I2 H2O2  2 H2O + O2
 ______ is a reaction intermediate
 ______ is the catalyst
It participates in forming the intermediate
It is returned to solution. (It is not used up.)
If there are 2 humps, there must be 2 transition states.
Question: If this mechanism is correct, what would the rate law
be?
Which step is rate limiting? Ans: ________
________________________________
Kinetics - 32
Step #1 is slow
because it has
the higher Ea
hump
B. How do Changes in Ea Affect the Rate?

 The lower the value of Ea, the more moleules have enough energy to react.

LP# 11.:Let’s assume that the only change is to the activation energy.
By what factor does the reaction rate change if we lower the activation energy from
75kJ/mol to 57kJ/mol?
Note:
 Both fractions of molecules w/ enough energy to react are very low.
 The bigger the value of Ea, the lower the fraction.
 When we lowered the value of Ea by only 30%,
the reaction rate went up by a factor of about 2000.
This means that there are almost 2000 times more particles with enough energy to react.
Kinetics - 33