Chemical Kinetics
The area of chemistry
that concerns reaction
rates.
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Collision Model
Key Idea: Molecules must collide to
react and must be oriented the correct
way.
However, only a small fraction of
collisions produces a reaction. Why?
Arrhenius: An activation energy must
be overcome.
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Reaction Rate
Change in concentration (conc) of a
reactant or product per unit time.
conc of A at time t2  conc of A at time t1
Rate =
t2  t1
A

Why does this formula
t
look so familiar?
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For example
2NO2  2NO + O2
As NO2 is used up more NO and O2
should appear.
How much O2 appears when 3.0 mol
of NO2 disappears?
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5
6
Factors affecting rate of rxn
Temperature- for every 10°C
increase, the rate of the reaction will
double (approximation)
[Reactants] – allows for more
collisions.
Catalyst – speeds up by lowering
activation energy.
Nature of reactants – Surface Area
mainly but also complexity of the
molecule.
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Rate Laws
Rate = k[NO2]n
k = rate constant
n = rate order – affect of changing
the [Reactant] on [Product]
k and n can only be found
experimentally.
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Method of Initial Rates
Initial Rate: the “instantaneous
rate” just after the reaction
begins. This is a cheat so we
don’t have to worry about the
reverse rxn.
The initial rate is determined in
several experiments using
different initial concentrations.
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You can figure out order of each
reactant and then figure out value of
k.
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Overall Reaction Order
Sum of the order of each
component in the rate law.
rate = k[H2SeO3][H+]2[I]3
The overall reaction order is 1 +
2 + 3 = 6.
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Types of Rate Laws
Differential Rate Law: expresses
how rate depends on concentration.
What we call the rate law.
Integrated Rate Law: expresses
how concentration depends on time.
Lets you figure out concentrations of
stuff after a certain amount of time.
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First-Order Rate Law
Find it in the equation sheet
For A  Products in a 1st-order reaction,
 A
Rate =
k A
t
Integrated first-order rate law is
ln[A]t = kt + ln[A]o
Notice this is slope intercept
formula
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Half-Life of a First-Order
Reaction
0.693
t1/2 
k
t1/2 = half-life of the reaction Half life
of a reaction is the time required for a
reactant to reach half its original
concentration
k = rate constant
For a first-order reaction, the half-life
does not depend on concentration.
(This is important)
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Second-Order Rate Law
For A  products in a second-order
reaction,
 A
Rate =
k A 2
t
Integrated rate law is (slope
intercept again)
1
1
 kt +
A
Ao
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Half-Life of a Second-Order
Reaction
t1/2
1

kA
o
t1/2 = half-life of the reaction
k = rate constant
Ao = initial concentration of A
The half-life is dependent upon the
initial concentration and changes
over time
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Zero order rxns
1. Rate is constant; it does not
change with changing concentration
2. Zero order sometimes happens
with catalysis
3. The integrate rate law is NOT
found on the formula sheet.
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Reaction Mechanism
 The
series of steps by which a
chemical reaction occurs.
A
chemical equation does not tell
us how reactants become
products - it is a summary of the
overall process.
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Reaction Mechanism
(continued)
 The
reaction
6CO 2  6H 2 O
light
C 6 H12 O 6  6O 2
has many steps in the reaction
mechanism.
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Often Used Terms
Intermediate: formed in one step and
used up in a subsequent step and so is
never seen as a product.
Molecularity: the number of species
that must collide to produce the
reaction indicated by that step.
Elementary Step: A reaction whose
rate law can be written from its
molecularity.
uni, bi and termolecular
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Elementary step example
2 NO2 + F2 2 NO2F
follows the mechanism,
NO2 + F2 NO2F + F
NO2 + F  NO2F
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Rate-Determining Step
In a multistep reaction, it
is the slowest step. It
therefore determines the
rate of reaction.
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Writing a rate law from an
elementary step
2 NO2 + F2  2 NO2F
follows the mechanism,
NO2 + F2  NO2F + F slow
NO2 + F  NO2F fast
The rate law can be written?????
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Tricks
If it says fast equilibrium then you
substitute
Overall reaction 2NO + O2  2NO2
Step 1 2 NO N2O2 (fast
equilibrium)
Step 2 N2O2 + O2  2NO2
slow
Rate law = k[NO]2 [O2]
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Tricks Part Dos
Remember all the elementary steps
must add up to the chemical
reaction.
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Molecularity Elementary step
Rate law
1
A -> products
rate = k [A]
2
A + A -> products
rate = k [A]2
2
A + B -> products
rate = k [A] [B]
3
A + A + A -> products
rate = k [A]3
3
A + 2 B -> products
rate = k [A] [B]2
3
A + B + C -> products
rate = k [A] [B] [C]
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Arrhenius Equation
 Collisions
must have enough energy
to produce the reaction (must equal
or exceed the activation energy).
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A graph showing two steps.
Blue is uncatalyzed
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Catalysis
Catalyst: A substance that speeds
up a reaction without being
consumed. It does so by lowering
the activation energy.
Enzyme: A large molecule (usually
a protein) that catalyzes biological
reactions.
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[NO2] what order?
time (s)
[]
ln
1/[]
0
0.5
-0.69315
2
1200
0.444
-0.81193
2.252252
3000
0.381
-0.96496
2.624672
4500
0.34
-1.07881
2.941176
9000
0.25
-1.38629
4
18000
0.174
-1.7487
5.747126
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[]
0.6
0.4
0.2
0
0
5000
10000 15000 20000
1/[]
8
6
4
2
0
0
ln
5000
10000
15000
20000
0
0
5000
10000
15000
20000
-1
-2
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Catalysts
Homogeneous (same phase) vs
Heterogenous (different phases)
The difference between an
intermediate and a catalsyt is?
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Think about your areas of
strength and weakness…
1. Figure out order from a set of data and
write a rate law.
2. Calculate the value of k including units.
3. Use and understand integrated rate laws.
4. Determine rate law from elementary
steps.
5. Know how catalysts work
6. Understand misc. vocabulary.
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