By: Olivia Bohnhoff & Zach Feldker
What is kinetics?
The study of the speed of reactions.
 It’s based on experiments.
 We did it first semester, so no one
probably remembers anything. (Right?)
 RATE LAWS.
 Mechanisms
 Half-lives
 Catalysts!
 And lots of graphs. Fun, right?
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5 things that affect a reaction
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Size of the atom of the reactants
- Larger molecules react more slowly than smaller
ones. It’s more dense, and therefore harder to get
to. The smaller they are, the easier to get to. Get
it?
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Temperature
- The higher the temperature is, the more quickly
things will react. The kinetic energy is greater and
more molecules are colliding and reacting than at
a lower temperature.
Five things continued
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Concentration
- The more concentrated something is, the more likely
it is to react.
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State of reactants
- Liquids react more quickly than solids, gases react
the most quickly because the molecules are the
tightest in a solid and less so in a liquid and even more
less so in a gas.
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Catalysts
- Catalysts speed up a reaction by reducing the
activation energy, the energy necessary to start the
reaction, or “getting over the hump.”
Fun graph(s)!
The reaction with and without a catalyst
RATE LAWS
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This is probably what we all failed on the
final, right?
Now you don’t have to!
Rate law is the equation involving k. Little
k. It’s a constant.
It tells us the rate of the reaction based on
the concentration.
Rate = k [A]m[B]n
There are three orders to the rate law: first,
second, and zero. Let’s take a look at them
individually, shall we?
First-Order Rate Laws
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It depends on only one reactant, and only
raised to the first power.
Rate = k[A]1 or Rate = k[A]1[B]1
If you want to look at it from a logarithmic point
of view (which is probably a good idea), the
equation looks like this.
ln[A]t-ln[A]0= -kt
[A]t = the concentration of A at a specific,
designated time.
[A]0 is the original concentration of A.
k is the rate constant.
t is the time elapsed.
First-Order Graph
This is what a first-order graph looks like. You can always recognize it
because it slopes downward and therefore has a negative slope.
Second-Order Rate Law
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Looks similar to the first, except squared.
Rate = k[A]2
The real (integrated) rate law:
1/[A]t – 1/[A]0 = kt
Again, [A]t = the concentration of A at time
t, your choice.
[A]0 is the original concentration.
k is still the rate constant.
And t is the amount of time.
Second-Order Graph
A second-order graph is characterized
by a straight line with a positive slope.
Zero-Order Rate Law
A little bit different than the first two in
that it is the exponent is 0, and therefore
leaves you with just the bare rate
constant.
 Rate = k
 The graph is a straight line with a
negative slope.
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Look! Mechanisms!
The overall idea of mechanisms is that
reactions do not occur in a single step,
but rather a bunch of little ones, called
elementary steps.
 It’s similar to Hess’s Law in that the
steps should add up to the overall
equation.
 It’s dissimilar to Hess’s Law in that it has
nothing to do with enthalpy.
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Key Things to know about
Mechanisms
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Intermediates- intermediates are substances
produced in one elementary step and then
used up in a later one, and so they won’t show
up in the overall equation.
Catalysts- catalysts are also included in
elementary steps. They show up as reactants
and then as products, opposite of
intermediates.
Types of Steps- Unimolecular (decompisition),
Bimolecular (pretty much everything else), and
Termolecular (pretty much nonexistent.)
More Mechanisms
In a mechanism, the slowest step is the
most important because it determines
the rate. Cleverly called the ratedetermining step.
 You can write rate laws for the individual
elementary steps, but you can’t put them
all together to write the overall equation.
You can only take that from the ratedetermining step.
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Mechanism Example
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A + 2B
E+F
A+B
C
 C+B
D
D
E+F
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What is the rate Determining step for the below reaction?
NO2(g) + CO(g)
NO(g) + CO2(g)
 Rate Law = k[NO2]2
 NO2(g) + NO2(g)
NO3(g) + NO(g)
 NO3(g) + CO(g)
NO2(g) + CO2(g)
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Multiple Choice!
A reaction follows the rate law: Rate =
k[A]2. Which of the following plots will
give a straight line?
a) 1/[A] versus 1/time
b) [A]2 versus time
c) 1/[A] versus time
d) Ln [A] versus time
e) [A] versus time
1.
Question # 2
2.
a)
b)
c)
d)
e)
For the following reaction: NO2(g) + CO(g)
NO(g) + CO2(g), the rate law is : Rate = k[NO2]2.
If a small amount of gaseous carbon monoxide
(CO) is added to a reaction mixture that was .10
molar NO2 and .20 molar CO, which of the
following statements is true?
Both k and the reaction rate remain the same
Both k and the reaction rate increase
Both k and the reaction rate decrease
Only k increases, the reaction rate remains the
same
Only the reaction rate increases; k remains the
same.
Question #3
The steps below represent a proposed mechanism for the
catalyzed oxidation of CO by O3
Step 1: NO2(g) + CO(g)
NO(g) + CO2
Step 2: NO(g) + O3(g)
NO2(g) + O2(g)
What are the overall products of the catalyzed reaction?
a)
CO2 and O2
b)
NO and CO2
c)
NO2 and O2
d)
NO and O2
e)
NO2 and CO2
3)
Question 4
4)
a)
b)
c)
d)
e)
The decomposition of ammonia to the element is
a first order reaction with a half-life of 200s at a
certain temperature. How long will it take the
partial pressure of ammonia to decrease from
0.100 atm to 0.00625 atm?
200s
400s
800s
1000s
1200s