AP Chemistry Notes
Unit 8: Chemical Kinetics
Reaction Rates



Chemical kinetics is the study of the “speed” or “rate” of a chemical reaction
The rate of a reaction is determined by measuring how the concentration
(mol/L) of either the reactants or the products changes every second, therefore,
the units used to report reaction rates are “moles per liter per second”
(mol/Ls).
Remember that brackets, [ ], symbolize concentrations in mol/L
Ex. [NO2] means “the concentration of NO2 in moles/L” or “the molarity of NO2”
o During a chemical reaction, the concentration of reactants, will decrease
over time because they are being consumed.
 So the moles/liter of reactant will decrease every second as
reactants get used up.
Rate =
 [reactant]
time
= (final concentration – initial concentration)
(final time – initial time)
*A negative sign is used in the formula to maintain a positive
numerical value for rate
o During a chemical reaction, the concentration of products, will increase
over time because they are being produced.
 So the moles/liter of product will increase every second as
products are formed
Rate =
[product]
time
= (final concentration – initial concentration)
(final time – initial time)
*A negative sign is NOT needed to obtain a positive rate value when
determining the rate by monitoring products
Ex. For the chemical reaction A  B:
Rate = [A]
t
and
Rate = [B]
t
1
Four Factors That Affect Reaction Rate
*Some reactions are very fast....


Combustion of finely powdered cornstarch
Precipitation of PbI2
Acid-Base neutralizations

These reactions may take only seconds or
fractions of seconds.
*While other reactions are very slow....


Rotting of meat
Trees dying (combustion of cellulose)
Decay of uranium

These reactions may take days, years,
or thousands of years.

The four factors that affect reaction rate are:
1. Surface area (of solids)
2. Concentration (of gases and solutions)
3. Temperature
4. Presence of a catalyst
Reaction Rate Demos
1. Increased surface area speeds up reactions:
a.) Cornstarch (or lycopodium powder) is flammable when finely
powdered
b.) 5g KMnO4 + 1mL glycerine (if ground up KMnO4, rxn starts
faster)
2. Increasing the concentration speeds up reactions:
a.) Mg ribbon in 2 test tubes......
add 1M HCl
add 12M HCl
3. Increasing the temperature speeds up reactions:
a.) 1mL NaClO3 + small scoop MnO2 in a test tube.....reacts after applying heat
from Bunsen burner
4. Adding a catalyst speeds up reactions:
a.) 10mL 30% H2O2 in 125mL flask + scoop of MnO2 as a catalyst
will produce large amount of steam quickly
2
Reaction Rates and Stoichiometry
Consider the graph below that illustrates the reaction:
2NO2(g)  2NO(g) + O2(g)
Conc (mol/L)
0.01
[NO2] decreases over time
0.0075
NO2
0.005
NO
O2
0.0025
0
0
100
200
300
400
[NO] increases over time
[O2] increases over time
The line representing NO has a steeper
initial slope than the line for O2. Why?
Time (s)
Because the stoichiometry of this balanced equation (look at the coefficients) shows
that 2 moles of NO are produced every time 1 mole of O2 is produced,
the rate of production of NO is twice that of O2.

This presents a problem because, depending on which substance you are
monitoring, you could report multiple rates for the same chemical reaction.
o To fix this, just divide the rate by the coefficient of the substance, then the
rate will always be the same numerical value, no matter which substance
was monitored to get it.
Rate =  1 [NO2]
2 t
= 1 [NO]
2 t
= 1 [O2]
1 t
3
Discussion Questions
The following reaction applies to all these discussion questions:
2H2(g) + O2(g)  2H2O(l)
1. Which concentrations would increase over time?
[H2O]
2. Which concentrations would decrease over time?
[H2] and [O2]
3. How does the rate of change of each substance compare?
[H2] would decrease at the same rate that [H2O] would increase, but the
[O2] would decrease only half as fast as the [H2] would decrease and half
as fast as the rate of [H2O] production.
4. If you were graphing the concentration of each substance vs. time, describe
the slope of the lines for each substance as it changes over time.
H2 and O2 would have negative slopes that were steep at first but level
off over time. The slope of H2 would be twice as steep as O2. The slope
of H2O would be positive with a steep slope initially, but leveling off over
time. The slope of H2O would be the opposite of the slope of H2.
4
Determining Rates Graphically

We can determine 2 kinds of rates from graphs
1. Average Rate = average rate over a long range of time
2. Instantaneous Rate = rate at a particular moment in time
1. Average Rate = -[reactant]
time
= -final concentration - initial concentration
final time - initial time
NOTE: we want the rates to be positive so we need a negative sign out in front
Example:
TIME
[NO2]
[NO]
[O2]
0
.0100
0
0
50
.0079
.0021
.0011
100
.0065
.0035
.0018
150
.0055
.0045
.0023
200
.0048
.0052
.0026
250
.0043
.0057
.0029
300
.0038
.0062
.0031
350
.0034
.0066
.0033
400
.0031
.0069
.0035
a.) Calculate the average rate at which NO2 changes during the first 50 seconds
of the reaction: 2NO2(g)  2NO(g) + O2(g)
Average Rate = - (.0079 - .0100) = 4.2 x 10-5 mol/Ls
50 s – 0 s
b.) Calculate the average rate at which [NO2] changes from 200 to 250 seconds.
Average Rate = -(0.0043 – 0.0048) = 1.0 x 10-5 mol/Ls
250 s – 200 s
5
Discussion Questions
The following discussion questions refer to the previous example problem and graph
1. What happens to the slope of every line as time goes by? What does this
mean about the change in concentration of each substance over time? What
does this mean about the rate of the reaction over time? Every slope flattens
out. This means the change in concentration every second eventually gets
smaller. As time goes by the rate of the reaction slows down.
2. When is the reaction fastest?
Initially when [reactants] is highest.
3. Why does the reaction rate change over time?
When the [reactants] decreases, there are fewer reactant particles to collide
with each other and cause a reaction.
4. What is the real rate of reaction if rates keep changing over time?
Average rates are reported for the “initial” reaction rate. The “real” rate is hard
to define other than initially…this is why it is sometimes necessary to know the
rate at a specific time instead of over a range of time.
2. Instantaneous Rate = - slope of a line tangent to the curve at any point
slope = -[reactant] =
t
-y
x
=
-(y2 - y1)
(x2 - x1)
*where t (change in time) is extremely small
What concept is this in calculus? A derivative
NOTE: The only reason we need a negative sign here is because we are basing
the rate on the decrease in concentration of reactants, yet we want a positive
value for the rate itself.
6
Example:
What is the instantaneous rate at t = 100 seconds?
Conc (mol/L)
0.01
0.0075
NO2
0.005
a.) On the curve representing the
reactant, draw a tangent line at
t=100 sec.
NO
O2
0.0025
0
0 50 100 150 200 250 300 350 400
Time (s)
b.) Calculate the slope of the
tangent line because the
instantaneous rate is equal to the
opposite of the slope of the
tangent line.
To help you calculate the slope of the tangent line, draw a right triangle such
that the tangent line is the hypotenuse.
Y1
tangent line
Y2
Slope = -(Y2 – Y1) = -(.005 - .0075)
(X2 – X1)
160 – 50
= 2.4 x 10-5 mol/Ls
X1 X2
Summary of Rates (so far)
1. Rate = speed of reaction
2. Rates are fastest initially, but decrease over time
3. During a reaction, the concentrations of reactants will decrease while the
concentrations of products will increase
4. We measure rates based on the rate of decrease in concentrations of reactants
5. Average Rate = general rate over a given time interval
6. Instantaneous Rate = specific rate at a single point in time
7
Determining Rates Using The Initial Rate Method

A simple method for calculating rates without graphing!

It is based on doing a series of reaction rate experiments and relating the
results to something called a RATE LAW

Once you establish a rate law for a reaction, you can use it to predict the rate
of a reaction without actually conducting the experiment.

A rate law relates the reaction rate to the concentration of reactants raised
to some exponent power. The exponent must be determined experimentally
Example: A + B  Products
o The generic rate law expression for this equation is: Rate = k[A]m[B]n
Where....
k = rate law constant
m = reaction order with respect to (w.r.t) A
n = reaction order w.r.t B
m + n = overall order of the reaction
o The size of "m" and "n" will show us a relation between [reactants]
and the rate, but “m” and “n” MUST be determined
EXPERIMENTALLY!!!
o NOTE: "m" and "n" do NOT correspond to the coefficients of the
equation whose rate we are studying

In order to obtain the values of the exponents in a rate law expression, we
must set up a series of experiments in which we vary the initial concentration
of one reactant (while holding the initial concentrations of all other reactants
constant) and observe if and how the initial rate changes.
8
Example: 2N2O5(g)  4NO2(g) + O2(g)
[N2O5]
1.0 x 10-2
2.0 x 10-2
Trial 1
Trial 2
Rate (mol/Ls)
4.8 x 10-6
9.6 x 10-6

The generic rate law for this equation is Rate = k[N2O5]m

To calculate the value of “m”, we need to compare two experiments in which the
concentration of N2O5 was changed and see how that affected the rate of the
reaction as shown below:
Rate2 = 9.6 x 10-6
Rate1 = 4.8 x 10-6
2
= k(2.0 x 10-2 )m
= k(1.0 x 10-2 )m
= 2m
So m = 1
In our example above, we doubled the concentration of reactant and
consequently the rate of the reaction doubled (went from 4.8x10-6 to
9.8x10-6). This is always the case if m = 1.
Because the exponent on [N2O5] was 1, we say:
“The rate is 1st order with respect to N2O5.”
9

Sometimes it’s hard to determine what the rate is doing (i.e. increasing 4x or 9x,
etc.) by simple inspection. To figure it out, just divide the two rates you are
comparing.
o From the previous example:
Trial 2 = 9.6 x10-6 = 2.0 (so the rate is increasing 2 fold or doubling)
Trial 1 4.8 x 10-6

To find the value of the rate law constant, k, simply pick any trial and substitute
those values into the rate law expression to solve for k.
Example 1:
A + B  Products
a.) Write the generic rate law
Rate = k[A]m[B]n
Exper. 1
Exper. 2
Exper. 3
[A]
2
6
6
[B]
4
4
8
Rate
4
108
432
b.) What is reaction order wrt A? Compare two experiments where [A] is changing
but [B] is not. (Like Experiments 1 and 2)
108 = k(6)m
so 27 = 3m and m = 3
4 = k(2)m
c.) What is reaction order wrt B? Compare two experiments where [B] is changing
but [A] is not. (Like Experiments 2 and 3)
so
4 = 2n
and n = 2
d.) What is the overall reaction order?
m+n = 3+2 = 5
e.) What is the value of k?
Trial 1: 4 M/s = k(2 M)3(4 M)2
4 M/s
(8 M3)(16 M2)
= k
0.03125 M-4s-1 = k
10
Example 2:
H2O2(aq) + 3I-(aq) + 2H+(aq)  I3-(aq) + 2H2O(l)
Exper. 1
Exper. 2
Exper. 3
Exper. 4
[H2O2]
[I-]
[H+]
Rate (mol/Ls)
.010
.020
.010
.010
.010
.010
.020
.010
.00050
.00050
.00050
.0010
1.15 x 10-6
2.30 x 10-6
2.30 x 10-6
1.15 x 10-6
a.) Generic Rate Law: Rate = k[H2O2]m[I-]n[H+]p
b.) Reaction order wrt each species:
Experiments:
1&2
2=2m
m=1
1&3
2=2n
n=1
1&4
1=2p
p=0
c.) Overall order of reaction: 1 + 1+ 0 = 2
"Reaction is 2nd order overall"
d.) Actual Rate Law: Rate = k[H2O2][I-]
*no [H+] since p = 0
e.) Rate constant:

k=
rate = 1.15 x 10-6 M/s
[H2O2][I-] (.010M)(.010M)
= 1.2 x 10-2 M-1s-1
Why should I care about “k”??????
o What does it mean if "k" is large? Small?
 A large k means fast reaction
 A small k means slower reaction
*Essentially, k is a multiplier in the rate equation
11
Integrated Rate Laws
 Using the Initial Rates Method to find the rate law is helpful if you want to know
what the rate is when you have a specific concentration of reactant "x" to start with.
Example
2H2O2(l)  2H2O(l) + O2(g)
Rate = k[H2O2]m
By performing just a few experimental
trials, you can determine the “Rates” at
specific [H2O2] and then solve for the
order of the reaction and finally, k.
This is the whole PURPOSE of
doing the initial rates method
Once you know "k" and the reaction
order you can find the Rate at any other
[H2O2] without further experiments.
But what if you want to know what the [H2O2] will be at t = 200 seconds?
The initial rate method can't answer this!
However, we can derive an equation from the rate law (by using Calculus and “taking
the integral of the rate law”) that WILL answer this.
The "integrated rate law" describes [H2O2] as a function of time
This equation depends on the order of the reaction
12
Integrated Rate Law for 1st Order Reaction

For AP Chem purposes, our discussion of integrated rate laws will be limited to
simple decomposition reactions that involve only one reactant. An example of such
a reaction might look something like this:
A  Products
o For a first order reaction, the rate law would be: Rate = k[A]1
o We can also define rate simply as the change in concentration of reactants
over the change in time: Rate = -[A]
t
o If we set these two definitions equal to each other we will get:
-[A]
= k[A]
t
o Which could be rearranged to:
[A]
= -kt
[A]
o Then, using Calculus, we could take the integral of each side:
conc at any t
[A]t

conc. at t=0
t
d[A]
[A]0 [A]
=

-k dt
rate constant
elapsed time
0
o The integrals above result in what we really need to know for AP Chemistry
which is the equation:
For Calculus fans only…
Know This
Equation!!!
What we did above is equivalent to:
b
1 dx = lnb - lna
a x
d[A]
1 d[A]
since
[A] =
[A]
=
ln[A]t - ln[A]0 = -kt
ln[A] t – ln[A]0
o This is really the equation of a line in "slope-intercept" form:
And be able to
calculate the
slope!!!
ln[A]t = -kt + ln[A]0
( y
= mx + b
)
where:
y = ln[A]t and x = t
b = y-intercept = ln[A]o
m = slope = -k
13

If you plot "ln[A]t vs. t" and the result is a straight line...
THE REACTION MUST BE FIRST ORDER!!!

If this plot is not a straight line, the reaction is NOT first order.
slope of line = -k
ln[A]t
T
Example
For the 1st order decomposition of H2O2, if k = 3.66 x 10-3 /s and [H2O2]0 = 0.882
M:
a.) Find the time at which [H2O2] = 0.600 M
ln[A]t – ln[A]0 = -kt
ln(.600) - ln(.882) = -3.66 x 10-3t
105 sec = t
b.) Find [H2O2] after 225 s
ln[H2O2]t – ln[H2O2]0 =
Don’t Freak!!!
Just hit “INV lnx”
of -.9496 on your
calculator or ex
-kt
= -(3.66 x 10-3)(225) = -.824
ln[H2O2]t
= (-.824) + ln(.882) = -.9496
[H2O2]t =
e-.9496
REMEMBER:
logN=x  N=10x
log210=3.332  10=23.322
[H2O2]t =
.387 M
lnN=x  N = ex
14
Half Life for First Order Reaction

A “half life” is the time required for half of a sample to decay
o For example, the time required for [A]t = 1/2[A]o

The formula used to calculate half life depends on the order of reaction
 The Formula for Half Life of 1st Order Reactions:
ln[A]t = -kt
[A]o
t1/2 = half life
ln1/2[A]o = -kt1/2
[A]o
ln(1/2) = -kt1/2
-.693 = -kt1/2
.693 = t1/2
k
For 1st order reactions, the half
life depends only on "k", not on
the concentration of reactant.
So the half life is constant
regardless of the initial
concentration!
Example: The thermal decomposition of N2O5(g) to form NO2(g) and O2(g) is a firstorder reaction. The rate constant for the reaction is 5.1 x 10-4 s-1 at 318K. What is the
half life for this process?
.693
= t1/2
-4
5.1 x 10
1.4 x 103 sec = t1/2
15
Integrated Rate Law for Zero Order Reaction

A  Products
For a zero order reaction such as:
o The rate law looks like:
Rate = k[A]0 = k
o As you can see, for zero order reactions, the rate doesn't depend on the
concentration of a reactant.
o The rate might depend on other factors such as intensity of light absorption,
surface area, etc.
Example
NH3(g)  1/2N2(g) + 3/2H2(g)
Rate = k[NH3]0 = k

Rate Law (See! It doesn't depend on
concentration of reactant)
For a zero order reaction, the rate law can be integrated to give:
[A]t – [A]0 = -kt


Integrated Rate Law
[A]t = -kt + [A]0
(in slope-intercept form)
(y
Again, slope = -k
=
mx + b)
If you plot "[A] vs. t" and the result is a straight line...
THE REACTION MUST BE ZERO ORDER!!!
If this plot is not a straight line, the reaction is NOT ZERO order.
slope of line = -k
[A]t
t
16
Half Life for Zero Order Reaction

The half life for a zero order reaction depends on the initial concentration AND the
rate constant
t1/2 = [A]0
2k
Integrated Rate Law for Second Order Reaction

A second order reaction is one where the rate depends on:
1. A single reactant concentration raised to the second power
OR
2. The concentrations of two different reactants each raised to powers that would
add together to be two.

Equations related to a situation like #2 above are beyond the scope of AP Chemistry
and we will focus only on situation #1 which might look like the following equation:
A  Products
o The rate law would be: Rate = k[A]2
o And the integrated rate law is:
1 - 1 = kt
[A]t [A]0
1 = kt + 1
[A]t
[A]0
in slope-intercept form
(y = mx + b)
17

If you plot " 1 vs. t" and the result is a straight line...
[A]
THE REACTION MUST BE SECOND ORDER!!!

If this plot is not a straight line, the reaction is NOT SECOND order.
1
[A]t
slope of line = k
t
Half Life for Second Order Reaction

The half life for a second order reaction also depends on the initial concentration
AND the rate constant
t1/2 =
1
k[A]0
18
Kinetics Cheat Sheet


Know EVERYTHING about this chart
Know how to look at a graph and identify the reaction order
Order
Rate Law
Integrated
Rate Law
Straight
Line Plot
Slope
=
Half Life
0
Rate = k
[A]t = -kt + [A]0
[A] vs. t
-k
[A]0
2k
1
Rate = k[A]
ln[A]t = -kt + ln[A]0
ln[A] vs. t
-k
0.693
k
2
Rate = k[A]2
1 = kt + 1
[A]t
[A]0
1 vs. t
[A]
k
1 .
k[A]0
Example
Identify the order of each reaction for each graph and give the slope.
[A]
ln[A]
t
0 order, slope = -k
1/[A]
t
1st order, slope = -k
t
2nd order, slope = k
19
Theories of Kinetics



Why are some reactions 0 order, 1st order, or 2nd order?
Why are some reactions fast while others are slow?
Why do reactions happen at all?
Attempts to answer these questions at the molecular level
has resulted in 2 coexisting theories:
1. Collision Theory
2. Transition State Theory
The Collision Theory

For particles to react, they must collide
o But only some collisions result in reactions
Rate of reaction is proportional to the frequency and effectiveness of the collisions!

An "effective" collision is one that results in a chemical reaction. Two conditions
must be met to have an “effective” collision.
1. The collision has to have enough energy to break bonds.
- the minimum amount of energy to break bonds = activation energy (Ea)
2. The particles must be oriented properly.
Example:
The diagrams on the left show the interaction
between I- and CH3Br.
I-
In figure (a), I- approaches the partially positive
carbon in CH3Br and a reaction DOES happen.
+
C
I-
In figure (b), I- approaches the partially
negative bromine in CH3Br and a reaction does
NOT happen. The I- and Br - particles repel
each other.
Br -
20
Transition State Theory

During reaction, some bonds are broken while some bonds are formed.
o There is sort of an intermediate stage during the reaction when new bonds
are starting to form but the old bonds aren't completely broken, yet
 This point in the reaction is called the "transition state"

During the transition state, the reaction could:
1. Proceed to products (reaction occurs)
2. Revert back to reactants (no reaction)
Example:
I- + CH3-Br  I---CH3---Br  I-CH3 + Br

Formation of this structure
occurs during the "transition state".
It is a very unstable intermediate
called the activated complex.
Perhaps an even better way to illustrate the Transition State Theory is by drawing
"reaction profiles"
Example:
O-C + O-N-O  O-C---O-N-O  O-C-O + N-O
A “Reaction Profile”
*You must be able to label the following items on a reaction profile for both an
exothermic and endothermic reaction:




Reactants
Products
Ea forward
Ea reverse



Transition state (activated
complex)
H (with correct sign)
Both axes
21
Exothermic Reaction
H is negative for
exothermic reactions
Endothermic Reaction
H is positive for
endothermic reactions

Ea = activation energy = minimum amount of energy needed for a reaction to occur
H = enthalpy = Ea forward - Ea reverse


In exothermic reactions, Ea forward < Ea reverse so the forward reaction is favored and
H is negative
In endothermic reactions, Ea forward > Ea reverse so the reverse reaction is favored and
H is positive
22
Effect of Temperature on Reaction Rate

Generally, higher temperatures speed up reactions
o Think about food spoiling (food spoils more slowly in the fridge)
o Think about cooking (stuff cooks faster at higher temps)

Increased temperatures mean particles have higher Kinetic Energies
o More Kinetic Energy leads to more frequent collisions and thus a higher % of
effective collisions
Collisions that have enough
energy and the proper orientation
to result in a chemical reaction

The Arrhenius Equation describes the effect of temperature on the rate constant, (k).
-Ea
k = AeRT
where:
e
Ea
=
=
base number for natural logarithms
activation energy
-Ea
eRT
=
fraction of collisions that are effective
R
A
=
=
gas constant (8.31 J/molK)
frequency factor = collision frequency x
probability of proper orientation
Don’t worry about “A”
23

The Arrhenius equation is more useful when you take the ln of both sides:
ln k = ln A + ln e –Ea/RT
ln k = ln A - Ea/RT
ln k = -Ea + ln A
RT
(y
= mx
+
We can use this equation to find the
Ea of a reaction….
b)
where the slope (m) = -Ea
R

With the Arhhenius equation in slope intercept form, you can graph “ln k” (your ycoordinates) and “1/T” (your x-coordinates)
o This technique provides a way for you to solve for Ea once you determine the
slope of the line
If you plot ln k vs. 1/T
The slope will be –Ea/R
slope = -Ea/R
ln k
-slope(R) = Ea
1/T (Kelvins-1)
Kinetics is
GGGRRRRATE!!
24

You can also use the Arrhenius Equation without graphing, but we need to get rid of
that ridiculous “ln A” term since A is so complex.
o Remember: the whole point of the Arrhenius Equation is to show the effect
temperature has on reaction rate.
o To get rid of the “ln A” term, let’s look at the Arrhenius Equation for a
reaction occurring at two different temperatures (T1 and T2) where the
corresponding rate constants are k1 and k2.
ln k2 = -Ea/RT2 + ln A
and
ln k1 = -Ea/RT1 + ln A
subtract ln k1 from ln k2
ln k2 - ln k1 = (-Ea/RT2 + ln A) - (-Ea/RT1 + ln A)
ln k2
k1
=
ln k2
k1
=
Ea - Ea + ln A - ln A
RT1
RT2
Ea
R
1 - 1
T 1 T2
Goodbye “A”
term…you were
such a complex
friend.
This is the most common way
to use the Arrhenius Equation
in problem solving!
25
Example:
The rate constant of a first-order reaction is 3.46 x 10-2 s-1 at 298 K. What is the
rate constant at 350. K if the activation energy for the reaction is 50.2 kJ/mol?
Substitute into the Arrhenius equation:
ln
k2
3.46 x 10-2/s
= (50,200 J/mol) (1/298K – 1/350K)
(8.31 J/mol K)
ln k2 – ln(3.46 x 10-2/s) = 3.01
ln k2 = 3.01 + ln(3.46 x 10-2/s)
ln k2 = -0.352
k2 = 0.703/s
*k2 got larger with larger temperature as it should
26
Reaction Mechanisms


Very few reactions occur as simply as is depicted by a balanced chemical equation
Chemical equations suggest that reactants go to products in one step, but more
often, reactions occur via a series of steps...
Example:
NO2(g) + CO(g)  NO(g) + CO2(g)
really occurs in 2 separate steps like this:
Step 1:
Step 2:
Net:
NO2 + NO2  NO3 + NO
NO3 + CO  NO2 + CO2
NO2(g) + CO(g)  NO(g) + CO2(g)

THE COMPLETE SERIES OF STEPS involved in converting reactants to products is
referred to as a reaction mechanism.

EACH INDIVIDUAL STEP in a reaction mechanism is called an elementary reaction.

Mechanisms explain why rate laws must be determined experimentally.
o Because some of the molecules are involved in various reactions (the
elementary reactions) that occur “behind the scenes” and these elementary
steps all have their own rates.
NOTE: Proposing a reaction mechanism is beyond the scope of AP
Chemistry. Our problems will focus on evaluating given mechanisms
to see if they are “plausible”.
27
Evaluating Reaction Mechanisms
To determine if a given mechanism is “plausible”, you must:
1. Write a rate law based on the slowest step in the mechanism
2. Make sure the rate law you write contains only substances that
are in the overall net reaction equation
3. Make sure the rate law you write matches the experimentally
determined rate law (which will probably be provided in the
problem)

Finding the slowest step in the mechanism
o The slowest step in the mechanism is called the “rate determining step” or
(RDS)
o Sometimes, the RDS will be noted by the word “slow” after that elementary
reaction
o If the words “slow” and “fast” are not noted in the reaction mechanism, use
molecularity to help determine the RDS:
o molecularity = the number of reactant particles that must collide in
an elementary reaction
o The more particles that must collide simultaneously (with enough
energy and proper orientation), the less likely it is to react
o This means that the elementary step with the most reactants (highest
molecularity) will generally be the RDS
o Once the RDS of the mechanism is determined, write a rate law based on
that elementary reaction.
o Since there are no other reactions taking place “behind the scenes” of
an elementary reaction, you can use the coefficients of the RDS to
serve as the exponents in the rate law you write.
Examples of determining molecularity :
O3  O2 + O
molecularity is “unimolecular” (a single molecule involved)
NO + Cl2  NOCl + Cl
molecularity is “bimolecular” (two molecules involved)
NO + NO + Br2  2 NOBr
molecularity is “termolecular” (three molecules involved)
*generally won’t see mechanisms proposed that require more than three molecules to
collide since that would be a rare occurrence
28

There are TWO mechanism scenarios:
1. The slow step is the first step in the mechanism (this is the easier problem)
2. The slow step is preceded by a fast step or steps (this is the harder problem)
Example of Scenario #1 (slow step is first)
Evaluate the proposed mechanism to determine if it is plausible.
Given Reaction:
2N2O(g)
 2N2(g) + O2(g)
Experimentally determined rate law: Rate = k[N2O]
Mechanism: N2O(g)
 N2(g) + O(g)
N2O(g) + O(g)  N2(g) + O2(g)
(slow step)
(fast step)
These are the 2
“elementary
reactions” in the
mechanism
Step 1: Write a rate law based on RDS: Rate = k[N2O]
Step 2: Make sure rate law only contains substances in the given equation.
Step 3: Make sure the rate law based on RDS matches the experimentally
determined rate law.
29
Example of Scenario #2 (slow step is preceded by fast step)
The experimentally determined rate law for the reaction below is
Rate = k[NO]2[O2]
Is the proposed mechanism valid?
Reaction:
2NO(g) + O2(g)  2NO2(g)
Possible Mechanism:
2NO
N2O2 + O2


N2O2
2NO2
Step 1: Write a rate law based on the RDS:
(fast step)
(slow step)
Rate = k[N2O2][O2]
Step 2: A valid rate law for this reaction can’t have [N2O2] in it because N2O2 is
NOT in the overall net equation for this reaction.
HOWEVER, in this scenario the fast step establishes an equilibrium. N2O2 starts to
accumulate because the slow step doesn’t use it very quickly, and the reaction starts
to proceed in the reverse direction. We can write an equilibrium expression for the
fast step which allows us to replace [N2O2] in the rate law with something else.
Remember Equilibrium Constants?
For an equation
aA + bB  cC + dD
Write an equilibrium expression for the fast step:
Keq = [C]c[D]d
[A]a[B]b
Keq = [N2O2]
[NO]2
Then rearrange it to show that [N2O2] is equal to something else:
[N2O2] = Keq[NO]2
Then substitute that new value for [N2O2] into the rate law for the slow step:
Rate = k[N2O2][O2] = k(Keq[NO]2)[O2] = k'[NO]2[O2]
Now the rate law only
contains substances that
are in the given equation
Step 3: The rate law based on the RSD matches the experimentally determined rate
law.
30
Another Example
A proposed mechanism for H2(g) + I2(g)  2HI(g) is:
I2

2I + H2 
2I
2HI
(fast)
(slow)
What is the order of the reaction according to the mechanism?
Rate Law from slow step: Rate = k[I]2[H2]
*a valid rate law can’t have [I] because that is not in the overall equation
Equilibrium Expression from fast step: Keq = [I]2
[I2]
Rearrange equilibrium expression to solve for [I]2:
[I]2 = Keq[I2]
Substitute in to rate law from slow step:
Rate = k[I]2[H2] = k(Keq[I2])[H2] = k'[I2][H2]
Reaction Order =
1+1=2
31
Catalysts

Speed up a reaction without being changed (they are consumed in one step of a
mechanism then produced again in a later step)

They work by changing the mechanism of a reaction to a process with a lower
activation energy (Ea)
These peaks indicate there were
two activated complexes
The graph with the higher peak is NOT catalyzed.
The graph with the lower peaks has been catalyzed.
A catalyst lowers the required activation energy.
This trough indicates one
reaction intermediate; they
are slightly more stable
hence the dip in energy

You must be able to define catalysts and intermediates and identify them within
reaction mechanisms:
Example:
Reaction:
2H2O2(aq)
Mechanism: H2O2 + I
H2O2 + OI
I

2H2O(l) + O2(g)


H2O + OI
H2O + O2 + I
(slow step)
(fast step)
o OI- is an intermediate (produced in one step, consumed in another);
intermediates are more stable (lower energy) than an activated complex, but
still very unstable
o I- is a catalyst (consumed in one step, produced in another); catalysts lower
the activation energy, Ea, barrier of a reaction
32

When writing a rate law for a slow step that contains a catalyst:
o Don’t include the catalyst because catalysts are never in the overall
chemical equation for a process
o Don’t include the catalyst because experiments confirm that changing the
concentration of a catalyst does not affect the rate of reaction (so if it
doesn’t affect the rate, why would it be in the rate law?)
33