AP Chemistry Chapter 14 Notes – Chemical Kinetics
Chemical kinetics is the study of the rates of chemical reactions, the factors that affect these rates, and the
mechanisms by which the reaction occurs.
The Rate of a Reaction
The rate of the reaction describes how fast the reactants are used up and the products form.
The rate of reactions are usually expressed in concentration per time, or in other words, moles per liter per time:
mol
or
mol
Ls
L min
Getting the measurements of the concentration while a chemical reaction is proceeding can be tricky. Perhaps for a
slow reaction involving an acid or base titration could be used. If the reaction involves a color change, then
equipment that can measure the intensity of the color by light absorption can be used. If a gas is given off or used
up, then a change in pressure can be used.
A balanced chemical reaction also works for the rate of change of the reactants or products.
Consider:
2N2O5(g) → 4NO2(g) + O2(g)
If NO2 is forming at the rate of 0.0072 mol L-1 s-1, then what is the rate of change for the other product and reactant?
What is the rate of the reaction?
It might be helpful to see in a graph what is happening:
Here’s the math that we are attempting to do, without needing calculus to take the derivative.
H2 + I2 → 2HI
Factors that Affect Reaction Rates
The rate of a chemical reaction depends on several factors, which further adds to the problem of finding ways to
measure the reaction rates.
A) Catalysts
B) Temperature
C) Nature of the reactants
D) Concentration of the reactants
The effects of catalysts
Catalysts are substances that increase the rate of a chemical reaction without being consumed. There are two basic
ways that catalysts can work.
1) Catalysts can lower the activation energy by reacting to form a reaction intermediate of lower potential
energy.
2) Catalysts can offer a surface for the reactants to “find” each other on.
A catalysts that increases the rate of a reaction by forming an intermediate takes an active role in the reaction
mechanism. Thus it must be in the same phase as the reactants, so it is sometimes called a homogeneous catalyst.
Consider this aqueous redox reaction:
2 Ce4+ + Tl+ → 2Ce3+ + Tl3+
Why is this redox? _____________________________________________________________________________
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What would it take for this reaction to occur as it is written? What would need to collide? What would need to
happen when they collide? _______________________________________________________________________
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Adding manganese (II) ions increases the rate of this reaction. It is thought to proceed like this:
Ce4+ + Mn2+ → Ce3+ + Mn3+
Ce4+ + Mn3+ → Ce3+ + Mn4+
Tl+ + Mn4+ → Tl3+ + Mn2+
2Ce4+ + Tl+ → 2Ce3+ + Tl3+
How is the manganese (II) ion involved? ___________________________________________________________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
How can it be involved yet not consumed? __________________________________________________________
_____________________________________________________________________________________________
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Why is this “easier” than before? ________________________________________________________________
_____________________________________________________________________________________________
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Because a catalyst increases the likelihood of a reaction to proceed, in chemical terms we would say it lowers the
activation energy.
Why is this referred to as tunneling? ______________________________________________________________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
Why is this referred to as an alternative pathway? ____________________________________________________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
Why does the activation energy decrease? __________________________________________________________
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Another type of catalyst does not take part in the reaction, but offers a location for the reaction to occur. This
catalyst is in a different phase than the reactants (usually the solid phase) so it is called a heterogeneous catalyst
(also sometimes a contact catalyst).
Why is the solid phase going to be different than the reactants’ phase? ____________________________________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
Enzymes in your body work as catalysts to increase the rate of the reactions in your body. Without them the
reactions in our body would happen at the rate of a tortoise.
Consider the catalyst bed in your catalytic converter.
The effects of temperature
Temperature is the measure of the average kinetic energy of the particles (relative to absolute zero) in the system.
It takes energy when atoms collide to cause the reactions (Ea), so if they do not collide with enough energy there will
not be a reaction.
So roughly temperature can measure the relative ability of the atoms to react. At low temperatures only a few atoms
will react even if a lot collide and at high temperatures a lot more will react when they collide.
Where is the Ea? ______________________________________________________________________________
_____________________________________________________________________________________________
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Why are the crests in different spots? ______________________________________________________________
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What is this really showing? _____________________________________________________________________
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Svante Arrhenius calculated the relationship between activation energy and temperature and the amount of possibly
productive collisions:
Ea
ln k = ln A RT
k = the specific rate constant of the reaction
A = a constant that represents the fraction of the collisions with the proper orientation when all concentrations are 1
M
R and T are as before
Notice, the larger the Ea the smaller the value of k, meaning the slower the reaction takes. Why? ______________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
What happens to Ea when a catalyst is present? ______________________________________________________
_____________________________________________________________________________________________
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What does this mean to the value of k and the rate of the reaction? _______________________________________
_____________________________________________________________________________________________
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It is possible to compare the rate at two different temperatures. We just calculate k at each temperature and just
subtract, or:
k2
ln
Ea
=
k1
1
(
R
1
-
T1
)
T2
Try this: what is the change in rate if the activation energy is 50 kJ mol -1 and the reaction is raised from 300K to
310K?
For a given reaction the rate constant, k, is 9.16 x 10 -3 s-1 at 0.0°C and the activation energy is 88.0 kJ mol-1. What
is the rate constant at 2.0°C?
At 600. K the value of k is 1.60 x 10-5 s-1. When the temperature is raised to 700. K, the value of k increases to 6.36
x 10-3 s-1. What is the Ea?
Determining the activation energy based on just two temperatures is not necessarily accurate.
But if the equation is rearranged a little, something interesting happens:
which can be: y =
m
x + b
How would this look to sketch it?
The effects of the nature of the reactants.
What does it take on the atomic level for reactants to react? ___________________________________________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
To facilitate a reaction atoms need to be mobile and as the reactions occur where different atoms can touch each
other it helps to have a high surface area.
What are the situations in which these conditions exist?
_____________________________________________________________________________________________
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Also most chemical reactions are redox reactions, so having low ionization energies would help. Why?
_____________________________________________________________________________________________
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What, then, is the “nature” part of the nature of reactants? _____________________________________________
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The effects of the concentration of the reactants.
Why does the concentration of the reactants matter to the rate of a reaction?
_____________________________________________________________________________________________
_____________________________________________________________________________________________
Because the concentration of the reactants has a HUGE effect on the rate of the reaction, it is common to describe
the rate of a reaction in terms of the concentrations and a rate constant, k.
rate = k [A]x[B]y[C]z…
What is the concentration measured in? ____________________________________________________________
_____________________________________________________________________________________________
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The superscripts are called the “order” of the substances, and the sum of the superscripts is called the “order of the
reaction” or “order overall”. The superscripts DO NOT come from the coefficients in front of the substance. If they
happen to match, it is purely coincidental.
Let’s look as some examples:
3NO(g) → N2O(g) + NO2(g)
the rate was determined to be = k[NO]2
This reaction is second order with respect to [NO] and second order overall. This means if we doubled [NO] the
reaction would increase by 4 times. This relationship is determined experimentally at first.
Why don’t the products appear in the rate law? ______________________________________________________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
Let’s look as some examples:
2NO2(g) + F2(g) → 2NO2F(g)
rate = k[NO2][F2]
This reaction is first order with respect to [NO2] and first order with respect to [F2], and second order overall.
What would happen to the rate if [NO2] was doubled? ________________________________________________
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What about both [NO2] and [F2]? _________________________________________________________________
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Let’s look as some examples:
H2O2(aq) + 3I-(aq) + 2H+(aq) → 2H2O(l) + I3-(aq)
rate = k[H2O2][I-]2
What is the order of each reactant and the order of the reaction? _________________________________________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
What does it mean to be zero in order? _____________________________________________________________
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What would happen to the rate if the [] of everything were doubled? _____________________________________
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Things you need to know about k:
1. k can only be determined experimentally, the balanced equation will not be useful
2. the units of k will change depending on the overall order of the reaction
3. k does NOT change if the [] changes or as time changes
4. k does depend on temperature, so often the temperature of the reaction is specified (or assumed 298 k)
5. k does change if a catalyst is added
Consider a simple reaction : A + 2B → C
From experiments we get:
Experiment
Initial [A]
Initial [B]
Initial Rate of
Formation of C
1
1.0 x 10-2 M
1.0 x 10-2 M
1.5 x 10-6 M s-1
2
1.0 x 10-2 M
2.0 x 10-2 M
3.0 x 10-6 M s-1
3
2.0 x 10-2 M
1.0 x 10-2 M
6.0 x 10-6 M s-1
let’s find the k and the orders:
rate = k [A] x[B]y
Look at experiments 1 and 2. What stays the same, and what changes? ___________________________________
_____________________________________________________________________________________________
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What is the result of doubling [B]? How would you say this on the AP test?
_____________________________________________________________________________________________
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Look at experiments 1 and 3. What stays the same, and what changes? ___________________________________
_____________________________________________________________________________________________
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What is the result of doubling [A]? How would you say this on the AP test?
____________________________________________________________________________________________
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What we know so far:
rate = k [A]2[B]
Why did we not compare 2 and 3? _______________________________________________________________
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How can we determine k? ______________________________________________________________________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
rate = k [A]2[B]
from experiment 1:
1.5 x 10-6 M s-1 = k (1.0 x 10-2 M)2(1.0 x 10-2 M)
1.5 x 10-6 M s-1
= 1.5 M-2 s-1
k=
1.0 x 10-6 M3
What determines the label for k? __________________________________________________________________
_____________________________________________________________________________________________
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What is another way to write the label of k? _________________________________________________________
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Your turn: 2A + B + C → D + E
From experiments we get:
Experiment
Initial
[A]
Initial
[B]
Initial
[C]
Initial Rate of Formation
of E
1
0.20 M
0.20 M
0.20 M
2.4 x 10-6 M min-1
2
0.40 M
0.30 M
0.20 M
9.6 x 10-6 M min-1
3
0.20 M
0.30 M
0.20 M
2.4 x 10-6 M min-1
4
0.20 M
0.40 M
0.60 M
7.2 x 10-6 M min-1
Determine the rate law for this reaction.
What is the general rate law form?
What is the order in respect to [A]?
What is the order in respect to [B]?
What is the order in respect to [C]?
What is the value of k?
What is the rate law?
What if you started a reaction and wanted to know the concentration of a reactant after some time, like “when would
half the reactants be left?”
To do this we use a different rate expression, called the integrated rate law (who loves calculus?), but the form of the
integrated rate law changes depending on the order of the reaction.
For our use, we will use the hypothetical equation of aA → products
aA → products: first order in respect to A and overall
ln ([A]0/[A]) = a k t
where [A]0 = initial [A]
[A] = [A] at the time of interest
a = coefficient of A
k = rate constant
t = time of interest
for example: a radioactive atom decays to form less energetic particles, the reaction is first order overall and has a
rate constant of 0.0450 s-1
What is the half life of this atom?
ln ([A]0/[A]) = a k t
aA → products: second order in respect to A and overall
1/[A] - 1/[A]0 = a k t
where [A]0 = initial [A]
[A] = [A] at the time of interest
a = coefficient of A
k = rate constant
t = time of interest
for example: a reaction of A and B is second order in respect to A and second order overall, and has a rate constant
of 0.622 M-1 s-1
What is the half life of 4.10x 10-2 M of A was initially present?
1/[A] - 1/[A]0 = a k t
aA → products: zero order in respect to A and overall
[A] = [A]0 - a k t
where [A]0 = initial [A]
[A] = [A] at the time of interest
a = coefficient of A
k = rate constant
t = time of interest
Now what would the units of k be?
Reaction Mechanisms
Chemists describe reactions occurring when reactants collide into each other. This is called collision theory.
However, not every collision guarantees a reaction. Kinetic energies, ionization energies, and orientation are
involved.
Some collisions are perfectly elastic and no reaction occurs, just deflection of the particles. Sometimes an inelastic
collision occurs long enough for a reaction to actually take place.
Consider CO2:
What shape is this molecule?
How would all three atoms need to collide?
Is that likely?
Consider this simple reaction: S + FeSe → Se + FeS
Consider this mechanism:
S + Fe - Se → S - Fe - Se → S - Fe + Se
reactants → transition state → products
What orientations would not work?
Consider I- + CH3Cl → CH3I + Cl-
What orientations would not work?
Would this be more or less likely to occur than the first equation?
The preceding examples are considered very simple because they only involve 1 step (1 collision between two
particles). Most reactions occur in many steps. These steps are called elementary steps and comprise the entire
reaction mechanism.
A reaction mechanism cannot occur faster than the slowest elementary step. This is called the rate determining
step.
The reaction order is equal to the coefficients of the rate determining elementary step.
if rate = k[A]x[B]y, then the rate determining step is xA + yB → products
This can help us determine a mechanism by giving a guide to the rate determining step.
Consider NO2(g) + CO(g) → NO(g) + CO2(g)
and rate = k [NO2]2
So the rate determining step is
2 NO2 → N2O4 (probably)
so what’s left?
N2O4 is called a reaction intermediate. What is characteristic of a reaction intermediate? _____________________
_____________________________________________________________________________________________
_____________________________________________________________________________________________
Consider NO2(g) + CO(g) → NO(g) + CO2(g)
and rate = k [NO2]2
Careful study has not detected N2O4, but has found NO3 in this reaction. How does that change things?
Consider 2NO(g) + Br2(g) → 2NOBr(g)
and rate = k [NO]2[Br2]
A collision involving only two particles seems more likely than one that involves three. This time the rate
determining step is the second.
NO + Br2 → NOBr2
NOBr2 + NO → 2NOBr (rate step)
2NO + Br2 → 2NOBr
How does this match the rate law? ________________________________________________________________
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Why does this seem better than only one step? ______________________________________________________
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The rate determine step was rate = k2[NOBr2][NO]
How does this become rate = k[NO]2[Br2]? _________________________________________________________
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In what way could a catalyst be involved? __________________________________________________________
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