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 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
Ls
or
mol
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
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)
B)
C)
D)
Catalysts
Temperature
Nature of the reactants
Concentration of the reactants
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?
Consider this aqueous redox reaction:
2 Ce4+ + Tl+ → 2Ce3+ + Tl3+
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?
Consider this aqueous redox reaction:
2 Ce4+ + Tl+ → 2Ce3+ + Tl3+
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+
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? Why is this “easier”
than before?
Because a catalyst increases the likelihood of a
reaction to proceed, in chemical terms we would
say it lowers the activation energy.
2 Ce4+ + Tl+ alone
with Mn2+
Why is this referred to as tunneling?
Why is this referred to as an alternative pathway?
Why does the activation energy decrease?
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.
CO
+ O2
CO
+O +O
CO2
+O
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?
Why are the crests in different spots?
What is this really showing?
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
Ea
ln k = ln A RT
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?
What does this mean to the value of k and the rate
of the reaction.
It is possible to compare the rate at two different
temperatures. We just calculate k at each
temperature and just subtract, or:
k2
Ea
1
1
ln
=
(
)
k1
R
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:
Ea
1
ln k = - (
)(
)+ ln A
R
T
which can be:
y=
m
x + b
Ea
1
ln k = - (
)(
)+ ln A
R
T
which can be: y =
m
x + b
How would this look to sketch it?
What does it take on the atomic level for reactants
to react?
Check this out!
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?
Also most chemical reactions are redox reactions,
so having low ionization energies would help. Why?
What, then, is the “nature” part of the nature of
reactants?
Why does the concentration of the reactants matter
to the rate of a reaction?
Check this out!
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?
rate = k [A]x[B]y[C]z…
The superscripts are called the “order” of the
substances, and the sum of the superscripts are
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? What about both [NO2] and [F2]?
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?
What would happen to the rate if the [] of
everything were doubled?
rate = k [A]x[B]y[C]z…
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?
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
What is the result of doubling [B]? How would you
say this on the AP test?
Between experiment 1 and 2 the [A] is held
constant and the [B] is doubled, and the rate of
formation of C doubles. So 1 x 2y = 2. So y = 1,
therefore the reaction is first order with respect to
[B].
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
Look at experiments 1 and 3. What stays the same,
and what changes?
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
What is the result of doubling [A]? How would you
say this on the AP test?
Between experiment 1 and 3 the [B] is held
constant and the [A] is doubled, and the rate of
formation of C increases by 4. So 2x x 1 = 4. So x = 2,
therefore the reaction is second order with respect
to [A].
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
What we know so far: rate = k [A]2[B]
Why did we not compare 2 and 3?
How can we determine k?
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
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
-2 s-1
k=
=
1.5
M
1.0 x 10-4 M3
What determines the label for k?
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
so, rate = 1.5 M-2 s-1 [A]2[B]
What is another way to write the label of k?
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
3
4
0.40 M
0.20 M
0.20 M
0.30 M
0.30 M
0.40 M
0.20 M
0.20 M
0.60 M
9.6 x 10-6 M min-1
2.4 x 10-6 M min-1
7.2 x 10-6 M min-1
Determine the rate law for this reaction.
Your turn: 2A + B + C → D + E
What is the general rate law form?
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
3
4
0.40 M
0.20 M
0.20 M
0.30 M
0.30 M
0.40 M
0.20 M
0.20 M
0.60 M
9.6 x 10-6 M min-1
2.4 x 10-6 M min-1
7.2 x 10-6 M min-1
Your turn: 2A + B + C → D + E
What is the order in respect to [A]?
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
3
4
0.40 M
0.20 M
0.20 M
0.30 M
0.30 M
0.40 M
0.20 M
0.20 M
0.60 M
9.6 x 10-6 M min-1
2.4 x 10-6 M min-1
7.2 x 10-6 M min-1
Your turn: 2A + B + C → D + E
What is the order in respect to [B]?
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
3
4
0.40 M
0.20 M
0.20 M
0.30 M
0.30 M
0.40 M
0.20 M
0.20 M
0.60 M
9.6 x 10-6 M min-1
2.4 x 10-6 M min-1
7.2 x 10-6 M min-1
Your turn: 2A + B + C → D + E
What is the order in respect to [C]?
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
3
4
0.40 M
0.20 M
0.20 M
0.30 M
0.30 M
0.40 M
0.20 M
0.20 M
0.60 M
9.6 x 10-6 M min-1
2.4 x 10-6 M min-1
7.2 x 10-6 M min-1
Your turn: 2A + B + C → D + E
What is the value of k?
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
3
4
0.40 M
0.20 M
0.20 M
0.30 M
0.30 M
0.40 M
0.20 M
0.20 M
0.60 M
9.6 x 10-6 M min-1
2.4 x 10-6 M min-1
7.2 x 10-6 M min-1
Your turn: 2A + B + C → D + E
What is the rate law?
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
3
4
0.40 M
0.20 M
0.20 M
0.30 M
0.30 M
0.40 M
0.20 M
0.20 M
0.60 M
9.6 x 10-6 M min-1
2.4 x 10-6 M min-1
7.2 x 10-6 M min-1
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?
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?
S + Fe - Se → S - Fe - Se → S - Fe + Se
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?
2 NO2 → N2O4
N2O4 + CO → NO + CO2 + NO2
-----------------------------------------NO2 + CO → NO + CO2
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?
Why does this seem better than only one step?
The rate determine step was rate = k2[NOBr2][NO]
How does this become rate = k[NO]2[Br2]?
In what way could a catalyst be involved?