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Collision Theory
and Rate of Reaction
6.4
Experimental observations and rate law equations describe what happens to the rate of
reaction when various factors such as initial concentration are changed. But how do reactions actually occur? Why do some reactions happen quickly and others slowly? For
example, why does potassium react so much more quickly than calcium, in water? Why can
butane gas from a lighter mix with air with no visible reaction until a spark is provided?
Rates of reaction can be explained with collision theory. Molecules are held together
with chemical bonds. According to collision theory, chemical reactions can occur only
if energy is provided to break those bonds; the source of that energy is the kinetic energy
of the molecules.
Concepts of the Collision Theory
•
A chemical system consists of particles (atoms, ions, or
molecules) that are in constant random motion at various
speeds. The average kinetic energy of the particles is
proportional to the temperature of the sample. Figure 1
shows the distribution of kinetic energies among particles
in a sample at two different temperatures.
•
A chemical reaction must involve collisions of particles
with each other or the walls of the container.
•
An effective collision is one that has sufficient energy
and correct orientation (alignment or positioning) of
the colliding particles so that bonds can be broken and
new bonds formed.
•
Ineffective collisions involve particles that rebound from
the collision, essentially unchanged in nature.
•
The rate of a given reaction depends on the frequency
of collisions and the fraction of those collisions that are
effective.
Number of Molecules with a
Particular Kinetic Energy
Kinetic Energy Distribution at Two Temperatures
T2
Kinetic Energy
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T 2 > T1
T1
Figure 1
Temperature is a measure of the
average kinetic energy of the particles. This graph shows how the
distribution of kinetic energies
changes when a substance is heated
or cooled. At any temperature there
are some particles with low kinetic
energy and some with high kinetic
energy. The higher the temperature,
the more particles there are with
higher kinetic energies.
Chemical Kinetics 383
The effects of these factors can be explained in terms of the last statement, which can
be expressed mathematically as
rate frequency of collisions fraction of collisions that are effective
For example, consider a system in which the collision frequency is 1000 collisions/s and
the fraction of effective collisions is 1/100 (meaning that only 1 collision of every 100 results
in reaction, and the other 99 collisions involve the molecules simply rebounding without
changing).
The reaction rate, expressed in terms of number of reactions/s, is
reactions per second
Figure 2
The reaction between nitrogen
and oxygen in air to produce
yellow-brown nitrogen dioxide
(N2 2 O2 → 2 NO2) is a very slow
reaction under normal conditions, but
occurs more quickly in the high temperatures of automobile engines. The
result can be a yellow haze of toxic
gas.
10 reactions/s
What happens if the collision frequency is increased to 5000 collisions/s?
reactions per second
5000 co llisions
1 reaction
1s
100 collisions
50 reactions/s
The rate becomes 50 reactions/s.
What happens if the fraction of effective collisions is increased to 1/20 from 1/100?
reactions per second
An Analogy for
Activation Energy
1000 col lisions
1 reaction
1s
100 collisions
1 reaction
1000 col lisions
20 collisions
1s
50 reactions/s
Potential Energy
The rate again becomes 50 reactions/s.
“activation”
energy
Eact
Clearly, increasing either the collision frequency or the fraction of effective collisions
will increase the rate.
net
energy
change
A
B
Progress of Trip
Figure 3
On a trip from A to B there is a net
decrease in overall (net) energy, but
there must be an initial increase in
potential energy (activation energy)
for the trip to be possible.
activation energy the minimum
increase in potential energy of a
system required for molecules to
react
384 Chapter 6
Activation Energy
Consider an empty soft drink can. In the nitrogen–oxygen air mixture in the can, an
estimated 1030 molecular collisions occur every second. This is an absolutely enormous
number: one thousand billion billion billion events per second. When nitrogen and
oxygen react, one of the products they form is a colourful, very toxic gas called nitrogen
dioxide, NO2(g) (Figure 2). However, our air remains relatively colourless so, although
nitrogen and oxygen can react at normal conditions, they must do so very slowly. How
can we use collision theory to explain this observation?
If each collision produced a reaction when reacting substances were combined, the rate
of any reaction would be extremely rapid, appearing essentially instantaneous. In the
pop can, the nitrogen and oxygen molecules would react completely to form nitrogen
dioxide in about 5 10–9 s (five-billionths of a second). Since the actual rate is too slow
to be measurable, we conclude that normally only an extremely tiny fraction of the collisions between oxygen and nitrogen molecules actually produce the new substances.
The collision frequency is high but the fraction of effective collisions is small.
The theoretical explanation for the empirical evidence involves the concept of
activation energy — the minimum energy with which particles must collide before
they can rearrange in structure, resulting in an effective collision. The concept of activation energy, Eact, can be illustrated by an analogy with gravitational potential energy.
Consider a billiard ball rolling (friction-free) on a smooth track shaped as shown in
Figure 3. The ball leaves point A moving toward the right. As it rises on the uphill portion of the track it slows down: Kinetic energy converts to potential energy. The ball can
only successfully overcome the rise of the track and proceed to point B if it has enough
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Section 6.4
initial speed (kinetic energy). We could call this situation an effective trip. The minimum kinetic energy required is analogous to the activation energy for a reaction. If the
ball does not have enough kinetic energy it will not reach the top of the track and will
just roll back to point A. This is analogous to two molecules colliding without enough
energy to rearrange their bonds: They just rebound. Since the activation energy sometimes seems to prevent reaction, it is often called an activation energy barrier.
Note that a ball that returns to point A will have the same potential and kinetic energy
it began with, but a ball that makes it to point B will have a different combination of
energies: It will have less potential energy because it is at a lower point on the diagram,
but it will have more kinetic energy because it will be moving faster. This billiard-ball
example is an analogy for the energy change taking place during an exothermic reaction,
in which energy is released to the surroundings. In kinetic theory terms, this means that
the energy is released to any other nearby molecules. These other molecules then move
faster, collide with more energy, and become more likely to react. The reaction, once
begun, is self-sustaining as long as enough molecules remain to make collisions likely.
For example, the energy released when wood burns allows the
reaction to proceed unaided by external sources of energy.
Exothermic reactions often drive themselves in this way once
begun, as shown in Figure 4.
The concept of activation energy fits well with the fundamental
idea from collision theory that rate is proportional to collision frequency and the proportion of effective collisions:
DID YOU
KNOW
?
Get in Line!
Some collisions appear to be
energetic enough for reaction to
occur but are still not effective
because the molecules do not
collide with the correct alignment.
The smaller and simpler in structure
the molecules, the more likely it is
that a collision will be effective.
rate frequency of collisions fraction of collisions that are effective
Increasing the concentration or the surface area of the reactants
increases the collision frequency — more collisions occur per unit
time. Changing the nature of the reactants or using a catalyst
changes the size of the activation energy barrier, making it easier
or more difficult for molecules to react. This has the effect of
changing the fraction of effective collisions. Increasing temperature
has a particularly dramatic effect on the reaction rate because it increases both collision
frequency and the fraction of effective collisions. Since the molecules are moving more
quickly, not only do they collide more often but they also collide with more energy, on
average, making it more likely that bonds will break.
SUMMARY
Figure 4
Once a fire has begun, the
exothermic combustion of wood is
self-sustaining and can destroy
large areas of forest.
Factors Affecting Rate and
Collision Theory
Each of the five factors that affect rate increases either collision frequency or the
fraction of collisions that are effective (or both) to increase rate of reaction.
rate collision frequency
concentration
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fraction effective
nature of reactant
surface area
catalyst
temperature
temperature
Chemical Kinetics 385
Potential Energy Changes During an Endothermic Reaction
Figure 5
Over the progress of an effective
collision between molecules in the
gas phase, the potential energy
increases to a maximum at the point
of closest approach, then decreases
to a final value higher than the initial
energy (as the reaction is
endothermic). The potential energy
gain of the molecules comes from
conversion of kinetic energy. The
overall reaction would lower the
temperature of the system and
surroundings.
activated complex
activation
energy
Ep
products
net potential energy change, H
reactants
Reaction Progress
Consider the reaction of hydrogen and iodine molecules, a single-step reaction at
high temperatures, plotted as potential energy of the molecules versus progress of the reaction (Figure 5).
H2(g) I2(g) → 2 HI(g)
activated complex an unstable
chemical species containing partially
broken and partially formed bonds
representing the maximum potential
energy point in the change; also
known as transition state
We can discuss the progress of this reaction in terms of molecular collisions, by moving
from left to right along the plot shown in Figure 5. Along the flat region to the left, the
molecules are moving toward each other, but are still distant from each other. As the
molecules approach more closely, they are affected by repulsion forces and begin to slow
down, as some of their kinetic energy is changed to potential energy (stored as a repelling
electric field between them). If the molecules have enough kinetic energy, they can
approach closely enough for their bond structures to rearrange to form an activated
complex. The activated complex is an unstable molecule with a particular geometry. It
is unstable because it possesses the maximum potential energy possible.
When the reacting system reaches the activated complex stage it may reverse to reactants, or it may continue to form product molecules. In either case, repulsion forces push
the molecules apart, converting potential energy to kinetic energy. Overall, there are
potential energy changes as bonds are broken and formed and products are formed. If the
energy difference is measured at constant pressure (the usual situation for a reaction
open to the atmosphere), it is called the enthalpy change or enthalpy of reaction, ∆H,
which is 53 kJ/mol for the endothermic formation of hydrogen iodide.
When the products of a reaction have higher potential energy than the reactants, they
will have lower kinetic energy (temperature). In their subsequent collisions with other
molecules in the system and in the surroundings, they will tend to decrease the speed of
molecules they collide with, resulting in a drop in the temperature of the system. This
is why endothermic reactions have the effect of cooling their surroundings.
In other types of reactions, the final potential energy may be lower than the initial
potential energy, meaning that the reaction is exothermic. In such cases the enthalpy of
the system decreases, so ∆H has a negative value. For example, the thermochemical
equation for the production of carbon dioxide and nitric oxide from carbon monoxide
and nitrogen dioxide is
CO(g) NO2(g) → CO2(g) NO(g)
∆H 227 kJ
During such reactions the temperature of the surroundings tends to rise, as heat is
released to the surroundings in the progress of the reaction (Figure 6).
386 Chapter 6
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Section 6.4
Potential Energy Changes During an Exothermic Reaction
activated complex
reactants
activation
energy
Ep
net potential
energy change,
H
products
Reaction Progress
Figure 6
Over the progress of this exothermic
reaction, the potential energy (or
enthalpy) increases to a maximum
as the activated complex forms, then
decreases to a final value lower than
the initial energy. The potential
energy lost by the molecules is converted to kinetic energy. The overall
reaction would raise the temperature of the surroundings.
Practice
Understanding Concepts
Potential Energy Diagram for System R P
1. (a) In your notebook sketch the graph shown in
Figure 7, and add labels for the axes.
(b) What does each curve represent?
(c) What type of reaction is occurring in terms of
energy flow to or from the surroundings?
(d) What does each number (i, ii, iii) represent?
P
(i) (ii)
(iii)
R
Figure 7
Reaction Mechanisms
How can you have a “collision” involving one particle? Consider the reaction in which
hydrogen peroxide decomposes to water and oxygen gas. How do some molecules obtain
enough energy to decompose? Some reaction mechanisms involve a step in which a
single molecule apparently hits container walls or any other particle in order to convert
enough energy from kinetic to potential for the molecule to decompose. Still other molecules absorb light energy in a reaction step to break a bond, resulting in two or more
atoms, such as the following reaction:
Cl2 light → 2 Cl•
Common sense, and calculations, indicates that collisions of three particles simultaneously must be much less frequent than two-particle collisions and that any collision
involving four or more particles is extremely unlikely indeed. Imagine a circle of students tossing Velcro-covered Ping-Pong balls toward the centre of the circle. The chances
of any two Ping-Pong balls colliding and adhering in the air is small, and the probability of three being in the same place at the same time is much smaller. Think how
incredibly unlikely it would be for four balls to collide simultaneously.
Scientists believe that most chemical reactions actually occur as a sequence of
elementary steps. This overall sequence is called the reaction mechanism.
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elementary step a step in a reaction mechanism that only involves
one-, two-, or three-particle collisions
reaction mechanism a series of
elementary steps that makes up an
overall reaction
Chemical Kinetics 387
DID YOU
KNOW
?
Uni, Bi, Tri ...
If only one reactant molecule is
involved in a reaction mechanism
(e.g., the decomposition of ozone:
O3(g) → O2(g) O(g)), the step is
called unimolecular.
If two reactant molecules are
involved (e.g., H2(g) I2(g) → 2 HI(g)),
the step is called bimolecular.
An automobile assembly line is a reasonable analogy to a reaction mechanism, because
a car is not built in a single, concerted step, but rather in a sequence of steps. Imagine a
car being assembled in a plant by a series of workers. One assembles the chassis, another
adds wheels, another the seats, and so on. The worker who controls the overall rate of production of cars will not necessarily be the first or the last but, rather, the slowest worker.
This worker could be called the “rate-determining worker.” If the slowest worker is at the
beginning of the line, other workers will wait for cars to arrive at their station; if the
slowest worker is at the end, partially assembled cars will stack up waiting to be finished. Adding workers—increasing concentration—at the fast steps will have no effect
on car production. However, increasing the “concentration” of workers at the slowest
step should increase the rate of production.
A chemical example of a reaction mechanism is the oxidation of hydrogen bromide,
which is rapid between 400°C and 600°C. This reaction has been studied extensively
because all substances are simple molecules and in the gas phase.
4 HBr(g) O2(g) → 2 H2O(g) 2 Br2(g)
rate-determining step the slowest
step in a reaction mechanism
reaction intermediates molecules
formed as short-lived products in
reaction mechanisms
It is highly unlikely that this reaction would occur in a single step, because a total of
five reactant molecules would have to collide simultaneously with the proper alignment
and sufficient energy to break and form new bonds. Experimental evidence shows that
increasing the concentration of oxygen increases the reaction rate, just as we would
expect. Since four molecules of HBr are involved for every molecule of O2, it seems logical to expect that a change in HBr concentration would have a much greater effect on
the rate, but measurement shows this is not the case. Quantitatively, the empirically
determined rate equation for this system is
r k[HBr][O2]
Relative Potential Energy Ep
Oxidation of HBr(g)
Eact
4 HBr + O2
H
2 H2O + 2 Br2
Reaction Progress
Figure 8
Over the progress of this reaction,
the potential energy increase necessary to reach the first activated
complex stage is the greatest
increase required, so this is the
rate-determining (slowest) step.
Energy released as kinetic energy
past this point is sufficient to quickly
carry the reaction mechanism to
completion.
388 Chapter 6
We explain by theorizing that the reaction occurs in the following elementary steps,
each of which involves a two-particle collision occurring at a different rate. (Note that
the steps sum to give the overall equation for the reaction.)
HBr(g) O2(g)
→ HOOBr(g)
(slow)
HOOBr(g) HBr(g)
→ 2 HOBr(g)
(fast)
2 {HOBr(g) HBr(g)
→ H2O(g) Br2(g)}
(fast)
4 HBr(g) O2(g)
→ 2 H2O(g) 2 Br2(g)
The theoretical interpretation is that the first elementary step is relatively slow because
it has a fairly high activation energy. The rate of the overall reaction is basically controlled
by this step, just as the slowest worker determined the overall rate at which cars were produced. The second step cannot use HOOBr any faster than the first step can produce
it, so the rate of the reaction overall is the same as the rate of the slowest step — in this
case, the first. The slowest reaction step in any reaction mechanism is called the ratedetermining step. Substances such as HOBr and HOOBr — which are formed during the
reaction but immediately react again and are not present when the reaction is complete
— are called intermediate products or reaction intermediates. On a potential energy
diagram like Figure 8, unstable activated complexes exist at the “peaks” and slightly more
stable reaction intermediates exist at the small “valleys” within a mechanism.
How do chemists determine a mechanism for a reaction? The first step is to perform
experiments and thereby determine a rate equation. There is a direct correlation between
the exponents in the rate equation and the equation coefficients in the rate-determining
step in the mechanism.
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Section 6.4
For example, in the hydrogen bromide–oxygen system, the empirically determined
rate equation has the exponent 1 on each of the concentrations of reactants.
r k[HBr]1[O2]1
Thus, the rate depends on the concentration of one molecule of each of the reactants.
The reaction coefficients of the rate-determining step are also 1 for each of the molecules.
1 HBr 1 O2 → reaction intermediate
In general, if the empirically determined rate equation is
r [molecule X]m[molecule Y]n
then the rate-determining step in the mechanism must be
m X n Y → products or reaction intermediates
Reaction mechanisms are only “best guesses” at the behaviour of molecules, but there
are three rules that must be followed in proposing a mechanism:
•
each step must be elementary, involving no more than three reactant (and more
usually only one or two) molecules;
•
the slowest or rate-determining step must be consistent with the rate equation; and
•
the elementary steps must add up to the overall equation.
It is often possible to create two or more mechanisms, each of which could account
for the empirically derived rate equation.
Finding the Rate-Determining Step
SAMPLE problem
Consider the decomposition of dinitrogen pentoxide
2 N2O5(g) → 2 N2O4(g) O2(g)
(a) What would the rate equation be if the reaction occurred in a single step?
(a) The exponent in the rate equation would be the same as the coefficient on the reactant. Therefore, the rate law equation would be
r k[N2O5]2
(b) The actual experimentally derived rate equation is
r k [N2O5 ]1
What is the rate-determining step?
(b) Because the coefficient on the reactant must be the same as the exponent in the rate
equation, the rate-determining step must be
1 N2O5(g) → some product or reaction intermediate
(c) Suggest a possible mechanism and indicate the slowest step.
(c) The only step that we are sure of is the rate-determining step; the others are guesses.
The following is a possibility.
N2O5 → N2O4 O
O N2O5 → N2O4 O2
(slow)
(fast)
The rate-determining step is the slow step, which could occur at any point in the
mechanism. Each step in the mechanism involves two or fewer reactant molecules,
and the steps sum up to the overall equation.
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Chemical Kinetics 389
Example
Consider the overall reaction involving three elements as reactants, and a compound as
the product:
X 2 Y 2 Z → XY2Z2
When a series of reactions is performed with different initial concentrations of reactants,
the results are as follows:
• doubling the concentration of X has no effect on the overall rate
• doubling the concentration of Y multiplies the overall rate by 4
• doubling the concentration of Z doubles the overall rate
State
(a)
(b)
(c)
(d)
the rate law equation for this system;
the rate-determining step;
a possible mechanism, indicating the slow step; and
a possible reaction intermediate in your mechanism.
Solution
(a) From the empirical information provided,
r [X]0; r [Y]2; and r [Z]1, giving a rate equation of
r k[Y]2[Z]1
(b) Therefore, the rate-determining step must be
2 Y 1 Z → some product(s)
(c) Any mechanism consistent with the above rules is acceptable. For example, one
possibility is
X Z
→ XZ
(fast)
2Y Z
→ Y2Z
(slow)
→ XY2Z2
__________________________________
XZ Y2Z
(fast)
X 2 Y 2 Z → XY2Z2
(d) Two possible reaction intermediates are XZ and Y2Z .
LAB EXERCISE 6.4.1
The Sulfur Clock (p. 405)
Analyze the evidence — how long
it takes for the “X” to disappear —
to find the order of the thiosulfate
reaction.
Practice
Understanding Concepts
2. Consider the overall reaction in which two elements combine to form a compound:
A2 2 B → 2 AB
When a series of reactions is performed with different initial concentrations of
reactants, the results are as shown in Table 1:
Table 1 Formation of AB
Trial
Initial [A2]
Initial [B]
Initial rate of production of AB
1
0.10
0.10
3 104
2
0.20
0.10
6 104
3
0.10
0.20
3 104
State (a) the rate law equation for this system;
(b) the rate-determining step; and
(c) a possible mechanism, indicating the slowest step.
390 Chapter 6
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Section 6.4
Section 6.4 Questions
Understanding Concepts
Making Connections
1. What is the difference between an elementary step and a
rate-determining step?
2. Consider the following mechanism, in which A, B, and E
may be elements or compounds, and C, D, and F are compounds:
(1) 2 A B → C
(2) C → D
(3) D E → F
(a) What is the overall equation?
(b) Which step is most likely to be the rate-determining
step? Explain.
3. (a) What is the (overall) activation energy for the following
reaction in the potential energy diagram in Figure 9?
reactants → products
(b)
(c)
(d)
(e)
(f)
What is the reaction enthalpy (∆H) for the reaction?
What is the rate-determining step for the reaction?
Is the reaction exothermic or endothermic?
Which letters represent activated complexes?
Which letters represent reaction intermediates?
4. (a) Consider the chemical reaction that happens in an
automobile engine when hydrocarbon molecules
(mostly octane) combine with oxygen molecules to
produce carbon dioxide and water vapour. Imagine
that you have an extraordinary microscope that
enables you to look at individual molecules. Describe
what you see as the chemical reaction occurs. Make
sure that you describe the positions and movements of
the molecules as specifically as possible.
(b) Compare your answer with what you gave for Reflect
on your Learning, question 2, at the beginning of this
chapter. How has your understanding changed?
5. Ozone is a molecule that is helpful in the upper atmosphere
but harmful at ground level. Research on the Internet to
answer the following questions:
(a) Why is ground-level ozone a problem?
(b) What are the mechanisms of the reactions that lead to
its production?
(c) What is the connection between the production of
ozone and kinetics?
GO
Relative Potential Energy Ep (kJ)
A Multistep Reaction
60
50
40
X
V
Z
W
Y
6. John Polanyi, a scientist at the University of Toronto, won
the 1986 Nobel Prize for his work in chemical kinetics.
Imagine that you are a newspaper reporter describing
Polanyi’s work in an article written the day after he received
the prize. Include in your article a description of his area of
research and the experiments that he performed.
GO
0
www.science.nelson.com
www.science.nelson.com
reactants
–35
products
Reaction Progress
Figure 9
Potential energy diagram
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Chemical Kinetics 391