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Rate of Reaction

By farisha9294 Nov 02, 2014 10674 Words
Kinetics

Kinetics
6.1

Rates of reaction

6.2

Collision theory

6

16.1 Rate Expression (AHL)
16.2 Reaction mechanism (AHL)
16.3 Activation energy (AHL)

6.1 Rates of reaction
6.1.1 Define the term rate of reaction.
6.1.2 Describe suitable experimental
procedures for measuring rates of
reactions.
6.1.3 Analyse data from rate experiments.
© IBO 2007

Figure 601  An explosion is a quick reaction

D

ifferent chemical reactions occur at different rates
(i.e. speeds). Some, such as the neutralisation of a
strong acid by a strong base in aqueous solution, take place very rapidly whilst others, such as the rusting of iron, take place far more slowly. Rates of reactions should not be
confused with how far a reaction goes - this is determined
by equilibrium.
The rate of a chemical reaction is a measure of the speed
at which products are formed, measured as the change in
concentration divided by the change in time, so reaction
rate has units of mol dm–3 s–1. This is equal to the rate at which the reactants are consumed, so for a reaction:

R

[ ]
∆​[ R  ]​ 
P, rate = ____
​ ∆​  P   ​ 
​ = –​ _____
​ 
∆t
∆t

Figure 602  Corrosion is a slow reaction

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Chapter 6

CORE

Note the minus sign for the reactants, which is necessary
as the concentrations of reactants decreases with time
whereas the concentrations of products increases. Rate is
always positive.
The numerical value will vary according to the amount of
the substance involved in the stoichiometric equation, so
that in the reaction:

concentration. It is most common to compare initial
rates, that is, the gradient of the tangent to the curve at
t = 0. At this time the concentrations of the reagents are
accurately known, as is the temperature of the system. It
is also easiest to draw tangents at this time as this section of the curve is the most linear. Typical curves obtained for the consumption of a reagent and formation of a product
are shown in Figure 603.

MnO4– (aq) + 8 H+ (aq) + 5 Fe2+ (aq)
Mn2+ (aq) + 4 H2O (l) + 5 Fe3+ (aq)
The rate of appearance of Fe3+ is five times as great as
the rate at which MnO4– is consumed. The rate is usually
considered to apply to a product that has a coefficient of
one as the equation is usually written:

Exercise
1.

∆​[ Mn​O4​–​ ]​ _1______
[ 3+ ]
Rate = – ________
​ 
   
 
​= ​ 5 ​​ ∆​  ​Fe​  ​  ​ 
​ 
∆t
∆t
Or more simply for a reaction: a A

The equation for a reaction is:

4 NO2 (g) + 2 H2O (g) + O2 (g)

Any property that differs between the reactants and the
products can be used to measure the rate of the reaction.
Refer to Section 6.2. Whichever property is chosen, a
graph is drawn of that property against time and the rate
of reaction is proportional to the gradient of the curve or
line ignoring the sign. Changes in the gradient of similar
graphs illustrate the effect of changing conditions on the
rate of reaction, without the need to convert the units to
mol dm–3 s–1.

Gradient = rate at time t = x

A

∆​[ N​O​2​ ]​
– __
​ 1 ​ _______
​ 
   
 

2 ∆t

B

∆​[ ​O​ ​ ]​
– __
​ 1 ​ _____
​   2  
​ 
2 ∆t

C

∆​[ ​H2​ ​O ]​
– _______
​ 
   
​ 
∆t

∆​[ HN​O3​ ​ ]​
1 ​ ________
D​ __
​ 
   
​ 
2
∆t
2.

Property

In most cases the rate of reaction decreases with time
because the concentration of the reactants decreases with
time and the reaction rate usually depends on the reactant

4 HNO3 (g)

Which one of the following is not numerically equal
to the others?

b B, then

[ ]
[ ]
1 ​_____
Rate = __
​ 1 ​_____
​ ∆​  B   ​ 
​ = –​ __
​ ∆​  A   ​ 
​ 
a
∆t
∆t
b

Property

6.1

Which of the curves on the following graph shows
the greatest initial reaction rate?

Gradient = – initial rate

Gradient = – rate at time t = x
Gradient = initial rate

x

Time

Product formation

x

Time

Reactant consumption

Figure 603  The variation of reaction rate with time

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Kinetics
A
B
Concentration

rates
D

Iodate(V) ions oxidise iodide ions in acidic solution
to form iodine and water according to the equation

IO3– (aq) + 5 I– (aq) + 6 H+(aq)
3 I2 (aq) + 3 H2O (l)
If the number of moles of each reactant consumed
after one minute was measured, which would have
been consumed least?
A
B
C
D
4.

IO3–
I–
H+
They would all have been consumed to
the same extent.

The rate of reaction between hydrogen peroxide and
the iodide ions was measured by monitoring the
absorption of blue light by the iodine. If the equation
for the reaction is

H2O2 (aq) + 2 I–(aq) + 2 H+(aq)
a)
b)

5.

I2 (aq) + 2 H2O (l)

If 0.005 mol dm-3 of iodine is produced in
the first 2 minutes, what is the initial reaction
rate in mol dm-3 s-1?
What is the rate at which

i hydrogen peroxide is consumed?

ii iodide ions are consumed?

Explain why these are different.

The rate of reaction between zinc and sulfuric acid
is measured by weighing a zinc plate, which is then
placed into a beaker of the acid. Every 10 minutes
it is removed, rinsed, dried and reweighed. This is
continued until all of the acid is consumed.
a)
b)
c)

Sketch the graph you would expect for the
mass of the zinc plate against time in the acid.
At what point is the reaction rate the
greatest? How can you tell?
Suggest another way that the rate of this
reaction could have been measured.

There are a variety of techniques that can be used to
measure the rate of a chemical reaction and some of the
more common are described below. Any property that
changes between the start and end of the reaction can in
principle be used. It is however best if this changes by a
large amount compared to the limits of accuracy of its
measurement. It is also simpler to use quantitatively if the characteristic is directly proportional to the concentration of one or more components. For these reasons monitoring
the rate of reaction by observing a pH change is generally
not to be recommended, because the pH, being a
logarithmic scale, will only change by 0.30 for a change of
[H+] by a factor of 2.

CORE

C

Time

3.

Some techniques for measuring

In some techniques the time taken for a particular event
to occur may be used to measure the reaction rate (e.g. the
time taken for a piece of magnesium ribbon to dissolve
in a dilute acid). In these techniques it is important to
remember that the greater the time the smaller the rate of
reaction, i.e. the rate of reaction is inversely proportional to the time taken:

Rate ∝ _____
​  1   ​ 
Time
If the purpose of the investigation is simply to observe
the effect of some variable, such as the concentration
of a particular species, on the rate of reaction, then a
graph of the property proportional to concentration
against time, will suffice. If however the reaction rate is
required in standard units (mol dm–3 s–1) then it will be necessary to calibrate the system so as to produce graphs
of concentration against time.
Whichever technique is being used, it is important to keep
the reaction mixture at a constant temperature during the
reaction, because temperature has a great effect on the
rate of reaction. For this reason it is usual to immerse the reaction vessel in a water bath at the required temperature. It is also preferable to immerse the reactants in the water
bath, to allow them to reach the required temperature
before mixing. One reason for prefering ‘initial rate’ data is that the effect of an endo- or exothermic reaction on the temperature of the system is minimised.

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Chapter 6

Titration

CORE

TOK  The empirical nature of chemistry
I used to think all magpies were black and white.
In both England and New Zealand (my two “home”
countries – whatever that concept implies) the
magpie (very different species in both countries
– the New Zealand version is a lot heavier and
doesn’t have the long tail) are black and white birds.
Then there’s the magpie robin, a much smaller bird,
and the magpie goose - both also black and white.
Indeed in England Newcastle United football club
are nicknamed ‘the magpies’ because of their black
and white shirts. My simple world was turned on its
head when I arrived in Hong Kong and encountered
the blue magpie. Definitely a magpie shape, but
with bright blue as the dominant colour (along with
black and white) and a gorgeous long tail. I’m afraid
I was another victim of the fundamental problem of
inductive logic - you can never be sure. However, in
case you’re thinking you will just rely on deductive
techniques, reflect for a few seconds on how you
establish your premises.
All empirical science suffers in the same way - it is
quite easy to prove a theory wrong, but it is never
possible to prove it correct. Tomorrow somebody
might come up with a piece of evidence that
contradicts your theory and then it’s back to the
drawing boards. Falsification is so fundamental to
science that the Austrian philosopher, Karl Popper,
said that if you could not think of an experiment
that would disprove your theory, then it was not a
scientific theory.
In kinetics, the balanced equation tells us nothing
about how changing the concentrations will
affect the rate and we have to do experiments to
determine the rate equation. We can then postulate
mechanisms that would account for the empirical rate
equation and certainly we can rule out some possible
mechanisms, but even if we have a mechanism that
explains the rate equation (and sometimes there can
be more than one mechanism that does this, read
this chapter) we can never be sure that tomorrow
somebody will not come up with an alternative.

This involves removing small samples from the reaction
mixture at different times and then titrating the sample to
determine the concentration of either one of the reactants
or one of the products at this time. The results can then be used directly to generate a graph of concentration against
time. In its simplest form this is only really suitable for
quite slow reactions, in which the time taken to titrate
the mixture is insignificant compared to the total time
taken for the reaction. One common variant that helps
to overcome this difficulty is to quench the reaction
before carrying out the titration. This means altering the
conditions so as to virtually stop the reaction. This can
be done by rapidly cooling the reaction mixture to a very
low temperature or by adding an excess of a compound
that rapidly reacts with one of the reactants. If for example the reaction was that of a halogenoalkane with an alkali,
it could be quenched by running the reaction mixture
into an excess of a strong acid. This means that the time at which the sample of the reaction mixture was quenched is
much easier to determine.
Another example of a reaction that can be readily measured
by this technique is the rate of reaction of hydrogen
peroxide with iodide ions in acidic solution to produce
iodine and water. The amount of iodine produced can be
measured by titrating the mixture with aqueous sodium
thiosulfate. The reaction mixture can be quenched by
adding excess of an insoluble solid base, such as powdered
calcium carbonate, to neutralise the acid required for
reaction.

H2O2 (aq) + 2 H+ (aq) + 2 I– (aq)

2 H2O (l) + I2 (aq)

Collection of an evolved gas/
increase in gas pressure
The gas produced in the reaction is collected either in a gas syringe, or in a graduated vessel over water. The volume
of gas collected at different times can be recorded. This
technique is obviously limited to reactions that produce a
gas. In addition, if the gas is to be collected over water, this gas must not be water soluble. An alternative technique is
to carry out the reaction in a vessel of fixed volume and
monitor the increase in the gas pressure. These techniques
would be suitable for measuring the rate of reaction
between a moderately reactive metal (such as zinc) and an
acid (such as hydrochloric acid), or reaction of a carbonate with acid:

Zn (s) + 2 H+ (aq)

Zn2+ (aq) + H2 (g)

Na2CO3 (s) + 2 HCl (aq)
2 NaCl (aq) + CO2 (g) + H2O (l)

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Measurement of the mass of the
reaction mixture
The total mass of the reaction mixture will only vary if a
gas is evolved. To be really effective, the gas being evolved should have a high molar mass (i.e. not hydrogen), so
that there is a significant change in mass, also the gas
should not be significantly soluble in the solvent used.
This technique would be suitable for measuring the rate
of reaction between a metal carbonate (such as calcium
carbonate, marble chips) and an acid (such as hydrochloric
acid), by measuring the rate of mass loss resulting from
the evolution of carbon dioxide.

CaCO3 (s) + 2 H+ (aq)

A reaction that is often studied by this technique is
the reaction between propanone and iodine to form
iodopropanone. The yellow–brown iodine is the only
coloured species involved and so the intensity of blue light (or light of wavelength ≈450 nm if a spectrophotometer is
used) passing through the solution will increase with time
as the concentration of the iodine falls. Most instruments,
however, give a direct reading of absorbance which has
an inverse relationship to the transmitted light, so that
absorbance decreases with time.

CORE

Kinetics

CH3COCH3 (aq) + I2 (aq)
CH3COCH2I (aq) + H+ (aq) + I– (aq)

Ca2+ (aq) + H2O (l) + CO2 (g)

Electrical conductivity
Light absorption
If a reaction produces a precipitate, then the time taken
for the precipitate to obscure a mark made on a piece of
paper under the reaction vessel can be used as a measure
of reaction rate. For simple work comparison of the times,
keeping the depth of the liquid constant will suffice; e.g.
if the time taken doubles then the reaction rate is halved.
A reaction that is often studied by this technique is the
reaction between aqueous thiosulfate ions and a dilute
acid which gives sulfur dioxide, water and a finely divided
precipitate of sulfur.

S2O32– (aq) + 2 H+ (aq)

H2O (l) + SO2 (g) + S (s)

A convenient way to follow this reaction is to place a black cross/mark on a white piece of paper under the reaction
mixture and to measure the time taken for the finely
suspended yellow sulfur to obscure the cross.
If the reaction involves a coloured reactant or product,
then the intensity of the colour can be used to monitor
the concentration of that species. In its simplest form
this can be done by comparing the colour by eye against
a set of standard solutions of known concentration.
The technique is far more precise if an instrument that
measures the absorbance (which is directly proportional
to concentration – refer to the Beer–Lambert law in
Section A8.6, Chapter 12) such as a colorimeter or
spectrophotometer is available. If a colorimeter is used
then a filter of the complementary colour to that of the
coloured species should be chosen – an aqueous solution of a copper(II) salt is blue because it absorbs red light, so that it is the intensity of transmitted red light not blue light that will vary with its concentration. If a spectrophotometer is

used, then a wavelength near to the absorption maximum
of the coloured species should be selected.

The presence of ions allows a solution to conduct, so if
there is a significant change in the concentration of ions
(especially hydrogen and hydroxide ions which have
an unusually high conductivity) during the course of
a reaction, then the reaction rate may be found from
the change in conductivity. This is usually found by
measuring the A.C. resistance between two electrodes
with a fixed geometry, immersed in the solution. A
reaction that is suitable for this technique would be the
hydrolysis of phosphorus(III) chloride that produces
dihydrogenphosphate(III) ions, hydrogen ions and
chloride ions from non–ionic reactants.

PCl3 (aq) + 3 H2O (l)
H2PO3– (aq) + 4 H+ (aq) + 3 Cl– (aq)

Clock techniques
There are some reactions in which the product can
be consumed by further reaction with another added
substance. When all of this substance is consumed then
an observable change will occur. The time taken for this
corresponds to the time for a certain amount of product
to have been formed and so is inversely proportional to
the rate of reaction. The classic reaction studied in this
way is the reaction between hydrogen peroxide and iodide
ions, in the presence of acid, to form iodine and water.
Thiosulfate ions are added to the system and these initially react rapidly with the iodine produced. When all of the
thiosulfate has been consumed, free iodine is liberated
and this colours the solution yellow, or more commonly
blue–black through the addition of starch solution (which
forms an intensely coloured complex with iodine) to the
system.

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Chapter 6
H2O2 (l) + 2 H+ (aq) + 2 I– (aq)

2 H2O (l) + I2 (aq)

CORE

2 S2O32– (aq) + I2 (aq)

S4O62– (aq) + 2 I– (aq)

The blue colour of the iodine-starch complex suddenly appears when all of the thiosulfate has been consumed. The time taken for this to occur is inversely proportional to the rate.

6.1

Exercise

1.

3. For which one of the following reactions would a
colorimeter be most suitable for monitoring the
reaction rate?
A

The rate of a chemical reaction can sometimes be
determined by measuring the change in mass of the
reaction flask and its contents with time. For which
of the following reactions would this technique be
most successful?
A
B
C
D

B

Magnesium oxide and
dilute sulfuric acid.
Aqueous sodium chloride and
aqueous silver nitrate.
Copper(II) carbonate and
dilute hydrochloric acid.
Zinc and
aqueous copper(II) sulfate.

2. You wish to carry out an investigation that involves
the use of a conductivity meter to monitor the rate
of a chemical reaction. Which of the reactions below
would be the least suitable for this?
A

H2O2 (aq) + 2 H+ (aq) + 2 I– (aq)
2 H2O (l) + I2 (aq)

B

Ba2+ (aq) + SO42– (aq)

C

POCl3 (l) + 3 H2O (l)
4 H+ (aq) + 3 Cl– (aq) + H2PO4– (aq)

D

2 H2O2 (aq)

BaSO4 (s)

2 H2O (l) + O2 (g)

C
D

The reaction of acidified permanganate
ions (manganate(VII)) ions with
ethanedioic acid (oxalic acid) to form
carbon dioxide, manganese(II) ions and
water.
The reaction of magnesium carbonate
with a dilute acid to form a soluble
magnesium salt, carbon dioxide and
water.
The reaction of bromobutane with
aqueous sodium hydroxide to form
butanol and aqueous sodium bromide.
The reaction of lithium with water to
form aqueous lithium hydroxide and
hydrogen.

4. You wish to measure the rate of reaction of acidified
dichromate(VI) ions with aqueous sulfur dioxide to
produce aqueous chromium(III) ions and aqueous
sulfate ions at 35 °C. This reaction involves a colour
change from orange to green. Discuss how you might
go about doing this, the measurements you would
need to take and the precautions required.
5. During your study of chemistry you will most likely
have studied the way in which altering certain
variables affected the rate of a chemical reaction.
a)
b)
c)
d)

What reaction did you study?
What technique did you use to study the
rate of this reaction? Why do you think this
method was appropriate?
What variable did you change? How was this
carried out?
How could you modify the investigation to
study another variable. State which variable
you are now going to study and outline how
you would carry this out along with any
precautions you would take.

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Reactants

Products

6.2.2 Define activation energy, Ea.
6.2.3 Describe the collision theory.

CORE

6.2.1 Describe the kinetic theory in terms
of the movement of particles whose
average energy is proportional to
temperature in kelvins.

Activation
Energy (E a)

6.2 Collision
theory

Potential Energy

Kinetics

Figure 605  The principle of activation energy

6.2.4 Predict and explain, using collision
theory, the qualitative effects of particle
size, temperature, concentration and
pressure on the rate of a reaction.
6.2.5 Sketch and explain qualitatively
the Maxwell–Boltzmann energy
distribution curve for a fixed amount
of gas at different temperatures and its
consequences for changes in reaction
rate.
6.2.6 Describe the effect of a catalyst on a
chemical reaction.
6.2.7 Sketch and explain Maxwell–Boltzmann
curves for reactions with and without
catalysts.
© IBO 2007

Collisions are vital for chemical change, both to provide
the energy required for a particle to change (for example, a bond to break), and to bring the reactants into contact.
As particles approach each other there is repulsion between
the electron clouds of the particles. In order for reaction
to occur, the collision must have sufficient kinetic energy
to overcome this repulsion. Frequently energy is also
required to break some of the bonds in the particles before
a reaction can take place. Hence not all collisions lead to
a reaction. This minimum amount of energy required for
reaction is known as the activation energy (Ea) for the
reaction (units of Ea: kJ mol–1). This is illustrated, for an exothermic reaction, in Figure 605.

The activation energy involved varies tremendously from
reaction to reaction. In some cases (such as the reaction
of the hydrogen ion and hydroxide ion) it is so low that
reaction occurs on almost every collision even at low
temperatures. In other cases (such as sugar and oxygen) it
is so high that reaction at room temperature is negligible.
In order to react, the two particles involved must:




collide with each other
the collision must be energetic enough to overcome
the activation energy of the reaction, (i.e. the
collision must have E > Ea)
the collision must occur with the correct geometrical
alignment, that is it must bring the reactive parts of
the molecule into contact in the correct way

This final factor, often called the steric factor, is particularly important with regard to reactions involving large organic
molecules.
If anything increases the collision rate, then the rate of
reaction increases. Similarly anything that increases the
proportion of the collisions that have an energy equal to
or greater than the activation energy will increase the rate of reaction. These factors are summarised in Figure 606.

Factors mainly
affecting the collision
rate

Factors mainly
affecting the
proportion with
required Ea

Concentration/pressure

Temperature

Surface area

Catalyst

Figure 606  Factors affecting the rate of reaction (table)

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Chapter 6

CORE

The Effect of Concentration

The rate at which particles collide is increased by increasing the concentration of the reactants. Thus marble chips
react faster with concentrated hydrochloric acid than they
do with the dilute acid. For reacting gases, increasing the
pressure is equivalent to increasing the concentration. The
effect of concentration on rate is illustrated in Figure 607.

The Effect of Temperature

Not all particles have the same energy so there is a
distribution of kinetic energy, and hence velocity,
amongst the particles of the gas, known as the MaxwellBoltzmann distribution, shown in Figure 609. As with cars on a freeway, some are moving more rapidly and
others more slowly. The mean speed of the particles is
proportional to the absolute temperature but, because the
curve is asymmetric, this does not coincide with the most
probable speed. The area under the curve represents the
total number of particles and hence, in a closed system,
this area must remain constant.

Figure 607  The effect of concentration

The Effect of Surface Area

If the reaction involves substances in phases that do not
mix (e.g. a solid and a liquid, or a liquid and a gas) then
an increase in the surface area in contact will increase the collision rate. As a result powdered calcium carbonate
reacts faster with hydrochloric acid than lumps of the solid with the same mass. This is illustrated in Figure 608.

Probability of this energy

Low temperature
High temperature

Ea
with catalyst

Ea
un-catalysed

Energy

Figure 609  The effects of temperature and catalyst
As previously stated, a collision must have a certain
minimum energy, the activation energy, before reaction
can occur. As can be seen in Figure 609, the number of
molecules with the required activation energy is much
greater at a higher temperature than at a lower one. This
means that marble chips react more rapidly with warm
hydrochloric acid than with cold hydrochloric acid.
Increasing the temperature also has a very slight effect on
the collision rate, but in most cases this is insignificant
compared to its effect on the proportion of collisions with
the required activation energy. For many reactions an
increase in temperature of 10 °C will approximately double
the rate of reaction (a crude generalization), but the increase in the collision rate that this rise in temperature causes
is only ≈2%. To summarise, increasing the temperature
increases the frequency of collisions but, more important
is the increase in the proportion of molecules with E > Ea.
The effect of temperature on reaction rate is discussed in
much more detail in Chapter 6.6.

Figure 608  The effect of surface area

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Reactants

With catalyst

CORE

Catalysed E a

Enthalpy

Without catalyst

Uncatalysed E a

Kinetics

∆H
Products
Figure 610  The effect of a catalyst

The Effect of Catalyst

A catalyst is a substance that is usually required in small
amounts and can increase the rate of a chemical reaction
without undergoing any overall change. Catalysts speed up
a reaction by providing an alternative reaction mechanism
or pathway (like a mountain pass) with a lower activation
energy by which the reaction can take place. This means
that a greater proportion of collisions will have the
required energy to react by the new mechanism and so
the reaction rate increases. Typically, the efficiency of a
catalyst decreases with time as it becomes inactive due to
impurities in the reaction mixture, side reactions, or if its surface becomes coated and unavailable for activity.
The way this affects the situation in terms of the MaxwellBoltzmann distribution and the energy diagram is shown in Figures 609 & 610.

Exercise

1.

The Effect of Light

Some chemical reactions are brought about by exposure
to light. Examples would be the darkening of silver halides
when exposed to sunlight (the basis of black and white
photography) and the reaction of alkanes with chlorine
or bromine (for example, the reaction of methane and
chlorine). This is because reactant particles absorb light
energy and this brings about either the excitation of one
of the reactants (as is the case with the silver halides) or the breaking of a bond (such as the halogen—halogen
bond in the halogenation of the alkanes) which initiates
the reaction. It is for this reason that many chemicals are
stored in brown glass containers.

Which one of the following factors does not affect
the rate of a chemical reaction?
A
B
C
D

2.

The amounts of the reagents.
The concentration of the reagents.
The temperature of the reagents.
The presence of a catalyst.

In most chemical reactions, the rate of reaction
decreases as the reaction proceeds. The usual reason
for this is that
A
B
C
D

3.

6.2

The energy for the reaction is running
out.
The concentrations of the reactants are
becoming lower.
The temperature is falling as the
reaction proceeds.
The activation energy becomes greater.

In which of the following situations would you
expect the rate of reaction between marble (calcium
carbonate) and nitric acid to be the greatest?
A
B
C
B

Powdered marble and 2 mol dm–3 acid
at 40 °C.
Powdered marble and 0.5 mol dm–3 acid
at 40 °C.
Powdered marble and 2 mol dm–3 acid
at 20 °C.
Marble chips and 0.5 mol dm–3 acid at
40 °C.

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Chapter 6
4.

In which one of the following reactions would surface
area not be a factor affecting the rate?
A
B
C
D

Which one of the following reactions must occur by
more than one reaction step?

AHL

5.

6.

A

H+ (aq) + OH– (aq)

B

2 H2O2 (aq)

C

2 H2 (g) + O2 (g)

D

H2 (g) + O3 (g)

H2O (l)

2 H2O (l) + O2 (g)
2 H2O (l)
H2O (l) + O2 (g)

Explain briefly why:
a)
b)
c)

7.

Zinc and sulfuric acid.
Carbon dioxide gas with limewater
(aqueous calcium hydroxide).
Vegetable oil and aqueous sodium
hydroxide.
Aqueous ethanedioic (oxalic) acid and
aqueous potassium permanganate.

Increasing the concentration of the reagents
usually increases the rate of a chemical
reaction.
A reaction does not occur every time the
reacting species collide.
Increasing the temperature increases the rate
of reaction.

The rate of decomposition of an aqueous solution of
hydrogen peroxide can be followed by recording the
volume of gas collected over water in a measuring
cylinder, against time.
a)
b)

c)

d)

Sketch the graph of volume against time you
would expect for the complete decomposition
of a sample of hydrogen peroxide.
On the same axes, use a dotted line to sketch
the curve that you would expect to find if the
experiment were repeated using a smaller
volume of a more concentrated solution of
hydrogen peroxide so that the amount of
hydrogen peroxide remains constant.
When lead(IV) oxide is added, the rate at which
oxygen is evolved suddenly increases, even
though at the end of the reaction, the lead(IV)
oxide remains unchanged. Explain this.
In what way, apart from altering the
concentration or adding another substance,
could the rate at which the hydrogen
peroxide decomposes be increased?

HIGHER LEVEL
16.1 Rate
Expression (aHL)
16.1.1 Distinguish between the terms rate
constant, overall order of reaction and
order of reaction with respect to a
particular reactant.
16.1.2 Deduce the rate expression for a reaction
from experimental data.
16.1.3 Solve problems involving the rate
expression.
16.1.4 Sketch, identify and analyse graphical
representations for zero-, first- and
second-order reactions.
© IBO 2007

Altering the concentration of the reactants usually affects
the rate of the reaction, but the way in which the rate is
affected is not the same for all substances, nor can it be
predicted from the balanced equation for the reaction.
The rate expression, which is a mathematical function
expressing the dependence of the rate on the concentrations
of the reactants, must be determined experimentally. This
is usually done by measuring the reaction rate whilst
varying the concentration of one species but holding
those of the other species constant. Consider a reaction
involving reactants A, B, etc. The rate expression for this
reaction takes the form:
[ ]
Rate of reaction = – ____
​ d​  A   ​ 
​ = k​​[ A ]m​​ ​​​[ B ]n​​ ​ etc dt

The order of reaction is said to be ‘m’ in substance A, ‘n’ in substance B etc. The overall order of the reaction is the sum of these powers, i.e. m + n etc. The constant ‘k’ in the rate expression is known as the rate constant.

Note that ‘k’ does not vary with concentration, but it varies greatly with temperature, so it is important to always state the temperature at which the rate constant was measured.
Note that where a solid is involved in a reaction, ‘k´
must also vary with particle size (since for example in
the reaction of acid with calcium carbonate at the same
concentration of acid and temperature, the rate changes as
particle size changes).

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Kinetics
[A]

[B]

[C]

Initial rate

mol dm–�3

mol dm–�3

mol dm–�3

mol dm–�3 s–�1

1

0.400

1.600

0.0600

4.86 × 10–3

2

0.800

1.600

0.0600

9.72 × 10–3

3

0.400

0.800

0.0600

4.86 × 10–3

4

0.800

1.600

0.1800

87.5 × 10–3

Experiment

Figure 611  The effect of concentration changes on the rate of a reaction If doubling the concentration of one species (say A), whilst the other conditions are held constant, has no effect on
the initial rate of reaction, then the reaction is zero order with respect to A (as 20 = 1). If doubling the concentration of A doubles the rate, then the reaction is first order with respect to A (as 21 = 2). If it increases by a factor of four it is second order with respect to A (as 22 = 4), by a factor of eight then third order with respect to A (as 23 = 8) etc. Similar considerations apply to altering the concentrations

by other factors (for example if the concentration was
decreased by a factor of 3, then the rate would also decrease by a factor of 3 if the reaction was first order with respect to this reagent).

units of mol dm–3 s–1, for known concentrations of the
reagents. Consider the data in Figure 611 above. The rate
constant can be calculated by substituting any set of data
in the rate expression. For example using the data from
experiment 1:

Figure 611 gives some data about the effect of varying
concentrations upon the rate of a chemical reaction
involving three species – A, B and C:

and using the data from experiment 2:

Comparing experiments 1 and 2, the only change is that the
concentration of A has been doubled. The data in the table
indicate that the rate has been doubled, so the reaction is
first order with respect to A. Comparing 1 and 3, the only
change is that the concentration of B has been halved, but
there is no effect on the reaction rate, indicating that the reaction is zero order with respect to B (if first order it
would be ½ the rate in 1, if second order, then ¼ of the rate in 1). Comparing 2 and 4 the only difference is that the
concentration of C has been increased by a factor of three.
The rate has increased by a factor of nine, so the reaction is second order in C (as 32 = 9 – if it had been first order in C, the rate would only have increased by a factor of 3.). This
means that the rate expression for this reaction is:

Rate = k.[A]1[B]0[C]2
or more simply

Rate = k.[A][C]2
Hence the reaction is third order (1 + 2) overall.
The rate constant for a reaction may be calculated provided
that the rate of reaction has been measured in standard

AHL

Rate = k.[A][C]2

4.86 × 10–3 = k[0.400][0.06]2

4.86 × ​10​–3​ 
k = ​ ___________
 ​ 
0.400 × ​0.06​2​

= 3.375 mol–2 dm6 s–1

Rate = k.[A][C]2

9.72 × 10–3 = k[0.800][0.06]2
9.72 × ​10​–3​ 
k = ​ ___________
 ​ 
0.800 × ​0.06​2​

= 3.375 mol–2 dm6 s–1

Note the units for the rate constant. These only hold
for a reaction that is third order overall and other order
reactions will have rate constants with different units.
The units can be calculated remembering that the units for
rates are mol dm–3 s–1 and the units for concentrations are mol dm–3. Hence the units of the rate constant are:
Zero order overall
First order overall
Second order overall
Third order overall

mol dm–3 s–1
s–1
mol–1 dm3 s–1
mol–2 dm6 s–1

In general (mol dm–3)q–1 s–1 where q is the overall order The order of reaction can also be found from a graph
showing the way in which the initial rate varies with the
initial concentration of the reactant, all other factors being equal. This is illustrated in Figure 612.

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Chapter 6
is halved, then the reaction is first order. If halving the
concentration causes the rate (gradient) to decrease by a
factor of 4, the reaction is second order (½)2 = ¼ etc. To be really useful, such experiments should have all but one
reagent in large excess, so that the order in the limiting
reagent is what causes the reaction rate to change.

Initial Rate

Zero order
First order
Second order

This is in fact the basis for a very powerful technique to
simplify the rate equation. Consider a reaction that is
second order overall in which the initial concentration of
A is 0.02 mol dm–3 and the initial concentration of B is
2 mol dm–3. When the reaction is complete (assuming
the stoichiometry is 1 mole of A reacts with 1 mole of
B), the final concentration of A is zero and that of B is
1.98 mol dm–3, i.e. the concentration of B remains virtually unchanged, hence:

Initial Concentration

AHL

Figure 612  The effect of concentration on rate
The order of a reaction can also be found from a graph of
concentration against time, which shows the effect of the
reactants being used up on the rate of reaction. The gradient of the graph at any point gives the rate of reaction. If this is constant (i.e. the graph is a straight line) then the reaction must be zero order in the reactants whose concentrations

are undergoing significant change, because the decrease in
concentration is not affecting the rate of reaction (which
we know from the constant gradient). If the reaction rate
(the gradient of the line) is halved when the concentration

Rate = k.[A].[B] ≈ k´.[A]

(where k´ = k.[B])

A reaction of this type is called a “pseudo first-order
reaction” because it obeys a first order rate law, but
the observed rate constant (k´) will depend on the
concentration of another species (if [B] is doubled then
the reaction rate will double and the half-life will decrease to half of its initial value). The same technique (making
the concentration of one reagent much less than that of the

Concentration

100
First order

75
50
25
0

t1/2

1

t1/2

3

4

Time

5

100
Z ero order

50
25
0

t1/2

2

Concentration

Concentration

100
75

Constant half–life = t1/2

5 Time

4
1
2
3
Each t1/2 is half preceding t1/2

Gradient constant

75

S econd order

50
25
0

4
5 Time
1
2
3
Each t1/2 is double preceding t1/2

Gradient after first half life = ¼ initial gradient

Figure 613 a, b and c.   Graphs showing the variation of concentration with time for reactions of different orders

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Kinetics
others) can be used to vary the observed order of reactions
with other rate expressions.

Exercise

It is easy to recognise a first order reaction from
graphs of concentration (or something proportional
to concentration) against time. This is because the
concentration shows an exponential decrease, that is the
time for the concentration to fall from its initial value to half its initial value, is equal to the time required for it to fall from half to one quarter of its initial value and from
one quarter to one eighth etc. This time is known as the
half-life t½ of the reaction and it is illustrated in Figure 613 which shows that the successive half lives of reactions
of other orders vary in characteristic ways.

16.1

Questions 1 to 4 refer to the rate expression for a chemical reaction given below:

Rate = k[A][B]2[H+]
Which one of the following statements is not true
about this reaction?
A
B
C
D
2.

The units of the rate constant (k) will be:
mol dm–3 s–1
mol s–1
dm3 mol–1 s–1
dm9 mol–3 s–1

A
B
C
D

t​_​ 1 ​​= ___
​ ln2   ​
2
k
For example, if the rate constant of a first order reaction is ln2
0.005 s-1, then the half-life will be ____
​ 0.005
  ​ = 139 s.

3.

This results from the integration of the first order
rate expression. If the half-life is known, say from a
concentration-time graph, then this equation may be
rearranged to find the rate constant. For example if the
half-life for a first order reaction is 4 minutes:

If the concentrations of A and B are both doubled,
but the concentration of H+ remains constant, the
rate would increase by a factor of:
A
B
C
D

k = _______
​ ( ln2  ) ​ 
= 0.0289 s-1
​  4 × 60  ​

4.

2
4
8
16

Which one of the following would lead to the greatest
increase in reaction rate?
A
B
C

Doubling the concentration of A only.
Doubling the concentration of B only.
Doubling the concentration of A and H+
only.
Doubling the concentration of B and H+
only.

D
5.

A

Which one of the graphs shown would indicate
that a reaction was zero order in the reactant whose
concentration was being varied?

B

C
Rate

Rate

Rate
Concentration

D

Concentration

Rate

The first order exponential decay is the same as that found
in radioactive decay. Because it remains constant, the
half-life is an important quantity for these systems and
it can be found from an appropriate graph (such as that
above) or it may be found from the rate constant (k) by
substituting in the equation:

It is first order in A.
It is second order in B.
It is first order in H+.
It is third order overall.

AHL

1.

Concentration

Concentration

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Chapter 6
6.

The following data refers to the acid catalysed iodination of propanone

CH3—CO—CH2—I (aq) + H+ (aq) + I– (aq)

CH3—CO—CH3 (aq) + I2 (aq)
[CH3—CO—CH3]

[I2]

[H+]

Initial Rate

mol dm–�3

mol dm–3

mol dm–3

mol dm–3 s–1

1

0.2

0.008

1

4 × 10–6

2

0.4

0.008

1

8 × 10–6

3

0.6

0.008

1

1.2 × 10–5

4

0.4

0.004

1

8 × 10–6

5

0.4

0.002

1

8 × 10–6

6

0.2

0.008

2

8 × 10–6

7

0.2

0.008

4

1.6 ���
����
10–�5

AHL

Solution

a)
b)
c)
7.

From the data in the table derive the rate expression for the reaction, explaining the evidence for the dependency on each of the species.
Give the order with respect to CH3—CO—CH3, I2 and H+, and the overall order.
Use the data from Solution 1 to calculate the value of the rate constant.

The data given below refer to the hydrolysis of a 0.002 mol dm–3 solution of an ester by 0.2 mol dm–3 aqueous sodium hydroxide.

Time (s)
[ester] (mmol dm–3)
a)
b)
c)
d)
e)

60

120

180

240

300

360

420

480

1.48

1.10

0.81

0.60

0.45

0.33

0.24

0.18

Plot a suitable graph to determine the order of the reaction with respect to the ester, explaining your method.
Use your graph to determine the half-life of the reaction and hence determine a value for the apparent rate constant, giving appropriate units. Why does this graph give no indication of the order with respect to the hydroxide ion?

How would you modify the experiment to determine the dependence on hydroxide ion?
Assuming that it is also first order in hydroxide ion, write a new rate expression.
Use this to calculate a value for the rate constant, giving appropriate units. Why does this differ from the value found in b) and how are the two related?

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Kinetics

16.2.1 Explain that reactions can occur by more
than one step and that the slowest step
determines the rate of reaction (ratedetermining step).
16.2.2 Describe the relationship between
reaction mechanism, order of reaction
and rate-determining step.
© IBO 2007

The chance of more than two particles colliding
simultaneously with the correct geometry and minimum
energy is very small. This means that if there are more
than two reactants, the reaction must occur by a number
of simpler reaction steps. In addition, many reactions that
have apparently simple equations do not occur in this
manner, but are the result of a number of steps. These
steps involve species that are the product of an earlier
step and are then completely consumed in a later step (so
they do not appear in the stoichiometric equation). These
species are known as “intermediates”. The simple stages
by which a chemical reaction occurs are known as the
mechanism of the reaction. The sum of the various steps
of the mechanism must equal the balanced equation for
the reaction.
The various steps in the reaction mechanism will have the
potential to occur at different rates. The products cannot
however be formed faster than the slowest of these steps
and so this is known as the rate determining step (rds). An
analogy would be that if people can come off a train at the
rate of 20 per second, can travel up the escalator at a rate

of 10 per second, and can pass through the ticket barrier
into the street at a rate of 50 per second, then they will
still only reach the street at a rate of 10 people per second. Making larger doors on the train or putting in an extra
ticket barrier will not make this any faster, only a change
affecting the escalator, the slowest step, will increase the rate.
In summary, a mechanism is a model of how a reaction
occurs. The rate of overall reaction is the rate of the slowest step. This slowest step is called the rate determining step. Species produced in earlier steps of the mechanism that
are consumed in later steps are called intermediates. A
mechanism must account for the overall stoichiometry
of the reaction, the observed rate expression and any
other available evidence (such as the effect of light or a
catalyst).
There are only two kinds of fundamental process that can
occur to bring about a chemical reaction. Firstly, a species can break up or undergo internal rearrangement to form
products, which is known as a unimolecular process.
As this only involves one species, a unimolecular step is
first order in that species. Radioactive decay, for example, is unimolecular. Secondly, two species can collide and
interact to form the product(s) and this is known as a
bimolecular process. As this involves the collision of the
two species then doubling the concentration of either will
double the collision rate. Hence, it is first order in each and second order overall. Both unimolecular and bimolecular
processes can be either reversible (lead to equilibrium) or
irreversible (lead to complete reaction) depending on the
relative stability of the reactants and products. Whether
a particular reaction step is unimolecular or bimolecular,
is known as the molecularity of that reaction step. In a
bimolecular process, the species collide to initially give a transition state (or activated complex), which then breaks
down to either form the products or reform the reactants.

A Unimolecular step

A Bimolecular step

A unimolecular step involves a single species as a reactant.

A bimolecular steps involves collision of two species (that
form a transition state or an activated complex that can
not be isolated).

A

/

Products

A+B

/

AHL

16.2 Reaction
mechanism (aHL)

Products

Its rate law is therefore 1st order with respect that reactant.

Its rate law is 1st order with respect to each of the colliding species and is therefore 2nd order overall.

rate ∝ [A]

rate ∝ [A][B]

Figure 617  Comparing uni- and bi-molecular steps

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Chapter 6
A transition state is an unstable arrangement in which
the bonds are in the process of being broken and formed.
It therefore occurs at the maximum point on a potential
energy diagram and cannot be isolated.
As outlined above, many chemical reactions occur by a
series of simple steps known as the mechanism of the
reaction. It is possible to write a number of mechanisms
(that is a series of fundamental processes by which the
reaction could occur) for any reaction and it is only possible to suggest which is the correct one by studying the kinetics of the reaction. Only species that are involved in the rate
determining step, or in an equilibrium preceding it, can
affect the overall rate of reaction. Hence, determining the
rate expression for a reaction will help to identify the rate determining step and this will eliminate many possible
mechanisms for the reaction.

AHL

Consider for example a reaction

2A+B

C+D

as an illustration of how the rate expression will depend
upon the mechanism and upon which step in the
mechanism is the rate determining step.
There are three particles involved in the reaction, so it
is most unlikely that this occurs as a single step. Many
mechanisms for the reaction could be written and these
would produce a variety of rate expressions. Some examples
are given in Figure 618. Note that adding together the
different steps always leads to the same overall equation
as shown for the first two possible mechanisms in Figure
618.
Some of these mechanisms involve equilibria, a topic that
is dealt with in greater detail in Chapter 7. For the present purposes it is enough to know that for the equilibrium,
A+B
X, the [X] will depend on both the [A] and [B].
Similarly in the equilibrium, A + A  
 A2, the [A2] will
2
be proportional to [A] . Note also that the concentration of the intermediate (X) never appears in the rate expression.
In I, the first bimolecular step is rate determining so that the rate will depend on the rate of collisions between A
and B, hence the rate will be proportional to [A].[B]. In
II the second bimolecular step is rate determining so that
the rate will depend on the rate of collisions between A
and X, hence the rate will be proportional to [A] [X],
but [X] will depend on both [A] and [B], so that taking
this into account the rate depends on [A]2 [B]. In V the
rate depends on the unimolecular conversion of B to an
intermediate X, so the rate only depends upon [B].

Rate
expression

Mechanism

I

A+B

X + C; Slow rds

A+X

D; Fast

(A + B + A + X
=2A+B

II

IV

X; Fast

A+X

C + D; Slow rds

(A + B + A + X

V

X+C+D

Rate ∝ [A]2 [B]

C + D overall)

A+A

A2; Fast

A2 + B

C + D; Slow rds

A+A

A2; Slow rds

A2 + B

C + D; Fast

B

Rate ∝ [A] [B]

C + D overall)

A+B

=2A+B
III

X+C+D

Rate ∝ [A]2 [B]

Rate ∝ [A]2

X; Slow rds

X+A

Y + C; Fast

Y+A

D; Fast

Rate ∝ [B]

Figure 618  Some possible mechanisms for the reaction;
2A+B
C+D

Note that III and IV only differ in which of the two steps
is the rate determining step. This is not necessarily fixed, for example at very low [B] the second step could be the
rate determining step (mechanism III), but at very high
[B] the second step will become much faster so that now
the first step might be rate determining (mechanism IV).
Because A2 will react with B as soon as it is formed, the
first step is now no longer an equilibrium. Note also that
both mechanism II and mechanism III lead to the same
rate expression and so some other means (such as trying
to get some information about the intermediate X) would
have to be used to decide which (if either.) was operating.
Consider as another example the reaction between
propanone and iodine:

CH3COCH3 (aq) + I2 (aq)
CH3COCH2I (aq) + H+ (aq) + I– (aq)
This would appear to be a simple bimolecular process,
but if this were the case, then the rate of reaction would
be expected to depend on the concentrations of both the

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Kinetics
A2

Enthalpy

A1

CH3COCH 3

A3

CH3C(OH+ )CH3
Intermediates

CH2=C(OH)CH3
∆H

CH 3COCH 2 I

propanone and the iodine hence the rate expression would
be:

Rate = k [CH3COCH3] [I2]
In practice it is found that the reaction is catalysed by acids and that the rate is independent of the concentration of
iodine hence the reaction is first order in both propanone
and hydrogen ions, but zero order in iodine. Hence the
rate expression is:

Rate = k [CH3COCH3] [H+]
This means that one molecule of propanone and one
hydrogen ion must be involved in the rate determining
step, or in equilibria occurring before this. The commonly
accepted mechanism for this reaction is:

CH3COCH3 + H+
Fast equilibrium

CH3C(OH+)CH3

CH3C(OH+)CH3
CH2=C(OH)CH3 + H+
Slow – rate determining step
CH2=C(OH)CH3 + I2
CH3COCH2I + H+ + I–
Fast (does not affect rate)
This mechanism agrees with the experimentally
determined rate expression. The rate expression can never
prove that a particular mechanism is correct, but it can
provide evidence that other possible mechanisms are
wrong.

The species CH3C(OH+)CH3 and CH2=C(OH)CH3 are
intermediates; they have a finite life and occur at a potential energy minimum on the reaction diagram. In this reaction
mechanism there would be a number of transition states,
firstly (A1) between CH3COCH3 and H+ before forming
CH3C(OH+)CH3 secondly (A2) when CH3C(OH+)CH3
starts to break up to form CH2=C(OH)CH3 and H+ and
finally (A3) in the reaction of this with iodine. These do
not have a finite life and occur at potential energy maxima
on the reaction diagram. This is illustrated in Figure 619
and the differences between intermediates and transition
states (activated complexes) are summarised in Figure
620.

Intermediates

Transition States

Exist for a finite time

Have only a transient
existence

Occur at a P.E. minimum

Occur at a P.E. maximum

Formed in one step of a
reaction and consumed in
a subsequent step

Exist part way through
every step of a reaction

AHL

Figure 619  P.E. diagram for the iodination of propanone

Figure 620  Differences between intermediates and
transition states
The differences between intermediates and transition
states can also be illustrated by the SN1 and SN2
mechanisms for the nucleophilic substitution reactions of
halogenoalkanes.

(Good TOK point – an example of a difficulty that besets all processes of inductive logic.)

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Chapter 6
Exercise

1.

Which one of these steps is unimolecular?
A
B
C
D

2.

AHL

B
C
D

+

NH3 + H
NH4
H+ + OH–
H2O
N2O4
2 NO2
H• + Cl•
HCl

An activated complex occurs at a
potential energy maximum and an
intermediate at a minimum.
An activated complex cannot take
part in bimolecular reactions, but an
intermediate can.
An activated complex does not exist for
a finite time but an intermediate does.
An activated complex can reform the
reactants, but an intermediate cannot.

The rate determining step of a mechanism is the one
which
A
B
C
D

4.

+

Which one of the following is not a difference
between an activated complex and an intermediate?
A

3.

5.

16.2

occurs most rapidly.
occurs most slowly.
gives out the most energy.
gives out the least energy.

Which one of the following mechanisms would give
a first order dependence on A and zero order on B for
the reaction below?

A+B

A reaction involves two reactants, A and B. The initial
reaction rate was measured with different starting
concentrations of A and B and the following results
were obtained at 25 °C:
[A]
mol dm3

[B]
mol dm3

Initial rate
mol dm3 s1

0.2

0.2

3.2 × 104

0.4

0.4

1.3 × 103

0.4

0.8

1.3 × 103

a)

Deduce the order of the reaction in A, in
B and the overall order. Hence write a rate
expression for the reaction.

b)

Calculate a value for the rate constant, giving
suitable units.

c)

What initial rate would you expect if the
initial concentrations of both reactants were
0.1 mol dm–3?

d)

If the overall equation for the reaction is
A+B
C + D, write a mechanism,
indicating which step is the rate determining
step, that is:
i
consistent with the rate expression
found.
ii inconsistent with the rate expression
found.

C

A

A+X
Y+B

Y (fast)
C + X (slow)

then

B

B
X (slow)
X+A
C (fast)

then

C

2A
A2 + B

A2 (slow)
C (fast)

then

D

A
X (slow)
X+B
C (fast)

then

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Kinetics

k = A​e​(​ –  RT  ​  )​​

16.3 Activation
energy (aHL)

​Ea​ ​
___

© IBO 2007

In a fluid, there is a distribution of energy amongst the
particles known as the Maxwell–Boltzmann distribution
(see Section 6.3). As the temperature increases, the
number of particles with a high energy increases, though
there are still some particles with very little energy. The
result is a flattening of the distribution curve, because the total area under it must remain constant (refer to figure
609, reproduced again as Figure 622 for convenience).
Probability of this energy

Low T
High T

Catalysed Ea

Uncatalysed Ea

Kinetic Energy

Figure 622  The distribution of kinetic energy at two
different temperatures
All chemical reactions require a certain minimum energy,
the activation energy (Ea), for the reaction to occur. As
can be seen from Figure 622, at the higher temperature
a greater proportion of the molecules have an energy
greater than the activation energy, i.e. there is a greater
proportion of the total area under the curve to the right
of the Ea line (either the catalysed or uncatalysed) on the
higher temperature curve than on the lower temperature
curve. This is usually the major reason why reactions
occur more rapidly at higher temperatures.

The expression indicates that the rate constant k depends
exponentially on temperature, which is why temperature
has such a large effect on reaction rate. Rather satisfyingly the expression for the area under the Maxwell–Boltzmann
distribution curve in excess of Ea also gives an exponential dependence if Ea>>RT. If we take logarithms to the base
e (natural logarithms (ln) - a mathematical procedure
you may possibly not have met) and then rearrange, the
Arrhenius equation is converted to:

(  )

​E​ ​ 1
lnk = lnA – ​ __
​  a ​   ​__
​   ​
R T

AHL

16.3.2 Determine activation energy (Ea) values
from the Arrhenius equation by a
graphical method.

This is the equation of a straight line so, as shown in Figure 623 below, a graph of ln k against __
​ 1 ​ will be linear with
T
​E​a​ and an intercept on the
gradient __
y–axis of ln A.
– ​   ​ 
R
The activation energy for a reaction can therefore be found
by measuring the rate of reaction at different temperatures, with all the other conditions unchanged (so that rate ∝ k), and then plotting ln(rate) against __
​ 1 ​ (Note again, T must
T
o
be in Kelvin, not Celsius).

ln k or ln(rate)

16.3.1 Describe qualitatively the relationship
between the rate constant (k) and
temperature (T).

where Ea is the activation energy, T the absolute
temperature (in Kelvin), R the gas constant (8.314 J K–1
mol–1) and A is called the Arrhenius constant (or the
pre–exponential factor). It is dependent on collision rate and steric factors, that is any requirements regarding the
geometry of the colliding particles. This equation is known
as the Arrhenius equation after the Swedish chemist,
Svante August Arrhenius, who first proposed it.

Intercept = ln A
E
Gradient = – -----a
R
1 –1
--T (K )
Figure 623  Determining the activation energy
graphically

For many reactions it is found that the effect of temperature on the rate constant for a reaction (k) is given by the
expression:

179
070809 Chem Chap 6-3 for correct179 179

7/12/2007 8:22:58 AM

Chapter 6
Exercise

1.

Which one of the following is not true about the
activation energy of a reaction?
A
B
C
D

AHL

2.

e)

f)

5.

the temperature.
the concentration of the reagents.
the concentration of the catalyst.
the surface area in contact.

2t
½t
t2
It would depend on the size of the
activation energy.

The activation energy for the reaction below is
112 kJ mol–1 and ∆H is +57 kJ mol–1.

a)

b)

c)

2 NO2 (g)

2 NO(g) + O2 (g)

Draw an energy level diagram illustrating
the energy changes for this reaction. Clearly
mark and label the activation energy Ea and
the enthalpy change (∆H).
On the same diagram, using dotted lines,
show the effect of a platinum catalyst, clearly
labelling the change ‘With catalyst’. Would
the platinum be acting as a homogeneous or
heterogeneous catalyst?
Is the reaction exothermic or endothermic?
When it occurs will the container become
hotter, or cooler?

If the temperature was increased, how would
this affect the rate of reaction? Explain this in
terms of the collision theory of reactions (a
diagram might help).
What other factor (i.e. not temperature or
catalyst) could be changed to increase the
rate of reaction – be precise, remembering
that these are all gases.
Nitrogen dioxide is a brown gas, whereas
nitrogen monoxide is colourless. Suggest
how you might be able to measure the rate of
reaction.

When aqueous solutions of benzenediazonium
chloride decompose, they evolve nitrogen gas. The
table below gives the volume of gas obtained at
different times for such a decomposition at 70 °C.

The temperature at which a reaction is carried out
is increased from 20 °C to 40 °C. If the half-life of
the reaction was initially t, the half-life at the higher
temperature will be:
A
B
C
D

4.

It is related to the enthalpy change (∆H)
of the reaction.
It is decreased by the addition of a
catalyst.
It is the minimum amount of energy
that the reactants must have in order to
form the products.
The greater the activation energy the
lower the rate of reaction.

The activation energy of a chemical reaction can be
determined by measuring the effect on reaction rate
of varying
A
B
C
D

3.

d)

16.3

a)

b)
c)

Time
min

Volume
cm3

1

5

2

9

3

13

4

17

5

21

7

28

9

33

12

40

16

48

20

54

If, when decomposition was complete, the
total volume of gas released was 70 cm3,
graphically determine the order of the
reaction. What further data, if any, would
you need to calculate a value for the rate
constant?
Draw the apparatus you could use to obtain
such data and state what precautions you
would take.
If you wanted to determine the activation
energy for this reaction, what further
experiments would you carry out? How
would you use the data from these to
determine the activation energy?

180
070809 Chem Chap 6-3 for correct180 180

7/12/2007 8:22:59 AM

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