L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.1
THE EFFECT OF TEMPERATURE CHANGE ON REACTION RATE
REVIEW
As the temperature of a reaction mixture increases, the reaction rate increases, it is generally found
that a temperature rise of
about 100C approximately doubles the rate of reaction. This can be explained in terms of
1. activation energy and
2. the collision theory.
II.
(A)
EFFECT OF TEMPERATURE CHANGE ON REACTION RATE IN TERMS OF
ACTIVATION ENERGY
Activation Energy
<1> During a reaction, bonds are first broken and others are then formed. Energy is required
to break certain bonds and start the process, whether the overall reaction is exothermic or
endothermic.
<2>
Particles will not always react when they collide because they may not have sufficient
energy for appropriate bonds to break.
If a reaction is to occur, the colliding reactant particles must process more than a certain
minimum amount of energy to overcome an energy barrier. This “minimum” amount of
energy is known as activation energy, Ea.
Conversely, if the reactant particles do not possess the activation energy required for a
reaction, this reaction will not occur.
If reactant particles possess energy equal to or greater than the activation energy,
they can cross the ‘energy barrier’. Reaction occur.
(B) Effect of Temperature change on Reaction Rate
<1> When temperature increases, the reactant particles will possess more energy.
<2> More reactant particles possess energy equal to or greater than the activation energy.
as a result, reaction rate increases.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
III.
(A)
Chpt. 15 : p.2
APPLICATION OF THE ARPHENIUS EQUATION TO DETERMINE THE ACTIVATION
ENERGY OF A REACTION
Arrhenius equation
The relationship between the rate constant and the temperature for a given reaction is given by
the Arrhenius equation, which is expressed as
k  A exp(
 Ea
)
RT
where
k = rate constant of the reaction
A = Arrhenius factor (of constant or coefficient)
exp = exponential constant = 2.718
Ea = activation energy of the reaction
T = temperature of the reaction mixture in the absolute scale (in K) i.e. ( 0C +273)
R = universal gas constant = 8.314 JK1molt
Note
 Ea
) suggest that
RT
increasing the temperature increases greatly the proportion of high kinetic energy
molecules.
<1> The exponential factor, exp(
<2> It is obvious that
Raising T makes Ea / RT smaller, k will increase and the reaction is faster.
Lowering T makes Ea / RT larger, k will decrease and the reaction is slower.
<3> Similarly,
If Ea is low, Ea / RT will be small. K will be large and the reaction will be fast.
If Ea is high, Ea / RT will be large. K will be small and the reaction will be slow.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
(B)
Chpt. 15 : p.3
Natural Log of The Arrhenius Equation
Taking natural log of the Arrhenius equation gives
ln k  ln A 
Ea
RT
log k  log A 
or
Ea
2.303RT
Measuring the rate constant at two different temperature provides enough information to evaluate the
activation energy.
The Activation energy and the rates or rate constants (k1 and k2) at two temperatures (T1 and T2) can be
related by the equation:
ln(
k1
Ea 1 1
)
x(  )
k2
R T1 T2
Exercise 1
If the rate constant of a reaction at 310K is double that a 298K, calculate the activation energy of the
reaction.
(Gas constant = 8.314 JK-1mol-1)
ANSWER
Exercise 2
In the gas reaction
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
doubling the initial concentration of NO made the initial rate four times as fast, doubling the initial
concentration of H2 made the reaction twice as rapid.
(a)
(b)
(c)
deduce the rate law.
what are the units of rate constant.
Calculate the activation energy of the reaction if the rate constant at 1115K is double the value at
1093K.
( Gas constant ,R = 8.314 JK-1mol-1)
ANSWER
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Exercise 3
At 300°C, the rate constant for the reaction
is 2.41x10-10 s-1. At 400°C. k equals to 1. 16x106s-1.
Determine the value of
(a) the activation energy (in kJ mol-1) and
(b) the Arrhenius factor A for this reaction.
(Gas constant, R = 8.314 J K-1 mol-1 )
ANSWER
Exercise 4
For the hypothetical reaction A  B + C,
the rate constant is 3.91x104 mol1dm3s-1 at 370°C and 4.05x102 mol-1dm3s-1 at 470°C.
calculate
(a) the activation energy, and
(b) the rate constant of the reaction at 450°C.
( Gas constant = 8.314 J K-1 mol-1 ) I
ANSWER
Chpt. 15 : p.4
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.5
(C) Arrhenius Plot
The activation energy is determined from the dependence of the natural logarithm of rate
constant on the reciprocal of temperature.
When the rate constant is known at more than two temperatures, the precision of the
determination of the activation energy is increased by the Arrhenius plot, in k against 1/T.
The equation
Ea
ln k  ln A 
RT
can be rewritten as
Ea 1
ln k  ln A 
( )
R T
to show the temperature dependence.
This expression is of the type y = mx + c, a plot of the values of ln k at different temperatures
against l/T will give a
Straight-line graph with a slope (Ea/RT) and an intercept of ln A. Hence the activation energy
and the Arrhenius Factor can be determined.
Example
This experiment is about the determination of the activation energy of the reaction between
bromide ion and bromate(V) ion in acid solution:
5Br(aq) + BrO3(aq) + 6H+(aq)  3Br2 (aq) + 3H2O(l)
The progress of the reaction is followed by adding a fixed amount of phenol together with some
methyl red indicator. The bromine produced during the reaction reacts very quickly with
phenol.
Once all the phenol is consumed, any further bromine produced bleaches the indicator
immediately.
(a) Investigation of the activation energy of the reaction
Experimental procedure
<1> Place phenol solution, bromide solution and bromate solution and methyl red indicator
into a boiling tube.
<2> place sulphuric acid into another boiling tube.
<3> Place both boiling tubes into a beaker of water (water bath)
which is maintained at 30°C. Allow the contents of both tubes
reach the temperature of the water bath.
<4> Mix the contents of the two tubes. Start the stop watch and swirl the tube gently. Keep the
tube in the water bath throughout the experiment.
<5> Record the time (t) taken for the complete disappearance of the red colour.
<6> Repeat the above steps, maintaining the reaction temperature
at different temperatures, e.g. 35°C, 40°C. 45°C, 50°C.
( Temperature must be made the only variable in this experiment. )
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.6
Principle of determination
The time for the reaction to proceed to a certain extent is determined. (t denotes the time a
definite quantity of bromine is produced at different temperatures.)
Reaction rate = (concentration change/time)
In general, the rate constant is related to the time of reaction by
where t = the time for methyl red to be bleached
k  1/t
For the Arrhenius equation k = exp(-Ea/RT), the following expression is obtained.
Ea 1
ln k  ln A 
( )
R T
Putting k = concentration change (c)/t, plot the graph
c
Ea 1
ln  ln A 
( )
t
R T
using t at different temperatures
Ea 1
ln t  ln c  ln A 
( )
R T
ln c and ln A are constants. The slope of the graph in t against 1/T is (Ea/R)
Knowing the value of the gas constant R, Ea can be determined.
(B)
Necessary apparatus required for the determination
Thermometer, stop watch, water bath, volume measuring device (e.g. dropper, burette,
measuring cylinder)
Exercise 5
The table below gives the rate constants obtained at different temperatures for the reaction
2N2O5(g)  2N2O4(g) + O2(g)
Temperature/ °C Rate constant / s-1
10
3.83 x 10-6
20
1.71 x 10-5
30
6.94 x 10-5
40
2.57 x 10-4
50
8.78 x 10-4
Determine
(a) the activation energy and
(b) the Arrhenius factor
for this reaction by plotting the graph for
Ea 1
ln k  ln A 
( )
[Gas constant, R = 8.314 J K-1 mol-1]
R T
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.7
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.8
THE INTERPRETATION OF RATES OF GASEOUS RACTIONS AT MOLECULAR LEVEL
I. DISTRIBUTION OF MOLECULAR SPEEDS IN A GAS
• At any given temperature, all gases have the same average kinetic energy.
• However, the motion of gas molecules is random. It follows that the collisions between
molecules also occur randomly and involve a transfer of energy.
• some collisions result in a gain of kinetic energy for one molecule and a loss of kinetic
energy for the other.
• If a molecule undergoes a series of collisions such that each collision adds to its kinetic
energy, it will end up with a kinetic energy higher than the average. Conversely, if a
molecule undergoes a series of collisions such that each collision results in a loss of kinetic
energy, it will end up with a kinetic energy lower than the average.
It can be concluded that
Since molecules undergo continual random collisions, the molecules of a gas at constant
temperature do not travel with the same speed. This results in a distribution of molecular
speeds in a gas.
II.
GRAPHICAL REPRESENTATION OF THE MAXWELL-BOLTZMANN DISTRIBUTION AND
ITS VARIATION WITH TEMPERATURE
Maxwell—Boltzmann distribution curve is a plot of distribution of molecular kinetic energies (or
speeds) different temperatures.
Interpretation
<1> The total area under the curve is proportional to the total number of molecules, the area under
any potion of the curve is proportional to the number of molecules with the energies in that
range.
<2>
The curve shows that some molecules have very low or very high speeds. However, most
molecules have intermediate speeds. This results a normal distribution curve.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
(A)
Chpt. 15 : p.9
Variation of Maxwell—Boltzmann curve with temperature
The effect on the Maxwell—Boltzmann distribution curve of increasing the temperature is
shown below:
The spread of the Maxwell-Boltzmann distribution increases with increasing temperature
It can be seen that at a higher temperature, the following changes occur:
<1> The peak of the curve moves to the right, so that the mean energy of the molecules
increases and the proportion of molecules having higher energy increases.
<2> The curve flattens so that there is a wider distribution of energies and the proportion of
molecules with the most probable speed decreases.
<3> The area under the curve is still the same as that of the one at lower temperature, as the
total number of molecules in the sample remains the same.
III. SIMPLE COLLISION THEORY
The collision theory explains chemical reactions at the molecular level. It is developed from the
kinetic theory of gases to account for the effects of concentration and temperature on reaction
rate. According to the collision theory
<1> Chemical reactions in the gas phase are due to collision of reactant particles.
<2> Not all collisions results in a reaction. For a collision to be effective such that a reaction
can occur, the following conditions must be necessary
(i)
The reactant particles must collide with kinetic energy greater than the activation
energy Ea (a certain threshold) to break the bonds that need to be broken, and
(ii)
The reactant particles must collide in the right direction (i.e. correct orientation or
collision geometry) so that new bonds can form.
As the number of effective collisions increases, the rate constant and the reaction rate increase.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.10
collision theory
In general, no of effective collisions =
collision frequency x
fraction of collision
x
with correct orientations
fraction of gas molecules with K.E.
greater than the Ea
where fraction of gas molecules with kinetic energy greater than the activation energy is given
by the Maxwell—Boltzmann energy distribution curve
nEa
 Ea
 exp[
]
ntotal
RT
Since rate constant is proportional to the neither of effective collisions, the Arrhenius equation
can be derived
 Ea
)
Rate constant k  A exp(
RT
(A) Temperature and the collision theory
When the temperature is raised, there is a greater proportion of molecules with kinetic energy
more than the activation energy than there is at the lower temperature.
Interpretation
<1> Ea represent the minimum collision energy necessary for the reaction to occur.
<2> The area under the curve to the right of the activation energy represents the proportion of
particles that collide with kinetic energies greater than the activation energy, Ea.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.11
<3> The increase in the reaction rate with temperature corresponds closely to the ratio of the
corresponding shaded areas.
<4> Therefore, as temperature changes, the area under the distribution curve for E > Ea changes. This
means that
At higher temperatures, more molecules collide with energy greater than the activation energy
and so the rate of reaction increase.
<5>
The fraction of effective collisions increases exponentially with temperature. Approximately.
10°C rise in temperature doubles the number of molecules enough to cross the activation
energy barrier.
Exercise 1
Discuss the effect of temperature on the rate of a reaction in terms of the Arrhenius equation.
ANSWER
(B)
Concentration and the Collision Theory
The increase in the rate of a reaction with the increase of concentration of one or more reactant
particles can be explained by the Collision Theory.
• As the concentration of the reactant increases, the frequency of collisions increases.
• The probability of effective collisions (collisions with correct orientations and sufficient
energy for the reaction to occur) also increases.
• Reaction rate therefore increases.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.12
ENERGY PROFILE
I.
ENERGY PROFILE
Most reaction occur in more than one step. Each intermediate product has its specific
potential energy.
Energy profile is a representation of the changes in potential energy (against time) during a reaction.
In an energy profile, the potential energies of the followings are shown:
1. reactants
2. products
3. intermediates
4. transition states.
II
TRANSITION STATE THEORY
The transition state theory is developed to explain the rate of the reaction. It is mainly
concerned with the events during collision.
In the transition state theory,
<1> As reactant particles collide and reaction takes place, they are temporarily in a less
stable than the reactants or products. The atoms are rearranging themselves.
<2>
As the atoms are separated, the potential energy of the system increases. This results in
an energy barrier between the reactants and products.
<3>
An activated complex in the transition state is formed.
Note
<1> The activated complex only exists for a fraction of a second.
It is in equilibrium with the reactants and the products:
[A.....B]
C
A+B
( Product )
activated complex
in transition state
<2>
The activated complex in the transition state possesses the maximum potential energy in
the energy profile. It may decompose to the product or reactant.
<3>
The products are formed only if the colliding particles have sufficient energy to
overcome the energy barrier. The energy required is referred to as the activation
energy.
<4>
The higher the activation energy, the lower the number of effective collisions.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
<5>
Chpt. 15 : p.13
Distinguish between an intermediate and an activated complex with the aid of an energy profile
An intermediate: It is an unstable species at a potential minimum which is relatively unstable
compared to the reactants or products. (It can be detected or even isolated).
An activated complex: It refers to a highly unstable species with the arrangement of nuclei and
electrons at the maximum potential energy. It exists in a state between the
reactants and products called the transition state.
III. SINGLE STATE AND MULTIPLE—STAGE REACTION
(A)
Single stage reaction
If a reaction between AB an C t es place in one—step process,
A—B + C
 [A….B….C]  A + B—C
transition state
the energy profile for the reaction can be represented as follows
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.14
Example : Alkaline hydrolysis of 1-bromobutane
C4H9Br + OH- C4H9OH + BrRate of hydrolysis = k[C4H9Br][OH-]
It can be deduced that
• both 1-bromobutane and hydroxide ion are involved in the rate determining step,
• the hydrolysis of 1-bromobutane is a single-stage reaction. The mechanism may be the
following:
C4HgBr + OH-  [HO••••C4H9•••Br]  C4H9OH + BrThe reaction is a bimolecular second order reaction.
(B)
Multi-stage reaction
Multi-stage reactions are more common than single-stage reactions.
For a reaction between DE and F, which takes place via a two-step or multi-step process, an
intermediate is formed.
D--E
intermediate
+
F
slow
intermediate
fast
Overall Reaction : D—E + F
D--F

D—F
+
E
+ E
The energy profile for this two-stage process can be represented as follows:
Example : hydrolysis of 2-bromo-2-methylpropane
(CH3)3CBr + OH  (CH3)3COH + Brrate of hydrolysis = k[(CH3)3CBr]
The reaction is independent of the alkali concentration, i.e. the reaction is zero order with
respect to OH-.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.15
It can be deduced that
•
only 2-bromo-2-methylpropane is involved in the rate determining step.
•
the hydrolysis is a two—step process, with the proposed mechanism as follows:
This hydrolysis is a unimolecular first order reaction.
IV.
MECHANISM
<1> Mechanism is a sequence of simple steps to account for the overall chemical reaction that takes
place (in terms of bond breaking and bond formation).
<2> Each of the steps in a mechanism is known as an elementary step or intermediate step.
<3> Molecularity refers to the number of particles (molecules, ions or atoms) involved in each
elementary step.
Example :
The rate determining step in the hydrolysis of 2-bromo-2-methyl propane
slow
(CH3)3CBr
 (CH3)3C+ + Bris unimolecular because only (CH3)3CBr is involved.
The hydrolysis of 1-bromobutane
(CH3)3CBr + OH-  (CH3)3COH + Bris bimolecular because the rate-determining step involves 2 reactant particles, (CH3)3Br and OH-.
V.
RATE DETERMINING STEP
The rate determining step in a multi-stage reaction is the slowest step in the reaction.
The rate-determining step control the overall rate of reaction.
Note:
<1>
<2>
The activation energy for the rate-determining step greatest.
In the hydrolysis of 2-bromo-2-methylpropane, the intermediate step
slow
(CH3)9CBr
 (CH3)3C+ + Br
is the slow step in the reaction because it requires the greatest activation energy.
The r.d.s. determines the overall rate of the hydrolysis because once this step has proceeded, the
following intermediate step will proceed very quickly.
fast
(CH3)3C+  (CH3)3COH
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.16
<3> The molecularity of the rate-determining step is usually the order of the reaction.
Exercise 2
Consider the energy profile shown below for a 3—step reaction:
(a)
(b)
ANSWER
Which is the rate-determining step in the reaction? Explain.
Is the reaction exothermic or endothermic? Explain.
Exercise 3
The following exothermic reaction:
A2 + 2B  2AB
proceeds in two steps:
1. A2 + B  intermediate
(slow)
2. intermediate + B  2AB (fast)
(a)
(b)
ANSWER
Sketch the energy profile against reaction coordinate for the two-step reaction.
Indicate clearly on your diagram the transition state(s), the intermediate, and the enthalpy
change of reaction for the overall reaction.
If A2 is both a reactant. and a solvent
for the reacting system, determine the order of
reaction for the overall reaction. Give reasons for your answer.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.17
Exercise 4
In acid solution, bromate ions slowly oxidize bromide ions to bromine:
BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2 + 3H2O(l)
To investigate the rate of this reaction, the following experimental data are obtained at 298K:
Mixture
A
B
C
D
(a)
(b)
(c)
(d)
Volume of
molar
bromate /cm3
50
50
100
50
Volume of
Volume of
molar
molar H+(aq)
bromide /cm3
/cm3
250
300
250
600
250
600
125
600
Volume of
water /cm3
400
100
50
225
Relative rate
of formation
of bromine
1
4
8
2
Suggest a titrimetric method for determining the rate of reaction at various time of the
experiment.
Name one physical method which can be used to determine the rate of reaction. (Details
are not required.)
What is the rate equation for the reaction? Work out the units of the rate constant.
A proposed mechanism for the reaction is
Is this mechanism consistent with the rate equation for the reaction in (c)?
Give your reasons and discuss each of the three reactants.
ANSWER
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.18
CATALYSTS AND THEIR EFFECT ON REACTION RATES
I.
CATALYSTS 催化劑
A catalyst is a substance which alters改變 the rate of a particular chemical reaction, but
itself is chemically unchanged at the end of the reaction.
Note:
<1> The catalyst may be changed physically or it may undergo temporary chemical change.
<2> negative catalyst— a catalyst which slows down a reaction.
positive catalyst — a catalyst which speeds up a reaction.
<3> Catalytic action often continues until all the reactants have been converted into products.
Reactants + catalyst
Products + catalyst
Catalyst goes back to catalyze more reactions.
II.
WORKING PRINCIPLE OF CATALYST AND THE EFFECT OF CATALYST ON
REACTION RATE 催化劑的工作原理及對反應速率的影響
Since the presence of catalyst does not change the reactant concentration, the catalyst must cause
the rate constant to increase or decrease. The working principles of a catalyst to increase the
reaction rate are listed as follows:
<1> A positive catalyst can increase the rate constant, and hence the reaction rate by providing
an alternative pathway from reactants to products which has a lower activation energy.
<2> In terms of collision theory, a catalyst increases reaction rate by
(i)
bringing the reactant molecules closer together, thereby increasing the collision
frequency 使反應物聚在一起，增加它們的撞擊頻率。
(ii)
orientating the reactant molecules so that they achieve the correct collision
geometry. 將反應物分子導向，使它們達到正確撞擊位置。
Exercise 1
Discuss, in terms of the Arrhenius equation, the effect on the reaction rate of lowering the
activation energy.
ANSWER
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.19
(A) Provision of an alternative pathway for a reaction 提供另一反應途徑
In a catalyzed reaction, the catalyst participates in the reaction by providing an alternative reaction
pathway (or mechanism) for the production of the products.
• positive catalyst — provides an alternative pathway with lower activation energy.
negative catalyst — provides an alternative pathway with higher activation energy.
• Energy profiles showing the effect of a positive catalyst on the activation energy.
•
The smaller Ea means that in the reaction mixture, there is a greater total fraction of
molecules possessing sufficient energy to react.
•
Thus, a positive catalyst can increase the chance of effective collisions, and hence greater
reaction rate.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
(B)
Chpt. 15 : p.20
Mechanism for the alternative pathway provided by the catalyst
<1> In general, a catalyst is chemically involved in the catalysed reaction. It is consumed in one
reaction step and regenerated in a subsequent step.
A + Catalyst
 A—catalyst
A—catalyst +
B 
A—B + Catalyst
---------------------------------------------------------------------------Overall reaction: A + B  A—B
<2> The involvement of a catalyst causes the reaction to take place via an alternative pathway (the
catalytic pathway).
<3> The uncatalysed reaction is replaced by two other reactions, both of which have lower
activation energies, as illustrated below:
In conclusion, in a catalysed reaction,
• The enthalpy of reaction, H, is unchanged. Only the activation energy is altered.
• The reaction rate is increased as the activation energy is lowered in the catalytic pathway.
Exercise 2
Does the use of a suitable catalyst in a reaction change
(a) the order of the reaction, and
(b) the enthalpy of reaction?
Explain your answer.
ANSWER
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.21
III. CHARACTERISTIC OF CATALYSTS 催化劑的特徵
Most important characteristics
<1> It does not alter the equilibrium position of a reversible reaction because it increases both
forward and backward rates. 不會改變可逆反應的平衡位置
of
<2> It can increase the rate of production in many industrial processes (e.g. Haber process and
contact process) but they do not increase the yield of product. It only increases the rate
attainment of the equilibrium position, and hence the reaction rate.
能加快生成產物的速度，但不能提高產物的數量
Other characteristics of catalyst:
<1>
<2>
<3>
the
<4>
<5>
<6>
<7>
<8>
<9>
A catalyst is chemically unchanged and are not used up at the end of a reaction.
The action of a catalyst is specific. A certain catalyst can only catalyze some particular
reactions, but have no effects on others. 一些催化劑只可對某反應起作用，其他的反
應並不適用
A catalyst must be able to bind結合to the reactant particles reversibly. If the bonding is
too weak, the catalyst will be less powerful; if the binding is too strong, the reactant
particles will be permanently bonded to the catalyst and cannot be converted into
product.
Its physical state may be altered after the reaction.
In general, small amount of catalyst is enough to increase the rate of reaction.
Adequate足夠amount of catalyst is enough. Further addition of catalyst cannot does not
change the reaction rate.
Many catalysts are transition metals and/or their compounds, e.g. palladium, nickel,
vanadium (V) oxide, manganese(IV) oxide. This is because the transition metal can use
low energy vacant d orbitals to form dative bond with the reactants.
Impurities may bind to a catalyst irreversibly and “poison” the catalyst. The catalyst is
destroyed and loses the catalytic function. (e.g. lead compounds in car exhaust would
poison the platinum catalyst in catalytic converter.)
Finely divided (solid) catalyst gives a larger catalytic surface and is therefore more
effective than a large catalyst.
IV. AUTOCATALYSIS 自動催化
For most catalytic reactions, the amount of catalyst remains constant with time. However, in
autocatalytic reaction, the amount of catalyst increases as the reaction proceeds.
In autocatalysis, the product of reaction becomes the catalyst for the reaction.
Example :Effect of Mn2+ on the reaction between MnO4- and C2O42- in acidic medium
The Mn2+ ion, formed in the oxidation of ethane-1,2-dioic acid (oxalic acid) by acidified
potassium manganate (VII) catalyses the reaction:
2MnO4- (aq) + 16H+(aq) + 5C2O42-  2Mn2+ (aq) + 8H2O(l) + 10CO2(g)
The following curve of plotted against time shows the variation of reaction rate in the course of
the autocatalytic reaction.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.22
Interpretation:
<1> At the beginning of the reaction, the rate
is slow because the collision of negatively
charged MnO4- and C2O42- is difficult.
<2> Once a certain amount of Mn2- has been formed,
the reaction rate increases because Mn2+ acts as the
catalyst for the reaction.
<3> Reaction rate falls as the concentrations of reactants
decreases.
Exercise 3
Potassium manganate(VII) will oxidize ethanedioic acid (oxalic acid). (CO2H)2 to carbon
dioxide and water in the presence of excess acid:
2MnO4-(aq) + 16H+ (aq) + 5C2O42-  2Mn2+(q) + 8H2O(l) + 10CO2(g)
2 experiments are devised to investigate the effect of Mn2+ ion on the reaction.
Solution
0.2 M ethanedioic acid
0.2 M manganese (II) sulphate
2 M sulphuric acid
water
Experiment 1
100 cm3
0
5 cm3
95 cm3
Experiment 2
100 cm3
15 cm3
5 cm3
80 cm3
50 cm3 of 0.02 M potassium manganate(VII) are added to each reaction mixture. 10 cm 3 portion
of the mixture is withdrawn and run up into a conical flask. This sample of reaction mixture is
then added to about 10 cm3 of 0.1 M potassium iodide solution.
This stops the reaction and releases iodine equivalent to the residual manganate(VII) ions.
The iodine liberated is titrated with 0.01 M sodium thiosulphate. and a little starch solution is
added near the end point. Further portions of the reaction mixture are removed and treated in
the same way.
The titration results are plotted for each experiment.
(a)
(b)
Describe how the reactant concentration at various time intervals is derived from the
titration results.
The titration results are given in the curve shown below:
(c)
Interpret curve(2) briefly.
Give an explanation for the abnormality observed in curve 1.
Name a physical method which can be used to follow the reaction.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.23
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
V.
Chpt. 15 : p.24
HOMOGENEOUS 均相催化 AND HETEROGENEOUS CATALYSIS多相催化
There are two main types of catalysts, i.e. homogeneous and heterogeneous catalysis.
Note:
Enzymes, which are very important biological catalysts, do not fit into either of the
above cataegories.
(A)
Homogeneous catalysis
In homogeneous catalysis, the catalyst and the reactants are in the same phase, i.e. physical state.
Note
<1> Usually, the catalyst forms an intermediate compound which decomposes at a later stage.
<2> Homogeneous catalysis in the liquid state may be illustrated by acid-catalyzed
esterification:
CH3COOH (l) + C2H5OH (l)  CH3COOC2H5 (l) + H2O (l)
This reaction between a carboxylic acid and an alcohol is catalysed by a little
concentration sulphuric acid.
Exercise 4
The reaction S2O82-aq + 2I-(aq)  2SO42-(aq) + I2(aq) is catalysed by Fe2+(aq), a transition
metal ion.
(a) Is this a homogeneous catalysis or heterogeneous catalysis? Explain.
(b) Outline the essential steps to show how the catalytic activity of Fe2+ (aq) can be
illustrated experimentally.
ANSWER
(B)
Heterogeneous catalysis
In heterogenous catalysis, the reactants and the catalyst are in two different phase.
Note:
<1> The behaviour of heterogeneous catalysts is explained by the Adsorption Theory
(a) The reactants come together by adsorption on the surface of the catalyst. The bonds
of the reactants are weakened.
(b) Reaction then occurs on the catalytic surface.
(c) After the product has been formed on the surface of the catalyst, the product
molecules are desorbed from the catalyst.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.25
<2> Adsorption Theory can be illustrated by the catalytic hydrogenation of ethene:
<3> Generally, heterogeneous catalysis involves the reaction of two gases on the surface of a catalyst.
Examples include:
1. the conversion of harmful pollutants from car exhausts into harmless substances in catalytic
converters
2. the hydrogenation of unsaturated oils (or alkenes) using nickel as catalyst.
3. the conversion of sulphur dioxide into sulphur trioxide in the Contact process
4. the production of ammonia from hydrogen and nitrogen in the Haber process
<4> In conclusion,
A heterogeneous catalyst provides a surface onto which reactants are adsorbed and from which
products of reaction are desorbed.
<5> Catalytic decomposition of hydrogen peroxide solution with manganese(IV)oxide (MnO2)
solid is an example of heterogeneous catalysis of a reaction in solution
MnO2
2H2O2(aq)
2H2O(l) + O2(g)
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.26
VI. APPLICATIONS OF CATALYSTS
Catalysts are widely used in many industrial processes, e.g. Haber process, contact process.
They increase the rate of production but they do not increase the yield of products.
(A)
Contact process (Production of sulphuric acid)
The major steps of the contact process are summarized in the table below:
Step
1
Description
Sulphur dioxide is produced by
roasting sulphur or heating
metal sulphides in air.
Equation
S(s) + O2(g)  SO2(g)
2ZnS(s) + 3O2(g)  2ZnO(s) + 2SO2(g)
4FeS2 + 11O2(g)  2Fe2O3(s) + 8SO2(g)
2
Sulphur dioxide and air are
purified to prevent poisoning
the catalyst in step (3).
3
Sulphur dioxide is oxidized by air
to sulphur trioxide. Conditions:
450 °C,
atmospheric pressure,1 atm
platinum or vanadium (V)oxide
catalyst.
4
Sulphur trioxide is dissolved in
concentrated sulphuric acid to
form oleum (fuming sulphuric
acid) in the absorber. Oleum is
dissolved in water to
SO3(g) + H2SO4(l)  H2S2O7(l)
5
Oleum is dissolved in water to
give sulphuric acid.
H2S2O7(l) + H2O(l)  2H2SO4(l)
2SO2(g) + O2(g)  2SO3(g)
Note
<1> The conversion of sulphur dioxide (SO2) to sulphur trioxide (SO3) involves the catalytic
action of platinum or vanadium (V) oxide.
<2> The flow diagram for the manufacture of sulphuric acid is shown below:
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt. 15 : p.27
(B) Haber process (Production of ammonia)
Catalyst used: finely divided iron, platinum or copper
In the Haber process. nitrogen and hydrogen combine directly to manufacture ammonia.
Procedure
<1> Initially, ammonia (1 part) and hydrogen (3 parts) are purified, dried and mixed. The
mixture is compressed to 200 atmospheres at 500°C and then passed over a catalytic
chamber containing the catalyst. The following reaction occurs: -
Fe
N2(g) + 3H2(g)
2NH3(g)
<2> The ammonia produced is cooled and condensed to liquid and collected.
<3> The unreacted nitrogen and hydrogen re—enter the catalyst chamber for recycling.
(C)
Hydrogenation of unsaturated oils
Margarine and other fats are made by passing hydrogen through a heated vegetable oil in the
presence of nickel catalyst. Under these conditions, hydrogen adds across some or all the double
bonds in the oil.
finely divided nickel,
—CH=CH—
unsaturated oil
+ H2
200 C
0
—CH2—CH2—
saturated oil
Note:
<1> The saturated oils forms high-melting point soft solid fats on cooling.
<2> Hydrogenation of unsaturated oils and alkenes using nickel, palladium or platinum catalyst
is an example of heterogeneous catalysis.
(D) Catalytic converters in car exhausts systems
A catalytic converter is used in some exhaust systems to convert some pollutants from car
exhausts (e.g. carbon dioxide, nitrogen monoxide and unburnt hydrocarbons) into relatively
harmless substances (e.g. carbon dioxide, nitrogen and water).
Working principles :
<1> The exhaust gases are passed through a “honeycomb” of small beads coated with metal
catalysts such as platinum (or palladium) and rhodium. The “honeycomb” increases the
surface area of the catalyst for better catalytic action.
<2> The converter is heated and extra air is pumped into it as shown.
A catalytic coverter
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 15
Chpt.15:p28
<3> Reactions:
(i) Nitrogen oxides react with carbon dioxide when they pass over the platinum catalyst
to form carbon dioxide and nitrogen:
2CO(g) + 2NO(g)  2CO2(g) + N2(g)
(ii)
Unburat hydrocarbons and carbon monoxide are oxidized by the air pumped into
the catalytic converter, forming water and carbon dioxide.
<4> Cars which include catalytic converters in their exhaust systems must use lead—free
petrol to prevent the catalyst from being coated with and poisoned by lead.
Conclusion
Many industrially important heterogeneous catalyst are d-block transition metals or their
compounds, e.g. nickel, vanadium, manganese, manganese(IV) oxide.
They use the 3d and 4s subshell to adsorb reactant molecules and lower the activation
energies by weakening the bonds in the adsorbed molecules.
VII. ENZYMES AS EXAMPLES OF BIOLOGICAL CATALYSTS
Enzymes are proteins that catalyze specific biochemical reactions in living systems. They are
often called biological catalysts. Without them, most biochemical reactions would be too
slow to sustain life.
For example, enzymes obtained from yeasts have long been used in the production of alcohol
by fermentation. The fermentation. of glucose to form ethanol can be represented by the
following equation:
C6H12O6
•
enzyme
2C2H5OH+ 2C02
Nowadays. enzymes have many uses, such as in the manufacture of biological washing
powders. These washing powders contain enzymes which can break down stains caused by
blood, egg, sweat and fats. They have the advantage of removing stains even at normal
temperature.