Download Unit 5 Chemical Kinetics Section 5.1 Rates of Chemical Reaction

Unit 5 - 1
Unit
5
Section
(1)
Chemical Kinetics
5.1
Rates of Chemical Reaction
The meaning of the rate of a chemical reaction
Chemical kinetics is the study of reaction rates, factors that affect the reaction rates and reaction
mechanisms.
Reaction mechanisms are suggested pathways of reactions.
The rate of reaction is the rate at which products are formed or the rate at which reactants are used
up in the reaction.
Rates vary immensely from reaction to reaction. Many ionic reactions take place instantaneously.
For example, the following neutralization reaction is instantaneous :
H+(aq) +
OH-(aq) → H2O(l)
In other cases, reactions keep on going for months or even years. The fermentation of grape juice
to form wine can take place many months to complete. And then, when the wine is stored, the complex
chemical processes which take place to bring out its full flavour may take years.
The average rate of a chemical reaction over a certain interval of time is equal to the change in the
concentration of a reactant or product that occurs during that time divided by the time.
Consider a reaction of the type
A →
B
where 1 mole of the reactant produces 1 mole of the product. A rate curve shows the change in
amount of either the product formed or the reactant used up with time.
concentration
t
time
The rate of reaction decreases as the reaction proceeds and the reactant is gradually used up.
At time t, the rate of reaction is the gradient of the tangent to the curve at that time.
d [ A]
d [ B]
=
Rate = dt
dt
The rate at the start of the reaction is called the initial rate. Since the reaction is fastest at the
beginning of the reaction, the slope for the initial rate is steepest. Gradually, as the reaction proceeds,
the slope becomes less steep. Finally, when there is no further reaction, the curve flattens out. The
tangent is a horizontal line and the reaction rate is zero.
Unit 5 - 2
For the general reaction :
aA
+
→ cC
bB
+
dD
The formal definition of the reaction rate is taken as the time derivative of a concentration divided by
the appropriate stoichiometric coefficient and converted to a positive value.
Rate
=
As concentration is expressed in mol dm-3
mol dm-3 s-1.
Example 1
In the reaction :
Rate
BrO3- (aq) +
5Br- (aq) +
6H+ (aq) →
3Br2 (aq)
+
3H2O (l)
=
Example 2
Consider the reaction :
(a)
and time in second, thus the unit of reaction rate is in
N2 (g)
+
3H2 (g)
→
2NH3 (g)
Write an expression for the rate of reaction.
(b) If the rate of formation of NH3 is 2.4 x 10-2 mol dm-3 s-1 at a particular moment, what are the rate of
disappearance of N2 and H2 ?
(c)
What is the rate of reaction ?
Unit 5 - 3
( 2 ) Following a reaction by chemical and physical methods
The rate of reaction is found by measuring some property of a reactant or a product at different times
after the start of the reaction.
1. Following the change in amount of reactant / product by titration
A reaction mixture is made up and small samples are withdrawn regularly from it, using pipette at
measured time intervals from the start of the reaction. In this way the main reaction is not interfered.
The reaction is then quenched (slowing it abruptly) by
rapid cooling in ice, or
removing the catalyst, or
diluting with cold water, thus lowering the reaction temperature and concentrations of the reactants at
the same time.
The concentrations of the reactants or products at the time the sample is being taken out can then be
determined by appropriate titrimetric methods.
Example : Hydrolysis of ethyl ethanoate
CH3COOC2H5(l) +
H2O(l)
CH3COOH(aq) +
C2H5OH(aq)
During the acid catalyzed hydrolysis of ethyl ethanoate, the concentration of ethanoic acid produced
at different times can be measured by taking samples of reaction mixture with a pipette, chilling them in
ice-cold distilled water and titrating them with aqueous sodium hydroxide solution of known
concentration.
2. Determining the volume of gas formed
Example : The reaction between magnesium and hydrochloric acid
Mg(s) + 2 HCl (aq) → MgCl2(aq) + H2(g)
The reaction between magnesium strips with hydrochloric acid involves the production of hydrogen
gas which does not dissolve readily in the reaction mixture. The gas can be collected and its volume
measured in a graduated gas syringe.
volume of H2(g)
0
time
The magnesium strips are in the form of fairly large pieces so that their surface area does not change
appreciably during the reaction The reaction is started by dropping the magnesium strips into the acid.
The air in the apparatus is displaced by the hydrogen produced. This pushes back the plunger of the gas
syringe. The volume of gas is taken every 30 seconds until the reaction is over and the volume no
longer changes. The plunger has to be rotated before taking each reading, to ensure that it has not
become jammed. This also ensures that the pressure inside the syringe is approximately atmospheric.
Unit 5 - 4
3. Following the change in mass
The method of following the change in mass is applicable to reactions involving the production of
gases. The gas formed is allowed to escape and the change in mass of the reaction mixture during the
course of the reaction is followed.
Example 1 : The reaction between marble chips and hydrochloric acid
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) +
Set-up :
CO2(g)
Rate curve :
Mass of CO2(g)
time
In some chemical reactions there may be some sudden observable changes after the reaction has
started for some time. The time interval to reach this easily seen stage of the reaction is therefore
inversely proportional to the average rate of reaction over that time interval.
Example 2 : The reaction between thiosulphate and hydrochloric acid
S2O32-(aq) + 2 H+(aq) → SO2(g) + S(s) + H2O(l)
The time required for the formation of a certain amount of sulphur can be measured. A simple way
to do this is to carry out the reaction in a beaker standing on a piece of paper on which a cross has been
marked. On looking vertically down onto the reaction mixture in the beaker, the cross will gradually
become fainter as more precipitate forms and will become invisible when a certain concentration of
sulphur is present.
Average rate of reaction
∝
Unit 5 - 5
4. Colorimetric measurement of light intensity
Colorimeter
A colorimeter is an instrument designed for measuring the intensity of light passing through
solutions. Inside this instrument, a narrow beam of light is allowed to pass through the solution under
investigation towards a sensitive photocell. A filter is usually inserted between the light source and the
solution, to select the most appropriate wavelength to be absorbed by the solution. The current
generated in the photocell is proportional to the amount of light transmitted by the solution, which in turn
depends on the colour intensity of the solution.
In a reaction, if one of the reactants or products has a colour, the intensity of this colour will change
during the reaction.
The colorimeter is calibrated to show the fraction of light absorbed, so that this will be directly
proportional to the concentration of the coloured species in the reaction mixture.
Absorbance (colour intensity) ∝ [coloured species]
Example : The reaction between bromine and methanoic acid
Br2(aq) + HCOOH(aq) → 2 Br-(aq) + 2 H+(aq) +
CO2(g)
The red-brown colour of bromine fades during the oxidation of methanoic acid by bromine. Such
change in colour intensity could be followed by a colorimeter. The concentration of bromine in the
reaction mixture at a particular instant can be read off from a calibration curve previously drawn up with
bromine solutions of known concentrations and colour intensity (absorbance). By plotting the
concentrations of bromine at various times during the course of the experiment, the rates of the reaction at
various times can be found by drawing tangents on various points on the curve, and determining values of
the gradients.
Calibration curve :
Rate curve :
[Br2(aq)]
Colour
intensity
[Br2(aq)]
time
Question : For the reaction given below, suggest methods for following the rate of reaction.
5 C2O42-(aq) + 2 MnO4-(aq) + 16 H+(aq) → 10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l)
1. Colorimetric method : MnO4- is the only coloured species with deep purple colour. By using a
colorimeter, the decrease in [MnO4-] can be followed by the decrease in colour intensity of light at
various time.
2. By using a graduated gas syringe, the volume of CO2 gas produced at various time can be found
under constant pressure and temperature.
Unit 5 - 6
Section
1.
2.
5.2
Factors influencing Reaction Rate
Basically for a chemical reaction to occur :
The reactants must collide together; and
The reactants must have the right amount of energy.
For a new substance to be made, some of the bonds in the original substance have to be broken
before new bonds can be formed. Bond breaking requires energy. The minimum energy that chemicals
must have before they can change to products is called the activation energy.
There are six factors which can affect the rates of chemical reactions.
(1)
Effect of concentration on reaction rate
An increase in concentration of reactants will give more particles per unit volume and hence more
collisions per unit time, thus increasing the chance of reaction.
Example : The reaction between magnesium and hydrochloric acid
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
The experiment is repeated with the same mass of large piece magnesium strips but different
concentrations of acid : 0.5 M. 1.0M, 1.5 M and 2.0 M. The reactions are followed by plotting the
volume of hydrogen gas collected with time of reaction. It is obvious that the initial rate is faster if the
acid is more concentrated, and the line steeper. Notice that as the magnesium strips are used up each
time, the total volume of H2(g) collected must be the same for these reactions.
Volume of
hydrogen gas
collected
Time
( 2 ) Effect of pressure on reaction rate
For reactions involving gases, an increase in pressure will have the same effect of increasing the
number of particles per unit volume. This again increases the number of collisions per unit time, and
hence the rate of the reaction.
Unit 5 - 7
( 3 ) Effect of surface area on reaction rate
Increasing the surface area of solid reactants is a convenient way of speeding up reactions. The
greater the surface area of a solid, the greater the area of contact open to reactants.
Example : The reaction between marble chips and hydrochloric acid
CaCO3(s) + 2 HCl(aq) → CO2(g) + H2O(l) + CaCl2(aq)
If equal masses of powdered marble and marble chips are added separately to equal volumes of
1 mol dm-3 hydrochloric acid at the same temperature, the ground up powder of marble can be seen to
react more readily with hydrochloric acid, since it takes a shorter time for the reaction to reach
completion.
Set-up :
Rate curve :
time
( 4 ) Effect of temperature on reaction rate
The most significant method to make reactions go faster is to heat the reactants. An increase in
temperature increases the number of reactant particles having energy greater than the activation energy of
the reaction, thus producing more fruitful collisions. Moreover, the increase in temperature also
increases the average kinetic energy of the particles. This will result in higher velocities of the particles
and more collisions per unit time, thus a greater chance of reaction. For many reactions, a rise in
temperature of 10℃ roughly doubles the reaction rate.
Example : The reaction between sodium thiosulphate and hydrochloric acid
S2O32-(aq) + 2 H+(aq) → S(s) + SO2(g) + H2O(l)
A beaker containing 25 cm3 of 0.1 mol dm-3 sodium thiosulphate solution is placed over a cross
drawn on a piece of paper. Then 5 cm3 of a dilute solution of hydrochloric acid is added. Timing is
started as soon as the acid is added. Eventually there is so much sulphur that it is no longer possible to
see the cross through the solution. The time taken for the cross to disappear ( t ) is noted. This
experiment is repeated at other higher temperatures, using the same volume and concentration of
reactants. Note that the reaction rate increases in an exponential way with temperature.
1
Average rate ∝
t
Temperature / oC
Unit 5 - 8
( 5 ) Effect of catalyst on reaction rate
A catalyst speeds up a reaction by providing an alternative mechanism (pathway) for a reaction.
The alternative mechanism has a lower activation energy than the original route, so that at a given
temperature, more molecules can react.
Example : Catalytic decomposition of hydrogen peroxide by MnO2
O2(g) + 2 H2O(l)
2 H2O2(aq)
Cat. MnO2(s)
Under normal conditions, the decomposition of hydrogen peroxide is very slow. When a fine,
black powdery form of manganese (IV) oxide is added to a cold and dilute hydrogen peroxide solution,
the hydrogen peroxide starts to decompose rapidly. The manganese (IV) oxide is not used up and when
the reaction finishes, the black powder is still there.
For a given amount of manganese (IV) oxide added, the concentration of hydrogen peroxide at
different times in the reaction mixture can be followed chemically by withdrawing a known volume of
reaction mixture, filtering off the MnO2 solids and adding to it excess dilute sulphuric acid. This
acidified mixture is then titrated against standard potassium manganate (VII) solution to obtain the
concentration of hydrogen peroxide :
concentration
of H2O2(aq)
time
( 6 ) Effect of light on reaction rate
Light affects the rates of certain chemical reactions by providing suitable frequencies of
electromagnetic radiation which can effect photochemical breakdown of covalent bonds, or excite the
outer electrons of atoms to result in their ionization.
Example : The reaction between bromine and hexane
UV light
C6H14(l) +
Br2(l)
C6H13Br(l) +
C6H12Br2(l) + ………… + HBr(g)
in CCl3CH3(l)
The reaction of bromine with hexane proceeds quite quickly under a lamp but much more slowly in
dark. One way of demonstrating this is to add several drops of 2% bromine in 1,1,1-trichloroethane into
10 cm3 of hexane in a test-tube. Loosely cork the tube and irradiate with a photoflood light. Prepare a
second similar tube and leave it in a dark place. Compare the colour intensity of bromine at intervals
using colorimetric method. Reaction in the irradiated tube is evidenced by the discharge of the bromine
colour and evolution of misty fumes of hydrogen bromide. The tube kept in dark has its bromine colour
relatively unchanged.
In the above reaction the Br-Br bond undergoes homolytic fission readily upon irradiation with light
and a chain reaction is initiated.
Unit 5 - 9
Section
5.3
Rate Equation and Order of Reaction
( 1 ) Rate equation and order of reaction
Rate Law
For any given reaction, the rate is depended on the concentration of the reactants. In fact, the rate is
directly proportional to the concentration of the reactants raised to some power.
Consider the general case of a reaction having the equation :
nB →
mA+
products
If the rate is found by experiment to be proportional to [A]x and [B]y , then
Rate
=
x
k [A] [B]
y
m ≠ x and n≠ y
This equation is called the rate equation for the reaction.
The constant k is known as the rate constant for a given reaction at a particular temperature.
It is found to vary with temperature.
The indices x and y are known as orders. The reaction is said to be of
x order with respect to A,
y order with respect to B and
(x + y) order overall.
The order of reaction is thus the sum of the indices of the concentration terms in the
experimentally-determined rate equation.
If the reactants are gases, the rate equation may be expressed in terms of partial pressures :
Rate
=
Example 1
For a general reaction :
x
k (PA) (PB)
y
bB →
a A+
cC +
dD
It is found experimentally that the reaction is zero order with respect to A and second order with
respect to B. Write the rate equation for this reaction.
The overall order for the reaction is
___.
Example 2
Some second order reactions are given in the table below:
Equation
3ClO-(aq) → ClO-3(aq)
H2(g) +
I2(g) →
Rate equation
+
2Cl-(aq)
2HI(g)
Unit 5 - 10
Example 3
A solution of Q , of concentration 0.20 mol dm-3 undergoes a first order reaction at an initial rate
of 3.0 x 10-4 mol dm-3 s-1. Calculate the rate constant.
The dimension of a first order rate constant is time-1 .
Example 4
A second order reaction takes between the reactants P and Q , which are both initially present at
concentration 0.20 mol dm-3 . If the initial rate of reaction is 1.6 x 10-4 mol dm-3 s-1 , what is the rate
constant ?
The dimension of a second order rate constant is concentration-1 time-1 .
Example 5
The order of reaction does not follow from its stoichiometric equation. The reaction between
bromate (V) ions, bromide ions and hydrogen ions to give bromine is represented by the equation :
BrO3-(aq) +
5 Br-(aq) +
6 H+(aq) → 3 Br2(aq) +
3 H2O(l)
From the results of kinetic measurements, the reaction is first order with respect to bromate(V), first
order with respect with bromide, second order with respect to hydrogen ion and fourth order overall.
Rate =
The dimension of rate constant is ___________________ .
Example 6
2HI(g) obeys overall second order kinetics and
The gas-phase reaction H2(g) + I2(g) →
is first order in each reactant. At 400oC the rate constant is 2.42 x 10-2 dm3 mol-1 s-1 . Calculate the
reaction rate when the concentration of each reactant is 0.50 mol dm-3 .
Unit 5 - 11
( 2 ) Order of reaction from initial rate
In a reaction : A →
X
the rate of reaction = k [A]n , where n = order of reaction.
The order of reaction may be found by comparing the initial rates of reactions at different
concentrations. Two experiments to find the rate of reaction, two ‘runs’, are done at different
concentrations of A.
[A0]1 = initial concentration of A in run 1, and R1 = the initial rate in run 1;
[A0]2 = initial concentration of A in run 2, and R2 = the initial rate in run 2.
n can be found by the ratio :
 [A ]
R1
=  0 1
R2
 [ A0 ] 2
log (



n
 [A ]
R1
) = n log  0 1
R2
 [ A0 ] 2



Example 1
The following results were obtained for a reaction between A and B :
Run
Concentrations
[A]
0.50
0.50
0.50
1.0
2.0
a
b
c
d
e
/ mol dm-3
[B]
1.0
2.0
3.0
3.0
3.0
/ mol dm-3 s-1
Initial rate
2.0
8.0
18.0
36.0
72.0
(a) What is the order of reaction with respect to A and with respect to B ?
Let the rate equation be rate = k [A]m [B]n
Compare runs d and e, in which [B] is constant.
m
rate e  2.0 
72
=2
∴m = 1
=
 =
rate d  1.0 
36
Compare runs a and b, in which [A] is constant.
n
rate b  2.0 
8
∴n = 2
=
 = =4
rate a  1.0 
2
The reaction is first order with respect to A and second order with respect to B.
(b) What is the rate equation for the reaction ?
rate = k [A] [B]2
(c) Calculate the rate constant.
From run c,
18 = k x 0.5 x (3.0)2
k = 4.0 mol-2 dm6 s-1
Unit 5 - 12
Example 2
For the homogeneous reaction
2X + 3Y → 2Z
the following experiment data at 37oC are available :
Concentration of X /mol
Concentration of Y /mol dm-3
dm-3
0.10
0.01
0.10
0.02
0.05
0.02
Initial rate
/mol dm-3 s-1
2.22 x 10-2
4.44 x 10-2
1.11 x 10-2
(a)
Deduce the rate equation for the reaction.
(b)
What is the order for the reaction ?
(c)
Calculate the initial rate if 20 cm3 of 0.10 mol dm-3 X is mixed with 30 cm3 of 0.20 mol dm-3 Y.
(d)
Calculate the initial rate of disappearance of Y and the initial rate of formation of Z for the same
conditions as in (c).
Example 3
In acid solution, chlorate (V) ions, ClO3-, slowly oxidize chloride ions to chlorine. The following
kinetic data are obtained at 25℃.
ClO-3(aq) + 5 Cl-(aq) + 6 H+(aq) → 3 Cl2(g) + 3 H2O(l)
Experiment
1
2
3
4
(a)
[ClO3-]
/ mol dm-3
0.08
0.08
0.16
0.08
[Cl-]
/ mol dm-3
0.15
0.15
0.15
0.30
[H+]
/ mol dm-3
0.20
0.40
0.40
0.20
Initial rate
/ mol dm-3 s-1
1.0 x 10-5
4.0 x 10-5
8.0 x 10-5
2.0 x 10-5
Determine the order of the reaction with respect to each reactant.
(b) Write the rate equation for the above reaction and calculate the rate constant.
Unit 5 - 13
( 3 ) First order reaction
If [A] is the concentration of the reactant at time t, the first order reaction rate is given by the
d [ A]
equation :
rate = = k [A]
dt
where k is the rate constant for the reaction.
If the initial concentration of A is [A]0, then at a later time t it will have fallen to [A]. Integrating of
equation gives :
-kt
[A] = [A]0 e
which shows that the concentration follows an exponential decay.
Figure 1 : A plot of rate against [A]
Figure 2 : A plot of [A] against t
gives :
[ A]
= −kt
ln [A] = ln [A]0 - kt
or ln
[ A]0
A reaction can be tested for first order character by plotting ln [A] against time. If the line obtained
is straight, then the reaction is first order and the slope gives the rate constant, -k.
Equation can be used to find the half-life ( t1 ) of a reactant : the time needed for its
concentration to fall to half its initial value.
On taking natural logarithms of equation
The half-life of a first order reaction is independent of the initial concentration.
initial concentration of A, it falls to half that value in a time 0.693/k.
Figure 3 : A plot of ln[A] against t
Whatever the
Figure 4 : The half-life of a first order reaction
is independent of the initial conc.
Unit 5 - 14
( 4 ) Second order reaction
If [A] is the concentration of the reactant at time t, the second order reaction rate is given by the
equation :
d [ A]
rate = = k [A]2
dt
where k is the rate constant for the reaction.
Figure 1 : A plot of rate against [A]
Figure 2 : A plot of rate against [A]2
If the initial concentration of A is [A]0, then at a later time t it will have fallen to [A]. Integrating of
equation gives :
1
1
=
+ kt
[ A] [ A]0
A reaction can be tested for second order character by plotting 1/[A] against time. If the line
obtained is straight, then the reaction is second order and the slope gives the rate constant, k.
Equation
can be used to find the half-life ( t1 ) of the reactant.
The half-life of a second order reaction depends on the initial concentration, and the higher that
concentration is, the more rapidly does [A] decrease to half its value.
Figure 3 : A plot of 1/[A] against t
Figure 4 : The half-life of a second order reaction
increases as the initial conc. of reactant is decreased.
Unit 5 - 15
( 5 ) Zeroth order reaction
In a zero order reaction the rate is independent of the concentration of the reactant.
equation for a zero order reaction is :
d [ A]
= k [A]0 = k
rate = dt
and the rate constant has the dimension concentration time-1 .
The rate
A plot of the concentration of the concentration of the reactant [A] against time has the form shown
in figure :
Since the rate is a constant during the reaction, a plot of rate against time or [A] has the form shown
in figure :
Pseudo zero order reactions
In some reactions, if one of the reactant is present in large excess, the concentration of that reactant
is practically constant, and the rate appears to be zero order with respect to it.
Example :
The acid-catalyzed hydrolysis of ethyl ethanoate is first order with respect to ester and first order
with respect to water :
CH3COOC2H5(l) + H2O(l) → CH3COOH(aq) + C2H5OH(aq)
H+(aq)
As water is present in large excess, only a small fraction of the water will be used up in the reaction.
The concentration of water remains unchanged, and the rate depends on the concentration of ester alone :
k’ = a first order rate constant.
The reaction seems to be zero order with respect to water.
Unit 5 - 16
( 6 ) Calculations involving rate equations and rate constants
Example 1
The hydrolysis of sucrose is a first order reaction, and the half-life is 80 min at 20℃.
(a) Calculate the rate constant at 20℃.
(b) Calculate the concentration of sucrose that remains after 240 min if the initial concentration of sucrose
is 0.10 mol dm-3.
(c) If 10% of sucrose is hydrolyzed in 30 min at 10℃, calculate the half-life at this temperature.
Example 2
The decomposition of nitrogen dioxide to nitrogen monoxide and oxygen at 300℃ is a second order
reaction :
2 NO2(g) → 2 NO(g) + O2(g)
(a) Write the rate equation for the reaction.
(b) If the initial concentration of NO2(g) is 0.10 mol dm-3 and it takes 10 s for [NO2] falls to half its initial
value, how long will it take for [NO2] falls to one-quarter its initial value ?
Example 3
On the iodination of propanone in acidified aqueous solution,
CH3COCH3(aq) + I2(aq)
CH3COCH2-I(aq) +
Cat. H+(aq)
the following results are observed :
(i) When the propanone concentration is doubled, the rate doubles.
(ii) When the iodine concentration is doubled, the rate remains unchanged.
(iii) When the pH of the solution is reduced from 1.0 to 0.70, the rate doubles.
Deduce the rate equation for the reaction.
HI(aq)
Unit 5 - 17
( 7 ) Simple rate equations determined from experimental results
Experiment 1 : Decomposition of NO2(g) to NO(g) and O2(g)
Object :
To determine the order and rate constant for the reaction
2 NO2(g) → 2 NO(g) + O2(g)
Results :
The following results were obtained at 300℃.
[NO2(g)]
/ mol dm-3
2.28 x 10-2
2.02 x 10-2
1.82 x 10-2
1.65 x 10-2
Rate of decomposition of NO2
/ mol dm-3 s-1
2.60 x 10-4
2.02 x 10-4
1.65 x 10-4
1.35 x 10-4
Interpretation :
The rate equation can be expressed as
where k is the rate constant and n is the order for the reaction.
On taking logarithms of equation gives :
A plot of log [rate] against log [NO2] would give the order n of the reaction as the slope of the
straight, and the intercept gives the value of log k.
Log [NO2(g)]
Log rate of decomposition of NO2
- 1.64
- 1.70
- 1.74
- 1.78
- 3.59
- 3.70
- 3.78
- 3.87
From the graph, slope of the straight line = 2. Therefore the order of the reaction is second order
with respect to NO2. Also from the graph, the y-intercept is – 0.32 and the rate constant k can be found :
Unit 5 - 18
Experiment 2 : Reaction between H2O2(aq) and
I-(aq)
under acidic condition
Object :
To investigate the rate of reaction
H2O2(aq) + 2 I-(aq) + 2 H+(aq) → I2(aq) +
Results :
The following results were obtained at 20℃.
Run
1
2
3
Volume of 1M
H2SO4 / cm3
10.0
10.0
10.0
Volume of 0.1M
H2O2 / cm3
5.0
5.0
5.0
2 H2O(l)
Volume of 0.1M
KI / cm3
5.0
10.0
15.0
Volume of
H2O / cm3
20.0
15.0
10.0
1
/ s-1
t
0.011
0.022
0.033
Interpretations :
(a) What method is used for finding t (the time for a small fixed quantity of I2 to be formed) for each
sample ? Explain how this is related to the initial rate of reaction for each run.
The method is to add a small but fixed quantity of sodium thiosulphate solution to each sample,
together with 1 cm3 starch indicator. The thiosulphate reacts with the iodine formed in the above
reaction :
2 S2O32-(aq) + I2(aq) → S4O62-(aq) + 2 I-(aq)
At the time ( t ) that all the thiosulphate has reacted, free iodine is produced and its presence is
shown by the appearance of the dark blue colour of the iodine-starch complex.
1
As long as the quantity of thiosulphate ions added for each run is the same,
can be taken as a
t
measure of the initial rate of reaction.
1
Initial rate ∝
t
(b) From the data above, what can you say about the order of the reaction with respect to iodide ion ?
As the concentrations of H+ and H2O2 are kept constant and
1
represents the rate :
t
(c) How would you find out the order of the above reaction with respect to H+(aq) ion ?
By keeping the concentrations of I-(aq) and H2O2(aq) constant and varying the concentration of H+(aq) in
1
different runs, order of reaction with respect to H+ ion can be determined by comparing the values of .
t
Unit 5 - 19
Experiment 3 : Decomposition of benzenediazonium chloride
Object :
To determine the rate equation for the reaction in which benzenediazonium chloride is hydrolyzed.
Introduction :
Benzenediazonium chloride is an unstable compound, which decomposes above 5℃ to give phenol,
nitrogen gas and hydrochloric acid :
C6H5N2Cl(aq) + H2O(l) → C6H5OH(aq) + N2(g) + HCl(aq)
Set-up and procedure :
Benzenediazonium chloride solution is poured into the side arm tube shown below. This side arm
tube is then immersed in a thermostat at 45℃.
The three way tap is then connected to the gas syringe and the volume of nitrogen collected is noted
at one minute intervals for about 25 minutes.
Results :
A plot of the volume of nitrogen gas collected against time is shown below :
Interpretations :
(a) Describe how the results could be used to show that the hydrolysis reaction is first order with respect
to benzenediazonium ion.
(b)
A benzenediazonium chloride solution gives 80 cm3 of nitrogen gas on complete decomposition.
It is found that, at 45℃, 40 cm3 of nitrogen are evolved in 16 min. How long after the start of the
decomposition will 70 cm3 of nitrogen have been evolved ?
Unit 5 - 20
Section
5.4
Arrhenius Equation
( 1 ) Explanation of the effect of temperature change on reaction rate in terms of
activation energy
In a chemical reaction, bonds are first broken and then others are made. Energy is therefore
required to start this process, whether the reaction is exothermic or endothermic overall. It is reasonable
to assume that reaction occurs only as a result of those collisions which occur between particles having a
certain minimum energy.
The minimum energy needed by the reacting particles so that a reaction may occur between is called
the activation energy.
Activation energy represents the ‘energy barrier’ which must be successfully overcome in order for
the reaction to occur. The energy required to overcome this energy barrier is provided by the kinetic
energies of colliding particles.
The potential energy – reaction coordination graphs for (a) exothermic reaction and (b) endothermic
reaction are shown below :
As the temperature rises, the kinetic energy of the reacting particles increases. Thus, the number
of particles possessing the required activation energy is increased. Consequently, there are more
effective collisions per unit time and a higher reaction rate is resulted.
(2)
The Arrhenius equation
As a general rule of thumb, the rate of reaction doubles for every 10℃ rise in temperature. This
would seem to suggest that there is an exponential relationship between rate and temperature.
The exact relationship was first proposed by a Swedish chemist called Arrhenius in 1889. It is
known as the Arrhenius equation :
where
k is the rate constant for the reaction;
A is a constant for the reaction (Arrhenius constant);
Ea is the activation energy for the reaction;
R is the gas constant; and
T is the temperature in Kelvin.
Note that the equation relates rate constant and not rate of reaction to temperature.
Unit 5 - 21
( 3 ) Application of the Arrhenius equation to determine the activation energy
of a reaction
The natural logarithmic form of the Arrhenius equation is
ln k = ln A -
Ea
RT
The activation energy of a reaction, Ea, and the Arrhenius constant, A, can be obtained by plotting
ln k against 1/T. The temperature T must be in kelvin.
The slope of the graph gives –Ea/R.
Moreover, if two rate constant at different temperatures are determined, Ea can also be found :
ln k2
-
ln k1
=-
Ea 1 1
( − )
R T2 T1
Example 1
Find the activation energy and the Arrhenius constant for the reaction
2 HI(g)
→
H2(g)
+
I2(g)
Values of the rate constant k at various temperatures are given below :
Temperature
/ K
Rate constant / dm3 mol-1 s-1
1/T
/ K-1
ln k
556
7.04 x 10-7
1.80 x 10-3
- 14.2
629
6.04 x 10-5
1.59 x 10-3
- 9.71
700
2.32 x 10-3
1.43 x 10-3
- 6.07
A plot of ln k against 1/T shows a straight with slope = - 2.35 x 104 K.
781
7.90 x 10-2
1.28 x 10-3
- 2.54
Unit 5 - 22
Example 2
The decomposition of dinitrogen pentoxide, N2O5, was followed at two different temperatures and
the rate constants for the reaction were found :
Temperature
Rate constant
300
3.38 x 10-5
/K
/ s-1
330
1.50 x 10-3
Calculate the activation energy of the reaction.
Example 3
Calculate the ratio of the rate constants at 30℃ and 20℃ for a reaction with activation energy
50 kJ mol-1.
Example 4
For the reaction
2 XY(g)
→
X(g)
+
Y2(g)
the rate constant is 3.91 x 10-4 mol-1 dm3 s-1 at 370℃ and 4.05 x 10-2 mol-1 dm3 s-1 at 470℃.
(a)
Calculate the activation energy.
(b)
Calculate the rate constant at 450℃.
Unit 5 - 23
Section
5.5
The interpretation of reaction rate at molecular level
( 1 ) Application of simple collision theory to gaseous systems
The collision theory is the most widely accepted theory of reaction rate. It is developed from the
kinetic theory of gases.
Major postulates of the collision theory :
1.
2.
3.
Reactant particles must collide in order to react.
Not every collision results in a reaction : there must be a certain minimum energy in the collision to
bring about the necessary reorganization of the bonds in the colliding particles.
The colliding particles must be correctly orientated with respect to each other in the collision,
otherwise the new bonds that are likely to be formed may not have the chance of forming.
Collision theory :
1. As reactant concentration increases, the frequency of collisions increases. Thus, the frequency of
properly orientated collisions having sufficient energy to cause reaction also increases.
2. As the temperature rises, the kinetic energy of the reacting particles increases. The number of
molecules possessing the required activation energy is increased. Consequently, there are more
effective collisions per unit time and a higher reaction rate is resulted.
Example :
For the reaction :
H2(g)
+
I2(g)
→
2 HI(g)
Collisions between molecules of hydrogen and iodine are illustrated below :
Unit 5 - 24
( 2 ) Maxwell-Boltzmann distribution curve and its variation with temperature
The particles in a gas are very far apart (between 1000 and 10000 times their own diameters) and are
moving at about the speed of a bullet. So it is possible to produce a theoretical model of a gas using
fairly simple mathematics. This mathematical treatment of gas particles is known as the kinetic theory
of gases and the hypothetical system it describes is called an ideal gas.
The kinetic theory makes certain assumptions about the behaviour of gas particles.
assumptions are :
These
1.
A gas is made up of small particles which are a very long way apart compared to their size. So the
total volume of the particles is very small compared to the volume of the gas as a whole.
2.
The particles do not attract each other at all.
3.
The gas particles are in continuous random motion. This means that there will be a range of
particle speeds. Even if they all started with the same speed, the results of collisions between
particle will slow some particles down while accelerating others.
The distribution of particle speeds can be calculated purely from the laws of probability. This was
first done by Maxwell and Boltzmann and it is often called a Maxwell-Boltzmann distribution.
The Maxwell-Boltzmann distribution curves at two different temperatures is shown below for
gaseous molecules :
Number of particles
with a given
velocity
Velocity v
Note that :
1. The increase in temperature has increased the average speed and the most probable speed.
2. The increase in temperature has increased the range of speeds in the system. The distribution curve
becomes flatter as the temperature rises. This means that there are fewer molecules with the most
probable speed, but there is a greater proportion of high speed molecules. As there are the same
number of molecules at two different temperatures, the area beneath the two curves must be
identical.
Unit 5 - 25
( 3 ) Distribution of kinetic energy of molecules in a gas
The average kinetic energy of the particles is proportional to the temperature of the system. The
particles have energy because they are moving, and this kinetic energy is a function of their mass and
speed.
Kinetic energy = 1 mv2
Because there is a range of particle speeds, there will be a range of particle energies. As energy is
proportional to the square of the particle speed, a plot of speed squared gives a distribution of the energies
in a gas system.
Number of particles
with a given
kinetic energy
Kinetic energy
As the temperature rises, the kinetic energy distribution curve shift to the right and becomes flatter.
There is a greater proportion of high energy molecules. The rate of reaction increases with temperature
because the fraction of particles with sufficient energy to overcome activation energy increases.
Unit 5 - 26
( 4 ) Reaction mechanism
A chemical equation does not reveal the pathway by which a reaction occurs. The reaction pathway
is called the reaction mechanism, which must be determined experimentally and cannot be deduced from
the equation.
If a chemical reaction occurs in a single step, it is called an elementary reaction.
The reaction rate will be equal to the rate of the single step.
For a chemical reaction occurs in a sequence of steps, it is known as the multi-step reaction.
A reaction intermediate is the product of a certain step in the multi-step reaction. It is used up as
soon as it is formed and so does not appear as the final products of the reaction. The sum of each step
must add up to the balanced chemical equation.
Example 1
A chemical reaction occurs in 2 steps :
1. A + B → C
2. C + B → D
Overall equation: A + 2 B → D
C is known as reaction intermediate.
The molecularity of a step is defined as the number of reactant participating in the process.
In general, the molecularity and the order are the same.
Thus a unimolecular step, involving only one reacting particle,
A → intermediate / product
rate = k [A]
A bimolecular step, involving two reacting particles,
A + B → intermediate / product
rate = k [A] [B]
A termolecular step, involving three reacting particles,
A + B + C → intermediate / product
rate = k [A] [B] [C]
In the sequence of steps making up a reaction mechanism, one step is usually slow compared to the
other steps. This slowest step is called the rate determining step (r.d.s.). The rate of the overall
reaction will be limited by, and be exactly equal to the rate of the rate determining step.
Example 2
For a chemical reaction :
2A + B + C → D
a two steps mechanism is found experimentally, the first step is
1. A + B → F
(a) What is the second step ?
2.
(b)
If the first step is the rate determining step, write the rate equation for the reaction.
Example 3
A reaction
A +
B + 2C
proceeds by a 3-steps mechanism:
1. A + C → D
+ F
2. F + B → G
3. G
+ C → E
What is the rate equation for the reaction ?
→
D
slow
fast
very fast
+
E
Unit 5 - 27
( 5 ) Energy profile of single step reaction
Consider a reaction between A-B and C that takes place in a single step mechanism.
When two molecules A-B and C approach closely on a collision course, their electron clouds begin
to repel. Unless they are moving very fast and have a lot of kinetic energy, this repulsion will push them
apart before they get close enough for new bonds to form. If they get sufficiently close, rearrangement
of the electrons in the outer shells can take place so that new bonds form and old ones break. The
kinetic energy of the collision is converted into potential energy. The highly energetic and unstable
chemical species which exist briefly at the point of maximum potential energy is known as the transition
state or activated complex.
Within the transition state, some of the old bonds are stretched to the point of breaking and the new
bonds are only partially formed. The dotted lines represent bonds in the process of breaking and
forming in the transition state.
The change in potential energy during a reaction can be represented by an energy profile.
The energy profile for the above reaction can be shown as follows :
The horizontal axis of the energy profile represents the course of the reaction and is called the
reaction coordinate.
During the course of reaction, as A-B and C come into contact, distort, and begin to exchange or
discard atoms, the potential energy rises to a maximum. It then falls as the products separate, eventually
attaining a value characteristic of the product. The transition state is therefore the highest point in the
energy profile. The energy gap between the reactants and the transition state is then the activation
energy, Ea, for the reaction.
Rate of reaction =
Unit 5 - 28
( 6 ) Energy profile of multi-step reaction
For a reaction between D-E and F, which takes place via a two-step mechanism,
Step 1
Step 2
____________________________________________________________
Overall reaction :
D-E +
F → D-F + E
The slow step is the rate determining step in which the activation energy is the greatest.
Rate of reaction =
The energy profile can be represented as follows :
Example :
The energy profile for a particular reaction is of the form shown in the following figure :
(a) Identify A, B, C and D.
(b) Identify L, M and N.
(c) Comment on the stabilities of L, M and N.
(d) What do the energy changes P, Q and R represent ?
(e) Which step do you expect to have the largest rate constant ?
Unit 5 - 29
Section
(1)
5.6
Catalysis
The action of catalyst
In general, a catalyst is chemically involved in a reaction. It is consumed in one reaction step and
regenerated in a subsequent step. The involvement of a catalyst causes the reaction to take place via an
alternative pathway (the catalytic pathway), and the reaction rate is increased as a result of the lower
activation energy ( Ea’ ) required in the catalytic pathway.
Example :
The decomposition of hydrogen peroxide is catalysed by iodide ions.
The intermediate ion in this case is IO-.
The steps are
1.
2.
Overall reaction :
Energy profile for the catalysed and uncatalysed reactions :
Unit 5 - 30
The characteristics of catalyst
1.
A catalyst provides a new, alternative reaction pathway of lower activation energy, thus enabling the
reaction pathway to proceed faster than the uncatalysed reaction. It is involved in one reaction step
and regenerated in a subsequent step. Although a catalyst does not undergo an overall chemical
change, it may well change its physical form.
2.
Since the catalyst is chemically unchanged at the end of the reaction, it does not contribute any
energy to the system. The enthalpy change if reactants and products are the same with or without a
catalyst, therefore a catalyst does not change the enthalpy change of reaction (∆H).
3.
Since a catalyst alters the activation energy for the reverse reaction by exactly the same amount as
for the forward reaction, thus a catalyst alters the rate of both forward and backward reactions to the
same extent. Thus a catalyst do not affect the position of chemical equilibrium, it merely alter the
rate at which equilibrium is achieved.
4.
A substance which decreases the rate of a chemical reaction is called a inhibitor.
Inhibitors are sometimes known as negative catalysts. They inhibit the normal course of a
chemical reaction by reacting with and removing reactive intermediates from the reaction sequence.
An example of an inhibitor is glycerine, which slows down the decomposition of hydrogen peroxide.
Dilute acid also inhibits this reaction.
5.
A promotor is a substance which enhance the activity of a catalyst.
Example: Aluminium oxide is the catalyst promotor in the Haber process.
N2(g) + 3H2(g)
2NH3(g)
It increases the efficiency of the catalyst finely divided iron.
6.
A poison is a substance which is harmful to the catalyst.
Example : Traces of arsenic poisons the platinum catalyst in the contact process:
2SO2(g)
+
O2(g)
2SO3(g)
Unit 5 - 31
( 2 ) Homogeneous catalysis
In homogenous catalysis the reactants and the catalyst are in the same phase.
Example 1 : Reaction of peroxodisulphate with iodide
S2O82-(aq) +
2 I-(aq) → 2 SO42-(aq) +
I2(aq)
The oxidation of iodide to iodine by peroxodisulphate is slow and may be catalysed by the presence
of iron(II) ions. Due to the strong repulsion between the negative ions, the uncatalysed
Reaction has high activation energy. This is not the case in the catalysed reaction :
1.
2.
Overall reaction :
Energy profile for the catalysed and uncatalysed reactions :
In catalytic reactions, the catalysts usually undergo alternative oxidation and reduction. This is
why transition elements, with their variable oxidation states, form so many compounds which are used as
catalysts.
Perhaps the commonest form of homogeneous catalysis is acid-base catalysis. In acid catalysis the
catalyst is H+, while in basic catalysis it is OH-. Acid catalysis is particularly common in organic
chemistry.
Unit 5 - 32
Example 2 : Iodination of propanone under condition of acid catalysis
Object :
To investigate the rate of reaction
CH3COCH3(aq) + I2(aq)
Cat. H+(aq)
CH3COCH2-I(aq) +
HI(aq)
Procedure :
50 cm3 of 0.01 mol dm-3 I2(aq) and 50 cm3 of acidified 0.125 mol dm-3 propanone were mixed. 10
3
cm portions of the reaction mixture were removed at 3 minute intervals and rapidly added to an excess of
0.25 mol dm-3 NaHCO3(aq) . The remaining iodine was titrated against standard sodium thiosulphate
solution.
Results :
Interpretations :
(a) Why are the 10 cm3 portions of reaction mixture added to aqueous sodium hydrogencarbonate
before titration with standard thiosulphate solution ?
The hydrogencarbonate reacts with the acid catalyst and effectively stops the reaction to enable
titration with thiosulphate to be carried out.
(b) How does the rate of change of iodine concentration vary during the experiment ?
Since the iodine concentration is proportional to the volume of thiosulphate used, the rate of change
of concentration of iodine is proportional to the slope of the graph. This is constant over the period
illustrated.
(c)
What is the order of reaction with respect to iodine ?
Since the concentrations of propanone and acid are much greater than the initial concentration of
iodine, they can therefore be assumed to remain effectively constant, i.e. pseudo zero order
throughout the experiment.
However, as in (b), when the concentration of iodine falls, the rate of reaction remains constant.
The reaction is therefore zero order with respect to iodine.
(d) Indicate by means of a sketch how the volume of S2O32-(aq) used would vary with time if no acid
catalyst was present in the reacting mixture of propanone and iodine.
The volume remains constant during the experiment because the reaction, in the absence of a catalyst,
proceeds too slowly to be measured.
Unit 5 - 33
( 3 ) Heterogeneous catalysis
Heterogeneous catalysis occurs when the catalyst is in a different phase from the reactants.
The catalyst is usually in the solid phase and the reactants in the gaseous or liquid phase.
A reaction between gaseous substances at a solid catalyst interface is prefaced by adsorption of the
gases on to the solid in such a way that reaction is encouraged.
Reaction occurs in the following stages :
1.
2.
3.
4.
5.
diffusion of the reactants to the solid catalyst;
adsorption on the catalyst surface;
reaction between suitably arranged molecules;
desorption from the catalyst;
diffusion of products away from the surface.
Since the position of the active centres are determined by the crystal structure of the catalyst and are
related to its chemical nature, catalysts are specific for given reactions.
Unit 5 - 34
Example 1
Of the same reactant ethanol, finely divided nickel is used for dehydrogenation and aluminium oxide
for dehydration.
(a)
Dehydrogenation by nickel :
(b) Dehydration by Al2O3
Example 2 : Decomposition of hydrogen peroxide
2 H2O2(aq)
2 H2O(l) +
O2(g)
Cat. MnO2(s)
The decomposition of hydrogen peroxide solution is slow at room temperature in the absence of
catalyst. Manganese(IV) oxide is a good catalyst for the reaction. To investigate this catalysed
reaction, a little granular manganese(IV) oxide (of known mass) is added to a certain volume of aqueous
hydrogen peroxide in the conical flask. The gas evolved is collected in the gas syringe. Its volume is
noted at certain time intervals. The experiment is repeated with a larger mass of the catalyst. A graph
of volume of oxygen evolved is plotted against time.
Unit 5 - 35
( 4 ) Autocatalysis
Autocatalysis is the catalysis of a reaction by one of the products of the reaction. The reaction is
initially slow but as products are formed the reaction speeds up.
Example : Oxidation of ethanedioic acid by potassium manganate(VII)
2 MnO4-(aq) 5 C2O42-(aq) + 16 H+(aq) → 2 Mn2+(aq) + 10 CO2(g) +
8 H2O(l)
The reaction is catalysed by Mn2+ ions. When the reaction is carried out at room temperature,
initially it is very slow. However, as Mn2+ ions are formed, the reaction speeds up. This can easily be
tested by adding some Mn2+ ions in the form of manganese(II) sulphate solution at the start of the reaction.
The reaction then proceeds at a reasonable rate right from the start.
(a) The rate of the above reaction may be followed with time by measuring the concentration of
manganese(VII) ions at regular intervals of time during the course of the reaction. Outline two
different ways to do this.
1. Titrimetric method
At regular time intervals small samples of the reaction mixture are withdraw by a pipette. This is
run into a conical flask into which excess potassium iodide solution is added. This stops the
reaction and release iodine quantitatively. The iodine is titrated with standard thiosulphate
solution, the titre gives the concentration of the manganate(VII) ion at that particular interval.
2. Colorimetric method
Adjust the meter of the colorimeter to 100% transmittance with a tube of water. Put the solution
of manganate(VII) ions into a test-tube that fits the colorimeter. Add the ethanedioic acid
solution to the manganate(VII) solution, shaking the mixture and start the clock. Take readings
of colour intensity (absorbance) from the colorimeter every 10 seconds or more frequently. From
a calibration curve of colour intensity with known concentrations of manganate(VII) solutions the
variation of manganate(VII) ions during the course of reaction can be worked out.
(b) The results obtained are shown in the graphs below :
What can you deduce from these results ? Draw a graph to show the change of [MnO4-] against
time if Mn2+(aq) ions were added right in the beginning,. Explain briefly.
The concentration of manganate(VII) ions stays relatively constant for about 20 seconds before
falling with time. The rate of reaction also stays low for the first 20 seconds before rising to a
maximum. This suggests that the rate of reaction must depend on the concentration of something
which produced in the reaction. Addition of Mn2+(aq) ions causes the reaction to proceed almost
immediately, as shown by the immediate fall of [MnO4-]. The reaction is said to undergo
autocatalysis of Mn2+(aq) ions.
(c) If the mechanism of the uncatalysed reaction were as shown :
1. Mn(VII) + C2O42- → Mn(IV) + CO2
2. Mn(IV) + C2O42- → Mn(II) + CO2
Suggest a possible mechanism for the reaction involving autocatalysis of Mn2+ ions.
Sketch an energy profile for both the catalysed and uncatalysed reactions.
Unit 5 - 36
( 5 ) Applications of catalysis
In the field of chemical industry, the problem of cost has always been one of the greatest concerns of
manufacturers. Competitors always try to find ways of making the same product, or another similar
substitute, at a suitably low cost. This has forced chemists to try to find ways of cutting down the cost
and at the same time enhancing the efficiency of catalysts involved in the industrial process.
Example 1 : Iron in the Haber Process
cat. Fe(s)
N2(g) + 3 H2(g)
2 NH3(g)
500℃, 200atm
The catalytic reaction involved in the synthesis of ammonia by Haber process occurs in a sequence
of steps :
1.
2.
N2(g) and H2(g) diffuse to the surface of the catalyst.
The gases are then adsorbed on the catalyst surface.
3.
Ammonia is formed by the stepwise hydrogenation of the adsorbed nitrogen.
4.
Ammonia molecules then desorb from the catalyst surface.
Transition metals like iron, platinum, ruthenium and osmium are all theoretically suitable catalyst
because they can chemisorb nitrogen to an adequate extent.
Iron is chosen as the industrial catalyst because its great abundance makes it economically attractive
for the large scale synthesis of ammonia.
Platinum has even a higher catalytic efficiency than iron. However, since the metal is much more
expensive to be used in a large scale and it is easily subjected to poisoning by impurities, its use in the
Haber Process is limited.
Ruthenium and osmium are too expensive to be used in a large scale.
Example 2 : Enzymes – the biological catalysts
Enzymes are proteins that catalyse specific biochemical reactions. They are often called
biological catalysts. Enzymes obtained from yeast have lone been used in the production of alcohol by
fermentation. The fermentation of glucose to form ethanol can be represented by the following
equation :
Nowadays, enzymes are used in various aspects such as in the manufacture of biological washing
powders. These washing powders contain enzymes which can break down stains caused by sweat, blood,
egg as well as fats. They have the advantage of removing stains even at normal temperature.
In the field of biotechnology, enzymes are tailored to catalyse particular processes with their usual
super-efficiency. The enzyme industry has grown enormously owing to this super-efficiency. Small
quantities of enzymes can convert large quantities of chemicals at relatively low temperature and normal
pressure. Many drugs are now synthesized by biotechnology, including the hormone insulin required by
many diabetic patients, and the new drugs ‘interferon’ and ‘AZT’ which are being used for their antiviral
effects.
Unit 5 - 37
Example 3 : Catalytic converters in car exhaust systems
In the catalytic converter, some pollutants from car exhaust (e.g. carbon monoxide, nitrogen
monoxide and unburnt hydrocarbons) are converted into relatively harmless substances (e.g. carbon
dioxide, nitrogen and water) with the use of metal catalysts such as platinum (or palladium) and rhodium.
1.
2.
3.
The catalysts are coated on a honeycomb support to increase the surface area for better action.
To use the catalytic converter the car must run on lead free petrol, otherwise the catalyst will be
poisoned by the lead. Unleaded petrol has been introduced in Hong Kong since 1991. The catalyst
system is only effective at temperatures over 400oC. With stop-start usage, which is typical of the busy,
congested traffic in Hong Kong, the converter will be inactive for much of the time as it would not reach
the operating temperature. Anyway, when in full gear such converters can convert 80 to 90 % of the
pollutant gases in exhaust fumes to carbon dioxide, water and nitrogen. From January 1st 1992 all new
cars in Hong Kong had to be fitted with these converters.