Form 5 Chemistry Notes – Ms. R. Buttigieg
Form 5 Chemistry Notes
What you shall be needing
A4 notebook for practicals
2 copybooks for HW and CW
A copy of the Chemistry syllabus
SEC exam papers 2004-2006
JL papers 2003 -2005
(can print out from internet by going on
http://www.curriculum.gov.mt/ then
choose Examination papers,
Select JL papers and the years you need)
Main topics to be discussed during the year:
1. Energetics
2. Organic Chemistry
3. Carbon chemistry and structures
Pg. 1
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 2
Topic 1 Energetics
Energy changes accompany both physical and chemical changes.
Principle of Conservation of Energy: Energy can be changed from one form to
another but it can neither be created nor destroyed.
Exothermic and endothermic reactions Chem 4 You pg. 190-195, GCSE Chem pg.190-191
When a chemical reaction happens, energy is transferred to or from the
surroundings and often there is a temperature change. For example, when
a bonfire burns it transfers heat energy to the surroundings. Objects near a
bonfire become warmer and the temperature rise can be measured with a
thermometer.
Exothermic reactions
These are reactions that transfer energy (liberate energy) to the surroundings. The
energy is usually transferred as heat energy, causing the reaction mixture and its
surroundings to get hotter. The temperature increase can be detected using a
thermometer. Some examples of exothermic reactions are:
•
•
•
burning
neutralisation reactions between acids and alkalis, and
the reaction between water and calcium oxide
These reactions can be represented by a thermochemical equation like the one below:
2 H2 (g) + O2 (g)
2H2O (l)
∆ H = -575 kJ
The symbol ∆ H shows the change in energy.
The minus (-) sign shows that energy is lost from the substance and an exothermic
reaction has occurred.
Endothermic reactions
These are reactions that take in energy from the surroundings. The energy is usually
transferred as heat energy, causing the reaction mixture and its surroundings to get
colder. The temperature decrease can be detected using a thermometer. Some examples
of endothermic reactions are:
•
•
•
electrolysis
the reaction between ethanoic acid and sodium carbonate, and
the thermal decomposition of calcium carbonate in a blast furnace
N2 (g) + O2 (g)
2NO (g)
∆ H = + 180.6 kJ (+ energy gained by substance)
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 3
The chemical energy stored in the bonds in a substance gives us a measure of a
chemical's energy level.
The higher the energy level of a substance, the more chemical energy is stored in its
bonds.
The reactants and products in a chemical reaction usually have different energy
levels, which are shown in a type of graph called an energy level diagram.
The vertical axis on this diagram represents the energy level and the horizontal axis
represents the progress of the reaction from reactants to products.
Energy level diagrams for exothermic reactions
In an exothermic reaction, reactants have a higher energy level than the products. The
difference between these two energy levels is the energy released to the surroundings in
the reaction, and an energy level diagram shows this as a vertical drop from a higher to a
lower level:
Usually some extra energy is
needed to get the reaction to start.
This is called the activation
energy and is drawn in energy level
diagrams as a hump.
Catalysts reduce the activation
energy needed for a reaction to
happen - this lower activation
energy is shown by the dotted line
in the diagram here.
Energy level diagrams for endothermic reactions
In endothermic reactions the
reactants have a lower energy level
than the products.
The difference between these two
energy levels is the energy gained
from the surroundings in the reaction,
represented in an energy level
diagram as a vertical jump from a
lower to a higher level - the bigger
the jump, the more energy is gained.
•
•
•
State symbols are important as the energy change also depends on the state
Energy must be supplied to break bonds.
Energy is released when bonds are formed.
1. STUDY CHEMISTRY FOR YOU PG. 194,
o
o
Work out pg. 198 nos 8,9.
Work out pg. 257 nos 25,26
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 4
In an exothermic reaction, the energy released from forming new bonds is greater than
the energy needed to break existing bonds.
In an endothermic reaction the energy needed to break existing bonds is greater than the
energy released from forming new bonds.
These energy changes can be shown in an energy level diagram.
The products have less energy than
the reactants.
30 kJ of energy has been given out.
The reaction is exothermic
The products have more energy than
the reactants.
15 kJ of energy has been taken in.
The reaction is endothermic.
The amount of energy transferred is found by comparing the energy in the reactants with
that in the products.
Reaction type:
exothermic
endothermic
heat is:
reaction vessel temperature:
enthalpy change is
net bond:
released
absorbed
rises
falls
negative
positive
formation
breaking
Answer the following:
1. When sodium carbonate and calcium chloride solution are mixed, the temperature of the
solution dropped by 4oC. Is the reaction exothermic or endothermic? ______________________
2. Are combustion reactions exothermic or endothermic? ______________________
3. Is energy needed or given out when bonds are broken? ______________________
4. Is energy needed or given out when bonds are formed? ______________________
5. Are ∆H values for an exothermic reaction positive or negative? ______________________
Now let’s have a look at some of the reactions accompanied by energy changes
See GCSE pg. 190-191
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 5
Energy changes are caused by bond breaking and forming
Breaking a bond - requires energy;
Forming a bond – releases energy
Energy Changes
Standard enthalpy of combustion – the heat change which occurs when 1 mole of the
substance burns completely in oxygen at STP (temperature of 25oC, pressure of 1 atm)
Example: When carbon burns completely it forms Carbon dioxide. If not completely
burnt it forms Carbon monoxide.
C (s) + O2 (g) CO2 (g)
∆ H = -394 kJ/mol
Standard enthalpy of formation – the heat change which occurs when 1 mole of the substance
is formed from its elements at STP
Example: When 1 mole of Carbon disulphide is formed from its elements.
C (s) + 2 S (g) CS2 (l)
∆ H = +117 kJ/mol
Standard enthalpy of precipitation – the heat change which occurs when 1 mole of the
substance is precipitated from solution at STP
Example: When 1 mole of Silver Chloride is precipitated
Ag+ (aq) + Cl- (aq) AgCl (s)
∆ H = - 50.2 kJ/mol
Standard enthalpy of neutralization – the heat change which occurs when 1 mole of acid is
neutralized by an alkali to form 1 mole of water at STP .
Example:
NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)
∆ H = - 57.5 kJ/mol
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 6
Standard enthalpy of solution– the heat change which occurs when 1 mole of the substance is
dissolved in a stated quantity of water (or till further dilution does not involve any energy
change).
The standard enthalpy of solution = lattice energy + hydration energy
The lattice energy is the energy needed to separate the ions and break down the lattice.
(Endothermic) The hydration energy is the energy released when the ions bond with water
molecules (exothermic)
Standard enthalpy of reaction – the energy change that occurs when the reactants react in the
number of moles as expressed by the equation.
Work these:
1.
2.
a)
b)
c)
d)
Give the meaning of:
Enthalpy
Exothermic
Principle of Conservation of energy
Standard enthalpy of combustion
STP
2 H2 (g) + O2 (g) 2H2O (l)
∆ H = -575 kJ
Explain the reaction in words
What is the standard enthalpy of reaction for the reaction above?
What is the standard enthalpy of combustion of Hydrogen?
What is the standard enthalpy of formation of water?
3a) The standard enthalpy of neutralization of hydrochloric acid and that for nitric acid is the
same. Why?
b) The standard enthalpy of neutralization of sulphuric acid is double that for hydrochloric acid.
Why?
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 7
Find the standard enthalpy change in each of the following and say whether each reaction
is exothermic or endothermic.
1. Hydrogen + Chlorine ==> Hydrogen Chloride
o
o
o
o
The symbol equation is: H2(g) + Cl2(g) ==> 2HCl(g)
but think of it as: H-H + Cl-Cl ==> H-Cl + H-Cl
(where - represents the chemical bonds to be broken or formed)
the bond energies in kJ/mol are: H-H 436; Cl-Cl 242; H-Cl 431
2. Hydrogen Bromide ==> Hydrogen + Bromine
o
o
o
The symbol equation is: 2HBr(g) ==> H2(g) + Br2(g)
but think of it as: H-Br + H-Br ==> H-H + Br-Br
the bond energies in kJ/mol are: H-Br 366; H-H 436; Br-Br 193
3. hydrogen + oxygen ==> water
o
o
o
2H2(g) + O2(g) ==> 2H2O(g)
or 2 H-H + O=O ==> 2 H-O-H (where - or = represent the covalent bonds)
bond energies in kJ/mol: H-H is 436, O=O is 496 and O-H is 463
4. nitrogen + hydrogen ==> ammonia
o
N2(g) + 3H2(g) ==> 2NH3(g)
o
o
or N
N + 3 H-H ==> 2
bond energies in kJ/mol: N N is 944, H-H is 436 and N-H is 388
SEC CHEMISTRY MAY 2002 – Paper I
5. Sulphur dioxide and oxygen react to form sulphur trioxide according to the equation:
2 SO2 (g) + O2 (g) -------- 2SO3 (g)
a)
b)
c)
d)
∆ H = -197 kJ
Is the reaction exothermic or endothermic?
What do you understand by the symbol ∆ H?
What is the value of ∆ H for the reverse reaction?
Use the information given in the equation above to draw a simple energy level diagram for
the formation of sulphur trioxide. Show ∆H on your diagram.
6. Consider the following reactions:
i.
C (s) + O2 (g) CO2 (g)
ii.
C (s) + 2S (s) CS2 (l)
∆ H = -393.5 kJ
∆ H = +117 kJ
a) Say whether the temperature increases, decreases or doesn’t change in each reaction.
b) Use the terms endothermic and exothermic to describe each reaction
c) Which one of the reactions would you use to heat a room? Give a reason for your answer.
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 8
Finding the Standard Enthalpy of Combustion in the
Lab
The diagram shows the apparatus to find the heat of
combustion in the laboratory. (draw complete setup)
The calorimeter has a copper spiral so that the hot gases
formed by the combustion of ethanol pass through the
spiral and so heat the surrounding water.
The calorimeter is well lagged on the outside.
The end of the spiral leads to a flask with water and
thermometer. This serves as a control to check whether
heat energy is lost from the calorimeter.
If all the heat energy is transferred to the water in the
calorimeter, the temperature in the flask should not rise.
METHOD:
1)
A lamp is filled with ethanol, capped and then the
lamp and cap is weighed.
The lamp is lit and put beneath the calorimeter.
The air-supply is controlled to get a blue flame.
Water is stirred continuously.
After some time, the flame is extinguished; the
lamp is immediately capped and weighed.
The maximum temperature of the water is noted.
The standard heat of combustion can then be found.
2)
3)
4)
5)
6)
7)
Readings you need to take
Mass of water
Mass of lamp after lighting
Initial temperature
Mass of Ethanol burnt
Temperature Rise
Mass of lamp and ethanol before burning
Final temperature
Mass of Copper Calorimeter
= Mass of lamp before
= Final
–
–
Mass of lamp after
Initial temperature
Calculation:
1. Heat energy liberated by ethanol = heat energy absorbed by water + copper calorimeter
=
mw x cw x ∆θ
+ mCu x cCu x ∆θ
= ___ joules
2. If x moles were burnt during the experiment,
x moles liberated ___ joules
1 mole liberates? Joules
Sources of Error:
Heat is lost to the surroundings
Some ethanol may have evaporated
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 9
A simpler method
This method is specifically for determining the heat energy released
(given out) for burning fuels. The burner is weighed before and after
combustion to get the mass of fuel burned. The thermometer records
the temperature rise of the known mass of water (1g = 1cm3).
You can use this system to compare the heat output from burning
various fuels. The bigger the temperature rise, the more heat energy is
released. See calculations below for expressing calorific values.
This is a very inaccurate method because of huge losses of heat eg
radiation from the flame and calorimeter, conduction through the
copper calorimeter, convection from the flame gases passing by the
calorimeter etc. BUT, at least using the same burner and set-up, you
can do a reasonable comparison of the heat output of different fuels.
•
You need to know the following:
o the mass of material reacting in the calorimeter (or their concentrations and
volume),
o the mass of water in the calorimeter,
o the temperature change
o the specific heat capacity of water, (shorthand is SHCwater), and this is 4.2J/g/oC,
this means it takes 4.2 J of heat energy to change the temperature of 1g of
water by 1oC (or K).
•
Example
o 100 cm3 of water (100g) was measured into the calorimeter.
o The spirit burner contained the fuel ethanol C2H5OH ('alcohol') and weighed 18.62g
o
o
o
o
o
at the start.
After burning it weighed 17.14g and the temperature of the water rose from 18
to 89oC.
The temperature rise = 89 - 18 = 71oC (exothermic, heat energy given out).
Mass of fuel burned = 18.62-17.14 = 1.48g.
Heat absorbed by the water = mass of water x SHCwater x temperature
= 100 x 4.2 x 71 = 29820 J (for 1.48g)
heat energy released per g = energy supplied in J / mass of fuel burned in g
heat energy released on burning = 29820 / 1.48 = 20149 J/g of C2H5OH
this energy change can be also expressed on a molar basis.
Relative atomic masses Ar: C = 12, H = 1, O = 16
Mr(C2H5OH) = (2 x 12) + (1 x 5) + 16 + 16 = 46, so 1 mole = 46g
Heat released (given out) by 1 mole of C2H5OH = 46 x 20149 = 926854
J/mole
At AS level this will be expressed as enthalpy of combustion
=
-1
Hcomb = -926.9 kJmol
The data book value for the heat of combustion of ethanol is -1371 kJmol-1,
showing lots of heat loss!
Form 5 Chemistry Notes – Ms. R. Buttigieg
Rates of reactions
How fast (or slow) a reaction occurs depends on 5 important factors:
Temperature
Surface area
Presence of a catalyst
Concentration
Pressure
Pg. 10
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 11
Factors, which increase the speed of a reaction, are those that increase the number of
collisions between particles and the energy of these collisions.
Concentration: So increasing the concentration of hydrochloric acid in its reaction with calcium
carbonate increases the speed of reaction because there are more acid particles per unit volume.
So they collide more with the calcium carbonate and the chance of them reacting is greater
Surface Area: Increasing the surface area of calcium carbonate in the same reaction also
increases the speed of reaction as you are exposing more calcium carbonate to be acted upon by
the acid. More collisions occur and the chance of reaction is greater.
Effect of temperature: Increasing the temperature increases the rate of reaction as raising the
temperature makes the particles move faster (have more kinetic energy). This means that the
particles collide more frequently with each other and the rate of the reaction increases.
Also, the faster the particles are traveling, the greater is the proportion of them, which
will have the required activation energy for the reaction to occur.
As a general guide, raising the temperature of a reaction by 10 °C will double the rate of
a reaction.
The gradient of the plot will be twice as steep
Pressure:
affects reactions involving gases, since increasing pressure increases the
concentration of gas.
Catalysts: Light catalyses certain reactions like the decomposition of silver chloride into silver
and chlorine – known as photoreduction of silver halides
In surface catalysts, the reaction takes place on the surface. Catalysts reduce the activation
energy needed for the reaction to begin.
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 12
Examples of catalysts
Light - A mixture of Chlorine, Cl2, and Hydrogen, H2, explodes when exposed to sunlight to
give Hydrogen Chloride, HCl. In the dark, no reaction occurs, so activation of the reaction by
light energy is required.
Cl2 (g) + H2 (g) ==> 2 HCl (g)
Catalyst
Reaction
Manganese Dioxide - MnO2
2 H2O2 (aq) 2 H2O (l) + O2 (g)
Iron - Fe
N2 (g) + 3 H2 (g) 2 NH3 (g)
Vanadium (V) Oxide
2 SO2 (g) + O2 (g) 2 SO3 (g)
In a Catalytic Converter
2 NO (g) + 2 CO (g) N2 (g) + 2 CO2 (g)
Platinum/Rhodium
4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (l)
Experiments to determine the Rate of Reaction:
In such reactions it is important that only one factor is changed each time. When the two
reactants are mixed the stopwatch is started and the volume/mass are noted at regular time
intervals. Graphs can then be plotted.
The effect of temperature on Reaction Rate
A crucible with zinc is placed floating on the dilute hydrochloric acid in the flask.
The flask is connected to an empty gas syringe.
The two reactants are mixed, while the stopwatch is started.
The volume of gas produced is measured at regular intervals till production stops.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
The temperature is then changed and the same experiment repeated (keeping mass, volume
and acid concentration constant).
Both graphs are plotted on the same axis.
It is noted that the reaction at a higher temperature has a steeper gradient.
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 13
The effect of surface area on Reaction Rate
The same set-up on page 12 is used.
Calcium carbonate chips are reacted with dilute hydrochloric acid.
Calcium carbonate powder is then reacted with dilute hydrochloric acid.
Both reactions are compared.
This reaction can also be done on a balance. When the two reactants are mixed, the stopwatch is
started and the mass taken at regular intervals.
The effect of concentration on Reaction Rate
A typical reaction used for such experiments is the reaction between sodium thiosulphate
solution and dilute hydrochloric acid. The experiment is repeated at different concentrations.
•
•
•
To follow this reaction in your investigation you can measure how long it takes for a certain
amount of sulphur to form.
You do this by observing the reaction down through a conical flask, viewing a black cross on
white paper (see diagram below).
The X is eventually obscured by the sulphur precipitate and the time noted.
Na2S2O3 (aq) + 2 HCl (aq) 2 NaCl (aq) + SO2 (g) + H2O (l) + S (s)
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 14
The effect of catalyst on Reaction Rate
Two test tubes are filled with some hydrogen peroxide.
In one flask some manganese dioxide powder, which has been
previously weighed, is placed in one test tube.
As soon as the manganese dioxide is added rapid effervescence
occurs.
The gas produced relights a glowing splint (oxygen)
When the manganese dioxide is filtered, washed, left to dry and
weighed, it is found that the reaction did not use it up. So MnO2
affects the speed of reaction without being used up.
The effect of light on silver chloride
Some silver nitrate is added onto sodium chloride solution.
A white precipitate of sodium chloride is formed.
One portion is placed in sunlight, the other in the dark.
After some time the one in sunlight goes black, the other remains white.
Silver nitrate ( ) + Sodium chloride ( ) Silver chloride ( ) + Sodium nitrate ( )
Answer the following:
1. Draw a labeled diagram of the set-up to follow the speed of reaction by measuring the volume
of gas produced when calcium carbonate is reacted with hydrochloric acid.
a.
If the experiment is done to find the effect of surface area you would carry out two
experiments. What must be kept the same in both experiments? What must be
changed from one experiment to the next?
b. Both graphs are plotted on the same graph paper, sketch the graphs that are expected.
i. Clearly label each graph
ii. Mark the point on the graph where the speed of reaction is the highest
iii. Mark the point where the reaction stops.
c. What can you conclude from this experiment?
d. Explain what happens in terms of particles.
2. Write the name of the catalyst used in the following reactions:
a. 2 H2O2 (aq) 2 H2O (l) + O2 (g)
b. 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (l)
3. Give an example and the equation of a reaction that is affected by light.
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 15
Reversible reactions and chemical equilibrium
If the direction of a reversible reaction is changed, the energy change is also reversed.
For example: the thermal decomposition of hydrated copper(II) sulphate.
On heating the blue solid, hydrated copper(II) sulphate, steam is given off and the white solid of
anhydrous copper(II) sulphate is formed.
o
This a thermal decomposition and is endothermic as heat is absorbed (taken in)
o
The energy is needed to break down the crystal structure and drive off the water.
When the white solid is cooled and water added, blue hydrated copper(II) sulphate is reformed.
o
The reverse reaction is exothermic as heat is given out.
o
i.e. on adding water to white anhydrous copper(II) sulphate the mixture heats up as
the blue crystals reform.
blue hydrated copper(II) sulphate + heat
CuSO4.5H2O(s)
white anhydrous copper(II) sulphate + water
CuSO4(s) + 5H2O(g)
Another example: is the chromate/dichromate equilibrium
Yellow chromate ion is turned orange by addition of acid. The orange dichromate is returned to
yellow by the addition of base.
2 CrO42- (aq) + 2 H+(aq)
Cr2O72- (aq) + H2O (l)
Chromate ions, CrO42- (aq), are yellow. Dichromate ions, Cr2O72- (aq), are orange. Since this is
an equilibrium, both ions are present in a solution at the same time. The color of the solution
indicates which of the ions is present in the greater amount.
Another example: is esterification
Ethanoic acid reacts with ethanol in the presence of concentrated sulphuric acid as a catalyst to
produce the ester, ethyl ethanoate. The reaction is slow and reversible. To reduce the chances of
the reverse reaction happening, the ester is distilled off as soon as it is formed
.
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 16
Kinetic picture of dynamic equilibrium
A dynamic equilibrium occurs when you have a reversible reaction in a closed system.
Nothing can be added to the system or taken away from it apart from energy.
At equilibrium, the quantities of everything present in the mixture remain constant, although
the reactions are still continuing. This is because the rates of the forward and the back
reactions are equal.
A dynamic equilibrium occurs when two reversible processes occur at the same rate. Many
processes (such as chemical reactions) are reversible.
An example of the process can be imagined if a bucket is filled with water and placed in a small
room. The water from the bucket will evaporate, and the air in the room will start to become
saturated with water vapor. Eventually, the air will be completely saturated with water, and the
level of water in the bucket will stop falling. However, water from the bucket is still evaporating.
What is happening is that molecules of water in the air will occasionally hit the surface of the
water and condense back into the liquid water, and this occurs at the same rate at which water
evaporates from the bucket. This is an example of dynamic equilibrium, because the rate of
evaporation equals the rate of condensation.
Water in a closed system, at constant temperature, is in a dynamic equilibrium.
H2O(l) H2O(g) forward reaction; H2O(g) H2O(l) reverse reaction
At equilibrium H2O(l)
H2O(g)
If you change the conditions in a way which changes the relative rates of the forward and back
reactions you will change the position of equilibrium - in other words, change the proportions of
the various substances present in the equilibrium mixture.
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 17
In a saturated solution of KCl (aq), at constant temperature, with excess KCl (s) the following
reactions occur.
KCl(s) K+(aq) + Cl-(aq) forward reaction (dissolution)
K+(aq) + Cl-(aq) KCl(s) reverse reaction (crystallization)
At equilibrium KCl(s)
K+(aq) + Cl-(aq)
The diagram shows what happens to the molecules when
there is a dynamic equilibrium.
If conditions are changed, the equilibrium shifts to one side
rather than the other, depending on which direction is
favoured.
Le Chatelier’s Principle says where the equilibrium
shifts when the conditions are changed.
The equilibrium shifts so as to
oppose the change disturbing it.
It states that:
E.g The following reaction is in equilibrium:
N2 (g) + 3 H2 (g)
2 NH3 (g)
If the reaction proceeds to the right , it decreases pressure as it reduces the number of particles
from 4 molecules (a nitrogen molecule and 3 hydrogen molecules) to 2.
If the reaction proceeds to the right , it produces heat; exothermic.
So if from outside the system the pressure is increased this will result in a higher percentage of
ammonia.
As according to Le Chateliers’ principle:
We increase pressure
The equilibrium tries to decrease it
Shift to right producing more ammonia
In industry a pressure of 300 atm is used to produce ammonia.
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 18
Temperature - If the mixture is heated from outside the equilibrium will reduce the heat so the
equilibrium shifts to the left , reducing the amount of ammonia.
So increasing the temperature reduces the amount of ammonia
In industry a balance is found:
o A low temperature produces more ammonia at equilibrium but this is formed very
slowly;
o A high temperature produces less ammonia at equilibrium but this is formed
quickly.
So a temperature of 450oC is used.
Less ammonia is formed at equilibrium but it is formed more quickly.
The effect of a catalyst on the rate of attainment of equilibrium and not on the
equilibrium position.
A catalyst does not effect the amounts of products at equilibrium as it catalyses both the
forward and backward reaction. It helps the equilibrium point to be reached more quickly.
In industry, iron is used in this reaction.
Form 5 Chemistry Notes – Ms. R. Buttigieg
Pg. 19
In industry ammonia is removed from the mixture as it becomes liquefied under that high
pressure.
This helps the equilibrium to keep shifting to the right to increase the amount of ammonia.
The following is another equilibrium reaction:
2 SO2 (g) + O2 (g)
2 SO3 (g)
Temperature
The forward reaction reduces pressure and is exothermic
If we perform the reaction at a low temperature, equilibrium shifts to increase
temperature; so it shift to the right and produces more sulphur trioxide.
In industry they don’t use low temperatures as this reduces rates.
So a temperature of 450oC is used.
Less sulphur trioxide is formed at equilibrium but is formed more quickly.
Oxygen
The oxygen used is 3 times as theoretically required.
Equilibrium shifts to reduce the amount of oxygen; so it shifts too the right and produces
more sulphur trioxide.
Pressure
If pressure is increased, the equilibrium shift to reduce pressure
So it shifts to the right to produce more sulphur trioxide
In industry high pressures aren’t used as at atmospheric pressure, 98% sulphur trioxide is
already formed at equilibrium
The catalyst vanadium (V) oxide doesn’t change the amount of sulphur trioxide produced
at equilibrium but it helps to reach the equilibrium point more quickly.
JL 2005 Annual
3. a) This question is about the effect of concentration and of temperature on an equilibrium
mixture. The equation for the system is:
i) What colour would the mixture turn to if some concentrated hydrochloric acid is added?
Give a reason for your answer using Le Chatelier’s principle.(3 marks)
ii) The forward reaction is endothermic. If a warm sample of the mixture is cooled, it turns pink.
Explain this observation in terms of Le Chatelier’s principle. (2 marks)
b) Cobalt chloride can act as a catalyst. What effect (if any) does a catalyst have on the
equilibrium position? Explain your answer. (3 marks)
c) Cobalt is a transition metal. Give two properties (apart from acting as a catalyst) that
cobalt would show. (2 marks)