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Pupil Number:
CfE Higher Chemistry
Unit 3
Chemistry in Society
Chemical Industry
Importance of the Chemical Industry

the Chemical Industry is one of the largest British industries.

its products are indispensable to many aspects of modern life and many are used for the
benefit of society

it is the only manufacturing industry to export more than it imports and so earns a trade
balance surplus from these exports for Britain

also invisible trade balance surplus from selling licenses to use British processes abroad

the chemical industry involves the investment of large sums of money but employs relatively
few people making it a capital intensive and not a labour intensive industry.
Stages in the manufacture of a new product
The manufacture of a new product is a step-wise process from its discovery.
research and
development
laboratory process
a new potentially useful chemical is prepared and patented. Some products
are discovered by accident, but others as a result of long and expensive
research.
small scale to review the production route
pilot study
the product is now required in larger amounts and will be manufactured in a
pilot study using the route identified by the research group but in kilogram
quantities. Product quality, health hazards, and production costs will be
discussed.
scaling up
planning the scaling up from lab quantities to full scale production will have
been going on from the pilot study stage.
Production
plant design, planning considerations, commissioning and start up
review
this will occur at each stage. All processes are reviewed and modifications
are made.
Raw Materials and Feedstocks
A feedstock is a chemical from which other chemicals are manufactured. Feedstocks are made from
raw materials; the basic resources that the earth supplies to us. They are:

Fossil fuels–coal, oil and natural gas

metallic ores–e.g. aluminium extracted from bauxite(Al2O3)

minerals–chlorine from sodium chloride

water and air – water in hydration of ethene to ethanol and nitrogen in the Haber Process,
oxygen in the catalytic oxidation of ammonia

organic materials–of plant and animal or vegetable oils and starch

Crude oil is a raw material from which naphtha is obtained by fractional distillation. Naphtha is
a feedstock that can be cracked to produce ethene.
2
Batch and Continuous Processes
In a batch process the chemicals are loaded into the reaction vessel. The reaction is monitored and
at the end of the reaction the product is separated and the reaction vessel cleaned out ready for the
next batch.
In a continuous process the reactants are continuously loaded at one end of the reaction vessel and
the products are removed at the other end. Each process has advantages and disadvantages.
Batch Process v Continuous Process
Process
Batch
Continuous
Advantages
Disadvantages

suited to smaller scale production
up to 100 tons per annum

possibility of contamination from
one batch to the next

more versatile than continuous as
they can be used form ore than
one reaction more suited for multi
step reactions or when reaction
time is long

filling and emptying takes time
during which no product, and
hence no money, is being made

suited to large scale production
>1000 tons per annum

very much higher capital cost
before any production can occur

suitable for fast single step
processes


more easily automated using
computer control
not versatile, can make only one
product not cost effective when
run below full capacity

smaller work force operates
round the clock, 365 days per
year

tend to operate with relatively
low volumes of reactants allowing
easy removal of excess heat
energy
In general products that are made on a very large scale will use a continuous process eg. sulfuric acid,
ammonia, iron, ethene, poly(ethene).
Products made on a smaller scale or when a continuous process would be difficult to devise or operate
will use a batch process
eg. pharmaceuticals, dyes, copper refining by electrolysis.
Factors Influencing the Choice of Synthetic Route
- Cost - availability of feedstocks
- Yield of the reaction – can unreacted starting materials be recycled, or sold?
- Difficulty and cost of waste disposal
- Energy consumption
- Emissions to the atmosphere
3
Economic Aspects

Operating Conditions
The conditions under which a chemical process operates are chosen to maximise economic efficiency.
We have considered these in other topics but examples are:
Raising the temperature may increase the rate of a reaction but it will increase energy costs so may
not be economic
Increasing the pressure may shift equilibrium in favour of the product but will mean using stronger
reaction vessels and more powerful compressors and may not be economic.

Costs in the Chemical Industry
Costs come under 3 main categories – capital, fixed and variable costs.

Capital Costs
These are incurred when building the plant. The life of a plant is assumed to be only about 10 years
after which it is written off. The cost of this depreciation is recovered under fixed costs.

Fixed Costs
These are costs that are the same whether 1 ton or 1000 tons of product are made. The effect of
the fixed cost decreases as the amount of product increases.
They include:
- Depreciation of the plant
- Labour
- Land purchase
- Variable Costs
- These are directly related to output and include:
- Raw materials and energy
- Packaging
- Waste disposal and effluent treatment
Location of the Chemical Industry
Many locations are for historical and practical reasons. They had to be near:

Water supply

Raw materials

Good communications; near ports, roads and rail

Reliable energy supplies

Available skilled labour
4
Safety and the Environment
The chemical industry is well aware of its environmental responsibilities and is acting
accordingly.
Power stations that burn fossil fuels must remove the sulfur dioxide from the gases before release
to the atmosphere. The SO2 is converted to H2SO4 , which is sold.
Waste used to be dumped in quarries, rivers, the sea or stored in containers from which it could leak
into streams. These methods are no longer acceptable and are increasingly becoming illegal. Waste
must be treated and discharged only when it is not harmful to the environment – it must meet
requirements of pH and metal ion content.
Water containing organic waste must not be discharged into rivers or canals if it will reduce
significantly the oxygen content of the water, causing fish to die.
Between 1990 and 1996 discharge of potentially harmful chemicals into UK rivers was reduced by
91%.
Plants have reduced accidents by 50% in the last decade.
Road and rail tankers that carry chemicals are constructed to withstand impact in accidents.
Plants have their own fire fighting teams on site.
Plants are designed with safety in mind. Chemicals are hazardous so the accident rate will never be
zero but the aim is to learn from mistakes and reduce the rate to a minimum.
It is essential at this stage to revise all of your National 5 calculations that were based on
moles and equations.
5
Calculations Based on Equations
A balanced equation is taken to give the relative number of moles of each reactant and product. Since
the mass of one mole of any substance is expressed in grams, the masses involved can then be
calculated as shown.
Worked Example
Calculate the mass of water produced on burning 1g of methane
CH4
1 mole
+
2O2
CO2
+
2H2O
2 moles
It is not necessary to calculate the masses of carbon dioxide and oxygen – these substances are not
included in the question.
Step 1
Convert moles to mass for the substances involved
CH4
2H2O
12 + (4 x 1)
16g
Step 2
2 [ (2 x 1) + 16 ]
36g
By proportion; determine the mass of water produced
16g
1g
36g
36 x 1
16g
= 2.25g
The Idea of Excess
Reactants are needed for a chemical reaction to take place. When one of the reactants is used up the
reaction will stop. Any reactant, which is left unreacted, is said to be “in excess”.
Worked Example 1
8g of methane is mixed with 16g of oxygen. A spark is applied to the mixture to start the reaction.
Calculate the mass of carbon dioxide produced.
CH4 + 2O2
1 mole 2 moles
Step 1
CO2 + 2H2O
1 mole
Calculate the number of moles of each reactant
No. moles CH4 =
No. moles O2
mass = 8
gfm
16
= 0.5moles
= mass = 16 = 0.5moles
gfm
32
6
Step 2
Use the balanced equation to determine the quantity of each reactant used in the
reaction.
CH4
1 mole
+
2O2
2 moles
Therefore, for every 0.5 moles of methane 1 mole of oxygen would be required. Looking at the
quantities of reactants from step 1 there is not enough oxygen to allow all of the methane to react
therefore some methane will be left over at the end. The methane is said to be in excess and the
oxygen will therefore determine the quantity of carbon dioxide produced.
Step 3
Convert moles to mass and carry out a proportion calculation to determine the mass of
carbon dioxide produced.
CH4 + 2O2
1 mole
2 moles
CO2 + 2H2O
1 mole
64g
16g
44g
16 x 44
64
= 11g
Worked Example 2
What mass of hydrogen gas is produced when 2.45g of magnesium is added to 100cm 3 of dilute
hydrochloric acid, concentration 1 moll-1?
Mg + 2HCl
1 mole 2 moles
Step 1
MgCl2 + H2
1 mole
Calculate the number of moles of each reactant
No. moles Mg =
mass =
gfm
No. moles HCl =
Step 2
So
2.45
24.5
= 0.1 moles
Cx V
1 x 0.1
= 0.1 moles
Use the balanced equation to determine the quantity of each reactant used in the
reaction.
Mg
+
1 mole
0.1 moles requires
2HCl
2 moles
0.2 moles
From step 1 it can be seen that there is not enough acid to allow all of the Mg to react hence the Mg
will be in excess and the acid will control the mass of hydrogen produced.
7
Experiment 3.1—Excess
Magnesium reacts with dilute sulphuric acid to produce hydrogen gas.
Mg + H2SO4 (aq)
H2 + MgSO4
If excess dilute sulphuric acid is used, the number of moles of unreacted acid can be found
experimentally by titration with sodium hydroxide solution using an indicator. This can then be
compared with the theoretical number of moles of unreacted acid.
Experiment

Accurately weigh out 0.4g of magnesium ribbon. Note the exact mass.

Add the magnesium to a 250cm3 volumetric flask.

Use a burette to add 50cm3 of dilute sulphuric acid, concentration 0.5 mol l -1 to the flask.

When the fizzing has completely stopped, add water to the flask up to the 250cm 3 calibration
point. Secure the lid and shake well.

Titrate 25cm3 portions of this solution with sodium hydroxide solution, concentration 0.1 mol l -1,
using phenolphthein as the indicator.

Repeat until two concordant results are obtained (within 0.1cm3)
Write Up
Complete a full write up for this experiment including the all appropriate calculations.
8
Molar Volume
The molar volume is the volume of which one mole of a gas occupies.
Worked Example 1
The volume of 8g of oxygen is 5.5 litres. Calculate the volume of 3 mol of oxygen.
gfm O2 = 2 x 16 = 32g
8g
96g
5.5
5.5 x 96
8
= 66 litres
Worked Example 2
Calculate the volume of 0.14g of nitrogen. (Assume the molar volume of the gas is 23 litres mol -1)
gfm N2 = 2 x 14 = 28g
28g
0.14g
23
23 x 0.14
28
= 0.115 litres or 115 cm3
Worked example 3
Calculate the volume of hydrogen produced when 0.2g of zinc reacts with excess sulfuric acid.
(Take the molar volume to be 22.2 litres mol-1)
Zn +
1 mole
H2SO4
1mole Zn
0.2 x 1
65
=0.003moles
ZnSO4
65g
0.2g
+
H2
1 mole
Therefore, we have 0.003 moles of H 2
1 mole
22.2 Litres
0.003moles
0.003 x 22.2
1
= 0.068 Litres
Reacting Volumes
1 mole of any gas occupies roughly the same volume (at the same temperature and pressure).
This means that equal volumes of gases contain the same number of moles.
=
=
1 litre of O2
1 litre of H2
1 litre of CO2
We already
equation gives the same number of moles of reactants and products.
N2(g) + 3H2(g)
1 mole 3 moles
So..
1 litre + 3 litres
9
2NH3(g)
2 moles
2 litres
know that a balanced
NOTE: liquid or solid reactants and products are not included in this
calculation!
C(s) +
1mole
O2(g)
1mole
CO2(g)
1mole
So.. not a gas 1 litre
1 litre
Worked Example 1
30cm3 of methane is completely burned in 100cm3 of oxygen. What is the volume and composition of
the gas at the end of the experiment? (All volumes are measured at atmospheric pressure and room
temperature)
CH4(g)
1mole
+
2O2(g)
2moles
CO2(g) + 2H2O(l)
1mole
Step 1: from the balanced equation determine which reactant is in excess
1 mole CH4 reacts with 2 moles of O2
30cm3
60cm3
Since there is 100cm3 of O2, O2 is in excess by 40cm3
Step 2: since the oxygen is in excess the methane will be used up first and hence will control the
volume of product.
CH4(g)
1mole
30cm3
+
2O2(g)
CO2(g) + 2H2O(l)
1mole
30cm3
So the total volume and composition of gas at the end of the experiment is:
Total Volume at the end will be: 70cm3
40cm of O2 left unreacted and 30cm3 of CO2 produced
3
NOTE: since the water is a liquid it is not considered.
10
Experiment 3.2—Synthesising sodium citrate—Percentage Yield Calculation.
Sodium citrate is also known as food additive E331. It is often used as an acidity regulator in drinks,
and as an emulsifier for oils when making cheese. It allows the cheeses to melt without becoming
greasy. Sodium citrate is also the main ingredient in over the counter cystitis medication. You are
going to synthesis sodium citrate and calculate the percentage yield and atom economy achieved.
Experiment

Weigh out 3.5g of citric acid crystals, add to a beaker

Measure out 50cm3 of 1.0 mol l-1 sodium hydroxide and add to the crystals

Stir the mixture until all the crystals have dissolved

Evaporate the sodium citrate solution in an evaporating basin

Weigh dried powder with an accurate balance and calculate the percentage yield.
Write up
Complete a full write up for this experiment. The calculation involving percentage yield can only be
completed once the next page is completed.
11
Percentage Yield Calculations
The yield in a chemical reaction is the quantity of product obtained. The actual yield can be
compared, as a percentage, with the theoretical.
Worked Example 1
5g of methanol reacts with excess ethanoic acid to produce 9.6g of methyl ethanoate. Calculate the
percentage yield.
Step 1:
determine the theoretical yield (the quantity expected from the balanced equation)
CH3OH + CH3COOH
CH3OOCCH3 + 2 H20
1 mol
74g
74 x 5
32
1 mol
32g
5g
= 11.56g
Theoretical Yield
Step 2:
= 11.56g
The actual yield is always given in the question.
Actual yield = 9.6g
Step 3:
Percentage yield =
=
9.6 x 100
11.56
83%
The percentage yield is a very important consideration for industrial chemists. They must take
account of cost of raw materials, plant-running costs etc. If the yield of product is not sufficient
enough to cover the costs of production then the process would not be considered to be economically
viable.
Atom Economy
Atom economy is a measure of the proportion of reactant atoms which are incorporated into the
desired product of a chemical reaction.
Calculation of atom economy therefore also gives an indication of the proportion of reactant atoms
forming waste products.
% atom economy =
Mass of desired product(s)
x 100
Total mass of reactants
In developing an atom economical reaction pathway the industrial chemist may well prefer
rearrangement and addition reactions over less environmental friendly substitution and elimination
reactions.
12
Example 1: Addition reaction – halogenation of an alkene
H
H
C
Br
+
C
H3C
H
Br2
CH3
Br
C
C
H3C
H
CH3
(Z)-but-2-ene
C4H8
1mol
(12 x 4) + (8 x 1)
Bromine
Br2
1mol
2 x 79.9
2,3-dibromobutane
C4H8Br2
1mol
(12 x 4) + (8 x 1) + (79.9 x 2)
= 56g
= 159.8g
= 215.8g
Total mass of reactants = 56 g + 159.8 g = 215.8 g
(Note: Product mass is also 215.8 g)
Mass of desired product (2,3-dibromobutane) = 215.8 g
% atom economy =
Mass of desired product(s)
x 100
Total mass HofHreactants
2
HO C
C Cl
H
% atom economy =
Ca(OH)2
+
H
215.8
2-chloroethanol
x 100 = 100%
215.8
C2H5OCl
2mol
2[(12 x 2) + (5 x 1) + 16 + 35.5]
Calcium hydroxide
Ca(OH)2
1mol
40 + 2(16 +1)
= 74g
This process is 100% atom efficient, with all the reactant
= 161g atoms included within the desired product.
Example 2: Elimination reaction
Desired Product
H H
2
HO C
H
C Cl
+
Ca(OH)2
2
H
2-chloroethanol
C2H5OCl
2mol
2[(12 x 2) + (5 x 1) + 16 + 35.5]
= 161g
Desired Product
O
C
H
H
C
+
CaCl2
+
2H2O
H
Calcium hydroxide
Ca(OH)2
1mol
40 + 2(16 +1)
ethylene oxide
C2H4O
2mol
2[(12 x 2) + (4 x 1) + 16]
Calcium chloride
CaCl2
1mol
40 + (2 x 35.5)
Water
H2O
2mol
2 [(2 x 1) + 16]
= 74g
= 88g
= 111g
= 36g
Waste Products
Total Omass of reactants = 161 g + 74 g = 235 +g
CaCl2
H
H
+
2 (Note:
product
mass = 235 g)
C Total
C
H
H
Waste Products
2H2atom
O
%
economy = Mass of desired product(s) x 100
Total Mass of reactants
H
Mass of desired product ethylene oxide = 88 g
ethylene oxide
C2H4O
2mol
2[(12 x 2) + (4 x 1) + 16]
Calcium chloride
CaCl2
1mol
40 + (2 x 35.5)
Water
H2O
% atom economy
2mol
2 [(2 x 1) + 16]
=
88
235
x 100 = 37.4%
This elimination reaction is therefore
= 36g efficient, with the remaining 62.6% in the
= 111g only 37.4% atom
form of unwanted waste products (calcium chloride and water).
= 88g
13
Equilibria
Reversible Reaction
Many reactions are reversible. A reversible reaction can reach equilibrium in a closed system. A
reaction reaches equilibrium when the rate of the forward reaction equals the rate of the reverse
reaction.
reactants
⇌
products
At equilibrium, the concentration of the products and the
reactants will remain constant.
The concentration of reactants will probably not equal the
concentration of the products.
Where an industrial process produces an equilibrium, costly reactants may not be completely
converted into products; chemists try to manipulate the equilibrium to achieve the best possible
conversion rate.
Catalysts increase the rate at which an equilibrium is formed but do not affect the equilibrium
position.
The Position of Equilibrium
The position of equilibrium means the extent to which the reaction has taken place.
The equilibrium in this case lies well to the right hand side. There is 80% conversion
of A into B. Sometimes this is indicated by the size of the arrows.
In this example, equilibrium would be said to lie to the left.
Can you explain what happens in the
case of water?
Le Chateliers Principle
An equilibrium will move to undo any change imposed upon it by temporarily favouring either the
forward or backward reaction until a new equilibrium position is reached.
reactants ⇌ products
If the forward reaction is favoured we say the equilibrium has moved to the right.
If the reverse reaction is favoured we say the equilibrium has moved to the left
14
Products of an Equilibrium Reaction with Time
At the start of the reaction there will be 100%
reactants. However, as the reaction proceeds the
reactants will be used up and the products will be
formed. Eventually the reaction will reach equilibrium.
At this stage the concentration of both the reactants
and products will remain constant.
100
100
75%
%
%
50
50
25%
0
0
Time
Time
100
75%
%
50
25%
0
Time
In a reversible reaction when the curve levels off this indicates that the reaction has reached
equilibrium. (it does not indicate that the reaction has stopped).
Shifting the Position of Equilibrium
If, once an equilibrium has been established the reaction conditions are changed then the position of
equilibrium will be altered. If a new equilibrium is established where there is an increase in products
we say the equilibrium has shifted to the right. If the new equilibrium results in an increased amount
of reactants then we say the equilibrium has shifted to the left.
Changing reaction conditions such as concentration, temperature and pressure will cause a shift in
equilibrium position.
Changing Concentration
Increasing the concentration of the reactants, or decreasing the concentration of the products
shifts the position of equilibrium to the right hand side
Decreasing the concentration of the reactants or increasing the concentration of the products shifts
the position of equilibrium to the left.
15
Changing Temperature
Increasing the temperature favours the endothermic reaction.
Decreasing the temperature favours the exothermic reaction.
Changing Pressure
In the gaseous state molecules have high energies and are fast moving. Pressure is the result of gas
molecules bombarding the walls of the vessel in which the gas is contained. The greater the number
of molecules in a given volume the greater the pressure.
Low pressure
High pressure
C(g)
2D(g)
The pressure on the right hand side is greater than the pressure on the left hand side because there
are more molecules.
NOTE:
1.
Pressure only affects the equilibrium of a system that involves gases.
2.
A pressure change will alter equilibrium only if there are different numbers of moles of
gases on each side.
A(g)
+
B(g)
C(g) +
D(g)
Changing pressure will not affect this equilibrium since there are the same number of moles of gas
on each side of the equation.
2A(g)
+
B(g)
C(g) +
D(g)
Changing pressure will affect this equilibrium since there are different numbers of moles of gases
on each side of the equation.
Increasing the pressure will favour the side with the lowest number of moles in the gas state.
Decreasing the pressure will favour the side with the highest number of moles in the gas state.
16
Experiment 3.3—Dehydration of hydrated cobalt chloride
Crystals of cobalt chloride which contain water molecules of crystallisation (hydrated cobalt chloride,
CoCl2.6H2O) are pink. Anhydrous cobalt chloride, CoCl2, on the other hand is blue. The conversion of
hydrated cobalt chloride to anhydrous cobalt chloride is an example of a reversible reaction, i.e. the
reaction can go forwards or backwards.
Experiment

Add a spatula of hydrated cobalt chloride to a test tube.

Heat the crystals until the water molecules are driven off. Observe the colour change.

Allow the anhydrous cobalt chloride to cool before adding a few drops of water. Observe the
colour change.
Write up
Complete a write up for this experiment to make suitable notes on your observations.
17
The equilibrium between nitrogen dioxide and dinitrogen tetroxide
In chemistry, Le Chatelier's principle, also called
Chatelier's principle or "The Equilibrium Law", can be used
to predict the effect of a change in conditions on a
chemical equilibrium. The principle states that:
If a chemical system at equilibrium experiences a
change in concentration, temperature, volume, or
pressure, then the equilibrium shifts to counteract the
imposed change and a new equilibrium is established.
Experiment
Pressure demo:
Collect a gas syringe filled with an equilibrium mixture of brown nitrogen dioxide and colourless
dinitrogen tetroxide. Press in the plunger of the syringe as far as possible. The colour of the gas
mixture will become darker as the concentration of the gas mixture increases. Within a few seconds
it will become paler, as the position of the equilibrium responds to the increased pressure, and brown
nitrogen dioxide is converted into colourless dinitrogen tetroxide. Release the plunger and the
changes will reverse themselves. Pull the plunger out to reduce the pressure and note the colour
changes.
Temperature demo:
Fill a second syringe with 50 cm3 of air and seal with a screw clip. Clamp both syringes vertically in a
2L beaker of water so that they are immersed up to the 100cm3 mark. Note the temperature and the
readings of both syringes (which should be the same). Heat the water gently with a Bunsen burner
and record the temperature and the volume of gases every 10 °C or so. Before taking each reading,
remove the Bunsen burner and stir the water for a couple of minutes to ensure that the
temperatures of the gases are the same as that of the water. Twist the plungers of the syringes
before taking each reading to ensure that they are not sticking. Continue taking readings until the
temperature is about 70 or 80 °C. Plot graphs of volume against temperature for both gases on the
same axes. The nitrogen dioxide/dinitrogen tetroxide mixture will expand more than air as the
equilibrium responds to the increase in temperature by producing more nitrogen dioxide. If there is
time, take further readings as the water cools to check for leaks.
Notes
A full description for this experiment can be found on the Royal Society of Chemistry web page
under classic chemistry demonstrations.
http://media.rsc.org/Classic%20Chem%20Demos/CCD-81.pdf
18
Effect of a Catalyst
Consider the reversible reaction at equilibrium:
If a catalyst is added the rates of both the forward and backward reaction will be increased.
without catalyst
with catalyst
Potential
energy
Ed
Ea
Eb
A + B
Ec
C+ D
Reaction pathway
There is the same decrease in the activation energies of the forward and backward reactions and
hence there is the same increase in their rates. Therefore, a catalyst does not alter the position
of equilibrium. It does however, speed up the rate at which equilibrium is attained.
Equilibrium in Industry
The manufacture of ammonia is a very important reaction. Ammonia is made from nitrogen and
hydrogen by the Haber Process. The reaction is reversible and if the conditions were kept constant
equilibrium would be attained.
At equilibrium only 15% yield of ammonia is achieved however, the reaction conditions can be altered
to increase the rate of the forward reaction to increase the yield of ammonia.
Use of a catalyst
An iron catalyst is used in the Haber Process. The catalyst does not alter the
position of equilibrium. However, it does allow the process to be carried out
more efficiently at a lower temperature.
The use of a catalyst does not change the equilibrium it only enables the
position of equilibrium to be reached more quickly.
19
(a)
Concentration
As a result of economic considerations, the Haber Process does not actually ever attain
equilibrium. In a condenser, the ammonia gas is cooled and the liquid ammonia piped off.
Constantly removing the ammonia gas decreases the rate of the backward reaction. In
addition, the unreacted nitrogen and hydrogen are recycled. The increase in the
concentrations of the reactant gases increases the rate of the forward reaction. The effect
of both of these factors is a more economical production of ammonia.
(b)
Pressure
Since the number of product molecules is less than the number of reactant molecules,
increasing the pressure increases the rate of the forward reaction. The pressure chosen is
about 200 atmospheres. Beyond this pressure, the relative increase in the rate cannot justify
the increased costs. SEE EXPERIMENT 3.4
(c)
Temperature
Since the reaction is exothermic, increasing the temperature decreases the percentage
conversion of reactants to ammonia. However, if the temperature is too low the reaction is
too slow. As a result, the Haber Process is carried out at a moderately high temperature of
about 400oC.
Experiment 3.4—Effect of pressure on equilibrium (DEMO)
Take a note of the
results/observations and
explain what you see.
Chemical Energy
Enthalpy of Combustion
When 1 mole of a substance burns it releases a certain amount of energy
which we call the enthalpy of combustion.
Definition of Enthalpy of Combustion:
The enthalpy of combustion is the energy released to the surroundings when one mole of a substance
burns completely in oxygen.
When writing balanced equations for enthalpies of combustion it is important to ensure only 1 mole of
the substance being burned is included in the equation:
C + O2 --> CO2
H2 + ½ O2 --> H20
S + O2 --> SO2
CO + ½ O2 --> CO2
C3H8 + 5O2 --> 3CO2 + 4H2O
20
Enthalpy Calculations
There are two types of combustion calculations. In the first type the temperature difference is not
mentioned and this calculation simply requires a proportion calculation to be carried out.
WORKED EXAMPLE 1
When 1g of ethanol, gram formula mass 46g, is burned 30kJ of energy is released. Calculate the
Enthalpy of Combustion for ethanol.
1g
30kJ
46g
30 x 46
1
= 1380kJ
This means that when 1 mole of ethanol burns it will produce 1380kJ. Since combustion is always an
-1
exothermic reaction the enthalpy of combustion is
- 1380 kJmol
WORKED EXAMPLE 2
0.16g of methanol, CH3OH is burned in a spirit burner. The heat from this combustion causes the
temperature of 100g of water to be raised from 20 oC to 27oC. Calculate the enthalpy of combustion
of methanol.
Step 1
Step 2
Eh = c m ΔT
= 4.18 x 0.1 x 7
= 2.926 kJ
0.16g
2.926kJ
32g
2.926 x 32
0.16
Remember mass is always
expressed in kg & 1cm3 of
water equals 1g
= 585.2kJ
So Enthalpy of Combustion, ΔH = - 585.2 kJmol-1
21
Experimental 3.5—Enthalpy of combustion
The enthalpy of combustion of a substance is the energy released when one mole of the substance is
completely burned in oxygen.
The aim of this experiment is to determine the enthalpy of combustion of ethanol i.e. the
enthalpy change for the reaction:
CH3CH2OH (l) + 3O2(g)
2CO2(g) + 3H2O(l)
A measured mass of ethanol is burned in a spirit burner and the heat
released is transferred to a copper can containing a known volume of
water. From the resulting temperature rise, the enthalpy of combustion of
ethanol can be calculated.
In this experiment we assume that all the heat released in the combustion
reaction is absorbed only by the water in the copper can.
Experiment

Weigh the spirit burner (already containing ~50cm3 ethanol) with its cap on and record its
mass. (The cap should be kept on to cut down the loss of ethanol through evaporation)

Using the measuring cylinder, measure out 100 cm3 of water into the copper can.

Set up the apparatus as directed by your teacher.

Measure and record the temperature of the water.

Remove the cap from the spirit burner and immediately light the burner.

Slowly and continuously stir the water with the thermometer. When the temperature has risen
by about 10 °C, recap the spirit burner and measure and record the maximum temperature of
the water.

Reweigh the spirit burner and record its mass.
Write up
Complete a full write up for this experiment. Within the experiment calculate the enthalpy of
combustion for ethanol.
22
Hess’s Law
The Law of Conservation of Energy states that energy can neither be created nor destroyed, but
that it can be changed from one form to another. This particular application of the conservation of
energy law to chemical reactions is known as Hess’s Law.
Hess’s Law:
the enthalpy change in converting reactants into products is the same regardless of
the route by which the reaction takes place.
1
A
B
∆H1
∆H2
2
C
∆H4
3
D
∆H5
∆H3
∆H6
E
The total enthalpy change for route 1 = ∆H1
The total enthalpy change for route 2 = ∆H2 + ∆H3
The total enthalpy change for route 3 = ∆H4 + ∆H5 + ∆H6
According to Hess’s Law the total enthalpy change for 1, 2 & 3 will be identical.
∆H1 = ∆H2 + ∆H3 = ∆H4 + ∆H5 + ∆H6
The reaction can take place by two methods:
∆H1
Route 1
NaOH(s)
∆H2
NaCl(aq)
+ HCl(aq)
+
Route 2
NaOH(aq)
According to Hess’s Law:
∆H1 = ∆H2 + ∆H3
23
+ HCl(aq)
∆H3
Applications of Hess’s Law
Hess’s Law can be used to calculate enthalpy changes, which are difficult or impossible to
determine, by experiment.
Enthalpy of Formation
Definition: The enthalpy of formation is the quantity of heat energy taken in or given out when 1
mole of a substance is formed from its elements in their normal state.
Calculations
The information you need to tackle a calculation will either be given in the question or can be found in
the data booklet.
Worked Example 1
Calculate the enthalpy of formation of ethane given that the enthalpies of combustion of carbon,
hydrogen and ethane are -394kJmoll-1, -286kJmoll-1 and -1561kJmol-1 respectively.
The first thing that you must do is write a balanced equation for the equation that you are required
to calculate the enthalpy change for. This is called the Target Equation.
Target Equation :
2C(s) +
3H2(g)
C2H6(g)
∆H = ?
Next you need to construct balanced equations for each of the substances in your target equation,
using the information given in the question.
(1)
(2)
(3)
C(s) +
H2(g) +
C2H6(g) +
O2(g)
½O2(g)
3½O2(g)
CO2(g)
H2O(l)
2CO2(g) +
3H2O(l)
∆H = -394
∆H = -286
∆H = -1561
These equations can now be rearranged to give the target equation. Note that whatever change you
make to the equation you must also make to the enthalpy value!
2 x (1)
2C(s) +
3 x (2)
3H2(g) +
reverse (3) 2CO2(g) +
2O2(g)
1½O2(g)
3H2O(l)
2CO2(g)
3H2O(l)
C2H6(g) +
3½O2(g)
∆H = -788
∆H = -858
∆H = +1561
Once everything that appears on both the reactant side and the product side have been cancelled you
should be left with your target equation!
2C(s) +
3H2(g)
C2H6(g)
24
∆H = -85kJmoll-1
Worked example 2
Use the enthalpies of combustion in your data booklet to find the enthalpy change for the reaction
between ethyne and hydrogen to produce ethane.
Target Equation :
C2H2
(1) C2H2 + 2½O2
(2) H2 +
½O2
(3) C2H6 +
3½O
(1)
2 x (2)
reverse (3)
+
2H2
2CO2 +
H 20
2CO +
C2H2 + 2½O2
2H2 +
O2
2CO2(g) +
3H2O(l)
C2H2
+
2H2
H 20
3H2O
C2H6
∆H = ?
∆H = -1301
∆H = -286
∆H = -1561
2CO2 + H20
∆H = -1301
2H20
∆H = -572
C2H6(g) +
3½O2(g) ∆H = +1561
∆H= -312kJmoll-1
C2H6
Bond Enthalpies
Bond enthalpies can be divided into two categories (see the Data Booklet).
The bond dissociation enthalpy is the energy required to break one mole of bonds and form two
separate atoms, all species being in the gaseous state. These values are accurately known but it is
only possible to calculate them for diatomic molecules, e.g.
H2(g)
→ 2H(g)
N2(g)
→ 2N(g)
HCl(g) →
H(g) + Cl(g)
+436 kJ mol–1
+945 kJ mol–1
+432 kJ mol–1
For molecules with more than two atoms, such as methane, it is only possible to calculate the mean
bond enthalpy since the situation is much more complex with the C–H bonds breaking off one after
the other to form a different fragment of the original molecule.
The removal of the first hydrogen atom from CH4 does not require the same energy as the removal
of the second hydrogen from the CH3 fragment and so on. The value obtained is the average or mean
bond enthalpy for C–H, i.e.
CH4(g)
→
C(g) + 4H(g)
∆H = 4 x C–H = 1648 kJ mol
Therefore the mean bond enthalpy of C–H = 1648 / 4 = 412kJ mol–1
Mean bond enthalpies are quoted in data books for bonds of that type in any molecule but can also be
calculated from other enthalpy changes, as shown in the following examples.
25
Experiment 3.6—Hess’s law
Solid potassium hydroxide can be converted into potassium chloride solution by two different routes:
Route 1 is the direct route whereby potassium chloride solution is made by adding solid potassium
hydroxide directly to hydrochloric acid. Let's suppose it has an enthalpy change of ΔH1.
KOH(s) + HCl(aq)
KCl(aq) + H2O(l) ΔH1
Route 2 is the indirect route and involves two steps. In the first of these solid potassium hydroxide
is dissolved in water:
KOH(s) + aq
KOH(aq) ΔH2a
The resulting potassium hydroxide solution is then added to hydrochloric acid to form potassium
chloride solution:
KOH(s) + HCl(aq)
KCl(aq) + H2O(l) ΔH2b
According to Hess's Law the overall enthalpy change involved in converting solid potassium hydroxide
into potassium chloride solution will be the same no matter whether the direct or indirect route is
taken.
The aim of this experiment is to confirm Hess's Law.
26
Experiment
ROUTE 1 (direct route)

Using the measuring cylinder, measure out 25cm3 of 1 mol l-1
hydrochloric acid into a plastic beaker or polystyrene cup.

Measure and record the temperature of the acid.

Weigh out accurately about 1.2g of potassium hydroxide into a
plastic beaker or polystyrene cup and record the mass. Make sure
the mass of potassium hydroxide does not exceed 1.4g.

Add the acid to the potassium hydroxide. Slowly and continuously stir the reaction mixture
with the thermometer until all the solid reacts.

Measure and record the highest temperature reached by the reaction mixture.
ROUTE 2 (indirect route)
Step A

The solution you prepare in this step is needed in step B - DON'T THROW ITAWAY!

Using the measuring cylinder, measure out 25 cm3 of water into a plastic beaker or polystyrene
cup.

Measure and record the temperature of the water.

Weigh out accurately about 1.2g of potassium hydroxide into a plastic beaker or polystyrene
cup and record the mass. Make sure the mass of potassium hydroxide does not exceed 1.4g.

Add the water to the potassium hydroxide. Slowly and continuously stir the reaction mixture
with the thermometer until all the solid dissolves.

Measure and record the highest temperature reached by the solution.

Keep the solution you have just prepared but allow it to cool down for some time before
proceeding to step B.
Step B

Using the measuring cylinder, measure out 25 cm3 of 1 mol l-1 hydrochloric acid into a plastic
beaker or polystyrene cup.

Measure and record the temperature of the acid.

Measure and record the temperature of the potassium hydroxide solution you prepared in step
A.

Add the acid to the potassium hydroxide solution and stir the reaction mixture slowly and
continuously with the thermometer.

Measure and record the highest temperature reached by the reaction mixture.
27
Example Question
Calculate the mean bond enthalpy of the C–H bond from the enthalpy of formation of methane
and any other required data from the Data Booklet.
Step 1: Write down the required equation:
CH4
→ C(g) +
∆H = 4 x C–H
4H(g)
Step 2: Express all the information given in equation form:
∆H = –75 kJ mol–1
∆H = + 716 kJ mol–1
∆H = +436 kJ mol–1
(1) C(s) + 2H2(g) → CH4(g)
(2) C(s) → C(g)
(3) H2(g) → 2H(g)
Step 3: Use the three equations to obtain the required balanced equation:
(1) Reverse CH4(g) → C(s) + 2H2(g)
(2) Nothing C(s) → C(g)
(3) x2
2H2(g) →4H(g)
∆H = +75 kJ mol–1
∆H = +716 kJ mol–1
∆H = +872 kJ mol–1
Step 4: Add the three equations to give the required equation and D H:
CH4(g) → C(g) + 4H(g)
∆H = 1663 kJ mol–1
1663 kJ =
C–H = 1663 / 4
4 x C–H
Mean bond enthalpy of C–H = 415.75kJ mol–1
REDOX Reactions
A REDOX reaction involves two half reactions - oxidation and reduction.
These half reactions can be written as ion-electron equations. Ion-electron equations are found on
page 12 of the Data Booklet.
Oxidation involves the LOSS of electrons (OIL):
Fe2+
Fe
+
2e
Mg2+
Mg
+
2e
The ion-electron equation for oxidation must be written in reverse.
Reduction involves the GAIN of electrons (RIG):
Cu2+
+
2e
Cu
Ag+
+
e
Ag
Remember, oxidation and reduction are two halves of the same chemical reaction. The combined
reaction is called a Redox reaction.
To form the overall redox reaction, the ion-electron equations for the oxidation and reduction must
be combined, ensuring that the electrons cancel.
28
Oxidation:
Mg(s)
Reduction:
Ag+
Mg2+
+
e
Mg2+
Oxidation:
Al(s)
Al3+
Reduction:
2H+
Mg(s) +
+
2e
2Ag+
2e
Ag(s)
2Ag+
Mg(s) +
+
(x2)
+ 2Ag(s)
+
3e (x2)
H2(g)
Mg2+
(x3)
+ 2Ag(s)
Oxidising and Reducing Agents
In a redox reaction the species that is oxidised is described as a reducing agent - a species that
allows reduction to occur.
Similarly, a species which is reduced is described as an oxidising agent - a species that allows
oxidation to occur.
eg.
Mg(s)
+
2Ag+
Mg2+
+
2Ag(s)
The elements with low electronegativities (metals) tend to form ions by losing electrons (oxidation)
and so can act as reducing agents. The strongest reducing agents are found in group 1.
The elements with high electronegativities (non-metals) tend to form ions by gaining electrons
(reduction) and so can act as oxidising agents. The strongest oxidising agents are found in group 7.
Using the Data Booklet
When writing redox equations you must first identify the oxidation and reduction half reactions and
then combine them to give the redox reaction. Use the Electrochemical Series on page 12 of the data
booklet.
Example:
Iron (II) sulfate solution is reacted with potassium permanganate solution until the first appearance
of a permanent pink colour.
Step 1:
What four ions are present in the two solutions?
Step 2:
Write the ion-electron equation for the oxidation reaction.
Step 3:
Write the ion-electron equation for the reduction reaction.
Step 4:
Add the two half equations together to give the redox equation (remember to balance
the electrons first)
29
Experiment 3.7—OXIDISING AND REDUCING AGENTS
Halogen displacement
Experiments with halogen displacement can be used to compare the strength of elements as oxidising
agents. In this experiment you will be using chlorine, bromine and iodine water firstly to test their
effect on pH paper, and secondly to test their reaction with potassium halides.
Experiment

Put a piece of Universal Indicator paper onto a white tile.

Use a glass stirring rod to transfer a few drops of the chlorine water onto the indicator paper.

Repeat this with a fresh piece of paper and the bromine water then the iodine water.

In a test-tube, add some chlorine water to a solution of potassium bromide.

Add some chlorine water to a solution of potassium iodide.

Add some bromine water to a solution of potassium chloride

Add some bromine water to a solution of potassium iodide

Add some iodine water to a solution of potassium chloride

Add some iodine water to a solution of potassium bromide

Complete the table noting your observations
Results table
Effect on
indicator paper
Reaction with
potassium
chloride solution
Chlorine water
Bromine water
Iodine water
30
Reaction with
potassium
bromide solution
Reaction with
potassium iodide
solution
Experiment 3.8—OXIDISING AND REDUCING AGENTS
Blue bottle experiment
The blue bottle reaction is a chemical reaction in which in a closed bottle an aqueous solution
containing dextrose (D-glucose), sodium hydroxide and a methylene blue and some air turns from
colorless to blue upon shaking and which then decolorizes again after a while. After shaking again the
blue color returns and this cycle can be repeated several times. The reaction will work with other
reducing sugars besides glucose and also with other reducing dyes. This experiment allows you to see
that dextrose acts as a reducing agent and oxygen acts as an oxidising agent.
Blue Bottle
NOT
Blue Bottle
Experiment
(Note the optimum temperature for this reaction is 25-300C)

Using a measuring cylinder, measure 30cm3, 0.5mol l-1 sodium hydroxide into a 100cm3 conical
flask, volumetric flask or bottle

Add 1g of D-glucose and swirl until the solids are dissolved.

Add 1cm3 of the methylene blue solution.

The blue colour will turn colourless after about one minute.

Stopper the flask or bottle. Shake the flask vigorously so that air dissolves in the solution. The
colour will change to blue. This will fade back to colourless over about 30 seconds. The more
shaking, the longer the blue colour will take to fade. The process can be repeated for over 20
cycles. After some hours, the solution will turn yellow and the colour changes will fail to occur.
Write up
Complete a full write up for this experiment. Make a note of any observations that you see.
31
More Complicated Half-Equations
There are some ion-electron equations that are not given in the data book. You must learn the rules
to work them out for yourself.
Worked example
Write an ion-electron equation for the following reaction:
Cr2O72Step 1:
If necessary, balance the central atom / ion.
Cr2O72-
Step 2:
+
7H2O
2Cr3+
+
7H2O
Calculate the total electrical charge on each side of the equation.
Cr2O72- + 14H+
(12+)
Step 5:
2Cr3+
Balance the hydrogen in the water by adding hydrogen ions.
Cr2O72- + 14H+
Step 4:
2Cr3+
Add water (H2O) if it is needed to balance the oxygen atoms.
Cr2O72-
Step 3:
Cr3+
2Cr3+
+ 7H2O
(6+)
Add electrons to balance the electrical charges.
Cr2O72- + 14H+ + 6e(12+) + (6-)
2Cr3+
+ 7H2O
(6+)
Chemical Analysis
Chromatography
Chromatography is a method of separating and analysing a mixture of
soluble chemical substances.
There are different types of Chromatography techniques. The most
common is paper chromatography.
Uses of Chromatography
Chromatography can be used to follow the course of a chemical reaction or to
establish the products of a chemical reaction.
It can also be used to identify additives in foods, analyse fibres found at a crime scene and
fingerprinting.
32
Paper Chromatography
During paper chromatography a small sample of the mixture being
tested is spotted onto the base line (a straight line usually drawn in
pencil above the level of the solvent) on the filter paper. The filter
paper is then placed in a solvent.
By capillary action the solvent moves up the paper. This is when
different components of the mixture are separated.
Components can move quickly or slowly up the paper depending on the solvent used.
There are two stages in paper chromatography; the Mobile Phase and the Stationary Phase.
The solvent is the mobile phase in paper chromatography and the wet paper is the stationary phase.
Why does separation occur?
Separation of the mixture occurs due to the polarity and size of the molecules within the mixture.
Depending on what solvent is used depends on how the molecules behave.
For example, if a mixture contains very polar molecules and non-polar molecules and a polar solvent,
such as ethanol, is used the polar molecules move much quicker up the filter paper. This is because
the less polar molecules will be more attracted to the mobile phase.
In terms of large and small molecules smaller molecules move much quicker and further up the filter
paper compared to larger molecules.
The Retention Factor (Rf) Values
The Rf factor can be used to compare the different components found in a sample. The Rf values of a
mixture being tested can be compared with known samples.
Note: if two substances on chromatography paper have the same Rf value it is likely that they are
the same compound. If they do not have the same Rf value then they are definitely different
compounds.
The Rf value of the red, green and
blue particles can be measured and
calculated to prove that certain
molecules are present in the mixture.
33
Experiment 3.10—CHEMICAL ANALYSIS
Wool chromatography
Chromatography is the collective term for a set of laboratory techniques for the separation of
mixtures. The mixture is dissolved in a fluid called the mobile phase, which carries it through a
structure holding another material called the stationary phase. The various constituents of the
mixture travel at different speeds, causing them to separate. This experiment uses white wool as the
stationary phase and ethanol as the mobile phase. You will be using this technique to separate the
colour constituents of a felt tip pen.
Experiment

Cut a length of white wool approximately 1m long

Fill a 100cm3 beaker with approximately 20cm3 of methylated spirits

Make a 1cm long mark on one end of the wool approximately 20cm from the end with your
chosen pen. Allow the ink to dry.

Attach a crocodile clip to each end of the wool and set the beaker of methylated spirits on the
edge of a table

Carefully lower the end of wool with the ink into the beaker making sure the ink is not
submerged, drape the wool over the spout of the beaker and allow it to hang towards the floor
with an empty beaker on the floor under the wool.

Watch as the alcohol is wicked up the wool and towards the floor under the influence of
gravity. The alcohol dissolves and separates the different colour components of the ink at
different rates resulting in a series of coloured band moving towards the floor.
34
Redox Titrations
As with volumetric titrations there has to be a way of determining the end-point of the reaction. For
many reactions an indicator must be used however, for some redox reactions the end-point can be
recognised from a colour change in one of the reactants, for example, when potassium permanganate
solution (purple) reacts with iron(II) sulfate solution, the permanganate ions are reduced to
colourless manganese ions. When a colour change involving one of the reactants is used to determine
the end-point the reaction is said to be self-indicating.
Worked example:
20cm3 of iron(II)sulphate were titrated with 0.01mol -1 potassium permanganate solution until a
permanent pink colour was observed. If the volume of potassium permanganate used was 25.6cm 3,
what is the concentration of the iron(II) sulfate solution?
Step 1:
write the ion-electron equations and combine to give the redox equation.
MnO4-
Fe2+
+ 8H+ + 5e-
Fe3+ + e(x5)
2+
Mn
+ 4H20
5Fe2+ + MnO4- + 8H+
Step 2:
5Fe3+ + Mn2+ + 4H2O
calculate the number of moles of the ‘known’ substance
Moles
MnO4-
=
c x v
0.01 x 0.0256
0.000256 moles
using the mole ratio:
Step 3:
calculate the concentration of the iron(II) sulfate solution
C =
=
moles / v
0.00128
0.02
0.064moll-1
35
Experiment 3.11—CHEMICAL ANALYSIS
Redox titration of Vitamin C
Vitamin C (ascorbic acid) is an important component of our diet. In its absence the protein, collagen,
cannot form fibres properly and this results in skin lesions and blood vessel fragility.
Although vitamin C occurs naturally in many fruits and vegetables many people take vitamin C tablets
to supplement their intake.
Vitamin C can undergo a redox reaction with iodine in which the vitamin C is oxidised and the iodine
molecules are reduced.
The aim of this experiment is to determine the mass of vitamin C in a tablet by carrying out a redox
titration using a solution of iodine of accurately known concentration and starch solution as an
indicator.
36
Experiment

Add a vitamin C tablet to the beaker.

Add some deionised water (approximately 50cm3) to the beaker and stir the mixture
until the tablet has dissolved.

Carefully add the resulting solution to the 250cm3 volumetric flask. Rinse out the
beaker several times with water and add the washings to the flask.

Add water to the standard flask to bring the volume of the solution up to the
graduation mark on the neck.

Stopper the flask and invert it several times to make sure the solution is thoroughly
mixed.
After rinsing the pipette with a little of the vitamin C solution, pipette 25cm 3 of it into
the conical flask.


Add a few drops of starch solution to the vitamin C solution in the conical flask.

After rinsing the burette with a little iodine solution, fill the burette with the iodine
solution.

Note the initial burette reading. Since the solution has a dark colour, it is difficult to
see the bottom of the meniscus. Take the burette reading from the top of the
meniscus.

Add the iodine solution slowly from the burette whilst gently swirling the solution in
the conical flask. Initially you will see a blue/black colour as the iodine reacts with
the starch but this will rapidly disappear as the iodine reacts with the vitamin C.

Near the end-point of the titration the colour disappears more slowly. At this point
add the iodine solution drop by drop until the solution just turns a blue/black colour
and remains so.

This is the end-point of the titration i.e. all the vitamin C has reacted. Note the final
burette reading.

Wash out the conical flask.

Repeat the titrations until concordant results are obtained.
Write up
Complete a full write up for this experiment. Show all calculations and determine the mass of Vitamin
C in the tablet.
37
Word
Meaning
Atom Economy
The proportion of reactant atoms which are incorporated into the
desired product of a chemical reaction.
% atom economy =
Mass of desired product(s)
x 100
Total mass of reactants
Batch processes
Bond Dissociation
Enthalpy
These involve mixing one batch of reactants in a vessel, allowing the
reaction to complete, then removal of the products, before reloading
with a fresh batch of reactants.
Energy required to break one mole of bonds to form two separate
atoms, all species being in the gaseous state.
Capital costs
Capital costs are incurred in the initial building of the plant and
associated infrastructure.
Cash flow
Cash flow is the movement of money into and out of a business, and is a
measure of profitability.
Continuous processes
These take place in a plant where reactants are fed in continuously at
one end, and products are removed continuously from the other.
Chromatography
A method of separating and analysing a mixture of soluble chemical
substances.
Dynamic equilibrium
A dynamic equilibrium is achieved when the rates of two opposing
processes become equal, so that no net change results
Electrodes
The electrodes (one negative and one positive) dip into the electrolyte
and form the connection to the electric wires. Any chemical reactions
occur at their surfaces.
Electrolysis
Electrolytes are liquids containing ions which are free to move, and are
therefore capable of conducting electricity.
Electrolyte
Electrolytes are liquids containing ions which are free to move, and are
therefore capable of conducting electricity.
Electrolytic cell
An electrolytic cell is the equipment used to carry out electrolysis –
oxidation occurs at the positive electrode and reduction at the negative
electrode.
38
Equilibrium
Chemical equilibrium is the state reached by a reaction mixture when
the rates of forward and reverse reactions have become equal
Excess
Reactant added which is not all used up in the reaction.
Feedstock
Feedstocks are the reactants from which other chemicals can be
extracted or synthesised.
Fixed costs
Fixed costs are incurred irrespective of whether a plant is operating at
maximum or only at partial capacity.
Hess’s law
The enthalpy change for a chemical reaction is independent of the route
taken, providing the starting point and finishing point is the same for
both routes.
Molar Volume
The volume, in litres, occupied by one mole of gas under standard
conditions of temperature and pressure.
Ion-electron equations
An ion-electron equation is a half-equation, either an oxidation or a
reduction, which in combination of the opposite type, can be part of a
complete redox equation.
Oxidation
An oxidation is a loss of electrons by a reactant in any reaction.
Oxidising agent
An oxidising agent is a substance which accepts electrons.
Pilot plant
A pilot plant is a small scale chemical plant, built with similar materials
to a fullscale one, to investigate the chemical and engineering aspects
of a process before a full-scale plant is built.
Raw materials
Raw materials are the initial starting materials from which the a
process’s feedstock is derived.
Reducing agent
A reducing agent is a substance which donates electrons.
Reduction
A reduction is a gain of electrons by a reactant in any reaction.
Stoichiometry
Stoichiometry means literally ’measuring elements’ and is concerned
with the quantities of substances reacting and the quantities of
substances produced as derived from the balanced equation, i.e. it is
concerned with mole relationships.
Variable costs
Variable costs are dependent on a plant’s output.
Yield
Quantity of product obtained in a reaction, usually given as % yield =
actual/theoretical x100.
39
The Chemical Industry
Exercise 1
1.
The chemical industry will design a process to
Maximise what?
Minimise what?
2.
Describe the stages in the manufacture of a new product.
3.
What does the term Feedstock mean?
4.
What are the main raw materials used in the chemical industry?
5.
Give examples of capital, fixed and variable costs.
6.
Describe a continuous process, giving advantages and disadvantages.
7.
Describe a batch process, giving advantages and disadvantages.
8.
Along with economic viability suggest other factors that will impact on the location and route
used to manufacture a new product.
9.
Aluminium is extracted from its purified oxide by molten electrolysis. Suggest two advantages
and disadvantages of siting the aluminium smelters in the Scottish Highlands.
The Chemical Industry
1.
Which element is NOT a raw material in the chemical Industry?
A
B
C
D
2.
Nitrogen
Oxygen
Sodium
Sulfur
The costs involved in the industrial production of a chemical are made up of fixed costs and
variable costs.
Which of the following is most likely to be classified as a variable cost?
A
B
C
D
3.
Exercise 2 Past Paper Questions
The cost of land rental
The cost of plant construction
The cost of labour
The cost of raw materials
Which of the following is produced by a batch process?
A
B
C
D
Sulfuric acid from sulfur and oxygen
Aspirin from salicylic acid
Iron from iron ore
Ammonia from nitrogen and hydrogen
40
The flow chart summarises some industrial processes involving ethene.
The feedstocks for ethene in these processes are
A
B
C
D
5.
A
B
C
D
ethane and glycol
ethane and ethanol
glycol and poly(ethene)
glycol, poly(ethene) and ethanol.
Which of the following compounds is a raw material in the chemical industry?
Ammonia
Calcium carbonate
Hexane
Nitric acid
6.
Cerium metal is extracted from the mineral monazite.
The flow diagram for the extraction of cerium from the
mineral is shown on the previous page.
Name the type of chemical reaction taking place in Step A.
In Step B, cerium hydroxide is heated to form cerium
oxide, Ce2O3, and compound Z.
Name compound Z.
In Step C, cerium metal is obtained by electrolysis
What feature of the electrolysis can be used to reduce the
cost of cerium production?
41
7.
About 2.5 million tonnes of sulfuric acid are produced each year in the UK. Sulfuric acid was
made industrially by the Chamber Process. The following chemical reactions are involved.
Sulfur is burned to produce sulfur dioxide.
Sulfur dioxide reacts with water to produce sulfurous acid.
Nitric oxide is produced by the catalytic oxidation of ammonia;
water is also a product of this reaction.
Nitric oxide reacts with oxygen to form nitrogen dioxide
Nitrogen dioxide reacts with sulfurous acid to form sulfuric acid and regenerate nitric oxide.
Copy and complete the flow diagram below of the Chamber process with the names of the chemicals
involved (2 Marks)
To show the name of a chemical
To Show a reaction
Sulfuric acid
42
The Mole Revision
Moles and mass
Exercise 3
1.)
a)
c)
Write the formula and then work out the gram formula mass (mass of 1 mole) of:
hydrogen sulfide
(b)
magnesium bromide
hydrogen sulfate
(d)
hydrogen nitrate
2.)
What is the mass of the following?
a)
c)
10 moles of calcium carbonate (b)
0.5 moles of lithium fluoride (d)
3.)
How many moles of compound are present in:
a)
c)
14 g of nitrogen gas
85 g of lithium chloride
3 moles of sodium chloride
0.2 moles of methane (CH4)
(b) 400 g of copper(I) oxide
(d) 117 g of sodium sulfide
Moles and concentration
1)
Calculate the number of moles of:
a)
sodium chloride in 1000 cm3 of 1 mol l-1 solution.
b)
sodium chloride in 500 cm3 of 0.5 mol l-1 solution.
c)
potassium nitrate in 250 cm3 of 0.5 mol l-1 solution.
2)
What volume of water is required to make:
a)
3 moles of potassium hydroxide into a 1.5 mol l -1 solution?
b)
0.5 moles of sodium chloride into a 2 mol l -1 solution?
c)
0.1 moles of lithium oxide into a 0.05 mol l -1 solution?
3)
What concentration of solution is obtained if we dissolve:
a)
2 moles of sodium hydroxide in 500 ml of water?
b)
1 mole of lithium chloride in 2 L of water?
c)
1.5 moles of magnesium bromide in 100 cm3 of water?
4)
What concentration of solution is obtained if we dissolve:
a)
b)
c)
5)
What volume of water is required to make:
a)
b)
c)
6)
40 g of sodium hydroxide in 1 litre of water?
58.5 g of sodium chloride in 100 cm3 of water?
585 g of sodium chloride in 10 litres of water?
56 g of potassium hydroxide into a 1.5 mol l -1 solution?
117 g of sodium chloride into a 2 mol l-1 solution?
15 g of lithium oxide into a 0.05 mol l-1 solution?
What mass of substance is required to make:
a)
b)
c)
500 cm3 of 2 mol l-1 aluminium nitrate solution?
2 L of 0.5 mol l-1 potassium bromide solution?
100 cm3 of 3 mol l-1 calcium sulfate solution?
43
Balancing Equations and Calculations from Balanced Equations Revision
1.
Balance the following equations:
a)
b)
c)
d)
N2
N2
Na
CH4
2.
2Mg + O2
+ H2
+ O2
+ H 2O
+ O2
NH3
NO
NaOH + H2
CO2 + H2O
MgO
Calculate the mass of oxygen required to burn 6 g of magnesium.
3.
CaCO3
CaO + CO2
What mass of CaCO3 is required to form 2.8 g of calcium oxide?
4.
N2 + 3H2
2NH3
What mass of ammonia (NH3) is produced from 7 g of nitrogen?
Excess Calculations
Exercise 4
In each of the following reactions calculate which reactant is in excess
1.
a)
Zn(s) +
H2SO4(aq)
ZnSO4(aq) +
H2(g)
6.5 g of Zinc added to 25 cm3 of dilute sulfuric acid,
concentration 2 mol l-1
b)
Mg
(s)
+
2HCl(aq)
MgCl2(aq)
+
H2(g)
2.4 g of magnesium added to 100 cm3 of dilute hydrochloric acid,
concentration 1 mol l-1
c)
Zn(s) +
2HCl(aq)
ZnCl2(aq)
+
H2(g)
3.6 g of Zinc added to 50 cm3 of dilute hydrochloric acid,
concentration 0.5 mol l-1
d)
Mg
(s)
+
H2SO4(aq)
MgSO4(aq) +
H2(g)
4.7 g of magnesium added to 25 cm3 of dilute sulfuric acid,
concentration 1 mol l-1
2.)
Iron(II) sulfide reacts with hydrochloric acid as follows:
FeS(s)
+ 2HCl(aq)
FeCl2(aq)
+
H2S(g)
If 4.4 g of iron(II) sulfide was added to 160 cm 3 of 0.5 mol l-1 hydrochloric acid, show by calculation
which substance is in excess.
44
3.
A student added 0.20 g of silver nitrate, AgNO3, to 25 cm3 of water. This solution was then
added to 20 cm3 of 0.0010 mol l-1 hydrochloric acid. The equation for the reaction is:
AgNO3(aq)
+
HCl(aq)
AgCl(s)
+
HNO3(aq)
Show by calculation which reactant is in excess.
4.
What mass of ammonia gas is produced when 1.32 g of ammonium sulfate is heated with 1 g of
sodium hydroxide?
(NH4)2SO4 +
2NaOH
2NH3 +
Na2SO4
+
2H2O
5.
What mass of sulfur dioxide is produced when 1.26 g of sodium sulphite is added to 50cm 3 of
dilute hydrochloric acid , concentration 2 mol l -1
Na2SO3
6.
+
2HCl
2NaCl +
H 2O +
SO2
Calcite is a very pure form of calcium carbonate which reacts with nitric acid as follows:
CaCO3(s)
+ 2HNO3(aq)
Ca(NO3)2(aq) + H2O(l) + CO2(g)
A 2.14 g piece of calcite was added to 50.0 cm3 of 0.200 mol l-1 nitric acid in a beaker. Calculate the
mass of calcite, in grams, left unreacted.
7.
Copper(II) oxide reacts with sulfuric acid as follows:
CuO(s)
+ H2SO4(aq)
CuSO4(aq)
+
H2O(l)
1.6 g of copper(II) oxide is added to a beaker containing 50 cm 3 of
0.25 mol l-1 sulfuric acid. Calculate the mass of copper(II) oxide remaining after the reaction was
complete.
8.
Lead reacts with hydrochloric acid as follows:
Pb(s) +
2HCl(aq)
PbCl2(aq)
+
H2(g)
If 6.22 g of lead was added to 50 cm3 of 1 mol l-1 hydrochloric acid, calculate the mass of lead left
unreacted.
9.
A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 10.00 g of
silver nitrate. The reaction that occurs is
Zn(s)
+
2AgNO3(aq)
2Ag(s)
+
(a) Determine which reactant is in excess.
(b) Calculate how many grams of silver will be formed.
45
Zn(NO3)2(aq)
10.
A piece of lithium with a mass of 1.50 g is placed in an aqueous solution containing 6.00 g of
copper (II) sulfate. The reaction that occurs is:
2Li(s)
+
CuSO4(aq)
Cu(s)
+
Li2SO4 (aq)
(a) Determine which reactant is in excess.
(b) Calculate how many grams of copper will be formed.
Molar Volume
1.
Exercise 5
Decide which has the greater volume, measurements made at the same temperature and
pressure
a)
b)
c)
d)
17 g of ammonia or 17 g of methane
20 g of neon or 20 g of nitrogen
4 g of hydrogen or 44 g of carbon dioxide
6.4 g of sulfur dioxide or 8 g of oxygen
2.
The volume of 0.22 g of propene C 3H6 is 118 cm3. Calculate the volume of 2 moles of propene
at this temperature and pressure .
3.
Calculate the volume of 2.4 g of ethane, C2H4. (take the molar volume to be 23.6 litres mol-1)
4.
The volume of 1 g of hydrogen is 11.6 litres. Calculate the volume of 4 moles of hydrogen at
this temperature and pressure.
5.
3 g of an alkane occupies a volume of 2.24 litres. What is the molecular formula of the alkane
(take the molar volume to be 22.4 litres mol-1)
6.
Using the densities in the data booklet calculate the volume of 10 g of
(a)
(b)
7.
Using the densities in the data booklet calculate the mass of 10 litres of
(a)
(b)
8.
Hydrogen
Argon
Helium
Nitrogen
From the following data calculate the approximate formula mass of the gas X.
Mass of plastic bottle empty
Mass of plastic bottle + Gas X
Capacity of plastic bottle
Molar volume of gas X
=
=
=
=
112.30 g
113.52 g
1 litre
23.6 litres
46
9.
A flask, capacity 600 cm3 was used to calculate the molar volume of sulfur dioxide.
The following data was obtained:
Mass of evacuated flask
Mass of flask + sulfur dioxide
10.
=
=
512.97 g
514.57 g
Some of liquid Z is injected into a large syringe and it evaporates. The following results were
recorded:
Mass of syringe before injection
Mass of syringe after injection
Syringe reading before injection
Syringe reading after injection
=
=
=
=
4.648 g
4.774 g
0 cm3
84 cm3
Calculate the relative formula mass of liquid Z
(Molar volume = 30.6 litre mol-1)
Balanced Equations Using Mass and Volumes
Exercise 6
Molar volume 24 litres unless stated otherwise
1.)
What volume (in l) of carbon dioxide would be produced by completely reacting 60 g of carbon
with oxygen?
C
2.)
4.)
O2
CO2
What volume (in l) of hydrogen would be produced by completely reacting 60 cm 3 of
hydrochloric acid of concentration 1.2 mol l –1 with zinc?
Zn
3.)
+
+
2HCl
ZnCl2 +
H2
What volume (in l) of carbon dioxide would be produced by completely reacting 10 g of calcium
carbonate with hydrochloric acid?
CaCO3
+
2HCl
Mg(s) +
2HCl(aq)
CaCl2 +
H 2O +
MgCl2(aq)
+
CO2
H2(g)
What volume of hydrogen gas is produced when 5 g of magnesium is added to 100 cm 3 of
dilute hydrochloric acid concentration 2 mol l -1 ( Molar volume =23.6 Litres mol-1)
5.)
What mass of magnesium oxide is obtained when 24 g of magnesium is ignited in 5 litres of
oxygen
Mg(s)
+
O2(g)
MgO(s)
(Molar volume is 22.8 litres mol-1)
47
6.
What mass of water is produced when 2 litres of hydrogen is burned in excess oxygen (use
the density of hydrogen given in the data book)
7.
Calculate the volume of oxygen required for the complete combustion of
1 g of ethane C2H4
(Molar volume is 32.2 litres mol-1)
8.
Chlorine gas can be produced by heating calcium hypochlorite, Ca(OCl) 2, in dilute hydrochloric
acid.
Ca(OCl)2(s) + 2HCl(aq)
Ca(OH)2(aq) + 2Cl2(g)
Calculate the mass of calcium hypochlorite that would be needed to produce 0·096 litres of
chlorine gas. (Take the molar volume of chlorine gas to be 24 litres mol –1.)
9.)
Hydrogen fluoride gas is manufactured by reacting calcium fluoride with concentrated
sulfuric acid.
CaF2 +
H2SO4
CaSO4
+
2HF
What volume of hydrogen fluoride gas is produced when 1.0 kg of calcium fluoride reacts
completely with concentrated sulfuric acid?
(Take the molar volume of hydrogen fluoride gas to be 24 litres mol –1.)
10.)
In the lab, nitrogen dioxide gas can be prepared by heating
copper(II) nitrate.
Cu(NO3)2(s)
CuO(s) + 2NO2(g) + ½O2(g)
Calculate the volume of nitrogen dioxide gas produced when 2.0 g of copper (II)
nitrate is completely decomposed on heating.
(Take the molar volume of nitrogen dioxide to be 24 litres mol –1.)
48
Reacting Volumes
1.
Exercise 7
In each of the following reactions decide the ratio of the volume of product(s) to the volume
of reactant(s).
a)
b)
c)
d)
2.
H2(g)
N2(g)
2C(s)
C2H4(g)
+
+
+
+
F2(g)
3H2(g)
O2(g)
3O2(g)
2HF(g)
2NH3(g)
2CO(g)
2CO2(g)
N2(g)
+
2O2(g)
2NO2(g)
+
2H2O(l)
What volume of nitrogen dioxide is produced when 100 cm3 of nitrogen is sparked in excess
oxygen?
3.
When the following gases are burned calculate the volume of oxygen required and the volume
of carbon dioxide produced
a)
b)
c)
4.
50 cm3 of methane
200 cm3 of ethane
250 cm3 of carbon monoxide
10 cm3 of butane gas is mixed with 75 cm3 of oxygen and the mixture exploded. Calculate the
volume and composition of the resulting gas mixture.
C4H10(g)
+
6.5 O2(g)
4CO2(g)
+
5H2O(l)
5.
If 100cm3 of propene is burned completely with 900 cm3 of oxygen what will be the volume
and composition of the resulting gas mixture?
6
50 cm3 of ethyne, C2H2 is burned completely in 220 cm3 of oxygen
a)
b)
What will be the volume and composition of the resulting gas mixture?
What will be the volume and composition of the resulting gas mixture if the
experiment was repeated at 200oC?
49
7.
What volume of oxygen (in litres) would be required for the complete combustion of a gaseous
mixture containing 1 litre of carbon monoxide and 3 litres of hydrogen?
8.
20 cm3 of ammonia gas reacted with an excess of heated
copper(II) oxide.
3CuO
+
2NH3
3Cu
+
3H2O
+
N2
Assuming all measurements were made at 200 °C, what would be the volume of gaseous
products?
9.
2C2H2(g)
ethyne
+
5O2(g)
4CO2(g) +
2H2O(l)
What volume of gas would be produced by the complete combustion of 100 cm 3 of ethyne gas?
All volumes were measured at atmospheric pressure and room temperature.
10.
The equation for the complete combustion of propane is:
C3H8(g)
+
5O2(g)
3CO2(g) +
4H2O(l)
30 cm3 of propane is mixed with 200 cm3 of oxygen and the mixture is ignited. What is the
volume of the resulting gas mixture? (All volumes are measured at the same temperature and
pressure.)
50
Percentage Yield and Atom Economy
1.)
CH3OH
Methanol
+
Exercise 8
C2H5COOH
propanoic acid
C2H5COOCH3
+
methyl propanoate
H 20
In preparation, 40.4 g of methyl propanoate is obtained from 18.3 g of methanol.
(a) Calculate the percentage yield
(b) Calculate the atom economy in regards to methyl propanoate
2.)
oxidation
C3H7OH
Propan-1-ol
C2H5CHO
Propanal
In a preparation, 3.2 g of propanal is obtained from 3.9g of propan-1-ol. Calculate the
percentage yield.
3.)
C3H6
+
Br2
C3H6Br2
20.4 g of 1,2-dibromopropane is obtained from 5.2 g of propene.
Calculate the percentage yield
4.)
N2
+
3H2
2NH3
Under test conditions, 2 kg of hydrogen reacts with excess nitrogen to produce 1.5 kg of
ammonia.
Calculate the percentage yield
5.)
2SO2
+
O2
2SO3
Under test conditions, 1 kg of sulfur dioxide reacts with excess oxygen to produce 0.8 kg of
sulfur trioxide.
Calculate the percentage yield
51
6.
2FeCl2
+
Cl2
2FeCl3
5.072 g of iron (II) chloride yields 4.869 g of iron (III) chloride, calculate the percentage
yield.
7.
8.
Calculate the atom economy for the production sulfur trioxide assuming that all the reactants
are converted into products.
H2SO4
H 2O +
SO3
Which reaction below has the highest atom economy for producing water?
2C2H6 +
7O2
4CO2 +
6H2O
C3H6
9.
4NH3
+
+
5O2
4½O2
3CO2
4NO +
+
3H2O
6H2O
a)
1.68 x 103 kg of NO was produced from1.36 x103 kg calculate the percentage yield.
b)
Calculate the atom economy in terms of the production of NO
10.
Fe2O3 +
3CO
2Fe
+
3CO2
6.65 x 104 kg of an iron ore which is impure iron (III) oxide is reacted with an excess of
carbon monoxide, producing 2.79 x 104 kg of iron.
a)
Calculate the mass of pure iron (III) oxide in the ore. (Assuming that all the iron
(III) oxide is reduced to iron and that the impurities do not take part in the reaction)
b)
Calculate the percentage by mass, of the iron (III) oxide in the ore.
52