Download CfE Higher Chemistry Unit 3: Chemistry in Society

SCHOLAR Study Guide
CfE Higher Chemistry
Unit 3: Chemistry in Society
Authored by:
Emma Maclean (George-Heriot’s School)
Reviewed by:
Diane Oldershaw (Menzieshill High School)
Previously authored by:
Peter Johnson
Brian T McKerchar
Arthur A Sandison
Heriot-Watt University
Edinburgh EH14 4AS, United Kingdom.
First published 2014 by Heriot-Watt University.
This edition published in 2016 by Heriot-Watt University SCHOLAR.
Copyright © 2016 SCHOLAR Forum.
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Distributed by the SCHOLAR Forum.
SCHOLAR Study Guide Unit 3: CfE Higher Chemistry
1. CfE Higher Chemistry Course Code: C713 76
ISBN 978-1-909633-22-3
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i
Contents
1 Getting the most from reactants
1.1 Prior knowledge . . . . . . . . . . . . . . . .
1.2 Introduction . . . . . . . . . . . . . . . . . .
1.3 The design of industrial chemical processes
1.4 Mole calculations . . . . . . . . . . . . . . .
1.5 Molar volume . . . . . . . . . . . . . . . . .
1.6 Reacting volumes . . . . . . . . . . . . . . .
1.7 Percentage yields . . . . . . . . . . . . . . .
1.8 The atom economy of a process . . . . . . .
1.9 Excess . . . . . . . . . . . . . . . . . . . . .
1.10 Summary . . . . . . . . . . . . . . . . . . . .
1.11 Resources . . . . . . . . . . . . . . . . . . .
1.12 End of topic test . . . . . . . . . . . . . . . .
2 Chemical equilibria
2.1 Prior knowledge . . . . . . . . .
2.2 Reversible reactions . . . . . . .
2.3 Dynamic equilibrium . . . . . .
2.4 Altering the equilibrium position
2.5 The Haber process . . . . . . .
2.6 Summary . . . . . . . . . . . . .
2.7 Resources . . . . . . . . . . . .
2.8 End of topic test . . . . . . . . .
3 Chemical energy
3.1 Prior knowledge . . . . . .
3.2 Introduction . . . . . . . .
3.3 Potential energy diagrams
3.4 Enthalpy changes . . . . .
3.5 Introduction to Hess's law
3.6 Bond enthalpies . . . . . .
3.7 Summary . . . . . . . . . .
3.8 Resources . . . . . . . . .
3.9 End of topic test . . . . . .
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4 Oxidising or reducing agents
4.1 Prior knowledge . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Elements as oxidising and reducing agents . . . . . . . . . .
4.4 Molecules and group ions as oxidising and reducing agents
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ii
CONTENTS
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
Uses for strong oxidising agents .
Ion-electron half equations . . . .
Combining ion-electron equations
Complex ion-electron equations .
Redox titrations . . . . . . . . . .
Summary cloze test . . . . . . . .
Summary . . . . . . . . . . . . . .
Resources . . . . . . . . . . . . .
End of topic test . . . . . . . . . .
5 Chromatography
5.1 Prior knowledge . . . . . . . . .
5.2 Introduction . . . . . . . . . . .
5.3 Paper chromatography . . . . .
5.4 Gas-liquid chromatography . . .
5.5 Size-exclusion chromatography
5.6 Summary . . . . . . . . . . . . .
5.7 Resources . . . . . . . . . . . .
5.8 End of topic test . . . . . . . . .
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6 Volumetric analysis
6.1 Prior knowledge . . . . . . . . . .
6.2 Introduction . . . . . . . . . . . .
6.3 Quality control . . . . . . . . . . .
6.4 Practical applications of titrations
6.5 Summary . . . . . . . . . . . . . .
6.6 Resources . . . . . . . . . . . . .
6.7 End of topic test . . . . . . . . . .
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189
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7 End of unit test
201
Glossary
216
Hints for activities
219
Answers to questions and activities
1
Getting the most from reactants
2
Chemical equilibria . . . . . . .
3
Chemical energy . . . . . . . .
4
Oxidising or reducing agents .
5
Chromatography . . . . . . . .
6
Volumetric analysis . . . . . . .
7
End of unit test . . . . . . . . .
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220
220
234
238
246
254
258
261
© H ERIOT-WATT U NIVERSITY
1
Topic 1
Getting the most from reactants
Contents
1.1
Prior knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.2
1.3
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The design of industrial chemical processes . . . . . . . . . . . . . . . . . . .
4
5
1.4
Mole calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 The mole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
10
1.5
1.4.2 Avogadro's constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Molar volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
22
1.6
1.7
Reacting volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Percentage yields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
32
1.8
1.7.1 Calculations involving percentage yields . . . . . . . . . . . . . . . . . .
The atom economy of a process . . . . . . . . . . . . . . . . . . . . . . . . . .
35
37
Excess . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.9.1 Calculating excess . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
43
1.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.11 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
47
1.12 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
1.9
Prerequisite knowledge
Before you begin this topic, you should know or be able to:
•
understand and explain the factors affecting rates of reaction (Higher, Unit 1);
•
to compare rates of chemical reactions, changes in mass, volume and other
quantities can be measured. Graphs can then be drawn to help this comparison
(National 4, Unit 1);
•
the gram formula mass is defined as the mass of one mole of a substance
(National 5, Unit 1);
•
use the chemical formula of any substance the gram formula mass can be
calculated using relative formula masses of its constituent elements (National 5,
Unit 1);
•
the concentration of solutions in moles per litre (National 5, Unit 1);
2
TOPIC 1. GETTING THE MOST FROM REACTANTS
•
calculations to determine the concentration and volume and the mass of a
substance through the number of moles present (National 5, Unit 1);
Learning objectives
At the end of this topic, you should know that:
The chemical industry
•
Industrial processes are designed to maximise profit and minimise the impact on
the environment.
•
Factors influencing process design include: availability, sustainability and cost of
feedstock(s); opportunities for recycling; energy requirements; marketability of byproducts; product yield.
•
Environmental considerations include: minimising waste; avoiding the use or
production of toxic substances; designing products which will biodegrade if
appropriate.
Chemical calculations
•
Balanced equations show the mole ratio(s) of reactants and products. Using
the balanced equation and the gram formula masses (GFM), mass to mass
calculations can be performed.
•
The quantity of a reactant or product can also be expressed in terms of moles.
The concentration of a solution can be expressed in mol l -1 .
•
Balanced equations can be used in conjunction with concentrations and volumes
of solutions and/or masses of solutes to determine quantities of reactants and/or
products.
Molar volume
•
The molar volume (in units of litres mol -1 ) is the same for all gases at the same
temperature and pressure. The volume of a gas can be calculated from the
number of moles and vice versa.
•
The volumes of reactant and product gases can be calculated from the number of
moles of each reactant and product.
Percentage yield
•
The efficiency with which reactants are converted into the desired product is
measured in terms of the percentage yield and atom economy.
•
Percentage yields can be calculated from mass of reactant(s) and product(s) using
a balanced equation.
•
Given costs for the reactants, a percentage yield can be used to calculate the
feedstock’s cost for producing a given mass of product.
© H ERIOT-WATT U NIVERSITY
TOPIC 1. GETTING THE MOST FROM REACTANTS
Atom economy
•
The atom economy measures the proportion of the total mass of all starting
materials successfully converted into the desired product.
•
It can be calculated using the formula shown below in which the masses of
products and reactants are those appearing in the balanced equation for the
reaction.
•
Atom Economy = (mass of desired product(s) / total mass of reactants) × 100.
•
Reactions which have a high percentage yield may have a low atom economy
value if large quantities of unwanted by-products are formed.
Excess
•
In order to ensure that costly reactant(s) are converted into product, an excess of
less expensive reactant(s) can be used.
•
By considering a balanced equation, the limiting reactant and the reactant(s) in
excess can be identified.
•
Whilst the use of excess reactants may help to increase percentage yields, this will
be at the expense of the atom economy so an economic / environmental balance
must be struck.
© H ERIOT-WATT U NIVERSITY
3
4
TOPIC 1. GETTING THE MOST FROM REACTANTS
1.1
Prior knowledge
Test your prior knowledge
Q1:
Go online
a)
b)
c)
d)
Which three variables can be altered to change the rate of a reaction?
Concentration
Particle Size
Temperature
All of the above
..........................................
Calculate the mass, in grams, of 0.25 moles of Butane (C 4 H10 ).
Q2:
a)
b)
c)
d)
58
56
29
14.5
..........................................
Q3: 10.0 cm3 of potassium hydroxide (KOH) of concentration 0.25 mol l -1 is titrated
with nitric acid. It takes exactly 12.5 cm3 of the acid to neutralise the alkali. Calculate
the concentration of the nitric acid.
0.1 mol l-1
0.2 mol l-1
1.0 mol l-1
2.0 mol l-1
a)
b)
c)
d)
..........................................
1.2
Introduction
The chemical industry is a major contributor to the British, and Scottish, economy.
This unit considers the principles governing chemical reactions. There are four major
sections:
•
Stoichiometry and equilibrium;
•
Kinetics;
•
Energy;
•
Chemical analysis.
Why is this topic important?
•
The oldest site for manufacturing chemicals in Scotland was established in 1871
and is still operating today.
•
More than 13,500 people are employed in the chemical industry in Scotland.
© H ERIOT-WATT U NIVERSITY
TOPIC 1. GETTING THE MOST FROM REACTANTS
•
The chemical industry generates a turnover of £3.4 billion annually.
•
Scotland is part of the EU - the world’s largest chemicals market both as a
producer and consumer.
After a look at general factors which influence the design of industrial processes, this
topic looks at stoichiometry, the relationship between the quantities of reactants and
products, and studies equilibrium in reversible reactions.
1.3
The design of industrial chemical processes
Key point
Industrial processes are designed to maximise profit and minimise the impact on
the environment.
"The Chemical Industry is not in existence to manufacture chemicals: like any other
industry it exists to create wealth and wealth can only be created if it can make profits."
So, the major determinant in any process design is how do we make the most profit from
our product, to enable investment in future products, and with due regard to the safety
of the workers, the local community and protecting the environment?
Until a few decades ago concern for the wider environment was not given much
consideration. Nowadays, the safety and environmental effects of the manufacture, use
and disposal of consumer products, including chemicals, is of paramount importance.
Many chemical processes are changing in response to the requirements of 'green
chemistry'. You will look at some of these later in this topic.
Process design economics
In order to maximise the profitability of a chemical process, several factors have to be
considered.
•
Raw materials and feedstocks have to be:
◦
readily available;
◦
preferably from sustainable sources;
◦
and at a suitable cost.
•
Recycling of materials should be able to reduce costs, compared with new
manufacture.
•
The energy requirements for a process should be minimised.
•
Can any unavoidable by-products also be sold at a profit or do they have to be
disposed of as costly waste?
•
The process should have the highest possible yield and the maximum atom
economy. (These will be discussed in detail later.)
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Sulfuric acid manufacture
The figure below shows a sulfuric acid manufacturing plant. The corresponding flow
chart is shown below.
Sulfuric acid plant at ICI Billingham, UK
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Q4: Look at the flow chart for the production of sulfuric acid. What are the feedstocks
for this process?
a)
b)
c)
d)
Sulfur, Air and Water.
Sulfur, Air, Sulfur Trioxide and Water.
Sulfur and Air.
Air and Water.
..........................................
Q5: What are the advantages of using these materials as feedstocks?
a) Readily available.
b) Sustainably sourced.
c) Available at a suitable cost.
d) No toxic by-products produced.
e) Waste by-products can be sold on.
..........................................
Q6: You can see that there are three passes of the SO 2 /O2 mixture through a series
of catalyst beds and coolers. Why do you think this is?
a) To make the reaction more economical.
b) To neutralise products.
c) Catalysts can be used over and over again.
d) So that no toxic by-products produced.
..........................................
Q7: Does this process generate wasteful by-products?
a) Yes
b) No
..........................................
Q8: Can you see anywhere where energy saving can be used?
..........................................
..........................................
Safety
The handling of chemicals is increasingly controlled by regulations such as COSHH
(Control of Substances Hazardous to Health Regulations, 1999), where the hazards
are clearly defined and measures to ensure safety are implemented before any contact
with the chemicals takes place. Part of this involves all chemicals being clearly labelled
with an EU Hazard symbol (some of these are shown below).
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!
Environmental considerations
These are largely achieved by designing processes which:
•
minimise the production of waste; either as unwanted products, or through the
use of materials (e.g. solvents) which require disposal at the end of the process;
•
and avoid the use or production of toxic materials, or those which will not
degrade in the environment.
DDT
The insecticide DDT was a very effective agent against malaria-carrying mosquitos. It
has several advantages:
•
it is cheap to produce;
•
it is effective against a wide variety of insect pests;
•
and it is relatively non-toxic to mammals.
The incidence of the cases of malaria dropped very significantly in the 1970s following
widespread spraying in countries such as India (with 500,000 deaths from malaria in
1960, but only 1000 each year in the early 1970s).
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Crop spraying in East Germany (http:/ / bit.ly/ 2bk47 V p, by http:/ / bit.ly/ 2bk24lk)
Deutsches Bundesarchiv (German Federal Archive), Bild 183-20820-0001. http:/ / cr e
ativecommons.or g/ licenses/ by-sa/ 3.0/ d e/ d eed .d e
Its stability, an advantage to its use, also meant that it would be able to persist for long
periods in the environment. DDT has even been carried to and detected in the
Antarctic, where it has never been used. It also increases in concentration along the
food chain. A consequence of this was the thinning of bird's egg shells, which broke
before hatching, resulting in alarming drops in the population of birds of prey like
sparrowhawks and peregrine falcons. Nowadays the use of DDT is banned in many
countries.
Use the internet to investigate new insecticides. Are they as effective at destroying
pests? What is their expected 'lifetime' in the environment?
..........................................
In many cases the problems associated with chemicals in the environment were not
anticipated when the chemical was first used.
Chemical companies can solve these problems by developing improved products, for
example:
•
Hydrogen-containing CFCs, called HCFCs, are refrigerants without the
ozone-damaging chlorine atoms. The importance of changing to alternative
chemicals is described in the Montreal Protocol on Substances that Deplete the
Ozone Layer, which came into effect in 1989.
•
Modern insecticides do not persist in the environment, so do not cause the
problems of DDT. The trade-off is that they are expensive and therefore not used
extensively by the poorer countries, where the incidence of malaria is increasing.
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•
Alternative and renewable sources of energy, which do not produce CO 2 (e.g.
hydrogen and wind or wave generated electricity) are replacing fossil fuels. The
Kyoto protocol, in 1997, was an attempt by world governments to reduce the
output of CO2 , and so reduce global warming effects.
1.4
Mole calculations
Whether you are carrying out a small scale preparation of a few grams of material in
a laboratory or producing hundreds of tonnes of a commercial product, it is vital that
you can calculate the amounts of reactants required and the amount of product(s) you
expect.
In order to do this you have to be able to balance chemical equations. All chemical
calculations depend on correctly balanced equations. If you are unsure about balancing
chemical equations you should practise now.
1.4.1
The mole
At National 5 level, one mole of a substance was defined as the formula mass expressed
in grams (i.e. the gram formula mass - GFM). The number of moles is therefore a
measure of the amount of substance.
Key point
The mole is fundamental to nearly all calculations in chemistry.
The equation below shows a balanced equation for the formation of carbon monoxide
from carbon and carbon dioxide, one of the reactions which occurs in a blast furnace,
used for making iron.
Formation of carbon monoxide
One interpretation of this equation is that one atom of carbon reacts with one molecule
of carbon dioxide to form two molecules of carbon monoxide. This is not much help to a
chemist since it is impossible to measure out individual atoms and molecules. A better
interpretation is that one mole of carbon and one mole of carbon dioxide react to form
two moles of carbon monoxide (see below).
Formation of CO with mole relationships
These are real quantities which can be measured. Notice that the total mass on the left
of the equation equals the total mass on the right. The mole provides a link between
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atoms and molecules reacting at the atomic level and chemical reactions occurring in
the 'real' world.
Mole display
The figure above shows one mole of each of four substances. Note that although the
amount of substance is the same (same number of moles), the mass of each substance
is different because the individual atoms and molecules making up each substance have
different masses.
Mass calculations
You should be familiar with calculations involving reacting masses from previous
chemistry courses, so here is a short exercise to check your knowledge. You will need
the SQA data booklet to obtain relative atomic masses.
Q9: 'Fools' gold' is a mineral with formula FeS 2 . A piece of the mineral contains 13.95
g of iron, how many grams of sulfur are present?
a)
b)
c)
d)
13.95
16.05
27.90
32.10
..........................................
Q10: Iron disulfide, FeS 2 can be used as a source of sulfur for sulfuric acid manufacture.
The ore is roasted in air to produce sulfur dioxide and iron(iii) oxide. Which is the
balanced equation for this reaction?
a)
b)
c)
d)
FeS2 + 3O2 → FeO2 + 2SO2
2FeS2 + 5O2 → Fe2 O3 + 4SO2
4FeS2 + 11O2 → 2Fe2 O3 + 8SO2
6FeS2 + 16O2 → 4Fe2 O3 + 16SO2
..........................................
Q11: 204 tonnes of iron disulfide ore was heated in air. How many tonnes of sulfur
dioxide were produced?
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a)
b)
c)
d)
204.0
217.9
256.4
512.8
..........................................
Q12: You need to make 100 kg of sulfuric acid. Assuming total conversion at each
stage, how many kg of FeS 2 would you need to start with?
a)
b)
c)
d)
244.65
122.32
100.00
61.16
..........................................
Q13: You are going to prepare 100 g of the ester methyl benzoate. Assuming complete
conversion, how many grams of methanol and benzoic acid would you require?
..........................................
Calculations involving solutions
Many chemical reactions are carried out in solution, so you have to be able to handle
calculations with solutions.
Chemists often express the concentration of solutions in mol -1 . You actually weigh out
grams of solutes, so you have to be able to convert mol -1 to g -1 and vice versa.
Here are some examples of calculation questions with solutions. You might like to try
to answer them yourself first, then check your solution (pun intended!). After these
examples there is a series of questions for you to try.
Example : Mass to concentration
What is the concentration, in mol -1 , of a solution of sodium chloride made by dissolving
9 g of NaCl in 1 litre of solution?
1. Calculate the GFM (gram formula mass) of sodium chloride:
NaCl is 23.0 + 35.5 = 58.5
2. Calculate the number of moles:
58.5 g is 1 mol
1 g is 1/58.5 mol
9 g is 9/58.5 mol
= 0.154 mol (to 3 decimal places)
3. Calculate the concentration:
moles
concentration = volume
(in litres)
concentration = 0.154
1
= 0.154 mol -1 to 3 decimal places
..........................................
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Examples
1. Mass to concentration
What is the concentration, in mol -1 , of a solution of silver nitrate made by dissolving
2.30 g of AgNO 3 in 100 ml of solution?
1. Calculate the GFM (gram formula mass) of silver nitrate:
AgNO3 is 107.9 + 14.0 + (3 × 16.0) = 169.9
2. Calculate the number of moles:
169.9 g is 1 mol
1 g is 1/169.9 mol
2.30 g is 2.3/169.9 mol
= 0.0135 mol (to 3 significant figures)
3. Change the volume to litres:
100 ml = 0.1 4. Calculate the concentration:
concentration = 0.0135
0.1
= 0.00135
= 0.00135 mol -1 to 3 significant figures
..........................................
2. Concentration to mass
How many moles and grams are in 250 ml of solution of calcium bromide of
concentration 0.15 mol -1 ?
1. Write the formula of calcium bromide, then calculate the GFM:
Formula is CaBr2
GRM is 40.0 + (2 ×79.9) = 199.8
2. Change volume to litres
250 ml = 0.25 3. Calculate number of moles of solute:
moles of solute = concentration (in mol -1 ) × volume (in litres)
moles = 0.15 × 0.25 = 0.0375 mol
4. Calculate number of grams of solute:
1 mol = 199.8 g
0.0375 mol = 0.0375 × 199.8 = 7.49 g to 3 sig. figures
..........................................
3. Reaction - masses and solutions
What mass of magnesium, in grams, can react completely with 100 ml of 0.25 mol -1
hydrochloric acid?
1. Write a balanced equation:
Mg + 2HCl → MgCl2 + H2
From which: 1 mol Mg reacts with 2 mol HCl
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2. Calculate the number of moles of HCl (what you know):
moles of solute = concentration (in mol -1 ) × volume (in litres)
moles = 0.25 × 0.10 = 0.025 mol
3. Calculate number of moles, and grams, of Mg reacting with this amount of HCl:
2 mol HCl reacts with 1 mol Mg
0.025 mol HCl reacts with 0.025/2 mol Mg = 0.0125 mol
4. Atomic mass of magnesium is 24.3
so mass of Mg:
0.0125 mol Mg = 0.0125 × 24.3 = 0.30375 g
..........................................
4. Reaction - masses and solutions
A chemist makes a solution containing copper(II) ion by dissolving 49.92 g hydrated
copper(II) sulfate (CuSO4 .5H2 O) in 500 ml of solution. What is the concentration of this
in mol -1 ?
Zinc metal, 3.27 g, was added to 200 ml of the copper sulfate solution. What mass, in
grams, of copper(II) ion was left in solution after reaction with the zinc?
Copper sulfate solution is 0.40 mol -1
200 ml of this contains 0.08 mol Cu 2+ ion
3.27 g Zn is 0.05 mol
from the equation 1 mol Zn reacts with 1 mol Cu 2+ , so 0.03 mol Cu2+ left in solution
1.905 g copper ion is left in solution
..........................................
5. Reaction - solutions
Insoluble barium sulfate can be prepared by adding solutions containing a soluble
barium salt and a soluble sulfate.
A solution of barium chloride (BaCl 2 ) contains 104.15 g in 1 litre. What volume of sodium
sulfate (Na2 SO4 ) containing 142.1 g in 1 litre, would be required to react completely with
50 ml of the barium chloride solution?
1. Write a balanced equation: BaCl 2 (aq) + Na2 SO4 (aq) → BaSO4 (s) + 2NaCl(aq)
2. Calculate the relative formula masses: BaCl 2 - 208.3; Na2 SO4 - 142.1
3. Calculate the concentrations of solutions: BaCl 2 - 104.15/208.3 = 0.5 mol -1 ;
Na2 SO4 - 142.1/142.1 = 1.0 mol -1
4. 50 ml of barium chloride contains 0.05 × 0.5 = 0.025 mol
5. 1 mol BaCl2 reacts with 1 mol Na2SO4, so 0.025 mol BaCl 2 reacts with 0.025 mol
Na2 SO4 .
6. Concentration of Na 2 SO4 is 1.0 mol -1 , so 0.025 mol is contained in 0.025 = 25
ml.
..........................................
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1.4.2
15
Avogadro's constant
Note: Avogadro’s Constant is a suggested learning activity, not a key area of the
course.
To understand what one mole of water and one mole of sodium chloride have in
common, we can use a simple analogy.
Q14: How many nails are on the right hand balance?
..........................................
In this example, the number of nails is small enough to be counted. However, this is not
feasible with very large numbers. Suppose you are a manufacturer of nuts, bolts and
washers. It is essential to manufacture the same number of nuts and bolts since one is
useless without the other.
Nuts and bolts
Figure 1.1: Nuts and bolts
Go online
..........................................
Q15: How many nuts are in the pile of nuts? (Note that one tonne = 1000 kg.)
..........................................
Q16: How many bolts are in the pile of bolts?
..........................................
Q17: The manufacturer wants the same number of washers. What mass in tonnes does
he need?
..........................................
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Key point
It is possible to count very large numbers of small objects by weighing them.
..........................................
A similar approach can be used with atoms as shown in the next activity.
Avogadro's constant calculations
Figure 1.2: Avogadro's constant
Go online
mass of one atom
mass of one atom
Number of atoms
He
4.0amu
4.0 g
C
12.0amu
12.0 g
L
?
Al
27.0amu
?
..........................................
Q18: On paper, write an expression for L, the number of atoms in one mole of helium.
(If you have difficulty, think back to the 'Nuts and bolts' activity. How did you work out the
number of nuts in the pile?)
..........................................
Q19: Using Figure 1.2, how many atoms of carbon are in one mole of carbon?
..........................................
Q20: What mass in grams of aluminium will contain the same number of atoms? Give
your answer to 1 decimal place.
..........................................
The number of atoms in one mole of any element is known as Avogadro's constant
which is given the symbol L. It has a value of 6.02 × 10 23 mol-1 . (Care must be taken
with elements such as oxygen and the halogens which are diatomic.)
This number is so enormous that it is very difficult to fully comprehend it. A number of
analogies have been developed to try to help, e.g.
•
In order to get 6.02 × 10 23 grains of sand, an area the size of the Sahara Desert
would have to be excavated to a depth of 2 metres.
•
Try to work the next one out for yourself.
Q21: If a 1p coin is 1 mm thick, how far would a column of 6.02 × 10 23 1p coins stretch?
Give your answer in light years, to three significant figures, where one light year is the
distance travelled by light in one year. (The speed of light = 3 × 10 8 m s-1 .)
..........................................
After the Sun, the nearest star to planet Earth is just over 4 light years away! Avogadro's
constant is a huge number which in turn means that individual atoms are incredibly
small.
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17
Key point
One mole of helium (4.0 g) contains the same number of atoms as one mole of
carbon (12.0 g) and one mole of aluminium (27.0 g). This number is known as
Avogadro's constant (symbol L) and has a value of 6.02 × 10 23 mol-1 .
..........................................
So, Avogadro's constant is the number of atoms in one mole of an element, except for
those elements which exist as diatomic molecules. What about compounds like sodium
chloride which contain ions rather than atoms?
Formula units
Figure 1.3: Formula units
Go online
..........................................
Q22: Complete the following sentence:
In diamond, one formula unit is a carbon .......
..........................................
Q23: Complete the following sentence:
In oxygen, one formula unit is an oxygen .......
..........................................
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Q24: In sodium chloride, one formula unit is:
a) a sodium ion.
b) a chloride ion.
c) one sodium ion and one chloride ion.
..........................................
Q25: In silicon dioxide, one formula unit is:
a)
b)
c)
d)
one silicon atom and one oxygen atom.
one silicon(IV) ion and one oxide ion.
one silicon atom and two oxygen atoms.
one silicon(IV) ion and two oxide ions.
..........................................
Key point
The term 'formula unit' is a general term which relates to the type of particles
which make up a substance. In general, it refers to the formula normally used for
the substance.
..........................................
In general:
•
The formula unit is an atom for all elements which do not exist as diatomic
molecules, e.g
4.0 g of helium
31.0 g of
phosphorus
55.8 g of iron
All contain L atoms (6.02 × 10 23 atoms)
•
The formula unit is a molecule for all diatomic elements and covalent molecular
compounds, e.g.
32.0 g of oxygen
44.0 g of carbon
dioxide
18.0 g of water
All contain L molecules (6.02 × 10 23 molecules)
•
For all ionic compounds and covalent network compounds, the formula unit is the
simplest ratio of atoms or ions in the formula of the compound, e.g. for sodium
chloride
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Molar quantity
58.5 g of Na+ Cl-
Number of formula units
L (6.02 × 1023 )
Number of Na+ ions
L
Number of
•
19
Cl-
ions
L
Total number of ions
2L
e.g. for silicon dioxide
Molar quantity
60.1 g of SiO2
Number of formula units
L (6.02 × 1023 )
Number of Si atoms
L
Number of O atoms
2L
Total number of atoms
3L
A more general definition of the mole and Avogadro's constant can now be given.
Key point
One mole of any substance contains 6.02 × 10 23 formula units. This number is
known as Avogadro's Constant (L). Equimolar quantities of substances contain
equal numbers of formula units.
It is very unusual to use exactly molar quantities of substances. However any mass of
substance can be quoted as a number of moles by using the relationship:
Number of moles =
mass in grams
mass in grams
or
GFM
mass of 1 mole
Since one mole of any substance contains L formula units, it follows that equimolar
quantities (quantities containing the same number of moles) must contain the same
number of formula units, e.g.
0.1 moles of each of the following substances will contain the same number of formula
units.
Substance
Mass of 0.1 mole
Number of formula units
oxygen
3.2 g
0.1 × L molecules
iron
5.58 g
0.1 × L atoms
sodium chloride
5.85 g
0.1 × L NaCl units
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The mole is being used as a counting unit. There are many other examples of names
being given to counting units, e.g. dozen, gross, ream (500 sheets of paper) etc.
Calculations using Avogadro's constant (optional)
Go online
At National 5 level, calculations involving the conversion of mass to moles and moles to
mass were covered. At Higher another stage is added:
Type A
Mass → number of moles → number of particles
Type B
Number of particles → number of moles → mass
Note that the number of moles is central to these calculations. These calculations are
much easier to manage if a standard routine is used each time a calculation is carried
out.
Examples
1. A - from mass to number of particles
A glass of water contains 210 g of water.
1. How many molecules of water are there in the glass?
2. What is the total number of atoms in the water?
The first step is to work out the direction of the calculation, i.e. what information have
you been given and what are you asked to find out. Always write the direction as shown
below with 'Given' to the left and 'Required' to the right. Then, below that write the 'Key
relationship' in the same order.
1. Now you can start to use the information from the question. The gram formula mass
of water is 18 g.
Given
Required
Mass of water
Number of molecules
Key
relationship
1 mole contains
L formula units
For water
18 g contains
6.02 × 1023 molecules
In this case
210 g contains
210/18 × 6.02 × 10 23
molecules
Answer
210 g water contains
7.02 × 1024 molecules
Here is the calculation:
18 g of water contains 6.02 × 10 23 molecules
By proportion,
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21
6.02 × 1023
molecules
18
23
210 × 6.02 × 10
molecules
18
1 g of water contains
210 g contains
= 70 × 1023 molecules
= 70 × 1024 molecules
2. One H2 O molecule contains two H atoms and one O atom, i.e. 3 atoms in one
molecule.
Total number of atoms
= 3 × 7.0 × 1024
= 2.1 × 1025
..........................................
2. B - Particles to mass
What mass of silicon dioxide would contain 1.204 × 10 21 atoms of silicon?
As before, work out the direction of the calculation and then write the key
relationship beneath it with the 'Given' to the left and the 'Required' to the right.
One formula unit of silicon dioxide is SiO 2 , i.e. one formula unit contains one Si atom.
Given
Required
Number of Si atoms
Mass of silicon dioxide
Key
relationship
L formula units
weigh 1 mole
For SiO2
6.02 × 1023 formula units
weigh 60.1 g
For SiO2
6.02 × 1023 Si atoms
are in 60.1 g SiO 2
In this case
1.204 × 1021 Si atoms
(≡ to 0.01204 × 10 23 )
are in 60.1 ×
0.01204/6.02 g
Answer
1.204 × 1021 Si atoms
are in 0.1202 g SiO 2
Here is the calculation:
6.02 × 1023 atoms of Si in one GFM
6.02 × 1023 atoms of Si in 60.1 g of SiO 2
By proportion,
1 atom in
60.1
g
6.02 × 1023
of SiO2
1.204 × 1021 atoms in
60.1 × 1.204
6.02
60.1 × 1.204 × 1021
6.02 × 1023
g
10−2 g
= 12.02 × 10-2 g
mass of SiO2 = 0.1202 g
..........................................
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Try the next five questions for yourself. Write your answers and working on paper
before revealing the answers. All have worked answers which you can use as a check if
you have difficulty.
Q26: How many molecules are there in 0.22 g of carbon dioxide?
..........................................
Q27: What mass of calcium carbonate would contain 3.612 × 10 20 formula units?
..........................................
Q28:
a) How many molecules are present in 0.112 g of methane?
b) How many atoms are present in 0.112 g of methane?
..........................................
Q29: How many ions are present in 37kg of magnesium nitrate?
..........................................
Q30: What mass of potassium sulfate would contain 9.03 × 10 21 ions?
..........................................
..........................................
1.5
Molar volume
When measuring out quantities of solid reactants or products, it is usual to weigh them
in a container of known mass. The mass of the solid can then be obtained by subtracting
the mass of the container. Pure liquids can be weighed in a similar way.
The masses obtained can be converted to molar quantities using the GFM of the
substance.
However, gases are very difficult to weigh since they have very low densities and tend
to spread easily, not to mention that most of them are invisible! It is more appropriate
to measure the volume of a gas. However, there are problems. The volume of a gas
depends on the temperature of the gas and also on the pressure. A balloon in front of
a fire expands as it gets hotter and hotter and the pressure inside increases until the
balloon bursts.
So the number of gas molecules in a one litre sample will depend on the conditions of
temperature and pressure.
e.g. at different temperatures,
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fewer molecules
23
more molecules
e.g. at different pressures,
more molecules
fewer molecules
A valid comparison between volumes of gases can only be made if the conditions of
temperature and pressure are the same. The following activity shows how the volume
occupied by one mole of a gas can be calculated. This volume is known as the molar
volume.
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Molar volume
Go online
Example : Calculating the molar volume
Using the apparatus shown, 1 of methane was collected. The initial reading on the
balance was 124.86 g and the final reading was 124.14 g. What is the molar volume of
methane under these conditions?
Results:
Volume of methane collected = 1.0 l
Initial mass = 124.86 g
Final mass = 124.14 g
Mass of methane released = 0.72 g
Step 1
The molar volume is the volume occupied by one mole of methane. Formula of methane
is CH4 . So one mole will weigh 16.0 g.
Step 2 Identify the direction of the calculation.
Direction of calculation is
mass ⇒ volume
0.72 g occupy 1 l
1
l
1.0 g occupy
0.72
16.0
l
16.0 g occupy
0.72
= 22.2 l
Molar volume of methane is 22.2 l mol −1
..........................................
Q31:
Under these conditions, using the same apparatus with the same initial mass, 1 of
hydrogen was collected. The final mass reading was 124.77 g. What is the molar
volume of hydrogen?
..........................................
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Q32:
Under these conditions, using the same apparatus with the same initial mass, 1 of
carbon dioxide was collected. The final mass reading was 122.88 g. What is the molar
volume of carbon dioxide?
..........................................
Q33: What do you notice about the molar volumes of methane, hydrogen and carbon
dioxide (under these conditions)?
..........................................
Key point
The molar volume is measured in units of mol -1 (also shown as dm3 mol-1 ) and
is the same for all three gases since the conditions of temperature and pressure
are the same.
..........................................
Density Calculations
The SQA data booklet shows the densities of selected elements including the gaseous
elements. Note that the density is quoted in g cm -3 and measured at s.t.p. (standard
temperature and pressure). Standard temperature for these measurements is 0 ◦ C and
standard pressure is one atmosphere. It is a simple matter to calculate the molar volume
of a gaseous element from its density.
Example : Molar volume from density
Calculate the molar volume of argon at s.t.p.
From the data booklet:
Density of argon = 0.0018 g cm -3
= 1.8 g l -1
(Since 1000 cm3 = 1 l)
1 mole of argon weighs 40.0 g
Direction of calculation:
mass ⇒ volume
1.8 g occupies 1.0 l
1.0 g occupies
1.0
1.8
40.0 g occupies
l
40.0
1.8
= 22.22 l mol-1
So the molar volume of argon at s.t.p. is 22.22 l mol -1
Note that this means that for any gas
molar volume =
GF M
density
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..........................................
Now you try. Calculate the molar volume for the following gases at s.t.p. using the values
in the data booklet. Give your answers in mol -1 to 2 decimal places.
Q34: Oxygen
..........................................
Q35: Krypton
..........................................
Q36: Chlorine
..........................................
The cube in the photograph opposite represents the molar volume of helium at s.t.p.
The density values in the data booklet have been rounded to 4 decimal places but even
so it can be seen that the molar volume is about the same for all gases under these
conditions (s.t.p.). This can be generalised further.
Key point
The molar volume is the same for all gases at the same temperature and
pressure.
Q37: Under particular conditions of temperature and pressure, the molar volume of
neon is 30.0 mol -1 . Under the same conditions of temperature and pressure, what will
be the molar volume of xenon in mol -1 ?
..........................................
This leads to a very useful statement often referred to as Avogadro's hypothesis:
'Equal volumes of different gases, under the same conditions of temperature and
pressure, contain the same number of molecules (atoms if the gas is a noble
gas).'
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So provided that the temperature and pressure remain the same the volume of a gas is
a measure of the number of moles of that gas. So the volume of a gas can be calculated
from the number of moles and vice versa. Work carefully through the two examples and
then try the questions which follow.
Examples
1. Volumes to moles
Under certain conditions, the molar volume of sulfur dioxide is 25.0 mol -1 . How many
moles are present in 100 cm 3 of sulfur dioxide gas?
Note that the units of molar volume are given in whereas the volume of SO 2 is given
in cm3 . Before starting the calculation, the units must be made compatible.
1000 cm3 = 1 l
So, 100 cm3
Direction of calculation:
Volume ⇒ Moles
25.0 l = 1.0 mol
1.0
mol
1.0 l =
25.0
0.1
0.1 l =
25.0
= 0.004 mol
..........................................
2. Moles to volumes
If the molar volume of carbon dioxide under certain conditions is 28.0 mol -1 , what
volume in cm3 would be occupied by 0.04 mol?
Direction of calculation:
Moles ⇒ Volume
1.0 mol = 28.0 l
0.04 mol = 28.0 × 0.04 l
= 1.120 l
= 1120 cm3
..........................................
Molar volume - Further practice
Click on Practice questions on molar volume to sit the test.
Go online
If you do not have access to the on-line version, try the following examples.
Q38: Calculate the molar volume of hydrogen sulfide, in l mol -1 , if 0.7g occupies 320
cm3 .
..........................................
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Q39: How many moles of oxygen are there in 900 cm 3 if the molar volume is 28 l mol-1 ?
..........................................
Q40: What volume is occupied by 0.59 moles of carbon dioxide if the molar volume is
28 l mol-1 ?
..........................................
1.6
Reacting volumes
Provided that the conditions of temperature and pressure do not change, the volume of
a gas can be used as a measure of the number of moles of the gas in the same way
as the mass of a solid or pure liquid. This enables the volume of a gaseous reactant or
product to be calculated from the appropriate balanced equation.
Consider the electrolysis of water.
The apparatus opposite can be used for the experiment. A small amount of sulfuric
acid is added to the water to increase the conductivity and speed up the process. This
does not affect the overall reaction. The gases produced at each electrode can be
collected and the volume measured.
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Example : Electrolysis of water
In an experiment, 50 cm 3 of hydrogen is collected. What volume of oxygen would be
collected at the same time?
Start with the balanced equation and mole relationship:
From the question, work out the direction of the calculation:
All measurements are made at the same time and under the same conditions of
temperature and pressure. The molar volumes of hydrogen and oxygen will be the
same - equal volumes of gases will contain the same number of moles.
2 volumes of H2
1 volume of O2
2 cm3
1 cm3
50 cm3
25 cm3
..........................................
The following activity looks at the combustion of methane, in which both reactants and
one of the products are gases. The combustion is carried out by mixing the methane
and oxygen and igniting the mixture using an electric spark. After the reaction, the
products and any unreacted reactants are allowed to cool back to room temperature.
Provided that all measurements are made under the same conditions of temperature
and pressure, the volumes of gases will be proportional to the number of moles
present.
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Note that the volumes of liquids and solids can be ignored. The volumes of solids and
liquids are negligible compared with equimolar volumes of gases as shown in the
photograph opposite.
Combustion of methane
Go online
Explanation
Initial volume of oxygen = 50 cm3
Initial volume of methane = 20 cm 3
Final volume = 30cm3
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Figure 1.4: Combustion of methane
..........................................
Almost always, one of the reactants will be in excess. Before continuing, it is essential
to establish which reactant is in excess. From the volume relationship (Figure 1.4), the
volume of oxygen required for complete reaction is twice the volume of methane, i.e.
20 cm3 of methane would require 40 cm 3 of oxygen. The initial volume of oxygen is 50
cm3 and so the oxygen is in excess.
From the equation,
20 cm3 of CH4 + 40 cm3 of O2 → 20 cm3 of CO2
volume of unreacted oxygen = 50 - 40
= 10 cm3
volume of CO2 produced = 20 cm3
So, the total volume of gas at the end = 30 cm 3
Note that if the temperature had been greater than 100 ◦ C, the water would have been
present as steam (i.e. gas not liquid). The volume of the steam would have to be taken
into account. From the mole relationship (Figure 1.4), 2 moles of H2 O are formed for
every 1 mole of CH4 used up. So, 2 volumes of steam would also have been produced,
an additional 40 cm 3 of gas. So the final volume would have been 70 cm 3 if all the
measurements had been made at a temperature above 100 ◦ C.
Questions of this type frequently arise in examinations. They are not as difficult as they
may appear, provided you set them out in a logical order as shown above.
..........................................
Key point
The volume of gaseous products or gaseous reactants can be calculated from a
balanced equation using the number of moles of a reactant or product.
Combustion of methane - Further practice
Try the next two questions for yourself before displaying the worked answers. A
randomised test is available online. Further questions will appear in the tutorial at the
end of the topic.
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Q41: 200 cm3 of propane was exploded with 1300 cm 3 of oxygen. Calculate the volume
and composition of the resulting gas mixture, assuming that all measurements were
made at the same room temperature and pressure.
..........................................
Q42: 400 cm3 of ethene was exploded with 900 cm 3 of oxygen. Calculate the volume
and composition of the resulting gas mixture, assuming that all measurements were
made at the same room temperature and pressure.
..........................................
1.7
Percentage yields
In the industrial manufacture of a chemical, it is important to ensure that the process
produces as much product as possible. The amount produced is known as the yield
and this is usually expressed as a percentage yield. A 100% yield means that the
maximum possible amount of product has been isolated. This amount is known as the
theoretical yield.
In practice, the theoretical yield is rarely, if ever, obtained for a variety of reasons:
•
some material is lost during purification;
•
there may be side reactions, i.e. more than one product;
•
the reaction may be reversible, i.e. only some of the reactants are converted into
products.
Balanced equations are used to calculate the theoretical yield.
Example : Calculating the theoretical yield
Consider the hydration of ethene to form ethanol.
If 14 tonnes of ethene are reacted, what is the theoretical yield of ethanol? (Steam is
assumed to be in excess.)
First write a balanced equation for the reaction.
Beneath the equation, write the mole relationship and beneath that insert the Gram
Formula Mass (use the SQA data booklet). You only need to do this for the reactant and
product stated in the question. Other reactants will be in excess.
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Theoretically,
28 g of ethene ⇒ 46 g of ethanol
28 tonnes ⇒ 46 tonnes
46
tonnes
1 tonne ⇒
28
46
× 14 tonnes
14 tonnes ⇒
28
= 23 tonnes
The theoretical yield of ethanol is 23 tonnes.
..........................................
If the actual mass of the product is known, then a percentage yield can be calculated.
Example : From theoretical yield to percentage yield
Following on from the last example, if only 20.7 tonnes of ethanol are produced, what is
the percentage yield?
The percentage yield is calculated by dividing the actual yield by the theoretical yield
and multiplying by 100.
percentage yield =
actual yield
× 100
theoretical yield
(1.1)
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..........................................
In this example:
percentage yield
=
20.7
23.0 x
100
= 90%
..........................................
If the percentage yield for a particular reaction is known, it is possible to predict the
mass of product formed.
Example : Predicting the mass of product
Predict the mass of ethyl ethanoate produced when 12.0 g of ethanoic acid reacts with
excess ethanol if the percentage yield is 60%.
As always, start with the balanced equation, mole relationship and Gram Formula Mass
(GFM).
60 g of ethanoic acid ⇒ 74 g of ethyl ethanoate
74
g of ethyl ethanoate
1 g of ethanoic acid ⇒
60
74
× 12.0
12.0 g of ethanoic acid ⇒
60
The theoretical yield is 14.8 g. The percentage yield was given as 60%.
Rearranging Equation 1.1 gives:
percentage yield × theoretical yield
100
60 × 14.8
=
100
= 8.88 g
actual yield =
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..........................................
1.7.1
Calculations involving percentage yields
This activity contains questions to give practice in calculations involving percentage
yields. The first two questions have worked solutions. In the on-line version, these
are followed by an on-line test which contains questions with random parameters which
can be tried over and over again to build confidence. These questions also contain hints
in the form of 'Steps'.
Q43:
Ethyne can be made in the laboratory by adding water to calcium carbide according to
the above equation. 2.56 g of calcium carbide produced 0.99 g of ethyne.
a) Calculate the theoretical yield of ethyne.
b) Calculate the percentage yield in the reaction.
..........................................
Q44:
1.3 g of ethyne reacts with an excess of chlorine to form 7.14 g of
1,1,2,2-tetrachloroethane.
a) Calculate the theoretical yield of 1,1,2,2-tetrachloroethane.
b) Calculate the percentage yield in the reaction.
..........................................
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TOPIC 1. GETTING THE MOST FROM REACTANTS
Calculations involving percentage yields - Further practice
Benzene is very important as an industrial feedstock.
Go online
Q45:
36 tonnes of benzene was converted into benzoic acid.
Calculate the theoretical yield. Give your answer to one decimal place.
..........................................
Q46:
The actual yield was 51 tonnes.
Calculate the percentage yield. Give your answer to one decimal place.
..........................................
Primary alcohols can be oxidised to carboxylic acids according to the following equation.
where R = the rest of the molecule.
The following table shows four primary alcohols.
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Compound
R
HCH3 -
1
2
3
C2 H5 C6 H5 -
4
Q47: If 11 g of compound 4 were oxidised, calculate the theoretical yield of acid. Give
your answer to 2 decimal places.
..........................................
Q48: If the percentage yield for the oxidation was 65%, calculate the actual yield of
acid. Give your answer to 2 decimal places.
..........................................
1.8
The atom economy of a process
Stoichiometry, equilibrium and percentage yield tell you a lot of vital information about
the chemistry of a process, but they give no help in determining how good the process is
to the environment. In the 1990s, Barry Trost of Stanford University in the USA devised
a simple measure of how 'green' a particular process might be.
The 'atom economy' of a process is defined as:
It is sometimes stated as: how much of what you put into your pot ends up in your
product.
Anything you produce but do not need is waste and will have to be disposed of and this
will add to the cost.
Q49:
Calculate the atom economy of the production of calcium oxide from calcium carbonate.
The equation is:
CaCO3 → CaO + CO2
..........................................
Q50: Calculate the atom economy of the production of sulfur trioxide from sulfur dioxide
and oxygen. The equation is:
SO2 + 1 /2 O2 → SO3
..........................................
Q51: What does an atom economy of 100% mean for the process?
..........................................
Q52: What about rearrangement reactions?
..........................................
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TOPIC 1. GETTING THE MOST FROM REACTANTS
Key point
This exemplifies a general principle that:
Addition reactions are preferable to elimination and substitution reactions, from
an environmental angle.
Ibuprofen
The analgesic (pain killer) ibuprofen has the structure shown in Figure 1.5. It is
illustrated in full structural formula and in skeletal form. In this activity you will consider
the atom economy of the original and modern synthetic routes. In order to simplify the
figures, skeletal structures will be used.
Figure 1.5: Ibuprofen structure
..........................................
Ibuprofen was invented by pharmacists at Boots in the 1960s. In order to allow
companies to benefit financially from their research, at that time a patent was granted
for several years during which time no other companies could manufacture or sell this
product.
BNC synthesis
In the 1980s, the Boots' patent on production and sales expired and ibuprofen became
licensed for over-the-counter sales. There was a rush to produce the pain-killer as
cheaply as possible, using the most environmentally acceptable methods. A new
company, BHC, now manufactures and sells ibuprofen. It uses an improved synthetic
route with only three stages, as shown (Figure 1.6).
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Figure 1.6: Ibuprofen synthesis
..........................................
Q53: The table below shows the reagents used in the new synthesis and how many of
these atoms are incorporated into ibuprofen. Complete the remaining columns showing
how many of these atoms are not used in the ibuprofen molecule and the masses used
and discarded.
Used in
ibuprofen
Reagent
Not used
Formula
Mr
Formula
Mr
Formula
Mr
C10 H14
134
C10 H13
133
?
?
C 4 H 6 O3
102
C2 H3 O
?
?
?
H2
?
H2
?
?
CO
?
CO
?
?
Ibuprofen
Total
?
C13 H18 O2
?
Waste
206
?
?
..........................................
Q54: Using the relevant M r data in the complete table, calculate the percentage atom
economy.
..........................................
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Q55: Look at the structures of the starting materials used to make ibuprofen in this
'green' method. Identify the atoms of these that end up in ibuprofen; circle the discarded
atoms.
..........................................
BOOTS' original process
In the original process the same starting material was employed, but the synthetic
process they devised had six stages.
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Figure 1.7: Boots original synthetic route
..........................................
Q56: One problem with syntheses that involve several steps is that the overall yield can
be quite low. Suppose that each stage of this six-stage synthesis has a 90% yield. What
is the overall yield in this case?
..........................................
Q57: How does this compare with the overall yield in the 3-stage BHC process?
(Assume that each stage has a 90% yield.)
..........................................
More importantly, the process has a low proportion of raw materials and reagents
actually ending up in the final product.
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The table below shows the reagents used in the synthesis, followed by a column showing
how many of these atoms are used in the ibuprofen molecule and how many atoms are
wasted.
Used in
ibuprofen
Reagent
Formula
C10 H14
Mr
Not used
Mr
27
13
134
Formula
C10 H13
C 4 H 6 O3
102
C2 H3
C4 H7 O2 Cl
122.5
CH
C2 H5 ONa
H3 O
68
0
19
0
NH3 O
33
0
H 4 O2
36
Total
C20 H42 NO10 ClNa 514.5
133
Mr
Formula
H
C 2 H 3 O3
1
75
C3 H6 O2 Cl
109.5
C2 H5 ONa
H3 O
68
33
HO2
Ibuprofen
33
NH3 O
H3
C13 H18 O2
206
Waste
C7 H24 NO8 ClNa
19
3
308.5
Q58: Using data in this table, calculate the percentage atom economy.
..........................................
About 3000 tonnes of ibuprofen are produced annually in the UK. (1 tonne = 1000 kg)
Q59:
A tablet of ibuprofen contains 200 mg. How many tablets of ibuprofen are taken each
year in the UK?
(Hint: take great care with the units mentioned. Manufactured amounts are in tonnes;
tablets are in mg. 200 mg is 0.2 g)
..........................................
Q60: The population of the UK is about 60 million. How many tablets is this per person
per year?
..........................................
Look at the two synthetic routes again (Figure 1.6 and Figure 1.7). You will have
noticed that the latter five stages of the original synthesis have been replaced by two
new methods for converting the -CO-CH 3 side chain to -CH(CH3 )-COOH in the BHC
synthesis.
Q61: What type of reactions are these new ones?
..........................................
Q62: Explain how this improved synthetic route has achieved a higher atom economy.
..........................................
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You can read more about ibuprofen synthesis at the http://www.rsc.org/learn-chemistry/
resources/chemistry-in-your-cupboard/nurofen/1.
http://www.rsc.org/learn-chemistry/resources/chemistry-in-your-cupboard/nurofen/1
..........................................
1.9
Excess
In chemical reactions, the reactants are often not present in quantities which react
exactly. Some of one or more of the reactants may remain at the end of the reaction they are said to be in excess. (A common analogy is making sandwiches.)
Q63: A loaf of bread contains 28 slices; a packet of cheese slices contains 12 slices.
How many sandwiches can you make from these and what will remain?
..........................................
1.9.1
Calculating excess
In the experiment on "Measuring rates" the reaction between zinc and hydrochloric acid
eventually stopped giving off hydrogen, and observation showed that some zinc was left
over. Revisit the simulation if you wish. The zinc is said to be "in excess" as all the
hydrochloric acid has been used up.
If the quantities present at the start of the reaction are known, the equation for the
reaction can be used to calculate the excess quantity of zinc.
Suppose the reaction started with 65.4 g of zinc and 500 cm 3 of 2 mol -1 hydrochloric
acid.
From the equation:
1 mole of zinc (65.4 g) reacts completely with 2 moles of hydrochloric acid.
But 500 cm3 of 2 mol -1 hydrochloric acid contains only 1 mole of acid.
Only half the zinc can react and 32.7 g would be left over (excess).
Not all examples are so straightforward, however, and the worked example here shows
a standard method of calculating which reactant is in excess and by how much.
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Problem
Magnesium with a mass of 9.72 g is dropped into 300 cm 3 of 1 mol -1 sulfuric acid.
Which reactant is in excess and by how much?
Solution
•
Write a balanced equation
•
Choose one reactant and calculate the number of moles involved
24.3 g of magnesium is 1 mol
1
mol
1 g of magnesium is
24.3
9.72 × 1
mol
9.72 g of magnesium is
24.3
= 0.40 mol
•
Calculate quantities of the other reactant involved
1000 cm3 1 mol l −1 sulfuric acid contains 1 mol
1
mol
1 cm3 1 mol l −1 sulfuric acid contains
1000
300 × 1
mol
300 cm3 1 mol l −1 sulfuric acid contains
1000
= 0.30 mol
•
Use the balanced equation to work out which reactant is in excess
In this case the magnesium and sulfuric acid react in a 1:1 ratio.
Magnesium (present as 0.40 mol) is in excess by 0.10 mol.
•
Calculate excess present so
1 mol magnesium is 24.3 g
0.10 mol magnesium is 2.43 g
(if required answer is in grams)
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45
Calculating excess - Tutorial examples
The questions in this set have the first example with a worked solution available at the
back of the book, the others have an answer given at the back of the book. Try the
questions on paper. Use your data booklet to find values for relative atomic masses.
Q64: The thermite process produces molten iron due to the heat given out in the
reaction. It can be used to weld railway tracks together.
An aluminium sample (50 g) is mixed with 100 g of iron (III) oxide and ignited. Calculate
which reactant is in excess and by how much.
..........................................
Q65: A 10 g piece of calcium carbonate is reacted with 500 cm 3 of 1 mol -1 hydrochloric
acid. Name the reactant in excess. (Hint: remember to write a balanced equation.)
..........................................
Q66: A 10g piece of calcium carbonate is reacted with 500 cm 3 1 mol -1 hydrochloric
acid. Calculate the number of moles of excess.
..........................................
Q67: The next three questions refer to this reaction.
Zinc reacts with hydrochloric acid, giving hydrogen gas and a solution of zinc (II)
chloride. When 1.962 g of zinc is used with 80 cm 3 1 mol -1 hydrochloric acid:
Name the reactant in excess.
..........................................
Q68: Write the number of moles of excess. (2 decimal places.)
..........................................
Q69: What mass (in grams) of hydrogen gas will have been given off? (2 decimal
places.)
..........................................
Q70: The next four questions refer to this reaction.
Copper (II) oxide powder with a mass of 1.693 grams is added to a beaker containing
50 cm3 of 0.25 mol -1 sulfuric acid.
Name the reactant in excess.
..........................................
Q71: The reaction beaker is now filtered. Which substance will be in the filter paper?
..........................................
Q72: What mass (in grams) of excess reactant will be left? (2 decimal places.)
..........................................
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TOPIC 1. GETTING THE MOST FROM REACTANTS
Q73: The remaining solution is evaporated and dried. What mass of solid would be
obtained? (1 decimal place.)
..........................................
1.10
Summary
Summary
The chemical industry
•
Industrial processes are designed to maximise profit and minimise the
impact on the environment.
•
Factors influencing process design include: availability, sustainability and
cost of feedstock(s); opportunities for recycling; energy requirements;
marketability of by-products; product yield.
•
Environmental considerations include: minimising waste; avoiding the use
or production of toxic substances; designing products which will biodegrade
if appropriate.
Chemical calculations
•
Balanced equations show the mole ratio(s) of reactants and products.
•
Using the balanced equation and the gram formula masses (GFM), mass to
mass calculations can be performed.
•
The quantity of a reactant or product can also be expressed in terms of
moles.
•
The concentration of a solution can be expressed in mol l -1 .
•
Balanced equations can be used in conjunction with concentrations and
volumes of solutions and/or masses of solutes to determine quantities of
reactants and/or products.
Molar volume
•
The molar volume (in units of litres mol-1) is the same for all gases at the
same temperature and pressure.
•
The volume of a gas can be calculated from the number of moles and vice
versa.
•
The molar volume is the same for all gases at the same temperature and
pressure. (approx. 24 l mol-1 ).
•
The volumes of reactant and product gases can be calculated from a
balanced equation using the number of moles of each reactant and product.
Percentage yield
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Summary continued
•
The efficiency with which reactants are converted into the desired product
is measured in terms of the percentage yield and atom economy.
•
Percentage yields can be calculated from mass of reactant(s) and
product(s) using a balanced equation.
•
Given costs for the reactants, a percentage yield can be used to calculate
the feedstock’s cost for producing a given mass of product.
Atom economy
•
The atom economy measures the proportion of the total mass of all starting
materials successfully converted into the desired product.
•
It can be calculated using the formula shown below in which the masses
of products and reactants are those appearing in the balanced equation for
the reaction.
•
Atom Economy = (mass of desired product(s) / total mass of reactants) ×
100%.
•
Reactions which have a high percentage yield may have a low atom
economy value if large quantities of unwanted by-products are formed.
Excess
•
In order to ensure that costly reactant(s) are converted into product, an
excess of less expensive reactant(s) can be used.
•
By considering a balanced equation, the limiting reactant and the
reactant(s) in excess can be identified.
•
Whilst the use of excess reactants may help to increase percentage
yields, this will be at the expense of the atom economy so an economic
/ environmental balance must be struck.
1.11
•
Resources
Higher Chemistry for CfE: J Anderson, E Allan and J Harris, Hodder Gibson,
ISBN 978-1444167528
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1.12
End of topic test
End of topic 1 test
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This end of topic test is available online. If you do not have access to the internet, here
is a paper version.
Q74:
The SQA data booklet contains information on the densities of elements at standard
temperature and pressure (s.t.p.). Using this data, calculate the molar volume of oxygen
in cm3 mol-1, giving your answer to 4 significant figures.
..........................................
Q75:
Identify the two gases which occupy the same volume.
(Assume all measurements are made under the same conditions of temperature and
pressure.)
A) 7 g CO
B) 32 g CH4
C) 4 g H2
D) 32 g SO2
E) 17 g NH3
..........................................
Q76:
In which of the following reactions do the products have a lower volume than the
reactants?
A) CaCO3 (s) + 2HCl(aq) → CaCl2 (aq) + CO2 (g) + H2 O(l)
B) C(s) + O2(g) → CO2 (g)
C) CH4(g) + 2O2 (g) → CO2 (g) + 2H2 O(l)
D) 2C(s) + O2 (g) → 2CO(g)
..........................................
Hydrazine, N2 H4 , is a hydride of nitrogen used in rocket fuel.
The balanced equation for the complete combustion of hydrazine is N 2 H4 (g) + 2O2 (g)
→ N2 (g) + 2H2 O(l)
Q77: What volume of oxygen would be required for the complete combustion of 50 cm 3
hydrazine?
..........................................
Q78:
What would the final volume if a mixture of 50 cm 3 of hydrazine and 150 cm 3 of oxygen
was ignited?
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Assume that all measurements were made at 25 ◦ C and 1 atmosphere pressure.
..........................................
Q79: Calculate the volume occupied by 0.80 g of hydrazine. Take the molar volume to
be 24.0 l mol-1 .
..........................................
Q80:
Nitrogen monoxide reacts spontaneously when mixed with oxygen.
2NO(g) + O2 (g) → 2NO2 (g)
How many litres of nitrogen dioxide could theoretically be obtained by mixing 5 litres of
nitrogen monoxide and 2 litres of oxygen gas?
(All volumes are measured under the same conditions of temperature and pressure.)
A)
B)
C)
D)
2
3
4
5
..........................................
Q81:
The equation shows the reaction between magnesium ribbon and dilute hydrochloric
acid.
Mg(s) + 2HCl(aq) → MgCl2 (aq) + H2 (g)
What volume of hydrogen will be produced when 1 g of magnesium is added to excess
acid?
(Take the molar volume of hydrogen to be 24.3 litres mol -1 .)
A)
B)
C)
D)
1.0 litre
2.0 litres
2.43 litres
24.3 litres
..........................................
Trichorophenol (TCP) is an antiseptic which is used to relieve throat infections. It is
prepared from phenol as follows:
C6 H5 OH + 3Cl2 → C6 H2 Cl3 OH + 3HCl
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Q82: Calculate the percentage yield if 537 g of TCP is obtained from 300 g of phenol.
Give your answer to the nearest unit.
..........................................
Q83: Calculate the atom economy of this process.
..........................................
Q84:
Benzene, phenol and related compounds are most important as:
A) face cream.
B) feedstocks.
C) fuels.
D) fish food.
E)
..........................................
Q85:
Excess marble chips (calcium carbonate) were added to 100 cm 3 of 1 mol l-1
hydrochloric acid. The experiment was repeated using the same mass of marble chips
and 100 cm3 of 1 mol l-1 ethanoic acid.
Which would have been the same for both experiments?
A) The mass of marble chips left over after the reaction had finished.
B) The time taken for the reaction to be completed.
C) The rate at which the first 10 cm3 of gas was produced.
D) The average rate of reaction.
..........................................
Q86:
Water from the Dead Sea contains 5 g l -1 of Br- ion. A bromine plant processes 1 million
litres of Dead Sea water each hour, producing 4 tonnes of bromine.
The percentage yield of the process is:
A) 100%
B) 80%
C) 40%
D) 8%
..........................................
The feedstock for making ammonia by the Haber process must be purified. For example,
carbon monoxide (CO) should not be present.
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Q87:
What are the features which can lead to improvements in the efficiency of a chemical
process?
A) Scrubbers
B) Heat exchangers
C) Improved catalysts
D) Continuous
E) Haber
F) batch
..........................................
Propanone, widely used as a solvent, is manufactured from cumene.
Cumene is oxidised by air and the cumene hydroperoxide product is then cleaved.
The mixture of propanone and phenol is separated by distillation.
Q88:
Complete the following flow chart to summarise the manufacture of propanone from
cumene.
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..........................................
Q89:
For every 10 tonnes of propanone produced in this industrial process, calculate the mass
of phenol (C6 H5 OH) produced.
Give your answer to 1 decimal place in tonnes.
..........................................
Q90: Calculate the atom economy of this process.
..........................................
Q91:
Propanone can also be manufactured by the catalytic oxidation of propan-2-ol with
oxygen.
Write a balanced equation for this reaction.
..........................................
Q92: Calculate the atom economy of this new process.
..........................................
Q93:
In industry, apart from atom economy, several factors influence the decision as to which
route might be employed.
Suggest three of these factors.
..........................................
..........................................
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Topic 2
Chemical equilibria
Contents
2.1 Prior knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
2.2 Reversible reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Dynamic equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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57
2.4 Altering the equilibrium position . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Changing temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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62
2.4.2 Changing pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.3 The influence of concentration . . . . . . . . . . . . . . . . . . . . . . .
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65
2.4.4 The influence of catalysts . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 The Haber process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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67
2.5.1 Adjusting the concentration . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.2 Temperature in the Haber process . . . . . . . . . . . . . . . . . . . . .
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69
2.5.3 Pressure in the Haber process . . . . . . . . . . . . . . . . . . . . . . .
2.5.4 Catalysts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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71
2.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74
Prerequisite knowledge
Before you begin this topic, you should know or be able to:
•
understand and explain the factors affecting rates of reaction (Higher, Unit 1);
•
compare rates of chemical reactions, changes in mass, volume and other
quantities can be measured. Graphs can then be drawn to help this comparison
(National 4, Unit 1);
•
the gram formula mass is defined as the mass of one mole of a substance
(National 5, Unit 1);
•
use the chemical formula of any substance the gram formula mass can be
calculated using relative formula masses of its constituent elements (National 5,
Unit 1);
•
the concentration of solutions in moles per litre (National 5, Unit 1);
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TOPIC 2. CHEMICAL EQUILIBRIA
•
calculations to determine the concentration and volume and the mass of a
substance through the number of moles present (National 5, Unit 1);
•
a very small proportion of water molecules will dissociate into an equal number of
hydrogen and hydroxide ions (National 5, Unit 1).
Learning objectives
At the end of this topic, you should know that:
Reversible Reactions
•
Many reactions are reversible, so products may be in equilibrium with reactants.
Equilibrium
•
At equilibrium, the concentrations of reactants and products remain constant, but
are rarely equal.
•
This may result in costly reactants failing to be completely converted into products.
•
In a closed system, reversible reactions attain a state of dynamic equilibrium when
the rates of forward and reverse reactions are equal.
Altering Position of Equilibrium
•
Le Chatelier's principle states that if the conditions of a chemical system at
equilibrium are changed, the system responds by minimising the effect of the
changes.
•
Changes in concentration, pressure and temperature can alter the position of
equilibrium.
The Haber Process
•
To maximise profits, chemists employ strategies to move the position of equilibrium
in favour of products.
•
A catalyst increases the rate of attainment of equilibrium but does not affect the
position of equilibrium. The effects of altering pressure, altering temperature, the
addition or removal of reactants or products can be predicted for a given reaction.
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2.1
55
Prior knowledge
Test your prior knowledge
Q1: What is equilibrium?
a) Equilibrium is a situation where forward and reverse reactions take place at the same
rate.
b) Equilibrium is a situation where forward and reverse reactions are at equal
proportions.
c) Equilibrium is a situation where forward reaction proceeds at a greater rate than the
reverse reaction.
d) Equilibrium is a situation where reverse reaction proceeds at a greater rate than the
forward reaction.
..........................................
Q2: What is a reversible reaction?
a) A situation where forward and reverse reactions take place at the same rate.
b) A reaction which proceeds in both directions at the same time.
c) A reaction where the reverse reaction always proceeds at a greater rate than the
forward reaction.
d) A reaction where the forward reaction always proceeds at a greater rate than the
reverse reaction.
..........................................
Q3: Which of the following is a correct representation of the water equilibrium?
a)
b)
c)
d)
H2 0(l) → H+ (aq) + OH- (ag)
H2 0(l) ← H+ (aq) + OH- (ag)
H2 0(l) = H+ (aq) + OH- (aq)
H2 0(l) H+ (aq) + OH- (aq)
..........................................
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2.2
Reversible reactions
Chemical reactions involve reactants changing into products and many reactions are
not reversible. Frying an egg cooks the white and it is impossible to change it back into
clear jelly (Figure 2.2). However some processes, including many chemical reactions
are reversible. The products can be turned back into reactants. Consider a mixture of
ice and water at 0◦ C (Figure 2.2). Some of the ice melts into water, while some of the
water freezes into ice. These two processes can be represented by equations (Figure
2.1).
Figure 2.1: Water freezing and melting
..........................................
Figure 2.2: Irreversible and reversible processes
Frying an egg is not a reversible process.
The reaction progresses in a forward
direction only.
The melting of ice and freezing of water
are reversible, and at 0◦ C are said to be
in an equilibrium situation.
..........................................
At 0◦ C the rates of melting and freezing become equal, and the ice and water are said
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57
to be in a state of equilibrium. The two equations in Figure 2.1 can be combined to
represent the equilibrium situation (Figure 2.3).
Figure 2.3: Water equilibrium
..........................................
The double arrow represents the equilibrium. The two-way sign shows that the process
can occur in either direction.
This equation shows the equilibrium in a chemical system. Hydrogen and iodine can
combine in the forward reaction to produce hydrogen iodide. At equilibrium, hydrogen
iodide breaks up at the same rate to form hydrogen and iodine.
Equilibrium of hydrogen iodide
The on-line version of this reaction is animated.
Go online
..........................................
2.3
Dynamic equilibrium
System in equilibrium
A system in equilibrium appears to be unchanging as far as an outside observer is
concerned. The bottle of soda water or lemonade shown below has carbon dioxide
dissolved in the water and also free carbon dioxide above the liquid. The system is
closed so that nothing can enter or leave the container.
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TOPIC 2. CHEMICAL EQUILIBRIA
As some carbon dioxide in the gas
dissolves, some carbon dioxide in the
solution leaves to become gas. So long as
the system remains closed there is a
balance between the rates of the
exchange. Notice that the concentrations
at equilibrium are not necessarily equal.
..........................................
The rates of forward and reverse reaction in this closed system are equal and the
reversible reaction (Figure 2.4) has attained a dynamic equilibrium.
This is the situation which applies in the closed container in the picture (left) but the
other container is an open system (right), which cannot reach equilibrium because the
CO2 is constantly escaping.
Figure 2.4: carbon dioxide equilibrium
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..........................................
The forward reaction is described as being from left to right and the reverse reaction
from right to left. Reversible reactions reach a point where they appear to stop and a
balance between reactant and product concentrations (not usually equal) is formed.
An example is the thermal (involves heat) breakdown of hydrated copper sulfate.
Figure 2.5
..........................................
The dynamic nature of equilibrium
..........................................
Go online
A balance can be achieved in this open system by careful control over the rate of the
forward and reverse reactions by control of the heating and the addition of water. An
open system allows material to be lost to or gained by the surroundings.
In a closed system reversible reactions reach a state of equilibrium when the rate of
the forward reaction is equal to the rate of the reverse reaction. In a closed system no
material is exchanged with the surroundings.
The on-line version of this topic contains a simulation experiment establishing the
reversible nature of the reaction. If you do not have access to the on-line material, this
paragraph will explain what the simulation illustrates.
Heating turns the blue copper sulfate crystals into white powder. The water is driven
away leaving anhydrous copper sulfate. Anhydrous means "without water".
The white anhydrous copper sulfate can be hydrated back in the reverse reaction by
adding water. Hydrated means "with water".
A balance can be achieved in this open system by careful control over the rate of the
forward and reverse reactions by control of the heating and the addition of water. An
open system allows material to be lost to or gained by the surroundings.
In a closed system reversible reactions reach a state of equilibrium when the rate of
the forward reaction is equal to the rate of the reverse reaction. In a closed system no
material is exchanged with the surroundings.
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TOPIC 2. CHEMICAL EQUILIBRIA
Establishing dynamic equilibrium
Go online
The on-line version of this topic contains an animated experiment which shows the
molecular collisions occurring as hydrogen and iodine react and form a chemical
equilibrium with hydrogen iodide. If you do not have access to the on-line material use
this diagram of the situation at equilibrium to answer the questions.
Q4: In the forward reaction, how many molecules of hydrogen and iodine (in total) are
present before reaction starts?
..........................................
Q5: In the forward reaction, how many reactant molecules of hydrogen and iodine are
present when equilibrium has been established?
..........................................
Q6: How many product molecules of hydrogen iodide are present when equilibrium
has been established?
..........................................
Q7: Which of these statements is correct?
products at equilibrium:
a)
b)
c)
d)
The concentrations of reactants and
change rapidly to different concentrations.
remain constant, but not necessarily equal.
change slowly to equal concentrations.
remain constant and equal.
..........................................
Q8: Now look closely at the reverse reaction. How many product molecules of
hydrogen iodide are present when equilibrium has been established?
..........................................
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Q9: Which of these statements is correct about the equilibrium mixture obtained when
starting with hydrogen iodide rather than hydrogen and iodine?
a)
b)
c)
d)
There is more HI than before.
There is less HI than before.
There is the same equilibrium mixture as before.
There is less H2 and I2 than before.
..........................................
Q10: Explain why this situation can be described as a dynamic equilibrium.
..........................................
Key point
A reversible reaction attains a state of dynamic equilibrium when the rates of
forward and reverse reactions are equal. At equilibrium, the concentrations
of reactants and products remain constant although not necessarily equal.
The same equilibrium position is reached irrespective of whether starting from
reactants or products.
2.4
Altering the equilibrium position
It is particularly important that in any industrial process as much product as possible is
produced. A process producing only small amounts of a required product in each cycle
would be regarded as very inefficient, and would need modification to product more.
This section looks at methods for influencing the equilibrium position.
In a reaction system which is in chemical equilibrium the rate of the forward reaction is
equal to the rate of the reverse reaction. At equilibrium, if there is a high proportion of
products to reactants, the position of equilibrium is said to lie well to the right. Chemists
often wish to influence the position of equilibrium to give a maximum yield of product.
How changes in conditions can influence equilibria can be summarised in Le Chatelier's
principle.
Key point
Le Chatelier's principle
If the conditions of a chemical system at equilibrium are changed, the system
responds by minimising the effect of the changes.
Factors which influence the rate of a reaction, such as temperature, pressure,
concentration and the addition of a catalyst might be expected to influence the position
of equilibrium.
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2.4.1
Changing temperature
The influence of temperature
The equilibrium between dinitrogen tetroxide and nitrogen dioxide is endothermic in
the forward direction (see below). The reaction is therefore exothermic in the reverse
direction.
Because N2 O4 is colourless and NO2 is brown, the colour of the equilibrium mixture
can give an indication of the equilibrium position.
Equilibrium and temperature
Go online
The on-line version of this activity contains a simulation experiment in which the effect
of temperature changes on the position of equilibrium can be investigated. If you do not
have access to the on-line version, look at this diagram before answering the questions.
..........................................
Q11: How many nitrogen dioxide molecules are present in the light brown equilibrium
mixture at room temperature?
..........................................
Q12: How many nitrogen dioxide molecules are present in the dark brown equilibrium
mixture at the increased temperature?
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..........................................
Q13: Has the equilibrium position shifted in the exothermic or endothermic direction?
..........................................
Q14: Use Le Chatelier's principle to explain why this occurs. Write an answer before
consulting the back of the text.
..........................................
Q15: How many dark brown nitrogen dioxide molecules are present in the equilibrium
mixture at the decreased temperature?
..........................................
Q16: Has the equilibrium position shifted in the exothermic or endothermic direction?
..........................................
Q17: Use Le Chatelier's principle to explain why this occurs. Write an answer before
consulting the back of the text.
..........................................
Key point
Le Chatelier's principle can be used to predict changes in the position of
equilibrium caused by a change in temperature in a reaction.
2.4.2
Changing pressure
The influence of pressure
The effect of changes in pressure on a system at equilibrium is only important for
reactions which involve gas molecules and have a volume change involved in the
chemical reaction.
The equilibrium between dinitrogen tetroxide and nitrogen dioxide involves a volume
change and is a suitable reaction to study.
Because N2 O4 is colourless and NO2 is brown, the colour of the equilibrium mixture
can give an indication of the equilibrium position.
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Equilibrium and pressure
Go online
The on-line version of this activity contains a simulation experiment in which the effect
of pressure changes on the position of equilibrium can be investigated. If you do not
have access to the on-line version, look at this diagram before answering the questions.
Consider the effect of increased pressure.
Q18: Has the total number of molecules present increased or decreased when the
pressure increased?
..........................................
Q19: Has the volume of gas present at the higher pressure increased or decreased?
..........................................
Q20: Has the equilibrium position shifted to the right or left?
..........................................
Q21: Explain how this could be predicted by Le Chatelier's principle.
..........................................
Now consider the effect of decreased pressure.
Q22: Use Le Chatelier's principle to explain fully the observations made. Write an
answer before consulting the back of the text.
..........................................
..........................................
Key point
Le Chatelier's principle can be used to predict changes in the position of
equilibrium in a reaction involving gas molecules due to pressure changes.
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2.4.3
65
The influence of concentration
The equilibrium reaction between nitrogen and hydrogen as reactants, and ammonia
as the product, is an extremely important industrial process. The Haber process (see
below) is aimed at manufacturing ammonia with the highest possible yield.
Before starting the next activity, consider Le Chatelier's principle in relation to the
concentrations of reactants or products in this system.
Key point
Le Chatelier's principle
If the conditions of a chemical system at equilibrium are changed, the system
responds by minimising the effect of the changes.
•
Try to predict what will happen to the position of equilibrium if the concentration of
a reactant is increased.
•
Try to predict what will happen to the position of equilibrium if the concentration of
a product is increased.
•
What would be the best strategy to yield the highest percentage of ammonia?
Changing concentration
The on-line version of this activity contains graphical simulations of changes in the
position of equilibrium which occur when the concentration of reactant or product are
altered in the Haber process. Look at these graphs before answering the questions.
Note: The concentration of substances, in mol -1 , can be represented by square
brackets around the formula. Thus [H 2 ] represents "the concentration of hydrogen".
The graphs show the basic equilibrium position (left) and the situation as the
concentration of one reactant is increased (right).
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Haber process graphs
Q23: Which reactant has been added to affect the equilibrium?
..........................................
Q24: In response to this, has the ammonia concentration increased or decreased?
..........................................
Q25: Has the equilibrium position shifted left or right?
..........................................
Q26: Explain how this could be predicted by Le Chatelier's principle.
..........................................
Q27: Use Le Chatelier's principle to explain the shape of the line tracing the hydrogen
concentration.
..........................................
In summary, Le Chatelier's principle predicts that high concentrations of both nitrogen
and hydrogen assist in shifting the position of equilibrium to the side which increases
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ammonia yield. Constant removal of the ammonia produced also encourages the
further production of the product.
..........................................
2.4.4
The influence of catalysts
Catalysts are very widely used in industry to speed up chemical reactions. Without
them modern chemical plants would not be able exist. Although they affect the rate of
reaction, they have no effect on the equilibrium position, since they affect the rates
of both the forward and reverse reactions similarly.
Key point
Changes in concentration, pressure and temperature can alter the position of
equilibrium. Le Chatelier's principle can be used to predict changes in the position
of equilibrium in a reaction.
A catalyst does not alter this position, merely changes the rate at which it is
attained.
2.5
The Haber process
Modern agriculture relies heavily on industrially manufactured fertilisers, particularly
nitrates. The Haber process produces ammonia (see below) which is a key chemical
for nitrate production and is a good example of how chemists can adjust the position of
equilibrium in an industrial process to provide a cost-effective yield.
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2.5.1
Adjusting the concentration
Changing concentration of ammonia
Go online
Investigate the effect that adjusting the concentration of ammonia has on the position
of equilibrium by examining the graphs obtained when you "increase ammonia" or
"decrease ammonia" as shown below.
..........................................
Q28: Changing the concentration of ammonia will affect the position of equilibrium.
Does shifting the position of equilibrium towards the product side occur if the
concentration of ammonia is increased or decreased?
..........................................
Q29: Explain how this could be predicted by Le Chatelier's principle.
..........................................
In the Haber process, industrial chemists cool the reaction mixture and the ammonia
produced condenses to a liquid form. The equilibrium responds by shifting to the right
and producing more ammonia. The liquid ammonia formed is easily run off into a
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69
storage tank and the nitrogen and hydrogen gases are recycled through the process
so there is almost no wastage.
2.5.2
Temperature in the Haber process
Temperature and pressure in the Haber process
The on-line version of this activity contains graphs which show how pressure and
temperature changes affect the yield of ammonia in the Haber process. If you do not
have access to the on-line version, look at these graphs before answering the
questions.
Look at the graph showing 50 atmospheres pressure.
Pressure and Temperature graphs
..........................................
Q30: At 50 atmospheres pressure, is the yield of ammonia highest at high or low
temperature?
..........................................
Q31: Explain how this could be predicted by application of Le Chatelier's principle.
..........................................
Q32: In the industrial process the temperature used is actually about 400 ◦ C. Which of
these could be the yield at 400 ◦ C and 50 atmospheres?
a) Less than 10% ammonia
b) Between 10% and 50%
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c) Between 50% and 100%
d) More than 100%
..........................................
Q33: If the Haber process was carried out at 50 atmospheres and room temperature
(25◦ C) the yield would be about 90% ammonia. In the industrial process the temperature
used is actually about 400 ◦ C. Explain why the higher temperature is used even though
it gives a lesser yield. (Hint: think about reaction rates).
..........................................
2.5.3
Pressure in the Haber process
Temperature and pressure in the Haber process
Go online
The on-line version of this activity contains graphs which show how pressure and
temperature changes affect the yield of ammonia in the Haber process. If you do not
have access to the on-line version, look at these graphs before answering the
questions.
Look at the graphs showing the yield of ammonia at 100, 200 and 400 atmospheres.
Pressure and Temperature graphs
..........................................
Q34: Is the yield of ammonia highest at high or low pressure?
..........................................
Q35: Explain how this could be predicted by application of Le Chatelier's principle.
..........................................
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Q36: In the industrial process the pressure used is actually about 200 atmospheres.
Explain why the lower pressure is used even though it gives a lesser yield. (Hint: think
about cost-effectiveness.)
..........................................
Key point
Pressure, temperature and catalyst can be controlled in the Haber process in
order to maximise a safe, cost-effective yield of ammonia.
The operating conditions in the Haber process, although easily predicted using Le
Chatelier's principle, are a series of compromises which have to take into account
building, maintenance, energy and labour costs, safety, catalyst activity, reaction rate
and yield. The industrial process has been developed to allow the chemical industry to
function in the most cost-effective way.
2.5.4
Catalysts
Key point
The effects of pressure and temperature, use of catalyst, removal of product
and recycling of unreacted gases can be considered in relation to the conditions
actually applied in the Haber process.
The catalyst used in the Haber process speeds up the rate of both forward and reverse
reaction and reduces the time required for equilibrium to be established. The finely
divided iron used as a catalyst in the reaction only operates between certain
temperatures, as illustrated in the graphs. The efficiency of the catalyst is increased by
the addition of small amounts of potassium oxide or aluminium oxide which act as
promoters.
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Summary of the Haber process
Go online
Q37: Copy this paragraph or photocopy the page and fill in the appropriate word from
the word bank.
..........................................
2.6
Summary
Summary
Reversible reactions
•
Many reactions are reversible, so products may be in equilibrium with
reactants.
Equilibrium
•
At equilibrium, the concentrations of reactants and products remain
constant, but are rarely equal.
•
This may result in costly reactants failing to be completely converted into
products.
•
The same equilibrium position is reached irrespective of whether starting
from reactants or products.
•
In a closed system, reversible reactions attain a state of dynamic equilibrium
when the rates of forward and reverse reactions are equal.
Altering position of equilibrium
•
Le Chatelier's principle states that “If the conditions of a chemical system at
equilibrium are changed, the system responds by minimising the effect of
the changes.”
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TOPIC 2. CHEMICAL EQUILIBRIA
Summary continued
•
Le Chatelier's principle can be used to predict changes in the position of
equilibrium caused by a change in temperature in a reaction.
•
Le Chatelier's principle can be used to predict changes in the position of
equilibrium in a reaction involving gas molecules due to pressure changes.
•
Le Chatelier's principle can be used to predict changes in the position of
equilibrium in a reaction when concentration of reactants or products are
altered.
•
Changes in concentration, pressure and temperature can alter the position
of equilibrium. Le Chatelier's principle can be used to predict changes in
the position of equilibrium in a reaction.
•
A catalyst does not alter this position, merely changes the rate at which it is
attained.
The Haber process
2.7
•
•
To maximise profits, chemists employ strategies to move the position of
equilibrium in favour of products.
•
Pressure, temperature and catalyst can be controlled in the Haber process
in order to maximise a safe, cost-effective yield of ammonia.
•
The effects of pressure and temperature, use of catalyst, removal of product
and recycling of unreacted gases can be considered in relation to the
conditions actually applied in the Haber process.
Resources
Higher Chemistry for CfE: J Anderson, E Allan and J Harris, Hodder Gibson,
ISBN 978-1444167528
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2.8
End of topic test
End of topic 2 test
Go online
This end of topic test is available online. If you do not have access to the internet, here
is a paper version.
Q38:
Chemical reactions are in a state of dynamic equilibrium only when the:
A) concentrations of reactants and products are equal.
B) rate of forward reaction equals the rate of reverse reaction.
C) reaction involves no enthalpy change.
D) activation energy of the forward and reverse reactions are equal.
..........................................
Q39:
At equilibrium, the concentrations of reactants and products always:
A) become zero.
B) reach their highest.
C) remain constant.
D) become equal.
..........................................
Q40:
A two way arrow is used in an equation to show that the reaction:
A) is completely finished.
B) is a reversible reaction.
C) has equal concentrations of reactants and products.
D) is changing direction.
..........................................
Q41:
In an industrial process, ethene and steam react together as follows.
C2 H4 (g) + H2 O(g) C2 H5 OH(g)
The reaction is exothermic.
Which set of conditions would give the best yield of ethanol at equilibrium?
A) Low temperature, high pressure
B) High temperature, high pressure
C) High temperature, low pressure
D) Low temperature, low pressure
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TOPIC 2. CHEMICAL EQUILIBRIA
..........................................
The questions below refer to this equilibrium reaction.
2SO2 (g) + O2 (g) 2SO3 (g)
The reaction is exothermic.
Q42:
Increasing the temperature would cause:
A) no change in the position of equilibrium.
B) both reactant gases to decrease concentration.
C) the proportion of sulfur trioxide to decrease.
D) the position of equilibrium to shift to the right.
..........................................
Q43:
Decreasing the pressure would cause:
A) the position of equilibrium to shift to the right.
B) no change in the position of equilibrium.
C) the proportion of sulfur trioxide to increase.
D) both reactant gases to increase concentration.
..........................................
Q44:
Using a catalyst would cause:
A) no change in the position of equilibrium.
B) the proportion of sulfur trioxide to decrease.
C) the position of equilibrium to shift to the right.
D) both reactant gases to increase concentration.
..........................................
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Chlorine gas dissolves in water, giving the following equilibrium:
Cl2 (aq) + H2 O(l) 2H+ (aq) + ClO- (aq) + Cl- (aq)
Q45:
Identify the compound which, if added to the equilibrium mixture, would shift the position
of equilibrium to the left.
A) KCl(s)
B) KOH(s)
C) Na2 SO4 (s)
D) AgNO3 (s)
E) KF(s)
F) NaNO3 (s)
..........................................
Q46:
Identify the compounds which, if added to the equilibrium mixture, would shift the
position of equilibrium to the right.
A) KCl(s)
B) KOH(s)
C) Na2 SO4 (s)
D) AgNO3 (s)
E) KF(s)
F) NaNO3 (s)
..........................................
Q47:
Identify the statements which can be applied to the role of a catalyst in a reversible
reaction.
A) It decreases the time required for equilibrium to be established.
B) It alters the equilibrium position.
C) It lowers the activation energy of the reverse reaction.
D) It decreases the enthalpy change for the reaction.
E) It increases the rate of the forward reaction more than the reverse reaction.
..........................................
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This equilibrium shows the relationship between dinitrogen tetroxide and nitrogen
dioxide.
N2 O4 (g) 2NO2 (g)
Q48: The amount of nitrogen dioxide formed by dissociation of dinitrogen tetroxide
increases as temperature rises.
The dissociation of dinitrogen tetroxide is:
..........................................
Q49: How would the equilibrium concentration of nitrogen dioxide be affected by a
decrease in pressure?
..........................................
The graph shows how the percentage of ammonia in the gas mixture at equilibrium
varies with pressure at different temperatures.
Q50: At what temperature would a pressure of 100 atmospheres give a yield of 20%?
..........................................
Q51:
In the industrial process the percentage yield of ammonia never rises above 15%.
Although the yield is low, the process is still profitable.
What is done to the excess nitrogen and hydrogen which allows the process to be
profitable?
..........................................
..........................................
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Topic 3
Chemical energy
Contents
3.1 Prior knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
3.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Potential energy diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
86
3.3.1 Enthalpy changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2 What produces this energy change? . . . . . . . . . . . . . . . . . . . .
87
90
3.3.3 The thermochemical equation . . . . . . . . . . . . . . . . . . . . . . .
3.4 Enthalpy changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
95
3.4.1 Enthalpy of combustion . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.2 Enthalpy of solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
99
3.4.3 Enthalpy of neutralisation . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Introduction to Hess's law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
102
106
3.5.1 Hess's law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.2 Verification of Hess's law . . . . . . . . . . . . . . . . . . . . . . . . . .
108
108
3.5.3 Applications of Hess's law . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Bond enthalpies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
112
116
3.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
118
3.8 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.9 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
120
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TOPIC 3. CHEMICAL ENERGY
Prerequisite knowledge
Before you begin this topic, you should know:
•
the gram formula mass is defined as the mass of one mole of a substance
(National 5, Unit 1);
•
using the chemical formula of any substance the gram formula mass can be
calculated using relative formula masses of its constituent elements (National 5,
Unit 1);
•
the concentration of solutions in moles per litre (National 5, Unit 1);
•
calculations to determine the concentration and volume and the mass of a
substance through the number of moles present (National 5, Unit 1);
•
define types of reaction (eg. neutralisation, combustion) (National 5);
•
alkanes and alcohols can be used as fuels (National 5, Unit 2);
•
combustion reactions are exothermic reactions (National 5, Unit 2);
•
the opposite of this is an endothermic reaction (National 5, Unit 2);
•
when a substance is combusted the reaction can be represented using a balanced
formulae equation (National 5, Unit 2);
•
the quantities of reactants and products in these reactions can be calculated
(National 5, Unit 2);
•
different fuels provide different quantities of energy and this can be measured
experimentally and calculated using Eh = cmΔT (National 5, Unit 2).
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TOPIC 3. CHEMICAL ENERGY
Learning objectives
At the end of this topic, you should be able to state that:
Chemical energy
•
For industrial processes, it is essential that chemists can predict the quantity of
heat energy taken in or given out.
•
If reactions are endothermic, costs will be incurred in supplying heat energy in
order to maintain the reaction rate.
•
If reactions are exothermic, the heat produced may need to be removed to prevent
the temperature rising.
Enthalpy
•
Chemical energy is also known as enthalpy.
•
The change in chemical energy associated with chemical reactions can be
measured. The specific heat capacity, mass and temperature can be used to
calculate the enthalpy change for a reaction.
•
The enthalpy of combustion of a substance is the enthalpy change when one mole
of the substance burns completely in oxygen.
•
These values can often be directly measured using a calorimeter and values for
common compounds are available from data books and online databases for use
in Hess’s law calculations.
Hess’s law
•
Hess’s law states that the enthalpy change for a chemical reaction is independent
of the route taken.
•
Enthalpy changes can be calculated by applying Hess’s law.
Bond enthalpies
•
For a diatomic molecule, XY, the molar bond enthalpy is the energy required to
break one mole of XY bonds.
•
Mean molar bond enthalpies are average values which are quoted for bonds which
occur in different molecular environments.
•
Bond enthalpies can be used to estimate the enthalpy change occurring for a gas
phase reaction by calculating the energy required to break bonds in the reactants
and the energy released when new bonds are formed in the products.
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Learning objective
At the end of this topic, you should be able to state that:
Chemical energy
•
For industrial processes, it is essential that chemists can predict the quantity
of heat energy taken in or given out.
•
If reactions are endothermic, costs will be incurred in supplying heat energy
in order to maintain the reaction rate.
•
If reactions are exothermic, the heat produced may need to be removed to
prevent the temperature rising.
Enthalpy
•
Chemical energy is also known as enthalpy.
•
The change in chemical energy associated with chemical reactions can be
measured. The specific heat capacity, mass and temperature can be used
to calculate the enthalpy change for a reaction.
•
The enthalpy of combustion of a substance is the enthalpy change when
one mole of the substance burns completely in oxygen.
•
These values can often be directly measured using a calorimeter and
values for common compounds are available from data books and online
databases for use in Hess’s law calculations.
Hess’s law
•
Hess’s law states that the enthalpy change for a chemical reaction is
independent of the route taken.
•
Enthalpy changes can be calculated by applying Hess’s law.
Bond enthalpies
•
For a diatomic molecule, XY, the molar bond enthalpy is the energy required
to break one mole of XY bonds.
•
Mean molar bond enthalpies are average values which are quoted for bonds
which occur in different molecular environments.
•
Bond enthalpies can be used to estimate the enthalpy change occurring for
a gas phase reaction by calculating the energy required to break bonds in
the reactants and the energy released when new bonds are formed in the
products.
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3.1
83
Prior knowledge
Test your prior knowledge
Q1: In which of the following reactions is oxygen used up?
a)
b)
c)
d)
Go online
Combustion
Neutralisation
Addition
Polymerisation
..........................................
Q2: Which of the following is a correctly balanced equation?
a)
b)
c)
d)
Li2 CO3 (s) + HCl(aq) → LiCl(aq) + CO2 (g) + H2 O(l)
Li2 CO3 (s) + HCl(aq) → 2LiCl(aq) + CO2 (g) + H2 O(l)
Li2 CO3 (s) + 2HCl(aq) → 2LiCl(aq) + CO2 (g) + H2 O(l)
Li2 CO3 (s) + HCl(aq) → 2LiCl(aq) + 2CO2 (g) + 2H2 O(l)
..........................................
Q3: What name is given to any chemical reaction which releases energy?
a)
b)
c)
d)
Exothermic
Endothermic
Combustion
Neutralisation
..........................................
Q4:
The student recorded the following data.
Mass of alkane burned
1g
Volume of water
Initial temperature of water
200 cm3
15◦ C
Final temperature of water
55◦ C
Specific heat capacity of water
4.18 kJ kg-1 ◦ C-1
How much energy was released, in kJ?
a)
b)
c)
d)
33440
33.44
167.2
45.98
..........................................
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Q5:
The reaction above is an example of:
a)
b)
c)
d)
hydration.
addition.
condensation.
saturation.
..........................................
3.2
Introduction
Much of the energy that we use in everyday activities is derived from the chemical
potential energy available from substances we call "fuels". Our cars are powered by
the energy released when petrol burns, many houses are heated by burning natural gas
(methane), electricity is mainly produced by burning coal or natural gas, and our bodies
are powered by using energy-rich foods like sugars and fats. The energy stored in these
sources (chemical potential energy) can be released when these substances undergo
chemical reactions, often involving oxidation. The energy can be released in various
forms such as heat, light, sound, electricity.
A chemical reaction producing heat, light and sound energy.
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TOPIC 3. CHEMICAL ENERGY
In fact many chemical reactions release heat to the surroundings as they proceed.
Examples: Metals reacting with acids; acids neutralising alkalis.
The "Thermite" reaction generates enough heat to melt iron, and is used to weld rails
together.
Reactions which release heat to the surroundings are called exothermic reactions.
Some reactions, however, absorb heat from the surroundings as they proceed.
Examples: Dissolving ammonium nitrate in water; solid ammonium thiocyanate and
barium hydroxide reacting.
The "cool packs" used for sports injuries contain chemicals which cool down the pack
when they react.
These types of reaction are called endothermic reactions.
Key point
Exothermic changes cause heat to be released to the surroundings; endothermic
changes cause absorption of heat from the surroundings.
The importance to industry
Any changes in the heat produced or absorbed by the reactions a in a chemical
production process have to be accounted for. Exothermic reactions may require cooling;
endothermic ones might need a supply of heat. Part of the energy economics of a
process plant is to try to use heat removed from one reaction to heat another. It will cost
if heat is wasted directly to the atmosphere, or if fuels have to be purchased specially.
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3.3
Potential energy diagrams
Consider the exothermic reaction between zinc and sulfuric acid.
After the reaction, the energy stored in the products is less than the energy originally
stored in the reactants - that is, some of the chemical potential energy stored in
reactants has been transferred to the surroundings. This transferred heat energy is
defined as the enthalpy change for this reaction.
Strictly, the change in energy which occurs during a chemical reaction consists of two
components: heat and work. We live on the surface of the Earth, at the bottom of a 'sea'
of atmosphere which exerts pressure. In the reaction above, the gaseous hydrogen
produced will have to 'work' to displace the atmosphere. This will require use of some
energy, so the heat energy released to the surroundings will be less than the total energy
change. In this reaction, for 1 mole of zinc, the heat (enthalpy) change is 152.4 kJ and
the work against the atmosphere 2.5 kJ, the overall energy change being 154.9 kJ.
Chemists often carry out reactions at constant pressure, hence the use of 'enthalpy
change', the heat change at constant pressure.
Let's consider another familiar exothermic reaction, the burning (oxidation) of methane
in a flame. You see this every time you use natural gas.
These changes are often shown as a potential energy diagram. An example is shown
in Figure 3.1 for methane oxidation.
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TOPIC 3. CHEMICAL ENERGY
Figure 3.1: Methane oxidation
..........................................
This diagram shows the potential energy of the reactants, at the start of the reaction. It
also shows the potential energy of the products, at the end of the reaction. The arrow
shows the energy given to the surroundings.
3.3.1
Enthalpy changes
The easiest measurement to make is the enthalpy change, ΔH, when the reaction
occurs. This is defined as the difference in enthalpy of the products and reactants by
the equation below. The units of energy and enthalpy are joules (J).
ΔH is negative for an exothermic reaction. The energy of the products is less than that
of the reactants.
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The potential energy diagram for an endothermic reaction, the decomposition of
calcium carbonate.
This is shown below.
In this case the energy of the products is greater than that of the reactants.
Q6:
So what will happen to the temperature of the surroundings?
a) It will increase.
b) It will not change.
c) It will decrease.
..........................................
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TOPIC 3. CHEMICAL ENERGY
Q7: What will the value of ΔH be?
a) Negative
b) Zero
c) Positive
..........................................
This is shown below.
In order to distinguish endothermic and exothermic reactions the + and - sign are
always used when giving ΔH values.
Q8:
The equation for the "water gas" reaction is:
Figure 3.2: Water gas reaction
..........................................
Note that this equation states that the carbon is in the solid state(s), and the other
materials are gases (g).
This reaction absorbs energy from its surroundings. What type of reaction is this?
a)
b)
c)
d)
Fast
Exothermic
Explosive
Endothermic
..........................................
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What will be the sign of ΔH?
Q9:
a)
b)
c)
d)
Positive.
Has no sign.
Negative.
Insufficient information to know.
..........................................
Q10: Draw a potential energy diagram for this reaction, and compare it with the answer
at the back of the book.
..........................................
But how much energy is involved in this reaction? In the case of the water gas reaction
(Figure 3.2) 121 kJ is absorbed for each mole of carbon and steam used. In other
words ΔH = +121 kJ mol-1 .
The full potential energy diagram for the water gas reaction is shown below.
Key point
3.3.2
•
A potential energy diagram can be used to show the energy changes for a
reaction.
•
The enthalpy change is the energy difference between products and
reactants.
•
The enthalpy change can be calculated from a potential energy diagram.
•
The enthalpy change has a negative value for exothermic reactions and a
positive value for endothermic reactions.
What produces this energy change?
The chemical energy in molecules is stored in the bonds. When a reaction occurs the
bonds in the reactant molecules are broken and the atoms rearrange to form new
bonds with different energies in the products. For example, think of making ammonia
from nitrogen and hydrogen.
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The balanced chemical equation is:
You can think of this reaction as first breaking the bonds in the nitrogen and hydrogen
molecules. This stage requires an input of energy.
Then these atoms rearrange and combine to form new N - H bonds in ammonia. This
stage releases energy.
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The enthalpy change will be given by the total energy in new bonds minus the energy
present in the original bonds.
It is important to remember that reactions do not normally proceed by such a route, but
the enthalpy change can be calculated as if they did.
3.3.3
The thermochemical equation
It is usual to express this information in a thermochemical equation, such as that for
the water gas reaction below (Figure 3.3).
Figure 3.3: Thermochemical equation for the water gas reaction
..........................................
Three points should be noted.
1. The enthalpy value quoted in the balanced equation is measured in kilojoules per
mole (kJ mol-1 ).
Look at Figure 3.3 again. The 121 kJ of energy is absorbed when one mole of solid
carbon and one mole of steam react; two moles of each reacting would absorb 242 kJ,
and so on. (This makes sense when you think that burning 2 kg of carbon would give
out twice the energy from burning 1 kg.)
2. The equation always contains the state symbols of reactants and products.
The enthalpy of reactions will change if the states are different. Look carefully at these
two equations for the reaction of hydrogen and oxygen
H2(g) + ½O2(g)
H2O(g)
ΔH = -242 kJ mol-1
(N.B. the equation has been written for forming 1 mole of water. The " 1 /2 " before the
oxygen means 1 /2 mole of reactant not 1 /2 a molecule of oxygen.)
but
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TOPIC 3. CHEMICAL ENERGY
The extra enthalpy in the second case is due to the heat released when the water
vapour condenses to liquid water.
3. Since the enthalpy change is independent of the route of the reaction, the enthalpy
change for the reverse reaction will be equal in value, but of opposite sign.
For the reaction below:
Q11:
So the enthalpy change for the reverse reaction:
CO2 (g) → CO(g) + 1 /2 O2 (g)
will be?
a)
b)
c)
d)
ΔH = -566 kJ mol-1
ΔH = -283 kJ mol-1
ΔH = 0 kJ mol-1
ΔH = +283 kJ mol-1
..........................................
Q12: The equation for calculating the enthalpy change (ΔH) for a reaction is:
a)
b)
c)
d)
ΔH = Hproducts + Hreactants
ΔH = Hproducts - Hreactants
ΔH = Hreactants + Hproducts
ΔH = Hreactants - Hproducts
..........................................
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Q13: A ΔH value of - 500 kJ mol -1 would indicate that the reaction is:
a)
b)
c)
d)
Fast
Reversible
Exothermic
Endothermic
..........................................
Q14: The potential energy diagram for the combustion of one mole of ethene is:
What is the enthalpy change for this reaction?
a)
b)
c)
d)
ΔH = +49 kJ mol-1
ΔH = -1313 kJ mol-1
ΔH = -1362 kJ mol-1
ΔH = -1411 kJ mol-1
..........................................
Q15:
The equation for the combustion of ethanol is:
How many kJ of energy will be released to the surroundings if 5 moles of ethanol are
burned?
..........................................
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TOPIC 3. CHEMICAL ENERGY
Q16:
Given that:
What would be the enthalpy change for the reaction below? (Hint: don't forget the sign.)
..........................................
Q17: The Ostwald process for producing nitric acid involves the oxidation of ammonia
(NH3 ) to form nitric oxide (NO) and water. Write a balanced equation for this reaction.
..........................................
Q18: Given that the enthalpy of the reactants (as written in the balanced equation) is
-184 kJ and the enthalpy of the products is -1090 kJ, calculate the enthalpy change for
this reaction in kJ. Remember the sign.
..........................................
Q19: What would be the enthalpy change per mole of ammonia oxidised? Give your
answer in kJ mol-1 to 1 decimal place.
..........................................
Q20: When 102 g of ammonia is oxidised, how many kJ of heat have to be removed to
maintain a constant temperature?
..........................................
3.4
Enthalpy changes
All chemical (and many physical) processes are associated with a particular enthalpy
change, but these are often difficult to measure.
Three particular enthalpy changes which can be measured in the laboratory are:
•
enthalpy of combustion;
•
enthalpy of solution, and;
•
enthalpy of neutralisation.
The methods of obtaining these values will be considered as well as their importance.
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The change in temperature during a reaction allows the heat energy change and ΔH to
be measured.
To obtain the most accurate result we also want to limit losses of heat from our system,
so that some form of insulation is normally employed to minimise these losses.
All the methods depend on measuring the temperature change of a known mass of
water (or dilute solution) during the course of a reaction.
The temperature change is related to the heat energy by using the equation shown in
Figure 3.4
Figure 3.4: Heat transfer equation
..........................................
The heat energy transferred is in kJ,
c is the specific heat capacity of water. The specific heat capacity relates the energy
change in a liquid to the change in temperature. For water it has a value of 4.18 kJ kg -1
◦ C-1 . In other words, when 1 kg of water absorbs 4.18 kJ of heat its temperature will
rise by 1◦ C.
m is the mass of water, or solution (in kg), and
ΔT is the temperature change (in ◦ C).
The use of this equation will be illustrated in the three particular cases described in the
next sections.
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97
Enthalpy of combustion
The enthalpy of combustion is defined as the enthalpy change that occurs when 1
mole of a substance is burned completely in oxygen. It is often called ΔH c .
To determine the heat given out from a measured quantity of a liquid fuel, say, ethanol,
the fuel is burned and the heat produced warms a known mass (or volume) of water.
This experiment and the calculation is described in the next activity.
Determination of the enthalpy of combustion of an alcohol
A interactive description of this experiment is available on-line. The equipment used is
shown below.
Enthalpy of combustion equipment
The spirit burner containing ethanol is weighed at the beginning and end of the
experiment to obtain the mass of ethanol used.
Before the experiment we should know the mass (or volume) of the water to be heated.
The temperature of the water in the can is measured at the start and finish, and the
temperature change noted.
We now have all the information to construct a table of data to determine the enthalpy
of combustion of ethanol. An example set of data is shown in the following table.
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Readings taken
Figure obtained
Value
Mass of burner at start &
end
Mass of ethanol burned
0.368 g
Volume of water heated
Mass of water (m)
0.200 kg
Temperature of water at
start & end
Temperature rise (ΔT)
+12.0 o C
-
Specific heat of water (c)
4.18 kJ o C-1 kg-1
Substituting these data into the equation in Figure 3.4
Energy transferred = c m ΔT
= 4.18 kJ kg - 1 o C - 1 × 0.200 kg × ( + 12.0 o C)
= 4.18 × 0.200 × 12.0 kJ (kg - 1 o C - 1 kg o C cancel out)
= 10.032 kJ
So 10.032 kJ of energy are released when 0.368 g of ethanol are burned. We need to
find how many kJ are released when 1 mole of ethanol burns.
Calculate the gram formula mass of ethanol.
Then work out the energy released when 1 mol of ethanol burns.
0.368 g of ethanol releases 10.032 kJ
0.368/46 mols of ethanol releases 10.032 kJ
0.008 mols of ethanol releases 10.032 kJ
1 mol of ethanol releases 10.032/0.008 kJ
1 mol of ethanol releases 1254 kJ
Remember that the heat has been measured in the water, i.e. energy is released to the
surroundings. The combustion of ethanol is exothermic, and the enthalpy change for
the reaction will be negative.
The enthalpy of combustion of ethanol (ΔH c ) is -1254 kJ mol-1 .
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The value from the data booklet is -1367 kJ mol -1 .
This experimental value is lower than the quoted one.
Q21: Can you think of some reasons for this?
..........................................
Enthalpy of combustion questions
Q22: Use the data booklet to find the enthalpy of combustion of propane and butane. If
you had a mole of each fuel, which would give you most heat energy?
a) Propane
b) Butane
..........................................
Q23: You are heating a pan containing 1.0 kg of water using a butane burner. If 0.1 mol
of butane is used, what would be the temperature rise of the water? Give your answer
to 1 decimal place.
..........................................
Q24: Your friend is using a methanol (CH 3 OH) burner to heat her water. She finds that
burning 6.4g of methanol raises the temperature of 0.50 kg of water by 50 ◦ C. From these
data what is the enthalpy of combustion of methanol? Give your answer in kJ mol -1 to 1
decimal place, and remember the sign.
..........................................
3.4.2
Enthalpy of solution
The enthalpy of solution is defined as the energy change (in kJ) when 1 mole of the
substance dissolves in water.
To determine the enthalpy of solution of a substance in water, a known weight of it is
dissolved in a known volume of water. The temperature change is noted when the
substance has dissolved. The experiment is carried out under conditions where the
system (substance plus water) is as thermally insulated from the surroundings as
possible. For practical purposes polystyrene cups are very useful - they have good
insulation and themselves have a very low specific heat capacity.
The enthalpy change in this case results from the difference between the enthalpy input
required to disrupt the crystal lattice (see ionic lattices) and the enthalpy obtained when
the ions are hydrated.
Enthalpies of solution can be +ve or -ve, unlike ΔH c which is always -ve.
An experiment to determine the enthalpy of solution of ammonium nitrate is described
in the next activity.
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An experiment to determine the enthalpy of solution of ammonium nitrate
Go online
Q25: In order to use the equation
Energy transferred = c m ΔT
what information do you need at the start of the experiment?
..........................................
Q26:
During the experiment, the temperature drops to a minimum, then slowly rises again.
Why do you think the temperature behaves like this?
..........................................
The lowest temperature reached was 20.4 ◦ C.
Q27: If the initial temperature was 22.0 ◦ C, what is the value of ΔT?
..........................................
Calculation of the enthalpy of solution of ammonium nitrate, using the
experimental data above
The balanced equation for the reaction is
NH4 NO3 (s) → NH4 + (aq) + NO3 - (aq)
Put the experimental data into a table:
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Readings taken
Figure obtained
Value
Mass of ammonium nitrate
-
1.00vg
Volume of water
Mass of water
0.050 kg
Temperature of water at
start
Temperature of solution at
end
Temperature change
- 1.6 ◦ C
-
Specific heat of water
4.18 kJ kg-1 ◦ C-1
N.B. Temperature change is -1.6 ◦ C (the temperature dropped, so ΔT is negative).
Once again, we use the equation in Figure 3.4, making the assumption that the
specific heat of the ammonium nitrate solution is the same as that of water.
Substituting into the equation gives,
Energy transferred = c m ΔT
= 4.18 kJ kg - 1 o C - 1 × 0.050 kg × ( - 1.6 o C)
= 4.18 × 0.050 × ( - 1.6) kJ
= - 0.334 kJ
0.334 kJ of energy was transferred from the surroundings (negative value) when 1g of
ammonium nitrate was dissolved in water.
The gram formula mass of NH 4 NO3 is
Calculate the number of kJ of energy absorbed when 1 mol of ammonium nitrate
dissolves.
1.00g of ammonium nitrate absorbed 0.334 kJ of energy
1.00/80 mol of ammonium nitrate absorbed 0.334 kJ of energy
0.0125 mol of ammonium nitrate absorbed 0.334 kJ of energy
1 mol of ammonium nitrate absorbed 0.334/0.0125 kJ of energy
1 mol of ammonium nitrate absorbed 26.7 kJ of energy
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The enthalpy of solution of ammonium nitrate is +26.7 kJ mol -1 .
NH4 NO3 (s) → NH4 + (aq) + NO3 - (aq) ΔH = +26.7 kJ mol-1
(Remember a final check, heat was absorbed as the temperature dropped, the reaction
is endothermic, ΔH will be +ve.)
..........................................
Questions on enthalpy of solution
Q28: When 1.20 g of sodium hydroxide is dissolved in 60 cm 3 of water the temperature
rises from 18.4◦ C initially to 23.2◦ C in the final solution. What is the enthalpy of solution
of sodium hydroxide in kJ mol-1 , to 1 decimal place?
..........................................
Q29:
Chemical heat packs, used to prevent hypothermia, release heat when calcium chloride
dissolves in water according to the equation:
How many kJ, to 3 significant figures, of energy will be obtained from a pack containing
500 g of calcium chloride?
..........................................
Key point
Enthalpy of solution is the energy change when one mole of a substance
dissolves in water. The units are kJ mol-1 .
3.4.3
Enthalpy of neutralisation
The enthalpy of neutralisation of an acid is defined as the energy change (in kJ)
when it is neutralised to form 1 mole of water.
Enthalpy changes for this reaction can be obtained by using polystyrene cups to
reduce heat loss to a minimum when a known volume of alkali is added to an
equivalent amount of acid and the temperature change noted.
The equipment to determine the enthalpy of neutralisation of hydrochloric acid is shown
below.
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The results for two different combinations of acid and alkali are described.
Calculation of the enthalpy of neutralisation of hydrochloric acid with sodium
hydroxide.
The temperatures taken were:
Temperature of the acid initially 20.5 ◦ C,
Temperature of the alkali initially 21.0 ◦ C,
Temperature of the final solution 27.5 ◦ C.
So, summarising the experimental data in a table.
Readings taken
Figure obtained
Value
Volume & concentration of
acid
Moles of acid reacting*
0.1 mol
Volume & concentration of
alkali
Moles of alkali reacting*
0.1 mol
Initial temperature of alkali
Initial temperature of acid
Average initial
temperature**
20.75◦C
Temperature of final
solution
Temperature change
ΔT = +6.75◦ C
-
Specific heat of water
4.18 kJ kg-1 ◦ C-1
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* Remember the equation
Moles
Volume (litres)
Which will rearrange to give
Molar concentration =
Moles = Concentration x Volume
In both these cases, the concentration is 1 mol l −1 and volume 0.1 l.
So Moles = 1 x 0.1 mol (l −1 l cancel out)
= 0.1 mols
** The initial temperature is calculated as the average of the temperatures of the acid
and alkali solutions. (It would be nice to mix them and take the initial temperature, but
they react on mixing, so the average initial temperature is obtained mathematically.)
Temperature of acid is 20.5 o C
Temperature of alkali is 21.0 o C
So average initial temperature is
20.5 + 21.0
= 20.75 o C
2
The temperature change (ΔT) is then
Final temperature of solution - average initial temperature
27.5 - 20.75 = + 6.75 o C
We are going to use the equation in Figure 3.4 to calculate the heat change in the
reaction.
Remember that m = 0.200 kg (100 cm 3 of acid + 100 cm3 of alkali)
Energy transferred = c m ΔT
= 4.18 kJ kg - 1 o C - 1 × 0.200 kg × ( + 6.75 o C)
= 4.18 × 0.200 × ( + 6.75) kJ (kg - 1 o C - 1 kg o C cancel out)
= 5.64 kJ
How many moles of water have been produced?
Q30: 0.1 mol of hydrochloric acid and 0.1 mol sodium hydroxide reacted, so how many
moles of water were produced?
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5.64 kJ of energy were released to the surroundings when 0.1 moles of water were
produced.
So 56.4 kJ of energy were produced for each mole of water.
Therefore ΔH for this neutralisation is -56.4 kJ mol -1 . Remember to check that you
have written a negative value of ΔH for an exothermic reaction.
The next questions concern the enthalpy of neutralisation of sulfuric acid by potassium
hydroxide, given the following data.
Sulfuric acid solution 50 cm3 of 1 mol -1 H2 SO4
Potassium hydroxide 50 cm3 of 2 mol -1 KOH
Initial temperature of acid 21.2 ◦ C
Initial temperature of alkali 21.6 ◦ C
Final temperature after mixing 35.0 ◦ C
Q31: Write a balanced equation for the reaction between sulfuric acid and potassium
hydroxide.
..........................................
Q32: What is the average initial temperature?
..........................................
Q33: What is the temperature change (ΔT)?
..........................................
Q34: What mass of solution was heated (m in kg)?
..........................................
Q35: How much energy (in kJ) has been transferred to the water? Give your answer to
3 decimal places.
..........................................
Q36: How many moles of water were produced when the sulfuric acid and potassium
hydroxide reacted?
..........................................
Q37: Calculate the enthalpy of neutralisation of sulfuric acid by potassium hydroxide, in
kJ mol-1 , to 2 decimal places (don't forget the sign).
..........................................
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Q38: What do you notice about the enthalpy of neutralisation of hydrochloric acid by
sodium hydroxide and of sulfuric acid by potassium hydroxide?
..........................................
Q39: Can you think of a reason for this similarity? Hint: All four reactants are completely
ionised in solution, so write ionic equations and think about the spectator ions.
..........................................
Key point
The enthalpy of neutralisation of an acid is the enthalpy change when the acid is
neutralised to form one mole of water.
Enthalpy changes can be calculated using the equation: energy transferred = c
m ΔT.
3.5
Introduction to Hess's law
You now know that the enthalpy change for a chemical reaction is described as the
energy transferred to the surroundings during an exothermic reaction or the energy
absorbed from the surroundings during an endothermic reaction. The enthalpy change
is therefore the difference in chemical potential energy between the reactants and
products, as shown in the potential energy diagrams (Figure 3.5).
Figure 3.5: Potential energy diagrams
..........................................
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TOPIC 3. CHEMICAL ENERGY
Q40: Which diagram (Figure 3.5) represents an exothermic reaction?
..........................................
Q41: Which of the following is the enthalpy change in reaction A?
a)
b)
c)
d)
-630
-570
+570
+630
..........................................
In the following diagram, the change in potential energy for the reaction between A 2
and B2 is shown, both in the presence, and in the absence, of a catalyst.
The numbers 1- 4 are pointing at energy changes and numbers 5 and 6 at key points
on each curve. Work out what each number is indicating (you might like to make a list)
before revealing the answers by moving the mouse pointer over each number in turn.
Potential Energy diagram for a catalysed reaction
Q42: Which number is pointing at the enthalpy change for the reaction?
..........................................
Q43: What is the significance of the energy changes indicated by 1 and 2?
..........................................
Key point
•
The use of a catalyst provides an alternative route for a reaction.
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Key point continued
3.5.1
•
The starting point and finishing point for both routes are the same and so
the enthalpy change is the same for both routes.
•
Chemists make use of this in Hess's law.
Hess's law
Hess's law can be stated in many different ways. One such statement is:
Key point
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.
This statement follows on from the law of Conservation of Energy - energy cannot be
created or destroyed.
Imagine a reaction which can take place by two different routes.
B can be made directly from A (Route 1) or via C (route 2). According to Hess's law,
ΔH1 = ΔH2 + ΔH3 . If this were not so, it would be possible to create energy, e.g.
Suppose ΔH1 = -50 kJ
but ΔH2 + ΔH3 = -60 kJ
It would then be possible to make B by Route 2 and release 60 kJ of energy. B could
then be converted back into A by reversing Route 1. This would use up 50 kJ of energy
and there would be a net gain of 10 kJ. Each time this cycle was repeated, 10 kJ
of energy would be created. This is impossible since it would contravene the law of
Conservation of Energy. Consequently the enthalpy change for Route 1 (ΔH 1 ) must
equal the enthalpy change for Route 2 (ΔH 2 + ΔH3 ), i.e. Hess's law must apply.
3.5.2
Verification of Hess's law
Hess's law can be verified experimentally. Sodium chloride solution can be made from
solid sodium hydroxide by two slightly different routes.
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Route A
directly by reacting solid sodium hydroxide with hydrochloric acid.
Route B
by first dissolving the solid sodium hydroxide in water and then
neutralising the resultant solution with hydrochloric acid.
All three steps involve reactions whose enthalpy changes are easily measured. The
reactions can be carried out in polystyrene cups using accurately measured quantities.
Route A
For Route A, the temperature of a known volume of 1.0 mol -1 hydrochloric acid
is measured before the addition of an accurately weighed sample of solid sodium
hydroxide. The temperature rises and the highest temperature is recorded as the final
temperature. From this information, the enthalpy change can be calculated.
The preceding figure contains all the information necessary for the calculation. The
following assumptions can be made:
•
assume there is no heat loss to the polystyrene cups or the surroundings;
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•
assume the solutions have the same specific heat capacity as water (data booklet
page 19);
•
convert masses to kilogram units.
Q44: What relationship is used to calculate the energy released from the temperature
rise?
..........................................
Q45: Calculate the enthalpy change in kJ mol -1 . Write your answer on paper, setting
out the calculation clearly, before displaying the worked answer.
..........................................
For Route B, step 1 involves dissolving sodium hydroxide in water and measuring the
temperature change as before.
Figure 3.6: Route B, Step 1
..........................................
Q46: Use the Figure 3.6 diagram to obtain the required data and then calculate the
enthalpy of solution of sodium hydroxide. If you are not confident about this or if you get
the answer wrong, try the further questions on the page noted below.
..........................................
Q47: What is the temperature rise (to one decimal place)?
..........................................
Q48: What mass of water is being heated?
..........................................
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Q49: How much energy is released when 0.80 g of NaOH dissolves?
..........................................
Q50: How much energy is released when 1 mole of NaOH dissolves?
..........................................
Q51: What is the enthalpy of solution?
..........................................
In step 2, the final solution from the first step (Figure 3.6) is cooled down and neutralised
by a known volume of hydrochloric acid and the temperature change is measured. The
initial temperature is obtained by measuring the temperature of each solution and taking
the average.
Figure 3.7: Route B, Step 2
..........................................
Q52: Use the diagram (Figure 3.7) to obtain the required data and then calculate the
enthalpy of neutralisation of sodium hydroxide with hydrochloric acid. If you are not
confident about this or if you get the answer wrong, try the further questions on the page
noted below.
..........................................
Q53: What is the temperature rise (to one decimal place)?
..........................................
Q54: What mass of water is being heated?
..........................................
Q55: How much energy is released when the two solutions are mixed?
..........................................
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Q56: How much energy is released when 1 mole of NaOH is neutralised?
..........................................
Q57: What is the enthalpy of neutralisation?
..........................................
Q58: Finally, calculate the total enthalpy change for Route B.
..........................................
The enthalpy changes for Route A and Route B are very similar and well within
experimental error and therefore Hess's law is obeyed. The assumptions made before
the calculation was done are important when considering the experimental errors.
3.5.3
Applications of Hess's law
Hess's law is important to chemists as they are often required to find values for enthalpy
changes which cannot be measured directly because the reaction does not occur under
normal conditions.
It is difficult to burn carbon and form carbon monoxide only (Figure 3.8 (b)). So the
value of ΔH2 for the formation of carbon monoxide would be difficult to determine by
experiment.
Figure 3.8
..........................................
According to Hess's law and the enthalpy cycles shown in Figure 3.8:
ΔH1 = ΔH2 + ΔH3
Hess's law can be applied in a situation like this to enable a determination of the value
of ΔH2 to be made. The first of the next two activities uses an enthalpy cycle diagram
and the second uses an algebraic method to calculate a value of ΔH 2 . You may be
directed to either or both methods of calculation.
Application of Hess's law by enthalpy cycle diagram
The Question
Go online
Calculate a value for the enthalpy change involved in the formation of one mole of
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carbon monoxide from carbon (graphite). The enthalpy of combustion of carbon to form
carbon dioxide is quoted in the data booklet and the enthalpy of combustion of carbon
monoxide is -283 kJ mol-1 .
The on-line version of this activity contains an interactivity showing the stages in the
calculation. If you do not have access to the on-line material use this energy cycle
diagram of the reactions, and work through the text.
Hess's law cycle for carbon monoxide formation
The overall enthalpy change is independent of the route taken and values of ΔH 1 and
ΔH3 are measurable.
Since: ΔH1 = ΔH2 + ΔH3
ΔH1 = enthalpy of combustion (C to CO2 ) = − 394 kJ mol−1
ΔH3 = enthalpy of combustion (CO to CO2 ) = − 283 kJ mol−1
ΔH1 =ΔH2 + ΔH3
by rearranging ΔH2 =ΔH1 − ΔH3
ΔH2 = − 394 − (−283) kJ mol−1
ΔH2 = − 111 kJ mol−1
So the enthalpy change involved in formation of carbon monoxide from carbon
(graphite) = -111 kJ mol -1
Notice that the enthalpy change represented by ΔH 3 had to be reversed in solving for
ΔH2 . The numerical value of the enthalpy change therefore changed from -283 kJ
mol-1 to +283 kJ mol-1 . In the same way, if an equation is multiplied or divided, the
value of ΔH should be treated in the same way.
Q59:
Since:
C(s) + 1 /2 O2 (g) → CO(g)
ΔH = -111 kJ
What would be the value (in kJ) for the enthalpy change for this reaction?
1 / C(s) + 1 / O (g) → 1 / CO(g)
ΔH = ? kJ
2
4 2
2
..........................................
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Q60:
Since:
C(s) + 1 /2 O2 (g) → CO(g)
ΔH = -111 kJ
What would be the value (in kJ) for the enthalpy change for this reaction?
2CO(g) → 2C(s) + O2 (g)
ΔH = ? kJ
..........................................
Key point
Enthalpy cycle diagrams can be used to calculate enthalpy changes for reactions
which are difficult to measure.
..........................................
Application of Hess's law by algebraic calculation
Go online
Some Hess's law problems are best solved by use of an algebraic treatment of the
balanced chemical equations. This activity sets out to solve the same problem as in the
last activity, but uses the equations rather than an enthalpy cycle diagram.
The Question
Calculate a value for the enthalpy change involved in the formation of one mole of
carbon monoxide from carbon (graphite). The enthalpy of combustion of carbon to form
carbon dioxide is quoted in the data booklet and the enthalpy of combustion of carbon
monoxide is -283 kJ mol-1 .
The on-line version of this activity contains an activity showing the stages in the
calculation. If you do not have access to the on-line material work your way through the
text.
The equation which represents the enthalpy change to be calculated is called the
"target" equation.
STEP 1
Write the 'target' equation.
C(s) + 1 /2 O2 (g) → CO(g)
ΔH = ?
STEP 2
Write equations for all the
information given to you.
equation (a)
C(s) + O2 (g) → CO2 (g)
ΔH = -394 kJ
(from data booklet)
equation (b)
CO(g) + 1 /2 O2 (g) → CO2 (g)
ΔH = -283 kJ
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To construct the "target" equation from (a) and (b),
leave equation (a) as written and reverse equation
(b). Hess's Law allows you to treat the ΔH values the
same way. The website animation shows the stages.
STEP 3
STEP 3
continued
equation (a)
C(s) + O2 (g) → CO2 (g)
ΔH = -394 kJ
Reverse
equation (b)
CO2 (g) → CO(g) + 1 /2 O2 (g)
ΔH = +283 kJ
C(s) + 1 /2 O2 (g) → CO(g)
ΔH = -111 kJ
add this to (a)
Notice that the enthalpy change represented by ΔH 3 (equation b) had to be reversed
in solving for ΔH2 (the target equation). The value of the enthalpy change therefore
changed from -283 kJ mol -1 to +283 kJ mol-1 . In the same way, if an equation is
multiplied or divided, the value of ΔH should be treated in the same way.
Q61:
Since:
C(s) + 1 /2 O2 (g) → CO(g)
ΔH = -111 kJ mol-1
What would be the value (in kJ) for the enthalpy change for this reaction? Remember
to include the sign.
1 / CO(g) → 1 / C(s) + 1 / O (g)
ΔH = ? kJ
2
2
4 2
..........................................
Q62:
Since:
C(s) + 1 /2 O2 (g) → CO(g)
ΔH = -111 kJ
What would be the value (in kJ) for the enthalpy change for this reaction? Remember
to include the sign.
3CO(g) → 3C(s) + 3 /2 O2 (g)
ΔH = ? kJ
..........................................
Key point
Algebraic calculations from balanced equations can be used to calculate enthalpy
changes for reactions which are difficult to measure.
..........................................
In both of the methods employed in calculating enthalpy changes by application of
Hess's law, the same rules have been used.
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•
If an equation is reversed, then the sign of ΔH also changes.
•
If an equation is multiplied the value of ΔH should be multiplied by the same
number.
•
If an equation is divided the value of ΔH should be divided by the same
number.
e.g.
Knowing the value for (1), we can write
equations like (2) and (3).
equation (1)
C(s) + O2 (g) → CO2 (g)
ΔH = -394 kJ
equation (2)
2C(s) + 2O2 (g) → 2CO2 (g)
ΔH = -798 kJ
equation (3)
1 / CO (g)
2
2
→ 1 /2 C(s) + 1 /2 O2 (g)
ΔH = +197 kJ
Key point
Enthalpy changes can be calculated by application of Hess's law. Calculations
can be carried out using enthalpy cycle diagrams and/or by using algebraic
calculations from balanced equations.
3.6
Bond enthalpies
You saw earlier that you can think of the reaction between nitrogen and hydrogen to
make ammonia as taking place by breaking all the bonds to produce atoms and then
recombining these in a new way to make ammonia. The reaction does not occur that
way; but now you know Hess's law, it is irrelevant how a reaction occurs; to calculate
a reaction enthalpy change you only need to consider the starting reactants and the
finishing products. So you can calculate the reaction enthalpy change in that way.
The overall enthalpy change of reaction will be given by the total enthalpy of new bonds
minus the enthalpy of the original bonds.
Because they are so useful in thermodynamic calculations, chemists have produced
tables of bond enthalpies. These can be found in your SQA data booklet. The enthalpy
change values in these tables are always quoted for breaking one mole of bonds for
gaseous reactants and products, and are always positive. (No bond will produce energy
when it breaks.)
For example, for a symmetrical bond such as H - H, the bond enthalpy is:
H2 (g) → 2H(g)
ΔH = +436 kJ mol-1
This bond enthalpy can only be for a hydrogen molecule, so the figure obtained should
© H ERIOT-WATT U NIVERSITY
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be exact. This will apply to other diatomic molecules where no other atoms are involved:
for example, Cl2 (and other halogens), O=O, N≡N. These bond enthalpies are listed in
one column of the SQA Data Booklet.
For bonds between different atoms, for example, C - H:
C - H(g) → C(g) + H(g)
ΔH = +412 kJ mol-1
In this case the values obtained differ slightly depending on the environment of the bond.
For example, the C - H bond enthalpy in methane (CH 4 ) will differ slightly from that in
ethene (CH2 = CH2 ). The bond enthalpies are given as average bond enthalpies from
several different reactants, in a separate table.
You will sometimes see bond enthalpies (or energies) written as:
E(X - Y)
for the bond enthalpy for a X - Y bond.
Bond enthalpies and the synthesis of ammonia
We can now apply bond enthalpy data to the synthesis of ammonia:
Go online
N2 (g) + 3H2 (g) → 2NH3(g)
Q63: Use the bond enthalpy data to construct a table showing enthalpy changes when
bonds are broken and when bonds are formed.
ΔH
Bonds broken
N≡N
?
3×H-H
?
Total enthalpy
change to break
bonds
?
Bonds formed
ΔH
6×N-H
?
Total enthalpy
change on making
bonds
?
Then calculate the overall enthalpy change for this reaction as the sum of these two
changes.
..........................................
The enthalpy of formation (ΔH f ) of a compound is the enthalpy change of reaction when
one mole of the compound is formed from its elements. In this reaction two moles of
ammonia have been made from its elements (nitrogen and hydrogen). The enthalpy of
formation of ammonia will be:
(-75)/2 = -37.5 kJ mol-1
A quoted value for ΔH f of ammonia is -46.1 kJ mol-1 . Although the enthaply changes
for breaking the H - H and N ≡ N bonds applied to the actual reactants (H 2 and N2 ), the
N - H bond value quoted is a mean and this has led to the value calculated using bond
enthalpies being lower than the actual value.
..........................................
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TOPIC 3. CHEMICAL ENERGY
The hydrogenation of ethene
The next series of questions concerns the hydrogenation of ethene to form ethane.
Go online
CH2 = CH2 + H2 → CH3 -CH3
The following bond enthalpies may be useful.
Bond
Bond enthalpy
/kJ mol-1
H-H
436 (for H2 )
C-H
412 (average)
C-C
348 (average)
C=C
612 (average)
Q64: Which type and number of bonds have to be broken?
a)
b)
c)
d)
1 x C=C only
1 x C=C and 1 x H - H
1 x C=C and 4 x C - H
1 x C=C, 4 x C - H and 1 x H - H
..........................................
Q65: Which type and number of bonds have to be made?
..........................................
Q66: Using the data provided, calculate (i) the enthalpy change to break the bonds, and
(ii) the enthalpy change from forming bonds. Hence, calculate the enthalpy change for
this reaction in kJ.
..........................................
Q67: The enthalpy change found directly for this reaction is -136.9 kJ. Why are these
results different?
..........................................
3.7
Summary
Summary
Chemical Energy
•
For industrial processes, it is essential that chemists can predict the quantity
of heat energy taken in or given out.
•
Exothermic changes cause heat to be released to the surroundings.
© H ERIOT-WATT U NIVERSITY
TOPIC 3. CHEMICAL ENERGY
Summary continued
•
Endothermic changes cause absorption of heat from the surroundings.
•
If reactions are endothermic, costs will be incurred in supplying heat energy
in order to maintain the reaction rate.
•
If reactions are exothermic, the heat produced may need to be removed to
prevent the temperature rising.
Enthalpy
•
Chemical energy is also known as enthalpy.
•
The change in chemical energy associated with chemical reactions can be
measured.
•
The enthalpy change is the energy difference between products and
reactants.
•
The enthalpy change can be calculated from a potential energy diagram.
•
The enthalpy change has a negative value for exothermic reactions and a
positive value for endothermic reactions.
•
The specific heat capacity, mass and temperature can be used to calculate
the enthalpy change for a reaction.
•
The enthalpy of combustion of a substance is the enthalpy change when
one mole of the substance burns completely in oxygen.
•
These values can often be directly measured using a calorimeter and
values for common compounds are available from data books and online
databases for use in Hess’s law calculations.
•
Enthalpy of solution is the energy change when one mole of a substance
dissolves in water. The units are kJ mol-1 .
•
The enthalpy of neutralisation of an acid is defined as the energy change
when it is neutralised to form 1 mole of water. The units are kJ mol -1 .
Hess’s Law
•
Hess's law is used to calculate enthalpy change values that would be difficult
to measure by experiment or cannot be measured directly because the
reaction does not occur under normal conditions.
•
Hess’s law states that 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.
•
Enthalpy changes can be calculated by applying Hess’s law.
•
Enthalpy changes can be calculated by application of Hess's law.
Calculations can be carried out using enthalpy cycle diagrams and/or by
using algebraic calculations from balanced equations.
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TOPIC 3. CHEMICAL ENERGY
Summary continued
•
Algebraic calculations from balanced equations can be used to calculate
enthalpy changes for reactions which are difficult to measure.
◦
If an equation is reversed, then the sign of ΔH also changes.
◦
If an equation is multiplied the value of ΔH should be multiplied by the
same number.
◦
If an equation is divided the value of ΔH should be divided by the same
number.
e.g. Knowing the value for (1), we can write equations like (2) and (3).
equation (1) C(s) + O2 (g) → CO2 (g) ΔH = -394 kJ
equation (2) 2C(s) + 2O2 (g) → 2CO2 (g) ΔH = -798 kJ
equation (3) 1/2 C(s) + 1/2O 2 (g) → 1/2CO2(g) ΔH = -197 kJ
Bond Enthalpies
3.8
•
•
For a diatomic molecule, XY, the molar bond enthalpy is the energy required
to break one mole of XY bonds.
•
Mean molar bond enthalpies are average values which are quoted for bonds
which occur in different molecular environments.
•
Bond enthalpies can be used to estimate the enthalpy change occurring for
a gas phase reaction by calculating the energy required to break bonds in
the reactants and the energy released when new bonds are formed in the
products.
Resources
Higher Chemistry for CfE: J Anderson, E Allan and J Harris, Hodder Gibson,
ISBN 978-1444167528
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End of topic test
End of topic 3 test
This end of topic test is available online. If you do not have access to the internet, here
is a paper version.
In a reaction, the enthalpy of the reactants is +50 kJ mol -1 and the enthalpy of the
products is -250 kJ mol-1 .
Q68:
Which equation should you use to calculate the enthalpy change for the reaction?
A) ΔH = H(products) - H(reactants)
B) ΔH = H(products) + H(reactants)
C) ΔH = H(reactants) - H(products)
D) ΔH = H(reactants) + H(products)
..........................................
Q69:
What is the value of ΔH for this reaction?
A) -300
B) -200
C) +200
D) +300
..........................................
Q70:
The reaction is:
A) endothermic
B) exothermic
..........................................
Q71:
Which equation illustrates an enthalpy of combustion?
A) C2 H5 OH(l) + O2 (g) → CH3 COOH(l) + H2 O(l)
B) C2 H6 (g) + 31 /2 O2 (g) → 2CO2 (g) + 3H2 O(l)
C) CH4 (g) + 11 /2 O2 (g) → CO(g) + 2H2 O(l)
D) CH3 CHO(l) + 1 /2 O2 (g) → CH3 COOH(l)
..........................................
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TOPIC 3. CHEMICAL ENERGY
Q72:
When 3.6 g of butanal (relative formula mass = 72) was burned, 134 kJ of energy was
released.
From this result, what is the enthalpy of combustion of butanal?
A) - 2680 kJ mol-1
B) - 6.7 kJ mol-1
C) + 6.7 kJ mol-1
D) + 2680 kJ mol-1
..........................................
Q73:
Ethanol (C2 H5 OH) has a different enthalpy of combustion from dimethyl ether
(CH3 OCH3 ).
This is because the compounds have different:
A) products of combustion.
B) bonds within the molecules.
C) molecular masses.
D) boiling points.
..........................................
Q74:
The potential energy diagram for the reaction CO(g) + NO 2 (g) → CO2 (g) +
NO(g) is shown.
What is the ΔH for the reaction?
a) - 361 kJ mol-1
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TOPIC 3. CHEMICAL ENERGY
b) - 227 kJ mol-1
c) - 93 kJ mol-1
d) + 361 kJ mol-1
..........................................
The enthalpies of combustion of methanol, ethanol and propan-1-ol are given below.
Methanol CH 3 OH ΔHc = -726 kJ mol-1
Ethanol C2 H5 OH ΔHc = -1367 kJ mol-1
Propan-1-ol C3 H7 OH ΔHc = -2021 kJ mol-1
Q75:
Why is there a regular increase in enthalpies of combustion from methanol to ethanol to
propan-1-ol?
..........................................
Q76:
What would you predict the enthalpy of combustion of butan-1-ol to be?
..........................................
Q77:
Identify the endothermic reactions from the thermochemical equations below.
A) Ba(OH)2 (aq) + 2HCl(aq) → BaCl2 (aq) + 2H2 O(l) ; ΔH = -113.5 kJ mol-1
B) 3O2 (g) → 2O3 (g) ; ΔH = +142.7 kJ mol-1
C) NaOH(s) → Na+ (aq) + OH- (aq) ; ΔH = -42.7 kJ mol-1
D) C3 H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2 O(g) ; ΔH = -2219 kJ mol-1
E) NH4 Cl (s) → NH4 + (aq) + Cl- (aq) ; ΔH = +15.0 kJ mol-1
F) 2Na(s) + Cl2 (g) → 2NaCl(s) ; ΔH = -411.1 kJ mol-1
..........................................
Q78:
Identify the equations which apply to an enthalpy of solution.
A) Ba(OH)2 (aq) + 2HCl(aq) → BaCl2 (aq) + 2H2 O(l) ; ΔH = -113.5 kJ mol-1
B) 3O2 (g) → 2O3 (g) ; ΔH = +142.7 kJ mol-1
C) NaOH(s) → Na+ (aq) + OH- (aq) ; ΔH = -42.7 kJ mol-1
D) C3 H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2 O(g) ; ΔH = -2219 kJ mol-1
E) NH4 Cl (s) → NH4 + (aq) + Cl- (aq) ; ΔH = +15.0 kJ mol-1
F) 2Na(s) + Cl2 (g) → 2NaCl(s) ; ΔH = -411.1 kJ mol-1
..........................................
Q79:
Identify the reaction which would be best able to provide heat to an environment.
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TOPIC 3. CHEMICAL ENERGY
A) Ba(OH)2 (aq) + 2HCl(aq) → BaCl2 (aq) + 2H2 O(l) ; ΔH = -113.5 kJ mol-1
B) 3O2 (g) → 2O3 (g) ; ΔH = +142.7 kJ mol-1
C) NaOH(s) → Na+ (aq) + OH- (aq) ; ΔH = -42.7 kJ mol-1
D) C3 H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2 O(g) ; ΔH = -2219 kJ mol-1
E) NH4 Cl (s) → NH4 + (aq) + Cl- (aq) ; ΔH = +15.0 kJ mol-1
F) 2Na(s) + Cl2 (g) → 2NaCl(s) ; ΔH = -411.1 kJ mol-1
..........................................
A pupil wanted to measure the enthalpy of combustion of butane. She used a camping
stove to heat a pot of water and found that 10 g of butane burned to give out 258.6 kJ of
energy.
Q80:
Write a balanced equation to show the reaction which corresponds to the enthalpy of
combustion of butane.
..........................................
Q81:
Apart from the mass of the butane cylinder at the start and the end of the experiment,
state three measurements which the pupil would have made.
..........................................
Q82:
Calculate the experimental value for the enthalpy of combustion of butane.
Give your answer to the nearest kJ mol-1 , and remember the sign.
..........................................
Q83:
Enthalpy changes for a reaction which can be carried out by different routes must be:
A) exothermic.
B) opposite.
C) the same.
D) endothermic.
..........................................
The enthalpy of combustion of carbon monoxide is -283 kJ mol -1 .
Q84:
What is the enthalpy change if 2.8 g of carbon monoxide is burned?
..........................................
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TOPIC 3. CHEMICAL ENERGY
Q85:
What is the enthalpy change when 88 g of carbon dioxide is converted to CO and
oxygen?
..........................................
Q86:
Look at the enthalpy cycle diagram below.
What is the relationship between the enthalpy changes?
a)
b)
c)
d)
ΔH3
ΔH3
ΔH2
ΔH1
= ΔH1
= ΔH2
= ΔH3
= ΔH2
- ΔH2
+ ΔH1
- ΔH1
- ΔH3
..........................................
Q87: What is the enthalpy change for ΔH 3 ?
..........................................
The questions below refer to the following reaction.
Si2 H6 (g) → 2Si(s) + 3H2 (g), ΔH = -78 kJ
Q88:
What is the enthalpy change for the reaction below?
2Si2 H6 (g) → 4Si(s) + 6H2 (g)
..........................................
Q89:
What is the enthalpy change for this reaction?
Si(s) + 3 /2 H2 (g) → 1 /2 Si2 H6 (g)
..........................................
Using the information below:
a) 2Fe(s) + 3 /2 O2 (g) → Fe2 O3 (s), ΔH = - 827 kJ
b) 2Al(s) + 3 /2 O2 (g) → Al2 O3 (s), ΔH = - 1675 kJ
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TOPIC 3. CHEMICAL ENERGY
Q90:
Calculate the enthalpy change for the reaction:
2Al(s) + Fe2 O3 (s) → Al2 O3 (s) + 2Fe(s)
..........................................
Q91:
Given the enthalpy changes for the three reactions below:
a) O2 (g) → 2O(g), ΔH = +496 kJ
b) C(s) → C(g), ΔH = + 715 kJ
c) C(s) + O2 (g) → CO2 (g), ΔH = - 394 kJ
Calculate the enthalpy change for the reaction:
C(g) + 2O(g) → CO2 (g)
..........................................
In this diagram ΔH is the enthalpy change involved in the formation of one mole of
ethanol from carbon, hydrogen and oxygen. Other enthalpy changes involved in the
cycle are the enthalpies of combustion of carbon, hydrogen and ethanol, which are
available from the data booklet.
Q92:
What is the enthalpy value, X, for the reaction 2C(s) + aO 2 (g) → 2CO2 (g)?
..........................................
Q93: What is the enthalpy value, Y, for the reaction 3H 2 (g) + bO2(g) → 3H2 O(l)?
..........................................
Q94: What is the value of c in the reaction 2CO 2 (g) + 3H2 O(l) - cO2 (g) → C2 H5 OH(l)?
..........................................
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Q95: What is the enthalpy value, Z, for the reaction 2CO 2 (g) + 3H2 O(l) - cO2 (g) →
C2 H5 OH(l)?
..........................................
Q96: Hence, what is the enthalpy value, ΔH, for the reaction 2C(s) + 3H 2 (g) + 1/2 O2 (g)
→ C2 H5 OH(l)?
..........................................
Q97: The mean bond enthalpy of the N-H bond is equal to one third of the value of ΔH
for which change?
a)
b)
c)
d)
N(g) + 3H(g) → NH3 (g)
N2 (g) + 3H2 (g) → 2NH3(g)
0.5N2 (g) + 1.5H2 (g) → NH3 (g)
NH3 (g) → 0.5N2 (g) + 1.5H2 (g)
..........................................
Q98: The table below contains information about some diatomic molecules.
Bond
H-H
◦
Boiling point C -253
Bond point (kJ
436
mol-1 )
H-CI
-85
CI-CI
-35
Br-Br
59
432
243
194
Which of the diatomic molecules listed has the strongest covalent bond?
a)
b)
c)
d)
H-H
H-CI
CI-CI
Br-Br
..........................................
Q99:
Bond
Enthalpy (kJ mol-1 )
CI-CI
C-CI
C-H
H-CI
243
338
412
432
Chloromethane can be produced by the reaction of methane with chlorine.
CH4 (g) + Cl2 (g) → CH3 Cl(g) + HCl(g)
Using the information in the table above, calculate the enthalpy change, in
kJ mol-1 , for this reaction.
a)
b)
c)
d)
-3897
-115
+115
+3897
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..........................................
Q100: Use bond enthalpy values from the data book to calculate the enthalpy change
for the following reaction. (Hint: remember the sign.)
CH4 (g) + Br2 (g) → CH3 Br (g) + HBr (g)
..........................................
Q101: The data book gives the enthalpy of combustion of methane as -891 kJ mol -1
Use bond enthalpies to calculate the enthalpy change for this reaction. (Hint: remember
the sign.)
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2 O (g)
..........................................
Q102: Using bond enthalpy values, calculate the enthalpy change for the following
addition reaction.
C2 H4 (g) + HBr (g) → C2 H5 Br (g)
..........................................
..........................................
© H ERIOT-WATT U NIVERSITY
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Topic 4
Oxidising or reducing agents
Contents
4.1
Prior knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
132
4.2
4.3
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Elements as oxidising and reducing agents . . . . . . . . . . . . . . . . . . . .
133
134
4.4
4.3.1 Displacement reactions . . . . . . . . . . . . . . . . . . . . . . . . . . .
Molecules and group ions as oxidising and reducing agents . . . . . . . . . . .
138
139
4.5
4.4.1 Oxygen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Uses for strong oxidising agents . . . . . . . . . . . . . . . . . . . . . . . . . .
140
142
4.6
4.7
Ion-electron half equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Combining ion-electron equations . . . . . . . . . . . . . . . . . . . . . . . . .
145
148
4.8
4.7.1 Tutorial - simple ion-electron equations . . . . . . . . . . . . . . . . . .
Complex ion-electron equations . . . . . . . . . . . . . . . . . . . . . . . . . .
149
150
4.9
4.8.1 Tutorial - complex ion-electron equations . . . . . . . . . . . . . . . . .
Redox titrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
153
155
4.10 Summary cloze test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
159
161
4.12 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
162
4.13 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
163
Prerequisite knowledge
You should already know that:
•
balanced ionic equations can be written to show the reaction of metals with water,
oxygen and acids (National 5, Unit 3);
•
ion-electron equations can be written for electrochemical cells including those
involving non-metals (National 5, Unit 3);
•
combinations of these reactions form redox equations (National 5, Unit 3);
•
spectator ions can be described and identified from equations (National 5, Unit 3);
•
students should be able to explain the principles of and carry out an acid / base
titration (National 5, Unit 3).
130
TOPIC 4. OXIDISING OR REDUCING AGENTS
Learning objectives
By the end of this topic you should be able to state that:
Elements as oxidising or reducing agents
•
A redox reaction is a reaction in which reduction and oxidation occur together,
reduction being the gain of electrons by a reactant and oxidation being the loss of
electrons by a reactant in a reaction.
•
An oxidising agent is a substance which accepts electrons.
•
A reducing agent is a substance which donates electrons.
•
Oxidising and reducing agents can be identified in redox reactions.
•
Elements with low electronegativities (metals) tend to form ions by losing electrons
(oxidation) and so can act as reducing agents.
•
Elements with high electronegativities (non-metals) tend to form ions by gaining
electrons (reduction) and so can act as oxidising agents.
•
The strongest reducing agents are found in Group 1.
•
The strongest oxidising agents come from Group.
•
The electrochemical series indicates the effectiveness of oxidising and reducing
agents.
Compounds as oxidising or reducing agents
•
Compounds can also act as oxidising or reducing agents.
•
Electrochemical series contain a number of ions and molecules.
•
The dichromate and permanganate ions are strong oxidising agents in acidic
solutions whilst hydrogen peroxide is an example of a molecule which is a strong
oxidising agent.
•
Carbon monoxide is an example of a gas that can be used as a reducing agent.
•
Oxidising and reducing agents can be selected using an electrochemical series
from a data booklet or can be identified in the equation showing a redox reaction.
Use of oxidising agents
•
Oxidising agents are widely employed because of the effectiveness with which
they can kill fungi and bacteria, and can inactivate viruses.
•
The oxidation process is also an effective means of breaking down coloured
compounds making oxidising agents ideal for use as ‘bleach’ for clothes and hair.
Ion-electron equations
© H ERIOT-WATT U NIVERSITY
TOPIC 4. OXIDISING OR REDUCING AGENTS
•
Oxidation and reduction reactions can be represented by ion-electron equations.
•
When molecules or group ions are involved, if the reactant and product species
are known, a balanced ion-electron equation can be written by adding appropriate
numbers of water molecules, hydrogen ions and electrons.
•
Ion-electron equations can be combined to produce redox equations.
Practical applications
•
Displacement reactions are example of redox reactions and oxidising and reducing
agents can be identified in these and other redox reactions.
•
The technique of titration can be applied to redox reactions, allowing the
concentration of a reactant to be calculated from results of volumetric titrations.
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4.1
Prior knowledge
Test your prior knowledge
Q1:
Go online
a)
b)
c)
d)
Reduction can be defined as:
gain of electrons.
loss of electrons.
loss in mass of a substance.
gain in mass of a substance.
..........................................
Q2: Potassium iodide will react with lead nitrate to produce a precipitate of lead iodide.
Which of the following best represents this?
a)
b)
c)
d)
2Pb+(aq) + 2NO3 - (aq) + 2K+ (aq) + 2l- (aq) → 2Pb+(l- )2 (s) + 2K+ (aq) + 2NO3 - (aq)
Pb+ (aq) + 2NO3 - (aq) + 2K+ (aq) + 2l- (aq) → Pb2+ (l- )2 (aq) + 2K+ (aq) + 2NO3 - (aq)
Pb2+ (aq) + 2NO3 - (aq) + 2K+ (aq) + 2l- (aq) → Pb2+ (l- )2 (s) + 2K+ (aq) + 2NO3 - (aq)
Pb2+ (NO3 - )2 (aq) + 2K+ l- (aq) → Pb2+ (l- )2 (s) + 2K+ NO3 - (aq)
..........................................
Q3:
Combine the following reactions to give the overall redox reaction.
Oxidation H2 O2 (l) → O2 (g) + 2H+ (aq) + 2eReduction 5e- + MnO4 - (aq) + 8H+ (aq) → Mn2+ (aq) + 4H2 O(l)
a) H2 O2 (l) + 5e- + MnO4 - (aq) + 8H+ (aq) → Mn2+ (aq) + 4H2 O(l) + O2 (g) + 2H+ (aq) + 2eb) H2 O2 (l) + MnO4 - (aq) + 8H+ (aq) → Mn2+ (aq) + 4H2 O(l) + O2 (g) + 2H+ (aq)
c) 5H2 O2 (l) + 10e- + 2MnO4- (aq) + 16H+ (aq) → 2Mn2+(aq) + 8H2 O(l) + 5O2 (g) +
10H+(aq) + 10ed) 5H2 O2 (l) + 2MnO4 - (aq) + 16H+ (aq) → 2Mn2+ (aq) + 8H2 O(l) + 5O2 (g) + 10H+(aq)
e) 5H2 O2 (l) + 2MnO4 - (aq) + 6H+ (aq) → 2Mn2+(aq) + 8H2 O(l) + 5O2 (g)
..........................................
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4.2
Introduction
The term 'redox' is an abbreviation of the chemical reactions commonly known as
reduction and oxidation. These processes occur together and include many common
reactions such as the burning of fossil fuels, the corrosion of metals and photosynthesis.
Rusting is a common redox reaction in
which iron is converted into iron oxides.
Photosynthesis is another common redox
reaction, involving both reduction and
oxidation stages.
An early definition of reduction describes the removal of oxygen from metal oxides. The
metal oxide is said to be reduced to the metal. An example is iron(III) oxide being
reduced to iron.
Iron oxide reduction
An early definition of oxidation describes the addition of oxygen to a metal forming a
metal oxide. A familiar example is the rusting of iron to give iron(III) oxide. In this case
the iron metal is being oxidised.
Iron oxidation
Iron(III) oxide consists of Fe3+ and O2- ions. Close analysis of the above reaction
shows that the iron atoms have had electrons removed from them, and these electrons
are transferred to the oxygen.
The iron has been oxidised and the oxygen has been reduced. The changes can be
shown clearly by writing ion-electron half equations for each reaction.
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Iron oxide half equations
Notice that the equations have been balanced by ensuring that the number of atoms on
each side of the arrow is the same and that the number of electrons exchanged
between the two half equations is the same.
A useful memory aid for the definitions of oxidation and reduction is provided by the
first letters in the phrase "OIL RIG" (see below).
Oxidation is Loss of electrons.
Reduction is Gain of electrons.
4.3
Elements as oxidising and reducing agents
The corrosion of iron results in the formation of iron(III) oxide (see below).
Iron oxidation 2
As already seen, this reaction can be represented by two ion-electron equations (see
below).
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Iron oxide half equations 2
The iron metal is being oxidised by the oxygen. Another way of expressing this is to say
that the oxygen is acting as an oxidising agent. The oxygen is receiving electrons
from the iron. An oxidising agent is therefore an electron acceptor.
In a similar way, the oxygen molecules are being reduced by the iron. The iron is acting
as a reducing agent. The iron is donating electrons to the oxygen. A reducing agent is
therefore an electron donor.
Sometimes there are ions present which do not react in the redox reaction. These ions
are present in the reactants and are still present in the same amount when the
products have formed. Such ions are known as spectator ions and are not normally
shown in the balanced redox equation.
The Electrochemical Series (also known as the Table of Standard Reduction Potentials)
can be found in your SQA Data Book and can give an indication of the strength of
oxidising and reducing agents. The easiest reductions occur at the foot of the series, left
to right. The best oxidising agents are therefore found at the bottom left side. Similarly
the easiest oxidations occur at the top of the series, right to left . This is summarised
below.
Reducing and oxidising agents
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Example exercise
When bromine is added to potassium iodide solution, iodine is displaced and
potassium bromide solution is formed (see below).
Potassium bromide formation
The two potassium ions are unchanged in this reaction. These are spectator ions, and
ignoring these ions results in the ion electron equations as shown below.
Iodide and bromine half equations
Q4: Is the first ion-electron equation (see above - Iodide and bromine half equations)
an oxidation or a reduction?
..........................................
Q5:
a)
b)
c)
d)
Which of these is acting as an oxidising agent?
Iodide
Iodine
Bromine
Bromide
..........................................
Q6:
Name the chemical species acting as a reducing agent.
..........................................
Q7: Where in the Electrochemical Series in the data booklet will you find the reaction
which is most likely to go in the opposite direction?
a) Top of the table.
b) Bottom of the table.
..........................................
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Q8: Use this equation showing a redox change in this and the next two questions.
Name the reactant being reduced.
..........................................
Q9: Name the reactant acting as an electron donor.
..........................................
Q10: Name the oxidising agent.
..........................................
Q11: Which of the following is the best oxidising agent?
a)
b)
c)
d)
e)
f)
Na (s)
Br2 (1)
Sn2+ (aq)
I- (aq)
Zn (s)
Cu2+ (aq)
..........................................
Q12: Which of the following is the best reducing agent?
a)
b)
c)
d)
e)
f)
Na (s)
Br2 (1)
Sn2+ (aq)
I- (aq)
Zn (s)
Cu2+ (aq)
..........................................
Q13: Which substance could be used as an oxidising agent or a reducing agent?
a)
b)
c)
d)
e)
f)
Na (s)
Br2 (1)
Sn2+ (aq)
I- (aq)
Zn (s)
Cu2+ (aq)
..........................................
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4.3.1
Displacement reactions
Displacement reactions
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A strip of zinc metal is placed into copper(II) sulfate solution. The zinc ionises and
passes into solution, and copper ions plate onto the surface of the strip as copper
atoms, replacing the zinc. The solution loses its blue colour since the resulting zinc
sulfate solution is colourless.
Zinc metal
Copper Sulfate solution
2+
2+
Cu
SO4
2+
SO4
Zn
2+
2+
SO
Zn
Zn
2+
Cu
Zn
2+
2+
SO4
Zn
Cu
2+
Cu
2+
SO4
2+
Zn
2+
Cu
Zn
Zn
Cu
2+
Cu
2+
4
SO
2+
Cu
SO4
Zn2+
SO4
2+
Zn
Cu
SO4
2+
Zn
Cu
2+
4
2+
Cu
2+
4
SO
Cu
2+
4
SO
2+
SO4
Zn2+
Zinc metal
Copper metal
Zinc Sulfate solution
Q14: Name the spectator ion in this reaction.
..........................................
Q15: Complete the ion-electron half equation for the oxidation taking place:
Zn →
..........................................
Q16: Complete the ion-electron half equation for the reduction taking place:
→ Cu
..........................................
Q17: Which of these is acting as an oxidising agent?
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a)
b)
c)
d)
Zinc atoms
Zinc ions
Copper atoms
Copper ions
..........................................
Q18: Name the reactant acting as an electron donor.
..........................................
Key point
An oxidising agent is a substance which accepts electrons; a reducing agent is
a substance which donates electrons. Oxidising and reducing agents can be
identified in redox reactions.
..........................................
4.4
Molecules and group ions as oxidising and reducing
agents
You might have noticed that as well as elements being listed in the table of Standard
Reduction Potentials, there are some compounds. Dichromate ion (Cr 2 O7 2- ) and
permanganate ion (MnO 4 - ) are commonly used as oxidising agents in chemical
reactions. You will study the reactions of permanganate and dichromate ions in more
detail in a later section.
Q19: Can you think of a practical reason why sodium dichromate or potassium
permanganate are frequently used as oxidising agents in chemistry as opposed to the
elements studied so far?
..........................................
Q20:
For example, a traditional method for producing chlorine is the reaction between
concentrated hydrochloric acid and potassium permanganate.
The formulae for the reactants and products are:
KMnO4 + HCl → KCl + MnCl2 + H2 O + Cl2
It is quite difficult, but can you balance this equation?
..........................................
Q21: Remembering that KMnO 4 consists of K+ and MnO4 - ions and HCl is also ionised
in solution, what is the oxidising agent in this reaction, and what are the reducing
agents?
..........................................
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Q22: Look carefully at the reactants and products again. What are the products from
these oxidising and reducing agents?
..........................................
Q23: Finally, are there any spectator ions in this reaction?
..........................................
4.4.1
Oxygen
Almost all life on Earth depends on oxygen to provide energy. When the Earth was
formed 4.5 billion (4.5 x 10 9 ) years ago* it was a very different place.
As the hot fireball cooled it formed a solid crust, volcanoes belched out gases: steam,
carbon dioxide and ammonia. These formed the oceans and an atmosphere with no
oxygen.
We now have an atmosphere consisting of about 21% oxygen. The rest is mainly
nitrogen. Here is our beautiful blue planet.
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The change was brought about somewhere around 2.7 to 2.2 billion years ago by vast
numbers of cyanobacteria (blue-green algae) in the oceans.
These organisms used the energy of sunlight in photosynthesis to convert the abundant
supplies of carbon dioxide and water into carbohydrates and oxygen. This dramatic
change in atmospheric composition fuelled the abundant numbers of plant and animal
species which now use oxygen to provide energy for life.
Q24: A simple equation for photosynthesis, making glucose, C 6 H12 O6 , is:
6CO2 + 6H2 O + light energy → C6 H12 O6 + 6O2
Can you see in this case what is oxidised and what is reduced?
..........................................
Q25: When we humans use glucose to provide energy for life, the equation is simply
reversed:
C6 H12 O6 + 6O2 → 6CO2 + 6H2 O + energy
The atmospheric concentration of carbon dioxide measured at Mauna Loa, Hawaii, has
risen from 315 ppmv in 1960 to 390 ppmv in 2010. ppmv is parts per million by volume;
μ CO2 per atmosphere. Assuming the 'effective volume' of Earth's atmosphere is 4 x
1020 , what volume of additional carbon dioxide has the atmosphere contained in going
from 1960 to 2010?
..........................................
[* Evidence for the age of the Earth comes mainly from studying the ages of rocks using
a variety of radioactive decay data.]
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4.5
Uses for strong oxidising agents
The main domestic and industrial uses of strong oxidising agents are as biocides and
bleaches.
Sodium chlorate(l) solution This weedkiller contains
is a powerful germicide and sodium chlorate(lll)
bleach.
The original washing
detergents contained
perborates, percarbonates
and silicates, from which
Persil derived its name.
Biocides
Biocides are used for killing fungi and bacteria, and inactivating viruses. The
main oxidising agents used are: chlorine, hypochlorite, chlorine dioxide, iodine, and
permanganate.
Disinfection during a foot-and-mouth
outbreak using hypochlorite
Disinfection with iodine during a bone
marrow donation
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Bleaches
These are used for bleaching textiles, paper and hair. Common oxidising agents include:
sulfite, chlorine and peroxides.
Perborate is used for adding extra whiteness to laundry and dibenzoyl peroxide for
treating spots and acne.
The original washing detergents contained
perborates, percarbonates and silicates,
from which Persil derived its name.
Gel for acne
Water purification
Water from reservoirs is purified considerably before it is pumped into buildings for
consumption. As well as removing particulate matter there are possible pathogens to be
removed. These comprise viruses, bacteria, including Escherichia coli, Campylobacter
and Shigella, and protozoa, including Giardia lamblia and other cryptosporidia.
The most common disinfection method involves some form of chlorine or its compounds
such as chloramine or chlorine dioxide. Chlorine dioxide is a faster-acting disinfectant
than elemental chlorine, however it is relatively rarely used, because in some
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circumstances it may create excessive amounts of chlorite.The use of chloramine is
becoming more common as a disinfectant. Although chloramine is not as strong an
oxidant, it does provide a longer-lasting residual than free chlorine and it won't form
undesirable by-products. Ozone (O 3 ) is an unstable molecule which readily gives up one
atom of oxygen providing a powerful oxidizing agent which is toxic to most waterborne
organisms. It is a very strong, broad spectrum disinfectant that is widely used in Europe.
For disinfection of small quantities of water when camping, for example, tablets which
produce hypochlorite (chlorate(I)) in situ can be used.
Propellants
A further use for strong oxidants is as propellants for space rockets. These include
hydrogen peroxide and liquid oxygen.
For example, the Apollo-Saturn V first stage uses kerosene-liquid oxygen, the upper
stages use liquid hydrogen-liquid oxygen.
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4.6
Ion-electron half equations
When sodium burns in chlorine the product, sodium chloride, is an ionic compound.
The reaction is a redox reaction and can be described by writing two ion-electron
equations. The first shows the sodium (see below).
Sodium ionising
This ion-electron equation can have its charge balanced by showing the two electrons
lost from the sodium atom on the right hand side of the equation (see below).
Sodium ionising 2
Similarly, the chlorine ion-electron half equation can be balanced by showing the gain
of two electrons by the chlorine on the left of the equation (see below).
Chlorine half equations
Practise writing ion-electron equations in this next activity.
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Writing ion-electron equations
Go online
Complete these oxidation and reduction half equations, then compare your answers
with those at the end.
Q26: Complete the ion electron half equation for the oxidation of potassium, forming a
K+ ion:
K→
..........................................
Q27: The oxidation of magnesium:
Mg →
..........................................
Q28: The oxidation of aluminium:
Al →
..........................................
Q29: Complete the ion-electron equation for the reduction of a fluorine atom to a
fluoride ion:
→F..........................................
Q30: The reduction of a nitrogen atom to nitride ion:
→ N3..........................................
Q31: The reduction of an oxygen atom to oxide ion:
→ O2..........................................
Many ion-electron equations do not have to be worked out as they can be found in a
table known as the "Electrochemical Series: Standard Reduction Potentials". This is
available in most chemistry texts or in data booklets.
The oxidation ion-electron equation which applies can be obtained by reversing the
reduction equation given in the table.
Simple rules
In a redox reaction a simple rule applies when determining which equation will be the
reduction and which the oxidation.
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The ion-electron equation appearing lowest in the electrochemical series will go as
written (i.e. as a reduction), and this reaction will be able to force any ion-electron
equation higher up in the table to go in reverse (i.e. as an oxidation). This can be
described as the 'Anticlockwise Rule' (see below).
Anticlockwise rule diagram
Use the diagram (above) and the Electrochemical Series in your SQA Data Booklet to
answer the following questions.
Q32: In the Electrochemical Series shown, where will the easiest reduction be found?
a) At the top of the table.
b) At the bottom of the table.
..........................................
Q33: In the Electrochemical Series shown, where will the easiest oxidation be found?
a)
b)
c)
d)
At the top of the table, going "as written".
At the bottom of the table, going "as written".
At the top of the table, in reverse.
At the bottom of the table, in reverse.
..........................................
Q34: Which of the following reductions is the easiest?
a)
b)
c)
d)
Na+ (aq) + e- → Na(s)
F2 (g) + 2e- → 2F- (aq)
Br2 (g) + 2e- → 2Br- (aq)
Cu2+ (aq) + 2e- → Cu(s)
..........................................
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Q35: Which of these reactions would be reversed by the following reduction?
2H+ (aq) + 2e- → H2 (g)
a)
b)
c)
d)
Na+ (aq) + e- → Na(s)
F2 (g) + 2e- → 2F- (aq)
Br2 (g) + 2e- → 2Br- (aq)
Cu2+ (aq) + 2e- → Cu(s)
..........................................
Key point
Ion-electron equations can be written for oxidation and reduction reactions. Many
ion-electron equations can be obtained from the Electrochemical Series available
in a suitable Data Booklet.
4.7
Combining ion-electron equations
The balanced redox equation for a chemical change can be worked out from the
ion-electron equations which make up the oxidation and reduction stages. Often these
can be found in the table of standard reduction potentials: the electrochemical series.
This is available in most chemistry texts or in data booklets.
The oxidation ion-electron equation which applies can be obtained by reversing the
reduction equation given in the table.
Writing the balanced redox equation can best be illustrated by use of an example.
Example Problem: Sodium reacting with chlorine
When sodium is placed into chlorine gas, the product, sodium chloride, is an ionic
compound. Write a balanced redox equation from the two ion-electron equations.
Solution
Consider the two reactants (sodium and chlorine). The ion-electron equation appearing
lowest in the electrochemical series will go as written (i.e. as a reduction). This is
described in the 'Anticlockwise Rule' (see above).
Chlorine to chloride
In this reaction (above) the chlorine is the oxidising agent and will be able to force any
of the ion-electron equations higher up in the table to go in reverse (i.e. as an
oxidation). In this case sodium will be oxidised (see below).
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149
Sodium oxidation
Combining the individual ion-electron equations into the redox equations requires a
number of "balancing" stages.
..........................................
Balancing redox equations
The diagram below shows the stages required.
Go online
Balancing redox equations
..........................................
4.7.1
Tutorial - simple ion-electron equations
Tutorial - simple ion-electron equations
Use the electrochemical series to try these examples. In each case:
Go online
•
write the ion-electron equation for the reduction.
•
write the ion-electron equation for the oxidation.
•
combine these to form the redox equation.
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Q36: When magnesium is placed into chlorine gas, the product, magnesium chloride,
is an ionic compound.
..........................................
Q37: When chlorine reacts with a solution containing iodide ions, iodine is one of the
products.
..........................................
Q38: Magnesium reacts with sulfuric acid to give hydrogen gas and magnesium sulfate
solution. The spectator ions can be ignored when writing the redox equation.
..........................................
Q39: Aluminium displaces silver ions from a solution of silver(I) nitrate, giving aluminium
nitrate solution and silver. The spectator ions can be ignored when writing the redox
equation.
..........................................
Key point
Ion-electron equations can be combined to produce redox equations.
4.8
Complex ion-electron equations
Many of the more complex ion-electron equations involve hydrogen ions and/or water
molecules in the chemical change. This is why reagents like "acidified" potassium
permanganate or "acidified" potassium dichromate (see below) require an acid solution
in which to act in redox reactions.
Complex ion-electron equations
Not all of these more complex ion-electron equations are listed in data booklets but,
given the reactant and product species, ion-electron equations involving H + (aq) and
H2 O(l) can be written.
Example
Acidified dichromate ions, Cr2 O7 2- , can be used to oxidise Fe 2+ ions. The Fe2+ ions are
changed to Fe 3+ ions and the dichromate to chromium(III), Cr 3+ .
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Write ion-electron equations for each reaction and combine these to give a redox
equation for the change.
Complex equations
Work out the balanced ion-electron equation for this next example using the steps
(below) and then answer the questions.
Complex equations
The oxidation step involves the Fe 2+ ions being changed into Fe 3+ ions.
The other steps in the diagram are:•
STEP 1 The reduction given in the question can be written and the number of
atoms in the given ions balanced.
•
STEP 2 Balance the number of oxygens by adding the correct number of water
molecules.
•
STEP 3 Balance the number of hydrogen atoms by adding the correct number of
hydrogen ions.
•
STEP 4 Balance the charge on each side of the equation by adding electrons to
the most positive side.
Writing the full redox equation requires the oxidation and reduction ion-electron
equations to be combined .
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Balance the number of electrons being transferred between the ion-electron equations
and add them to give the redox equation.
..........................................
Example
Fe2+ ions can be changed into Fe 3+ ions in reaction with hypochlorite ions. The
hypochlorite ion is converted to chloride (Figure 4.1).
Figure 4.1: hypochlorite reduction
..........................................
Write a balanced ion-electron equation for this change (Figure 4.1) using the steps
shown above then answer the questions.
Q40: How many water molecules have to be added to the product side?
..........................................
Q41: How many hydrogen ions have to be added to the reactant side?
..........................................
Q42: How many electrons have to be added to balance the charge?
..........................................
Q43: On which side do the electrons get added?
a) Reactant
b) Product
..........................................
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Q44: Is the completed ion-electron equation an oxidation or a reduction?
a) Oxidation
b) Reduction
..........................................
Q45: Write a complete balanced redox equation for the reaction between hypochlorite
and Fe2+ .
..........................................
4.8.1
Tutorial - complex ion-electron equations
Tutorial - complex ion-electron equations
In the first example, write a balanced ion-electron equation for the given change and
then answer the questions which take the process step by step.
Example 1
V3+ ions can be changed into VO 3- ions as part of a redox reaction. Write a balanced
ion-electron equation for the change and then answer the questions which take the
process step by step.
Q46: How many water molecules have to be added to the reactant side?
..........................................
Q47: How many hydrogen ions have to be added to the product side?
..........................................
Q48: How many electrons have to be added to balance the charge?
..........................................
Q49: On which side do the electrons get added?
a) Reactant
b) Product
..........................................
Q50: Is the completed ion-electron equation an oxidation or a reduction?
a) Oxidation
b) Reduction
..........................................
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TOPIC 4. OXIDISING OR REDUCING AGENTS
In the second example, write a balanced ion-electron equation for the given change and
then check your answer at the end.
Example 2
Q51:
ClO3 - ions can be changed into Cl 2 molecules as part of a redox reaction. Write a
balanced ion-electron equation for the change. (Hint: balance the number of chlorine
atoms first).
..........................................
Use the electrochemical series in the data booklet to help with these examples. In each
case:
•
write the ion-electron equation for the reduction.
•
write the ion-electron equation for the oxidation.
•
combine these to form the redox equation.
Q52: When zinc is placed into sulfuric acid, the products are hydrogen gas and zinc
sulfate solution.
..........................................
Q53: In acid solution, potassium permanganate oxidises iron(II) to iron(III).
..........................................
Q54: In dilute nitric acid solution, copper metal is oxidised to copper(II) ions. The nitrate
ion is reduced to nitrogen monoxide gas.
..........................................
..........................................
Key point
Given reactant and product species, ion-electron equations which include H + (aq)
and H2 O(l) can be combined to produce redox equations.
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4.9
Redox titrations
Any balanced chemical equation can be used to calculate quantities of chemicals
involved in the reaction. Knowing the quantity of one reactant can allow calculation of the
quantity of another, as long as the end-point of the reaction can be observed. Titration
of a solution of an oxidising agent against a reducing agent is known as a redox titration
and, if a suitable indicator is available, can be used to determine the concentration of
unknown reagents.
For example: When acidified dichromate ions, Cr 2 O7 2- , are used to oxidise Fe2+ ions.
The Fe2+ ions are changed to Fe 3+ ions and the dichromate to chromium(III), Cr 3+ . The
redox equation is given below. This equation was fully developed in the example given
in the previous page.
Iron(III) dichromate redox
The balanced redox equation (above) shows that 1 mole of dichromate ions (Cr 2 O7 2- )
is able to oxidise 6 moles of iron(II) ions (Fe2+ ).
Example : Redox titration example
Individual 20 cm3 portions of acidified Fe2+ solution of unknown concentration are
titrated with 0.02 mol l -1 potassium dichromate solution, using a suitable indicator.
Ignoring the first rough titration, the average volume of dichromate solution used was
found to be 14.5 cm 3 .
The results of the experiment and the balanced redox equation can be used to work
out the concentration of the unknown Fe 2+ solution.
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..........................................
This type of calculation will be used in the next activity and in the tutorial examples.
Estimation of vitamin C in fruit juices
Go online
Vitamin C (formula C6 H8 O6 ) is a vital dietary component found in fruit and vegetables.
A lack of this vitamin in the diet leads to scurvy, a potentially fatal disease. The
concentration of vitamin C in a sample can be determined in a redox titration using
an iodine solution of known concentration and starch solution as indicator (see below).
Vitamin C titration equations
The on-line version of this activity contains a simulation experiment titrating three
different fruit juices against a known concentration of iodine. If you do not have access
to the on-line material, study this diagram (above) and text before trying the questions.
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A 20 cm3 portion of orange juice is pipetted into a flask and a few drops of starch
solution added. Iodine solution of known concentration is added from the burette until,
at the end point, the mixture turns a permanent blue/black colour. The first titration
value (rough) is discarded and two or more further titrations are done until two volumes
within 0.2 cm3 of each other are obtained. An average of these concordant titrations is
noted and the process repeated for the other two types of fruit juice. The table of
results is shown.
GRAPE
FRUIT
Vitamin C titration
Q55: Which fruit juice required the largest volume of iodine solution and has the highest
concentration of vitamin C?
..........................................
Q56: What volume of iodine solution (to 1 decimal place) was required to oxidise the
vitamin C in each titration of lemon juice?
..........................................
Q57: Given that the concentration of iodine solution used was 0.005 mol l -1 , calculate
the number of moles of iodine used in each titration of the lemon juice.
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..........................................
Q58: Using the balanced redox equation for the reaction (see 'Vitamin C titration
equations' above), which of these ratios shows the ratio of vitamin C to iodine?
a)
b)
c)
d)
1:1
1:2
2:1
6:2
..........................................
Q59: Given that the volume of iodine solution used in each titration was 20.0 cm 3 , use
your answer to the last question to calculate the concentration of vitamin C in the lemon
juice in mol l -1 .
..........................................
Q60: One mole of vitamin C has a molar mass of 176 g mol -1 .
concentration of the lemon juice solution in mg l -1 .
Calculate the
..........................................
Q61: The recommended daily allowance (RDA) of vitamin C suggested as being
necessary to maintain good health, is 60 mg. What volume of lemon juice would be
required to meet this requirement?
..........................................
Tutorial - calculations in redox titrations
Go online
Q62: Return to the titration diagram and calculate the concentration of vitamin C in the
orange juice in mol l -1
..........................................
Q63: One mole of vitamin C has a molar mass of 176 g mol -1 .
concentration of the orange juice solution in mg l -1 .
Calculate the
..........................................
Q64: The recommended daily allowance (RDA) of vitamin C suggested as being
necessary to maintain good health, is 60 mg. What volume of orange juice would be
required to meet this requirement?
..........................................
A worked solution to the next three questions can be accessed at the back of the book.
Try to work out your answers on paper before accessing the answer at the end.
Q65: The water in swimming pools can be kept sterile by adding chlorine, which kills
micro-organisms. Chlorine levels in swimming pools can be monitored by titration with
iron(II) sulfate. The related redox equation is shown below.
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Iron(II) chlorine redox
A 1000 cm3 sample of water from a swimming pool which is being heavily chlorinated
before use by the public required 15 cm 3 of iron(II) sulfate of concentration 0.6 mol l -1
to react completely.
Calculate the number of moles of iron(II) sulfate used in the titration. Give your answer
to three decimal places.
..........................................
Q66: Calculate the concentration of chlorine in the swimming pool in mol l -1 . Give your
answer to four decimal places.
..........................................
Q67: Calculate the concentration of chlorine in the swimming pool in mg l -1 . Give your
answer to one decimal place.
..........................................
Q68: Present your working in a tidy fashion. An example is shown at the end.
..........................................
..........................................
Key point
The concentration of a reactant can be calculated from results of volumetric
titrations.
4.10
Summary cloze test
Summary exercise
Online there is an interactive exercise summarising this topic. If you do not have access
to the web, here is a paper version.
Q69: The thermit reaction is used to generate molten iron, for example to weld railway
lines in situ.
The reactants are iron(iii) oxide and aluminium powder. The equation for the reaction is:
a) Fe2 O3 + Al → Fe + Al2 O3
b) Fe2 O3 + 2Al → 2Fe + Al2 O3
c) 3FeO + 2Al →3 Fe + Al 2 O3
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..........................................
Q70: What word can be used to describe the oxide ion (O 2- )?
a) Common
b) Watching
c) Spectator
..........................................
Q71: The ionic equation can be written as:
a) Fe + Al → Fe + Al
b) Fe+ + Al → Fe + Al+
c) Fe3+ + Al → Fe + Al3+
..........................................
Q72: Oxidation is defined as . . . . . . of electrons.
a) change
b) loss
c) gain
..........................................
Q73: In this reaction what is oxidised?
a)
b)
c)
d)
Al
Fe
Al3+
Fe3+
..........................................
Q74: What is acting as the oxidising agent?
a)
b)
c)
d)
Al
Fe
Al3+
Fe3+
..........................................
Q75: Using the electrochemical series, which metal could not be used to reduce Fe 3+
ion to Fe?
a) Copper
b) Magnesium
c) Manganese
..........................................
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TOPIC 4. OXIDISING OR REDUCING AGENTS
Q76: You are required to make a mixture of aluminium (Al) and iron(III) oxide (Fe 2 O3 )
to use for this reaction. Use your data booklet to calculate how much Fe 2 O3 is required
to add to 1 kg of Al to make a reaction mixture.
a) 1 kg
b) 2 kg
c) 3 kg
..........................................
4.11
Summary
Summary
You should be able to state that:
Elements as oxidising or reducing agents
•
A redox reaction is a reaction in which reduction and oxidation occur
together, reduction being the gain of electrons by a reactant and oxidation
being the loss of electrons by a reactant in a reaction.
•
An oxidising agent is a substance which accepts electrons.
•
A reducing agent is a substance which donates electrons.
•
Oxidising and reducing agents can be identified in redox reactions.
•
Elements with low electronegativities (metals) tend to form ions by losing
electrons (oxidation) and so can act as reducing agents.
•
Elements with high electronegativities (non-metals) tend to form ions by
gaining electrons (reduction) and so can act as oxidising agents.
•
The strongest reducing agents are found in Group 1.
•
The strongest oxidising agents come from Group 7.
•
The electrochemical series indicates the effectiveness of oxidising and
reducing agents.
Compounds as oxidising or reducing agents
•
Compounds can also act as oxidising or reducing agents.
•
Electrochemical series contain a number of ions and molecules.
•
The dichromate and permanganate ions are strong oxidising agents in
acidic solutions whilst hydrogen peroxide is an example of a molecule which
is a strong oxidising agent.
•
Carbon monoxide is an example of a gas that can be used as a reducing
agent.
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Summary continued
•
Oxidising and reducing agents can be selected using an electrochemical
series from a data booklet or can be identified in the equation showing a
redox reaction.
Use of oxidising agents
•
Oxidising agents are widely employed because of the effectiveness with
which they can kill fungi and bacteria, and can inactivate viruses.
•
The oxidation process is also an effective means of breaking down coloured
compounds making oxidising agents ideal for use as ‘bleach’ for clothes and
hair.
Ion-electron equations
•
Oxidation and reduction reactions can be represented by ion-electron
equations.
•
When molecules or group ions are involved, if the reactant and product
species are known, a balanced ion-electron equation can be written
by adding appropriate numbers of water molecules, hydrogen ions and
electrons.
•
Ion-electron equations can be combined to produce redox equations.
Practical applications
•
Displacement reactions are example of redox reactions and oxidising and
reducing agents can be identified in these and other redox reactions.
•
The technique of titration can be applied to redox reactions, allowing the
concentration of a reactant to be calculated from results of volumetric
titrations.
4.12
Resources
Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN
978-1444167528
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4.13
163
End of topic test
End of topic 4 test
This end of topic test is available online. If you do not have access to the internet, here
is a paper version.
Q77: Which of the following reductions is the easiest?
a)
b)
c)
d)
Cl2 + 2e- → 2Cl2H+ + 2e- → H2
Fe3+ + e- → Fe2+
Na+ + e- → Na
..........................................
Q78: An oxidising agent is a substance which
a)
b)
c)
d)
is oxidised in a reaction.
accepts electrons.
donates electrons.
causes reduction.
..........................................
Q79: Which of these is the equation for oxidation of bromide ions?
a)
b)
c)
d)
2Br- + 2e- → Br2
2Br- → Br2 + 2eBr2 + 2e- → 2BrBr2 → 2Br- + 2e..........................................
Q80: In which of these reactions is the reactant a reducing agent?
a)
b)
c)
d)
Fe3+ + 3e- → Fe
Fe2+ → Fe3+ + eH+ + HO- → H2 O
2H2 O + 2e- → H2 + 2OH..........................................
Q81: Which of these is the equation for the reduction of copper ion?
a)
b)
c)
d)
Cu2+ + 2e- → Cu
Cu2+ → Cu + 2eCu + 2e- → Cu2+
Cu → Cu2+ + 2e..........................................
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Q82: Consider the following two reactions.
Cl2 + 2e- → 2ClI2 + 2e- → 2IWhich of these reactions will take place in solution?
a)
b)
c)
d)
Chlorine will react with iodine.
Iodide ions will react with chloride ions.
Iodide ions will react with chlorine.
Chloride ions will react with iodine.
..........................................
Q83: The following reduction fits into the electrochemical series: standard reduction
potentials at a value of E o = + 0.68 V.
O2 + 2H+ + 2e- H2 O2
Which of these chemicals would convert H 2 O2 into O2 ?
a)
b)
c)
d)
Sn4+
Fe2+
Sn2+
Fe3+
..........................................
Q84: How many electrons have to be added to balance this incomplete ion-electron
equation?
Hg2 Cl2 → 2Hg + 2Cla)
b)
c)
d)
Add 1 electron to the left hand side.
Add 2 electrons to the right hand side.
Add 2 electrons to the left hand side.
Add 1 electron to the right hand side.
..........................................
Q85: Consider the following two oxidations.
Fe2+ → Fe3+ + eSn2+ → Sn4+ + 2eWhich of these reactions is likely to take place in solution?
a)
b)
c)
d)
Tin(IV) ions will react with iron(II) ions.
Tin(IV) ions will react with iron(III) ions.
Iron(II) ions will react with tin(II) ions.
Iron(III) ions will react with tin(II) ions.
..........................................
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Q86: In the equation below, highlight the chemical acting as a reducing agent.
..........................................
Q87: In the equation below, highlight the chemicals which are necessary to act as an
oxidising agent.
..........................................
Q88: Which of the following contains the best oxidising agent?
a)
b)
c)
d)
e)
f)
Cs+ (aq)
Br- (aq)
Ca (s)
Cl2 (g)
Fe2+ (aq)
Sn (s)
..........................................
Q89: Which of the following contains the best reducing agent?
a)
b)
c)
d)
e)
f)
Cs+ (aq)
Br- (aq)
Ca (s)
Cl2 (g)
Fe2+ (aq)
Sn (s)
..........................................
Q90: Which of the following contains a substance which could be used as an oxidising
agent or a reducing agent?
a)
b)
c)
d)
e)
f)
Cs+ (aq)
Br- (aq)
Ca (s)
Cl2 (g)
Fe2+ (aq)
Sn (s)
..........................................
Look at this incomplete ion-electron equation.
XeO3 → Xe
Q91: How many water molecules have to be added to the product side?
..........................................
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Q92: How many hydrogen ions have to be added to the reactant side?
..........................................
Q93: How many electrons have to be added to balance the charge?
..........................................
Q94: On which side do the electrons get added?
a) Reactant
b) Product
..........................................
Q95: The completed ion-electron equation is
a) an oxidation.
b) a reduction.
..........................................
Hydrogen peroxide (H 2 O2 ) is used as a hospital antiseptic.
A technician checking the concentration of peroxide titrated 25.0 cm 3 portions against
0.125 mol l-1 potassium permanganate and found an average titration value of 12.8 cm 3 .
The redox reaction is 2MnO 4- + 6H+ + 5H2 O2 → 2Mn2+ + 8H2 O + 5O2
Q96: Calculate the number of moles of potassium permanganate used in the titration.
Give your answer to 4 decimal places.
..........................................
Q97: What ratio of permanganate to hydrogen peroxide should be used in the
calculation of the concentration of the peroxide solution?
a)
b)
c)
d)
1:2.5
1:5
1:1
2:1
..........................................
Q98: Calculate the concentration of the hydrogen peroxide solution in mol l -1 to 2
decimal places.
..........................................
Q99: Calculate the concentration of H 2 O2 in g l-1 . Give your answer to 2 decimal places.
..........................................
..........................................
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Topic 5
Chromatography
Contents
5.1 Prior knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169
5.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Paper chromatography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169
172
5.3.1 Thin layer chromatography . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.2 Chromatography of protein hydrolysates . . . . . . . . . . . . . . . . . .
174
175
5.4 Gas-liquid chromatography . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5 Size-exclusion chromatography . . . . . . . . . . . . . . . . . . . . . . . . . . .
178
180
5.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
183
183
5.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
184
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TOPIC 5. CHROMATOGRAPHY
Prerequisite knowledge
You should already know that:
•
chemical analysis permeates all aspects of chemistry. It is important that you
understand the significance of analysis and are aware of how to carry out simple
analytical techniques (National 4, Unit 3);
•
analytical techniques could include:
◦
chromatography;
◦
flame tests;
◦
pH measurement using indicators / pH meters;
◦
separation techniques (National 4, Unit 3);
•
titration is an analytical technique used to determine the accurate volumes involved
in chemical reactions such as neutralisation (National 5, Unit 1);
•
an indicator is used to show the end-point of the reaction (National 5, Unit 1);
•
chemists play an important role in society by monitoring our environment to ensure
that it remains healthy and safe and that pollution is tackled as it arises (National
5, Unit 3);
•
a variety of methods exist which enable chemists to monitor the environment both
qualitatively and quantitatively, such as acid/base titration, precipitation and flame
testing (National 5, Unit 3).
Learning objectives
By the end of this topic you should be able to state that:
•
in chromatography, differences in the polarity / size of molecules are exploited to
separate the components present within a mixture;
•
depending on the type of chromatography in use, the identity of a component can
be indicated either by the distance it has travelled or by the time it has taken to
travel through the apparatus (retention time);
•
the results of a chromatography experiment can sometimes be presented
graphically showing an indication of the quantity of substance present on the yaxis and retention time on the x-axis.
Note: Learners are not required to know the details of any specific chromatographic
method or experiment.
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169
Prior knowledge
Test your prior knowledge
Q1: Which of the following techniques would you use to separate a mixture of water
and alcohol?
a)
b)
c)
d)
Distillation
Filtration
Chromatography
Titration
..........................................
Q2: Which of the following techniques would you use to separate a mixture of water
and sand?
a)
b)
c)
d)
Distillation
Filtration
Chromatography
Titration
..........................................
Q3: A precipitation reaction is one where:
a)
b)
c)
d)
the pH of a reaction moves closer to 7.
two solutions react to form a solid.
two small molecules react to produce a larger molecule and water.
a metal higher up in the electrochemical series displaces one lower down from its
ions in solution.
..........................................
5.2
Introduction
Much of chemistry is concerned with producing materials. These could be either a small
amount of a novel substance produced in a laboratory to enable further research into
new, unknown compounds or, alternatively, the manufacture of a large quantity of an
industrial product.
The other main strand of chemistry is concerned with analysis; finding out the structure
or composition of a natural or artificial material.
Chemical analysis has importance in:
•
environmental protection;
•
chemical manufacturing, e.g. pharmaceuticals;
•
consumer protection, e.g. food analysis;
•
forensics;
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TOPIC 5. CHROMATOGRAPHY
•
bioanalysis and clinical chemistry;
•
experimental design, including controlling chemical processes.
Analytical methods that are widely used to separate and identify the components of
mixtures come under the heading of chromatography.
There are several different types of chromatography, but the name derives from the early
methods used by the Russian, Michael Tswett, to separate mixtures of plant pigments
into a pattern of coloured components. The Greek words chroma (colour) and graphein
(to write) were chosen to name the method.
All chromatographic methods involve a mobile phase moving over a stationary phase.
Separation of the components in the sample occurs because they have different
solubilities (strictly, partition coefficients) between the stationary and mobile phases.
Substances present in the initial mixture which dissolve more in the stationary phase
will move more slowly than materials which dissolve more in the mobile phase.
Hornets versus bees
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To illustrate this separation process, follow the images below of some bees and hornets
moving over a flower bed. The bees spend some time collecting nectar from the
(stationary) flowers, and so arrive at the edge of the bed later than the hornets.
Hornets versus bees - start
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TOPIC 5. CHROMATOGRAPHY
The bees land on the flowers to collect nectar
The hornets finish first
The bees and hornets are separated
..........................................
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5.3
Paper chromatography
In paper chromatography the mixture of components to be separated is placed as a
small spot close to the bottom of a rectangular piece of absorbent paper (like filter
paper). The bottom of the paper is placed in a shallow pool of solvent in a tank. An
example of the solvent would be an alcohol. Owing to capillary attraction, the solvent is
drawn up the paper, becoming the mobile phase. The solvent front is clearly visible as
the chromatography progresses.
When the paper is removed from the solvent, the various components in the initial spot
have moved different distances up the paper.
Chromatography of blue and black inks
The start and final positions of a chromatographic analysis are shown below.
Go online
Start of chromatographic analysis
End of chromatographic analysis
This separation depends on the different polarities of the various components in the
sample. The components are dissolved in the moving solvent and the water trapped in
the paper. Substances which are less polar will dissolve mainly in the solvent mobile
phase and so will move further up the paper than more polar substances which dissolve
more in the aqueous stationary phase.
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TOPIC 5. CHROMATOGRAPHY
Q4: By observing the chromatography simulation above, which ink (blue or black) has
the most components?
a) Blue
b) Black
c) Both the same.
..........................................
Q5: Which material in the black ink has stayed longest on the stationary phase?
a)
b)
c)
d)
e)
Red
Yellow
Dark blue
Blue
Green
..........................................
Q6: What can be deduced about the polarity of the green component in black ink?
a) It is more polar than the red component.
b) It is less polar than the red component.
c) It is a non-polar substance.
..........................................
Q7: The red coloured spot has moved furthest, this would indicate that the red
material:
a) has the highest solubility in the solvent.
b) has the lowest solubility in the solvent.
c) has the lowest molecular mass.
..........................................
Q8: The movement of materials on paper chromatography is often described by an R f
(relative to the front) value, which is the distance travelled by the spot divided by the
distance travelled by the solvent front. As long as the conditions of chromatography
remain the same, a compound will have a constant R f value.
Which colour in the black ink could have an R f value of 0.4?
a)
b)
c)
d)
e)
Red
Yellow
Dark blue
Blue
Green
..........................................
Q9: Both the blue and dark blue spots from both the original inks have moved similar
distances. What might you conclude from this?
..........................................
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TOPIC 5. CHROMATOGRAPHY
Q10: You set up a paper chromatography experiment using butanol as the mobile
phase. All the dyes in your ink hardly move from the initial position; all have low R f
values. What does this tell you about your choice of mobile phase? What could you do
about it?
..........................................
5.3.1
Thin layer chromatography
Although paper chromatography is still used today, it has been largely replaced by thin
layer chromatography (TLC). In this method, a support plate of glass or aluminium is
coated, usually with a thin layer of silica, alumina or cellulose. The processing is identical
to paper chromatography but TLC allows a more rapid separation (which prevents the
spots spreading too far) and makes detection of the spots easier.
Most materials are not coloured, but can still be separated by chromatography. An
additional stage is required with non-visible components.
The invisible spots on paper or thin layer chromatography are revealed by use of a
locating reagent. These are applied once the chromatography is complete and react with
the compounds in the spots to produce a coloured derivative. For example, ninhydrin
solution can be sprayed onto chromatograms to reveal amino-acids as purple spots.
Separation of black ink on a TLC plate
(https:/ / commons.w ikimed ia.or g/ w iki/ F ile:T LC_black_ink.jpg by http:/ / en.w iki
ped ia.or g/ w iki/ User :Natr ij is licensed under http:/ / cr eativecommons.or g/ license
s/ by-sa/ 3.0/ )
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175
Further developments
In another TLC detection system, the silica stationary phase is mixed with a fluorescent
dye, so that at the end of the process, viewing the plate under an ultraviolet lamp will
cause the background to glow (often an eerie green) except where there are spots,
which appear black.
TLC is a rapid method for indicating qualitatively what materials are present in a sample
mixture. When carrying out an experimental synthesis, for example, putting a small spot
of your reaction mixture onto a TLC plate and developing it will show you whether your
reactants have turned into products.
It is not so effective at giving quantitative information about how much of a certain
material is in the sample. To obtain this information different systems are employed.
5.3.2
Chromatography of protein hydrolysates
Chromatography of protein hydrolysates
The diagrams below show the results of a simulated experiment on the chromatography
of protein hydrolysates. These results show the chromatographic separation of the
amino acids present in a section of protein. Three different proteins X, Y and Z have
been hydrolysed.
Protein X results
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TOPIC 5. CHROMATOGRAPHY
Protein Y results
Protein Z results
Q11: Look at the diagram showing the results of a chromatographic separation for the
hydrolysed protein X.
How many amino acids are present ?
..........................................
Q12: Name any amino acid not present in X.
..........................................
Q13: It is possible to conclude from the results that hydrolysed protein X contains:
a)
b)
c)
d)
B and C
A and B
A, B and C
A, B and D
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TOPIC 5. CHROMATOGRAPHY
..........................................
Q14: Referring to protein sample X, draw your prediction of the chromatogram which
would result if this protein section was hydrolysed and separated alongside samples of
A, B, C and D.
..........................................
Q15: Look at the diagram showing the results of a chromatographic separation for the
hydrolysed protein Y.
Explain fully your conclusions about the composition of Y from the results of the
experiment. There could be three marks for this answer in an exam setting. Try to
give three pieces of information. Write an answer before revealing the possible model
answer.
..........................................
Q16: Look at the diagram showing the results of a chromatographic separation for the
hydrolysed protein Z.
How many amino acids are present in the hydrolysed protein Z ?
..........................................
Q17: Name an amino acid which is not present in Z.
..........................................
Q18: Which amino acid seems to be present in the greatest amount?
..........................................
Q19: Look at the structures of the four standard amino acids used in this simulation and
the position of these after chromatography. Decide whether the following statement is
true or false:
•
The separation in this chromatography depends on molecular mass.
a) True
b) False
..........................................
Q20: Give a piece of evidence to support your answer to the previous question.
..........................................
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TOPIC 5. CHROMATOGRAPHY
Q21: If you examined the products of hydrolysis of a complex protein, say the enzyme
pepsin, what do you think the chromatogram would look like?
..........................................
Q22: How might you suggest this problem could be resolved?
..........................................
5.4
Gas-liquid chromatography
In gas-liquid chromatography (GLC) the stationary phase is a high boiling point liquid
held on an inert, finely-powdered support material, and the mobile phase is a gas (often
called the carrier gas).
The stationary phase is packed into a tubular column usually of glass or metal, with a
length of 1 to 3 metres and internal diameter about 2 mm. One end of the column is
connected to a gas supply (often nitrogen or helium) via a device which enables a small
volume of liquid sample, containing the mixture to be analysed, to be injected into the
gas stream. The other end of the column is connected to a device which can detect the
presence of compounds in the gas stream.
The column is housed in an oven to enable the temperature to be controlled throughout
the chromatographic analysis. One reason for this is that the materials to be analysed
must be gases during the analysis, so that gas-liquid chromatography is often carried
out at elevated temperatures.
GLC apparatus
A mixture of the material to be analysed is injected into the gas stream at zero time.
The individual components travel through the packed column at rates which depend on
their partition coefficients between the liquid stationary phase and the gaseous mobile
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TOPIC 5. CHROMATOGRAPHY
phase.
The detector is set to measure some change in the carrier gas that signals the presence
of material coming from the end of the column. Some detectors measure the thermal
conductivity of the gas, others burn the material from the column in a hydrogen-air flame
and measure the presence of ions in the flame.
The signal from the detector is recorded and plotted against time to give a series of
peaks each with an individual retention time.
A typical trace, chromatogram, for a complex mixture, premium grade petrol, is shown
below.
Chromatogram of premium grade petrol
As long as the conditions remain constant (same column and stationary phase, same
pressure of gas, and same temperature) the retention time is the same for a given
compound. By chromatographing known materials (standards) the identity of many
peaks in a GLC trace can be established.
The other important feature of many chromatograms is that the area under the peak is
directly proportional to the amount of material present.
The composition of petrol is changed with the seasons by the suppliers. In the
cold winter months petrol needs a higher proportion of more volatile, low-boiling
components; in summer too much of this composition would evaporate, so more highboiling compounds are present. GLC analysis is used by the oil companies to ensure
consumers are getting the correct grade of petrol.
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TOPIC 5. CHROMATOGRAPHY
Gas-liquid chromatography
Go online
Q23: The retention times, under the same conditions as the chromatogram figure
above, for four compounds are shown in the table below:
Hydrocarbon
Structure
Retention time
Butane
C4 H10
5.5 min
Pentane
C5 H12
7.4 min
Benzene
C6 H6
12.7 min
Toluene
C7 H8
15.1 min
Which of these hydrocarbons is present in the greatest amount in petrol?
a)
b)
c)
d)
Butane
Pentane
Benzene
Toluene
..........................................
Q24: How are retention time and molecular mass related for these four hydrocarbons?
a) The retention time increases as mass increases.
b) The retention time decreases as mass increases.
c) They are not related.
..........................................
Q25: Using the answer to the previous question, suggest a possible hydrocarbon for
the peak with retention time 20.8 min.
a)
b)
c)
d)
Hexane (C6 H14 )
Heptane (C7 H16 )
Xylene (C8 H10 )
Propane (C3 H8 )
..........................................
5.5
Size-exclusion chromatography
The previous separations depended mainly on the differences in polarities between
the components in a sample. Size-exclusion chromatography, sometimes called gel
filtration, separates on the basis of molecular size. It is widely used to characterise
polymers, including biological material.
A column is packed with porous beads and the sample containing a range of differently
sized molecules added to the top in a suitable solvent. A flow of solvent then moves
down the column. The sample components are detected as they leave the column and
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181
a graph of detector response against time is constructed to reveal the peaks of eluted
compounds.
There is a range of materials with different pore sizes in the beads, so that a wide range
of sizes of analytes (materials to be separated) can be chromatographed. In order to
quantify the sizes of analytes there are standards containing molecules of a defined
range of sizes which can be used in the system.
Only the smaller molecules can enter the pores in the stationary beads; the larger
molecules are excluded and remain in the mobile phase solvent. You can see this in
the diagram below.
Gel filtration
Go online
Q26: Larger molecules in the sample will leave the column (be eluted) first. Why does
this happen?
a) The smaller molecules adhere more strongly to the beads.
b) The larger molecules are excluded form the pores in the stationary beads.
c) The larger molecules are heavier and move to the bottom faster.
..........................................
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TOPIC 5. CHROMATOGRAPHY
Q27: If the pores in the beads were too small to allow any solute molecules to enter the
beads, what would happen?
a) The sample molecules would remain on the top of the column and never be eluted.
b) The sample molecules would be slowly eluted from the column in many peaks.
c) The sample molecules would elute from the column in a single peak.
..........................................
Q28: A size-exclusion column was constructed and a standard containing three proteins
A, B and C with molecular masses of 1,000, 3,000 and 6,000 respectively was applied
to the column and eluted with solvent. The pattern of elution is shown below.
Proteins A, B and C
Starting with the earliest, what is the order of elution?
a)
b)
c)
d)
A, B then C
A, C then B
C, B then A
B, A then C
..........................................
Q29: A sample, X, containing unknown proteins was applied to the same column and
eluted under identical conditions to the previous question. The pattern is shown below.
Unknown proteins
What can you tell about the proteins in sample X?
..........................................
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TOPIC 5. CHROMATOGRAPHY
Q30: Think very carefully. Is the answer to the question above absolutely correct?
..........................................
Q31: Another experiment was performed on this system by mixing the standard mixture
and the unknown sample then eluting this mixture. What would you expect the elution
pattern to look like?
..........................................
Q32: What could you do to try and establish whether there was more than one type of
protein with the same size in the sample?
..........................................
..........................................
5.6
Summary
Summary
You should now be able to state that:
•
in chromatography, differences in the polarity / size of molecules are
exploited to separate the components present within a mixture;
•
depending on the type of chromatography in use, the identity of a
component can be indicated either by the distance it has travelled or by
the time it has taken to travel through the apparatus (retention time);
•
the results of a chromatography experiment can sometimes be presented
graphically showing an indication of the quantity of substance present on
the y-axis and retention time on the x-axis.
Note: Learners are not required to know the details of any specific
chromatographic method or experiment.
5.7
•
Resources
Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson,
ISBN 978-1444167528
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184
TOPIC 5. CHROMATOGRAPHY
5.8
End of topic test
End of topic 6 test
Go online
Q33: This question refers to the chromatographic analysis of hydrolysed protein section
P.
Name the amino acids which are present in the hydrolysed protein section P.
..........................................
Q34: How many amino acids are present in hydrolysed protein section P?
..........................................
Q35: From the results, it can be concluded that the hydrolysed protein section P
contains:
a)
b)
c)
d)
A and B
A and B only
A, B and C
A, B, C and D
..........................................
Q36: Which of these protein sections could be P?
a)
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TOPIC 5. CHROMATOGRAPHY
b)
c)
d)
..........................................
Q37: If protein section Q, shown below, were hydrolysed and run against A, B, C and
D, which spots would be present?
a)
b)
c)
d)
A only
A and D
A, C and D
A, B, C and D
..........................................
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TOPIC 5. CHROMATOGRAPHY
Q38: Gas chromatograms of the hydrocarbons found in two sediment samples a few
miles from an oil production platform are shown below.
The numbers refer to the number of carbon atoms in the straight chain n-alkanes. Pr
and Ph refer to pristane and phytane, two terpenoid hydrocarbons.
Gas chromatography separates volatile compounds in order of boiling point. Pristane
and phytane are highly branched hydrocarbons containing 19 and 20 carbon atoms
respectively.
Looking at the chromatograms, which of the following statements is correct?
a)
b)
c)
d)
You can tell nothing about boiling points from these chromatograms.
Boiling points of hydrocarbons with the same number of carbon atoms are the same.
All hydrocarbons are present in these samples.
Branched alkanes have lower boiling points than n-alkanes with the same number
of carbon atoms.
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TOPIC 5. CHROMATOGRAPHY
..........................................
Q39: Samples of hydrocarbons which have been derived from crude oil spills and those
from natural plant waxes differ in their carbon preference index (CPI). Plant waxes
contain a preponderance of hydrocarbons with odd numbers of carbon atoms over those
with even numbers (CPI > 1); crude oil hydrocarbons show no such pattern (CPI = 1).
Look particularly at the hydrocarbons from 27 to 36. What do you think was the source
of the sample?
a) Plant waxes
b) Cannot tell
c) Crude oil
..........................................
Q40: Sample A was taken from a place where marine currents supply new material on
a regular basis. Sample B was from a region where hydrocarbon breakdown is more
rapid than addition of new material.
Comparing these two samples which of the following statements is correct?
a) All hydrocarbons degrade at the same rate.
b) Alkanes with odd numbers of carbon atoms are degraded faster than those with
even numbers.
c) Shorter hydrocarbons are degraded more slowly than longer ones.
d) Branched alkanes, such as pristane and phytane, are degraded more slowly than
n-alkanes.
..........................................
Q41: What happens to hydrocarbons when they are degraded in the environment?
..........................................
Q42: From your knowledge of chemistry describe why marine oil spills are frequently
treated with detergents/emulsifiers in attempts to clear them more rapidly.
..........................................
..........................................
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TOPIC 5. CHROMATOGRAPHY
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Topic 6
Volumetric analysis
Contents
6.1 Prior knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
191
6.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2.1 Titration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
192
193
6.3 Quality control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4 Practical applications of titrations . . . . . . . . . . . . . . . . . . . . . . . . . .
194
194
6.4.1 The purity of aspirin tablets . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.2 Antacid tablet CaCO3 content (back titration) . . . . . . . . . . . . . . .
194
195
6.4.3 The chloride ion content of seawater (precipitation) . . . . . . . . . . . .
6.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
196
198
6.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
198
199
190
TOPIC 6. VOLUMETRIC ANALYSIS
Prerequisite knowledge
Before you begin this topic, you should already know that:
•
a neutralisation reaction is one in which an acid reacts with a base to form water.
A salt is also formed in this reaction (National 4, Unit 1);
•
titration is an analytical technique used to determine the accurate volumes involved
in chemical reactions such as neutralisation (National 5, Unit 1);
•
an indicator is used to show the end-point of the reaction (National 5, Unit 1);
•
combinations of oxidation and reduction half reactions form redox equations
(Higher, Unit 3, Topic 5);
•
the technique of titration can be applied to redox reactions, allowing the
concentration of a reactant to be calculated from results of volumetric titrations
(Higher, Unit 3, Topic 5).
Learning objectives
By the end of this topic you should be able to state that:
•
volumetric analysis involves using a solution of accurately known concentration in
a quantitative reaction to determine the concentration of another substance;
•
a solution of accurately known concentration is known as a standard solution;
•
the volume of reactant solution required to complete the reaction is determined by
titration;
•
calculations from balanced equations can then be carried out to calculate the
concentration of the unknown solution;
•
redox titrations are based on redox reactions;
•
substances such as potassium permanganate(VII), which can act as their own
indicators, are very useful reactants in redox titrations;
•
the concentration of a substance can be calculated from experimental results by
use of a balanced equation.
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6.1
191
Prior knowledge
Test your prior knowledge
Q1: What is the name of the technique which separates substances by their
solubilities?
a)
b)
c)
d)
Chromatography
Distillation
Titration
Precipitation
..........................................
Q2: What is the term given to when a titration has just been neutralised and no more?
a)
b)
c)
d)
End point
Titre
Neutralisation
Burette
..........................................
Q3: Reaction of an acid and an alkali will produce which products?
a)
b)
c)
d)
A salt and hydrogen
A salt and water
A salt, water and carbon dioxide
A salt and oxygen
..........................................
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TOPIC 6. VOLUMETRIC ANALYSIS
6.2
Introduction
You should be familiar with volumetric analysis as a means of obtaining an accurate
concentration of a substance in solution. A solution of accurately known concentration,
a standard solution, is used to titrate the unknown solution until an end-point is
obtained. Calculation then enables the unknown concentration to be found.
Volumetric analysis equipment
Go online
The diagram below shows the equipment that you would use to carry out volumetric
analysis.
Q4:
a)
b)
c)
d)
e)
Identify the burette from the diagram above.
A
B
C
D
E
..........................................
Q5:
a)
b)
c)
d)
e)
Identify the pipette from the diagram above.
A
B
C
D
E
..........................................
Q6:
a)
b)
c)
d)
e)
Identify the indicator from the diagram above.
A
B
C
D
E
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193
..........................................
Q7: Identify the volumetric flask from the diagram above.
a)
b)
c)
d)
e)
A
B
C
D
E
..........................................
6.2.1
Titration
The following example describes an acid-based titration.
A metal plating plant produces waste nitric acid which is discharged into a holding tank
before disposal. The process chemists need to know the concentration of the acid in the
tank, so take regular samples to estimate the nitric acid concentration by titration with
standard sodium hydroxide solution.
A 25 ml sample from the tank was titrated with 0.100 mol -1 NaOH and required 28.5
ml for complete neutralisation. What is the concentration of nitric acid in mol -1 and g
-1 ?
1. Write a balanced equation: HNO 3 + NaOH → NaNO 3 + H2 O
The mol relationship: 1 mol HNO 3 reacts with 1 mol NaOH
2. Moles of NaOH used = concentration × volume = 0.100 × 0.0285 = 0.00285 mol
3. Moles of HNO3 in sample = 0.00285
Concentration = moles/volume = 0.00285/0.025 = 0.114 mol -1
4. Molecular mass of HNO3 = 1 + 14 + 3 × 16 = 63 (g mol -1 )
0.114 mol -1 HNO3 is 0.114 × 63 = 7.182 g -1
Titration
Q8: A soap factory produces an alkaline waste containing potassium hydroxide
solution.
A 25 ml sample from the waste was titrated with 0.150 mol -1 hydrochloric acid and
required 34.5 ml for complete neutralisation. What is the concentration of potassium
hydroxide in mol -1 and g -1 ?
..........................................
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TOPIC 6. VOLUMETRIC ANALYSIS
6.3
Quality control
Quality assurance is often applied to an end-product or service. The consumer should
be able to be assured that whatever is being supplied will meet certain defined criteria.
Quality control is used internally by manufacturers, say a pharmaceutical manufacturer,
to assess the progress of the production. At various critical stages, samples are taken
and analysed to ensure that "all is well" with the process and to correct any problems
before they lead to further difficulties.
Quality control
Go online
Q9: An antiseptic healing cream used for mild skin irritation such as nappy rash,
sunburn or abrasions is made by taking a slurry of zinc oxide (ZnO) in water and blending
this with oils and waxes to make an emollient cream. Before this blending, the zinc
oxide suspension is sampled and titrated with standard hydrochloric acid to determine
the concentration of zinc oxide. It should contain between 400 and 450 g ZnO -1 of
slurry.
The sample volume of 1.00 ml is neutralised completely by 20.64 ml of 0.500 mol -1
hydrochloric acid. Is the batch within the requirements?
..........................................
6.4
Practical applications of titrations
You have already met redox titrations in a previous topic. The following pages contain
some useful titration experiments for you to try.
There are suggested results, so that you can practise the calculations.
6.4.1
The purity of aspirin tablets
The purity of aspirin tablets
The chemical structure of the analgesic aspirin (acetylsalicylic acid) is shown below.
Go online
Aspirin
Q10: Look at the chemical structure. What method would you suggest to quantify the
mass of aspirin in a tablet?
© H ERIOT-WATT U NIVERSITY
TOPIC 6. VOLUMETRIC ANALYSIS
..........................................
Q11: Each aspirin tablet should have 300 mg of acetylsalicylic acid.
Calculate the relative formula mass of aspirin and then calculate how many moles of
aspirin should be in one tablet.
..........................................
Q12: A sample tablet of aspirin was dissolved in water and titrated with 0.300 mol -1
sodium hydroxide solution, using phenolphthalein as indicator.
If the tablet was 100% aspirin, what volume of sodium hydroxide solution would be
required?
..........................................
Q13: As a method for checking the aspirin content of tablets this experiment could be
improved in several ways. What might you suggest?
..........................................
6.4.2
Antacid tablet CaCO3 content (back titration)
Many indigestion remedies contain 'antacids', compounds which react with the
hydrochloric acid in stomach juices and so reduce the excess acidity. One of the
commonest of these is calcium carbonate.
Antacid tablets
(Antacid -L47 8 by http:/ / commons.w ikimed ia.or g/ w iki/ User :Mid nightcomm is
licensed under http:/ / cr eativecommons.or g/ licenses/ by-sa/ 3.0/ )
You cannot estimate the amount of antacid in an indigestion tablet directly because the
material is insoluble in water. Consequently, you estimate the calcium carbonate content
of antacid tablets by performing a "back titration".
A tablet is added to a known volume of standard hydrochloric acid and the acid remaining
is measured by titration with standardised sodium hydroxide solution. From this figure,
the amount of acid neutralised by the tablet can be calculated. The method has the
added advantage that you don't need to know the composition of the tablet to obtain a
value for its effectiveness.
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TOPIC 6. VOLUMETRIC ANALYSIS
Method of back titration
An antacid tablet is placed in a conical flask with about 25 ml of water and 5.00 ml of
5.00 mol -1 hydrochloric acid is added. The tablet effervesces and the solution is gently
boiled for a minute to ensure complete reaction. Any cloudiness remaining will be due
to other ingredients, e.g. starch, in the tablet.
The solution is then cooled, a few drops of indicator solution, phenolphthalein, is added
and the solution titrated with 0.500 mol -1 sodium hydroxide solution, until a faint pink
colour remains.
Back titration
Go online
Q14: Assuming the volume of sodium hydroxide solution needed to neutralise the
remaining HCl was 22.0 ml, what mass of calcium carbonate is in one tablet?
Try this calculation yourself before looking at the answer.
..........................................
6.4.3
The chloride ion content of seawater (precipitation)
The salinity (saltiness) of sea water is very important in determining which animals
and plants can survive there and in driving the major ocean currents. The density of
sea water depends on its salinity (the more saline the denser) and temperature (higher
temperatures expand water, making it less dense).
Gulf stream
You may have heard of the Gulf Stream, see image above, which is a surface water
current that carries warm water from the tropics north-east across the Atlantic to bathe
western Scotland and allow palm trees to grow in Plockton! It is part of a circulation that
involves this warm surface water cooling in the Northern hemisphere, becoming more
dense and sinking to form a return sub-sea current flowing south-west. There is some
worry that, as the Earth generally becomes warmer, the Gulf Stream that keeps Britain
warm might stop.
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197
Since the predominant salt in seawater is sodium chloride, a good estimate of salinity
can be obtained by determining the chloride ion content. The chloride ion content can
be calculated by titrating the seawater with standard silver nitrate solution until excess
silver ions can be seen. This is done by adding potassium chromate(VI). In solution
chromate(VI) ion is yellow, but with silver ion it forms an intense brick red precipitate.
The vast amount of chloride ion in seawater is thought to have originated from the
'outgassing' of the mantle as volcanos in the very early history of the Earth. The negative
ions that are carried in rivers due to weather erosion of rocks contain very little chloride
ion; they are mainly hydrogencarbonate (HCO 3 - ) owing to the action of carbonic acid
(H2 CO3 ) on rock minerals. Any chloride ion present has probably been recycled from
rain produced when clouds form over the salty oceans.
The chloride ion content of seawater (precipitation)
Q15: Seven ions constitute more than 99% of the ions in seawater and the ratios of
these ions is constant throughout the world's oceans, so estimating one (e.g. chloride)
can lead to estimates of the others and hence overall salinity.
Which seven ions do you think are the major ones in seawater?
..........................................
Q16: A 10 ml sample of seawater is carefully measured into a conical flask and diluted
with about 20 ml of water. A few drops of a solution of potassium chromate(VI), KCrO 4 ,
is added as indicator. The solution is titrated with a standard solution of silver nitrate,
AgNO3, containing 0.50 mol -1 , until a trace of red precipitate remains.
What do you think the red precipitate might be?
..........................................
Q17: Assuming the volume of silver nitrate solution needed to react with the chloride
ion in seawater was 12.0 ml, what is the salinity of seawater in g NaCl -1 ?
Try this calculation before looking at the answer.
..........................................
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TOPIC 6. VOLUMETRIC ANALYSIS
..........................................
6.5
Summary
Summary
You should now be able to state that:
6.6
•
•
volumetric analysis involves using a solution of accurately known
concentration in a quantitative reaction to determine the concentration of
another substance;
•
a solution of accurately known concentration is known as a standard
solution;
•
the volume of reactant solution required to complete the reaction is
determined by titration;
•
calculations from balanced equations can then be carried out to calculate
the concentration of the unknown solution;
•
redox titrations are based on redox reactions;
•
substances such as potassium permanganate(VII), which can act as their
own indicators, are very useful reactants in redox titrations;
•
the concentration of a substance can be calculated from experimental
results by use of a balanced equation;
•
quality control of chemical processes requires analysis to ensure that the
process requirements are being met.
Resources
Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson,
ISBN 978-1444167528
© H ERIOT-WATT U NIVERSITY
TOPIC 6. VOLUMETRIC ANALYSIS
6.7
199
End of topic test
End of Topic 6 test
Q18: As vegetable oils age they produce fatty acids. It is, therefore, important to be
able to estimate the concentration of 'free fatty acids' in samples of vegetable oils.
This is usually done by an acid-base titration. A sample of the oil is titrated with standard
potassium hydroxide solution.
Explain why the oil cannot be titrated directly with aqueous potassium hydroxide.
..........................................
Q19: The acid-based titration reaction is carried out in a solvent mixture, often a mixture
of ethanol and ether, which dissolves both vegetable oils and aqueous potassium
hydroxide.
An example procedure is as follows:
1. A sample of the oil weighing 5.53 g is dissolved in about 50 ml of the solvent in a
250 ml conical flask.
2. A few drops of indicator solution (phenolphthalein) are added.
3. The solution is titrated with 0.1 mol l -1 potassium hydroxide solution until a faint pink
colour persists, indicating the end-point.
4. The titration value for this sample is 12.32 ml.
Calculate the mass, in grammes, of oleic acid (GFM = 282) in the flask.
..........................................
Q20: Calculate the percentage of oleic acid in the oil sample.
..........................................
Q21: When making biodiesel from vegetable oil it important to start with a neutral oil,
so any free fatty acid is neutralised with sodium hydroxide.
What mass of solid sodium hydroxide, in kg, would need to be added to 100 kg of this
oil sample to neutralise it exactly?
..........................................
Q22: The water in swimming pools can be kept sterile by adding chlorine, which kills
micro-organisms. Chlorine levels in swimming pools can be monitored by titration with
iron(II) sulfate.
The redox equation for this is:
2Fe2+ + Cl2 → 2Fe3+ + 2ClA 1000 cm3 sample of water from a swimming pool which is being heavily chlorinated
before use by the public required 15 cm3 of iron(II) sulfate of concentration 0.6 mol l-1
to react completely.
Calculate the number of moles of iron(II) sulfate used in the titration. Give your answer
to three decimal places.
..........................................
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TOPIC 6. VOLUMETRIC ANALYSIS
Q23: Calculate the concentration of chlorine in the swimming pool in mol l -1 . Give your
answer to four decimal places.
..........................................
Q24: Calculate the concentration of chlorine in the swimming pool in mg l -1 . Give your
answer to one decimal place.
..........................................
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Topic 7
End of unit test
202
TOPIC 7. END OF UNIT TEST
End of unit 3 test
Q1:
Go online
On going down a group in the Periodic Table, the first ionisation energy:
a) increases.
b) decreases.
..........................................
Q2:
a)
b)
c)
d)
When the enthalpy change has a positive sign, the reaction is:
exothermic.
endothermic.
large.
small.
..........................................
Q3: Ionic bonding is most likely when the electronegativity difference between the
elements is:
a)
b)
c)
d)
exothermic.
endothermic.
large.
small.
..........................................
Q4: The same reaction was carried out at four different temperatures. The table below
shows the time taken for the reaction to occur.
Temp (◦ C)
20
30
40
50
Time
(seconds)
60
30
14
5
The results show that:
a)
b)
c)
d)
a small rise in temperature results in a large increase in reaction rate.
the reaction is endothermic.
the activation energy increases with increasing temperature.
the rate of the reaction is directly proportional to the temperature.
..........................................
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TOPIC 7. END OF UNIT TEST
Q5: The potential energy diagram below refers to the reversible reaction involving
reactants R and products P.
What is the value of the activation energy for the forward reaction (reactants to
products)?
a)
b)
c)
d)
10 kJ mol-1
20 kJ mol-1
40 kJ mol-1
60 kJ mol-1
..........................................
Q6: What is the enthalpy change for the forward reaction (reactants to products)?
a)
b)
c)
d)
-30 kJ mol-1
-20 kJ mol-1
+20 kJ mol-1
+30 kJ mol-1
..........................................
Q7: Which of the following molecules may be described as polar?
a)
b)
c)
d)
..........................................
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TOPIC 7. END OF UNIT TEST
Q8: The melting points of Group 7 elements increase on descending the group
because the . . . . . . . . . . . . . . . . . . increase.
a)
b)
c)
d)
mean bond enthalpies
nuclear charges
covalent bond lengths
London dispersion forces
..........................................
Q9: The difference between the covalent radii of lithium and carbon is mainly due to
the difference in the:
a)
b)
c)
d)
number of neutrons.
mass of each atom.
number of electrons.
number of protons.
..........................................
Q10: In general, covalent substances have lower melting points than ionic substances
because:
a)
b)
c)
d)
covalent bonds have no electrostatic attractive forces.
covalent compounds are composed of non-metals which have low melting points.
bonds between molecules are weaker than bonds between ions.
ionic bonds are stronger than covalent bonds.
..........................................
Q11:
The product of the oxidation of the above compound is:
a)
b)
c)
d)
4-methylpentan-2-one.
2-methylpentanal.
2-methylpentan-4-one.
4-methylpentanal.
..........................................
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TOPIC 7. END OF UNIT TEST
Q12:
Rum flavouring is based on the compound with the formula shown above.
It can be made from:
a)
b)
c)
d)
propanol and ethanoic acid.
ethanol and butanoic acid.
butanol and methanoic acid.
ethanol and propanoic acid.
..........................................
Q13:
What product(s) would be expected upon dehydration of the alcohol shown above?
a)
b)
c)
d)
2-methylbut-2-ene only
2-methylbut-2-ene and 2-methylbut-1-ene
3-methylbut-1-ene and 2-methylbut-2-ene
2-methylbut-1-ene only
..........................................
Q14: The enthalpy changes (ΔH values) for two reactions are given below.
S(s) + O2 (g) → SO2 (g) ΔH = -297 kJ
SO2 (g) + 1 /2 O2 (g) → SO3 (g) ΔH = -101 kJ
What is ΔH for the following reaction?
S(s) + 3 /2 O2 (g) → SO3 (g)
A)
B)
C)
D)
-398 kJ
-196 kJ
+196 kJ
+398 kJ
..........................................
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Q15: For the reaction:
PCl3 (g) + Cl2 (g) PCl5 (s)
ΔH = -126 kJ
Which of the following changes will cause the greatest increase in the percentage of
PCl5 in the above equilibrium mixture?
a)
b)
c)
d)
Decrease in temperature, decrease in pressure
Decrease in temperature, increase in pressure
Increase in temperature, decrease in pressure
Increase in temperature, increase in pressure
..........................................
Sulfur burns in oxygen according to the equation S(s) + O 2 (g) → SO2 (g)
0.642 g of sulfur is placed in a vessel containing 0.25 l of oxygen. Under these conditions
the molar volume of oxygen is 25.0 l mol -1 .
Q16: How many moles of sulfur are present? Give your answer to 2 decimal places.
..........................................
Q17: By calculating the number of moles of oxygen, which element is in excess?
..........................................
Graph A shows the volume of hydrogen gas produced against time when an excess of
magnesium is added to 50 cm 3 of hydrochloric acid of concentration 1 mol l-1 at 20◦ C.
Graph B was obtained when the reaction was repeated with excess magnesium and
hydrochloric acid of the same concentration but at a different temperature.
Q18: Which reaction was at the higher temperature?
a) A
b) B
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..........................................
Q19: How many cm3 of acid was required to produce graph B?
..........................................
The apparatus shown can be used to prepare iron(III) chloride.
Q20: On the diagram, highlight a substance which contains metallic bonding.
..........................................
Q21: By considering the method of production, predict the type of bonding in iron(III)
chloride.
..........................................
Q22: Which of the following describes the bonding in the chlorides across Period 3 from
NaCl to SCl2 ?
a)
b)
c)
d)
Ionic → polar covalent
Ionic → polar covalent → pure covalent
Polar covalent → ionic
Ionic → pure covalent → polar covalent
..........................................
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The potential energy diagram below represents the decomposition of hydrogen iodide:
2HI(g) → H2 (g) + I2 (g)
Q23: What is the value for the activation energy (E A ) for the above reaction?
..........................................
Q24: What would be the enthalpy change (ΔH) for the reaction?
H2 (g) + I2 (g) → 2HI(g)
..........................................
Q25: The use of a catalyst would . . . . . . the activation energy in part 1.
..........................................
Q26: The use of a catalyst would . . . . . . the enthalpy change in part 2.
..........................................
Q27: The enthalpy change for the reaction
1 / H (g) + 1 / I (s) → HI(g)
2 2
2 2
is known as the enthalpy of formation of hydrogen iodide. It differs from the equation
above in two respects:
1. it involves the formation of one mole of HI,
2. and the reactants are in their standard states, so iodine is a solid.
Given that the enthalpy of sublimation of iodine
I2 (s) → I2 (g) is ΔH = + 62 kJ
calculate the enthalpy of formation of hydrogen iodide.
..........................................
Q28: If the enthalpy of formation of hydrogen chloride is -92 kJ mol -1 , predict a value
for the enthalpy of formation of hydrogen bromide.
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..........................................
Q29: Identify a peptide link on the following diagram.
..........................................
Q30: Identify the phenyl group on the following diagram.
..........................................
Q31: On hydrolysis, this section of protein produces three amino acids, none of which
can be made by the human body.
What name is given to such amino acids?
..........................................
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Q32: Which of the following amino acids could NOT be produced on hydrolysis of this
section of protein?
a)
b)
c)
d)
..........................................
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The following diagram shows some reactions of ethene and ethyne.
Q33: Compounds A and C are isomers. Which is the structural formula of compound
C?
a)
b)
c)
d)
..........................................
Q34: Ethanol can be made industrially by the catalytic hydration of ethene as shown in
the diagram.
If the process is 85% efficient, how many kg of ethanol will be produced from 112 kg of
ethene?
Give your answer to one decimal place.
..........................................
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The enthalpy changes for three reactions are shown below.
C(s) + O2 (g) → CO2 (g) ΔH1
H2 (g) + O2 (g) → H2 O(l) ΔH2
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2 O(l)
ΔH3
Q35:
What is the ΔH value for the following reaction?
C(s) + 2H2 (g) → CH4 (g)
A) ΔH1 + 2ΔH2 - ΔH3
B) ΔH3 - 2ΔH2 - ΔH1
C) ΔH1 + ΔH2 + ΔH3
D) ΔH1 + ΔH2 - ΔH3
..........................................
Methanol is manufactured by the reaction of carbon monoxide and hydrogen in synthesis
gas.
CO(g) + 2H2 (g) CH3 OH(g) ΔH = - 91 kJ mol-1
Q36: What is the atom economy of this reaction?
..........................................
Q37: In the manufacture of methanol, which way would the equilibrium move if the
pressure was increased?
..........................................
Q38: Would this increase or decrease the yield of methanol?
..........................................
Q39: In the manufacture of methanol, what would be the effect on the yield of product
of decreasing the temperature?
..........................................
Q40: The process used to be operated at 300 atmospheres pressure, but improved
catalysts allow an efficient process at 100 atmospheres.
Give one reason why 100 atm. is better for industry than 300 atm.
..........................................
Q41: What effect would the improved catalyst have on the position of equilibrium?.
..........................................
Q42: HgCl2 (aq) + SnCl2 (aq) → Hg(l) + SnCl4 (aq)
What ion is oxidised in the above redox reaction?
a) Cl- (aq)
b) Sn2+ (aq)
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TOPIC 7. END OF UNIT TEST
c) Sn4+ (aq)
d) Hg2+ (aq)
..........................................
Q43: Which of the following reactions can be classified as reduction?
a)
b)
c)
d)
CH3 CH2 COCH3 → CH3 CH2 CH(OH)CH3
CH3 CH(OH)CH3 → CH3 COCH3
CH3 CH2 OH → CH3 COOH
CH3 CH2 CHO → CH3 CH2 COOH
..........................................
Q44: Which of the following is a redox reaction?
a)
b)
c)
d)
ZnO + 2HCl → ZnCl 2 + H2 O
ZnCO3 + 2HCl → ZnCl2 + H2 O + CO2
Zn(OH)2 + 2HCl → ZnCl2 + 2H2 O
Zn + 2HCl → ZnCl 2 + H2
..........................................
Q45: In which of the following reactions is a positive ion reduced?
a)
b)
c)
d)
Sulfate → sulfite
Gold(II) → gold(III)
Iron(III) → iron(II)
Bromide → bromine
..........................................
Q46: In which of the following reactions is the hydrogen ion acting as an oxidising
agent?
a)
b)
c)
d)
Fe + 2HCl → FeCl 2 + H2
KOH + HNO3 → KNO3 + H2 O
MgCO3 + H2 SO4 → MgSO4 + H2 O + CO2
CH3 COONa + HCl → NaCl + CH 3 COOH
..........................................
The water in swimming pools is often disinfected by adding "chlorinated lime". This
produces chlorate(I) ions (OCl- ) in solution, which effectively kill pathogens, as long as
their concentration is sufficiently high.
Q47: One method of estimating the OCl- concentration is to take a known volume
of pool water and add an excess of potassium iodide solution. The reactions below
produce iodine which can then be measured.
Fill in the gaps to complete the ion-electron equations below.
..........................................
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Q48:
..........................................
Q49: Which ionic equation is correct for the whole reaction?
a)
b)
c)
d)
OCl- + 2H+ + 2I- → Cl- + H2 O + I2
OCl- + H+ + 2I- → Cl- + H2 O + I2
OCl- + 2H+ + I- → Cl- + H2 O + I2
2OCl- + 2H+ + I- → 2Cl- + H2 O + I2
..........................................
Q50: If 100 cm3 of the pool water produced 0.002 moles of iodine, what was the
concentration of OCl- in mol l-1 ?
..........................................
Q51: In order to be effective the concentration of chlorate(I) should be between 0.4 and
1.5 mg l-1 . What is the concentration of OCl- in mg l-1 ?
..........................................
Q52: Using the table of standard reduction potentials, which of these redox reactions
will take place?
a) Cl2 (g) + 2Br- (aq) → 2Cl- (aq) + Br2 (aq)
b) Br2 (l) + 2Fe2+(aq) → 2Br- (aq) + 2Fe3+(aq)
c) Ca2+ (aq) + Cu(s) → Ca(s) + Cu2+ (aq)
d) Zn2+ (aq) + Cu(s) → Zn(s) + Cu2+ (aq)
e) Mg2+ (aq) + H2 (g) → Mg(s) + 2H+ (aq)
f) Cu(s) + Cu2+ (aq) → 2Cu+ (aq)
..........................................
Q53: The iodate ion, IO 3 - , can be converted to iodine. Which is the correct ion-electron
equation for the reaction?
a)
b)
c)
d)
2IO3 - (aq) + 12H+ (aq) + 11e- → I2 (aq) + 6H2 O(l)
2IO3 - (aq) + 12H+ (aq) → 2I- (aq) + 6H2 O(l)
2IO3 - (aq) + 12H+ (aq) → 2I- (aq) + 6H2 O(l)
IO3 - (aq) + 6H+ (aq) → I- (aq) + H2 O(l)
..........................................
Q54: A major source of iodine is Caliche, mined in Chile.
The mass of iodine in a 10 g sample of caliche can be determined by dissolving the
sample in water and adding hydrogen peroxide solution to oxidise the iodide to iodine
molecules. The ion-electron equation for the reduction reaction is shown.
H2 O2 (aq) + 2H+ + 2e- → 2H2 O(l)
Write a balanced redox equation for the reaction of hydrogen peroxide with iodide ion.
..........................................
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215
Q55: Using starch solution as indicator, the iodine solution is then titrated with sodium
thiosulfate solution to determine the mass of iodine in the sample. The balanced
equation for the reaction is shown.
2Na2 S2 O3 (aq) + I2 (aq) → 2NaI(aq) + Na2 S4 O6 (aq)
In an analysis of a sample, 13.25 cm 3 of 0.0100 mol l-1 sodium thiosulfate solution was
required to reach the end-point.
Calculate the mass of iodine present in the sample of caliche.
..........................................
Q56: Hydrogen peroxide is used in gels to whiten teeth. The ion-electron equation for
the oxidation of hydrogen peroxide is:
H2 O2 → O2 + 2H+ + 2eUsing your knowledge of chemistry, comment on possible methods for measuring and
comparing the concentration of hydrogen peroxide present in two different gels.
(This question is worth 3 marks)
..........................................
Q57: Concentrated solutions of hydrogen peroxide are used in the propulsion systems
of torpedoes. Hydrogen peroxide decomposes naturally to form water and oxygen:
2H2 O2 (aq) → 2H2 O() + O2 (g)
ΔH = -196.4 kJ mol-1
Transition metal oxides act as catalysts in the decomposition of the hydrogen peroxide.
Unfortunately, there are hazards associated with the use of hydrogen peroxide as a fuel
in torpedoes. It is possible that a leak of hydrogen peroxide solution from a rusty torpedo
may trigger an explosion.
Using your knowledge of chemistry, comment on why this could happen.
(This question is worth 3 marks)
..........................................
..........................................
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GLOSSARY
Glossary
Avogadro's constant
this is the number of atoms in one mole of an element. In particular, it is the
number of atoms in 12.0g of the isotope carbon-12. This number is given the
symbol L and has a value of 6.02 × 10 23
Bond enthalpies
bond enthalpy is the amount of energy needed to break one mole of a bond in a
gaseous molecule
Chromatography
an analytical method where mixtures are separated into their components by
partitioning between a stationary and mobile phase. The stationary/mobile phases
are solid/liquid in paper and thin layer chromatography, and liquid/gas in gas-liquid
chromatography.
Dynamic equilibrium
a dynamic equilibrium is achieved when the rates of two opposing processes
become equal, so that no net change results
Endothermic reactions
absorb heat energy from the surroundings
End-point
the point at which the reaction is just complete
Enthalpy change
the name given to the amount of heat evolved or absorbed in a reaction carried
out at constant pressure. It is given the symbol ΔH, read as "delta H"
Enthalpy of combustion
is the enthalpy change that occurs when 1 mole of a substance is burned
completely in oxygen
Enthalpy of neutralisation
is the energy change (in kJ) when an acid is neutralised to form 1 mole of water
Enthalpy of solution
is the energy change (in kJ) when 1 mole of the substance dissolves in water
Equilibrium
chemical equilibrium is the state reached by a reaction mixture when the rates of
forward and reverse reactions have become equal
Exothermic reactions
release heat energy, which is given up to the surroundings
© H ERIOT-WATT U NIVERSITY
GLOSSARY
Feedstocks
a feedstock is a reactant from which other chemicals can be extracted or
synthesised. Feedstocks are themselves derived from raw materials either by
physical separation or by chemical reaction
Formula unit
the term 'formula unit' is a general term. A formula unit may be an atom (for all
elements which do not exist as diatomic molecules), a molecule (for all covalent
molecular substances) or the simplest ratio of atoms or ions (for network or lattice
substances).
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
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
Molar volume
the molar volume is the volume occupied by one mole of a substance. For gases,
the units used are mol-1 . (Note that some texts will quote the molar volume in
units of decimetres cubed per mole (dm 3 mol-1 )
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
Potential energy diagram
shows the enthalpy of reactants and products, and the enthalpy change during a
chemical reaction
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
Retention time
the time taken for an individual peak to traverse the gas-liquid chromatographic
column after the injection time
Specific heat capacity
relates the energy change in a liquid to the change in temperature. For water it
has a value of 4.18 kJ kg-1 ◦ C-1 . In other words, when 1 kg of water absorbs 4.18
kJ of heat its temperature will rise by 1◦ C.
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GLOSSARY
Standard solution
a solution of accurately known concentration
Theoretical yield
the theoretical yield is the maximum possible amount of product in a reaction, i.e.
all of the reactant(s) have been converted into product
Thermochemical equation
states the enthalpy change for the reaction defined, with reactants and products in
the states shown
Titration
determines the volume of reactant solution required to react completely with the
test solution
Volumetric analysis
involves analysis using a solution of accurately known concentration in a
quantitative reaction to determine the concentration of another substance
© H ERIOT-WATT U NIVERSITY
HINTS
Hints for activities
Topic 1: Getting the most from reactants
Test your prior knowledge
Hint 1:
1. Write a balanced equation.
2. Identify mole ratio.
3. Calculate number of moles of potassium hydroxide (n = cv).
4. Use mole ratio to determine number of moles of nitric acid.
5. Calculate concentration of nitric acid (c = n/v).
Topic 3: Chemical energy
Test your prior knowledge
Hint 1: Question 4
•
Identify equation to use the equation Eh = cmΔT.
•
Convert mass of water to kg.
•
Calculate ΔT. (Final Temperature - Initial Temperature).
•
Use equation to solve.
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ANSWERS: TOPIC 1
Answers to questions and activities
1 Getting the most from reactants
Test your prior knowledge (page 4)
Q1:
d) All of the above
Q2:
d) 14.5
Q3:
b) 0.2 mol l-1
Sulfuric acid manufacture (page 6)
Q4:
a) Sulfur, Air and Water.
Q5:
a) Readily available and c) Available at a suitable cost.
Clearly air and water are very abundant and free. (Why do you think that major sulfuric
acid plants are built near rivers?) Sulfur is available as a mineral deposit, but it is not
sustainable, although it is not thought to be in short supply.
Q6:
a) To make the reaction more economical.
A single catalyst bed converts only part of the sulfur dioxide to sulfur trioxide. This is
because the reaction releases heat which inhibits further reaction. So the mixture is
cooled and passed through further catalysts, so that a large proportion of the sulfur
dioxide is oxidised to trioxide. This makes the process more economic.
Q7:
a) Yes
There are no products other than H 2 SO4 . However, the catalyst will eventually become
ineffective and have to be disposed of.
Q8: The burning of sulfur and conversion in the catalyst beds generate heat which can
be used to heat up other processes that require heat, or to heat business premises.
Answers from page 11.
Q9:
b) 16.05
Q10: c) 4FeS2 + 11O2 → 2Fe2 O3 + 8SO2
Q11: b) 217.9
Q12: d) 61.16
Q13: 23.53 g CH 3 OH; 89.71 g C6 H5 COOH
1. Write the formulae for methanol, benzoic acid and methyl benzoate then calculate
the relative formula masses.
(CH3 OH - 32; C6 H5 COOH - 122; C6 H5 COOCH3 - 136)
2. Write a balanced equation.
(CH3 OH + C6 H5 COOH → C6 H5 COOCH3 + H2 O)
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221
3. From the balanced equation, write down the mole ratios, then the ratio of masses.
(1 mol CH3 OH reacts with 1 mol C6 H5 COOH to make 1 mol C 6 H5 COOCH3
32g CH3 OH reacts with 122 g C6 H5 COOH to make 136 g C 6 H5 COOCH3)
4. Use mathematical ratios to calculate the required quantities of alcohol and acid to
make 100 g ester.
Answers from page 15.
Q14:
We know the total mass of the nails and we know the mass of one nail.
total mass
mass of one
10.75
=
0.43
= 25
The number of nails =
Nuts and bolts (page 15)
Q15:
The number of nuts =
=
However, 1 tonne =
So, the number of nuts =
=
total mass
mass of one
3 tonnes
3g
1000 kg = 1000000 g
3000000
3
1000000 (one million)
Q16: 1000000
Q17: 2
Avogadro's constant calculations (page 16)
Q18:
total mass of helium
mass of one He atom
4.0 g
=
4.0 amu
1.0 g
=
1.0 amu
Number of atoms (L) =
Q19: L
Q20: 27.0
Q21: 63600
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ANSWERS: TOPIC 1
Formula units (page 17)
Q22: atom
Q23: molecule
Q24: c) one sodium ion and one chloride ion.
Q25: c) one silicon atom and two oxygen atoms.
Calculations using Avogadro's constant (optional) (page 20)
Q26:
The GFM for carbon dioxide is 44.0 g.
Given
Required
Mass of CO2
Number of molecules
Key
relationship
1 mole contains
L formula units
For CO2
44 g contains
6.02 × 1023 molecules
In this case
0.22 g contains
0.22/44 × 6.02 × 10 23
molecules
Answer
0.22 g CO2 contains
3.01 × 1021 molecules
Here is the calculation:
44.0 g contains 6.02 × 10 23 molecules
1.0 g contains
6.02 × 1023
44.0
molecules
0.22 × 6.02 × 1023
44.0
10 23 molecules
0.22 g contains
= 0.0301 ×
molecules
= 3.01 × 1021 molecules
Q27:
Direction
Key relationship
Given
→
Required
number of particles
→
mass
6.02 × 1023 formula units are in one GFM
The GFM for calcium carbonate is 100.0 g.
6.02 × 1023 formula units in 1 GFM
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ANSWERS: TOPIC 1
223
6.02 × 1023 formula units in 100.0 g
1 formula unit in
100.0
g
6.02 × 1023
3.612 × 1020 formula units in
= 3.612
6.02
3.612 × 1020 × 100.0
6.02 × 1023
g
× 10-1 g
= 0.6 × 10-1 g
Q28:
a)
Direction
Key relationship
Given
→
Required
mass
→
number of
molecules
one GFM contains 6.02 × 10 23 molecules
The GFM for methane is 16.0 g.
16.0 g contains 6.02 × 10 23 molecules
6.02 × 1023
molecules
16.0
6.02 × 1023
molecules
0.112 g contains 0.112 × 16.0
23
= 0.04214 × 10 molecules
1.0 g contains
= 4.214 × 1021 molecules
b) Methane has the formula CH 4 and so there are five atoms in each molecule.
The number of atoms
= 5 × 4.214 × 10 21
= 2.107 × 1022
Q29:
Magnesium nitrate has the formula Mg(NO 3 )2 . Each formula unit contains three ions
(one Mg2+ and two NO3 - ).
The GFM of magnesium nitrate is 148.3 g.
148.3g contains 6.02 × 10 23 formula units
6.02 × 1023
formula units
148.3
37000 × 6.02 × 1023
formula
37000g contains
148.3
= 1502 × 1023 formula units
1.0g contains
= 1.502 × 1026 formula units
Number of ions
= 3 × 1.502 × 10 26
= 4.506 × 1026 ions
© H ERIOT-WATT U NIVERSITY
units
224
ANSWERS: TOPIC 1
Q30:
Potassium sulfate has the formula K 2 SO4 and a GFM of 174.3 g. One formula unit
contains three ions (two K+ and one SO 4 2- )
9.03 × 1021 ions = 3.01 × 1021 formula units
6.02 × 1023 formula units in GFM
6.02 × 1023 formula units in 174.3 g
1 formula unit in
174.3
6.02 × 1023
g
3.01 × 1021 formula units in
3.01 × 1021 × 174.3
6.02 × 1023
g
= 87.15 × 10-2 g
= 0.8715 g
Molar volume (page 24)
Q31:
Initial mass of container and gas = 124.86 g
Final mass of container and gas = 124.77 g
Mass of hydrogen = 0.09 g
GFM of hydrogen = 2.0 g
Direction of calculation is mass ⇒ volume
0.09 g occupy 1 l
1.0 g occupy
2.0 g occupy
1
0.09
2.0
0.09
l
l
= 22.2
Molar volume of hydrogen is 22.2 l mol -1
Q32:
Initial mass of container and gas = 124.86 g
Final mass of container and gas = 122.88 g
Mass of carbon dioxide = 1.98 g
GMF of carbon dioxide = 44.0 g
Direction of calculation is mass ⇒ volume
1.98 g occupy 1 l
1
1.98 l
occupy 44.0
1.98
1.0 g occupy
44.0 g
l
= 22.2
Molar volume of carbon dioxide is 22.2 mol -1
Q33: They are all the same
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 1
Answers from page 26.
Q34: 22.86
Q35: 22.65
Q36: 22.19
Answers from page 26.
Q37: 30.0
Molar volume - Further practice (page 27)
Q38: 16 l mol-1
Q39: 0.0321 moles
Q40: 16500cm 3
Combustion of methane - Further practice (page 31)
Q41:
Start by writing the balanced equation, mole relationship and volume relationship.
Work out which reactant is in excess:
1 volume of propane requires 5 volumes of oxygen
200 cm3 of propane require exactly 1000 cm 3 of oxygen
There are 1300 cm3 of oxygen and so oxygen is in excess.
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226
ANSWERS: TOPIC 1
1000 cm3 of the oxygen is used up. So 300 cm 3 remains unreacted.
600 cm3 of carbon dioxide is formed. So the total volume of the resulting gas mixture
is 900 cm3 .
Q42:
Start by writing the balanced equation, mole relationship and volume relationship.
Work out which reactant is in excess:
1 volume of ethene requires 3 volumes of oxygen
400 cm3 of ethene require exactly 1200 cm3 of oxygen
There are only 900 cm3 of oxygen and so ethene is in excess.
300 cm3 of the ethene is used up. So 100 cm 3 of ethene remains unreacted.
600 cm3 of carbon dioxide is formed. So the total volume of the resulting gas mixture
is 700 cm3 .
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 1
227
Answers from page 35.
Q43:
a)
64 g ⇒ 26 g
26
g
1g ⇒
64
26
× 2.56 g
2.56 g ⇒
64
= 1.04 g
Theoretical yield is 1.04 g.
b)
actual
× 100
theoretical
0.99
× 100
=
1.04
= 95.2%
Percentage yield =
Q44:
a)
26 g ⇒ 168 g
168
g
1g ⇒
26
168
× 1.3 g
1.3 g ⇒
26
= 8.4 g
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ANSWERS: TOPIC 1
Theoretical yield is 8.4 g.
b)
actual
× 100
theoretical
7.14
× 100
=
8.4
= 85%
Percentage yield =
Calculations involving percentage yields - Further practice (page 36)
Q45: 56.3 tonnes
Q46: 90.6%
Q47: 12.43 grams
Q48: 8.08 grams
Answers from page 37.
Q49:
Using relative atomic masses from p. 1 of the data booklet, mass of CaO is 56, mass of
CaCO3 is 100.
substituting into the equation: atom economy = (56/100) × 100 = 56%
The atom economy of the production of calcium oxide from calcium carbonate is
56%
Q50:
Using relative atomic masses from p. 1 of the data booklet, mass of SO 3 is 80.1, mass
of SO2 is 64.1, mass of 1 /2 O2 is 16.
substituting into the equation: atom economy = (80.1/80.1) × 100 = 100%
The atom economy of the production of sulfur trioxide from sulfur dioxide and
oxygen is 100%
Q51: All the materials go to produce the product. There is no waste from the process.
Q52: Rearrangements have an identical mass of reactant and product, so have 100%
atom economy.
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229
Ibuprofen (page 38)
Q53:
Used in
ibuprofen
Reagent
Not used
Formula
Mr
Formula
Mr
Formula
Mr
C10 H14
134
C10 H13
133
H
1
C 4 H 6 O3
102
C2 H3 O
43
C 2 H 3 O2
59
H2
2
H2
2
0
CO
28
CO
28
0
Ibuprofen
Total
C15 H22 O4
266
C13 H18 O2
Waste
206
C 2 H 4 O2
60
Q54: atom economy = (206/266)100 = 77%
Q55:
Q56:
Step 1 - 90% yield = 0.9 (so for each 100 g of product you should theoretically get, there
is only 0.9 i.e. 90 g produced);
Step 2 - 0.9 × 0.9 = 0.81 = 81% (should get 90 g but only 0.9 of this, so 81 g) ;
Step 3 - 0.9 × 0.9 × 0.9 = 0.73 = 73%
... Step 6 - (0.9)6 = 0.53 = 53%
Q57: 73%
Q58:
atom economy = mass of product / mass of starting materials
percentage atom economy = (206/514.5)100 = 40%
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ANSWERS: TOPIC 1
Q59:
Converting to grams, 3000 tonnes is 3 × 10 9 g
Number of tablets taken is 3 × 10 9 /0.2 = 1.5 × 1010
Q60: 250
Q61: Addition reactions
Q62: The two new stages are addition reactions which have a 100% atom economy.
These replace the wasteful previous five stages.
Answers from page 43.
Q63: Each sandwich needs two slices of bread and one slice of cheese, so you can
make 12 sandwiches. This will use all 12 cheese slices and 24 slices of bread. What
remains? 4 slices of bread. The bread can be said to be 'in excess'. The cheese is
'limiting', since the number of cheese slices limits the number of sandwiches you can
prepare.
Calculating excess - Tutorial examples (page 45)
Q64:
Solution
•
Write a balanced equation, including the mole ratios
•
Choosing aluminium and calculating quantities involved
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ANSWERS: TOPIC 1
•
Calculate quantities of the other reactant involved
•
Use the balanced equation to work out which reactant is in excess
In this case the aluminium to iron (III) oxide react in a 2:1 ratio.
So 1.85 mol of aluminium would require 0.925 mol iron (III) oxide.
(There is not enough iron (III) oxide, aluminium is in excess.)
So 0.63 mol iron (III) oxide would require 1.26 mol of aluminium.
Excess aluminium = 1.85 - 1.26 = 0.59 mol
•
Calculate excess present
The aluminium is in excess by 0.59 mol.
1 mol aluminium = 27 g
So 0.59 mol aluminium = 0.59 × 27 g
0.59 mol aluminium = 15.93 g
Excess aluminium = 15.93 g
Q65: hydrochloric acid
Q66: 0.3
Q67: hydrochloric acid
Q68: 0.02
Q69: 0.06
Q70: copper (II) oxide
Q71: copper (II) oxide
Q72: 0.70
Q73: 2.0
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ANSWERS: TOPIC 1
End of topic 1 test (page 48)
Q74: 228.6 cm3 mol-1
Q75: B and C. 32g CH 4 and 4 g H2
Q76: C. CH4(g) + 2O2 (g) → CO2 (g) + 2H2 O(l)
Q77: 100 cm3
Q78: 100 cm3
Q79: 600 cm3
Q80: C. 4
Q81: A. 1.0 litre
Q82: 85%
Q83: 64%
Q84: b) feedstocks.
Q85: A: The mass of marble chips left over after the reaction had finished.
Q86: B: 80%
Q87: b) heat exchangers and c) improved catalysts
Q88:
Q89: 16.2 tonnes
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ANSWERS: TOPIC 1
Q90: 38%
Q91: 2CH3 CH(OH)CH3 + O2 → 2CH3 COCH3 + 2H2 O
Q92: 76%
Q93: Any three from:
•
percentage yield;
•
availability of raw materials;
•
potential sales of co-product (phenol);
•
energy efficiency of the process;
•
costs of building and running the plant.
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ANSWERS: TOPIC 2
2 Chemical equilibria
Test your prior knowledge (page 55)
Q1: a) Equilibrium is a situation where forward and reverse reactions take place at the
same rate.
Q2:
b) A reaction which proceeds in both directions at the same time.
Q3:
d) H2 0(l) H+ (aq) + OH- (aq)
Establishing dynamic equilibrium (page 60)
Q4:
6
Q5:
4
Q6:
2
Q7:
b) remain constant, but not necessarily equal.
Q8:
2
Q9:
c) There is the same equilibrium mixture as before.
Q10: This is an equilibrium because the rate of forward reaction (left to right) is equal
to the rate of backward reaction (right to left). It is dynamic because both reactions
continue and although it looks, to an external observer, as if no change is occurring,
products continue to be formed and undergo further reaction to reform reactants. (This
answer could gain 2 marks.)
Answers from page 62.
Q11: 4
Q12: 6
Q13: endothermic
Q14: This is to be expected as Le Chatelier's principle would predict that an increase
in temperature will favour the reaction which absorbs heat energy, the endothermic
reaction, which is the forward reaction. This is shown in the colour of the gas, which
is dark brown. This answer could gain 2 marks.
Q15: 2
Q16: exothermic
Q17: This is to be expected as Le Chatelier's principle would predict that a decrease
in temperature will favour the reaction which gives out heat energy, the exothermic
reaction, which is the reverse reaction. This is shown in the colour of the gas, which
is very pale, almost colourless. This answer could gain 2 marks.
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ANSWERS: TOPIC 2
Equilibrium and pressure (page 64)
Q18: decreased
Q19: decreased
Q20: left
Q21: Le Chatelier's principle states that if the conditions of a chemical system at
equilibrium are changed, the system responds by minimising the effect of the changes.
Increasing the pressure moves the equilibrium to the side of the equation with fewer gas
molecules as this reduces the pressure. The position of equilibrium shifts to the side
with the N2 O4(g) (the left). The colour lightens. A full answer like this would be worth two
or three marks in an examination situation.
Q22: Le Chatelier's principle states that if the conditions of a chemical system at
equilibrium are changed, the system responds by minimising the effect of the changes.
Decreasing the pressure shifts the equilibrium to the side of the equation with a larger
volume and more gas molecules as this increases the pressure. The system shifts to
the side with the NO2 (g) (the right). This is shown in the observation that the colour
darkens due to the increase in the number of NO 2 (g) molecules.
Changing concentration (page 65)
Q23: nitrogen
Q24: increased
Q25: right
Q26: Le Chatelier's principle states that if the conditions of a chemical system at
equilibrium are changed, the system responds by minimising the effect of the changes.
Increasing the concentration of nitrogen shifts the position of equilibrium to the side
of the equation which uses up the nitrogen, thus producing more ammonia. The
equilibrium position shifts to the right.
Q27: Le Chatelier's principle suggests that increasing the concentration of nitrogen
shifts the position of equilibrium to the side of the equation which uses up the nitrogen.
As this happens, hydrogen has to be used up to react with the nitrogen and form
ammonia, thus the hydrogen line drops after nitrogen is added.
Answers from page 68.
Q28: decreased
Q29: Le Chatelier's principle states that if the conditions of a chemical system at
equilibrium are changed, the system responds by minimising the effect of the changes.
Decreasing the concentration of ammonia causes an initial dip in the concentration,
but the system then responds. The equilibrium shifts to the side of the equation which
minimises the change, by producing more ammonia. The equilibrium shifts right.
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ANSWERS: TOPIC 2
Answers from page 69.
Q30: low
Q31: Is the forward reaction exothermic or endothermic? According to Le Chatelier's
principle will high or low temperature favour the forward reaction? (Remember: Le
Chatelier's principle states that if the conditions of a chemical system at equilibrium
are changed, the system responds by minimising the effect of the changes). Since the
forward reaction is exothermic, low temperature favours the forward reaction.
Q32: a) Less than 10% ammonia
Q33: At low temperatures the reaction rate is very slow and it takes a long time to
reach equilibrium. At 400 ◦ C the reaction reaches equilibrium faster and although the
yield is lower, separating off the ammonia by cooling and condensing, and recycling the
unreacted gases gives a continuous stream of ammonia at a reasonable rate.
Answers from page 70.
Q34: high
Q35: Is there an increase or decrease in volume in the forward reaction? According
to Le Chatelier's principle will high or low pressure favour the forward reaction?
(Remember: Le Chatelier's principle states that if the conditions of a chemical system at
equilibrium are changed, the system responds by minimising the effect of the changes).
The forward reaction occurs with a decrease in volume. High pressure favours this
decrease in volume.
Q36: High pressure processes are very expensive to maintain, have high construction
and energy costs and are less safe. At 200 atmospheres there is still a reasonable yield
of ammonia, the pressure is high enough to force the equilibrium to the ammonia side
and the separating of ammonia and recycling of unreacted gases ensures an economic
process.
Summary of the Haber process (page 72)
Q37:
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ANSWERS: TOPIC 2
End of topic 2 test (page 74)
Q38: B. rate of forward reaction equals the rate of reverse reaction.
Q39: C. remain constant.
Q40: B. is a reversible reaction.
Q41: A. Low temperature, high pressure
Q42: C. the proportion of sulfur trioxide to decrease.
Q43: D. both reactant gases to increase concentration.
Q44: A. no change in the position of equilibrium.
Q45: A. KCl(s)
Q46: B and D: KOH(s) and AgNO 3 (s)
Q47: A and C: It decreases the time required for equilibrium to be established. and It
lowers the activation energy of the reverse reaction.
Q48: endothermic
Q49: The concentration of nitrogen dioxide would increase.
Q50: 450◦ C
Q51: They are recycled.
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ANSWERS: TOPIC 3
3 Chemical energy
Test your prior knowledge (page 83)
Q1:
a) Combustion
Q2:
c) Li2 CO3 (s) + 2HCl(aq) → 2LiCl(aq) + CO2 (g) + H2 O(l)
Q3:
a) Exothermic
Q4:
b) 33.44
Q5:
b) addition.
Answers from page 88.
Q6:
c) It will decrease.
Q7:
c) Positive
Answers from page 89.
Q8:
d) Endothermic
Q9:
a) Positive.
Q10:
Answers from page 93.
Q11: d) ΔH = +283 kJ mol-1
Q12: b) ΔH = Hproducts - Hreactants
Q13: c) Exothermic
Q14: d) ΔH = -1411 kJ mol-1
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ANSWERS: TOPIC 3
Q15: -6835 kJ
Q16: +150 kJ mol-1
Q17: 4NH3 (g) + 5O2 (g) → 4NO(g) + 6H2 O(g)
Q18: -906 kJ
Q19: -226.5 kJ mol-1
Q20: 1359 kJ
Determination of the enthalpy of combustion of an alcohol (page 97)
Q21: Some of the heat from the burning will have been lost completely from the
equipment, e.g. in heating the air.
You took account only of the heat given to the water, but the container and thermometer
would also take some heat to warm up.
Perhaps the alcohol did not burn completely to water and CO 2 .
All these effects mean that the temperature rise noted was lower than it should have
been, resulting in a smaller value for ΔH c .
Enthalpy of combustion questions (page 99)
Q22: b) Butane
Q23: 68.9 ◦ C
Q24: -522.5 kJ mol-1
An experiment to determine the enthalpy of solution of ammonium nitrate (page
100)
Q25: You need the mass of water being heated (m), and the initial temperature. The
value of c is a constant, obtained from the data booklet.
Q26: The reaction occurring cools the solution. Once it is below the ambient
temperature it will slowly absorb heat from the surroundings and rise in temperature.
Q27: -1.6 ◦ C
Answers from page 102.
Q28: -40.1 kJ mol-1
Q29: 369 kJ
Answers from page 104.
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ANSWERS: TOPIC 3
Q30: 0.1 mol of water
Answers from page 105.
Q31: H2 SO4 (aq) + 2KOH(aq) → K2 SO4 (aq) + 2H2 O(l)
Q32: 21.4 ◦ C
Q33: +13.6 ◦ C
Q34: 0.10 kg
Q35: 5.685 kJ
Q36: 0.1
Q37: -56.85 kJ mol-1
Q38: They are almost the same
Q39:
Sulfuric acid and potassium hydroxide are almost completely ionised in solution, so if
you write ionic equations for these two neutralisation reactions then remove spectator
ions, they both reduce to the same equation. The same reactions must have the same
ΔH value.
which are both
Answers from page 106.
Q40: B
Q41: c) +570
Answers from page 107.
Q42: 4
Q43: They are activation energies for the reaction. Energy 2 is the large activation
energy for the uncatalysed reaction; energy 1 shows the reduced activation energy in
the presence of a catalyst.
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ANSWERS: TOPIC 3
241
Answers from page 110.
Q44:
Q45:
E = c × m × ΔT
= 4.18 × 0.025 × 18.1
= 1.89 kJ
This is the energy released when 0.80 g of sodium hydroxide reacts.
1 mole of NaOH = 40 g
0.80 g releases 1.89 kJ
1.89
kJ
1.0 g releases
0.80
1.89
kJ
40.0 g releases 40 ×
0.80
The enthalpy change = −94.5 kJ mol −1
(Note the negative sign since energy is released.)
Answers from page 110.
Q46: -41.8 kJ mol-1
Q47: 8.0 o C
Q48: 0.025 kg
Q49: 0.836 kJ
Q50:
1 mole of NaOH = 40 g
0.80 g releases 0.836 kJ
0.836
kJ
1.0 g releases
0.80
0.836
kJ
40.0 g releases 40.0 ×
0.80
= 41.8 kJ
Q51: -41.8 kJ mol-1
Answers from page 111.
Q52: -52.25 kJ mol-1
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ANSWERS: TOPIC 3
Q53: 5.0 ◦ C
Q54: 0.05 kg
Q55: 1.045 kJ
Q56:
The solution from step 1 contains 0.80 g of NaOH
1 mole of NaOH = 40 g
0.80 g releases 1.045 kJ
1.045
kJ
1.0 g releases
0.80
0.836
kJ
40.0 g releases 40.0 ×
0.80
= 52.25 kJ
Q57: -52.25 kJ mol-1
Q58: -94.05 kJ mol-1
Application of Hess's law by enthalpy cycle diagram (page 112)
Q59: -55.5 kJ
Q60: +222 kJ
Application of Hess's law by algebraic calculation (page 114)
Q61: +55.5 kJ
Q62: +333 kJ
Bond enthalpies and the synthesis of ammonia (page 117)
Q63:
Bonds broken
N≡N
3×H-H
Total enthalpy
change to break
bonds
ΔH
945
Bonds formed
ΔH
6×N-H
6 × 388
Total enthalpy
change on making
bonds
-2328 kJ
3 × 436
+2253 kJ
ΔH to break bonds + ΔH to make bonds:
2253 + (-2328) = -75 kJ
(You may have noticed earlier in the topic that the reaction enthalpy change was
described as the enthalpy in the product bonds minus the enthalpy in the reactant bonds.
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ANSWERS: TOPIC 3
The equation above is for enthalpy changes and, therefore, includes an appropriate sign
for exo- or endothermic changes. The overall change is then the sum of the changes.)
The hydrogenation of ethene (page 118)
Q64: d) 1 x C=C, 4 x C - H and 1 x H - H
Q65: 1 × C - C and 6 × C - H
Q66: Enthalpy change to break bonds = 612 + (4 × 412) + 436 = +2696
Enthalpy change to form bonds = -348 + (6 × -412) = -2820
Enthalpy change for reaction is +2696 + (-2820) = -124 kJ
Q67: In this case average bond enthalpies were used, and since bond enthalpies are
slightly different in different environments, the results will differ slightly.
End of topic 3 test (page 121)
Q68: A: ΔH = H(products) - H(reactants)
Q69: A: -300
Q70: B: exothermic
Q71: B: C2 H6 (g) + 31 /2 O2 (g) → 2CO2 (g) + 3H2 O(l)
Q72: A: - 2680 kJ mol-1
Q73: B: bonds within the molecules.
Q74: a) - 361 kJ mol-1
Q75:
As molecules are getting larger, there are greater London Dispersion Forces between
them which require more energy to be overcome."
Q76: - 2688 kJ mol-1
Q77: B and E: 3O 2 (g) → 2O3 (g) ; ΔH = +142.7 kJ mol-1
and NH4 Cl (s) → NH4 + (aq) + Cl- (aq) ; ΔH = +15.0 kJ mol-1
Q78: C and E: NaOH(s) → Na + (aq) + OH- (aq) ; ΔH = -42.7 kJ mol-1
and NH4 Cl (s) → NH4 + (aq) + Cl- (aq) ; ΔH = +15.0 kJ mol-1
Q79: D: C3 H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2 O(g) ; ΔH = -2220 kJ mol-1
Q80: C4 H10 (g) + 61 /2 O2 (g) → 4CO2 (g) + 5H2 O(g)
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ANSWERS: TOPIC 3
Q81:
1. The mass of water in the pot.
2. The temperature of the water at the start of the experiment.
3. The temperature of the water at the end of the experiment.
Q82: - 1500 kJ mol-1
Q83: C: the same.
Q84: - 28.3 kJ
Q85: + 566 kJ
Q86: a) ΔH3 = ΔH1 - ΔH2
Q87: 400 kJ
Q88: - 156 kJ
Q89: + 39 kJ
Q90: - 848 kJ
Q91: - 1605 kJ
Q92: -788 kJ
Q93: -858 kJ
Q94: 3 moles
Q95: +1367 kJ
Q96: -279 kJ
Q97: a) N(g) + 3H(g) → NH3 (g)
Q98: a) H-H
Q99: b) -115
Q100:
Worked Solution
Bond breaking
Bond making
4 mol C-H = 4 × 412 = 1648
1 mol Br-Br = 194
3 mol C-H = 3 × 412 = 1236
1 mol C-Br = 276
1mol H-Br = 366
Total energy given out = -1878 kJ
Total energy put in = +1842 kJ
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ANSWERS: TOPIC 3
245
ΔH = 1842 - 1878
= -36 kJ mol-1
Q101:
Worked Solution
Bond breaking
Bond making
4 mol C-H = 4 × 412 = 1648
2 mol O=O = 2 × 498 = 996
Total energy put in = +2644 kJ
2 mol C=O = 2 × 743 = 1486
4 mol H-O = 4 × 463 = 1852
Total energy given out = -3338 kJ
ΔH = 2644 - 3338
= -694 kJ mol-1
Q102:
Worked Solution
Bond breaking
Bond making
1 mol C=C = 612
4 mol C-H = 4 × 412 = 1648
1 mol H-Br = 366
Total energy put in = +2626 kJ
1 mol C-C = 348
5 mol C-H = 5 × 412 = 2060
1 mol C-Br = 276
Total energy given out = -2684 kJ
ΔH = 2626 - 2684
= -58 kJ mol-1
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ANSWERS: TOPIC 4
4 Oxidising or reducing agents
Test your prior knowledge (page 132)
Q1:
a) gain of electrons.
Q2:
c) Pb2+ (aq) + 2NO3 - (aq) + 2K+ (aq) + 2l- (aq) → Pb2+ (l- )2 (s) + 2K+ (aq) + 2NO3 - (aq)
Q3:
e) 5H2 O2 (l) + 2MnO4 - (aq) + 6H+ (aq) → 2Mn2+(aq) + 8H2 O(l) + 5O2 (g)
Answers from page 136.
Q4:
oxidation
Q5:
c) Bromine
Q6:
iodide
Q7:
a) Top of the table.
Q8:
sulfur
Q9:
magnesium
Q10: sulfur
Q11: b) Br2 (1)
Q12: a) Na (s)
Q13: c) Sn2+ (aq)
Displacement reactions (page 138)
Q14: sulfate
Q15: Zn → Zn2+ +2eQ16: Cu2+ +2e- → Cu
Q17: d) Copper ions
Q18: zinc atoms
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ANSWERS: TOPIC 4
Answers from page 139.
Q19: These salts are soluble in water so can be used in reactions in solution - the
commonest ones in chemistry.
Q20: 2KMnO4 + 16HCl → 2KCl + 2MnCl 2 + 8H2 O + 5Cl2
Q21: The permanganate ion (MnO 4 - ) is the oxidising agent and the reducing agents are
chloride ion (Cl- ) and hydrogen ion (H + ).
Q22: MnO4 - is reduced to manganese(II) ion (Mn 2+ ). Chloride ion is oxidised to chlorine
and hydrogen ion oxidised to water.
Q23: K+ is a spectator ion.
Answers from page 141.
Q24: Looking at the reverse reaction it is much easier to see that glucose is oxidised by
oxygen to carbon dioxide and water. In the case of photosynthesis, the carbon dioxide
is reduced to glucose, and the water is oxidised to molecular oxygen.
Q25: Increase in ppmv 390 - 315 = 75 ppmv CO 2
So volume of CO2 is: Total volume of atmosphere x ppmv x 10 -6
4 x 1020 x 75 x 10-6 = 3 x 1016 litres of CO2
Writing ion-electron equations (page 146)
Q26: K → K+ + 1eQ27: Mg → Mg2+ + 2eQ28: Al → Al3+ + 3eQ29: F + 1e- → FQ30: N + 3e- → N3Q31: O + 2e- → O2-
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ANSWERS: TOPIC 4
Answers from page 147.
Q32: b) At the bottom of the table.
Q33: c) At the top of the table, in reverse.
Q34: b) F2 (g) + 2e- → 2F- (aq)
Q35: a) Na+ (aq) + e- → Na(s)
Tutorial - simple ion-electron equations (page 149)
Q36:
Q37:
Q38:
Q39:
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ANSWERS: TOPIC 4
Answers from page 152.
Q40: 1
Q41: 2
Q42: 2
Q43: a) Reactant
Q44: b) Reduction
Q45: The oxidation equation for Fe 2+ has to be multiplied by 2 and then added to the
reduction equation:
Tutorial - complex ion-electron equations (page 153)
Q46: 3
Q47: 6
Q48: 2
Q49: b) Product
Q50: a) Oxidation
Q51:
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ANSWERS: TOPIC 4
Q52:
Q53:
Q54:
The reduction ion-electron equation can be developed first:
Combining this with the oxidation:
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ANSWERS: TOPIC 4
Estimation of vitamin C in fruit juices (page 156)
Q55: orange juice
Q56: 4.0 cm3
Q57: Number of moles of iodine used = 0.005 x 0.004 = 2.0 x 10 -5 mol
Q58: a) 1:1
Q59: Number of moles of iodine used = Number of moles of vitamin C = 2.0 x 10 -5 mol
This is present in 20.0 cm3 , therefore in one litre:
Vitamin C concentration = 2.0 x 10 -5 mol divided by 0.02 litres (20.0 cm3 )
This is equal to 0.001 mol l -1
Q60: Vitamin C concentration = 0.001 mol l -1
In mg l -1 = 0.001 x 176 = 0.176 g l -1 = 176 mg l -1 .
Q61: 340.9 cm3
Tutorial - calculations in redox titrations (page 158)
Q62: 0.00225 mol l -1
Q63: Vitamin C concentration = 0.00225 mol l -1
In mg l -1 = 0.00225 x 176 = 0.396 g l -1 = 396 mg l -1 .
Q64: 151.5 cm3
Q65: 0.009 moles
Q66: 0.0045 mol l -1
Q67: 319.5 mg l -1
Q68: The working for the last three questions can be shown thus:
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ANSWERS: TOPIC 4
=
Summary exercise (page 159)
Q69: b) Fe2 O3 + 2Al → 2Fe + Al2 O3
Q70: c) Spectator
Q71: c) Fe3+ + Al → Fe + Al3+
Q72: b) loss
Q73: a) Al
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ANSWERS: TOPIC 4
Q74: d) Fe3+
Q75: a) Copper
Q76: c) 3 kg
End of topic 4 test (page 163)
Q77: a) Cl2 + 2e- → 2ClQ78: b) accepts electrons.
Q79: b) 2Br- → Br2 + 2eQ80: b) Fe2+ → Fe3+ + eQ81: b) Cu2+ → Cu + 2eQ82: c) Iodide ions will react with chlorine.
Q83: d) Fe3+
Q84: c) Add 2 electrons to the left hand side.
Q85: d) Iron(III) ions will react with tin(II) ions.
Q86:
Q87:
Q88: d) Cl2 (g)
Q89: c) Ca (s)
Q90: e) Fe2+ (aq)
Q91: 3
Q92: 6
Q93: 6
Q94: a) Reactant
Q95: b) a reduction.
Q96: 0.0016
Q97: a) 1:2.5
Q98: 0.16
Q99: 5.44
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ANSWERS: TOPIC 5
5 Chromatography
Test your prior knowledge (page 169)
Q1:
a) Distillation
Q2:
b) Filtration
Q3:
b) two solutions react to form a solid.
Chromatography of blue and black inks (page 172)
Q4:
b) Black
Q5:
e) Green
Q6:
a) It is more polar than the red component.
Q7:
a) has the highest solubility in the solvent.
Q8:
c) Dark blue
Q9: They are probably the same dyes in both cases since the R f values would be
the same. If they were different materials, they would probably have moved different
distances.
Q10: If the sample components are hardly moving with the mobile phase, the polarity of
the phase is too high, too similar to the stationary water. Try and use a less polar mobile
phase, for example, octanol.
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ANSWERS: TOPIC 5
Chromatography of protein hydrolysates (page 175)
Q11: 3
Q12: cysteine
Q13: d) A, B and D
Q14:
Q15: Hydrolysed protein Y contains two amino acids (one mark). It contains cysteine
because it has a spot which travelled the same as spot "C" (one mark). It contains an
amino acid which cannot be named but it is not A, B or D (one mark).
Q16: 3
Q17: phenylalanine
Q18: alanine
Q19: b) False
Q20: Glycine with the lowest molecular mass travels furthest; but phenylalanine, with
the highest molecular mass is not far behind, so mass does not determine the distance
travelled.
Q21: There are more than 20 amino acids found in proteins. Each one will move a
different distance, so the developed chromatogram will have a large number of spots
merging into a streak from the origin to the solvent front.
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ANSWERS: TOPIC 5
Q22: You need a very much longer distance for the amino acids to travel so that they
can eventually separate. The problem with longer and longer plates is that i) as the spots
travel diffusion causes them to spread out, and they also form a longer trail and ii) the
solvent front slows down considerably as it travels further, so TLC is usually performed
only on short plates.
To separate many amino acids a technique called high pressure liquid chromatography
(HPLC) is employed. The stationary phase is just like TLC (often silica), but the material
is packed into a long column, so allowing time and distance for separating, and the
mobile solvent is forced through the column with a high pressure pump.
Variations of HPLC are very widely employed in a vast range of analytical systems
from clinical chemistry to the petrochemical industry. Many of these are considerably
automated.
Gas-liquid chromatography (page 180)
Q23: b) Pentane
Q24: a) The retention time increases as mass increases.
Q25: c) Xylene (C8 H10 )
Gel filtration (page 181)
Q26: b) The larger molecules are excluded form the pores in the stationary beads.
Q27: c) The sample molecules would elute from the column in a single peak.
Q28: c) C, B then A
Q29: There are two proteins, one has a molecular mass between 1,000 and 3,000
(probably about 2,000); the other has a molecular mass close to 6,000.
Q30: No. There are only two molecular masses shown, but there could be more than
one type of protein, with very similar molecular masses, under each peak.
Q31:
Note the increased area of the first (6,000) peak owing to the presence of standard and
sample at that position.
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ANSWERS: TOPIC 5
Q32: You need a method that does not depend on size, but on some other property.
You could try paper chromatography, which depends on polarity. If sample X produced
more than two 'spots' there are more than two proteins.
End of topic 6 test (page 184)
Q33: Glycine and alanine
Q34: There are three spots and so three amino acids.
Q35: a) A and B
Q36: c)
Q37: b) A and D
Q38: d) Branched alkanes have lower boiling points than n-alkanes with the same
number of carbon atoms.
Q39: a) Plant waxes
Q40: d) Branched alkanes, such as pristane and phytane, are degraded more slowly
than n-alkanes.
Q41: They are oxidised (1 mark) to carbon dioxide and water (1 mark).
Q42: The oxidation process required for total removal of an oil spill is mainly carried out
by bacteria and fungi. These exist in the water, as does the oxygen. In order to allow
these organisms to oxidise the hydrocarbons, they should be in as small a droplets as
possible, with the largest area in contact with water. Detergents and emulsifiers, with
the churning action of the sea, will break the oil into these small droplets.
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ANSWERS: TOPIC 6
6 Volumetric analysis
Test your prior knowledge (page 191)
Q1:
a) Chromatography
Q2:
a) End point
Q3:
b) A salt and water
Volumetric analysis equipment (page 192)
Q4:
a) A
Q5:
e) E
Q6:
d) D
Q7:
c) C
Titration (page 193)
Q8:
1. Write a balanced equation: HCl + KOH → KCl + H 2 O
The mol relationship: 1 mol HCl reacts with 1 mol KOH
2. Moles of HCl used = concentration volume = 0.150 × 0.0345 = 0.005175 mol
3. Moles of KOH in sample = 0.005175
Concentration = moles/volume = 0.005175/0.025 = 0.207 mol -1
4. Molecular mass of KOH = 39 + 16 + 1 = 56 (g mol -1 )
0.207 mol -1 KOH is 0.207 × 56 = 11.592 g -1
Quality control (page 194)
Q9: Equation: ZnO + 2HCl → ZnCl 2 + H2 O
Mole ratios: 1 mol ZnO reacts with 2 mol HCl
Moles of HCl used: 0.500 × 0.02064 = 0.01032 mol
Moles of ZnO: 0.01032/2 = 0.00516 mol
GFM of ZnO: 65.4 + 16 = 81.4
0.00516 mol = 0.00516 × 81.4 = 0.42 g
0.42 g ml-1 ZnO = 420 g -1 (1 = 1000 ml)
So the slurry contains 420 g ZnO -1 and is within requirements.
The purity of aspirin tablets (page 194)
Q10: Aspirin is a carboxylic acid (acetylsalicylic acid), so it can be estimated by titration
with standard sodium hydroxide solution.
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ANSWERS: TOPIC 6
Q11: The formula of aspirin is: C 9 H8 O4 : RFM = 9×12 + 8×1 + 4×16 = 180
180 g aspirin is 1 mole
1 g aspirin is 1/180 mol
0.300 g aspirin (1 tablet) is 0.300/180 = 0.001667 mol
Q12: From the equation 1 mol aspirin reacts with 1 mol NaOH
So 0.001667 mol aspirin reacts with 0.001667 mol NaOH
For the NaOH solution: 0.300 mol is contained in 1000 ml
1 mol is in 1000/0.300 ml
0.01667 mol is in (1000 × 0.001667)/0.300 = 5.556 ml
Q13:
•
Taking one tablet is not representative, it could have a chip missing, for example. It
would be better to take, say, ten, dissolve them in water and take an aliquot (part)
of this solution for analysis. For example, you could make the solution accurately
to 250 ml in a volumetric flask and take a 25 ml aliquot (1/10 th ) for analysis.
•
The titre of more than 50 ml is quite high. If you make a solution of several tablets
(as above) you could take 1/20 th to get a better titre value.
•
This will also let you carry out a rough titration and as many accurate ones as you
need to feel confident in the result.
Back titration (page 196)
Q14:
1. Moles of HCl remaining: m × v = 0.500 × 0.022 = 0.0110 mol
2. Moles of HCl originally: m × v = 5.00 × 0.005 = 0.0250 mol
3. Moles of HCl neutralised: 0.0250 - 0.0110 = 0.0140 mol
4. Balanced equation: CaCO 3 + 2HCl → CaCl2 + H2 O + CO2
5. Mole ratios: 1 mol CaCO 3 reacts with 2 mol HCl
6. 0.014 mol HCl reacts with 0.007 mol CaCO 3
7. Relative formula mass of CaCO 3: 40.0 + 12.0 + (3 × 16.0) = 100.0
8. 0.007 mol CaCO 3 is 0.7 g
9. One antacid tablet contains 0.7 g CaCO 3 .
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ANSWERS: TOPIC 6
The chloride ion content of seawater (precipitation) (page 197)
Q15:
•
Na+
•
Mg2+
•
K+
•
Ca2+
•
Cl-
•
SO4 2-
•
HCO3 -
Q16: Silver chromate(VI), AgCrO4
Q17:
1. Balanced equation: Ag + (aq) + Cl- (aq) → AgCl(s)
2. Mole ratios: 1 mol Ag+ reacts with 1 mol Cl3. Moles of Ag+ used: m × v = 0.50 × 0.012 = 0.006 mol
4. Moles of Cl- in 10 ml seawater = 0.006 mol
5. Relative formula mass of NaCl: 23.0 + 35.5 = 58.5
6. Mass of NaCl in 10 ml seawater: 0.006 × 58.5 = 0.351 g ≡ 35.1 g NaCl -1
End of Topic 6 test (page 199)
Q18: The oil and aqueous solution will not mix so will be unable to react well. You might
think of a way round this problem, before looking at the procedure (below) for a solution.
Q19: 0.347424 g
Q20: 6.28%
Q21: 0.89 kg
Q22: 0.009 moles
Q23: 0.0045 mol l -1
Q24: 319.5 mg l -1
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ANSWERS: TOPIC 7
261
7 End of unit test
End of unit 3 test (page 202)
Q1: b) decreases.
Q2: b) endothermic.
Q3: c) large.
Q4: a) a small rise in temperature results in a large increase in reaction rate.
Q5: b) 20 kJ mol-1
Q6: a) -30 kJ mol-1
Q7: a)
Q8: d) London dispersion forces
Q9: d) number of protons.
Q10: c) bonds between molecules are weaker than bonds between ions.
Q11: a) 4-methylpentan-2-one.
Q12: b) ethanol and butanoic acid.
Q13: b) 2-methylbut-2-ene and 2-methylbut-1-ene
Q14: A: -398 kJ
Q15: b) Decrease in temperature, increase in pressure
Q16: 0.02 mol
Q17: sulfur
Q18: b) B
Q19: 100cm3
Q20:
Fe(s)
Cl2(g)
Heat
Iced waterr
FeCl3(g)
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ANSWERS: TOPIC 7
Q21: Polar covalent
Q22: b) Ionic → polar covalent → pure covalent
Q23: 190 kJ
Q24: - 10 kJ
Q25: decrease
Q26: have no effect on
Q27: + 26.5 kJ mol-1
Q28: - 32.75 kJ mol-1
Q29:
Q30:
Q31: Essential amino acids
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ANSWERS: TOPIC 7
Q32: a)
Q33: a)
Q34: 156.4 kg
Q35: A: ΔH1 + 2ΔH2 - ΔH3
Q36: 100 %
Q37: Increasing pressure would move the equilibrium to the right (towards product),
since there are fewer molecules on the product side of the equation.
Q38: Increasing pressure would increase the yield.
Q39: Decreasing the temperature would increase the yield since the forward reaction
is exothermic and lowering the temperature would favour the reaction which produces
heat.
Q40: The lower pressure is cheaper to use and maintain and it is safer.
Q41: None. Catalysts affect the rate of a reaction, not the position of the equilibrium.
Q42: b) Sn2+ (aq)
Q43: a) CH3 CH2 COCH3 → CH3 CH2 CH(OH)CH3
Q44: d) Zn + 2HCl → ZnCl 2 + H2
Q45: c) Iron(III) → iron(II)
Q46: a) Fe + 2HCl → FeCl 2 + H2
Q47:
Q48:
Q49: a) OCl- + 2H+ + 2I- → Cl- + H2 O + I2
Q50: 0.02 mol l-1
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Q51: 1.03 mg l-1
Q52: a) Cl2 (g) + 2Br- (aq) → 2Cl- (aq) + Br2 (aq) and b) Br2 (l) + 2Fe2+ (aq) → 2Br- (aq) +
2Fe3+ (aq)
Q53: a) 2IO3 - (aq) + 12H+ (aq) + 11e- → I2 (aq) + 6H2 O(l)
Q54: H2 O2 (aq) + 2H+ (aq) + 2I- (aq) → 2H2 O(l) + I2 (aq)
Q55: 0.0168 g
Q56: As a general rule, award yourself a mark (up to a maximum of three marks in
total) for each point you made. [NB this is a general rule only, remember there are no
half marks awarded at Higher]
•
Measure the rate of reaction of the decomposition of hydrogen peroxide with
manganese peroxide and therefore the faster the reaction, the higher the
concentration of hydrogen peroxide.
•
You could also measure the conductivity as electrons are produced. The greater
the number of electrons released the more hydrogen peroxide has decomposed.
Therefore the higher the conductivity, the more hydrogen peroxide present in the
gel.
•
You could measure the pH of the products as the lower the pH (higher H +
concentration), the more hydrogen peroxide present in the gel.
Q57: As a general rule, award yourself a mark (up to a maximum of three marks in
total) for each point you made. [NB this is a general rule only, remember there are no
half marks awarded at Higher]
•
If hydrogen peroxide was to leak from a rusty torpedo then it would decompose to
form water and oxygen.
•
This reaction is exothermic and so releases energy in the form of heat. The oxygen
gas which is released would combust with any fuel it came into contact with and
this could cause an explosion.
•
The decomposition of hydrogen peroxide is slow at room temperature and pressure
but the presence of the iron (in the rust) would act as a catalyst and greatly increase
the rate of decomposition.
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