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ECZ GRADE 11 CHEMISTRY SUMMARISED
NOTES (FOR 5070 AND 5124) WITH QUESTIONS
AND ANSWERS.
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ESKULU ZM
5/30/18
0
G11 CHEMISTRY SUMMARIZED
NOTES(5070 & 5124)
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This document summarizes Chemistry (5070 & 5124)) notes according to the ECZ
(Examinations Council Syllabus).
The questions and answers are adapted from actual international past exam
papers.
Prepared by William N for eskulu.com
Contact +260978031524
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1. CLASSIFYING BY GROUPS ................................................................................... 4
1.1 Classifying by Periods ................................................................................... 6
2. Rules of Behavior ............................................................................................. 10
2.1 Empirical Formulae .................................................................................... 12
2.2 The Mole .................................................................................................... 16
2.3 Titrations .................................................................................................... 19
challenging questions I......................................................................................... 22
3. THE BEHAVIOUR OF GASES .............................................................................. 24
3.1 The Particles in a Gas ................................................................................. 24
3.2 Avogadro’s law ........................................................................................... 25
3.3 Boyle’s law and Charles’ law ...................................................................... 26
Molar Gas Volume ........................................................................................... 29
3.4 The Ideal Gas .............................................................................................. 30
Using Molar Volumes in Calculations ............................................................... 32
4. ENERGY CHANGES IN CHEMICAL REACTIONS ................................................... 34
Energy Diagrams .............................................................................................. 34
Energy Changes and Heat Content ................................................................... 36
Hess’s Law........................................................................................................ 37
4.2 Endothermic and Exothermic Reactions ..................................................... 38
4.3 Activation Energy ....................................................................................... 39
Challenging Questions II....................................................................................... 40
5. RATE OF REACTION .......................................................................................... 42
5.1 Factors Affecting the Rate of Reaction ....................................................... 42
5.2 Catalysts in Industry ................................................................................... 45
5.3 Reversible Reactions .................................................................................. 46
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6. DYNAMIC EQUILIBRIUM ................................................................................... 47
6.1 What is Dynamic Equilibrium? ................................................................... 47
6.2 Industrial Applications................................................................................ 49
7. FUELS ............................................................................................................... 51
7.1 What is a Fuel? ........................................................................................... 51
7.2 Fossil Fuels ................................................................................................. 52
7.3 Fractional Distillation of Petroleum ............................................................ 52
Fractional Distillation in the Laboratory ........................................................... 53
7.4 Alternative Energy Sources ........................................................................ 55
7.5 Chemicals from Petroleum ......................................................................... 56
8. OXIDATION AND REDUCTION ........................................................................... 58
8.2 Electron Loss and Electron Gain ................................................................. 58
8.3 Reactions Involving Redox.......................................................................... 63
9. Electrolysis ....................................................................................................... 66
9.1 Electrolytes and non-electrolytes ............................................................... 66
9.2 Mechanism of Electrical Transfer ............................................................... 67
9.3 Examples in the laboratory ........................................................................ 69
9.4 The Faraday Constant ................................................................................ 70
10. ACIDS, BASES AND SALTS ............................................................................... 71
10.1 Proton Donors and Acceptors .................................................................. 71
10.2 Salts and their Preparation ....................................................................... 71
10.3 Titrations .................................................................................................. 74
10.4 Hydrolysis of Salts .................................................................................... 76
10.5 Identifying anions ..................................................................................... 78
10.6 Identifying Cations ................................................................................... 79
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1. CLASSIFYING BY GROUPS
A vertical column of elements in the periodic table (Appendix I) is called a group.
GROUP I – the alkali metals
An example is the column containing lithium, sodium and potassium. If you look
again at table 1.5, you can see that each of these elements has one electron in its
outer most shell. This is true of all the elements in the group. The group is called
Group 1.
The reactions of these elements are controlled by this one outermost electron; it is
called the valency electron. For example all these elements in the group react with
water to form hydrogen gas. Here is the equation for the reaction of sodium with
water:
2Na + 2H2O
H2 + 2NaOH
The compounds of Sodium are ionic, not covalent, so it is better to write:
2Na + 2H2O
H2 + 2Na+ + 2OH-
From this ionic equation you can see that the atoms of sodium have each lost one
electron.
These are the characteristics of group behavior. Elements in a group have the
same reactions, but show a trend in their reactions. This may be a trend in vigour,
in speed, or in reaction type.
Group 7 – the halogens
From the table below we see that the element fluorine has seven electrons in its
outermost (L) shell. Counting from the left-hand side of the Periodic Table, fluorine
heads group 7.
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The other common members of the group are chlorine, bromine, iodine. Like the
Group 1 elements, all these have properties in common. They all:






Are coloured,
Have seven valency electrons,
Form ionic, in preference to covalent, compounds,
Form ions with a charge of minus one,
Form colourless compounds,
Act as oxidizing agents (in the right circumstances)
The elements show several trends in their properties. For example:
 Fluorine if the most reactive, iodine the least,
 Fluorine is very pale yellow; chlorine is yellow-green; bromine gas is orange;
iodine gas is deep purple,
 Fluorine and chlorine are gases; bromine is a liquid at RTP; iodine is a solid.
Here is the equation for the reaction between chlorine and bromide ions:
Cl2 + 2Br- (aq)
2Cl- (aq) + Br2 (g)
Group 8 (0) the Noble gases
In Grade 10 you learned that the elements helium, neon, argon, krypton and xenon
are unreactive because they have full electron shells. They form a clearly defined
group of the Periodic Table. They have these properties:
 They form almost no compounds, either with each other or with other
elements.
 All are colourless.
 All are gases at RTP.
Because they have no chemical reactions, we find trends only in their
physical properties. For example the melting points and boiling points of
these elements increase as the relative atomic mass increases.
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1.1 Classifying by Periods
A Horizontal row in the Periodic table is called a period. The elements in any one
period all have different electron structures, so the chemical properties of any
one element differ from those of its neighbours. The electron structures change
regularly, by one electron for each element. All the elements in one period have
an electron core, which is the electron arrangement in the noble gas which
completed the previous period, with outer electrons in addition. So there are
some properties which show a clear trend as we go across a period. This time,
though, there are some which change sharply instead.
Changes in Period 2
The table below shows the second period of the Periodic table with two properties
of the elements.
changes in properties of the elements of period 2 in the Periodic Table
State at RTP
Usual valency
Li
(s)
1
Be
(s)
2
Elements in Period 2
B
C
N
O
(s)
(s)
(g)
(g)
3
4
3
2
F
(g)
1
Ne
(g)
0
Hydrogen
In our Periodic Table we have placed hydrogen at the top of group 1. This is because
hydrogen, like the Group 1 metals, has one electron in its outermost shell (the K
shell). It forms, as do the metals, ionic compounds containing an ion with a single
positive charge.
Some people put hydrogen at the top of Group 7 because it is one electron short
of a noble gas structure (He) and it can also form an ion with a single negative
charge. To get round this problem, we can draw the position of hydrogen as shown
in figure 1.3.
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H
Li
F
There is no right or wrong way to place hydrogen; it is a matter of your own
preference! But if you place hydrogen above lithium, it is usual to show a small gap
between the two.
 Elements with similar outer shell electron structures form groups in the
Periodic Table. They have similar chemical properties. Each property shows a
trend from one element to the next
 Elements with the same electron core form a period in the Periodic Table.
Across a period a property may change sharply or make up a trend
 The number of shells containing electrons is equal to the number of the
period in which the element is placed.
 The number of electrons in the outermost shell is equal to the number of the
group in which the element is placed.
The metals make up most of the elements. The non-metals are clustered together
in the top right-hand corner of the Periodic Table. Some elements for example
silicon (Si), seem to have properties between those of metals and those of nonmetals.
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They are often called metalloids and lie on the dividing line between the two
extremes. Place an element in the Periodic Table and you know at once what its
character will be
Metal oxides, nitrates and carbonates
A similar pattern is found when we examine the oxides of the elements. Oxides of
metals are basic; they react with acids to give a salt and water. Here is the equation
for the reaction between magnesium oxide and hydrochloric acid:
MgO + 2HCl
MgCl2 + H2O
The oxides of non-metals are acidic; in water they react with bases to give a salt
and water. Here is the equation for the reaction between carbon dioxide and
sodium hydroxide:
2NaOH + CO2
Na2CO3 + H2O
Some oxides will react with acids and react with bases. Fresh aluminium oxide
reacts with hydrochloric acid to give aluminium chloride
Al2O3 + 6HCl
2alCl3 + 3H2O
But it also reacts with sodium hydroxide solution to give a soluble compound,
sodium aluminate:
Al2O3 + 2NaOH
2NaAlO2 + H2O
Such oxides are called amphoteric oxides.
A few common oxides have neither acidic nor basic properties. They do not react
with either bases or acids to form a salt. They are called neutral oxides. Three
common examples are water (H2O), nitrogen monoxide (NO) and carbon monoxide
(CO). Nitrogen and carbon both form other oxides which are acidic.
Summary
 The elements are arranged in order of their atomic number (Z) to form the
Periodic Table.
 Vertical columns in the Periodic Table are called groups.
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 Horizontal rows in the Periodic Table are called periods.
 Each element in a group has the same arrangement of electrons in its
outermost shell hence elements in a group have similar chemical properties.
 Properties of elements in a group frequently show a trend
 Properties of elements in a period may show a trend or change abruptly.
 The position of an element in the Periodic Table can be used to predict its
properties and those of its compounds.
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2. RULES OF BEHAVIOR
If matter is made of atoms, then compounds are formed when atoms join together.
But they can only do this in a limited number of ways. Think about copper oxide.
We can find out by experiment that this compound contains only copper and
oxygen. The atoms might join together in various ways, for example:
Cu O or
Cu O Cu
or
O Cu O
When we get a bottle of copper oxide from the shelf perhaps it contains a mixture
of all these different materials. Below is a picture of two bottles of copper oxide.
One compound is black and the other is red. It seems likely that each compound is
not a mixture, but just one of the possible atomic combinations.
Dalton said, “All samples of a pure compound contain the same elements in the
same proportions by mass.” Remembering that all atoms of one element have
the same mass, this means that all samples of a pure compound, however they
are made, contain the same elements in the same atomic ratio. This is called the
law of constant composition.
Experiment: To find the composition of black copper oxide
Black copper oxide can be reduced to metallic copper by warm ammonia gas. We
can weigh a container, weigh the same container with some dry copper oxide, pass
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ammonia gas over the warmed copper and weigh the tube after the reaction is
over.
Requirements
Flat-bottomed or round-bottomed flack, drying tower, glass tubing, perforated test
tube, small length of rubber tubing, clip.
Caution
There is little risk in this experiment if you do it carefully. However ammonia gas is
very irritating to the eyes and lungs. Only do this experiment where there is good
ventilation.
Method
Set up the apparatus as below. The container for the copper oxide can be a test
tube with a hole blown in it or a length of glass tubing. It should have a loose plug
of fibre near one end. When the apparatus is set up follow these steps:
1. Weigh the empty, dry container.
2. Add a little dry copper oxide and reweigh. Insert the loose fibre plug
3. Warm the flask to start the production of ammonia and after one minute
begin to warm the copper oxide in the tube. When the ammonia reaches the
oxide the reaction will begin. Notice the colour changes.
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Explanation
The ammonia gas converted the copper oxide into copper. The oxygen from the
oxide was lost as water from the heated tube. Heat was produced in the reaction.
Through the early years of the nineteenth century chemists did a large number of
experiments like this one and found that the law of constant composition was
indeed true. This in turn meant that Dalton’s atomic theory was also conformed.
2.1 Empirical Formulae
Percentage composition
To find out the formula of a substance we often begin by finding out which
elements are present and in what quantities. It is useful to express the composition
in percentage terms. We imagine that we start with 100 mass units of the
compound and find out how many of those 100 parts come from each element.
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For example a compound contains only carbon, hydrogen and oxygen. We discover
that 20g of the compound contained 10.4g of carbon and 2.61g of hydrogen. To
find the percentage composition of the compound we follow these steps.
1. 20g contained 10.4g carbon therefore 1g contained (10.4g/20g) x 1g carbon
and 100g contained 100g carbon x (10.4/20) that is 52g carbon. Therefore
the compound contains 52 percent of carbon by mass.
2. Similarly we find that the percentage of hydrogen is 100g x (2.61/20) namely
13.05g, or 13.05 percent hydrogen by mass.
3. Therefore since there is only carbon, hydrogen and oxygen present, the
percentage of oxygen is 100 – (52 + 13.05) namely 34.95 percent by mass.
From Percentage Mass to Formulae
If we know the relative atomic masses of the atoms in a compound and its
percentage composition by weight, we can work out how many atoms of each type
are present. This is how to do it.
1. Divide each percentage weight by the relative atomic mass of the atom.
2. Divide each of these answer by the smallest of them
3. The numbers you get are the numbers of each type of atom in the
compound.
Example
In an analysis of black copper oxide, the percentage of copper was found to be 79
percent therefore the percentage of oxygen was (100 - 79) = 21 percent. From
appendix 1 we see that Ar for oxygen is 16 and Ar, for copper is 63.5.
Atoms present are
Cu
O
Percent of each
79
21
1. Divide by Ar:
79/64 = 1.24
21/16 = 1.31
2. Divide by the smaller:
1.24/1.24 = 1.00 1.31/1.24 = 1.06
3. But atoms must be there whole numbers! The nearest whole numbers are:
Copper = 1,
Oxygen = 1.
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So the atoms are present in the ratio 1:1. Now we can write a formula for the
compound; it is Cu1O1 or simply CuO.
The idea of molecules; molecular formulae
A molecule is the smallest particle of the substance capable of independent
existence.
Now we know that many materials, not just gases, exist as molecules. Any covalent
compound is made up of molecules. The molecules are made up of atoms tightly
bound together. The molecules thus formed are usually less tightly bound to each
other.
When we heat some alcohol molecules come apart from each (vaporize) without
the individual atoms coming apart (decomposing). To decompose the molecules (to
make the atoms separate from each other) we have to make the vapour much
hotter.
In the last section we saw how to calculate the ratio of the number of atoms in a
compound. In the example, the formula was CuO. Figure 2.4 shows several
different molecules all having this atom ratio.
All the molecules in figure 2.4 have the same atom ratio. The formula CuO, which
shows this ratio, is called the empirical formula. The formula which gives a picture
of the real molecule is called the molecular formula.
The empirical formula shows the simplest whole number ratio of the atoms
present. The molecular formula shows the actual number of each atom present.
In black copper oxide the empirical formula and the molecule formula and the
molecule formula are the same. But in ethane they are not. The molecular formula
of ethane is C2H4. The empirical formula is CH2.
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Percentage Composition
Sometimes we need we need to state, in quantitative terms, how a substance is
made up. A convenient way of doing this is to use the percentage composition by
mass.
To find the percentage composition by mass of a compound we follow these steps.
1. We need to know the formula and the values of the relative atomic mass (Ar)
for each element.
2. We then work out the relative molecular mass (Mr) by multiplying the value
of Ar for each element by the number of atoms of that element in the formula
(X) and adding together the values X for each element.
3. Express the value X for each element as a percentage of Mr.
Example
To calculate the percentage composition by mass of sodium nitrate we follow these
steps
1. Sodium nitrate has a formula NaNO3 and values of Ar are NA 23, N 14, O 16.
2. Mr = 23 + 14 + (3 x 16) = 85
3. Na = (23/85) x 100 = 27 percent; N = (14/85) x 100 = 16.5 percent
O = ((3 x 6)/85) x 100 = 56.5 percent
Example
To calculate the percentage by mass of aluminium in aluminium nitrate:
1. Aluminium nitrate has the formula Al (NO3)3 and values of Ar are Al 27, N 14,
o 16.
2. Mr = 27 + (3 x 14) + (3 x 16)) = 213.
3. Al = (27/213) x 100 = 12.7 percent
When we go to the market to buy eggs, we do not buy one egg at a time. Instead
we buy a unit of eggs. The word “unit” is used to mean a quantity of eggs, in fact a
bag of ten eggs. In chemistry we use the word “mole” to mean a quantity of a
chemical substance. One mole is a very large number, it is:
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602 200 000 000 000 000 000 000 (6.022 x 1023) particles
A mole is a quality of a substance. One mole of a substance is 6.022 x 1023 particles
of that substance. There are 6.022 x 1023 particles per mole. The symbol for a
mole is mol.
Although the number is large, because the particles – atoms and molecules – are
very small, one mole of a chemical is about a small handful, smaller in size than a
unit of eggs. The mole is a very useful quantity. Most quantitative work in chemistry
uses moles as we will discover later on in this unit.
Examples
1 mol of copper atoms contains 6.022 x 1023 copper atoms.
1 mol of sugar molecules contains 6.022 x 1023 sugar molecules.
1 mol of eggs would contains 6.022 x 1023 eggs.





6.022 x 103 is called the Avogadro number or Avogadro constant
The symbol for the Avogadro constant is L
In calculations we often use the approximate value 6 x 103
The number of particles equals the number of moles multiplied by 6 x 103
When using moles, always state which particles are involved.
2.2 The Mole
Using Moles
Before the mole was thought of, the manufacture of chemicals was a hit-and miss
affair. Raw materials were mixed in quantities which gave a good result but not
necessarily the best or cheapest result.
To make cement, we mix one bucketful of cement with four bucketsful of sand and
add enough water to make a slippery paste. Water does not cost very much so if
we add a little too much no harm is done. But suppose that we want to make
copper nitrate to sell. We need to be more accurate.
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Example
Suppose we have some copper, how much nitric acid will we need? Nitric acid is
expensive so we don’t want to waste it. First we write down a balanced equation
to represent the reaction.
Cu + 4HNO3
Cu (NO3)2 + 2H2O + 2NO2
We see from this that one atom of copper reacts exactly with four molecules of
nitric acid. Two atoms of copper would react exactly with eight molecules of nitric
acid and ten molecules of copper would react exactly with 10 x 4 molecules of
nitric acid. 6 x 1023 atoms of copper would react exactly with 4 x (6 x 1023)
molecules of nitric acid.
One mole of copper atoms would react exactly with four moles of nitric acid
molecules. The highlighted numbers in the equation tell us how many atoms and
molecules are involved.
They also tell us how many moles of each reactant are needed.
Converting Moles to Mass
Avogradro’s number (L), 6.023 x 1023 , is the number of atoms of hydrogen
which weigh 1.000g. So L hydrogen atoms, have a mass of (weigh) one gram
To convert from moles to grams, multiply the number of moles by the relative
particle mass.
Grams = moles x relative particle mass
Converting Mass to Moles
Two moles of helium weigh 8g. Looking at this from another angle, if we know
that the relative mass of an atom is four, and we know that we have 8g of that
atom, then we can see that we have two moles of the atom
𝑚𝑜𝑙𝑒𝑠 =
𝑔𝑟𝑎𝑚𝑠
𝒓𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆 𝒎𝒂𝒔𝒔
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Calculating Reaction Quantities
Examples
The following reaction starts off with 1000 tonnes of copper. Work out exactly
how much nitric acid is required.
Cu + 4HNO3  Cu(NO3)2 + 2H2O + 2NO2
Example. Method – 1
From the equation we see that 1 mol of copper reacts with 4 mol nitric acid
1000 tonnes = 1000 x 106 g
Ar for copper = 63.5, therefore copper weighs 63.5g/mol
𝟏𝟎𝟗 𝒈
𝟏𝟎𝟎𝟎 𝒕𝒐𝒏𝒏𝒆𝒔 =
𝟔𝟑. 𝟓𝒈/𝒎𝒐𝒍
= 𝟏. 𝟓𝟗𝟓 × 𝟏𝟎𝟕 𝒎𝒐𝒍 𝒄𝒐𝒑𝒑𝒆𝒓
There is 1.575 x 107 mol copper so he will need:
4 x 1.575 x 107 mol of nitric acid = 6.30 x 107 mol of nitric acid
Mr for nitric acid = (1 + 14 + 48) = 63, therefore nitric acid weighs 63g/mol
𝒎𝒂𝒔𝒔 𝒏𝒆𝒅𝒆𝒅 = 𝟔𝟑
= 𝟑𝟗𝟔𝟗𝒕𝒐𝒏𝒏𝒆𝒔
𝒈
× (𝟔. 𝟑𝟎 × 𝟏𝟎𝟕 )
𝒎𝒐𝒍
Example. Method – 2
1 mol reacts with 4 mol nitric acid.
1 x 63.5g copper reacts with 4 x 63g nitric acid
63.5 parts by mass of copper react with 4 x 63 (252) parts by mass nitric acid
63.5 tonnes of copper react with 252 tonnes nitric acid
Therefore:
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1 tonne copper reacts with 252/63.5 tonnes nitric acid**
1000 tonnes copper react with 1000 x (252/63.5) tonnes nitric acid
= 3968 tonnes
The two answers are not exactly the same because the figures have been rounded
up throughout the calculations.
NB: It is usually better to use moles
Quantity and Concentration
The concentration of a solution is either the mass of solute it contains or the
number of moles of solute it contains divided by the volume of the solution:
concentration = mass/volume (usually g/dm3) or
concentration = moles/volume (usually mol/dm3)
2.3 Titrations
You can use indicators to find out whether a substance is acid or alkaline.
Now we will use indicators to find the concentration of a solution.
Experiment: To Find the Concentration of a Solution
Aim
To find the concentration of an alkali.
You have to react a fixed volume of the alkali with an acid solution whose
concentration you know exactly. Then the concentration of the alkali is calculated.
This is called titration.
Requirements
Burette and stand, pipette (10, 20 or 25cm3), an acid/base indicator (methyl
orange, phenolphthalein or bromothymol blue, preferably not litmus or any
universal indicator)
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Procedure
1. Use a small funnel to fill the burette with acid; there is no need to fill to the
0.0 line; anywhere in the graduated part will do.
2. Use the pipette to transfer 25.0cm3 of alkali solution into the beaker
3. Use the glass rod to put one drop of indicator into the alkali. If the colour is
not deep enough, add another drop.
4. Copy the table below into your note book.
5. Read the level of the acid in the burette. Read the bottom of the liquid
meniscus as shown above.
6. Add acid to the alkali roughly 1 cm3 at a time, swirling the liquid in the
beaker to mix it after every addition, until the indicator changes colour.
7. Read the level of the liquid in the burette.
8. Work out what volume of acid was used.
9. Start again! But this time add 1cm3 less acid than was needed, all in one go.
Mix the liquid very thoroughly.
10. Now add acid from the burette one drop at a time until the indicator just
changes. This is called the end-point.
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11. Read the burette again.
12. Work out the exact volume of acid needed for the reaction.
13.Make sure that you have noted your results as shown below.
Concentration of acid (mol/dm3)
Volume of alkali taken (cm3)
First burette reading (cm3)
Second burette reading (cm3)
Third burette reading (cm3)
Fourth burette reading (cm3)
Accurate volume of acid needed is
(cm3)
25.0
14. Write down the equation for the reaction, for example:
NaOH + HCl  NaCl + H2O
15. From the concentration of the acid work out how many moles of acid you
added (call this A moles)
16. From the equation, work out how many moles of alkalis are needed to
react with A moles of acid (B moles)
17. B moles of alkali must have been in the 25cm3 of alkali with which you
began.
18. How many moles of alkali would be in 1.0cm3 of solution?
19. How many moles of acid would be in 1000cm3 of solution (in 1dm3 of
solution).
20. Write down the concentration of the alkali in mol/dm3
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CHALLENGING QUESTIONS I
Instructions: try to attempt the following before checking the answers.
1. 33 mL of 3 M Hydrochloric acid is titrated with sodium hydroxide to form water
and sodium chloride. How many mols of sodium hydroxide are consumed in this
reaction?
a. 3 mol
b. 10 mol
c. 33 mol
d. 100mol
answer: d. 100mol
2. 50 mL of 0.5M barium hydroxide are required to fully titrate a 100 mL solution
of sulfuric acid. What is the initial concentration of the acid?
a. 50 M
b. 5 M
c. 100 M
d. 25M
answer: d. 25M
3. An experiment was done on the reaction of copper oxide (CuO) with methane
(CH4).
(a) The equation for this reaction is shown below.
4CuO(s) + CH4(g) → 4Cu(s) + 2H2O(g) + CO2(g)
The water and carbon dioxide produced escapes from the test tube. Use
information from the equation to explain why.
........................................................................................................................ [1]
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(b) (i) Calculate the relative formula mass (Mr) of copper oxide (CuO).
Relative atomic masses (Ar):
O = 16; Cu = 64.
Relative formula mass (Mr) =.......................................................................... [2]
(ii) Calculate the percentage of copper in copper oxide.
Percentage of copper = .................................. %
(iii) Calculate the mass of copper that could be made from 4.0 g of copper oxide
Mass of copper = ............................................. g
answers:
3. (a) because they are gases ignore vapours / evaporate / (g) allow it is a gas 1
(b) (i) 80 / 79.5 correct answer with or without working = 2 marks ignore
units if no answer or incorrect answer then evidence of 64 / 63.5 + 16 gains 1
mark 2
(ii) 80 / 79.87 / 79.9 / 79.375 / 79.38 / 79.4 correct answer with or without
working = 2 marks if no answer or incorrect answer.
(iii) 3.2 correct answer with or without working = 1 mark
allow (ecf)
4 x ((b)(ii)/100) for 1 mark if correctly calculated
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3. THE BEHAVIOUR OF GASES
3.1 The Particles in a Gas
Experiment: To estimate the separation between the particles in a gas.
Requirements
A polythene bag (sandwich bag), alcohol, electric kettle, teat pipette, sellotape or
clip, tongs.
Inside the hot kettle the liquid alcohol changes state. The particles move further
apart as they vaporise. Alcohol has a boiling point of 78C so it condensed back to
liquid as soon as you took it out of the hot steam. You might find the result of
your calculation surprising.
Despite the huge change in the volume the
particles only move apart by less than ten
times their own size. But there are some
equally surprising consequences!
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3.2 Avogadro’s law
When Avogadro suggested that elements could exist as molecules, rather than
single atoms, he was trying to support an idea which he had put forward several
years before but which no one believed.
Avogadro had suggested that equal volumes of all gases contained equal numbers
of particles. It was well known that when gases reacted together, they did so in
volumes which were either equal or simple multiples of each other. For example
hydrogen and chlorine react together in equal volumes to make hydrogen
chloride. It seemed obvious that:
one hydrogen particle + one chloride particle
= one hydrogen chloride particle
So if there were L particles in one volume, this would be the same as saying that:
one volume of hydrogen + one volume of chlorine
= one volume of hydrogen chloride
But we know they make two volumes of hydrogen chloride! So each particle of
hydrogen and each particle of chlorine must contain two atoms each to give two
particles of hydrogen chloride. The particles are molecules, not atoms. So now we
can say that:
Equal volumes of gases, at the same temperature and pressure, contain equal
numbers of molecules. This is called Avogadro’s Law.
If we use L (the Avogadro constant) as the number of particles whose volume we
measured, then, because L is the number of particles in one mole, we can say that
one mole of any gas, under the same conditions of temperature and pressure, will
occupy the same volume. This volume is called the molar volume Vm.
When we have considered what happens to the particles of a gas as it is heated,
cooled, or squeezed, we shall be in a position to ask what that volume is.
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3.3 Boyle’s law and Charles’ law
The gas particles in the box on the left-hand side of Figure 3.3 are all moving. They
move in straight lines unless they hit each other or they hit the wall of the box.
When they hit the wall they bounce off. It is as though each wall is being
continuously struck by thousands of tiny hammers. The walls are being
continuously pushed outwards by the gas molecules. We call this the pressure of
the gas.
If we were to squirt more gas into the box, so that the number of gas molecules
was doubled (as in the box on the right hand side of Figure 3.3) then the walls
would be hammered twice as often – the pressure would be doubled.
This is the same as saying that the concentration of gas particles in the box has
been doubled.
Instead of doubling the amount of gas in the box, we could have halved the
volume of the box as shown in Figure 3.4. the pressure of the gas in the small box
in Figure 3.4 is exactly the same as it was in the large box on the right-hand side of
Figure 3.3 with double the number of particles because the concentration of
particles is the same in both.
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So when the volume was halved, the pressure doubled. If the volume had been
made one tenth of the original, the pressure would have gone up by a factor of
ten. When the volume increases, the pressure decreases and vice-versa.
The volume and the pressure of the gas are inversely proportional to each other.
This was first pointed out by Robert Boyle. It is called Boyle’s law.
Boyle’s law states that if the temperature is constant, the volume and pressure
of a gas are inversely proportional to one another, in short form:
P 1/V
Increasing the concentration of a gas in a box is not the only way of increasing its
pressure. If we could increase the speed of the molecules v, then they would hit
the box walls harder, because their kinetic energy (mv2) would be greater. We can
do this by heating the gas to a higher temperature.
Nearly one hundred years after Boyle stated his law, Charles discovered that
when a gas is heated, the change in pressure is related to the change in
temperature in a simple way. If the temperature increases, the pressure also
increases.
Charles’ law states if the pressure of a gas is constant, the volume of the gas is
directly proportional to the absolute temperature, in short form:
VT
Notice that in Charles’ law, it is the absolute temperature which matters.
Absolute temperature is measured in Kelvin (K) where:
T(K) = T() + 273
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Example
The temperature in a room is 25. The atmospheric pressure on that day is 1.0 x
105 Pa. a chemist collects 600cm3 of a gas in a balloon and puts it in a refrigerator
at 4. What will be the volume of the gas?
The pressure in the refrigerator will be the same as in the room. Only the
temperature has changed.
Original temperature = (25 + 273) K = 298 K
Final temperature = (4 + 273) K = 277 K
The gas is colder, the temperature has gone down, and so the volume will go
down; the new volume is:
(600cm3 x 277 K/298 K) = 557.7cm3
Example
A chemist collects two samples of gas. Each has a volume of 1000cm3. He stores
one in a glass gas jar with a sealed lid. He stores the other in a balloon. The
atmospheric pressure on that day is 0.98 x 105 Pa. The next day the temperature
has not changed but the pressure has risen to 1.01 x 105 Pa. What will be the
volume of gas in each of the containers?
The volume of the gas jar will not change because the jar cannot change and a
gas takes the volume of its (rigid) container. The volume of the gas in the balloon
changes because the container is not rigid; the pressure has gone up so the
volume will go down. The new volume is
(1000 cm3 x 0.98 x 105) Pa/(1.01 x 105)Pa = 970.3cm3
Exercise
1. A gas has a volume of 2.00dm3 at a pressure of 1 x 105Pa. what will be its
volume at the same temperature, at a pressure of 3 x 105 Pa?
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2. 150cm3 of oxygen are heated from 25 to 45, at atmospheric pressure. What is
the new volume of the gas?
3. A certain mass of gas is found to have a volume of 15.00dm3 at a temperature
of 25 and a pressure of 1.00 x 105 Pa. what is the volume of the gas at 0 and 5 x
104 Pa?
Hint: Ask yourself each time “will the volume get larger or smaller?” Then you will
know if you have got the numbers the right way up!
Molar Gas Volume
Because the volume of a fixed mass of gas varies with its pressure and
temperature we cannot use Avogadro’s law to say “One mole of any gas will have
the same volume”. One worker might measure that mole under conditions
different from those chosen by someone else.
So scientists have chosen a set of standard conditions which everyone can use.
Standard conditions for the measurement of gas volumes are a pressure of 105 Pa
and a temperature of 273 K. Under these conditions one mole of any gas occupies
22.4dm3. These conditions are called standard temperature and pressure (STP).
STP:
Standard pressure
= 105 Pa
Standard temperature = 273 K
Molar gas volume
= 22.4 dm3
Very few measurements are actually taken at 273 K so for many purposes we use
not STP but RTP (room temperature and pressure). At RTP the molar volume of a
gas is taken as 24.0dm3.
RTP:
Room pressure
= 105 Pa
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Room temperature = 298 K
Molar gas volume = 24.0 dm3
3.4 The Ideal Gas
If it is true that for a fixed mass of gas:
p 1/V
then it would follow that:
p = k/V
where k is a constant and so:
pV = constant
A graph of pV against p for any gas at constant temperature should be a
horizontal straight line. Instead these lines are curved, as shown below.(Note
that the curves are exaggerated to show the effect.)
For some gases pV is usually smaller than Boyle’s law would predict (the gas is
taking up less space than the law suggests). The differences are not great but they
are very important in practice. What can be the explanation?
If Boyle’s law was accurate, if the pressure of a gas was increased to infinity the
volume would be zero. But the volume could never be zero, because the
molecules take up some space.
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So when a gas takes up more space than we expect, it is because of the volume of
the molecules themselves. The space between them obeys Boyle’s law, but not
the gas itself!
When two molecules come together, they usually attract one another. The
molecules of polythene, for example, must have something holding them
together, or polythene would be a gas.
When sulphur dioxide gas is compressed it turns into a liquid because the
molecules, when they get close enough, attract each other so strongly that the
kinetic energy (mv2) of the particles is not enough to make them separate again.
Although these forces (usually called van der Waals’ forces) are only weak even
when the molecules are touching each other, they do still have some effect in the
gas. They pull the molecules of the gas just a little closer together than would
otherwise be the case. So the gas has a slightly smaller volume than Boyle’s law
suggests. As the pressure goes up the volume decreases, the molecules get closer
together and this effect increases.
These two effects work against one another. Which one wins depends on the gas
and the pressure. So sometimes the product pV is smaller than we expect and
sometimes larger. If the gas is at low pressure – the molecules are far apart from
each other – then these effects are not noticeable. We say that the gas is
behaving like an ideal gas. For an ideal gas we can combine Boyle’s and Charles’
laws together:
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If pV = constant,
and p = constant x T
then pV = constant x T
As usual, it is sensible to work with one mole of gas. When we do, the constant in
the equation is called the molar gas constant. The molar gas constant has the
symbol R and its value is 8.314 J K-1 mol-1. The gas law becomes, for one mole of
an ideal gas:
pVm = RT
Using Molar Volumes in Calculations
From the equation for any reaction we know how to work out the quantities
involved by mass or by moles and how to convert from moles to volumes. This
means that we can work out what volumes of gas are produced, or used up, in
any reaction.
Example
Nitrogen and hydrogen react to make ammonia. Ammonia is a gas at RTP. The
equation for the reaction is:
N2 + 3H2 2NH3
From this equation, using values of Ar from Appendix I, we see that:
 28g of nitrogen reacts with 6g of hydrogen to give 34g of ammonia.
 One molecule of nitrogen reacts with three molecules of hydrogen to give two
molecules of ammonia.
 One mole of nitrogen reacts with three moles of hydrogen to give two moles
of ammonia.
 At RTP, 24dm3 of nitrogen reacts with 72dm3 of hydrogen to give 48dm3 of
ammonia.
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We can use this information to answer several different types of questions. For
example, suppose that we wanted to know what is the maximum volume of
ammonia which can be made, at RTP, from 56g of nitrogen, here is how we would
work it out:
56g of nitrogen = 56g/28g/mol = 2 mol
1 mol nitrogen yields 2 mol ammonia. (1 x 2)
2 mol nitrogen yields 4 mol ammonia. (2 x 2)
1 mol gas at RTP occupies 24dm3.
4 mol occupies (4 x 24) dm3 = 96dm3
Notice that this tells us nothing about the conditions needed to do the reaction,
how fast the reaction will happen, or how much gas might be lost.
Example
Supposing 6dm3 of ammonia were lost and only 90dm3 of ammonia were
collected, what was the yield of the reaction? In this case, as we expected to
collect 96dm3, the percentage yield is (90dm3/96dm3) x 100 = 93.75%
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4. ENERGY CHANGES IN
CHEMICAL REACTIONS
In every chemical change there is a change of temperature and a flow of heat into
or out of the reacting system.
Experiment:
To observe changes in temperature in chemical reactions,
Requirements:
Anhydrous calcium chloride, ammonium chloride, test tubes.
Both the compounds used are ionic solids. When the solids dissolve in the water,
firstly, the ions of the solid become separated from each other. Bonds are broken.
Secondly, the water molecules form bonds with the cations forming new
chemicals Ca(H2O)n2+ and NH4(H2O)m+.
Energy Diagrams
In the diagram below, the height above the baseline represents the energy
content of the system. The higher the system is in the diagram, the more energy it
contains. The separate atoms are high up the scale; the bonded atoms are low
down. The separate atoms contain more energy than the bonded atoms.
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If the atoms are to end up in exactly the same state as they began, the loss of
energy from the atoms as the bond forms must exactly equal their gain in energy
as the bond is broken.
Energy is released into the surroundings when bonds form. Energy is gained from
the surroundings when bonds are broken.
Overall changes
In a chemical change, the starting materials (reactants) are used up and new
substances (products) are formed. Bonds in the reactants are broken, and bonds
in the products are formed.
Example
Think about the reaction between hydrogen and oxygen. Showing the bonds, we
can write:
2(H–H)(g) + OO(g) 2(H–O–H)(g)
As the reaction goes on, H–H and O=O bonds are broken and bonds between
oxygen atoms and hydrogen atoms are formed (H–O–H). energy has to be put
into the system to break the old bonds, but energy is given out of the system as
the new bonds are formed.
Measurements of the energies involved show that 436kJ of energy are needed
when one mole of hydrogen molecules is broken into atoms and 498kJ of energy
are needed when one mole of oxygen molecules is broken into atoms. When one
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mole of water is formed from hydrogen and oxygen atoms, 463kJ of energy are
liberated. We can draw an energy diagram for these changes.
The diagram below shows that when one mole of water (in the gaseous state) is
formed from its elements, more energy is given out than has to be put in.
Energy Changes and Heat Content
Energy changes in chemical reactions affect heat content. That is why we measure
them by observing changes in temperature. To show the energy changes, in an
equation, we use the symbol H where means “the change in” and H means “heat
content”.
Note that the H is in italics to distinguish it from the symbol for hydrogen (H).
(Another word for “heat content” is enthalpy, but you do not have to remember
this word at this grade.)
When heat energy is added to a system, to break bonds, H is positive (+ve). When
heat energy is lost from a system because bonds are formed, H is negative (-ve).
Example
In the reaction between hydrogen and oxygen, it shows that 482kJ are lost from
the system for every mole of oxygen used. So we can write either:
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2H2(g) + O2(g) 2H2O(g) – 482 kJ
or:
2H2(g) + O2(g) 2H2O(g) + H; H = – 482 kJ/mol
H = heat content of products minus heat content of reactants. When the heat
content goes up, H is positive. When it goes down, H is negative.
Hess’s Law
It states that the overall energy change in converting reactants to products is the
same no matter what route the change takes.
The value of H for a reaction is a useful figure; it is measured per mole of reactant
under standard conditions, and is called the standard molar heat of the reaction.
The standard conditions are usually 298 K and 1.01 x 105 Pa (1 atmosphere).
The standard molar heat of a reaction is the heat change per mole of the reaction
as specified by a balanced chemical equation. The heat change must be measured
under standard conditions.
For example we can make copper(II) oxide in several ways; we can heat copper
metal in oxygen gas:
Cu + O2 CuO + H1
or we can dissolve copper metal in nitric acid to make copper nitrate, then heat
he copper nitrate strongly:
Cu + 4HNO3 Cu(NO3)2 + 2H2O + 2NO2 + H2
Cu(NO3)2 CuO + O2 + 2NO2 + H3
also:
4HNO3 2H2O + 4NO2 + O2 + H4
Using Hess’s law, we can say that:
H1 = H 2 + H3 – H4
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4.2 Endothermic and Exothermic Reactions
Exothermic reaction is one in which heat is given out from the chemicals to the
surroundings. (The prefix exo means outside).
Example
Consider the reaction between solid carbon and carbon dioxide to form carbon
monoxide:
C(s) + CO2(g) 2CO(g)
The energy formed in breaking and forming the bonds is shown below.
Step
Energy (kJ/mol)
Vaporise C(s)
715
Break each (of two) carbon–oxygen bonds (in
804
CO2)
Form carbon–oxygen
-1076
More energy is required on the left-hand side of the reaction equation, namely;
{715 kJ/mol + (2 x 804 kJ/mol)} = 2323 kJ/mol
Than is liberated by forming two moles of carbon monoxide:
(2 x 1076 kJ/mol) = 2152 kJ/mol
H is +171 kJ/mol. We can show this in the equation:
C(s) + CO2(g) 2CO(g);
H = +171 kJ/mol
Heat has to flow into the system for the reaction to proceed. This is an
endothermic reaction. This particular reaction occurs in the special conditions of
the blast furnace. (The prefix endo means within).
Note: Negative H implies an exothermic reaction. Positive H implies an
endothermic reaction.
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Spontaneous and non-spontaneous changes
If something is hotter than its surroundings, it will lose heat. This happens without
us having to do anything about it. Processes like this are called spontaneous
processes. These processes happen without any external help or influence. Most
exothermic reactions are spontaneous. Once they begin, they go on.
Most endothermic processes are not spontaneous. They cannot go on unless
something drives them. Photosynthesis is an example of an endothermic process
because it needs sunlight to occur.
Some endothermic processes are spontaneous. For example ammonium chloride
dissolves spontaneously in water.
4.3 Activation Energy
The figure below shows how the heat content of a reacting system varies with
time, from start to finish. Part (a) shows an exothermic reaction, in which the
overall heat content falls. Part (b) shows an endothermic reaction in which the
overall heat content goes up.
In both cases the
reacting system
needs to gain
energy first,
before the final
products can be
formed. It is as
though the
system has to climb up an energy hill before it can slide down the other side. The
additional energy is called the energy of activation for the reaction.
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CHALLENGING QUESTIONS II
Instructions: attempt the following questions before checking the answers.
1 (a) Marble chips react with hydrochloric acid to produce carbon dioxide. The
equation for the reaction is CaCO3 + 2HCl o CaCl2 + H2 O + CO2 Which one of
these changes would decrease the rate of this reaction?
A. use hydrochloric acid which is more dilute.
B. use smaller sized marble chips.
C. use marble chips which have a larger surface area.
D. use a larger volume of the hydrochloric acid.
(b) Explain why increasing the temperature of a reaction increases the rate of
the reaction ………………………………………………………………………………………………………
(c)(i) The rate of decomposition of hydrogen peroxide can be increased by
adding a catalyst. Which of these graphs shows the mass of the catalyst during
the reaction?
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(ii) The decomposition of hydrogen peroxide, H2 O2 , produces oxygen and water.
Give the balanced equation for this reaction.
....................................................................................................................................
(d) Explain, in terms of the energy involved in the breaking of bonds and in the
making of bonds, why some reactions are exothermic.
Answers:
3(a) A use hydrochloric acid which is more dilute
3(b) An explanation linking two of M1 {particles/reactants/collisions} have more
energy (1 mark) M2 more frequent collisions (1 mark) M3 more
{productive/successful/effective} collisions (1 mark)
(c)(i) C
(c)(ii) 2H2O2→ 2H2O + O2 (2 marks) all formulae correct (1 mark) balancing
correct formulae(1 marks)
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5. RATE OF REACTION
The rate of a reaction is stated in terms of grams per second (g/s), or moles per
second (mol/s) of reactant used up or products produced.
Because the rate of a reaction often changes as time goes on, we may plot a
graph of amount of substance against time. The slope of this graph gives the rate
of the reaction, just as the slope of a distance against time graph gives the rate
(speed) of a moving object.
5.1 Factors Affecting the Rate of Reaction
Two chemicals can react together when their particles hit one another. Anything
which changes the rate at which particles collide will change the rate at which
they react. Speeding up the collision rate will speed up the reaction, and slowing
down the collision rate will slow the reaction down.
Not every collision will result in a reaction. Unless the particles collide there will
be no products. In some collisions there will not be enough energy to break the
original chemical bonds and so no new substances can be made. Anything which
changes the amount of energy needed to activate the reaction will also change
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the rate of reaction, because a different fraction of the collisions will be
successful.
We can illustrate this by thinking about a reaction between two compounds A–B
and C–D which react according to the equation:
A–B + C–D A–C + B–D
In this reaction energy will be needed to break the bonds between A and B and
those between C and D are formed. Any excess energy will be given off as heat.
This is shown below.
Not all the colliding particles have the same energy. Some particles are always
travelling more quickly than others. In a gentle collision the total energy is not
enough to force A–B and C–D into a state from which A–C and B–D can form.
The diagram below shows how the energies of the collisions are distributed. A
few have only small energies; most have energies near the average; and some
have very large energies indeed.
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In the diagram above, the shaded area shows the proportion of collisions which
have energies above the energy E shown by the vertical line. Diagram (a) is at a
low temperature, and diagram (b) is at a higher temperature. The value of E is the
same in both. As the temperature goes up, the proportion of collisions whose
total energy is above E goes up.
If E is the energy needed to allow the reaction to continue, then at the higher
temperature, more reactions become possible. E is called the energy of
activation.
When the temperature goes up, there are more collisions in a fixed time. When
the temperature goes up, a larger fraction of the collisions give a reaction.
The rate of reaction is controlled by these factors.
a) The concentration of the reactants: The more concentrated the solution is,
the closer the reacting particles are. The closer together the particles are, the
more often they will collide. So the more concentrated the solution, the faster
the rate. The rate of reaction is directly proportional to the concentration of
the reactant. This is true for every reactant. In general we can say that the rate
of a reaction is proportional to the product of the concentration of the
reacting substances.
b) Surface area (particle size): The more finely divided a solid reactant is, the
more surface area it exposes and the more quickly it reacts.
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c) Temperature: The rate of a chemical reaction increases with increasing
temperature.
d) Gas pressure.
e) Presence of a catalyst: A catalyst is a substance which changes the speed of a
chemical reaction but is unchanged in mass at the end of the reaction. Most
catalysts speed up the rate of reactions. Some however, slow down the rate of
reaction.
5.2 Catalysts in Industry
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A catalyst provides an alternative reaction path with a lower activation energy.
5.3 Reversible Reactions
Example: Copper (II) Sulphate crystals contain water chemically bonded to the
copper ions. The formula of the crystalline material is CuSO45H2O. When you heat
the crystals gently, the bonds between the water and the copper ions are broken
and the water is driven off as water vapour.
This vapour condenses on the cold parts of the test tube and appears as a
colourless liquid. If you collected enough of it you could boil it and measure the
boiling point to show that it was indeed water. The solid residue is still copper (II)
sulphate (unless you have heated it too strongly) but it is white, not blue. It is
called anhydrous copper (II) sulphate because it now has no water in it.
When the water mixes once more with the anhydrous copper (II) sulphate, the two
recombine, giving the original blue hydrated material. Because this is the reverse
of the first reaction and the first one was endothermic (took in heat), this is an
exothermic reaction (it gives out heat).
The reactions you have done are two parts of one reversible reaction.
The direction of change of a reversible reaction can be controlled by changing the
conditions.
We can write one equation which shows both forward and reverse reactions by
using a special symbol.
CuSO4H2O ⇌ CuSO4 + 5H2O
All reactions are reversible to some extent.
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6. DYNAMIC EQUILIBRIUM
6.1 What is Dynamic Equilibrium?
Look at the graphs above. If, after time t, the rate of the forward reaction (A) is
higher than the rate of the backward reaction (B), more products are being
formed than are being converted back into limestone. So the amount (and the
surface area), of the limestone must go on falling, and the rate of the forward
reaction falls with it.
But less quicklime and carbon dioxide are being reconverted than are being
formed, so the rate of the backward reaction must still go up. This is true as long
as A is greater than B. Therefore as the forward rate falls and the backward rate
goes up the two rates must get closer and closer together.
Eventually the two rates will become equal. When this happens, the products are
being formed at the same rate as they are being reconverted. The reactants are
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being used up at the same rate as they are being re-formed. So the
concentrations of the reactants and the products no longer change. It is as
though the reaction has stopped, though in reality the forward and the backward
reactions are still going on.
When this happens, and no further change can be detected in the system a state
of dynamic equilibrium has been reached.
A reacting system is said to be in dynamic equilibrium when the rate of the
forward reaction is equal to that of the backward reaction. In a reacting system
which is in dynamic equilibrium no further change can be detected.
This does not mean that when a system is in dynamic equilibrium the products
and reactants are present in equal amounts or concentrations. Look at the
diagram below which shows graphs of rate against time for three separate types
of reaction.
The first case (a) shows a reaction in which the forward reaction goes on very
quickly but the backward reaction goes on slowly. In this case the reactants are
used up almost entirely long before the reverse reaction can have much effect.
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The system comes to equilibrium but the reactants have almost gone and only
products are present in any concentration.
The second case (b) shows a reaction in which the forward and backward
reactions have nearly the same rate throughout. This time there will be similar
concentrations of reactants and products in the equilibrium mixture.
The third case (c) shows the opposite of (a). The backward reaction is fast, and so
the reactants are re-formed almost as soon as they are used up. When
equilibrium is reached, there will only be small concentrations of products in the
mixture.
An equilibrium mixture may contain mostly reactants, or a mixture of both.
6.2 Industrial Applications
Sulphuric Acid Production – The Contact Process
Sulphuric acid is made from sulphur. Sulphur dioxide SO2 is made by burning the
sulphur or by roasting a sulphide mineral such as iron pyrites in air. The sulphur
dioxide is then oxidised to form sulphur (VI) oxide SO3. Sulphur (VI) oxide is a white
solid at room temperature but at the temperature of the Contact process it is a
gas.
The gas can be dissolved in water but the reaction is so violent that instead it is
dissolved in concentrated sulphuric acid to give an oily, super-concentrated
product called oleum H2S2O7. Water is then added to the oleum giving sulphuric
acid.
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The equations for the Contact process reactions are:
1)
2)
3)
4)
S(s) + O2(g) SO2(g)
2SO2(g) + O2(g) 2SO3(g);
H-ve
H2O(l) + 2SO(g) H2S2O7(l)
H2S2O7(l) + H2O(l) 2H2SO4(l)
It is exothermic. It must use a catalyst that passes through a converter containing
vanadium (v) oxide V2O5. In this process, a high yield is produced by high pressure
and low temperature.
Ammonia Production – The Haber Process
Nitric acid and nitrates are important in the manufacture of fertilisers and
explosives. It can be made from ammonia which in turn can be made from
nitrogen and hydrogen.
The Haber process for ammonia manufacture is based on the reaction:
N2(g) + 3H2(g) 2NH3(g); H-ve
This reaction is also exothermic – it gives out heat as it goes.
Nitrogen is obtained from the air, while Hydrogen is obtained from methane
(natural gas).
The three conditions needed for this process are;
 A catalyst of pure iron with a trace of added aluminium oxide.
 A temperature of about 400C
 Pressure of 250 atmospheres.
In the Haber process a high equilibrium yield is produced by low temperature and
high pressure.
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7. FUELS
7.1 What is a Fuel?
It is a substance which burns in air to provide us with useful energy.
Carbon burns in air to form either carbon monoxide (if the air supply is limited,
which is known as incomplete combustion) and carbon dioxide (if the air is
plentiful, which is known as complete combustion). The equations are:
2C + O2 2CO
incomplete combustion
C + O2 CO2
complete combustion
What makes a Fuel Good?
 It’s availability.
 Yield as much heat as possible.
 Avoidance of unpleasant fumes when it is burnt.
 Easy to handle.
 Be cheap.
Heat of Combustion
The heat of combustion of a fuel is the total amount of heat released when one
mole of the fuel is completely burned in air or oxygen.
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Heats of combustion
Substance
Carbon (s)
Coal (s)*
Ethanol (l)
Hydrogen (g)
Propane (g)
Wood (s)*
*Average values.
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Heat of
combustion
(kJ/mol)
394
280
1370
286
2200
540
Heat of combustion
(kJ/g)
32.8
23
29.8
143.0
50.0
18
7.2 Fossil Fuels
These are fuels that were formed a long time ago and cannot be replaced. They
are finite resources.
Burning fossil fuels adds carbon dioxide to the air. Carbon dioxide is a greenhouse
gas – it helps to trap heat which would otherwise leak out into space by radiation.
The use of fossil fuels causes climatic change.
7.3 Fractional Distillation of Petroleum
Crude oil is a complex mixture of hydrocarbons with small amounts of other
substances. Crude oil cannot be used in its natural state as it comes from the
Earth. It must be separated into fractions each containing a much smaller range
of compounds. The range chosen depends on the intended use of the fuel. The
separation is done by fractional distillation. A fraction consists of a mixture of
chemicals with similar boiling points.
In distillation, a liquid is heated to its boiling point to turn it into a vapour. The
vapour is collected and cooled to turn it back into a liquid. In the process any
material, dissolved in the liquid, which cannot easily be vaporised, is left behind.
Simple distillation is used to purify a liquid by removing non-valuable impurities.
The liquid being purified turns into vapour; the impurities do not.
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In fractional distillation we heat a mixture of two or more volatile liquids, each
with its own boiling point and then cool the vapour progressively. The liquids with
high boiling points condense before the liquids with low boiling points. The liquids
are separated.
Fractional Distillation in the Laboratory
The diagram above shows the laboratory apparatus for fractional distillation. A
long tube, packed with small pieces of glass, is fitted vertically between the top of
the flask and the condenser. This is called the fractionating column. As the flask is
heated the vapour rises up the column and condenses.
If the flask is not heated very strongly the temperature of the top of the column
(heated by the condensing vapour) never reaches the boiling point of the liquid
and no vapour gets to the outlet.
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As the flask is heated more strongly more vapour condenses in the fractionating
column, which heats up from the bottom upwards. The part near the flask is
always hotter than the part near the top. A temperature gradient has been set up
in the column.
Eventually the temperature at the top of the column rises enough that light
molecule vapour flows out and down to the condenser. However the temperature
in the lower part of the column is below the boiling point of the heavier molecules
so they condense there and drop back into the flask as shown in the diagram.
When all the light molecules have been distilled over nothing more happens until
extra heat is supplied to the flask when the column heats up further and the next
liquid can be distilled off. Eventually only one liquid is left in the flask.
Fractional Distillation in Industry
Fractional distillation is also used in the petroleum industry to separate the main
constituents of crude oil, on a larger scale.
The oil is vaporised and the vapour passed into a fractionating column. This is like
a laboratory column but is made in such a way that liquids of different boiling
points collect in trays in the column and the gas continually bubbles through
these liquid fractions. The arrangement of trays and the bubble caps which cover
them.
The gas from the top of the column contains ethane, propane and butane. It is
used as a fuel either directly or liquefied. Some is used to produce ethene (C2H4)
which is a valuable synthetic chemical.
The naphtha is mainly used to produce synthetic chemicals.
The petrol contains molecules with five to ten carbon atoms. It is used directly as
a fuel for cars and vans and also after the process of cracking to make other
chemicals.
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The kerosene (paraffin) fraction contains hydrocarbons with eleven and twelve
carbon atoms and is used as aviation fuel and for domestic heating systems.
The residue is used to make bitumen which is used to surface roads and make
waterproofing materials.
7.4 Alternative Energy Sources
As resources diminish and people become more aware of the effects on their
environment which burning fossil fuels can have, other sources of energy are
being developed.
Nuclear Energy
Nuclear fission is now widely used. The light isotope of uranium (235U) is unusual.
If it is struck by a neutron it breaks up into smaller nuclei and gives out more
neutrons. This is called fission. Each of the neutrons can then go on to strike more
nuclei and cause more fissions. In every fission a tiny amount of mass is lost. This
mass is actually turned into energy. This differs from a conventional chemical
reaction in which the total mass remains the same.
The law of conversation of mass states that in a chemical reaction matter is
neither created nor destroyed. However, when a heavy nucleus splits into lighter
nuclei some matter is converted into energy.
Every time a 235U nucleus breaks up energy is released because the bonds which
hold the components of the fission fragment nuclei together are stronger than
the bonds which held the components of the uranium nucleus together.
The amounts of energy which are released are enormous because the relationship
between mass lost and energy produced is:
E = mc2
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where E is the energy released (in joules), m is the mass lost (in kilograms) and c is
the velocity of light (in metres per second).
Chemical Energy Sources
Chemically, several alternative sources are available.
Coal can be gasified. In one process the coal is heated, in the absence of air, to
produce useful chemicals, hydrogen gas and methane. The charred residue – coke
– is then heated with steam at a high temperature, to form a mixture of carbon
monoxide and hydrogen:
C + H2O CO + H2
Some countries now produce fuels from biomass. Biomass includes all plant and
animal material. For example sugar can be grown as an energy crop. It can be
fermented to give ethanol:
C6H12O6 2C2H5OH + 2CO2
Animal waste can be digested, in the absence of air and in the presence of
bacteria which decompose it, to form methane. Produced in this way methane is
called biogas.
7.5 Chemicals from Petroleum
Some of the products from the fractional distillation of crude oil are not directly
marketable. They can be changed into different more marketable molecules by
cracking. In steam cracking the unwanted fraction is heated and mixed with
steam. The mixture is passed into a furnace. At high temperatures the molecules
decompose and a mixture, consisting mostly of ethene and propene, is produced.
Ethene and propene are valuable synthetic chemicals. For example they form the
basis (the feedstock) for making plastics by polymerisation.
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Catalytic cracking is a process applied to the heavier fractions. The large
molecules are heated strongly. They break up into fragments. In the presence of a
suitable catalyst the fragments join up again into smaller, more valuable
molecules. Heavy oils, for example, can be changed into petrols by this process.
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8. OXIDATION AND
REDUCTION
8.1 Oxygen and Hydrogen
Oxygen is a reactive element. Most metals and most non-metals, except the noble
gases, combine with oxygen to form one or more oxides.
A substance is oxidised when oxygen is combined with it or when hydrogen is
removed from it. A substance is reduced when hydrogen is combined with it or
when oxygen is removed from it.
Examples of oxidation and reduction include:
Magnesium is oxidised since it has combined with oxygen.
2Mg + O2 2MgO
Copper (II) oxide is reduced since it has lost oxygen.
CuO + H2 Cu + H2O
Sulphur is reduced since it has combined with hydrogen
8H2 + S8 8H2S
In the decomposition of ammonia, ammonia is oxidised since it has lost hydrogen.
2NH3 N2 + 3H2
8.2 Electron Loss and Electron Gain
According to the definition a substance is oxidised when it reacts with oxygen to
form its oxide, for example when magnesium burns in oxygen:
2Mg(s) + O2(g) 2MgO(s)
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the product, magnesium oxide, is ionic. We could write an ionic equation (without
state symbols, to be clearer):
2Mg + O2 2Mg2+ + 2O2-
(1)
Magnesium will burn in chlorine. The product is magnesium chloride. It is also
ionic. The ionic equation for the reaction (again leaving out the state symbols) is:
Mg + Cl2 Mg2+ + 2Cl-
(2)
In each of the reactions the magnesium atoms have lost electrons and become
positively charged ions. If the first case is called oxidation then the second must
be called oxidation also.
The electrons which have been lost from the magnesium atoms have been
transferred to the other reagent (the oxygen in one case, the chlorine in the
other).
It is as though the two reagents competed for the electrons and the magnesium
lost. Losing electrons is called being oxidised; gaining electrons is called being
reduced.
This leads to a new definition of oxidation and reduction. A substance is oxidised
when it loses one or more electrons. A substance is reduced when it gains one or
more electrons.
Silver metal reacts with sulphur to form black silver sulphide. (This type of
reaction, which makes the silver lose its polish, is called tarnishing.) the equation
for the reaction is:
2Ag(s) + S(s) Ag2S(s)
or ionically:
2Ag + S 2Ag+ + S2-
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The silver atoms have been oxidised because they have lost electrons. The
sulphur atoms have been reduced because they have gained electrons.
In these reactions there is no overall loss or gain of electrons. The electrons
removed from one reagent must be added to another one.
There cannot be an oxidation reaction unless something is reduced at the same
time, and vice-versa. So this type of reaction is often called a redox reaction.
Redox is short for reduction – oxidation. Oxidation and reduction must take place
together.
In a redox reaction the reagent which causes the oxidation is called the oxidising
agent. The reagent which causes the reduction is called the reducing agent. Note
that because the oxidising agent causes the oxidation causes the oxidation
(removes electrons) it itself is reduced because it accepts them. Similarly the
reducing agent gives up electrons and is itself oxidised.
Oxidation
Loss of electrons
Oxidising agent
Gains electrons, is reduced
Reduction
Gain of electrons
Reducing agent
Loses electrons, is oxidised
Halogens can react with each other’s ions. These reactions can be done by
dissolving the reagents in water and mixing the solutions.
Fluoride ion
Chloride ion
Bromide ion
Iodide ion
Fluorine



Chlorine
x


Bromine
x
x

Iodine
x
x
x
-
Look at the reaction between chlorine gas and bromide ions. We can write an
ionic equation for this reaction:
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Cl2(g) + 2Br-(aq) 2Cl-(aq) + Br2(aq)
The chlorine gas has gained electrons from the bromide ions. The bromide ions
have lost electrons and become bromine. This means that the chlorine gas acted
as an oxidising agent.
But when bromine gas was tested against chloride ions, there was no reaction.
Bromine, although a strong enough oxidising agent to oxidise iodide ions, was not
a strong enough oxidising agent to take electrons from the chloride ions. From the
table above you can see that only fluorine can do that. Therefore fluorine is a
stronger oxidising agent than bromine. Some oxidising agents are stronger than
others.
The halogen order of their oxidising power is as follows:
Iodine
Bromine
Chlorine
Fluorine
Least powerful  Oxidising agent  Most powerful
Investigating the reducing power of some metals
Three metals are used namely; copper, iron and magnesium
Adding electrons to hydrogen ions produces hydrogen atoms which bond
together to make hydrogen gas. The hydrogen ions are reduced by the metal
atoms.
Magnesium and iron reduce the hydrogen ions to hydrogen. The equations for
the reactions are:
Fe(s) + 2H+(aq) Fe2+(aq) + H2(aq)
Mg(s) + 2H+(aq) Mg2+(aq) + H2(aq)
With iron and with magnesium a colourless gas is given off. This gas is hydrogen.
With copper metal no gas is given off. There is no reaction. Therefore copper is a
worse reducing agent than either iron or magnesium.
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The reaction with magnesium is much more vigorous than the reaction with the
iron. This suggests that magnesium is a more powerful reducing agent than iron.
The three metals can be written in order of their reducing power, as follows:
Magnesium
Iron
Copper
Most powerful  Reducing agent  Least powerful
The reactivity series of the metals towards water
Metal
Reducing power
Sodium
Best reducing agent
Calcium
Magnesium
Aluminium
Zinc
Iron
Tin
Lead
Hydrogen
Copper
Silver
Worst reducing agent
Metals and Their ores
Name of ore
Metal Contained
Cinnabar
Mercury
Zinc blende
Zinc
Haematite
Iron
Chalcopyrite
Copper
Bauxite
Aluminium
Metal Compound
Mercury (II) sulphide
Zinc sulphide
Iron (II) oxide
Copper (II) iron (II)
sulphide
Aluminium oxide
Metal Ion(s)
Hg2+
Zn2+
Fe3+
Cu2+Fe2+
Al3+
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Redox Processes in Metal Production
Metal compound Reducing
Process
agent
Mercury (II)
None
Heat in air
sulphide
Zinc sulphide
Carbon
Heat
Iron (III) oxide
Carbon
monoxide
Copper (II)
Sulphur in
sulphide
the ore
Aluminium oxide Electrons
(from oxide
ions)
Reduced
product
Mercury
Zinc
Heat
Iron
Heat with
oxygen
Electrolysis
Copper
Aluminium
Oxidised
product
Sulphur
dioxide
Carbon
dioxide
Carbon
dioxide
Sulphur
dioxide
Oxygen
(becomes
CO2)
8.3 Reactions Involving Redox
Corrosion
Corrosion occurs when a metal is oxidised. In some cases the corrosion is only a
nuisance when silver tarnishes in an atmosphere containing sulphur.
2Ag(s) + S(g) Ag2S(s)
The silver sulphide is black but can be removed by polishing the metal with a mild
abrasive.
In other cases corrosion leads to severe weakening of the metal and is both
difficult and expensive to prevent. An example is the rusting of iron. When iron
rusts it forms a compound which is a mixed oxide-hydroxide.
Experiment: To investigate the rusting of iron
Requirements
Clean iron plate about 10cm square, dilute salt solution, ferroxyl indicator, dilute
solutions of sodium hydroxide and iron (II) sulphate.
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After a while the plate looks like the
one shown above. The red areas
indicate the presence of an alkali
namely the presence of excess
hydroxide ions in the solution. The
2+
blue areas indicate the presence of iron (II) (Fe ) ions in the solution. The brown
ring formed last. It is rust, formed by the other two ions coming together.
Fe2+ + 2OH- Fe(OH)2
this is then oxidised further:
Fe(OH)2 + OH- Fe(OH)3
Brown solid
Rusting of iron needs moisture and oxygen. At first iron atoms pass into solution
as Fe2+ ions quite randomly. Random blue areas are seen. The electrons are
liberated from the iron atoms:
Fe Fe2+ + 2e-
(1)
The electrons travel to the outside of the drop where oxygen from the air is
available. They react with water and oxygen to make negatively-charged
hydroxide ions.
2H2O + O2 4e- 4OH-
(2)
After a short time reaction 1 only goes on in the centre of the drop of solution, in
places away from the oxygen. Reaction 2 can only go on at the edges of the drop,
where oxygen is available.
The production of rust goes on between these two other areas. It needs both Fe2+
and OH- ions. Anything which will stop either reaction 1 or reaction 2 will prevent
the rust forming. Rusting needs oxygen (to form OH- ions), a clean iron surface (to
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release iron ions) and water (to transport the ions). Removing any one of these
will stop rusting.
Corrosion Prevention
1. Painting – paint prevents the access of water and oxygen as long as the paint is
not scratched. If the paint is scratched, rusting is rapid because the rust forces
its way under the exposed edges of the paint and pushes it off.
2. Oil or grease.
3. Rust preventer – usually a solution containing phosphate ions which form
insoluble iron phosphates on the metal surface and stop iron ions passing into
solution. This can be used as a primer under paint, or where paint has become
scratched and is being repaired.
4. Galvanising – this means that the iron is dipped into molten zinc so that a thin
layer of zinc is deposited on the iron surface. Zinc is more easily oxidised than
iron so the electrons needed to form the hydroxide ions come from the zinc,
not the iron.
5. Sacrificial protection- using another metal to protect the other from rusting.
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9. ELECTROLYSIS
9.1 Electrolytes and non-electrolytes
Electrolytes are very often salts. A salt is formed when some or all of the
hydrogen atoms of an acid are replaced by a metal. Salts are made up of an array
of positively charged metal ions (cations) and negatively charged ions (anions)
arranged in a crystal lattice. The relative numbers of cations and anions make the
lattice electrically neutral.
For example sodium chloride, NaCl is made up of sodium ions, Na+ and chloride
ions, Cl-. In sodium chloride the ions are present in equal numbers (1:1).
Sodium sulphate, Na2SO4 is made up of sodium ions, Na+ and sulphate ions,SO42-.
In sodium sulphate there are two sodium ions to one sulphate ion (2:1).
The ions in a solid salt cannot move. If the salt is melted or made into a solution,
the ions separate from each other. In molten salt the ions stay close together but
in solution they become separated by large numbers of solvent molecules. (See
diagram below.)
When an electric current flows through a liquid, producing a chemical change, the
process is called electrolysis. The electrode connected to the positive side of the
DC supply is called the positive electrode or anode; the electrode connected to
the negative side of the DC supply is called the negative electrode or cathode.
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9.2 Mechanism of Electrical Transfer
When two electrodes are placed in an electrolyte and connected to a DC source,
an electric field is produced. The ions in the liquid interact with the electrical field
and experience a force accelerating them towards the electrode of the opposite
sign and repelled by the electrode of the same sign. The current pushes the
electrons around the circuit.
As the ions move towards the electrodes they have to push through the solvent
molecules which surround them. Small ions can move faster through the solution
than big ions.
When a positive ion reaches the negative electrode (cathode) it takes electrons
away from the electrode. When a negative ion reaches the positive electrode it
gives up electrons into the electron holes on the positive electrode (anode). In
both cases the ions lose their overall charge and become neutral. As many
electrons are taken from the negative electrode (cathode) as are given to the
positive electrode (anode).(See the next figure)
When the ions have lost their charge at the electrodes they may become atoms
which are stable, like the copper atom below. They may become atoms (or groups
of atoms) which are not stable and react further. The chlorine atoms shown
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forming below are an example. Chlorine atoms cannot exist alone. They combine
to form chlorine gas.
Each half of an electrolytic cell can be regarded as separate from the other. The
reaction which takes place at an electrode is called a half-cell reaction.
The half-cell reaction which occurs at the negative electrode (cathode) involves
the donation of electrons from the cathode to the ions reacting. This is a
reduction reaction. The ion reacting at the cathode is reduced. At the positive
electrode (anode) the ion gives up electrons to the electrode. This is an oxidation
reaction. The ion reacting at the anode is oxidised.
The overall chemical changes which take place during electrolysis can be found by
adding together the reactions which go on at the separate electrodes. So for the
example above:
At the negative electrode (cathode):
Cu2+ + 2e- Cu(s)
At the positive electrode (anode):
2Cl- 2e- Cl2(g)
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Overall:
Cu2+ + 2Cl- Cu(s) + Cl2(g)
Electrolytic and metallic conduction
In an electrolytic cell the current is carried through the cell by ions of both
charges. At the electrodes, chemical changes take place. In metallic conduction
the current is carried everywhere in the circuit by electrons. There are no
chemical changes in the metallic conductor. In both cases there is heating effect.
9.3 Examples in the laboratory
Experiment 1: To observe the products of the electrolysis of water
At the negative electrode, hydrogen gas was produced. When ignited the gas
burns with an almost invisible flame. In this experiment there is always some air
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in with the hydrogen and the gas mixture burns very quickly with a shrill “pop”.
This is used as a test for hydrogen.
At the positive electrode, oxygen gas was produced. Air contains only 20 per cent
of oxygen – enough to allow a charred splint to glow. In the pure gas the higher
concentration of oxygen makes the glowing splint burn much more quickly and it
burns into flame again. This is used as a test for oxygen.
The water has been electrolysed into hydrogen and oxygen:
2H2O 2H2O2
9.4 The Faraday Constant
The coulomb (C) is the unit of electrical charge. When an electrical current of one
ampere (A) flows for one second, one coulomb of charge flows round the circuit:
Coulombs = amperes x seconds
1C = 1As
Just as we can have one mole of atoms or ions or molecules so we can have one
mole of electrons. Knowing the charge on one electron (1.602 x 10-19C) and the
value of the Avogadro constant we can calculate the charge on one mole of
electrons.
The charge on one mole of electrons is equal to the charge on one electron
multiplied by the Avogadro constant, namely:
1.602 x 10-19 C x 6.022 x 1023 = 9.65 x 104C
This quantity, the amount of charge carried by one mole of electrons, is called the
Faraday constant (F).
Electrolysis is used in aluminium production and in electroplating.
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10. ACIDS, BASES AND SALTS
10.1 Proton Donors and Acceptors
Acids react with bases to give a salt and water only. Salts are substances formed
when the hydrogen ions of an acid are replaced by metal ions or by ammonium
ions from a base:
acid + base salt + water
An acid can only exist in aqueous solution. It is the solution which is acidic rather
than the parent compound even though we speak loosely of nitric acid or
sulphuric acid.
An acid is a proton donor. A base is a proton acceptor.
10.2 Salts and their Preparation
Salts
A salt is a compound formed when the hydrogen ions in an acid are replaced by
metal ions or by ammonium ions.
Each acid gives rise to a series of salts named by the anion which they contain.
 Hydrochloric acid gives chlorides (sodium chloride, ammonium chloride).
 Nitric acid gives nitrates (barium nitrate, copper nitrate).
 Sulphuric acid gives sulphates (silver sulphate, iron (II) sulphate).
 Phosphoric acid gives phosphates (sodium phosphate, ammonium phosphate).
Salts which contain no replaceable hydrogen are called normal salts. Salts which
contain replaceable hydrogen are called acid salts (because when that hydrogen
is replaced they are acting like an acid).
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Methods of salt preparation
There are a number of different methods of salt preparation. Most (but not all)
involve the reaction between an acid and a base. These are shown below.
Methods of salt preparation from acids and bases
Acid
Dilute mineral acid (*)
Dilute mineral acid (*)
Dilute mineral acid (**)
Base
Metal
Insoluble metal oxide or
hydroxide
Alkali
Mineral acid (**)
Metal carbonate
Concentrated mineral
acid
Any of the above
Notes
Not all metals react
Includes ammonium
salts
Gives off carbon
dioxide
May be vigorous, not
normally used
The steps involved depend largely on the solubilities of the reagents and
products.
In general terms they are as follows.
1. (a) Neutralise the acid with an excess of the other reactant and filter off any
excess solid reagent (marked * in the table),
(b) exactly neutralise the acid with the other reagent, (** in the table).
2. Evaporate the solution to the crystallisation point.
3. Cool to produce crystals of the salt.
4. Filter, wash and dry the crystals.
Salt Preparation by Neutralising an Acid with a Metal
When a metal reacts with an acid the atoms of the metal become ions, and the
hydrogen ions from the acid become hydrogen atoms which join together to form
hydrogen molecules. The hydrogen comes off as a gas. Most metals have a
valency of two; we can use the symbol M for any metal and write a general
equation:
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M(s) + 2H+(aq) M 2+(aq) + 2H
H + H H2
Preparation of a Crystaline Salt by Neutralising an Acid with Excess Base
Information
Copper (II) oxide reacts with dilute sulphuric acid according to the rquation:
CuO(s) + H2SO4(aq)  CuSO4(aq) + H2O(l)
Explanation
As the oxide dissolves, the solution changes colour. The blue colour of the Cu2+(aq)
ion appears. If the crystals, which contain water of crystallisation, had been
evapourated to dryness, the water would have been driven off and some
decomposition would have taken place. The copper(II) oxide had all been
removed in the filter and the only other product of the reaction was water so
there was no need to rinse the crystals.
Salt Preparation by Precipitation (Double Decomposition)
Insoluble salts can be made by mixing solutions containing their separate ions.
They process is called double decompostiotion.
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Pb(NO3)2 + H2SO4  PbSO4 + 2HNO3
Pb+ + 2NO3- + 2H+ + SO42-  PbSO4(s) + 2H+ + 2NO3-
10.3 Titrations
To get more accurate results of a quantity of a salt we perform a titration.
Using a Titration to Prepare a Salt
Requirements:
Hydrochloric acid (1.0 mol/dm3), sodium hydroxide solution (about 2 mol/dm3)
bromothymol blue or other suitable indicator, pipette of suitable volume (10, 20,
or 25cm3), burette and stand, dropper, conical flask or beaker (about 100cm3).
Method: (Refer to page XXXX)
1. Titrate 20cm3 of base solution with acid.
2. Note the volume of acid needed.
3. Work out the concentration of sodium hydroxide solution.
4. Work out the volume of base solution you will need to give the desired mass of
salt.
5. From your titration value, work out what volume of acid will be needed exactly
to neutralize this volume of base.
6. Mix the two solutions and evaporate until crystals begin to form
7. Set the dish aside to crystalise, then filter off the product.
*Use titration to prepare a salt when both acid and base are soluble in water.*
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Calculation Method for Titrations
1. Calculate the number of moles of acid added from the volume used and known
concentration.
2. From the equation for the reaction work out how many moles of alkali will
react with this number of moles of acid.
3. Work out the concentration of alkali solution.
Example
This is based on using a 20cm3 pipette. The volume of acid required is shown in
the table below.
Rough titration
Starting volume
(cm3)
Final volume
(cm3)
Added
Volume(cm3)
0.4
First accurate
titration
22.4
Second accurate
titration
0.6
22.4
42.0
20.2
20.0
19.6
19.6
1. Volume of acid needed is 19.6cm3. Concentration of acid is 1.0mol/ dm3.
Therefore moles of acid are
= 0.0196𝑚𝑜𝑙
19.6
1000𝑑𝑚3
× 1.0𝑚𝑜𝑙/𝑑𝑚3
2. The equation is:
2NaOH + H2SO4  Na2SO4 + 2H2O
so one mole of acid reacts with two moles of alkali so (0.0196 x 2) mol of
alkali were present
= 0.0392 mol alkali
3. The volume of alkali was 20cm3 so 1 cm3 of alkali contained (0.0392/20) mol,
therefore
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1000cm3 (1dm3) contained
(0.0392/20) mol x 1000
= 1.96mol
The value of Mr for NaOH is
(23 + 16 + 1) = 40
Therefore 1 mol NaOH contains 40g.
1.96mol contains 1.96g X 40g/mol
= 78.4g
and the concentration is 78.4g/dm3
10.4 Hydrolysis of Salts
Reactions with Water
Water is a reactive chemical. A reaction involving splitting a chemical has a name
ending –lysis; when this is done by water the process is called hydrolysis.
Acids and bases react to form salts and water. This process can be reversed.
When this happens we say that the salt has been hydrolysed.
Hydrolysis is the reaction of an ion or a molecule with water. When crystals of a
salt are dissolved in water the salt may be hydrolysed. As a result, the solution
may be acidic, neutral or alkaline.
Equilibrium in Hydrolysis Reactions
Weak acids and alkalis ionise to only a small extent when in solution in water.
Most of the dissolved material is in the form of molecules.
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Weak
CH3COOH(aq)
Ethanoic Acid
Strong
HCl(aq)
Hydrochloric acid
in solution
Alkali
CH3COOH
NH3(aq)
Ammonia solution
H+, ClNaOH(aq)
Sodium Hydroxide Solution
in solution
NH4OH
Na+, OH-
Acid
Strong acid, weak base:
H+ + A- + BOH ⇌ B+ + A- + H2O
Weak acid, strong base:
HA + B+ + OH- ⇌ B+ + A- + H2O
Strong acid, strong base:
H+ + OH- ⇌ H2O
Weak acid, weak base:
HA + BOH ⇌ BA + H2O
These are reversible reactions. To see the effect of dissolving the salt in water look
at each one in turn from right to left. (in reverse)
The salt of a strong acid and weak base reacts with water to give an acidic
solution because of the ionised acid and the unionised base.
The salt of a weak acid and strong base reacts with water to give an alkaline
solution because of the ionised base and the unionised acid.
The salt of a strong acid and strong base reacts with water to give a neutral
solution because of the equal number of H+ and OH- ions.
We cannot give any predictions about the salt of a weak acid and a weak base.
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10.5 Identifying anions
Test
Carbonate (CO3 2-)
add dilute hydrochloric
acid to solution of CO3
Chloride (Cl-)
Add an equal volume of
dilute nitric acid
Nitrate (NO3-)
To solid salt add sodium
hydroxide solution and
heat gently
Sulphate (SO42-)
Add an equal volume of
dilute nitric acid (to
prevent precipitation of
barium cabonate)
Reagent
Positive observation
Dilute hydrochloric acid
Colourless gas which first
turns lime water milky
and then clear again
Silver nitrate
solution(about 0.1
mol/dm3)
White precipitate (which
turns purple in the light)
Devarda’s alloy (or
aluminium powder)
Colourless gas which
turns damp red litmus
paper blue (the gas is
ammonia)
Barium chloride solution
(about 0.1 mol/dm3)
White precipitate
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10.6 Identifying Cations
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References
pixabay.com
physicsandmathstutor.com
Curriculum development Centre’s chemistry 10 – macmillan Zambia
Scullion Family chemistry material
Wikimedia Commons
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