Chemistry
Edexcel International GCSE in Chemistry (4CH0)
First examination June 2013
1

Principles of chemistry
–
States of matter
– Atoms
–
Atomic structure
– Relative formula masses and molar volumes of gases
–
Chemical formulae and chemical equations
– Ionic compounds
– Covalent substances
–
Metallic crystals
– Electrolysis
2

States of matter
1.1 understand the arrangement, movement and energy of the particles in each of the three states of matter: solid, liquid and gas
– Solid: Particles in fixed positions, regular repeating pattern, strong inter-particle forces.
– Liquid: Particles not in fixed positions, can flow, irregular pattern, weaker inter-particle forces.
–
Gas: Particles not in fixed positions, can flow, irregular pattern, non-existent interparticle forces.
1.2 understand how the interconversions of solids, liquids and gases are achieved and recall the names used for these interconversions
– Solid  Liquid is MELTING.
Heat the solid until it melts.
–
Liquid

Sold is FREEZING. Cool the liquid until it freezes.
– Liquid  Gas is EVAPORATION. Heat the liquid until it boils.
–
Liquid

Gas is CONDENSATION.
Cool the gas until it condenses.
– Solid  Gas is SUBLIMATION. Heat the solid until it sublimes.
3

States of matter
1.3 explain the changes in arrangement, movement and energy of particles during these interconversions.
– Melting: The particles gain kinetic energy as they are heated and vibrate faster and faster. This allows the particles to overcome the forces of attraction between them.
This allows the particles to break the regular pattern and to flow over one another.
– Freezing: The particles lose kinetic energy as they are cooled and this allows the forces of attraction to hold them together. The particles arrange themselves into a regular pattern and are no longer able to flow over one another.
– Evaporation: The particles gain kinetic energy as they are heated and move further apart. Eventually the forces of attraction between the particles are completely broken and they are able to escape.
–
Condensation: The particles lose kinetic energy as they are cooled and this allows the forces of attraction to bring them closer together. The particles gradually clump together to form a liquid.
–
Sublimation: The particles gain kinetic energy as they are heated and vibrate faster and faster. Eventually the forces of attraction between the particles are completely broken and they are able to escape.
4

Atoms
1.4 describe and explain experiments to investigate the small size of particles and their movement including: i dilution of colored solutions
–
When potassium manganate (VII) crystals are dissolved in water, a purple solution is formed. When this solution is diluted with water several times, the color becomes less intense. However it takes many dilutions for the color to completely fade.
– This indicates that there is a large number of particles in a small solid, denoting that potassium manganate particles are very small.
ii diffusion experiments
– The bromine gas in the flask on one side will diffuse quickly into the flask on the other side. This is because of large gaps between particles, allowing the air and bromine to easily mix.
– The ammonium chloride experiment shows that particles in different gases travel at different speeds. The while layer of ammonium chloride is closer to the hydrochloric acid because the ammonia gas travels faster because it is lighter.
5

Atoms
1.5 understand the terms atom and molecule
– Atoms are made up of smaller, sub-atomic particles called protons, neutrons and electrons.
–
Molecules are made up of two ore more atoms covalently bonded together.
1.6 understand the differences between elements, compounds and mixtures
– Elements are substances that cannot be further broken down.
–
Compounds are substances that are created through a chemical reaction from different elements. They have different properties to their elements.
– Mixtures are substances that are created through a physical action of mixing from different elements. They have properties from each of the elements used.
6

Atoms
1.7 describe experimental techniques for the separation of mixtures, including simple distillation, fractional distillation, filtration, crystallization and paper chromatography
– Filtration is to separate an undissolved solid from a mixture of the solid and a liquid.
– Pass the mixture through a filter, thus filtering out the solid.
– Evaporation is to separate a dissolved solid from a solution, when the solid has similar solubilities in both cold and hot solvent.
– Heat the solution until the liquid evaporates and only the solid remains.
– Crystallization is to separate a dissolved solid from a solution, when the solid is much more soluble in hot solvent than in cold.
– Dissolve the solution in a hot solvent, then allow it to rest and crystals will appear in the flask. Using filtration, extract the crystals, then rinse it with cold solution to remove impurities.
–
Simple Distillation is to separate a liquid from a solution.
– Heat the solution until the liquid evaporates, then pass it through a condenser to condense the gas back into liquid.
7

Atoms
–
Fractional Distillation is to separate two or more liquids that are miscible with one another.
– Heat the mixture up to a certain boiling point that only one of the liquids evaporates, then pass it through a condenser to condense the gas back into liquid.
– Paper Chromatography is to separate substances that have different solubilities in a given solvent.
– The different substances travel up the paper at different speeds due to their different solubilities.
1.8 explain how information from chromatograms can be used to identify the composition of a mixture.
–
The distance they travel on the paper denotes their solubility.
8

Atomic Structure
1.9 understand that atoms consist of a central nucleus, composed of protons and neutrons, surrounded by electrons, orbiting in shells
– Neutrons
–
Electrons
– Protons
9

Atomic Structure
1.10 recall the relative mass and relative charge of a proton, neutron and electron
– Protons: +1 Charge, 1 Mass
–
Electrons: -1 Charge, 1/1836 (Basically 0) Mass
– Neutrons: 0 Charge, 1 Mass
1.11 understand the terms atomic number, mass number, isotopes and relative atomic mass ( A r
)
–
Atomic Number is the number of protons.
– Mass Number is the number of protons + the number of neutrons.
– Isotopes occur when atoms have the same number of protons but different number of neutrons.
– Relative Atomic Mass is calculated from the masses and relative abundancies of all the isotopes of an element. Given the symbol A r
.
10

Atomic Structure
1.12 calculate the relative atomic mass of an element from the relative abundances of its isotopes
– Chlorine has two isotopes : chlorine-35 and chlorine-37.
–
In a typical sample of chlorine, 75% is chlorine-35 and 25% is chlorine-37.
– (0.75 X 35) + (0.25 X 37) = 35.5 = A r of chlorine
1.13 understand that the Periodic Table is an arrangement of elements in order of atomic number
–
The periodic table is arranged by the number of protons in the element. The element is defined by its number of protons.
1.14 deduce the electronic configurations of the first 20 elements from their positions in the Periodic Table
– The period (row) in the periodic table defines the number of shells it has.
11

Atomic Structure
1.15 deduce the number of outer electrons in a main group element from its position in the Periodic Table.
– The group (column) 1-7 in the periodic table defines the number of electrons on the outer shell.
12

Relative formula masses and molar volumes of gases
1.16 calculate relative formula masses ( M r
) from relative atomic masses ( A r
)
– Hydrogen has an A r
– H
2 has a M r of 1.
of 2 X 1 = 2
– Oxygen has an A r
– H
2
O has a M r of 16.
of 2 X 1 + 16 = 18
– Likewise, (NH
4
)
2
SO
4 has an M r of 132
– (2 X 14) + (8 X 1) + 32 + (4 X 16) = 132
1.17 understand the use of the term mole to represent the amount of substance
–
Mole is a measure of the amount of substance.
1.18 understand the term mole as the Avogadro number of particles (atoms, molecules, formulae, ions or electrons) in a substance
– 6.023 X 10 23 is known as the Avogrado number.
–
1 mol of an substance contains 6.023 X 10 23 of the substance.
13

Relative formula masses and molar volumes of gases
1.19 carry out mole calculations using relative atomic mass ( A r
) and relative formula mass ( M r
)
– Mole calculations can be done with A r
– e.g H
2 has a M r of 2 and M r
..
– 1 mol of H
2 is 2g.
– Likewise, 8 mols of H
2
– e.g. Mg has a A r of 24
– 0.5 mols of Mg is 12g is 16g
– Mass of Substance (g) = Amount (mols) X M r
1.20 understand the term molar volume of a gas and use its values (24 dm
3
24,000 cm
3
) at room temperature and pressure (rtp) in calculations.
–
One mole of any gas has a volume of 24 dm 3 .
and
–
To calculate the volume of gas at rtp: Volume of gas (dm 3 ) = Amount (mols) X 24
– To calculate the amount of gas at rtp: Amount of gas (mols) = Volume of gas (dm 3 ) /
24
14

Chemical formulae and chemical equations
1.21 write word equations and balanced chemical equations to represent the reactions studied in this specification
– Unbalanced Equation: Al + CuO  Al
2
O
3
– Left side: 1 Al, 1 Cu, 1 O
+ Cu
– Right side: 2 Al, 1 Cu, 3 O
–
Balanced Equation: 2Al + 3CuO

Al
2
O
3
– Left side: 2 Al, 3 Cu, 3 O
– Right side: 2 Al, 3 Cu, 3 O
+ 3Cu
– Word Equation: Aluminum + Copper Oxide  Aluminum Oxide + Copper
1.22 use the state symbols (s), (l), (g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectively
–
Zn(s) + CuSO
4
(aq)  ZnSO
4
(aq) + Cu(s)
–
Tells us that when solid zinc is added to an aqueous Copper Sulfate solution, it forms an aqueous Zinc Sulfate solution and solid Copper.
15

Chemical formulae and chemical equations
1.23 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallization
– e.g. When 20g of hydrated copper sulfate salts are heated, and 7.2g of water evaporates.
– The mass of the anhydrous salt copper sulfate is 12.8g.
– The mass of the water is 7.2g.
– To deduce the mole ratio divide the masses by their respective
M r
..
– Copper Sulfate: 12.8 / 160 = 0.08
– Water: 7.2 / 18 = 0.4
– 0.08 : 0.4 = 1 : 5
– The empirical formula of hydrated copper sulfate salts is CuSO
4
.5H
2
O
16

Chemical formulae and chemical equations
1.24 calculate empirical and molecular formulae from experimental data
– e.g. To find the empirical formula of aluminum oxide:
– From an experiment it is deduced that there is 52.94g of aluminum and 47.06g of oxygen in a sample of aluminum oxide.
– Aluminum: 52.94 / 27 = 1.96
– Oxygen: 47.06 / 16 = 2.94
– 1.96 : 2.94 = 2 : 3
– The empirical formula of aluminum oxide is Al
2
O
3
– e.g. To find the molecular formula of benzene:
– Empirical formula of benzene is CH.
– The
M r of Benzene is 78.
– The empirical formula mass of CH = 13
– 78 / 13 = 6
– Therefore, the molecular formula of Benzene is C
6
H
6
17

Chemical formulae and chemical equations
1.25 calculate reacting masses using experimental data and chemical equations
– e.g. Calculate the mass of magnesium oxide made by completely burning 6g of magnesium in oxygen. (2Mg + O
2
 2MgO)
–
A r of magnesium is 24. Therefore, 6 / 24 = 0.25 mols.
– The equation states that 2 mols of Mg makes 2 mols of MgO, thus 0.25 mols of
MgO is created from 0.25 mols of Mg.
–
M r of MgO is 40
– 0.25 X 40 = 10g of MgO formed.
18

Chemical formulae and chemical equations
1.26 calculate percentage yield
– Percentage yield = Yield obtained / theoretical yield
– e.g. To displace copper from copper sulfate, 6.5g of zinc was added to an excess of copper sulfate solution. The mass of copper obtained was 4.8g.
– The equation for the reaction is Zn(s) + CuSO
4
–
A r of zinc is 65. Therefore, 6.5 / 65 = 0.10 mols.
(aq)  ZnSO
4
(aq) + Cu(s)
– Amount of copper = 0.10 mols. Mass of copper = (0.10 X 64) = 6.4g
– Yield obtained = 4.8g
– Theoretical yield = 6.4g
– 4.8 / 6.4 X 100 = 75%
19

Chemical formulae and chemical equations
1.27 carry out mole calculations using volumes and molar concentrations.
– e.g. Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm 3 that is required to react completely with 2.5g of calcium carbonate.
–
M r of calcium carbonate is 100. Therefore, 2.5 / 100 = 0.25 mols.
– The equation states that 1 mol of CaCO
3 of CaCo
3 requires 0.05 mols of HCl.
requires 2 mols of HCl, thus 0.025 mols
– Volume = (Amount (cm 3 ) X 1000) / concentration (mol / dm 3 )
– (0.05 X 1000) / 1.0 = 50 cm 3 of hydrochloric acid needed.
20

Ionic compounds
1.28 describe the formation of ions by the gain or loss of electrons
– An ion is an electrically charged atom or group of atoms. Ions are formed by the loss or gain of electrons.
1.29 understand oxidation as the loss of electrons and reduction as the gain of electrons
–
Oxidation: Loss of electrons
– Reduction: Gain of electrons
1.30 recall the charges of common ions in this specification
Ag +
NH
4
+
Cr 3+
Zn 2+
OH -
NO
3
-
SO
4
2-
CO
3
2-
21

Ionic compounds
1.31 deduce the charge of an ion from the electronic configuration of the atom from which the ion is formed
– An atom seeks to achieve a full shell by oxidation or reduction.
– When oxidation occurs, an atom loses electrons and thus gains a positive charge.
– When reduction occurs, an atom gains electrons and thus gains a negative charge.
– e.g. Magnesium is in Group 2, thus has 2 electrons in its outer shell.
– To achieve a full shell, it has to lose 2 electrons.
– Magnesium gains a charge of +2, since each electron lost has a -1 charge.
22

Ionic compounds
1.32 explain, using dot and cross diagrams, the formation of ionic compounds by electron transfer, limited to combinations of elements from Groups 1, 2, 3 and 5, 6, 7
– e.g. In NaCl, sodium chloride:
– Na has 1 electron in its outer shell.
– Cl has 7 electrons in its outer shell.
– To achieve full shells, Na donates its extra electron to Cl.
– Now that Na has one less electron, it has a charge of +1.
– Now that Cl has one more electron, it has a charge of -1.
Na
[ ] +
Cl
[ ] -
23

Ionic compounds
1.33 understand ionic bonding as a strong electrostatic attraction between oppositely charged ions
– An ionic compound forms when a oppositely charged (positive / negative) atoms are attracted to each other. It is a bond between a metal and a non-metal.
1.35 understand the relationship between ionic charge and the melting point and boiling point of an ionic compound
– Strong ionic charge: Very high melting and boiling point (MgO: 2800 degrees)
–
Weak ionic charge: High melting and boiling point (NaCl: 800 degrees)
1.36 describe an ionic crystal as a giant three-dimensional lattice structure held together by the attraction between oppositely charged ions
– When many ions arrange themselves in a three dimensional structure called an giant ionic lattice. Ionic compounds often form crystals.
24

Ionic compounds
1.37 draw a diagram to represent the positions of the ions in a crystal of sodium chloride.
25

Covalent substances
1.38 describe the formation of a covalent bond by the sharing of a pair of electrons between two atoms
– When atoms of non-metallic elements combine together they share electrons between them to make full shells. This electron bonds them together.
1.39 understand covalent bonding as a strong attraction between the bonding pair of electrons and the nuclei of the atoms involved in the bond
– When the electrons are shared, the shared region of the electron shells bond the two nuclei together. This forms a covalent compound.
26

Covalent substances
1.40 explain, using dot and cross diagrams, the formation of covalent compounds by electron sharing for the following substances:
H H
Cl Cl
H
H C H
H
H
H C
H
H
C H
H
H
C
H
H
C
H
H Cl
27

Covalent substances
1.41 understand that substances with simple molecular structures are gases or liquids, or solids with low melting points
– Simple molecular structures are formed from covalent bonds because it consists of individual molecules, these structures have low melting points.
– Often gases (e.g. hydrogen, oxygen, nitrogen, methane).
–
Low boiling point liquids (e.g. water, bromine)
–
Low melting point solids (e.g. iodine)
1.42 explain why substances with simple molecular structures have low melting and boiling points in terms of the relatively weak forces between the molecules
– Forces of attraction between molecules are relatively weak compared to ionic bonds, therefore, very little energy is required to overcome them.
– e.g. If it is a diatomic chlorine molecule, it is breaking the intermolecular bond between the chlorine molecules not the covalent bond.
28

Covalent structures
1.44 draw diagrams representing the positions of the atoms in diamond and graphite
Graphite
Diamond
29

Covalent substances
1.45 explain how the uses of diamond and graphite depend on their structures, limited to graphite as a lubricant and diamond in cutting.
– Diamond can be used in cutting because covalent bonds are strong, and there are many of them in the giant covalent structure. There are no weak forces in diamond.
– Graphite can be used as a lubricant because the weak intermolecular forces of attraction between the layers are weak so the layers easily slide over one another and can easily be separated. This gives graphite the soft and slippery property.
30

Metallic crystals
1.46 understand that a metal can be described as a giant structure of positive ions surrounded by a sea of delocalized electrons
– Metals have a giant, three dimensional lattice structure in which positive ions are arranged in a regular pattern in a sea of electrons. The outer shell electrons are detached from the atoms and are delocalized throughout the entire structure.
1.47 explain the electrical conductivity and malleability of a metal in terms of its structure and bonding.
– Metallic bonds are strong and there are many of them in a giant structure, hence a lot of heat energy is required, so melting points are high.
– Metals can conduct electricity because the delocalized electrons are free to move when a potential difference is applied across the metal.
–
Metals are malleable and ductile because the layers of positive ions can easily slide over one another and take up different positions. The electrons move with them so no bonds are broken.
31

Electrolysis
1.48 understand that an electric current is a flow of electrons or ions
– An electric current is a flow of electrons or ions within a substance, this is why metals are such good conductors.
1.49 understand why covalent compounds do not conduct electricity
–
Covalent compounds do not conduct electricity because they do not have free delocalized electrons that enables this property.
1.50 understand why ionic compounds conduct electricity only when molten or in solution
– Ionic compounds conduct electricity when molten or in aqueous form because the ions are free to move in these forms and thus allow it to conduct a charge.
32

Electrolysis
1.51 describe experiments to distinguish between electrolytes and nonelectrolytes
– Electrolytes are substances that conduct electricity in the molten state or when dissolved in water.
–
Nonelectrolytes are substances that don’t conduct electricity in these states
– To distinguish between these two, simply place two electrodes in the water connected to a power source and a light bulb, and if the bulb lights up, electrolytes are present.
1.52 understand that electrolysis involves the formation of new substances when ionic compounds conduct electricity
– Passing a charge through the electrolytes cause chemical changes, thus the chemical reactions produce new products.
33

Electrolysis
1.53 describe experiments to investigate electrolysis, using inert electrodes, of molten salts such as lead(II) bromide and predict the products
– The electrodes are inert which means they do not react.
–
Cathode: The negative electrode, when ions get to the electrode, they gain electrons.
– Anode: The positive electrode, when ions get to the electrode, the lose electrons.
– e.g. In the electrolysis of lead(II) bromide:
– Pb 2+ is attracted towards the cathode, where they gain electrons to become Pb.
– Pb 2+ + 2e  Pb
– Br is attracted towards the anode, where they lose electrons to become Br.
– 2Br  Br
2
+ 2e-
34

Electrolysis
1.54 describe experiments to investigate electrolysis, using inert electrodes, of aqueous solutions such as sodium chloride, copper(II) sulfate and dilute sulfuric acid and predict the products
–
Selective discharge occurs:
– At the cathode, hydrogen is preferably discharged over all cations unless the cation is less reactive than it.
– At the anode, oxygen is preferably discharged over all anions unless there is a high concentration of anions.
– e.g. in the electrolysis of sodium chloride:
– There are three particles present in the solution, Na + , Cl , H
2
O.
– At the cathode, the water gains electrons to produce hydrogen gas and hydroxide ions.
– 2H
2
O + 2e 
H
2
+ 2OH -
– At the anode, the chloride loses electrons to produce chlorine gas.
– 2Cl 
Cl
2
+ 2e -
– 2NaCl(aq) + H
2
O(l)  Cl
2
(g) + H
2
(g) + 2NaOH(aq)
35

Electrolysis
– e.g. in the electrolysis of copper(II) sulfate:
– There are three particles present in the solution, Cu 2+ , SO
4
2, H
2
O.
– At the cathode, copper ions gain electrons to produce copper deposits. (Copper is lower than hydrogen on the reactivity series).
– Cu 2+ + 2e  Cu
– At the anode, water loses electrons to produce oxygen gas and hydrogen ions.
– 2H
2
O  O
2
+ 4H + +4e -
– 2CuSO
4
(aq) + 2H
2
O(l)  2Cu(s) + O
2
(g) + 2H
2
SO
4
(aq)
– e.g. in the electrolysis of dilute sulfuric acid:
– There are three particles present in the solution, H + , SO
4
2, H
2
O.
– At the cathode, hydrogen ions gain electrons to produce hydrogen gas.
– 2H + + 2e  H
2
– At the anode, water loses electrons to produce oxygen gas and hydrogen ions.
– 2H
2
O  O
2
+ 4H + 4e and then 2H
2
O  O
2
+ 2H
2
O
– The water is decomposed leaving a more concentrated solution of sulfuric acid.
36

Electrolysis
1.55 write ionic half-equations representing the reactions at the electrodes during electrolysis
– Typically ionic half equations show the gain and loss of electrons at the electrodes:
– At the cathode: cation + e 
[hydrogen or metal]
– At the anion: anion  [non-metal] + e -
1.56 recall that one faraday represents one mole of electrons
–
One faraday is one mole of electrons, which equals 96500 coulombs of charge.
37

Electrolysis
1.57 calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions.
– e.g. When copper is deposited, the cathode reaction is Cu 2+ + 2e  Cu
– Thus, for every 1 mol of copper deposited, 2 mols of electrons are needed.
– e.g. Calculate the masses of metals deposited when one faraday of electricity flows through molten lead bromide.
– Pb 2+ + 2e 
Pb
– 1 mol of electrons deposit 0.5 mols of lead = 103.5g
– e.g. A current of 0.010 A passes for 4 hours through a solution of gold (III) nitrate.
What mass of metal is deposited.
– Charge = Current X Time
– 0.010 X (4 X 60 X 60) = 144 C
– Au 3+ + 3e  Au
– 3 x 96500C deposit 1 mol (197g) of gold.
– 144C deposits (197 / (3 X 96500)) X 144 = 0.098g of gold.
38

Electrolysis
– e.g. A metal of A r
48.0 is deposited by electrolysis. If 0.239g of the metal is deposited when 0.100 A flow for 4 hours, what is the charge on the ion of this element.
– 0.100 X (4 X 60 X 60) = 1440 C
– 1440 C deposits 0.239g of metal.
– 48g of metal is deposited by 1440/0.239 X 48 = 289205 C
– 289205 / 96500 = 3 mols of electrons.
– 3 mols of electrons produces 1 mol of metal.
– The metal has the charge of 3+.
39

Chemistry of the elements
–
The Periodic Table
– Group 1 elements – lithium, sodium and potassium
–
Group 7 elements
– chlorine, bromine and iodine
– Oxygen and oxides
–
Hydrogen and water
– Reactivity series
– Tests for ions and gases
40

The Periodic Table
2.1 understand the terms group and period
– Groups are the vertical columns of the periodic table. The elements have the same number of electrons in their outermost shell.
–
Periods are the horizontal columns of the periodic table. The elements have the same number of electron shells.
2.2 recall the positions of metals and non-metals in the Periodic Table
– The metals are on the left and the righter-most metals are Beryllium, Aluminum,
Gallium, Tin, Bismuth.
2.3 explain the classification of elements as metals or non-metals on the basis of their electrical conductivity and the acid-base character of their oxides
– Metals conduct electricity, are malleable and ductile, often shiny, solid at room temperature and form oxides that are basic.
–
Non metals do not conduct electricity, are brittle and form oxides that are acidic.
41

The Periodic Table
2.4 understand why elements in the same group of the Periodic Table have similar chemical properties
– They have similar chemical properties because they have the same number of electrons in their outer shell.
2.5 understand that the noble gases (Group 0) are a family of inert gases and explain their lack of reactivity in terms of their electronic configurations.
– Noble gases have full outer shells, thus they do not lose or gain electrons easily, because doing so would either require breaking a full shell, or creating a new one.
Therefore, they are inert.
42

Group 1 elements – lithium, sodium and potassium
2.6 describe the reactions of these elements with water and understand that the reactions provide a basis for their recognition as a family of elements
– The alkali metals are recognized as a family due to their reactions with water:
– Lithium: Moves around the surface of the water, hissing sound, effervescence, deteriorates over time.
– Sodium: All of the above, melts into a shiny ball.
– Potassium: All of the above, burns with a lilac-colored flame.
2.7 describe the relative reactivities of the elements in Group 1
– Increases down the group: Potassium  Sodium  Lithium
2.8 explain the relative reactivities of the elements in Group 1 in terms of distance between the outer electrons and the nucleus.
–
The reactivity goes up going down the group because less energy is required to remove the one electron on the outer shell. This is because the electron is further away from the nucleus, thus it is less strongly attached.
43

Group 7 elements – chlorine, bromine and iodine
2.9 recall the colors and physical states of the elements at room temperature
Chlorine Gas
Bromine Liquid
Iodine Solid
Pale green
Red-brown
Black
2.10 make predictions about the properties of other halogens in this group
– Fluorine is a pale-yellow gas.
–
Astatine is a black solid.
2.11 understand the difference between hydrogen chloride gas and hydrochloric acid
– Hydrochloric acid is hydrogen chloride dissolved in water, as a solution. It forms H + ions which makes it acidic.
44

Group 7 elements – chlorine, bromine and iodine
2.12 explain, in terms of dissociation, why hydrogen chloride is acidic in water but not in methylbenzene
– When hydrogen chloride is dissolved in water, it dissociates (splits up) into H + ions and Cl ions. The H + ions are the ones responsible for its acidic properties.
– When hydrogen chloride is dissolved in methylbenzene, it only dissolves and does not dissociate, thus not forming ions. This means it only exists as HCl molecules, not
H + and Cl ions.
2.13 describe the relative reactivities of the elements in Group 7
–
Decreases down the group: Fluorine

Chlorine

Bromine

Iodine

Astatine
45

Group 7 elements – chlorine, bromine and iodine
2.14 describe experiments to demonstrate that a more reactive halogen will displace a less reactive halogen from a solution of one of its salts
– e.g. Adding chlorine to a potassium bromide solution will displace the bromide from the potassium.
– Cl
2
+ 2KBr  2KCl + Br
2
–
The color of aqueous chlorine is very pale green.
–
The color of aqueous bromine is orange.
– The color of aqueous iodine is brown.
2.15 understand these displacement reactions as redox reactions.
–
In the above reaction, chlorine oxidizes bromine and gains an electron each, this makes Cl which is attracted to the K + ion. There is now two bromine atoms, and thus forms a covalent bond to make the diatomic molecule Br
2
.
– Cl
2
+ 2e 
–
2Br 
Br
2
2Cl
+ 2e
-
-
46

Oxygen and oxides
2.16 recall the gases present in air and their approximate percentage by volume
– Composition of air is 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% carbon dioxide.
2.17 explain how experiments involving the reactions of elements such as copper, iron and phosphorus with air can be used to investigate the percentage by volume of oxygen in air
–
In the reaction of copper and oxygen: 2Cu + O
2

2CuO
– Using syringes to pass air over copper over a Bunsen burner. Measure how much air there is left after it stops decreasing. The difference is the volume of oxygen.
– Placing wet iron filings in the end of the burette, and allowing the water to rise up the burette to show the volume of oxygen that reacted with the iron. The volume of oxygen used divided by the initial volume of gas is the percentage of oxygen in air.
47

Oxygen and oxides
2.18 describe the laboratory preparation of oxygen from hydrogen peroxide, using manganese(IV) oxide as a catalyst
– 2H
2
O
2
(aq)  2H
2
O(l) + O
2
(g)
– Oxygen is not very soluble in water, so very little is lost.
– The speed of hydrogen peroxide decomposition can be catalyzed by MnO
2
.
2.19 describe the reactions of magnesium, carbon and sulfur with oxygen in air, and the acid-base character of the oxides produced
–
Magnesium burns with a bright, white flame to form a white powder.
– In solution it forms Mg(OH)
2
, a basic solution with a pH of 10.
–
Carbon burns with a yellow-orange flame to form a colorless gas.
– In solution it forms H
2
CO
3 a acidic solution with a pH of 5-6.
–
Sulfur burns with a blue flame to form a colorless gas.
– In solution it forms H
2
SO
3
(sulfurous acid), a acidic solution with a pH of 3-4.
48

Oxygen and oxides
2.20 describe the laboratory preparation of carbon dioxide from calcium carbonate and dilute hydrochloric acid
– CaCO
3
(s) + 2HCl(aq)  CaCl
2
(aq) + H
2
O(l) + CO
2
(g)
– Calcium carbonate is used in the form of marble chips. They are easy to handle and the rate of reaction is not too fast.
2.21 describe the formation of carbon dioxide from the thermal decomposition of metal carbonates such as copper(II) carbonate
–
CuCO
3
(s)

CuO(s) + CO
2
(s)
– Green copper carbonate  Black copper oxide
–
MgCO
3
(s)  MgO(s) + CO
2
(g)
– CaCO
3
(s)  CaO(s) + CO
2
(g)
–
ZnCO
3
(s)  ZnO(s) + CO
2
(g)
– White to yellow when hot, then yellow again when cold.
49

Oxygen and oxides
2.22 describe the properties of carbon dioxide, limited to its solubility and density
– Carbon dioxide is slightly soluble in water.
–
It is more dense than air.
2.23 explain the use of carbon dioxide in carbonating drinks and in fire extinguishers, in terms of its solubility and density
– It is used as a fire extinguisher as it sinks below other gases onto the fire and prevents oxygen from getting to it.
–
Under high pressure it becomes much more soluble, hence its use to make carbonated drinks.
2.24 understand that carbon dioxide is a greenhouse gas and may contribute to climate change.
– As a greenhouse gas, it traps heat inside the atmosphere, thus heating up the earth.
50

Hydrogen and water
2.25 describe the reactions of dilute hydrochloric and dilute sulfuric acids with magnesium, aluminum, zinc and iron
– Metals above hydrogen in the reactivity series will react with acids to form a salt.
– Metal + hydrochloric acid  metal chloride + hydrogen
– Metal + sulfuric acid  metal sulfate + hydrogen
–
Common observations:
– Metal disappears
– Effervescence
– Colorless solution formed (except green iron)
2.26 describe the combustion of hydrogen
– Hydrogen combusts violently to give water.
–
To test for hydrogen, place the flame of a lighted splint at the mouth of the test tube, a squeaky pop will be heard.
51

Hydrogen and water
2.27 describe the use of anhydrous copper(II) sulfate in the chemical test for water
– Add the liquid to anhydrous copper(II) sulfate. The white power will turn blue if water is present.
– CuSO
4
(s) + 5H
2
O(l)

CuSO
4.
5H
2
O(l)
– Anhydrous copper(II) sulfate is white .
– Hydrated copper (II) sulfate is blue .
2.28 describe a physical test to show whether water is pure.
– Boil the water and measure its boiling point.
– If it is 100 degrees, then the water is pure.
52

Reactivity series
2.29 understand that metals can be arranged in a reactivity series based on the reactions of the metals and their compounds: potassium, sodium, lithium, calcium, magnesium, aluminum, zinc, iron, copper, silver and gold
–
Potassium  Sodium  Lithium  Calcium  Magnesium  Aluminum  Carbon

Zinc

Iron

Tin

Lead

Copper

Silver

Gold

Platinum
– Hydrogen is between lead and copper.
2.30 describe how reactions with water and dilute acids can be used to deduce the following order of reactivity: potassium, sodium, lithium, calcium, magnesium, zinc, iron and copper
– First four react vigorously with cold water to form a solution of metal hydroxide and hydrogen. Magnesium reacts very slowly with cold water but burns in steam. Zinc, iron do not react with cold water but react with steam, without burning. They react with steam to form metal oxide and hydrogen, and react with decreasing vigor with dilute acids to form the metal salt and hydrogen.
– Copper does not react with anything
53

Reactivity series
2.31 deduce the position of a metal within the reactivity series using displacement reactions between metals and their oxides, and between metals and their salts in aqueous solutions
–
A metal oxide is a compound of metal and oxygen. When the oxygen is removed from, the oxide is reduced . The substance that carries out the reduction is the reducing agent .
– A metal that is higher in the reactivity series will displace a metal from its oxide.
– e.g. CuO(s) + Mg(s) 
Cu(s) + MgO(s)
– In this case, magnesium is the reducing agent.
–
A metal that is higher in the reactivity series will displace a metal from a solution of its salts.
– e.g. Mg(s) + FeSO
4
(aq)  MgSO
4
(aq) + Fe(s)
– In this case, magnesium is the reducing agent.
– Pale green of FeSO
4 replaced by colorless of MgSO
4.
54

Reactivity series
2.32 understand oxidation and reduction as the addition and removal of oxygen respectively
– Oxidation: addition of oxygen.
–
Reduction: removal of oxygen.
2.33 understand the terms redox, oxidizing agent, reducing agent
– Redox reaction: A reaction in which both reduction and oxidation are taking place.
–
Oxidizing agent: A substance that is capable of oxidizing another substance.
– Reducing agent: A substance that is capable of reducing another substance.
2.34 describe the conditions under which iron rusts
– Iron rusts in the presence of oxygen and water.
–
It is catalyzed by the presence of electrolytes such as salt.
55

Reactivity series
2.35 describe how the rusting of iron may be prevented by grease, oil, paint, plastic and galvanizing
– One can stop rusting by preventing contact with oxygen and water, this is done by:
– Covering it with grease / oil.
– Painting the iron.
– Covering it with plastic.
– However, one it is scratched, the iron will rust very quickly.
–
Galvanizing .
– Coating it with a more reactive metal.
2.36 understand the sacrificial protection of iron in terms of the reactivity series.
– Zinc is more reactive than iron, thus, the zinc will react with the air before it can corrode the iron. Therefore, the zinc sacrifices itself to stop iron from rusting.
Galvanization is only used to describe being coated by zinc, for other metals it is sacrificial protection .
56

Tests for ions and gases
2.37 describe tests for the cations: i Li + , Na + , K + , Ca 2+ using flame tests
–
Clean the end of a piece of platinum wire by dipping it into clean HCl and then placing it into a roaring flame until no color is produced.
– The end of the wire should be dipped into hydrochloric acid and then into the sample.
– Lithium (Li + ) burns red.
– Sodium (Na + ) burns yellow / orange.
– Potassium (K + ) burns lilac.
– Calcium (Ca 2+ ) burns brick red.
ii NH4 + , using sodium hydroxide solution and identifying the ammonia evolved
– Add aqueous sodium hydroxide to the solid, or solution, under test and warm it.
–
If ammonium is present, pungent-smelling gas is produced. The gas produced turns damp red litmus paper blue. It is ammonia (NH
3
).
– NH
4
+ (aq) + OH  NH
3
+ H
2
O
57

Tests for ions and gases iii Cu2+, Fe2+ and Fe3+, using sodium hydroxide solution
– Most metal hydroxides are insoluble and hence can be precipitated by adding sodium hydroxide solution drop by drop.
– Copper(II) (Cu 2+ ): Blue precipitate
– Cu 2+ (aq) + 2OH (aq)  Cu(OH)
2
(s)
– Iron(II) (Fe 2+ ): Green precipitate
– Fe 2+ (aq) + 2OH (aq)  Fe(OH)
2
(s)
– Iron(III) (Fe 3+ ): Brown precipitate
– Fe 3+ (aq) + 3OH (aq)  Fe(OH)
3
(s)
58

Tests for ions and gases
2.38 describe tests for the anions: i Cl , Br and I , using dilute nitric acid and silver nitrate solution
–
To an aqueous solution of the solid under test, add some dilute nitric acid followed by a few drops of silver nitrate solution.
– Chloride (Cl ): White precipitate
– Ag + (aq) + Cl (aq)

AgCl(s)
– Bromide (Br ): Cream precipitate
– Ag + (aq) + Br (aq)  AgBr(s)
– Iodide (I ): Yellow precipitate
– Ag + (aq) + I (aq)  AgI(s)
–
The precipitate goes from white to yellow down the halogen group.
59

Tests for ions and gases ii SO
4
2, using dilute hydrochloric acid and barium chloride solution
– To an aqueous solution of the solid under test, add dilute hydrochloric acid followed by a few drops of barium chloride solution.
– Sulfate (SO
4
2): White precipitate
– Ba 2+ (aq) + SO
4
2(aq)  BaSO
4
(s) iii CO
3
2, using dilute hydrochloric acid and identifying the carbon dioxide evolved
–
To either the solid, or an aqueous solution of it, under test add dilute hydrochloric acid.
– Carbonate (CO
3
): Bubbles of gas. The gas produced turns limewater milky.
– CO
3
2(aq) + 2H + (aq)  CO
2
(g) + H
2
O(l)
2.39 describe tests for the gases: i hydrogen
– Mix with air and ignite. If positive, burns with a squeaky pop.
– 2H
2
(g) + O
2
(g)

2H
2
O(l)
60

Tests for ions and gases ii oxygen
– Insert glowing splint into test tube of oxygen. If positive, split relights.
iii carbon dioxide
–
Bubble through limewater. If positive, limewater turns milky colored.
– Ca(OH)
2
(aq) + CO
2
(g)  CaCO
3
(s) + H
2
O(l)
– The CaCO
3 solid turns limewater milky.
iv ammonia
–
Damp red litmus paper. If positive, red litmus paper turns blue.
– NH
3
(g) + H
2
O(l)  NH
4
+ (aq) + OH (aq)
–
OH responsible for basic properties. OH present.
– H + responsible for acidic properties. H + not present.
61

Tests for ions and gases v chlorine
– Damp litmus paper. If positive, bleaches litmus paper white.
62

–
Introduction
– Alkanes
–
Alkenes
– Ethanol
Organic chemistry
63

Introduction
3.1 explain the terms homologous series, hydrocarbon, saturated, unsaturated, general formula and isomerism.
– Homologous series: a series of organic compounds that have the same general formula , similar chemical reactions, and where each member differs from the next by a -CH
2
- group.
– Hydrocarbon: a compound containing only the elements hydrogen and carbon.
–
Saturated: an organic compound in which all the bonds are single bonds.
– Unsaturated: an organic compound that contains a carbon-carbon double bond.
–
General formula: a formula that states the ratio of atoms of element in the formula of every compound in a particular homologous series.
– Isomerism: compounds that have the same molecular formula but different displayed formulae. The different compounds are called isomers .
64

Alkanes
3.2 recall that alkanes have the general formula C n
H
2 n +2
– The general formula C n
H
2 n +2 produces CH
4,
C
2
H
6
, etc…
3.3 draw displayed formulae for alkanes with up to five carbon atoms in a molecule, and name the straight-chain isomers
Butane Methan e
Ethane Propane
Pentane
65

Alkanes
3.4 recall the products of the complete and incomplete combustion of alkanes
– In complete combustion, the products are CO
2
– CH
4
(g) + 2O
2
(g)

CO
2
(g) + 2H
2
O(l) and water:
– C
3
H
8
(g) + 5O
2
(g)  3CO
2
(g) + 4H
2
O(l)
–
In incomplete combustion, the products are CO and water:
– CH
4
(g) + 1.5O
2
(g)  CO(g) + 2H
2
O(l)
3.5 describe the substitution reaction of methane with bromine to form bromomethane in the presence of UV light.
– Methane and bromine react together in the presence of ultra-violet radiation to form bromomethane.
– CH
4
(g) + Br
2
(g)  CH
3
Br(g) + HBr(g)
–
One hydrogen atom in methane has been replaced by a bromine atom, this reaction is called substitution reaction.
66

Alkenes
3.6 recall that alkenes have the general formula C n
H
2 n
– The general formula C n
H
2 n produces C
2
H
4
, C
3
H
6
, etc…
3.7 draw displayed formulae for alkenes with up to four carbon atoms in a molecule, and name the straight-chain isomers (knowledge of cis- and trans- isomers is not required)
Butene Ethene Propene
But-1ene
But-2ene
Methylpropene
67

Alkenes
3.8 describe the addition reaction of alkenes with bromine, including the decolorizing of bromine water as a test for alkenes.
– A bromine molecule will perform an addition reaction across the double bond of ethene to form 1,2-dibromoethane:
– 1.2-dibromoethane is colorless, so when bromine or bromine is shaken with any alkenes, it will decolorize bromine because of the carbon-carbon double bond.
–
This reaction can be used as a test for unsaturated compounds (hydrocarbons).
68

Ethanol
3.9 describe the manufacture of ethanol by passing ethene and steam over a phosphoric acid catalyst at a temperature of about 300 degrees and a pressure of about 60 –70 (atm)
–
Direct hydration of ethene:
– C
2
H
4
(g) + H
2
O(g)  C
2
H
5
OH(g)
– The ethanol is condensed as a liquid.
–
Process described in statement.
3.10 describe the manufacture of ethanol by the fermentation of sugars, for example glucose, at a temperature of about 30 degrees
– Fermentation:
C
6
H
12
O
6
(aq)  2C
2
H
5
OH(aq) + 2CO
2
(g)
– Dissolve sugar in water and add yeast.
– Leave mixture to ferment at 25-40 degrees for several days in the absence of air.
–
Filter off the excess yeast to obtain a dilute solution of ethanol.
69

Ethanol
3.11 evaluate the factors relevant to the choice of method used in the manufacture of ethanol, for example the relative availability of sugar cane and crude oil
Method Fermentation Hydration of ethene
Raw Materials Uses renewable resources Uses non-renewable resources
Type Batch process Continuous process
Rate
Quality
Conditions
Slow (several days)
Dilute ethanol solution
Low temp.
Fast
Pure ethanol
High temp. and pressure
3.12 describe the dehydration of ethanol to ethene, using aluminum oxide.
–
C
2
H
5
OH(g)

C
2
H
4
(g) + H
2
O(g)
– Aluminum oxide is used as a catalyst to the reaction.
70

Physical chemistry
–
Acids, alkalis and salts
– Energetics
–
Rates of reaction
– Equilibria
71

Acids, alkalis and salts
4.1 describe the use of the indicators litmus, phenolphthalein and methyl orange to distinguish between acidic and alkaline solutions
– Litmus: 1-5 red, 5-8 purple, 8-14 blue
–
Methyl orange: 0-3 red, 3-5 orange, 5-14 yellow
– Phenolphthalein: 0-10 colorless, 10-14 red
4.2 understand how the pH scale, from 0 –14, can be used to classify solutions as strongly acidic, weakly acidic, neutral, weakly alkaline or strongly alkaline
–
0-3 pH is strongly acidic.
– 4-6 pH is weakly acidic.
–
7 is neutral.
– 8-10 is weakly basic.
–
11-14 is strongly basic.
72

Acids, alkalis and salts
4.3 describe the use of universal indicator to measure the approximate pH value of a solution
– From strongly acidic to neutral to strongly alkali: Red  Yellow  Green  Blue 
Purple.
4.4 define acids as sources of hydrogen ions, H+, and alkalis as sources of hydroxide ions, OH
¯
– Hydrogen (H + ) ions give an acidic nature.
–
Hydroxide (OH ) ions give an alkaline nature.
73

Acids, alkalis and salts
4.5 predict the products of reactions between dilute hydrochloric, nitric and sulfuric acids; and metals, metal oxides and metal carbonates (excluding the reactions between nitric acid and metals)
–
Metal + acid  salt + hydrogen
– Mg(s) + 2HCl(aq)  MgCl
2
(aq) + H
2
(g)
–
Acids + base  salt + water
– NaOH(aq) + HCl(aq) 
NaCl (aq) + H
2
O
– Acids + metal carbonates  salt + carbon dioxide + water
– Na
2
CO
3
(s) + 2HCl(aq)

2NaCl(aq) + CO
2
(g) + H
2
O(l)
74

Acids, alkalis and salts
4.6 understand the general rules for predicting the solubility of salts in water:
– Soluble
– All common sodium, potassium and ammonium salts (includes carbonates)
– All nitrates
– All common chlorides (except silver and lead(II) chloride)
– All common sulfates (except barium, calcium and lead(II) sulfate)
– Insoluble
– All carbonates (except sodium, potassium and ammonium carbonate)
4.7 describe experiments to prepare soluble salts from acids
– Acid + insoluble base
– Put some dilute acids into a beaker and heat it using a Bunsen burner flame.
– Add the insoluble base, a little at the time and stir until there is excess.
– Filter the mixture into an evaporation basin to remove the excess base.
– Leave the filtrate for the water to evaporate and to form the crystals.
75

Acids, alkalis and salts
–
Acid + soluble base
– Put an aqueous solution of the alkali into a conical flask and add a suitable indicator. Add dilute acid from a burette until the indicator just changes color.
– Add powdered charcoal and shake the mixture to remove the color of the indicator.
– Filter to remove the charcoal and then obtain crystals from the filtrate in through crystallization.
– Dilute acid + metal carbonate
– For most carbonates, follow acid + insoluble base, except for sodium and potassium carbonates where following the acid + soluble technique is suitable.
– Dilute acid + metal
– Follow acid + insoluble base procedure.
76

Acids, alkalis and salts
4.8 describe experiments to prepare insoluble salts using precipitation reactions
– AgNO
3
(aq) + NaCl(aq)  AgCl(s) + NaNO
3
(aq)
– Silver nitrate provides Ag +
– Sodium chloride provides Cl -
– Ag + + Cl  AgCl
– The precipitate of silver chloride is removed by filtration, washed with a little distilled water, and left to dry in a warm place.
77

Acids, alkalis and salts
4.9 describe experiments to carry out acid-alkali titrations.
– Titrations are used to find the volume of acids required to react exactly with a given volume of alkali.
– Using a pipette, put 25.0 cm 3 of the alkali solution into a conical flask.
– Add a few drops of an indicator, such as methyl orange.
– Put the acid into a burette and note the initial reading.
– Add the acid into the alkali until the indicator just changes color.
– Methyl orange turns from yellow to orange
– Note the final reading of acid in the burette.
– Subtract the initial reading from the final reading to obtain the volume of acid added.
78

Energetics
4.10 understand that chemical reactions in which heat energy is given out are described as exothermic and those in which heat energy is taken in are endothermic
– Exothermic: Energy is evolved. (Rise in temperature of surroundings)
– Reactants have more energy than products.
– Water  Ice
– Combustion of fuels
– Negative heat energy change
–
Endothermic: Energy is absorbed. (Fall in temperature of surroundings)
– Reactants have less energy than products.
– Ice  Water
– Photosynthesis
– Positive heat energy change
79

Energetics
4.11 describe simple calorimetry experiments for reactions such as combustion, displacement, dissolving and neutralization in which heat energy changes can be calculated from measured temperature changes
– e.g. for reactions in solution
– Using a measuring cylinder, place 50 cm 3 of 0.1 mol / dm 3 HCl into a polystyrene cup supported in a beaker. Measure and record the temperature of the acid.
– Tip approximately 0.15g of magnesium power into the acid and stir the mixture.
– Measure and record the highest temperature reached by the mixture.
– Heat given out = mass of solution x heat capacity x temperature rise
– Temp. rise = 10 degrees
– Moles of acid = 50 / 1000 X 0.1 = 0.005 mols
– Heat given out = 50 X 4.2 X 10J = 2.1 kJ (Density = 1g / cm 3 , capacity = 4.2)
– 0.0005 mol of acid produces 2.1 kJ
– 1 mol of acid produces = 420 kJ
– Molar enthalpy change = ▲H = -420 kJ / mol
80

Energetics
– e.g. combustion reactions
– Using a measuring cylinder, put 100 cm 3 of water into a copper can.
– Measure and record the initial temperature of the water.
– Fill the spirit burner with the substance that is being measured and record its mass.
– Place the burner under the copper can and light the wick.
– Stir the water constantly with the thermometer and measure the highest temperature of the water.
– Measure and record the mass of the empty spirit burner.
– Temp. rise = 24.5 degrees
– Mass of ethanol burnt = 0.46g
– Heat given out = 100 X 4.2 X 24.5J = 10.29 kJ
– 0.46g of ethanol produces 10.29 kJ
– 1 mol of ethanol = 46g
– 10.29 X 46 / 0.46 = 1029 kJ
– Molar enthalpy change = ▲H = -1029 kJ / mol
81

Energetics
4.16 use average bond energies to calculate the enthalpy change during a simple chemical reaction.
– Breaking bonds takes in energy. (endothermic / positive)
–
Making bonds gives out energy. (exothermic / negative)
– e.g. H-H = 436, Cl-Cl = 242, H-Cl = 432
– In the reaction H
2
+ Cl
– Breaking H-H = +436
2
– Breaking Cl-Cl = +242
= 2HCl:
– Breaking bonds total = +678
– Forming 2 X H-Cl = 2 X -432 = -864
– ▲H = +678 – 864 = -186kJ / mol
82

Rates of reaction
4.17 describe experiments to investigate the effects of changes in surface area of a solid, concentration of solutions, temperature and the use of a catalyst on the rate of a reaction
– e.g. In the laboratory preparation of CO
2
:
– CaCO
3
(s) + 2HCl(aq)

CaCl
2
(aq) + H
2
O(l) + CO
2
(g)
– The size of the marble chips (surface area).
– Concentration of HCl.
– Temp.
– e.g. In the laboratory preparation of O
2
:
– 2H
2
O
2
(aq)  2H
2
O(l) + O
2
(g)
– Use of manganese(IV) oxide as a catalyst.
83

Rates of reaction
4.18 describe the effects of changes in surface area of a solid, concentration of solutions, pressure of gases, temperature and the use of a catalyst on the rate of a reaction
– Increase concentration: Increase rate.
–
Increase temperature: Increase rate.
– Increase surface area: Increase rate.
–
Catalysts: Increase rate.
84

Rate of reaction
4.19 understand the term activation energy and represent it on a reaction profile
– With catalyst
–
Without catalyst
Activation Energy
Reactants
Products
85

Rate of reaction
4.20 explain the effects of changes in surface area of a solid, concentration of solutions, pressure of gases and temperature on the rate of a reaction in terms of particle collision theory
–
Concentration: More particles in a given volume. Particles collide more often, more successful collisions per second.
– Temperature: Particles have more energy. More collisions have required activation energy, more successful collisions per second. Particles move faster, particles collide more often.
– Surface area: More particles in contact with other reactant. Particles collide more often, more successful collisions per second.
4.21 explain that a catalyst speeds up a reaction by providing an alternative pathway with lower activation energy.
–
Catalysts provide an alternate pathway for the reaction. This pathway has a lower activation energy. This means that although there are the same number of collisions, there are more collisions that have enough activation energy.
86

Equilibria
4.22 understand that some reactions are reversible and are indicated by the symbol ⇌ in equations
– The symbol ⇌ indicates that the reaction occurs in both directions. Such reactions are said to be reversible.
4.23 describe reversible reactions such as the dehydration of hydrated copper(II) sulfate and the effect of heat on ammonium chloride
– When hydrated copper(II) sulfate crystals are heated, it turns from blue to white because the water of crystallization is evaporated.
– CuSO
4
.5H
2
O(s)

CuSO
4
(s) + 5H
2
O
– When water is added to anhydrous copper(II) sulfate, it turns from white to blue as it turns back into hydrated copper(II) sulfate.’
– CuSO
4
(s) + 5H
2
O

CuSO
4
.5H
2
O(s)
– These two reactions can be represented by a single equation:
– CuSO
4
.5H
2
O(s) ⇌ CuSO
4
(s) + 5H
2
O
87

Equilibria
4.24 understand the concept of dynamic equilibrium
– If a reversible reaction is carried out in a closed reaction container, then it is possible for the reaction to reach a position of dynamic equilibrium .
– e.g. 3H
2
(g) + N
2
(g) ⇌ 2NH
3
– At the beginning, when hydrogen and nitrogen are in high concentrations, the rate of the forward reaction is fast.
– However, as N
2 and H
2 concentration decreases, the rate of reaction decreases.
– Meanwhile, as the concentration of NH
3 increases, the rate increases.
– This occurs until the rate of forward reaction and rate of backward reaction becomes equal. Thus the reactions reach a dynamic equilibrium.
– The reaction looks to have stopped, but it is just that the rates are the same.
– Amount of reactants remain the same.
– Any property, such as color or pressure, which depends on the amounts of reactions and products that are present remains constant.
88

Equilibria
4.25 predict the effects of changing the pressure and temperature on the equilibrium position in reversible reactions.
– e.g. for 3H
2
(g) + N
2
(g) ⇌ 2NH
3
– If the equilibrium is altered so that more ammonia is produced, and less hydrogen and nitrogen, the equilibrium shifts to the right.
– If the equilibrium is altered so that less ammonia is produced, and more hydrogen and nitrogen, the equilibrium shifts to the left.
– Increase in pressure: Equilibrium to the right, more ammonia is produced.
– Smaller number of molecules of gas.
– Decrease in pressure: Equilibrium to the left, less ammonia is produced.
– Greater number of molecules of gas.
– Increase in temperature: Equilibrium to the left, less ammonia is produced.
– Shifts towards the endothermic (backward reaction) direction.
– Decrease in temperature: Equilibrium to the right, more ammonia produced
– Shifts towards the exothermic (forward reaction) direction.
89

Chemistry in industry
–
Extraction and uses of metals
– Crude oil
–
Synthetic polymers
– The industrial manufacture of chemicals
90

Extraction and uses of metals
5.1 explain how the methods of extraction of the metals in this section are related to their positions in the reactivity series
– The more reactive the element, the more stable their compounds.
–
All metals above carbon have to be extracted through electrolysis of the molten chloride or oxide. This is because it cannot use carbon as a reducing agent.
–
All metals below carbon can use heat with a reducing agent such as carbon or carbon monoxide.
– Silver and gold occur naturally.
5.2 describe and explain the extraction of aluminum from purified aluminum oxide by electrolysis, including: i the use of molten cryolite as a solvent and to decrease the required operating temperature
– Purified aluminum oxide (after purification from bauxite) has a very high melting point and hence it is dissolved in molten cryolite to make the electrolyte. This mixture has a much lower melting point and is also a better conduct of electricity than molten aluminum oxide.
91

Extraction and uses of metals ii the need to replace the positive electrodes
– Positive electrodes need to be replaced because the oxygen produced at the anode reacts with the carbon used as the material. The anode gradually burns away and needs to be replaced.
iii the cost of the electricity as a major factor
–
Electricity is required in high amounts in the extraction of aluminum oxide, thus molten cryolite is needed to lower the melting temperature to reduce operating costs.
5.3 write ionic half-equations for the reactions at the electrodes in aluminum extraction
– At the cathode: Al 3+ + 3e  Al
–
At the anode: 2O 2
O
2
+ 4e -
– The O
2 reacts with the carbon in anode to form CO
2
:
– C(s) + O
2
(g)

CO
2
(g)
92

Extraction and uses of metals
5.4 describe and explain the main reactions involved in the extraction of iron from iron ore
(hematite), using coke, limestone and air in a blast furnace
– The raw materials are iron ore (hematite), coke (carbon), limestone (calcium carbonate) and air .
– The materials are mixed together and fed into the top of the blast furnace.
–
Hot air is blasted into the bottom of the furnace.
– Oxygen in the air reacts with the coke to form carbon dioxide:
– C(s) + O
2
(g)  CO
2
(g)
– Carbon dioxide reacts with coke to form carbon monoxide:
– CO
2
(g) + C(s)  2CO(g)
– Carbon monoxide reduces the iron(III)oxide in the iron ore, which melts:
– Fe
2
O
3
(s) + 3CO(g)  2Fe(l) + 3CO
2
(g)
– The molten iron collects at the bottom and is tapped off.
– Calcium carbonate in the limestone decomposes to form calcium oxide:
– CaCO
3
(s)  CaO(s) + CO
2
(g)
93

Extraction and uses of metals
– The calcium oxide reacts with silicon dioxide, an impurity in the ore to form calcium silicate, which melts:
– CaO(s) + SiO
2
(s)  CaSiO
3
(l)
– The molten slag collects on top of the molten iron, and is tapped off.
5.5 explain the uses of aluminum and iron, in terms of their properties.
– Aluminum can be used for:
– Aeroplane bodies: high strength to weight ratio
– Overhead power cables: good conductor of electricity
– Saucepans: Good conductor of heat
– Food cans: Non-toxic
– Windows frames: Corrosion-resistant
–
Iron is used in car bodies, iron nails, ships and bridges because of its strength.
94

Crude oil
5.7 describe and explain how the industrial process of fractional distillation separates crude oil into fractions
– Crude oil is refined through fractional distillation, this is done in a fractionating column.
– Crude oil is split into various fractions , each of these fractions have different boiling points.
– Crude oil is heated to vapor and fed into the bottom of a fractionating column.
– The hydrocarbons with very high boiling points (bitumen, fuel oil) immediately turn back into liquids and are tapped off at the bottom of the column.
– The hydrocarbons that have higher boiling points lower than 400 degrees remains as gases and rise up the column, as they rise they cool down.
– The hydrocarbons condense at different heights and are tapped off individually.
– The fraction with the lowest boiling point remains as a gas.
95

Refinery Gases
40 degrees
Crude oil
Gasoline
Kerosene
Diesel oil
Fuel oil 400 degrees
Bitumen
96

Crude oil
5.8 recall the names and uses of the main fractions obtained from crude oil: refinery gases, gasoline, kerosene, diesel, fuel oil and bitumen
– Refinery gases: bottled gas for camping
–
Gasoline: petrol for cars
– Kerosene: fuel for airplanes, paraffin for small heaters and lamps
–
Diesel oil: diesel fuel for buses, lorries, trains and cars.
– Fuel oil: fuel for ships and industrial heating
–
Bitumen: road surfaces
5.6 understand that crude oil is a mixture of hydrocarbons
–
Crude oil is a thick, sticky black liquid that is found under the ground and sea, it is a mixture of hydrocarbons.
97

Crude oil
5.9 describe the trend in boiling point and viscosity of the main fractions
– As boiling point increases, viscosity increases as well.
– Refinery gases have the lowest boiling point, and the lowest viscosity, which means it flows well.
– Bitumen gases have the highest boiling point, and the highest viscosity, which means it is sticky and cannot flow well.
5.10 understand that incomplete combustion of fuels may produce carbon monoxide and explain that carbon monoxide is poisonous because it reduces the capacity of the blood to carry oxygen
– Incomplete combustion of hydrocarbons produces carbon monoxide, water and soot.
Carbon monoxide combines with the hemoglobin in the red blood cells, hindering its capability to carry oxygen.
98

Crude oil
5.11 understand that, in car engines, the temperature reached is high enough to allow nitrogen and oxygen from air to react, forming nitrogen oxides
–
In high temperatures such as car engines, the nitrogen and oxygen from the air react, forming nitrogen oxides.
5.12 understand that nitrogen oxides and sulfur dioxide are pollutant gases which contribute to acid rain, and describe the problems caused by acid rain
– Nitrogen dioxide dissolves in the water to become  nitric acid
–
Sulfur dioxide dissolves in the water to become
 sulfuric acid
– The problems caused by acid rain are:
– Stunted and sickly tree growth
– Release of heavy metals into the food chain
– Lakes are acidic and they can no longer support aquatic life
99

Crude oil
5.13 understand that fractional distillation of crude oil produces more long-chain hydrocarbons than can be used directly and fewer short-chain hydrocarbons than required and explain why this makes cracking necessary
–
The amount of long-chain hydrocarbons (high boiling point) is more than necessary, and a lot of shorter-chain hydrocarbons such as gasoline is needed. Therefore cracking is necessary to convert long-chain into short chain.
5.14 describe how long-chain alkanes are converted to alkenes and shorter-chain alkanes by catalytic cracking, using silica or alumina as the catalyst and a temperature in the range of 600 –700 degrees.
–
The long-chain hydrocarbon molecules are passed over a catalyst (silica or aluminum oxide) and heated to about 600-700 degrees. The long-chain hydrocarbons break down into short-chain alkane molecules and at least one alkene molecule.
– C
10
H
22
(g)  C
8
H
18
(g) + C
2
H
4
– C
8
H
18
– C
2
H
4
(octane) is used to make petrol.
is the alkene that can be used to make poly(ethene).
100

Synthetic polymers
5.15 understand that an addition polymer is formed by joining up many small molecules called monomers
– An addition polymer is a long-chain molecule that has been formed when many small molecules (monomers) have been joined together.
– The process of making this is called addition polymerization.
– The monomer molecule contains a carbon-carbon double bond.
– One of the bonds breaks during addition polymerization and this allows the monomer molecules to join together to form a long chain of carbon atoms.
– For example, when ethene undergoes addition polymerization it forms poly(ethene):
– CH
2
= CH
2
 -(-CH
2
-CH
2
-) n
-
– The letter n represents how many times this structure repeats itself in the molecule.
101

Synthetic polymers
5.16 draw the repeat unit of addition polymers, including poly(ethene), poly(propene) and poly(chloroethene)
H H CH
3
H H Cl
C C C C C C
H H poly(ethene)
H H poly(propene)
H H poly(chloroethene)
102

Synthetic polymers
5.17 deduce the structure of a monomer from the repeat unit of an addition polymer
Ethene Propene Ethene
103

Synthetic polymers
5.18 describe some uses for polymers, including poly(ethene), poly(propene) and poly(chloroethene)
– Low density poly(ethene) is used for plastic bags. High density poly(ethene) is used for plastic bottles.
– Poly(propene) is used to make ropes and crates.
–
Poly(chloroethene) or PVC is strong and rigid, used in pipes and electric insulation.
5.19 explain that addition polymers are hard to dispose of as their inertness means that they do not easily biodegrade
– Their inertness means that they do not easily react with other substances. This means that they are hard to corrode.
104

Synthetic polymers
5.20 understand that some polymers, such as nylon, form by a different process called condensation polymerization
– Not all polymers are made with addition polymerization, some are formed by a different process called condensation polymerization.
– This is because they are made from monomers which are saturated as they do not have double bonds between the carbon atoms. This means that an addition polymerization is not possible.
5.21 understand that condensation polymerization produces a small molecule, such as water, as well as the polymer.
– Groups of atoms need to be removed from the monomer structures to free up electrons to allow it to connect to other monomers.
–
The atoms which have been removed make a new small molecule such as water or hydrogen chloride.
– This is called condensation polymerization.
105

Synthetic polymers
H H H
H N C N
H diamine monomer
H
H
N
H
C
H
H
N
H
H H water
H
O H O
O C C C
H dicarboxylic acid monomer
O H
O
O
C
H
C
H
O
C O H
106

Synthetic polymers
–
In the end, a section of the nylon polymer would have the following structure:
O H O H H H O H O H H
C C N N C C N C
H
C
H
C
H
C
H
H
N
107

The industrial manufacture of chemicals
5.22 understand that nitrogen from air, and hydrogen from natural gas or the cracking of hydrocarbons, are used in the manufacture of ammonia
– 3H
2
(g) + N
2
(g) ⇌ 2NH
3
(g) ▲H = -98kJ / mol
5.23 describe the manufacture of ammonia by the Haber process, including the essential conditions:
–
The Haber process reaction is reversible and can reach a position of equilibrium.
The equilibrium position can be shifted to the right to achieve a high yield of ammonia by using high pressures (less ammonia molecules) and low temperatures
(exothermic).
i a temperature of about 450 degrees
– The rate of reaction is very slow at low temperatures, so a compromise of 450 degrees is used. This is a compromise between rate and yield.
ii a pressure of about 200 atmospheres
–
Carrying out a reaction at very high pressures is expensive since more energy is used to compress the gases. Therefore a pressure of 200 atmospheres is used.
108

The industrial manufacture of chemicals iii an iron catalyst
– The reaction is still fairly slow at 450 degrees unless a catalyst is present. There are a number of metals that will catalyze this reaction, but the most economical one is iron.
5.24 understand how the cooling of the reaction mixture liquefies the ammonia produced and allows the unused hydrogen and nitrogen to be recirculated
–
Under these conditions about 15% of the hydrogen and nitrogen is converted into ammonia. The mixture leaves the reaction vessel and is then cooled. The ammonia liquefies and is tapped off.
–
The hydrogen and nitrogen remain as gases and are mixed with more hydrogen and nitrogen and passed again into the reaction chamber. It is recycled.
5.25 describe the use of ammonia in the manufacture of nitric acid and fertilizers
– Manufacture of fertilizers
– Manufacture of nitric acid
–
Manufacture of nylon
109

The industrial manufacture of chemicals
5.26 recall the raw materials used in the manufacture of sulfuric acid
– Sulfur
–
Air
– Water
5.27 describe the manufacture of sulfuric acid by the contact process, including the essential conditions:
–
Stage 1: Producing sulfur dioxide
– S + O
2
 SO
2
– Stage 2: Producing sulfur trioxide
– 2SO
2
+ O
2
⇌ 2SO
3
▲H = -196 kJ / mol
– Stage 3: Making sulfuric acid
– H
2
SO
4
(l) + SO
3
(g)

H
2
S
2
O
7
(l)
– H
2
S
2
O
7
(l) + H
2
O(l)  2H
2
SO
4
(aq)
–
Final product is a concentrated solution of sulfuric acid.
110

The industrial manufacture of chemicals
–
The reaction is reversible and can reach a position of equilibrium. The equilibrium position can be shifted to the right to achieve a high yield of ammonia by using high pressure (less sulfur trioxide molecules) and low temperatures (exothermic).
i a temperature of about 450 degrees
– The rate of reaction is very slow at low temperatures, so a compromise of 450 degrees is used. This is a compromise between rate and yield.
ii a pressure of about 2 atmospheres
–
The yield would be increase with high pressure but the yield is already very high
(98%) at a pressure of 2 atmospheres, so using higher pressures would be uneconomical.
iii a vanadium(V) oxide catalyst
– Vanadium(V) oxide is used to speed up the reaction by giving a alternate pathway with lower activation energy. This is because at 450 degrees, the reaction is still not quite slow.
111

The industrial manufacture of chemicals
5.28 describe the use of sulfuric acid in the manufacture of detergents, fertilizers and paints
– Manufacture of detergents
–
Manufacture of fertilizers
– Manufacture of paints
5.29 describe the manufacture of sodium hydroxide and chlorine by the electrolysis of concentrated sodium chloride solution (brine) in a diaphragm cell
–
Sodium hydroxide and chlorine are manufactured by the electrolysis of a concentrated solution of sodium chloride.
– At the anode the chloride ions lose electrons to form chlorine molecules.
– This is because the chloride ions are in high concentration, thus preferably electrolyzed over oxygen.
– At the cathode, water molecules gain electrons to form hydroxide ions and hydrogen molecules.
– The diaphragm is important to keep apart the sodium hydroxide and chlorine.
112

The industrial manufacture of chemicals
5.30 write ionic half-equations for the reactions at the electrodes in the diaphragm cell
– At the cathode: 2H
2
O(l) + 2e  2OH (aq) + H
2
(g)
–
At the anode: 2Cl (aq)

Cl
2
(g) + 2e -
– Overall equation: 2NaCl(aq) + 2H
2
O(l)  NaOH(aq) + H
2
(g) + Cl
2
(g)
5.31 describe important uses of sodium hydroxide, including the manufacture of bleach, paper and soap; and of chlorine, including sterilizing water supplies and in the manufacture of bleach and hydrochloric acid.
– Sodium hydroxide
– Manufacture of bleach (If sodium hydroxide and chlorine mix)
– Manufacture of paper
– Manufacture of soap
– Manufacture of detergents
113

The industrial manufacture of chemicals
–
Chlorine
– Sterilizing water supplies
– Manufacture of bleach
– Manufacture of hydrochloric acid
– Manufacture of PVC
114