Chemistry

3:16 practical: investigate the effect of different solids on the catalytic decomposition of hydrogen peroxide solution

Oxygen (O2) is made in the lab from hydrogen peroxide (H2O2) using manganese(IV) oxide (MnO2) as a catalyst.

 

 

Different catalysts could be used to investigate which is the most effective in decomposing hydrogen peroxide. Examples of other substances which could be tested are:

  • Manganese dioxide
  • Liver
  • Potato
  • Potassium iodide
  • Copper oxide
  • Sodium chloride

Only some of these are effective catalysts when used with hydrogen peroxide. If a substance is not a catalyst, there will be no bubbles of oxygen produced. For other substances, such as liver which is a very effective catalyst in the decomposition of hydrogen peroxide, bubbles of oxygen will be produced quickly.

 

 

 

3:18 describe reversible reactions such as the dehydration of hydrated copper(II) sulfate and the effect of heat on ammonium chloride

Dehydration of copper(II) sulfate

 

Heating ammonium chloride

On heating, white solid ammonium chloride decomposes forming ammonia and hydrogen chloride gas. On cooling, ammonia and hydrogen chloride react to form a white solid of ammonium chloride:

 

3:20 (Triple only) know that the characteristics of a reaction at dynamic equilibrium are: the forward and reverse reactions occur at the same rate, and the concentrations of reactants and products remain constant

Features of a reaction mixture that is in dynamic equilibrium:

  1.   the concentrations of reactants and products remain constant
  2.   rate of forward reaction = rate of backward reaction

3:21 (Triple only) understand why a catalyst does not affect the position of equilibrium in a reversible reaction

A catalyst is a substance which increases the rate of reaction without being chemically changed at the end of the reaction.

A reversible reaction is one where the forward reaction and the backward reaction happen simultaneously. For example:

3H₂ + N₂ ⇋ 2NH₃

In such a reaction a catalyst speeds up both the forward and the backward reactions. Hence, although the system will reach dynamic equilibrium more quickly, the addition of a catalyst will not affect the position of equilibrium.

3:22 (Triple only) predict, with reasons, the effect of changing either pressure or temperature on the position of equilibrium in a reversible reaction (references to Le Chatelier’s principle are not required)

In a reversible reaction the position of the equilibrium (the relative amounts of reactants and products) is dependent on the temperature and pressure of the reactants.

If the conditions of an equilibrium reaction are changed, the reaction moves to counteract that change.

Therefore by altering the temperature or pressure the position of the equilibrium will change to give more or less products.

Adding a catalyst does not affect the position of the equilibrium.

 

If a change in conditions moves equilibrium to the right, the yield of the substances on the right is increased.

 

Changing the temperature:

All reactions are exothermic in one direction and endothermic in the other way.

For this reaction the enthalpy change, ΔH is negative therefore the forward reaction is exothermic:

     CO(g)     +             2H2(g)                    ⇌            CH3OH(g)              ΔH = –91 kJ/mol

If temperature is decreased the position of the equilibrium will shift to the right because it is an exothermic reaction.

For this reaction the enthalpy change, ΔH is positive therefore the forward reaction is endothermic:

     CH4(g)                     +              H2O(g)                    ⇌            CO(g)      +              3H2(g)                     ΔH = +210 kJ mol–1

If temperature is increased the position of the equilibrium will shift to the right because it is an endothermic reaction.

Key point: an increase (or decrease) in temperature shifts the position of equilibrium in the direction of the endothermic (or exothermic) reaction

 

Changing the pressure:

Reactions may have more molecules of gas on one side than on the other.

For this reaction there are 2 molecules on the left and 4 molecules on the right:

     CH4(g)                     +              H2O(g)                    ⇌            CO(g)      +              3H2(g)                     ΔH = +210 kJ mol–1

If the pressure is increased the position of the equilibrium will shift to the left because there are fewer molecules on the left-hand side.

For this reaction there are 3 molecules on the left and 1 molecule on the right

     CO(g)     +             2H2(g)                    ⇌            CH3OH(g)              ΔH = –91 kJ/mol

If the pressure is decreased the position of the equilibrium will shift to the left because there are more molecules on the left-hand side.

Key point: an increase (or decrease) in pressure shifts the position of equilibrium in the direction that produces fewer (or more) moles of gas

4:02a understand how to represent organic molecules using molecular formulae, general formulae, structural formulae and displayed formulae

The molecular formula shows the actual number of atoms of each element in a molecule.

The general formula shows the relationship between the number of atoms of one element to another within a molecule. Members of a homologous series share the same general formula. The general formula for alkanes is CnH2n+2 and the general formula for alkenes is CnH2n.

A structural formula shows how the atoms in a molecule are joined together.

The displayed formula is a full structural formula which shows all the bonds in a molecule as individual lines.

 

The terms above are demonstrated with the example of butane.

Image result for butane

  • Displayed formula:
  • Molecular formula: C₄H₁₀
  • General formula (alkanes): CnH2n+2
  • Structural formula: CH₃ – CH₂ – CH₂ – CH₃

 

The terms above are demonstrated with the example of ethene, which contains a double bond.

Image result for ethene

  • Displayed formula:
  • Molecular formula: C₂H₄
  • General formula (alkenes): CnH2n
  • Structural formula: CH₂ = CH₂

4:02 understand how to represent organic molecules using empirical formulae, molecular formulae, general formulae, structural formulae and displayed formulae

The molecular formula shows the actual number of atoms of each element in a molecule.

The empirical formula shows the simplest whole number ratio of atoms present in a compound. So the molecular formula is a multiple of the empirical formula.

The general formula shows the relationship between the number of atoms of one element to another within a molecule. Members of a homologous series share the same general formula. The general formula for alkanes is CnH2n+2 and the general formula for alkenes is CnH2n.

A structural formula shows how the atoms in a molecule are joined together.

The displayed formula is a full structural formula which shows all the bonds in a molecule as individual lines.

 

The terms above are demonstrated with the example of butane.

Image result for butane

  • Displayed formula:
  • Molecular formula: C₄H₁₀
  • Empirical formula: C₂H₅
  • General formula (alkanes): CnH2n+2
  • Structural formula: CH₃ – CH₂ – CH₂ – CH₃

 

The terms above are demonstrated with the example of ethene, which contains a double bond.

Image result for ethene

  • Displayed formula:
  • Molecular formula: C₂H₄
  • Empirical formula: CH₂
  • General formula (alkenes): CnH2n
  • Structural formula: CH₂ = CH₂

4:03 know what is meant by the terms homologous series, functional group and isomerism

A functional group is an atom or a group of atoms that determine the chemical properties of a compound.

For example the functional group of an alcohol is the -OH group and that of alkenes is the C=C carbon to carbon double bond.

 

A Homologous series is a group of substances with:

  • the same general formula
  • similar chemical properties because they have the same functional group
  • a trend (graduation) in physical properties

 

 

Isomers are molecules with the same molecular formula but with a different structure.

4:04 understand how to name compounds relevant to this specification using the rules of International Union of Pure and Applied Chemistry (IUPAC) nomenclature. Students will be expected to name compounds containing up to six carbon atoms

The names of organic molecules are based on the number of carbon atoms in the longest chain. This chain is the longest consecutive line of carbon atoms, even if this line bends. 

The name is based on the number of carbon atoms in the longest chain
1 Meth-
2 Eth-
3 Prop-
4 But-
5 Pent-
6 Hex-
7 Hept-
8 Oct-
9 Non-
10 Dec-

Hydrocarbons are molecules which contain only hydrogen and carbon.

 

Naming straight-chain alkanes

The simplest hydrocarbons are alkanes. They contain only single bonds, and have “-ane” in the name.

For example, the displayed formula of ethane (C₂H₆) is:

The name “ethane” contains “eth-” because there are 2 carbon atoms in the longest chain, and the name contains “-ane” because the molecule only has single bonds so is an alkane.

 

Another example is pentane (C₅H₁₂) which has the displayed formula:

The name “pentane” contains “pent-” because there are 5 carbon atoms in the longest chain, and the name contains “-ane” because the molecule only has single bonds, so is an alkane.

 

Remember, it does not matter if the longest consecutive line of carbons bends around. For example the displayed formula below still shows a very normal molecule of pentane (5 carbons in a row). Pentane is not normally drawn with the longest chain of carbons bent around because it could be confusing.

 

You might also see the bonds drawn at angles. Don’t worry, the displayed formula below is still pentane, as can be seen by the fact there are 5 carbon atoms in the longest chain, surrounded by hydrogen atoms bonded to the carbon atoms by single bonds.

Image result for displayed formula pentane

 

A shorter way to express the detailed structure of an organic molecule is the structural formula. The structural formula for pentane is CH₃-CH₂-CH₂-CH₂-CH₃, which tells us the same information about the molecule as does the displayed formula, without the hassle of having to draw all the bonds or all the hydrogen atoms.

 

Naming straight-chain alkenes

Another simple group of hydrocarbons is the alkenes. They contain a carbon-to-carbon double bond, which also means they have two fewer hydrogen atoms than their corresponding alkane. An alkene has “-ene” in its name.

For example, the displayed formula for ethene (C₂H₄) is:

ethene has 2 carbon atoms and 4 hydrogen atoms

 

and the displayed formula of propene (C₃H₆) is:

propene has 3 carbon atoms and 6 hydrogen atoms

 

With longer alkene molecules the double bond might appear in different locations of the carbon chain, so the name needs to be a little bit more complicated to be able to describe these differences clearly. A number is added in the middle of the name to indicate at which carbon the double bond starts.

So the displayed formula of pent-1-ene is:

 

and this is the displayed formula of pent-2-ene:

 

However, take care that when counting which carbon has the double bond. The numbers start from the end that produces the smallest numbers in the name. For example, this is the displayed formula for pent-1-ene again, but just drawn the other way round. It is still pent-1-ene (you can’t get pent-4-ene):

 

Naming straight-chain alcohols

We get the same pattern all over again with the group of organic molecules called alcohols, which are recognised by an -OH functional group. For example here is the displayed formula for ethanol, which has 2 carbon atoms in the longest chain:

Image result for displayed formula ethanol

 

and here is the displayed formula for butanol:

Image result for displayed formula butanol

 

Summary of naming simple straight-chain organic molecules

The following table summarises the naming of some of the straight-chain alkanes, alkenes and alcohols, giving a name and a molecular formula for each:

Carbons in longest chainAlkanesAlkenesAlcohols
1methane, CH₄-methanol, CH₄O
2ethane, C₂H₆ethene, C₂H₄ethanol, C₂H₆O
3propane, C₃H₈propene, C₃H₆propanol, C₃H₈O
4butane, C₄H₁₀butene, C₄H₈butanol, C₄H₁₀O

 

Naming branched alkanes and alkenes

The naming conventions for organic molecules cover more than the straight chain molecules. Branched molecules are named depending on the number of carbon atoms in the branch. A branch with 1 carbon is called “methyl” and a branch with 2 carbons is called “ethyl”. This is similar to the conventions covered above, plus the “-yl-” bit just says it is a branch.

For example, this is the displayed formula for 2-methyl hexane:

In the name 2-methyl hexane, the number 2 indicates that when counting along the longest carbon chain the methyl branch comes off the second carbon atom. The “methyl” bit of the name says there is one branch of 1 carbon. The “hex” bit of the name says the longest consecutive chain of carbon atoms is 6. The “ane” bit says the molecule has only single bonds.

 

When counting along the carbon atoms of the longest chain to work out the name, the numbering of carbon atoms starts from the end nearest to the branch. Another way to put this is that the name is given such that the numbers in the name are as low as possible. For example, here is the displayed formula for 4-ethyl octane:

 

Another example, this time with 2 methyl branches coming off the second and third carbons of the chain, is 2,3-dimethyl hexane. The “di” in the name indicates there are two methyl groups. This is the same way in which “di” indicates there are two oxygen atoms in carbon dioxide.

 

Another example of how the naming convention works for branches is 2,2-dimethyl hexane:

 

 

This naming of branches also applies to alkenes. Here is the displayed formula of 4-methylpent-1-ene:

4:05 understand how to write the possible structural and displayed formulae of an organic molecule given its molecular formula

The molecular formula describes the actual number of each type of each atom in a molecule.

For example, a molecular formula of C₆H₁₂ tells us that in each molecule there are 6 carbon atoms and 12 hydrogen atoms.

However, the molecular formula tells us nothing about how those atoms are arranged. For example, it does not tell us if there any branches of carbon atoms coming off the main carbon-carbon chain, nor how long or how many there might be.

On the other hand, the structural formula and displayed formula of a molecule tell us clearly how the atoms are arranged in that molecule.

This means that if we are given a molecular formula only, there may be several possible structural and displayed formulae all of which could apply for that molecule.

When trying to work out possible structural or displayed formulae from a molecular formula there are several clues:

  • If the molecular formula only has carbon and hydrogen in it, then of course the structural formula will only have atoms of these 2 elements.
  • If the molecular formula has twice as many carbons as hydrogens (CnH2n) then the molecule is an alkene and has a double bond somewhere between 2 of the carbon atoms.
  • If the molecular formula has two more than twice as many carbons as hydrogens (CnH2n+2) then the molecule is an alkane and only has single bonds.
  • If the molecular formula has two more than twice as many carbons as hydrogens and also has an oxygen atom (CnH2n+2O), then the molecule is an alcohol, and somewhere in the displayed formula will be a carbon single-bonded to an oxygen which itself is then single-bonded to a hydrogen.

4:06 understand how to classify reactions of organic compounds as substitution, addition and combustion. Knowledge of reaction mechanisms is not required

In a substitution reaction an atom or group of atoms is replaced by a different atom or group of atoms. For example when ethane reacts with bromine gas one of the hydrogen atoms in ethane is substituted by one of the atoms of bromine from within the bromine molecule:

CH₃-CH₃         +         Br-Br         →         CH₃-CH₂Br         +         H-Br

ethane         +         bromine         →         bromoethane         +         hydrogen bromide

 

 

An addition reaction occurs when an atom or group of atoms is added to a molecule without taking anything away. For example when ethene reacts with bromine gas, the product is simply the addition of the two molecules:

CH₂=CH₂         +         Br-Br         →         CH₂Br-CH₂Br

Image result for ethene + bromine displayed formula

 

 

A combustion reaction is another way to say ‘burning’ and is a reaction with oxygen. Combustion of hydrocarbons with excess oxygen gives the products water and carbon dioxide, and also releases heat energy (exothermic reaction). Two examples the combustion of propane and the combustion of butene:

C₃H₈         +         5O₂         →         3CO₂         +         4H₂O

C₄H₈         +         6O₂         →         4CO₂         +         4H₂O

4:08 describe how the industrial process of fractional distillation separates crude oil into fractions

  • Crude oil is separated by fractional distillation.
  • Crude oil is heated and the oil evaporates.
  • The gas goes into the fractional distillation tower. As the gas rises the temperature falls.
  • Fractions with higher boiling points condense and are collected nearer the bottom of the tower.

 

Fractional distillation – videos

This video does not quite use the right language for the various fractions as appropriate to the Edexcel iGCSE, but it is nevertheless a good description of the process. Make sure you use the notes on tutorMyself.com to get the exact language you will need for your exam.

And another somewhat older video showing the industrial process of fractional distillation:

4:09 know the names and uses of the main fractions obtained from crude oil: refinery gases, gasoline, kerosene, diesel, fuel oil and bitumen

Crude oil is separated into fractions by the process of fractional distillation.

FractionUse
Refinery gasesBottled gas
GasolineFuel for cars
KeroseneFuel for aeroplanes
Diesel OilFuel for lorries
Fuel OilFuel for ships
BitumenRoad Surfacing

4:10 know the trend in colour, boiling point and viscosity of the main fractions

The boiling point increases as the number of carbon atoms (chain length) increases.

The viscosity increases as the number of carbon atoms (chain length) increases.

The greater the number of carbon atoms (chain length), the darker in colour that fraction is.

The viscosity of a fluid describes how easily it flows. Water has a low viscosity, it flows very easily. Crude oil has a higher viscosity than water, it does not flow very easily.

Fractions (in order)Properties
Refinery gasesSmallest molecules. Lowest boiling point. Lowest viscosity. Lightest in colour.
Gasoline
Kerosene
Diesel
Fuel oil
BitumenLargest molecules. Highest boiling point. Highest viscosity. Darkest in colour.

4:12 know the possible products of complete and incomplete combustion of hydrocarbons with oxygen in the air

Complete Combustion happens when there is enough oxygen available, producing carbon dioxide (CO2) and water (H2O)

 

Incomplete Combustion happens when there is not enough oxygen available, with possible products being carbon monoxide (CO), carbon (C, soot), carbon dioxide (CO2) and water (H2O)

4:13 understand why carbon monoxide is poisonous, in terms of its effect on the capacity of blood to transport oxygen references to haemoglobin are not required

Carbon monoxide may be produced from the incomplete combustion of fuels:

Carbon monoxide is poisonous because it reduces the capacity of the blood to carry oxygen.

4:14 know that, in car engines, the temperature reached is high enough to allow nitrogen and oxygen from air to react, forming oxides of nitrogen

When fuels are burned in vehicle engines, high temperatures are reached.

At these high temperatures nitrogen and oxygen from the air react to produce nitrogen oxides:

          nitrogen          +          oxygen        →          nitrogen oxides

eg

          N2 (g)              +          O2 (g)          →          2NO (g)

In the atmosphere these nitrogen oxides can combine with water to produce nitric acid (HNO3).

4:15 explain how the combustion of some impurities in hydrocarbon fuels results in the formation of sulfur dioxide

Fossil fuels such as coal, gas and oil are derived from crude oil.

These fuels are hydrocarbons, but also include impurities such as sulfur.

When the fuels are burned, sulfur dioxide is produced which can escape into the atmosphere:

S (s)         +         O₂ (g)         →         SO₂ (g)

 

4:16 understand how sulfur dioxide and oxides of nitrogen oxides contribute to acid rain

Acids formed in the atmosphere can fall as acid rain. This can be a major problem, killing trees and fish in lakes. The acid rain also corrodes limestone buildings and marble statues since these are both made of calcium carbonate (CaCO₃). Some metals such as iron are also attacked by acid rain.

 

Sulfur dioxide released into the atmosphere from the burning of fossil fuels can react with water and oxygen to make sulfuric acid (H₂SO₄):

2SO₂ (g)         +         2H₂O (l)         +         O₂ (g)         →         2H₂SO₄ (aq)

 

Also, if sulfur dioxide in the atmosphere reacts with just water, a weaker acid called sulfurous acid (H₂SO₃) is formed:

SO₂ (g)         +         H₂O (l)         →         H₂SO₃ (aq)

 

In car engines the temperature is high enough for the nitrogen in the air to react with oxygen to produce oxides of nitrogen, e.g:

N₂ (g)         +         O₂ (g)         →         NO₂ (g)

In the atmosphere these nitrogen oxides can produce nitric acid (HNO₃).

4:17 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 600–700⁰C)

Cracking involves the thermal decomposition of long-chain alkanes into shorter-chain alkanes and alkenes:

Conditions

Temperature: 600oC

Catalyst: aluminium oxide, Al2O3

4:18 explain why cracking is necessary, in terms of the balance between supply and demand for different fractions

Cracking converts long chain hydrocarbons into short chain hydrocarbons.

Long-chain alkanes are broken down into alkanes and alkenes of shorter length.

Crude oil contains a surplus long chains.

Shorter chain hydrocarbons are in greater demand, e.g. petrol.

Cracking also produces alkenes which are used in making polymers and ethanol.

4:20 explain why alkanes are classified as saturated hydrocarbons

 

Saturated:   A molecule containing only single bonds between carbon atoms. For example, alkanes as described as saturated molecules.

Unsaturated:   A molecule containing a carbon-carbon double or triple bond. For example, alkenes as described as unsaturated molecules.

 

4:21 understand how to draw the structural and displayed formulae for alkanes with up to five carbon atoms in the molecule, and to name the unbranched-chain isomers

The displayed formulae show all the atoms and bonds drawn out.

The molecular formulae just show the number of each type of atom in the molecule.

4:22 describe the reactions of alkanes with halogens in the presence of ultraviolet radiation, limited to mono-substitution knowledge of reaction mechanisms is not required

Alkanes react with bromine in the presence of UV light, e.g. sunlight.

A hydrogen atom in the alkane is replaced by a bromine atom.

This is known as substitution.

 

4:23 know that alkenes contain the functional group >C=C<

Alkenes are a homologous series of hydrocarbons which contain a carbon-carbon double bond. This double bond is shown in formulae as a double line.

The names of alkenes end with “ene”.

An example is ethene, the structural formula for which is CH₂ = CH₂

For a molecule with more than two carbon atoms, the position of the double bond within the molecule can vary as indicated by the name and the structural formula.

4:25 explain why alkenes are classified as unsaturated hydrocarbons

Saturated:   A molecule containing only single bonds between carbon atoms. For example, alkanes as described as saturated molecules.

Unsaturated:   A molecule containing a carbon-carbon double or triple bond. For example, alkenes as described as unsaturated molecules.

4:26 understand how to draw the structural and displayed formulae for alkenes with up to four carbon atoms in the molecule, and name the unbranched-chain isomers. Knowledge of cis/trans or E/Z notation is not required

The displayed formulae show all the atoms and bonds drawn out.

The molecular formulae just show the number of each type of atom in the molecule.

4:27 describe the reactions of alkenes with bromine, to produce dibromoalkanes

Alkenes react with bromine water. UV light is not required for this reaction.

The double bond is broken and the bromine atoms are added. This is an addition reaction.

During this reaction there is a colour change from orange to colourless.

For example:

This is how we can test for the presence of an alkene or another type of unsaturated molecule.

 

 

4:28 describe how bromine water can be used to distinguish between an alkane and an alkene

In the absence of UV light an alkane added to bromine water will not react: the bromine water will stay orange.

However, alkenes react with bromine water even without UV light. There will be a colour change of orange to colourless.

 

 

 

4:29 (Triple only) know that alcohols contain the functional group −OH

The member of the homologous series called Alcohols have names which end in “ol”. Examples are methanol, ethanol and propanol.

Alcohols all contain an -OH functional group attached to a hydrocarbon chain.

4:30 (Triple only) understand how to draw structural and displayed formulae for methanol, ethanol, propanol (propan-1-ol only) and butanol (butan-1-ol only), and name each compound, the names propanol and butanol are acceptable

Structural formula and displayed formula for methanol:

CH₃-OH

 

Structural formula and displayed formula for ethanol:

CH₃-CH₂-OH                    (or simply C₂H₅OH)

Image result for ethanol

 

Structural formula and displayed formula for propan-1-ol:

CH₃-CH₂-CH₂-OH

 

Structural formula and displayed formula for butan-1-ol:

CH₃-CH₂-CH₂-CH₂-OH

 

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     Skills and equipment

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Section 1: Principles of chemistry

      a) States of matter

      b) Atoms

      c) Atomic structure

     d) Relative formula masses and molar volumes of gases

     e) Chemical formulae and chemical equations

     f) Ionic compounds

     g) Covalent substances

     h) Metallic crystals

     i) Electrolysis

 Section 2: Chemistry of the elements

     a) The Periodic Table

     b) Group 1 elements: lithium, sodium and potassium

     c) Group 7 elements: chlorine, bromine and iodine

     d) Oxygen and oxides

     e) Hydrogen and water

     f) Reactivity series

     g) Tests for ions and gases

Section 3: Organic chemistry

     a) Introduction

     b) Alkanes

     c) Alkenes

     d) Ethanol

Section 4: Physical chemistry

     a) Acids, alkalis and salts

     b) Energetics

     c) Rates of reaction

     d) Equilibria

Section 5: Chemistry in industry

     a) Extraction and uses of metals

     b) Crude oil

     c) Synthetic polymers

     d) The industrial manufacture of chemicals

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