SpecPoint

3:07 (Triple only) use bond energies to calculate the enthalpy change during a chemical reaction

Each type of chemical bond has a particular bond energy. The bond energy can vary slightly depending what compound the bond is in, therefore average bond energies are used to calculate the change in heat (enthalpy change, ΔH) of a reaction.

Example: dehydration of ethanol

Note: bond energy tables will always be given in the exam, e.g:

BondAverage bond energy in kJ/mol
H-C412
C-C348
O-H463
C-O360
C=C612

So the enthalpy change in this example can be calculated as follows:

Breaking bondsMaking bonds
BondsEnergy (kJ/mol)BondsEnergy (kJ/mol)
H-C x 5(412 x 5) = 2060C-H x 4(412 x 4) = 1648
C-C348C=C612
C-O360O-H x 2(463 x 2) = 926
O-H463
Energy needed to break all the bonds3231Energy released to make all the new bonds3186

Enthalpy change, ΔH = Energy needed to break all the bonds - Energy released to make all the new bonds

ΔH = 3231 – 3186 = +45 kJ/mol (ΔH is positive so the reaction is endothermic)

3:08 practical: investigate temperature changes accompanying some of the following types of change: salts dissolving in water, neutralisation reactions, displacement reactions and combustion reactions

Calorimetry allows for the measurement of the amount of energy transferred in a chemical reaction to be calculated.

 

EXPERIMENT1: Displacement, dissolving and neutralisation reactions

Example: magnesium displacing copper from copper(II) sulfate

Method:

  1. 50 cm3 of copper(II) sulfate is measured and transferred into a polystyrene cup.
  2. The initial temperature of the copper sulfate solution is measured and recorded.
  3. Magnesium is added and the maximum temperature is measured and recorded.
  4. The temperature rise is then calculated. For example:
Initial temp. of solution (oC)Maximium temp. of solution (oC)Temperature rise (oC)
24.256.732.5

Note:  mass of 50 cm3 of solution is 50 g

 

EXPERIMENT2: Combustion reactions

To measure the amount of energy produced when a fuel is burnt, the fuel is burnt and the flame is used to heat up some water in a copper container

Example: ethanol is burnt in a small spirit burner

Method:

  1. The initial mass of the ethanol and spirit burner is measured and recorded.
  2. 100cm3 of water is transferred into a copper container and the initial temperature is measured and recorded.
  3. The burner is placed under of copper container and then lit.
  4. The water is stirred constantly with the thermometer until the temperature rises by, say, 30 oC
  5. The flame is extinguished and the maximum temperature of the water is measured and recorded.
  6. The burner and the remaining ethanol is reweighed. For example:
Mass of water (g)Initial temp of water (oC)Maximum temp of water (oC)Temperature rise (oC)Initial mass of spirit burner + ethanol (g)Final mass of spirit burner + ethanol (g)Mass of ethanol burnt (g)
10024.254.230.034.4633.680.78

The amount of energy produced per gram of ethanol burnt can also be calculated:

3:09 describe experiments to investigate the effects of changes in surface area of a solid, concentration of a solution, temperature and the use of a catalyst on the rate of a reaction

The rate of a chemical reaction can be measured either by how quickly reactants are used up or how quickly the products are formed.

The rate of reaction can be calculated using the following equation:

The units for rate of reaction will usually be grams per min (g/min)

 

An investigation of the reaction between marble chips and hydrochloric acid:

Marble chips, calcium carbonate (CaCO3) react with hydrochloric acid (HCl) to produce carbon dioxide gas. Calcium chloride solution is also formed.

Using the apparatus shown the change in mass of carbon dioxide can be measure with time.

As the marble chips react with the acid, carbon dioxide is given off.

The purpose of the cotton wool is to allow carbon dioxide to escape, but to stop any acid from spraying out.

The mass of carbon dioxide lost is measured at intervals, and a graph is plotted:

 

Experiment to investigate the effects of changes in surface area of solid on the rate of a reaction:

The experiment is repeated using the same mass of chips, but this time the chips are larger, i.e. have a smaller surface area.

Since the surface area is smaller, the rate of reaction is less.

Both sets of results are plotted on the same graph.

If instead the chips were smashed into powder (and again same mass of chips used) the surface area would be much larger and so the rate of reaction higher (steeper line on graph).

 

Experiment to investigate the effects of changes in concentration of solutions on the rate of a reaction:

The experiment is again repeated using the exact same quantities of everything but this time with half the concentration of acid. The marble chips must however be in excess. The reaction with the half the concentration of acid happens slower and produces half the amount of carbon dioxide.

 

Experiment to investigate the effects of changes in temperature on the rate of a reaction:

The experiment is once again repeated using the exact same quantities of everything but this time at a higher temperature. The reaction with the higher temperature happens faster.

 

Experiment to investigate the effects of the use of a catalyst on the rate of a reaction:

Hydrogen peroxide naturally decomposes slowly producing water and oxygen gas.

Manganese (IV) oxide can be used as a catalyst to speed up the rate of reaction.

The rate of reaction can be measured by measuring the volume of oxygen produced at regular intervals using a gas syringe.

Both sets of results are plotted on the same graph.

 

 

 

 

 

 

 

Experiment to investigate the reaction between varying concentrations of sodium thiosulfate and hydrochloric acid

Sodium thiosulfate (Na2S2O3) and hydrochloric acid (HCl) are both colourless solutions. They react to form a yellow precipitate of sulfur.

     sodium thiosulfate   +   hydrochloric acid    →     sodium chloride   +   sulfur dioxide   +   sulfur   +   water

    Na2S2O3(aq)         +         2HCl(aq)           →           2NaCl(aq)         +         SO2(g)         +         S(s)         +         H2O(l)

 

To investigate the effects of changes in concentration of sodium thiosulfate on the rate of a reaction, the conical flask is placed above a cross. The reaction mixture is observed from directly above and the time for a cross to disappear is measured. The cross disappears because a precipitate of sulfur is formed.

In order to change the concentration of sodium thiosulfate, the volumes of sodium thiosulfate and water are varied (see results table). However the total volume of solution must always be kept the same as to ensure that the depth of the solution remains constant.

In this reaction, sulfur dioxide gas (SO2), which is poisonous is produced therefore the experiment must be carried out in a well ventilated room.

The results are recorded in the table below and then plotted onto a graph.

Volume of Na2S2O3(aq) (cm3)Volume of water (cm3)Concentration of Na2S2O3(aq) (mol/dm3)Time taken for cross to disappear (s)Rate of reaction (s-1) (1/time)
5000.10450.0222
40100.08600.0167
30200.06800.0125
20300.04130.0769
10400.022550.0039

The graph shows that the rate of reaction is directly proportional to the concentration.

The experiment can also be repeated to show how temperature affects the rate of reaction.

In this experiment the concentration of sodium thiosulfate is kept constant but heated to range of different temperatures.

As a rough approximation, the rate of reaction doubles for every 10oC temperature rise.

 

3:10 describe the effects of changes in surface area of a solid, concentration of a solution, pressure of a gas, temperature and the use of a catalyst on the rate of a reaction

Increasing the surface area of a solid increases the rate of a reaction.

Increasing the concentration of a solution increases the rate of a reaction.

Increasing the pressure of a gas increases the rate of a reaction.

Increasing the temperature increases the rate of a reaction.

Using a catalyst increases the rate of a reaction.

3:11 explain the effects of changes in surface area of a solid, concentration of a solution, pressure of a gas and temperature on the rate of a reaction in terms of particle collision theory

Increasing the surface area of a solid:

  • more particles exposed
  • more frequent collisions
  • increase the rate of a reaction

 

Increasing the concentration of a solution or pressure of a gas:

  • more particles in same space
  • more frequent collisions
  • increase rate of reaction

 

Increasing the temperature:

  • particles have more kinetic energy
  • more frequent collisions
  • and a higher proportion of those collisions are successful because the collision energy is greater or equal to the activation energy
  • increase rate of reaction

 

3:12 know that a catalyst is a substance that increases the rate of a reaction, but is chemically unchanged at the end of the reaction

A catalyst is a substance that increases the rate of a reaction, but is chemically unchanged at the end of the reaction.

3:13 know that a catalyst works by providing an alternative pathway with lower activation energy

Catalyst: A substance that speeds up a chemical reaction while remaining chemically unchanged at the end of
the reaction.

A catalyst is not used up in a reaction.

A catalyst speeds up a reaction by providing an alternative pathway with lower activation energy.

3:14 (Triple only) draw and explain reaction profile diagrams showing ΔH and activation energy

Below is a diagram showing the reaction profile for the reaction of hydrogen with oxygen, which is EXOTHERMIC:

The activation energy is the minimum amount of energy required to start the reaction.

For an exothermic reaction, the products have less energy than the reactants. The difference between these energy levels is ΔH.

For an exothermic reaction, more energy is released when bonds are formed than taken in when bonds are broken.

 

Below is a diagram showing the reaction profile for the thermal decomposition of calcium carbonate, which is ENDOTHERMIC:

The activation energy is the minimum amount of energy required to start the reaction.

For an endothermic reaction, the products have more energy than the reactants. The difference between these energy levels is ΔH.

For an endothermic reaction, more energy is taken in to break bonds than is released when new bonds are formed.

3:15 practical: investigate the effect of changing the surface area of marble chips and of changing the concentration of hydrochloric acid on the rate of reaction between marble chips and dilute hydrochloric acid

The rate of a chemical reaction can be measured either by how quickly reactants are used up or how quickly the products are formed.

The rate of reaction can be calculated using the following equation:

The units for rate of reaction will usually be grams per min (g/min)

 

An investigation of the reaction between marble chips and hydrochloric acid:

Marble chips, calcium carbonate (CaCO3) react with hydrochloric acid (HCl) to produce carbon dioxide gas. Calcium chloride solution is also formed.

Using the apparatus shown the change in mass of carbon dioxide can be measure with time.

As the marble chips react with the acid, carbon dioxide is given off.

The purpose of the cotton wool is to allow carbon dioxide to escape, but to stop any acid from spraying out.

The mass of carbon dioxide lost is measured at intervals, and a graph is plotted:

 

Experiment to investigate the effects of changes in surface area of solid on the rate of a reaction:

The experiment is repeated using the exact same quantities of everything but using larger chips. For a given quantity, if the chips are larger then the surface area is lesson. So reaction with the larger chips happens more slowly.

Both sets of results are plotted on the same graph.

 

Experiment to investigate the effects of changes in concentration of solutions on the rate of a reaction:

The experiment is again repeated using the exact same quantities of everything but this time with half the concentration of acid. The marble chips must however be in excess. The reaction with the half the concentration of acid happens slower and produces half the amount of carbon dioxide.

 

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.

 

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

 

4:31 (Triple only) know that ethanol can be oxidised by: burning in air or oxygen (complete combustion), reaction with oxygen in the air to form ethanoic acid (microbial oxidation), heating with potassium dichromate(VI) in dilute sulfuric acid to form ethanoic acid

1) Ethanol can be oxidised by complete combustion. With excess oxygen the complete combustion of ethanol (C₂H₅OH) in air produces carbon dioxide and water:

C₂H₅OH (l)         +         3O₂ (g)         →         2CO₂ (g)         +         3H₂O (l)

 

2) Ethanol can be oxidised in air in the presence of microorganisms (‘microbial oxidation’) to form ethanoic acid (CH₃COOH).

 

3) Ethanol can be oxidised by heating with the oxidising agent potassium dichromate(VI) (K₂Cr₂O₇) in dilute sulfuric acid (H₂SO₄).

In the equation below, [O] means oxygen from an oxidising agent.

CH₃CH₂OH         +         2[O]         →         CH₃COOH         +         H₂O

This mixture starts orange but when the reaction happens turns green which indicates the presence of Cr³⁺ ions which are formed when the potassium dichromate(VI) is reduced.

4:32 (Triple only) know that ethanol can be manufactured by: 1) reacting ethene with steam in the presence of a phosphoric acid catalyst at a temperature of about 300⁰C and a pressure of about 60–70atm; and 2) the fermentation of glucose, in the absence of air, at an optimum temperature of about 30⁰C and using the enzymes in yeast

In the hydration of ethene, ethanol is made by passing ethene and steam over a catalyst.

Water is added to ethene, this is known as hydration.

Conditions

Catalyst: Phosphoric acid (H3PO4)

Temperature: 300°C

Pressure: 60 atm

 

 

Fermentation is the conversion of sugar, e.g. glucose into ethanol by enzymes from yeast.

 

Conditions

Catalyst: Zymase (enzyme found in yeast)

Temperature: 30°C  –  The process is carried out at low temperatures as not to denature the enzymes.

Other: Anaerobic (no oxygen present) – if oxygen were present, the yeast produce carbon dioxide and water instead of ethanol.

 

Select a set of flashcards to study:

     Terminology

     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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