GCSE_Chemistry_SpecPoint

2:36 understand that an acid is a proton donor and a base is a proton acceptor

An acid is a proton donor.

A base is a proton acceptor.

 

A proton is the same as a hydrogen ion. A good way to think about that is to realise that a hydrogen atom is just one proton and zero neutrons surrounded by only one electron. If that atom becomes an ion by the removal of the electron, then only one proton is left.

2:37 describe the reactions of hydrochloric acid, sulfuric acid and nitric acid with metals, bases and metal carbonates (excluding the reactions between nitric acid and metals) to form salts

Acid reactions summary

         alkali      +      acid      →      water      +      salt

         base      +      acid      →      water      +      salt

         carbonate      +      acid      →      water      +      salt      +      carbon dioxide

         metal   +   acid   →   salt   +   hydrogen

To assist remembering this list, many pupils find it useful to remember this horrid looking but very effective mnemonic:

         AAWS

         BAWS

         CAWS CoD

         MASH

Acids are a source of hydrogen ions (H⁺) when in solution. When the hydrogen in an acid is replaced by a metal, the compound is called a salt. The name of the salt depends on the acid used. For example if sulfuric acid is used then a sulfate salt will be formed.

Parent acidFormulaSaltFormula ion
sulfuric acidH2SO4sulfateSO42-
hydrochloric acidHClchlorideCl-
nitric acidHNO3nitrateNO3-

 

Acid + Alkali   and   Acid + Base

A base is a substance that can neutralise an acid, forming a salt and water only.

Alkalis are soluble bases. When they react with acids, a salt and water is formed. The salt formed is often as a colourless solution. Alkalis are a source of hydroxide ions (OH⁻) when in solution.

         alkali      +      acid      →      water      +      salt

         base      +      acid      →      water      +      salt

Examples of acid + alkali reactions:

  •          sodium hydroxide   +   hydrochloric acid   →   sodium chloride   +   water
  •          NaOH (aq)         +         HCl (aq)         →         NaCl (aq)         +         H₂O (l)
  •          potassium hydroxide   +   sulfuric acid   →   potassium sulfate   +   water
  •          2KOH (aq)         +         H₂SO₄ (aq)         →         K₂SO₄ (aq)         +         2H₂O (l)

Example of an acid + base reaction:

         CuO (s)         +         H₂SO₄ (aq)         →         CuSO₄ (aq)         +         H₂O (l)

 

Acid + Carbonate

         carbonate      +      acid      →      water      +      salt      +      carbon dioxide

A carbonate is a compound made up of metal ions and carbonate ions. Examples of metal carbonates are sodium carbonate, copper carbonate and magnesium carbonate.

When carbonates react with acids, bubbling is observed which is the carbon dioxide being produced. If the acid is in excess the carbonate will disappear.

Examples of acid + carbonate reactions:

  •          calcium carbonate   +   hydrochloric acid   →   calcium chloride   +   water   +   carbon dioxide
  •          CaCO₃ (s)         +         2HCl (aq)         →         CaCl₂ (aq)         +         H₂O (l)         +         CO₂ (g)
  •          potassium carbonate   +   hydrochloric acid   →   potassium chloride   +   water   +   carbon dioxide
  •          K₂CO₃ (aq)         +         2HCl (aq)         →         2KCl (aq)         +         H₂O (l)         +         CO₂ (g)

 

Acid + Metal

         metal   +   acid   →   salt   +   hydrogen

Metals will react with an acid if the metal is above hydrogen in the reactivity series.

When metals react with acids, bubbling is observed which is the hydrogen being produced. If the acid is in excess the metal will disappear.

Examples of acid + metal reactions:

  •          magnesium   +   sulfuric acid   →   magnesium sulfate   +   hydrogen
  •          Mg (s)         +         H₂SO₄ (aq)         →         MgSO₄ (aq)         +         H₂ (g)
  •          aluminium   +   hydrochloric acid   →   aluminium chloride   +   hydrogen
  •          2Al (s)         +         6HCl (aq)         →         2AlCl₃ (aq)         +         3H₂ (g)
  •          copper   +   hydrochloric acid   →   no reaction (since copper is below hydrogen in the reactivity series)

2:38 know that metal oxides, metal hydroxides and ammonia can act as bases, and that alkalis are bases that are soluble in water

A base is a substance that neutralises an acid by combining with the hydrogen ions in them to produce water.

A base usually means a metal oxide, a metal hydroxide or ammonia.

Alkalis are bases which are soluble in water.

 

Some metal oxides are soluble in water and react with it to form solutions of metal hydroxides. For example:

Na₂O (s)         +         H₂O (l)         →         2NaOH (aq)

Apart from this and other group 1 oxides (such as potassium oxide) most other metal oxides are not soluble in water.

One exception is calcium oxide which does dissolve slightly in water to form calcium hydroxide (known as limewater):

CaO (s)         +         H₂O (l)         →         Ca(OH)₂ (aq)

 

Ammonia is another base. Ammonia reacts with water to form ammonium ions and hydroxide ions:

NH₃ (aq)         +         H₂O (l)         ⇋         NH₄⁺ (aq)         +         OH⁻ (aq)

 

All the solutions produced here contain hydroxide ions (OH⁻) so they are all alkalis.

 

 

2:39 describe an experiment to prepare a pure, dry sample of a soluble salt, starting from an insoluble reactant

Excess Solid Method:

Preparing pure dry crystals of copper sulfate (CuSO4) from copper oxide (CuO) and sulfuric acid (H2SO4)

StepExplanation
Heat acid (H2SO4) in a beakerSpeeds up the rate of reaction
Add base (CuO) until in excess (no more copper oxide dissolves) and stir with glass rodNeutralises all the acid
Filter the mixture using filter paper and funnelRemoves any excess copper oxide
Gently heat the filtered solution (CuSO4)To evaporate some of the water
until crystals form on a glass rodShows a hot saturated solution formed
Allow the solution to cool so that hydrated crystals formCopper sulfate less soluble in cold solution
Remove the crystals by filtrationRemoves crystals
Dry by leaving in a warm placeEvaporates the water

 

2:40 (Triple only) describe an experiment to prepare a pure, dry sample of a soluble salt, starting from an acid and alkali

Titration Method:

Preparing pure dry crystals of sodium chloride (NaCl) from hydrochloric acid (HCl) and sodium hydroxide (NaOH)

Before the salt preparation is carried out using the below method, the volume of acid that exactly reacts with 25cm3 of the alkali is found by titration using methyl orange indicator.

StepExplanation
Pipette 25cm3 of alkali (NaOH) into a conical flaskAccurately measures the alkali (NaOH)
Do not add indicatorPrevents contamination of the pure crystals with indicator
Using the titration values, titrate the known volume acid (HCl) into conical flask containing alkaliExactly neutralises all of the alkali (NaOH)
Transfer to an evaporating basin & heat the solutionForms a hot saturated solution (NaCl(aq))
Allow the solution to cool so that hydrated crystals formSodium chloride is less soluble in cold water
Remove the crystals by filtration and wash with distilled waterRemoves any impurities
Dry by leaving in a warm placeEvaporates the water

(Note – This process could be reversed with the acid in the pipette and the alkali in the burette)

How to select the right method for preparing a salt:

2:41 (Triple only) describe an experiment to prepare a pure, dry sample of an insoluble salt, starting from two soluble reactants

Precipitation Method:

Preparing pure dry crystals of silver chloride (AgCl) from silver nitrate solution (AgNO3) and potassium chloride solution (KCl)

StepExplanation
Mix the two salt solutions together in a beakerForms a precipitate of an insoluble salt (AgCl)
Stir with glass rodMake sure all reactants have reacted
Filter using filter paper and funnelCollect the precipitate (AgCl)
Wash with distilled waterRemoves any the other soluble salts (KNO3)
Dry by leaving in a warm placeEvaporates the water

2:42 practical: prepare a sample of pure, dry hydrated copper(II) sulfate crystals starting from copper(II) oxide

Excess Solid Method:

Preparing pure dry crystals of copper sulfate (CuSO4) from copper oxide (CuO) and sulfuric acid (H2SO4)

StepExplanation
Heat acid (H2SO4) in a beakerSpeeds up the rate of reaction
Add base (CuO) until in excess (no more copper oxide dissolves) and stir with glass rodNeutralises all the acid
Filter the mixture using filter paper and funnelRemoves any excess copper oxide
Gently heat the filtered solution (CuSO4)To evaporate some of the water
until crystals form on a glass rodShows a hot saturated solution formed
Allow the solution to cool so that hydrated crystals formCopper sulfate less soluble in cold solution
Remove the crystals by filtrationRemoves crystals
Dry by leaving in a warm placeEvaporates the water

 

2:43 (Triple only) practical: prepare a sample of pure, dry lead(II) sulfate

Objective: prepare a pure, dry sample of lead (II) sulfate (PbSO₄).

Preparing a pure, dry sample of lead (II) sulfate (PbSO₄) from lead (II) nitrate solution (Pb(NO₃)₂) and sodium sulfate solution (Na₂SO₄).

      Pb(NO₃)₂ (aq)      +      Na₂SO₄ (aq)      →        PbSO₄ (s)      +      2NaNO₃ (aq)

  1. Mix similar volumes lead nitrate solution and sodium sulfate solution in a beaker. The precise volumes do not matter since any excess will be removed later.
  2. A white precipitate of lead (II) sulfate will form.
  3. The reaction mixture is filtered.
  4. The residue left on the filter paper is washed with distilled water several times to remove impurities.
  5. The residue is then moved to a warm oven to dry.

 

2:44 describe tests for these gases: hydrogen, oxygen, carbon dioxide, ammonia, chlorine

Tests for gases

GasTestResult if gas present
hydrogen (H2)Use a lit splintGas pops
oxygen (O2)Use a glowing splintGlowing splint relights
carbon dioxide (CO2)Bubble the gas through limewaterLimewater turns cloudy
ammonia (NH3)Use red litmus paperTurns damp red litmus paper blue
chlorine (Cl2)Use damp litmus paperTurns damp litmus paper white (bleaches)

2:45 describe how to carry out a flame test

A flame test is used to show the presence of certain metal ions (cations) in a compound.

  • A platinum or nichrome wire is dipped into concentrated hydrochloric acid to remove any impurities.
  • The wire is dipped into the salt being tested so some salt sticks to the end.
  • The wire and salt are held in a non-luminous (roaring) bunsen burner flame.
  • The colour is observed.

Properties of the platinum or nichrome wire is:

  • Inert
  • High melting point

2:46 know the colours formed in flame tests for these cations: Li⁺ is red, Na⁺ is yellow, K⁺ is lilac, Ca²⁺ is orange-red, Cu²⁺ is blue-green

When put into a roaring bunsen burner flame on a nichrome wire, compounds containing certain cations will give specific colours as follows.

IonColour in flame test
lithium (Li⁺)red
sodium (Na⁺)yellow
potassium (K⁺)lilac
calcium (Ca²⁺)orange-red
copper (II) (Cu²⁺)blue-green

2:47 describe tests for these cations: NH₄⁺ using sodium hydroxide solution and identifying the gas evolved, Cu²⁺, Fe²⁺ and Fe³⁺ using sodium hydroxide solution

Underneath are the tests for : ammonium ion test, copper (II) ion test, iron (II) ion test, iron (III) ion test
 
Describe tests for the cation NH4+, using sodium hydroxide solution and identifying the ammonia evolved

 

Describe tests for the cations Cu2+, Fe2+ and Fe3+, using sodium hydroxide solution

First, add sodium hydroxide (NaOH), then observe the colour:

2:48 describe tests for these anions: Cl⁻, Br⁻ and I⁻ using acidified silver nitrate solution, SO₄²⁻ using acidified barium chloride solution, CO₃²⁻ using hydrochloric acid and identifying the gas evolved

Describe tests for anions: Halide ions (Cl, Br and I)

Underneath are the tests for:

chloride ion test, bromide ion test, iodide ion test, sulfate ion test, carbonate ion test

 

Describe tests for anions: Sulfate ions (SO42)

 

Describe tests for anions: Carbonate ions (CO32-)

2:49 describe a test for the presence of water using anhydrous copper(II) sulfate

Add anhydrous copper (II) sulfate (CuSO4) to a sample.

If water is present the anhydrous copper (II) sulfate will change from white to blue.

3:01 know that chemical reactions in which heat energy is given out are described as exothermic, and those in which heat energy is taken in are described as endothermic

Exothermic: chemical reaction in which heat energy is given out.

Endothermic: chemical reaction in which heat energy is taken in.

 

(So, in an exothermic reaction the heat exits from the chemicals so temperature rises)

 

3:02 describe simple calorimetry experiments for reactions such as combustion, displacement, dissolving and neutralisation

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

 

The cup used is polystyrene because:

polystyrene is an insulator which reduces heats loss

 

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:

3:03 calculate the heat energy change from a measured temperature change using the expression Q = mcΔT

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:

3:05 (Triple only) draw and explain energy level diagrams to represent exothermic and endothermic reactions

The symbol ΔH is used to represent the change in heat (or enthalpy change) of a reaction.

ΔH is measured in kJ/mol (kilojoules per mole).

The change in heat (enthalpy change) can be represented on an energy level diagram. ΔH must also labelled.

 

In an exothermic reaction, the reactants have more energy than the products.

Energy is given out in the form of heat which warms the surroundings.

ΔH is given a negative sign, because the reactants are losing energy as heat, e.g  ΔH = -211 kJ/mol.

 

 

 

 

In an endothermic reaction, the reactants have less energy than the products.

Energy is taken in which cools the surroundings.

ΔH is given a positive sign, because the reactants are gaining energy, e.g  ΔH = +211 kJ/mol.

 

 

 

3:06 (Triple only) know that bond-breaking is an endothermic process and that bond-making is an exothermic process

During chemical reactions, the bonds in the reactants must be broken, and new ones formed to make the products.

Breaking bonds need energy and therefore is described as endothermic.

Energy is released when new bonds are made and therefore is described as exothermic.

 

If bonds are both broken and made during chemical reactions, why can a reaction overall be describe as either exothermic or endothermic?

Example: hydrogen reacts with oxygen producing water. Overall energy is released and therefore the reaction is exothermic.

         

The reaction is exothermic because the energy needed to break the bonds is less than the energy released in making new bonds.

If a reaction is endothermic then the energy needed to break the bonds is more than the energy released in making new bonds.

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

4:02 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.

Select a set of flashcards to study:

     Terminology

     Skills and equipment

     Remove Flashcards

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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