5.31 describe important uses of sodium hydroxide, including the manufacture of bleach, paper and soap; and of chlorine, including sterilising water supplies and in the manufacture of bleach and hydrochloric acid.

Uses of Sodium Hydroxide and Chlorine:

Sodium Hydroxide	Chlorine
Manufacture of Bleach	Sterilising water supplies
Manufacture of Paper	Manufacture of Bleach
Manufacture of Soap	Manufacture of hydrochloric acid
Manufacture of Detergents	Manufacture of PVC

5.30 write ionic half-equations for the reactions at the electrodes in the diaphragm cell

Positive electrode (anode):
Chloride ions lose electrons to form chlorine molecules

2Cl^- à Cl₂ + 2e^-

Negative electrode (cathode):

Water molecules gain electrons to form hydroxide ions and hydrogen molecules

2H₂O + 2e^- à 2OH^- + H₂

5.29 describe the manufacture of sodium hydroxide and chlorine by the electrolysis of concentrated sodium chloride solution (brine) in a diaphragm cell

Positive electrode (anode):
Chloride ions lose electrons to form chlorine molecules

2Cl^- à Cl₂ + 2e^-

Negative electrode (cathode):

Water molecules gain electrons to form hydroxide ions and hydrogen molecules

2H₂O + 2e^- à 2OH^- + H₂

Overall equation:

2NaCl + 2H­₂O à 2NaOH + H₂ + Cl₂

The diaphragm keeps the sodium hydroxide and chlorine apart to stop them forming bleach.

5.28 describe the use of sulfuric acid in the manufacture of detergents, fertilisers and paints

The 3 main uses of sulfuric acid are:

1. Manufacture of detergents
2. Manufacture of fertilisers
3. Manufacture of paints

5.27 describe the manufacture of sulfuric acid by the contact process

1. Sulfur Dioxide is made by burning sulfur in air
2. Sulfur dioxide is reacted with oxygen to form sulfur trioxide
3. Sulfur trioxide is absorbed into concentrated sulfuric acid to make oleum:

H₂SO₄(l) + SO₃(g) à H₂S₂O₇(l)

(Take note of the state symbols)

The oleum is then diluted to form a solution of sulfuric acid:

H₂S₂O₇(l) + H₂O(l) à 2H₂SO₄ (aq)

Process:

1. Air and Sulfur Dioxide are inputted 
2. The mixture of gases is cleaned, dried and heated to 400-500C and compressed to 2 atm. 
3. The mixture of gases is passed over the catalyst vanadium(V) oxide. Oxygen and sulfur combine to form sulfur dioxide and the yield is 98%.
4. Any unreacted gases are recycled. Sulfur trioxide is dissolved in concentrated sulfuric acid to form oleum. 
5. The oleum is diluted to form sulfuric acid.

5.26 recall the raw materials used in the manufacture of sulfuric acid

The raw materials of the contact process (manufacture of sulfuric acid):

• Sulfur dioxide
• Air

5.21 understand that condensation polymerisation produces a small molecule, such as water, as well as the polymer.

Condensation reactions takes two molecules and reacts them together to form a larger molecule and the elimination of a smaller molecule like water.
e.g.
Reacting a carboxylic acid and alcohol together forms an Ester with the elimination of a water molecule.

5.20 understand that some polymers, such as nylon, form by a different process called condensation polymerisation

Nylon is a product of a process called condensation polymerisation.

5.18 describe some uses for poly(chloroethene)

• Poly(chloroethene) or PVC for short is tougher than poly(ethene), very hard wearing and more stable to heat
• It's used for plastic sheets, drainpipes and gutters, insulation for wires and casing for electrical plugs

5.16 draw the repeat unit of poly(chloroethene)

This is the monomer of poly(chloroethene) and is called chloroethene.

This is the repeat unit of poly(chloroethene).

4.16 use average bond energies to calculate the enthalpy change during a simple chemical reaction.

Bond Energies

The energy required to break a bond is called bond energy. It's measured in kJ/mol (of bonds).

Breaking bonds takes in energy i.e. is endothermic

Making bonds gives out energy i.e. is exothermic

To calculate enthalpy change using bond energies:

1. Calculate the sum of energies for the bonds broken, Σ (bonds broken)
2. Calculate the sum of energy for bonds made, Σ (bonds made)
3. Calculate H by using the formulaH = Σ (bonds broken) - Σ (bonds made)

e.g. 

H₂ + O₂ à 2H₂O

Σ (bonds broken) = (H-H) + (O-O) = 436 + 496 = 932

Σ (bonds made) = 4(O-H) = (4 x 463) = 1852

Σ (bonds broken) – Σ (bonds made) = 932 – 1852

H = 932 – 1852 = -920 kJ/mol

4.12 calculate molar enthalpy change from heat energy change

Enthalpy Calculations

For reactions carried out at constant pressure the heat energy change is called the enthalpy change.

To standardise this we find the molar enthalpy change which is referred to asH. 

Heat Given Out = Mass x Specific Heat Capacity x Temperature Rise

To convert the heat given out into molar enthalpy change you have to find out how energy 1 mol of the substance produces.

e.g.

Reaction between Magnesium and Hydrochloric Acid

Temp Rise = 10 degrees

Moles of acid = 0.005

Heat given out = 50 x 4.2 (specific heat capacity of water) x 10J = 2100J = 2.10kJ

0.005 mol of acid produces 2.10kJ

1 mol of acid produces = 2.10/0.005 = 420kJ

H = -420kj/mol 

3.12 describe the dehydration of ethanol to ethene, using aluminium oxide.

Dehydration of Ethanol

Ethanol can be dehydrated by passing ethanol vapour over hot aluminium oxide (catalyst):

C₂H_5­OH à C₂H₄ + H₂O

3.11 evaluate the factors relevant to the choice of method used in the manufacture of ethanol, for example the relative availability of sugar cane and crude oil

Fermentation vs Hydration of Ethene

	Fermentation	Hydration of Ethene
Raw Materials	Use renewable sources	Uses non-renewable crude oil
Type of process	Batch process	Continuous process
Rate of reaction	Very slow	Fast
Quality of Product	Produces a dilute solution of ethanol that needs processing for pure ethanol	Produces pure ethanol
Reaction conditions	Low temperatures required	High temps and pressures means increased cost

The method chosen depends on many factors:

• If you have a climate suited to growing sugar cane then fermentation's better but if crude oil is more widely available then hydration of ethene is better.
• If a dilute solution of ethanol is required for the production of wine or vinegar then fermentation is better
• If pure ethanol is required is required for a solvent then the hydration of ethene would be better.

3.10 describe the manufacture of ethanol by the fermentation of sugars, for example glucose, at a temperature of about 30°C

Production of Ethanol

Ethanol can be produced in two ways - fermentation or direct hydration of ethene.

Fermentation:

1. Dissolve sugar or starch in water and add yeast
2. Leave the mixture to ferment at 25-40°C for several days - anaerobically
3. Filter off the excess yeast to obtain a dilute solution of ethanol

When the ethanol content reaches around 15% then the yeast is killed.

If you need a more concentrated solution then this must be fractionally distilled.

NB - the enzymes in the sugar produce glucose and then convert it into ethanol.

Sugar / Starch à C₆H₁₂O₆

C₆H₁₂O_{6 ­}à 2C₂O₅OH + 2CO₂

3.9 describe the manufacture of ethanol by passing ethene and steam over a phosphoric acid catalyst at a temperature of about 300°C and a pressure of about 60–70 atm

Production of Ethanol

Ethanol can be produced in two ways - fermentation and direct hydration of ethene.

Hydration of Ethene:

1. A mixture of ethene and steam is passed over phosphoric acid catalyst at a temperature of 300°C and 60-70 atmospheres pressure. 

C₂H₄ + H₂O à C₂H₅OH

1. The ethanol is condensed as a liquid
2. The ethene required for this reaction is obtained from crude oil.

2.8 explain the relative reactivities of the elements in Group 1 in terms of distance between the outer electrons and the nucleus

Reactivity of Group 1 metals

The metals in Group 1 increase in reactivity as you descend down the group. This is because of the distance between the outer electrons and the nucleus. 

In Group 1 all the atoms have 1 outer electrons. They don't want this electron so are trying to get rid of it to get a full outer shell. 

If the distance between the nucleus and the electron is small then there is a strong force of attraction so it's hard for it to give away its electron.

Similarly if the distance between the nucleus and the electron is great then there is a weaker force so it can give away its electron easily, thus being able to react easier with other atoms. It's more reactive.

1.57 calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions.

Electrolysis Calculations

Using moles = mass / RFM

The charge on 1 mol of electrons is 96500C.
96,500C/mol is called Faradays Constant.

When copper is deposited during electrolysis:

Cu²⁺ + 2e^-  à Cu

So for every mol of copper 2 mols of electrons are needed to discharge it. 

(The number before the symbol represents the number of moles e.g 2e^-)

96,500C x 2 is the quantity of charge required for discharge. 

To find the amounts of products of electrolysis:

Use moles = mass/RFM

e.g.

Find the mass of metal deposited when 1F of electrons flows through Silver Nitrate:

Ag⁺ + e^- à Ag

So 1 mol of electrons is required to discharge 1 mol of Ag ions

mass = moles x RFM

mass = 1 x 108 = 108g

Find the amount of metal deposited when 1F of electrons flows through 5g of Nickel (II) Nitrate:

Ni²⁺ +2e^- à Ni

So 1 mols of electrons are required to discharge 0.5 mole of Ni ions 

(always use 1 mol of electrons and adjust the other figures accordingly)

mass = moles x RFM

mass = 0.5 x 59 = 29.5g of Ni

Using C = I x t

e.g.

A current of 0.01 amperes passes for 4 hours through a solution of gold (III) nitrate. What mass of metal is deposited.

Total Quantity of Charge = 0.010 x (4 x 60 x 60)

C = 144C

 

Equation: Au³⁺ + 3e^- à Au

So 3 moles of electrons are required to discharge 3 moles of gold

3 x 96,500C deposits 1 mol (197g) of gold

144C deposits 197(mass) / 3 x 96,500(quantity of charge required to discharge) x 144g of gold  (because of the coulombs)

144C = (197/3x96500) x 144g = 0.098g

NB - a good way of tackling these questions is to look at the question, see what it's asking you, see what its given you and to see which formulas you should use e.g. moles = mass / RFM

1.56 recall that one faraday represents one mole of electrons

1 Faraday = 1F

Coulombs = Quantity of Charge

Coulombs (C) = Current (A) x Time (s)

1F = 96,500 Coulombs (C)

1 mol of electrons = 1F or 96,500C

1.54 describe experiments to investigate electrolysis, using inert electrodes, of aqueous solutions such as sodium chloride, copper(II) sulfate and dilute sulfuric acid and predict the products

Electrolysis

Aqueous Solutions

1. Use inert electrodes like platinum or carbon
2. Place them the aqueous solution in a beaker
3. Put a current through the solution
4. Watch as the products are formed

Predicting the products

The positive ions will always go to the negative electrode and vice versa. 

From that we can see what we will see at each electrode.

e.g.

2Cl^- --> Cl₂ + 2e- 

This means that 2 Chloride ions have lost electrons (oxidation) and formed Cl₂ and 2 electrons which will go and be used at the other electrode.