1. Patterns of Chemical Change

2. Rates of Reaction
Chemical reactions occur when different atoms or molecules collide:
For the reaction to happen the particles must have a certain amount of energy – this is called the ACTIVATION ENERGY.
The rate at which the reaction happens depends on four things:
The temperature of the reactants,
Their concentration
Their surface area
Whether or not a catalyst is used

3. Measuring rate of reaction
Two common ways:
1) Measure how fast the products are formed
2) Measure how fast the reactants are used up

4. Rate of reaction graph Amount of product formed
Slower rate of reaction here due to reactants being used up
Fast rate of reaction here
Slower reaction
Time

5. Biological catalysts
Our bodies contain catalysts for all the reactions that go on inside them – they are called ENZYMES. Enzymes work best at certain temperatures:
Optimum temperature (roughly 40OC)
Enzyme activity
Enzymes are denatured beyond 40OC
Increasing activity due to more collisions
Temperature
Enzymes are used in the food industry. Two examples:
Fermentation: Glucose Ethanol + Carbon dioxide
Ethanol is used for alcohol, carbon dioxide for making bread rise
Yoghurt production – enzymes in bacteria produce yoghurt from milk by converting sugar (lactose) into lactic acid, which tastes sour

6. Endothermic and exothermic reactions
Step 1: Energy must be SUPPLIED to break bonds:
Step 2: Energy is RELEASED when new bonds are made:
A reaction is EXOTHERMIC if more energy is RELEASED then SUPPLIED. If more energy is SUPPLIED then is RELEASED then the reaction is ENDOTHERMIC

7. Energy level diagrams Energy level Activation energy
Using a catalyst might lower the activation energy
Energy given out by reaction
Reaction progress

8. Exothermic vs endothermic:
EXOTHERMIC – more energy is given out than is taken in (e.g. burning, respiration)
ENDOTHERMIC – energy is taken in but not necessarily given out (e.g. photosynthesis)

9. Burning Methane CH4 + 2O2 2H2O + CO2
To burn methane you have to break all of these bonds:
And then you have to make these ones:

10. Bond energies C-H = 435 Kj O=O = 497 Kj
Total for breaking bonds = 4x x497 = 2734 Kj
H-O = 464 Kj
C=O = 803 Kj
Total for making bonds = 2x x464 = 3462 Kj

11. Drawing this on an energy diagram:
3462 Kj
2734 Kj
-728 Kj
More energy is given out (3462) than is given in (2734) – the reaction is EXOTHERMIC. The total (“nett”) energy change is –728 Kj. An endothermic reaction would have a positive energy change.

12. PROTON NUMBER = number of protons (obviously)
Atomic mass
RELATIVE ATOMIC MASS, Ar (“Mass number”) = number of protons + number of neutrons He 2 4
SYMBOL
PROTON NUMBER = number of protons (obviously)

13. Relative formula mass, Mr
The relative formula mass of a compound is blatantly the relative atomic masses of all the elements in the compound added together.
Relative atomic mass of O = 16
E.g. water H2O:
Relative atomic mass of H = 1
Therefore Mr for water = 16 + (2x1) = 18
Work out Mr for the following compounds:
HCl
NaOH
MgCl2
H2SO4
K2CO3
H=1, Cl=35 so Mr = 36
Na=23, O=16, H=1 so Mr = 40
Mg=24, Cl=35 so Mr = 24+(2x35) = 94
H=1, S=32, O=16 so Mr = (2x1)+32+(4x16) = 98
K=39, C=12, O=16 so Mr = (2x39)+12+(3x16) = 138

14. More examples CaCO3 40 + 12 + 3x16 100 HNO3 1 + 14 + 3x16 2MgO80
3H2O
3 x ((2x1) + 16)
4NH3
2KMnO4
3C2H5OH
4Ca(OH)2

15. Calculating percentage mass
If you can work out Mr then this bit is easy…
Percentage mass (%) =
Mass of element Ar
Relative formula mass Mr
x100%
Calculate the percentage mass of magnesium in magnesium oxide, MgO:
Ar for magnesium = 24 Ar for oxygen = 16
Mr for magnesium oxide = = 40
Therefore percentage mass = 24/40 x 100% = 60%
Calculate the percentage mass of the following:
Hydrogen in hydrochloric acid, HCl
Potassium in potassium chloride, KCl
Calcium in calcium chloride, CaCl2
Oxygen in water, H2O

16. Calculating the mass of a product
E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in air?
IGNORE the oxygen in step 2 – the question doesn’t ask for it
Step 1: READ the equation:
2Mg + O MgO
Step 2: WORK OUT the relative formula masses (Mr):
2Mg = 2 x 24 = MgO = 2 x (24+16) = 80
Step 3: LEARN and APPLY the following 2 points:
48g of Mg makes 80g of MgO
60g of Mg makes 60/48 x 80 = 100g of MgO

17. When water is electrolysed it breaks down into hydrogen and oxygen:
2H2O H2 + O2
What mass of hydrogen is produced by the electrolysis of 6g of water?
Work out Mr: 2H2O = 2 x ((2x1)+16) = H2 = 2x2 = 4
36g of water produces 4g of hydrogen,
6g of water will produce (6/36) x 4 = 0.66g of hydrogen
2) What mass of calcium oxide is produced when 10g of calcium burns?
2Ca + O CaO
Mr: 2Ca = 2x40 = CaO = 2 x (40+16) = 112
80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO
3) What mass of aluminium is produced from 100g of aluminium oxide?
2Al2O Al + 3O2
Mr: 2Al2O3 = 2x((2x27)+(3x16)) = Al = 4x27 = 108
204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3

18. So mass of product = (4/36) x 6g = 0.66g of hydrogen
Another method
Try using this equation:
Mass of product IN GRAMMES
Mass of reactant IN GRAMMES
Mr of product
Mr of reactant
Q. When water is electrolysed it breaks down into hydrogen and oxygen:
2H2O H2 + O2
What mass of hydrogen is produced by the electrolysis of 6g of water?
Mass of product IN GRAMMES 4 6g 36
So mass of product = (4/36) x 6g = 0.66g of hydrogen

19. Calculating the volume of a product
At normal temperature and pressure the Relative Formula Mass (Mr) of a gas will occupy a volume of 24 litres
e.g. 2g of H2 has a volume of 24 litres
32g of O2 has a volume of 24 litres
44g of CO2 has a volume of 24 litres etc
Q. When water is electrolysed it breaks down into hydrogen and oxygen:
2H2O H2 + O2
What VOLUME of hydrogen is produced by the electrolysis of 6g of water?
On the previous page we said that the MASS of hydrogen produced was 0.66g
2g of hydrogen (H2) will occupy 24 litres (from the red box above),
So 0.66g will occupy 0.66/2 x 24 = 8 litres

20. Example questions
What volume of hydrogen is produced when 18g of water is electrolysed?
2H H2 + O2
Marble chips are made of calcium carbonate (CaCO3). What volume of carbon dioxide will be released when 500g of CaCO3 is reacted with dilute hydrochloric acid?
CaCO3 + 2HCl CaCl2 + H2O + CO2
In your coursework you reacted magnesium with hydrochloric acid. What volume of hydrogen would be produced if you reacted 1g of magnesium with excess acid?
Mg + 2HCl MgCl2 + H2

21. A classic exam question:
Empirical formulae
Empirical formulae is simply a way of showing how many atoms are in a molecule (like a chemical formula). For example, CaO, CaCO3, H20 and KMnO4 are all empirical formulae. Here’s how to work them out using masses:
A classic exam question:
Find the simplest formula of 2.24g of iron reacting with 0.96g of oxygen.
Step 1: Divide both masses by the relative atomic mass:
For iron 2.24/56 = For oxygen 0.96/16 = 0.06
Step 2: Write this as an equation and simplify:
0.04:0.06 is equivalent to 2:3
Step 3: Write the formula:
2 iron atoms for 3 oxygen atoms means the formula is Fe2O3

22. Example questions
Find the empirical formula of magnesium oxide which contains 24g of magnesium and 16g of oxygen.
Find the empirical formula of a compound that contains 42g of nitrogen and 9g of hydrogen.
Find the empirical formula of a compound containing 20g of calcium, 6g of carbon and 24g of oxygen.

23. Electrolysis Molecule of solid copper chloride
Molecule of solid copper chloride after being dissolved
Chloride ion
Copper ion

24. Electrolysis Electrolysis is used to extract a HIGHLY REACTIVE metal.
= chloride ion
= copper ion
When we electrolysed copper chloride the negative chloride ions moved to the positive electrode and the positive copper ions moved to the negative electrode – OPPOSITES ATTRACT!!!

25. Electrolysis equations
We need to be able to write “half equations” to show what happens during electrolysis (e.g. for copper chloride):
At the negative electrode the positive ions GAIN electrons to become neutral copper ATOMS. The half equation is:
Cu e Cu 2
At the positive electrode the negative ions LOSE electrons to become neutral chlorine MOLECULES. The half equation is:
Cl e Cl2 2 2

26. Calculating masses and volumes from electrolysis
Example question: How much chlorine is released at the positive electrode if 2g of copper is collected at the negative electrode?
Consider those two half equations again:
Cu e Cu 2Cl e Cl2
1) Write down the relative atomic mass:
Copper = 63
2 molecules of chlorine (Cl2) = 2x35 = 70
Actual mass of chlorine
Actual mass of copper
Relative mass of chlorine
Relative mass of copper
2) Now use this equation:
3) Write down the answer: _______g of chlorine is liberated, which would occupy a volume of ________litres