1. Patterns of Chemical Change
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Patterns of Chemical Change
R. Cozier

2. Rates of Reaction
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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
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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
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Amount of product formed
Slower rate of reaction here due to reactants being used up
Fast rate of reaction here
Slower reaction
Time

5. Enzymes
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Enzymes are biological catalysts. They help the reactions that occur in our bodies by controlling the rate of reaction.
Yeast is an example of an enzyme. It is used to help a process called fermentation:
Sugar Alcohol + carbon dioxide
The alcohol from this process is used in making drinks and the carbon dioxide can be used to make bread rise.
Enzymes work best in certain conditions:
Enzymes are denatured beyond 40OC
Could be protease (found in the stomach)
Could be amylase (found in the intestine)
Enzyme activity
Temp pH
400C

6. Uses of enzymes
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Daz
1) Enzymes are used in washing powders to help digest food stains. Biological washing powders will only work on 400C or lower.
2) Enzymes are used in baby foods to “pre-digest” the proteins.
3) Enzymes are used to convert starch into sugar which can then be used in food.
4) Conversion of glucose into fructose – glucose and fructose are “isomers” (they have the same chemical formula), but fructose is sweeter.

7. Endothermic and exothermic reactions
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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

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

9. Exothermic vs endothermic:
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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)

10. Reversible Reactions
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Some chemical reactions are reversible. In other words, they can go in either direction: A + B C D
e.g. Ammonium chloride
Ammonia + hydrogen chloride
NH4Cl
NH3 + HCl
If a reaction is EXOTHERMIC in one direction what must it be in the opposite direction?
For example, consider copper sulphate:
Hydrated copper sulphate (blue)
Anhydrous copper sulphate (white)
+ Heat
+ Water
CuSO4 + H2O
CuSO4.5H2O

11. Endothermic reactions
Reversible Reactions
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When a reversible reaction occurs in a CLOSED SYSTEM (i.e. no reactants are added or taken away) an EQUILIBRIUM is achieved – in other words, the reaction goes at the same rate in both directions: A + B C D
Endothermic reactions
Increased temperature:
Decreased temperature:
Exothermic reactions
Increased temperature:
Decreased temperature: A + B C D A + B C D
More products
Less products A + B C D A + B C D
Less products
More products

12. Making Ammonia Nitrogen + hydrogen Ammonia N2 + 3H2 2NH3
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Fritz Haber,
Guten Tag. My name is Fritz Haber and I won the Nobel Prize for chemistry. I am going to tell you how to use a reversible reaction to produce ammonia, a very important chemical. This is called the Haber Process.
Nitrogen + hydrogen Ammonia
N2 + 3H NH3
To produce ammonia from nitrogen and hydrogen you have to use three conditions:
High pressure
450O C
Iron catalyst
Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy.
Nitrogen
Hydrogen
Recycled H2 and N2

13. Ammonia + nitric acid Ammonium nitrate
Uses of Ammonia
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Ammonia is a very important chemical as it can be used to make plant fertilisers and nitric acid:
Water and oxygen
Nitrogen monoxide
Ammonia gas
Oxygen
Hot platinum catalyst
Nitrogen monoxide
Nitric acid
Cooled
More ammonia can then be used to neutralise the nitric acid to produce AMMONIUM NITRATE (a fertiliser rich in nitrogen).
Ammonia + nitric acid Ammonium nitrate
NH3 + HNO NH4NO3
The trouble with nitrogen based fertilisers is that they can also create problems – they could contaminate our drinking water.

14. Haber Process: The economics
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A while ago we looked at reversible reactions: A + B C D
Endothermic, increased temperature A + B C D
Exothermic, increase temperature
Nitrogen + hydrogen Ammonia
N2 + 3H NH3
Endothermic
Exothermic
1) If temperature was DECREASED the amount of ammonia formed would __________...
However, if temperature was INCREASED the rate of reaction in both directions would ________ causing the ammonia to form faster
If pressure was INCREASED the amount of ammonia formed would INCREASE because there are less molecules on the right hand side of the equation

15. Haber Process Summary
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A low temperature increases the yield of ammonia but is too slow
A high temperature improves the rate of reaction but decreases the yield too much
A high pressure increases the yield of ammonia but costs a lot of money
To compromise all of these factors, these conditions are used:
200 atm pressure
450O C
Iron catalyst
Recycled H2 and N2
Nitrogen
Hydrogen
Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy.

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

17. Burning Methane CH4 + 2O2 2H2O + CO2 Methane Oxygen Water
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CH4 + 2O H2O + CO2
Methane
Oxygen
Water
Carbon dioxide

18. Bond energies C-H = 435 Kj O=O = 497 Kj
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C-H = 435 Kj
O=O = 497 Kj
Total for breaking bonds = 4x x497 = 2734 KJ/mol
H-O = 464 Kj
C=O = 803 Kj
Total for making bonds = 2x x464 = 3462 KJ/mol
Total energy change = = KJ/mol

19. Drawing this on an energy diagram:
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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.

20. Bond energy values C-H = 435 KJ/mol O-H = 464 KJ/mol O=O = 497 KJ/mol
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C-H = 435 KJ/mol
O-H = 464 KJ/mol
O=O = 497 KJ/mol
C=O = 803 KJ/mol
C-O = 360 KJ/mol
C-C = 346 KJ/mol

21. Burning Methanol 2CH3OH + 3O2 2CO2 + 4H2O
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2CH3OH + 3O CO2 + 4H2O
Total for breaking bonds = 6x435 (C-H) + 2x360 (C-O) + 2x464 (O-H) + 3x497 (O=O) = 5749 KJ/mol
Total for making bonds = 4x803 (C=O) + 8x464 (O-H) = 6924 KJ/mol
Energy change = (divide this by two as we are dealing with two molecules of methanol) = KJ/mol

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

23. Relative formula mass, Mr
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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

24. More examples CaCO3 40 + 12 + 3x16 100 HNO3 1 + 14 + 3x16 2MgO
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CaCO3
x16
100
HNO3
x16
2MgO
2 x ( ) 80
3H2O
3 x ((2x1) + 16)
4NH3
2KMnO4
3C2H5OH
4Ca(OH)2

25. Calculating percentage mass
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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

26. Calculating the mass of a product
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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 3 points:
48g of Mg makes 80g of MgO
1g of Mg makes 80/48 = 1.66g of MgO
60g of Mg makes 1.66 x 60 = 100g of MgO

27. Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108
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?
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Work out Mr: 2H2O = 2 x ((2x1)+16) = H2 = 2x2 = 4
36g of water produces 4g of hydrogen
So 1g of water produces 4/36 = 0.11g of hydrogen
6g of water will produce (4/36) x 6 = 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

28. So mass of product = (4/36) x 6g = 0.66g of hydrogen
Another method
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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

29. Calculating the volume of a product
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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

30. Example questions
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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

31. A classic exam question:
Empirical formulae
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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:
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 a ratio 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

32. Example questions
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Find the empirical formula of magnesium oxide which contains 48g of magnesium and 32g 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.

33. Electrolysis Molecule of solid copper chloride CuCl2 (s)
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Molecule of solid copper chloride
CuCl2 (s)
Molecule of solid copper chloride after being dissolved
CuCl2 (aq)
Chloride ion
Copper ion

34. Electrolysis
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Electrolysis is used to separate a metal from its compund.
= chloride ion
= copper ion
When we electrolysed copper chloride the _____ chloride ions moved to the ______ electrode and the ______ copper ions moved to the ______ electrode – OPPOSITES ATTRACT!!!

35. Electrolysis equations
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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

36. Calculating masses and volumes from electrolysis
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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
Follow the steps:
63g of copper makes 70g of chlorine…
…so 1g of copper would make (70/63) g of chlorine… (=________g)
…so 2g of copper would make (70/63) x 2g of chlorine (=________g)
3) Work out the volume: 70g of chlorine would occupy 24 litres, so 2.22g of chlorine would occupy a volume of (2.22/70) x 24 = ______ litres