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Patterns of Chemical Change

Patterns of Chemical Change. 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.

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Patterns of Chemical Change

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

  5. Biological catalysts Optimum temperature (roughly 40OC) Enzymes are denatured beyond 40OC Increasing activity due to more collisions • 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 Our bodies contain catalysts for all the reactions that go on inside them – they are called ENZYMES. Enzymes work best at certain temperatures: Enzyme activity Temperature

  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 Activation energy Energy given out by reaction Using a catalyst might lower the activation energy Energy level 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 H-O = 464 Kj C=O = 803 Kj Total for breaking bonds = 4x435 + 2x497 = 2734 Kj Total for making bonds = 2x803 + 4x464 = 3462 Kj

  11. Drawing this on an energy diagram: 2734 Kj 3462 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. Atomic mass RELATIVE ATOMIC MASS, Ar (“Mass number”) = number of protons + number of neutrons 4 He SYMBOL 2 PROTON NUMBER = number of protons (obviously)

  13. Relative formula mass, Mr Relative atomic mass of O = 16 Relative atomic mass of H = 1 The relative formula mass of a compound is blatantly the relative atomic masses of all the elements in the compound added together. E.g. water H2O: 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

  15. Calculating percentage mass Mass of element Ar x100% Percentage mass (%) = Relative formula mass Mr If you can work out Mr then this bit is easy… Calculate the percentage mass of magnesium in magnesium oxide, MgO: Ar for magnesium = 24 Ar for oxygen = 16 Mr for magnesium oxide = 24 + 16 = 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 IGNORE the oxygen in step 2 – the question doesn’t ask for it Step 1: READ the equation: 2Mg + O2 2MgO E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in air? Step 2: WORK OUT the relative formula masses (Mr): 2Mg = 2 x 24 = 48 2MgO = 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 2H2 + O2 • What mass of hydrogen is produced by the electrolysis of 6g of water? 2) What mass of calcium oxide is produced when 10g of calcium burns? 2Ca + O2 2CaO 3) What mass of aluminium is produced from 100g of aluminium oxide? 2Al2O3 4Al + 3O2 • Work out Mr: 2H2O = 2 x ((2x1)+16) = 36 2H2 = 2x2 = 4 • 36g of water produces 4g of hydrogen, • 6g of water will produce (6/36) x 4 = 0.66g of hydrogen Mr: 2Ca = 2x40 = 80 2CaO = 2 x (40+16) = 112 80g produces 112g so 10g produces (112/80) x 10 =14g of CaO Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108 204g produces 108g so 100g produces (108/204) x 100 =52.9g of Al2O3

  18. Another method Q. When water is electrolysed it breaks down into hydrogen and oxygen: 2H2O 2H2 + O2 What mass of hydrogen is produced by the electrolysis of 6g of water? Mass of product IN GRAMMES Mr of product Mass of reactant IN GRAMMES Mr of reactant Try using this equation: 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 Q. When water is electrolysed it breaks down into hydrogen and oxygen: 2H2O 2H2 + O2 What VOLUME of hydrogen is produced by the electrolysis of 6g of water? 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 • 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? • 2H20 2H2 + 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. 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 = 0.04 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 Copper ion Chloride ion

  24. Electrolysis = chloride ion = copper ion Electrolysis is used to extract a HIGHLY REACTIVE metal. 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 At the negative electrode the positive ions GAIN electrons to become neutral copper ATOMS. The half equation is: Cu2+ + e- Cu At the positive electrode the negative ions LOSE electrons to become neutral chlorine MOLECULES. The half equation is: Cl- - e- Cl2 We need to be able to write “half equations” to show what happens during electrolysis (e.g. for copper chloride): 2 2 2

  26. Calculating masses and volumes from electrolysis Consider those two half equations again: Cu2+ + 2e- Cu2Cl- - 2e- Cl2 2) Now use this equation: Actual mass of chlorine Actual mass of copper Relative mass of chlorine Relative mass of copper Example question: How much chlorine is released at the positive electrode if 2g of copper is collected at the negative electrode? 1) Write down the relative atomic mass: Copper = 63 2 molecules of chlorine (Cl2) = 2x35 = 70 3) Write down the answer: _______g of chlorine is liberated, which would occupy a volume of ________litres

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