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. 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

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

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

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

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

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

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)

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:

Bond energies C-H = 435 Kj O=O = 497 Kj Total for breaking bonds = 4x435 + 2x497 = 2734 Kj H-O = 464 Kj C=O = 803 Kj Total for making bonds = 2x803 + 4x464 = 3462 Kj

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.

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)

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

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

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 = 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

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 + O2 2MgO 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

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? 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 2) What mass of calcium oxide is produced when 10g of calcium burns? 2Ca + O2 2CaO Mr: 2Ca = 2x40 = 80 2CaO = 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? 2Al2O3 4Al + 3O2 Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108 204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3

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 2H2 + 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

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 2H2 + 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

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

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 = 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

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.

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

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!!!

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: Cu2+ + 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

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: Cu2+ + 2e- Cu 2Cl- - 2e- 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