GR11 Support pack for Chemical change

Study & Master Support Pack | Grade 11 CAPS Physical Sciences Chemical change This support pack for the Chemical change strand of the Physical Sciences Grade 11 CAPS curriculum provides revision summaries on the topic to help prepare for the examinations. Learners can work through these individually at home or these could form the basis of a catch-up class or online lesson. You have permission to print or photocopy this document or distribute it electronically via email or WhatsApp. Cambridge University Press Africa is a proudly South African publisher. For more information on our Study & Master CAPS-approved textbooks and valuable resource materials, visit www.cambridge.org Brighter thinking I Better learning www.cambridge.org

Unit 1: Quantitative aspects of chemical change Summary • Relative formula mass or relative molecular mass: the sum of the atomic masses of all the atoms in the formula or molecule. • Mole refers to the amount of substance that contains the same number of elementary units as the number of atoms in 12,0 g of carbon-12. • Avogadro’s number = 6,02 × 1023 particles. • 1 mole = relative formula mass in g = 22,4 dm3 for gasses at STP = 6,02 × 1023 particles. m • To calculate mole: n = M • Concentration is the amount of solute present in a given quantity of solvent n (c = V ) • An empirical formula is the simplest whole-number ratio of elements in a compound. • A molecular formula is the total number of atoms of each element present in a compound. • The limiting reactant is the substance that will be used up totally during a reaction. The limiting reactant determines the amount of product formed. • Theoretical percentage yield refers to the amount of product formed when the limiting reactant has reacted completely. • Actual yield is the amount of product that is in practice obtained from a chemical reaction. actual yield – percentage yield = theoretical yield × 100% mass of pure product – percentage purity = mass of impure product × 100% 30 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Revision exercises: Unit 1 – Quantitative aspects of chemical change 1. Give one term for the following descriptions. a) Determine the empirical formula of the a) This formula shows the lowest whole number two substances. ratio of the elements in the compound b) If 0,1 mol of lactic acid has a mass of 9 b) The reactant that is entirely used up before g, determine the molecular formula of any of the other reactants, determining the lactic acid. maximum yield of the product c) If the molecular mass of ribose is c) The amount of product that will be 150 g·mol−1, what is the molecular formula synthesised in ideal conditions of ribose? d) In a chemical reaction, matter can be changed from one form into another but 5. The concentration of a solution depends on the total mass remains constant the amount present in a solution. e) 273 K and 1,013 3 105 Pa a) Calculate the concentration of a solution 2. In an experiment, James mixed 50 cm3 of prepared by dissolving 0,085 moles of 0,6 mol·dm−3 lead(II) nitrate solution with KNO3 in water and diluting it to a total an excess of hydrochloric acid. A white volume of 250 cm3. precipitate formed, which he filtered and b) What is the concentration of a 800 cm3 solution that contains 196 g of H2SO4? dried. He determined the mass of the dry c) How many grams of the solid are needed precipitate to be 5,2 g. to prepare 300 ml of a 1,25 mol·dm−3 Pb(NO3)2(s) 1 2HCl(aq) PbCl2(s) 1 Mg(NO3)2 solution? 2HNO3(aq) d) 4 litres of 0,4 mol·dm−3 HNO3 is added to a) What is the name of the precipitate that 1 litre of 0,2 mol·dm−3 HNO3. What is the formed? concentration of the HNO3 mixture? b) What is the theoretical yield of precipitate? e) 150 ml of water are added to 250 ml of a c) Calculate the percentage yield. 0,6 mol·dm−3 KMnO4 solution. Determine the final concentration after the dilution. 3. Ammonium nitrate decomposes according to the equation: 6. Many samples of chemicals are not pure. NH4NO3(s) N2O(g) 1 2H2O(g) We can define percentage purity as: mass of pure compound in the impure sample A 2,8 g sample of ammonium nitrate is sealed total mass of impure sample 3 100 and then heated in a 1 dm3 (1 litre) flask. The Consider the reaction between magnesium ammonium nitrate decomposes completely. hydroxide powder and a phosphoric acid a) Calculate the total number of moles of solution. gaseous products that form. 3Mg(OH)2 1 2H3PO4 Mg3(PO4)2 1 6H2O b) If the temperature in the flask is 344 K, what is the pressure in the flask in kPa? a) Calculate the percentage purity of a 4. Lactic acid is a chemical produced in sample of Mg(OH)2 if 2,57 g of the sample your muscles during exercise. Ribose is a reacts completely with 40,45 cm3 of a component of vitamin B2. When these two 0,64 mol·dm−3 H3PO4 solution. substances are analysed, they are found to b) Determine the amount of moles of water have the same percentage composition: 40,0% that was formed after completion of the carbon, 6,7% hydrogen and 53,3% oxygen reaction. by mass. CHEMICAL CHANGE 31

Unit 1: Answers 1. a) Empirical formula Molecular formula subscripts 5 3 3 b) Limiting reactant empirical formula subscripts c) Theoretical yield Empirical formula: CH2O d) Principle of conservation of matter Therefore molecular formula: C3H6O3 e) STP (standard temperature and pressure) 150 g·mol−1 c) 30 g·mol−1 5 5 2. a) Lead chloride Molecular formula subscripts 5 5 3 b) nlead nitrate 5 cV 5 0,6 mol·dm−3 3 0,05 dm3  empirical formula subscripts 5 0,03 mol Therefore molecular formula: C5H10O5 nlead chloride 5 nlead nitrate 5 0,03 mol mlead chloride5 nM 5 0,03 mol 3 278 g·mol−1 n 5. a) c 5 V 5 8,34 g 0,085 mol actual yield 5 0,25 dm3 c) Percentage yield 5 theoretical yield 3 100 5,2 g 5 0,34 mol·dm−3 5 8,34 g 3 100 5 62,35% m b) c 5 MV m 2,8 g 196 g 3. a) nNH NO 5 M 5 80 g·mol−1 5 0,035 mol 5 98 g·mol−1 3 0,8 dm3 4 3 nN O 5 0,035; nH O 5 0,07 5 2,5 mol·dm−3 2 2 ntotal 5 0,105 mol c) m 5 McV b) PV 5 nRT 5 148 g·mol−1 3 1,25 mol·dm−3 3 0,3 dm3 nRT 5 55,5 g ⇒P5 V d) nT 5 n1 1 n2 0,105 mol 3 8,31 J·K−1·mol−1 3 344 K 5 cV1 1 cV2 5 1 3 10−3 m3 5 (0,4 mol·dm−3 3 4 dm3) 1 5 300 157,2 Pa ø 300,16 kPa (0,2 mol·dm−3 3 1 dm3) 5 1,8 mol 4. a) n c5V Element g/ m Simplest Empirical 1,8 mol n5 M 5 5 dm3 100g ratio formula C 40,0 40,0 3,33 CH2O 5 0,36 mol·dm−3 12,0 5 3,33 3,33 5 1,00 e) ni 5 cV H 6,7 6,7 6,70 1,0 5 6,70 3,33 5 2,01 5 0,6 mol·dm−3 3 0,25 dm3 O 53,3 53,3 3,33 5 0,15 mol 16,0 5 3,33 3,33 5 1,00 ni 5 nf 5 0,15 mol n cKMnO 5 V 4 b) The molar mass of the empirical formula 0,15 mol 5 (0,25 + 0,15) dm3 CH2O: (12 1 2 1 16) g·mol−1 5 30 g·mol−1 The molar mass of the molecular formula: 5 0,375 mol·dm−3 m 9g M 5 n 5 0,1 mol 5 90 g·mol−1 90 g·mol−1 30 g·mol−1 5 3 32 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

6. a) nH PO 5 cV 3 4 5 0,64 mol·dm−3 3 40,45 3 10−3 dm3 5 0,0259 mol nMg(OH) that react with 0,0259 mol 2 3 5 0,0259 mol 3 2 5 0,0389 mol mMg(OH) 5 nM 2 5 0,0389 mol 3 58 g·mol−1 5 2,26 g % purity 5 mass of pure compound in the impure sample total mass of impure sample 3 100 2,26 g 5 2,57 g 3 100 5 87,94% b) nH O 5 3nH PO 2 3 4 5 3 3 0,0259 mol 5 0,078 mol CHEMICAL CHANGE 33

Unit 2: Energy and chemical change Summary • Enthalpy change is the energy change that occurs when energy is transferred from or to a system at constant pressure (ΔH = Hproducts – Hreactants) • Exothermic reactions give out energy (ΔH < 0); endothermic reactions take heat in (ΔH > 0) • Bond energy refers to the amount of energy required to break a particular type of bond in a molecule; the higher the energy, the stronger the bond. • Activation energy is the minimum amount of energy needed to start a reaction. • We show energy changes for a reaction in an energy profile or energy level diagram. 34 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Revision exercises: Unit 2 – Energy and chemical change 1. Study the simple illustration that shows the b) State what the diagram suggests about the difference between the bond breaking and sum of the bond enthalpies (energies) for bond making of a water molecule. the reactant molecules compared with the product molecules. bond breaking c) In terms of K and L, identify the following + energy for the reaction: molecule separate atoms i) the value for the activation energy ii) the value for the overall enthalpy bond making change. + energy 3. Disposable hand warmers are very popular in separate atoms molecule countries with very cold winters. They heat up by means of an exothermic reaction. These a) Is energy absorbed or released to separate packaged warmers contain iron particles, salt the atoms in a water molecule? water, carbon and other chemicals that act as b) When the separate atoms combine to form insulators. When the pouch is removed from a water molecule, is the process exo- or its outer packaging, outside air penetrates the endothermic? chemicals inside. Oxygen reacts chemically c) Explain your answer in b). with the iron, causing the iron to oxidise and d) How will the strength of a chemical bond release heat. The overall reaction occurring influence the amount of energy involved in in many disposable hand warmers can be a bond being broken or formed. represented by: e) Water reacts with calcium: 4Fe(s) 1 3O2(g) 2Fe2O3(s) 1 1 652 kJ Ca 1 2H2O Ca(OH)2 1 H2 Are the bonds in the water molecules being broken or formed? 2. An equation for the reaction between A and B to form C is given: A 1 B C. The curve in the diagram below illustrates the reaction profile for the reaction. Enthalpy K A+B L C a) Is the formation of C exothermic or endothermic? CHEMICAL CHANGE 35

a) Complete the potential energy curve for f) Does the enthalpy diagram below the above reaction on the given set of axes. represent the energy changes during Indicate the 1 652 kJ value on the graph. photosynthesis or respiration? Potential energy products Ep (kJ) Progress ΔH b) What is the general term given to the value 1 652 kJ? reactants c) Determine the energy released when 1 mol Fe2O3 is produced. d) Calculate the mass of Fe that would be required to release 33,04 kJ of energy. Reaction progress g) Give the numerical value for ΔH in the 4. Green plants have the remarkable ability to diagram. produce glucose from carbon dioxide and water. The process is called photosynthesis: 6CO2(g) 1 6H2O(ℓ) 1 light energy C6H12O6(s) 1 6O2(g) a) Is photosynthesis an exo- or endothermic process? b) Explain the answer in a). c) Name the main source of light energy for this reaction. d) Animal and plant cells have the equally remarkable ability to metabolise glucose and use the energy released to do biological work, such as contracting muscles or thinking. During cellular respiration glucose is broken down as shown: C6H12O6(s) 1 6O2(g) 6CO2(g) 1 6H2O(ℓ) ΔH 5 −2 880 kJ Calculate the enthalpy change if 6,3 g of glucose is metabolised. e) Explain the importance of regular exercise. 36 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Unit 2: Answers 1. a) Absorbed 4. a) Endothermic b) Exothermic b) There is a net absorption of energy during c) Energy is released when the bonds are the reaction. formed. c) The sun m d) Stronger bonds will release more energy d) nglucose 5 M when formed and absorb more energy 6,3 g 5 180 g·mol−1 when broken. e) Broken 5 0,035 mol 1 mol glucose liberates 2 880 kJ 2. a) Exothermic 0,035 mol liberates b) The sum of the bond enthalpies of 0,035 mol 3 2 880 kJ·mol−1 5 100,8 kJ the reactants is smaller than the bond e) Exercise uses energy; so more glucose is enthalpies of the product. broken down by the body. This reduces c) i) K weight gain and keeps the body and ii) K – L metabolism healthy. f) Photosynthesis 3. a) g) 12 880 kJ Potential energy –1 652 kJ Progress b) Enthalpy change / heat of reaction / heat of solution c) When 2 mol of Fe2O3 form, 1 652 kJ are released. When 1 mol of Fe2O3 form, 826 kJ are released. d) 1 652 kJ of energy is released when 4 mol of Fe react. 33,04 kJ of energy is released when 33,04 kJ 4 3 1 652 kJ 5 0,08 mol Fe react mFe 5 nM 5 0,08 mol 3 56 g·mol−1 5 4,48 g CHEMICAL CHANGE 37

Unit 3: Types of reactions Summary • Brønsted acid is a proton donor. Brønsted base: a proton acceptor. • An acid forms its conjugate base when it has donated a proton; a base forms its conjugate acid after it has accepted a proton. • An ampholyte is a substance that can act as an acid or a base, e.g. water. • Neutralisation is a reaction between and acid and a base; the products can include a salt, water and carbon dioxide. • Indicators are coloured substances that change colour at different pH values. • The oxidation number signifies the number of charges the atom would have in a molecule if electrons were transferred completely. • Electrons are transferred during redox reactions. • During an oxidation half-reaction, electrons are lost. A reduction half-reaction refers to a gain of electrons. • An oxidation half-reaction is loss of electrons. A reduction half-reaction refers to gain of electrons. • The reactant that is reduced is the oxidising agent. The reactant that is oxidised is the reducing agent. • During oxidation, the oxidation number increases. During reduction, the oxidation number decreases. 38 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Revision exercises: Unit 3 – Types of reactions 1. Give one term for the following descriptions. 3. Gunpowder is a mixture of potassium nitrate, carbon and sulfur. When these substances a) Reactions where electrons are transferred react, large volumes of gas are produced from from one species to another a small mass of solid. b) The oxidation number (or oxidation state) of an element 2KNO3(s) 1 S(s) 1 3C(s) K2S(s) 1 c) This happens to the oxidation number of N2(g) 1 3CO2(g) an element during reduction a) Explain the term oxidation in terms of d) An ion that is present in a redox reaction electron transfer. but does not undergo a change in b) Explain, in terms of oxidation numbers, oxidation state why the above reaction is a redox reaction. e) Chemical change in which an element c) Which substance is oxidised? gains electrons d) Explain your answer in terms of its change f) The species that remains when a proton is in oxidation number. removed from the acid e) Which substance is reduced? g) Colourants that change their colour at f) Explain your answer in terms of its change different pH values in oxidation number. h) Can act as both an acid and base g) In the 15th century, potassium nitrate was i) Reactions in which H+ ions are transferred obtained from urine. Today this salt can be from one reactant to the other manufactured by neutralisation. j) An acid and base react to form a salt i) What is the meaning of the term ‘salt’? and water ii) Give the names of three pairs of compounds that could react to form 2. Copper reacts with silver nitrate solution. The potassium nitrate by neutralisation. reaction is described as a redox reaction. The balanced equation and ionic equation for the 4. Work out the net ionic equation for each of reaction are given below. the pairs of half-equations. Balanced equation: a) The reaction between chlorine gas and Cu 1 2AgNO3 2Ag 1 Cu(NO3)2 iodide ions Ionic equation: Cl2 1 2e− 2Cl− Cu 1 2Ag+ Cu2+ 1 2Ag 2I– I2 1 2e− a) What is a redox reaction? b) The reaction between copper and dilute b) Which ion does not undergo any change? nitric acid to give nitrogen monoxide c) Which substance is oxidised? Cu Cu2+ 1 2e− d) Write the reduction half-reaction. e) Which substance is formed as a result of NO3− 1 4H+ 1 3e− NO 1 2H2O oxidation? c) The reaction between copper and f) Two observations are made during the concentrated nitric acid to give nitrogen reaction. A shiny grey solid forms and the dioxide solution turns light blue. Which product is responsible for each observation? Cu Cu2+ 1 2e− NO3− 1 2H+ 1 e− NO2 1 H2O CHEMICAL CHANGE 39

d) The reaction between iron and chlorine gas b) What is the household name for NaOH? to give iron chloride c) Calculate the number of moles of hydrochloric acid used in the titration. Fe Fe3+ 1 3e− d) Write the balanced equation for the Cl2 1 2e– 2Cl− reaction. e) Aluminium bromide, AlBr3, which is made e) How many moles of sodium hydroxide by passing gaseous bromine over hot were required? aluminium f) What is the concentration of the sodium hydroxide? Al Al3+ 1 3e− Br2 1 2e− 2Br− 5. Determine the oxidation number of: a) S in H2SO4 b) P in HPO32 − c) P in H4P2O7 d) O in H2O2 e) Cr in Cr2O3 f) N in NO2− 6. Give the conjugate base for each of the following acids. a) HCl b) HCO− c) H2PO3− 7. Give the conjugate acid for each of the following bases. a) HSO4− b) ClO− c) OH− 8. Complete the following equations by writing the formulae for the missing acid and base that could form the given products. Finally balance the equation. a) metal hydroxide 1 acid H2O 1 CaCl2 b) metal oxide 1 acid H2O 1 MgSO4 c) _____ 1 _____ H2O 1 CO2 1 KNO3 d) metal oxide 1 acid H2O 1NaNO3 9. 25 cm3 of a sodium hydroxide (NaOH) solution was titrated with a standard solution of hydrochloric acid of concentration 0,125 mol·dm−3 using phenolphthalein indicator. It was found that exactly 14,5 cm3 of the standard solution was required to neutralise the NaOH solution. a) What is the role of the indicator in this reaction? 40 S T U DY & M A S T E R P H Y S I C A L S C I E N C E S S U P P O R T PA C K G R A D E 1 1

Unit 3: Answers 1. a) Redox reaction (reduction–oxidation 4. a) Cl2 1 2I− 2Cl− 1 I2 reaction) b) 3Cu 1 2NO3− 1 8H+ 3Cu2+ 1 2NO 1 b) Zero 4H2O c) Decreases c) Cu 1 2NO3− 1 4H+ Cu2+ 1 2NO2 1 2H2O d) Spectator ion d) 2Fe 1 3Cl2 2Fe3+ 1 6Cl− e) Reduction e) 2Al 1 3Br2 2Al3+ 1 6Br− f) Conjugate base g) Acid–base indicator 5. a) 16 d) −1 h) Ampholyte b) 13 e) 13 i) Acid–base reactions c) 15 f) 13 j) Neutralisation 6. a) Cl− 2. a) During a redox reaction electron transfer b) CO2− takes place between two species that react. c) HPO32 − b) Nitrate ion (NO3− ) c) Copper (Cu) 7. a) H2SO4 d) Ag+ 1 e− Ag b) HClO e) Copper ion (Cu2+) or Cu(NO3)2 c) H2O f) The shiny grey solid is due to the formation of silver metal. 8. a) Ca(OH)2 1 2HCl 2H2O 1 CaCl2 The blue solution is due to the formation b) MgO 1 H2SO4 H2O 1 MgSO4 of copper ions in solution (Cu2+). c) K2CO3 1 2HNO3 H2O 1 CO2 1 2KNO3 d) Na2O 1 2HNO3 H2O 1 2NaNO3 3. a) Oxidation is the loss of electron(s) by a substance. 9. a) The indicator was used to determine the b) Carbon shows an increase in oxidation end point of the titration. number and sulfur and nitrogen both show b) Caustic soda a decrease in oxidation number. c) n 5 cV c) Carbon (C) 5 0,125 mol·dm−3 3 0,0145 dm3 d) Oxidation number increases from 0 14 5 1,8125 3 10−3 mol (indicates loss of electrons) d) NaOH 1 HCl NaCl 1 H2O e) Sulfur (S); nitrogen (N) e) 1,8125 3 10−3 mol n f) Oxidation number for sulfur decreases f) c 5 V from 0 −2 ; oxidation number for 1,8125 3 10−3 mol 5 0,025 dm3 nitrogen decreases from 15 0 (these indicate the gaining of electrons) 5 0,0725 mol·dm−3 g) i) Salt is an ionic compound formed from a neutralisation reaction between an acid and a base. ii) K2CO3 and HNO3; K2O and HNO3; KOH and HNO3 CHEMICAL CHANGE 41