Physical Sciences Grade 12 IEB Paper 2 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2016 PHYSICAL SCIENCES: PAPER II MARKING GUIDELINES Time: 3 hours 200 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 2 of 9 QUESTION 1 MULTIPLE-CHOICE 1.1 D 1.2 B 1.3 C 1.4 D 1.5 B 1.6 A 1.7 C 1.8 A 1.9 D 1.10 B QUESTION 2 CHEMICAL BONDING 2.1 2.1.1 It is a bond occurring between atoms within molecules. 2.1.2 Polar/covalent/bond. 2.1.3 A (weak) force of attraction between molecules or between atoms of noble gases. 2.1.4 Hydrogen bond (Hydrogen bonding intermolecular force). 2.1.5 It is a (small) atom with a high electronegativity and it has at least one lone pair of electrons. 2.1.6 δ+ (allow positive; +; α+) 2.2 2.2.1 A transfer of electrons and subsequent electrostatic attraction. 2.2.2* • Large number of STRONG electrostatic forces (ionic bonds, coulombic forces) of attraction between ions in giant network structure. • Large amount of energy required to break/overcome these forces. 2.2.3 Ion-dipole force 2.2.4 Sodium ion (Na+). The water molecules are orientated in such a way that the (slightly) negative oxygen atoms (δ − ) are closest to the positive ion. c.o.e. from 2.1.6 Cℓ- The water molecules are orientated in such a way that the (slightly) positive oxygen atoms are closest to the positive ion. If 2.1.6 is wrong no mark is given in 2.2.4 for either Na+ or Cℓ- without justification. 2.2.5* The electrostatic forces (ionic bonds) holding the ions in the crystal lattice are much stronger than the ion-dipole forces between the ions and the water molecules. Large numbers of water molecules need to surround each ion in order to overcome the strong electrostatic forces and "pull" the ion out of the lattice. * Penalise wrong use of terminology (e.g. atoms instead of ions) once in total in Question 2.2.2 and Question 2.2.5 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 3 of 9 QUESTION 3 ENERGY CHANGE AND REACTION RATES 3.1 (Sulphur) is insoluble/solid/precipitate/murky/opaque/turbid and blocks the light. Don't give a mark for saying there is a decrease in light intensity as that is stated in the question. 3.2 3.2.2 Scale • If scale is wrong no marks can be awarded for plotting 180 points. • No marks for best-fit line which is 'dot-to-dot/line 160 segments' or 'sketched/hairy' but a wobbly curve is acceptable. 140 3.2.4 best-fit line 120 Time (s) 100 3.2.3 plot point 80 60 40 3.2.3 plot point 20 0 0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 3.2.1 label concentration mol dm–3 or M Graph 3.3 Time is the dependent variable OR time depends on concentration. No marks for stating 'time is measured'. 3.4 Straight line, which passes through the origin (0). Correct sketch of graph (one mark only). Ignore reference to positive or negative slope. 3.5 The reaction rate increases as there are more particles (of Na2S2O3) per unit volume therefore more effective (correct orientation and sufficient activation energy) collisions per unit time. IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 4 of 9 3.6 Place the flask containing the sodium thiosulphate solution on top of a large black cross drawn on a piece of white paper. Carry out step 2 of the original method but stop the watch when the cross is no longer visible when viewed through the solution from above. OR Measure change in mass/pressure/volume of gas collected/pH with time. m OR 3.7.1* nsulphur = 3.7 3.7.1 n = c.V M = 0,2 × 0,1 (conversion) 0,18 = 0,02 mol of HCℓ = 32 = 0,005625 mol 3.7.2 mol ratio HCℓ : S mol ratio HCℓ : S 2:1 2 :1 0,02 : 0,01 0,01 mol (2d.p.) 0,01125 : 0,005625 OR 3.7.2* If do alternative given above then m= n × M n = c.V IF one error = 0,01 × (32) = 0,2 × 0,1 (conversion) = 0,02 mol of HCℓ only (eg = 0,32 g 0 , 01125 wrong molar % yield = × 100 0 ,18 0 , 02 mass of S or % yield = × 100 (method) = 56,25% wrong ratio) 0 ,32 OR n = m/M = 0,18/32 = 0,005625 mol of S gets 3/5 = 56,25% 0 , 005625 % yield = × 100 (method)= 56,25% 0 , 01 QUESTION 4 CHEMICAL EQUILIBRIUM 4.1 As A2B decomposes in the forward reaction, the concentration of A2 and B2 (or product) increases OR more particles of A2and B2. 4.2 4.2.1 There has been an increase in concentration (or pressure) of both reactants and products. 4.2.2 Reverse 4.2.3 The reverse reaction is favoured since it leads to the formation of fewer moles of gas, which relieves the stress of high pressure (or lowers the pressure). (NO c.o.e. from 4.2.2) 4.3 Exothermic. • Stress: Decrease in temperature. • Response: Reverse reaction rate decreases more than forward reaction rate, therefore forward reaction favoured. • Reason: Forward reaction produces heat (exothermic) in order to relieve the stress (raise the temperature). 4.4 A catalyst is added. 4.5 No change in [A2B]. The rates of BOTH forward and reverse reactions have increased. EQUALLY or whilst remaining in equilibrium OR A2B being broken down and produced at the same rate. 4.6 4.6.1 No effect. 4.6.2 Increases. Must be consistent with 4.3 If state ENDO in 4.3 then must state decrease in 4.6.2 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 5 of 9 n 4.7 4.7.1 V= c 3, 6 = 1, 2 = 3 dm3 [A 2 ]2 ⋅ [B2 ] ( ) -1 mark 4.7.2 Kc = [A 2 B]2 4.7.3 [A2B] = n/V [B2] = n/V c.o.e. (5,1 − 3, 6) method 1,5 1,8 volume from = or = 3 3 3 Question 4.7.1 = 0,5 mol ⋅ dm–3 ÷ volume = 0,6 mol ⋅ dm–3 OR TABLE MOLE table 2A2B(g) 2A2(g) + B2(g) *Mole ratio 2 2 1 Moles at start 5,1 0 0 *Moles used/formed 3,6 3,6 1,8 Moles at equilibrium (5,1 – 3,6) = 1,5 3,6 1,8 Conc. at equilibrium (mol ⋅ dm–3) ÷ vol (1,5/3) = 0,5 1,2 1,8/3 = 0,6 (c = n/V) (V = 3 dm3) CONC table 2A2B(g) 2A2(g) + B2(g) *Mole ratio 2 2 1 Conc at start 5,1/3 = 1,7 0 0 *Conc change 1,2 1,2 0,6 Conc. at equilibrium (mol ⋅ dm–3) (1,7 – 1,2) = 0,5 1,2 0,6 (c = n/V) (V = 3 dm3) [A 2 ]2 ⋅ [B2 ] c.o.e. Kc = Kc from [A 2 B]2 Question 4.7.2 1, 22 × 0, 6 subst = 0,52 = 3,46 QUESTION 5 QUANTITATIVE CHEMISTRY AND ACIDS AND BASES 5.1 5.1.1 A solution of known concentration. 5.1.2 Error & correction (any TWO of the following) • Use of tap water. She should have used distilled water. • Use of a beaker. She should have used a volumetric flask. • Use of a glass rod to stir (as could transfer solute to rod). Mix by shaking the solution (in a sealed volumetric flask). • She did not weigh the filter paper before and after transferring the solute to the beaker in order to ascertain (by subtraction) exactly how much solute was transferred. She should have done this (implied). IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 6 of 9 • She added the solute to 250 cm3 of water, which would have resulted in a greater total volume than 250 cm3. She should have added water to the solute to make up the total volume to 250 cm3. 5.1.3 sodium carbonate (Na2CO3) 5.2 5.2.1 It is the point where an acid and base have reacted, so neither is in excess OR where equivalent (NOT same or equal) number of moles of acid and base have reacted. 5.2.2 Colour: Orange since more HIn is formed. Stress: Increase [H3O+] Response: Favour reverse reaction Reason: Reverse reaction uses up H3O+ or decreases [H3O+] therefore relieves stress. V 5.3 5.3.1 n= Vm = 0,56 ÷ 22,4 = 0,025 mol of CO2 (OR 0,03 mol to 2 dp) 5.3.2 Mol ratio HCℓ : CO2 2:1 0,05 : 0,025 (OR 0,06 : 0,03) n V= c 0 , 05 0,06 = 0,25 0 , 25 V = 0,2 dm3 of HCℓ = 0,24 dm3 5.4 5.4.1 CH3COOH + H2O H3O+ + CH3COO– Don't penalise single arrow. Reactants/Products. 5.4.2 GREATER THAN Nitric acid is a stronger acid than ethanoic acid therefore it will ionise more/produce more ions/donates H+ more readily OR stronger acids have higher Ka values. 5.4.3 Nitric acid (c.o.e.) (Allow ethanoic acid if they have identified it as the stronger acid in Question 5.4.2.) It would have a higher concentration of ions (more ions) as it is a stronger acid. 5.4.4 Ethanoic acid (Allow HNO3 if c.o.e. from Question 5.4.2.) It is a weaker acid, therefore it would have a lower concentration of H3O+ ions. Kw = [H3O+] i [OH–] = constant OR [OH-] inversely proportional to [H3O+] + – ∴ if [H3O ] is lower, then [OH ] must be higher. IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 7 of 9 5.5 5.5.1 A salt is a substance in which the hydrogen of an acid has been replaced by a cation. 5.5.2 pH=7,1 – 11 It is a weak base/basic salt/salt of a strong base and a weak acid as it causes an increase in [OH–]. 5.5.3 2CH3COOH + Na2O 2CH3COONa + H2O Reactants/Products/Balanced QUESTION 6 GALVANIC CELLS Diagram 6.1 Electrons Salt bridge V If identify Pb as cathode then (anode to cathode) Electrodes in solutions coe for rest of diagram, Voltmeter & wires Cu(NO3)2 as electrolyte in If show symbol for cell anode half-cell and electron instead on voltmeter give from Cu to Pb - Don't penalise + max of 4/6 if rest of Q (Anode half-cell) no cotton wool, correct ripples on Allow any lead salt or Copper cathode surface of Pb2+ solution. electrolytes. Lead nitrate or Pb(NO3)2(aq) State symbols not 6.2 Pb(s)|Pb2+ (aq, 1 mol ⋅ dm–3)||Cu2+ (aq, 1 mol ⋅ dm–3)|Cu(s) at 25º C required. Anode/Salt bridge/Cathode/Standard conditions NO c.o.e from 6.1 6.3 Eº cell = Eº cathode – Eº anode NO c.o.e from 6.1 = 0,34 – (–0,13) = 0,47 V 6.4 6.4.1 Completes circuit/allows ions through maintains electrical neutrality/ neutralises electrolyte/acts as an ion pump. 6.4.2 As a result of oxidation of the lead anode the concentration of positive lead ions (cations) in the lead half-cell increases therefore the negative iodide ions move into the lead half-cell in order to maintain electrical neutrality (maintain balance between positive and negative ions). 6.4.3 Increases emf. 6.5 6.5.1 Increases 6.5.2 No effect IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 8 of 9 QUESTION 7 ELECTROLYTIC CELLS 7.1 Electrical energy to chemical (potential) energy. 7.2 Aℓ3+ + 3e– Aℓ (–1 per error) –1 in total for double arrows 7.3 2O2– O2 + 4e– (–1 per error) C + 2O2– CO2 + 4e– (–1 per error) 7.4 The oxygen gas produced at the anode reacts with the graphite (carbon) to form carbon dioxide. C + O2 CO2 Reactants/Product 2Aℓ2O3 + 3C 3CO2 + 4Aℓ 7.5 Less coal is burnt to produce electricity therefore fewer CO2 (greenhouse gas) emissions OR conserving non-renewable fossil fuel OR less poisonous gas emissions (from burning coal to make electricity). (ONE reason only which links to the environment.) Do NOT accept reduces electricity/energy consumption without a link to the environment. 7.6 Cryolite is poisonous or toxic fluoride (PFC) emissions (ONE reason only) Cryolite is mined which damages environment. 7.7 Aℓ3+ ions are a weaker oxidising agent (have a more negative Eo) than H2O molecules, therefore H2O will be reduced at the cathode in preference to Aℓ3+ ions. 2H2O + 2e– H2 + 2OH– (–1 per error) QUESTION 8 ORGANIC 8.1 8.1.1 A compound containing only carbon and hydrogen atoms in which there is at least one double and/or triple bond between carbon atoms. 8.1.2 2-methylbut-2-ene Mark allocation: 2-methyl but -2- ene The first 2- is redundant and need not be shown. -1 if formatting error (gap between methyl and but- or no hyphens) 8.2 8.2.1 An atom or group of atoms that forms the centre of chemical activity in the molecule OR an atom or group of atoms that gives a compound its physical and chemical properties. 8.2.2 3 C's Structure correct Functional group correctly identified If give semi-structural or condensed structural formula –1 per question. 8.2.3 If give molecular formula then give 0 Ethyl & methyl correct IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 9 of 9 8.2.4 They have the same molecular formula (but different structural formulae) and different functional groups/homologous series. 8.2.5 Methanol 8.2.6 Ethyl methanoate 8.2.7 Propanoic acid 8.2.8 • Propanoic acid has hydrogen bonds between molecules. • Methyl ethanoate has dipole-dipole forces between molecules. • Hydrogen bonds are stronger than dipole-dipole forces. • More energy needed to overcome the stronger intermolecular forces. Do NOT give final mark if they say more energy …. to break bonds. Wrong format of formula -1 8.3 8.3.1 (a) CH3CH3 or CH3-CH3 2C 6H (structural or molecular) Gap between CH3 and CH3 -1 (a) and (b) penalise each error once only (b) CH3CH2OH CH3CH2 OH 8.3.2 (a) chloroethane chloro ethane Don't penalise gap in name if already (b) ethene eth ene penalised in 8.1.2 (c.o.e.) Don't penalise redundancy unless wrong 8.3.3 haloalkane (alkyl halide) eg take off a mark for 2-chloroethane and eth-2-ene 8.3.4 alkene 8.3.5 (a) substitution (free radical substitution/halogenation/chlorination) (b) substitution (hydrolysis) (c) elimination (dehydration) Mark first answer if more than one is given. (d) addition (hydrogenation) (e) combustion (redox/oxidation) Total: 200 marks IEB Copyright © 2016