Physical Sciences Grade 12 IEB Paper 2 2017 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2017 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 © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 2 of 12 QUESTION 1 MULTIPLE CHOICE 1.1 A 1.2 D 1.3 B 1.4 C 1.5 B 1.6 C 1.7 B 1.8 C 1.9 A 1.10 D QUESTION 2 CHEMICAL BONDING 2.1 A covalent bond is a sharing of at least one pair of electrons by two (non- metal) atoms. 2.2 A non-polar covalent bond is an EQUAL sharing of electrons. A polar covalent bond is an UNEQUAL sharing of electrons. 2.3 2.3.1 hydrogen OR chlorine 2.3.2 hydrogen chloride OR hydrogen fluoride 2.3.3 aluminium oxide 2.3.4 hydrogen fluoride 2.3.5 hydrogen chloride 2.3.6 argon 2.3.7 magnesium 2.3.8 graphite IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 3 of 12 QUESTION 3 ENERGY CHANGE 3.1.1 Activation energy is the MINIMUM energy required to start a chemical reaction. 3.1.2 A catalyst is a (chemical) substance that increases the rate of a chemical reaction but remains unchanged at the end of the reaction 3.2 EA = (242,6 + 92,4) = 335 kJmol–1 (-1 if no unit) 3.3 Error: There will NOT be more collisions taking place per second (per unit time). The number of EFFECTIVE (successful) collisions (per second) increases (since more of the molecules will have the required activation energy). 3.4 Platinum. It lowers the activation energy the most, (has the lowest activation energy) which means that more of the molecules will have the activation energy required for an effective collision, therefore there will be a greater number of effective collisions per unit time, therefore a faster reaction rate. or more product formed per unit time. Give mark for EITHER point NOT both. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 4 of 12 QUESTION 4 RATES OF REACTION 4.1 Various options each out of 3 marks … (apparatus measure time) Place the flask on a scale/weighing apparatus/digital balance and measure the loss in mass over a set time period. Attach a syringe/gas measuring apparatus to the flask and measure the volume/amount/how much (of) gas produced in a set time period. Collect the gas (by the downward displacement of water) using an inverted measuring cylinder (or a burette) filled with water and measure the volume/amount/how much (of) gas produced in a set time period. 4.2 The particle size/physical state (lump vs granules)/surface area of the calcium carbonate. 4.3 Flask X. The smaller the particle size, the greater the surface area exposed to the acid. There will be more effective (correct orientation and Ek ≥ Ea) collisions per unit time (frequency/chance) and therefore a faster reaction rate. No marks for flask W (Apply principle of negative marking.) 4.4 n = c.V (c = n/V) = 0,1 × 0,1 (conversion) = 0,01 mol of HCℓ Mol ratio HCℓ : CO2 2: 1 0,01 : 0,005 V = n.Vm = 0,005 × 22,4 = 0,11 dm3 of CO2 (or 112 cm3) Award marks for the following skills: Method to calculate mols of HCℓ (n = c.V) Unit conversion (cm3 to dm3) Applying mol ratio (HCℓ : CO2 = 2 : 1) Multiplying mols of CO2 × 22,4 dm3 Answer (0,11 dm3) 4.5 4.5.1 Equal to 4.5.2 There are more molecules /moles/quantity of acid to react with the excess calcium carbonate therefore producing more carbon dioxide. OR More (or bigger volume of) (molecules/moles/quantity) acid which is the limiting reactant/reagent. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 5 of 12 QUESTION 5 CHEMICAL EQUILIBRIUM 5.1 The reaction is in (dynamic chemical) equilibrium or the rates of the forward and reverse reactions are equal or the rate of formation of products is equal to the rate of formation of reactants. 5.2 1,6 Number of moles 1,2 H2(g) H2O(g) 0,8 CH4(g ) 0,4 CO(g ) Time t0 t1 t2 5.2.1 Graph levels out at t1 at 1,2 moles Between t0 and t1 line must be curved (decreasing gradient) as shown. (-1 if H2 not started at zero) 5.2.2 Graph levels out BEFORE t1 at 0,8 moles Between t0 and t1 line must be curved (decreasing gradient) as shown (–1) if not. (Allow c.o.e. from 5.2.1.) (-1 if not started at 1,2 mol) CO × H2  3 5.3 Kc = (–1 for use of round brackets) CH4  × H2O  Correct moles for reactants  0,4 / 2  × 1,2 / 2  3 Kc =  Correct moles for products (c.o.e. from 5.2.1 if  0,8 / 2  × 1/ 2  incorrect number of moles used for H2) = 0,22  Divide by volume 5.4 Low yield of products (c.o.e. from 5.3) (Answer MUST link to 5.3) 5.5 When an external stress is applied to a system in chemical equilibrium, the equilibrium point will change in such a way as to counteract the stress 5.6 Explanation using Le Chatelier’s Principle Decrease or low yield of hydrogen. High pressures favours the reverse reaction which produces less moles of gas and relieves the stress of high pressure or reduces the pressure). OR Explanation using rates Decrease or low yield of hydrogen. A high pressure increases the rates of both the forward and reverse reactions but the reverse rate increases more as it involves more gas particles, therefore favours the reverse reaction. 5.7 No change 5.8 Faster reaction rate. (Products produced faster.) Higher yield of products. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 6 of 12 QUESTION 6 EXPERIMENTAL SKILLS AND THE AUTO-IONISATION OF WATER 6.1 Kw = (1,71 × 10–7)2 = 2,92 × 10–14 6.2 ENDOTHERMIC since Kw or [H3O+] increases (or forward reaction is favoured) with an increase in temperature. Award marks for the following skills: Linking temperature change to correct shift in equilibrium (either increase temperature favours forward or decrease temperature favours reverse) No marks for EXOTHERMIC. 6.3 Graph to show the relationship between [H3O+] and the temperature (of water) (order of [H3O+] and temperature not important) OR Graph of [H3O+] vs temperature (of water) (order of [H3O+] and temperature IS important) Heading Label x-axis with unit 2,50 Suitable scale x-axis Must use > ½ blocks 1st point at 0 °C correct 2,25 Points at 30 °C and 50 °C both correct Smooth curve (6) 2,00 1,75 [H3O+] (x 10–7) (moldm–3) 1,50 1,25 1,00 0,75 0,50 0,25 0 5 10 15 20 25 30 35 40 45 50 0 o Temperature ( C) 6.4 [H3O+] at 45 oC = 2 (+ 0,05) × 10–7 moldm–3 (c.o.e. from graph) IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 7 of 12 If lines indicating where graph was read are not shown (−1) 6.5 It is not a straight line or it does not have a constant gradient (y/x ≠ k) or it is a curve or it is exponential. It does not pass through the origin. 6.6 6.6.1 Remains the same 6.6.2 Decreases 6.7 Decreases IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 8 of 12 QUESTION 7 ACIDS & BASES 7.1 An acid is a proton (H+) donor. 7.2 A concentrated acid has a large amount of acid per unit volume of water. (High proportion of acid to water.) A strong acid ionises (almost) completely in water 7.3 7.3.1 LiOH 7.3.2 H3PO4 7.3.3 HNO3 7.3.4 HCOOH 7.3.5 NaHCO3 7.4 H2SO4 + 2H2O  2H3O+ + SO42- Reactants Products Balanced OR H2SO4 + H2O  H3O+ + HSO4- 7.5 7.5.1 n = c.V (c=n/V) = 0,8 × 0,05 = 0,04 mol of HCℓ 7.5.2 n = c.V = 0,5 × 0,02 = 0,01 mol of NaOH Therefore 0,01 mol of HCℓ neutralised (excess). Mols of HCℓ that reacted with MgO = (0,04 – 0,01) (c.o.e from 7.5.1) = 0,03 mol of HCℓ Mol ratio: MgO : HCℓ 1 : 2 0,015 : 0,03 Mass of MgO m=n×M = 0,015 × 40,3 = 0,6045 g of MgO % of MgO in tablet = 0,6045/0,96 × 100 = 62,97% OR Assume 100% purity n= m/M = 0,96/40,3 = 0,02382134 mol of MgO But only 0,015 mol of MgO in tablet % of MgO in tablet = 0,015/0,02382134 × 100 = 62,97% IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 9 of 12 QUESTION 8 GALVANIC CELLS 8.1 8.1.1 The electrode where oxidation takes place. 8.1.2 A substance that donates electrons (electron donor) 8.2 Chemical (potential) energy to electrical energy. 8.3 Temperature = 25 oC (298 K) Concentration of electrolyte = 1 moldm–3 If give 1 atm pressure then give a maximum of 1 mark for either one or both of the other 2 conditions correct. 8.4 3X + 2Y3+ 3X2+ + 2Y Reactants Products Balanced (–1 for double arrow) 8.5 Y3+ 8.6 Z|Z+||X2+|X Anode Salt bridge Cathode Ignore balancing numbers. 8.7 X2+ 8.8 8.8.1 Z strongest 8.8.2 Y weakest 8.9 Cell C: EoCell = Eo cathode – Eo anode Cell A: EoCell A = EoY – EoX 0,84 = EoY – EoX (1) Cell B: EoCell B = EoX – EoZ 0,68 = EoX – EoZ (2) (1) + (2) 0,84 + 0,68 = EoY – EoX + EoX – EoZ 1,52 V = EoY – EoZ Cell C:  EoCell C = EoY – EoZ  EoCell C = 1,52 V IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 10 of 12 QUESTION 9 ELECTROLYTIC CELLS 9.1 In order to free the ions from the crystal lattice so that it would conduct electricity. 9.2 Good electrical conductor Inert/unreactive/not readily oxidised or reduced ANY 2 High melting point (OR solid at high temp) 9.3 9.3.1 2Cℓ–  Cℓ2 +2e– 9.3.2 Na+ + e–  Na 9.3.3 2H2O + 2e–  H2 + 2OH– 9.4 Na+ 9.5 NaOH or sodium hydroxide or caustic soda 9.6 In the dilute solution there is relatively more water. Either of these points Water is a stronger reducing agent than chloride ions. Water is oxidised to oxygen. 2H2O  O2 + 4H+ + 4e– No marks for showing oxidation of OH− ions / H2O2 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 11 of 12 QUESTION 10 ORGANIC CHEMISTRY (1) 10.1 10.1.1 Esters 10.1.2 butyl ethanoate 10.1.3 ethanoic acid c.o.e. from 10.1.2 (must make link) 10.1.4 C6H12O2 + 8O2  6CO2 + 6H2O Reactants Products Balancing 10.1.5 Isomers are compounds having the same molecular formula but different structural formulae. 10.1.6 (a) ethyl butanoate or propyl propanoate or methyl pentanoate or pentyl methanoate (b) hexanoic acid hex anoic acid no c.o.e. from 10.1.2 10.2 10.2.1 substitution (hydrolysis) 10.2.2 CH3CHCℓCH3 + NaOH  CH3CHOHCH3 + NaCℓ (structural or molecular formulae –1) Award marks for the following: 2-chloropropane 3C's rest of formula correct Propan-2-ol OH group (knowing it's an alcohol) rest of formula NaCℓ 10.2.3 H H C=C–C–H C = C rest of formula correct H H H -1 if use condensed structural formula IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 12 of 12 QUESTION 11 ORGANIC CHEMISTRY (2) 11.1 Cracking 11.2 A saturated hydrocarbon is a compound containing only carbon and hydrogen atoms in which all of the bonds between carbon atoms are single bonds. 11.3 Alkanes 11.4 The hydrocarbons have induced dipole forces or London forces between their molecules. Water has hydrogen bonds between its molecules. Hydrogen bonds are STRONGER than London forces or the intermolecular forces of the hydrocarbons are not similar in STRENGTH to those of water. (either point for 1 mark) therefore insoluble. OR Dipole induced-dipole forces between the hydrocarbon molecules and the water solvent are weaker than the hydrogen bonds between the water molecules therefore insoluble. 11.5 11.5.1 Ethene 11.5.2 Addition (halogenation, bromination) 11.6 2,2–dimethylbutane 2,2 – dimethyl but ane 11.7 11.7.1 2 molecules of X = (12C + 26H) – (4C + 10H) – (2C + 4H) (OR C12H26 – C4H10 – C2H4) = 6C + 12H or C6H12 Award marks for the following (÷ 2) skills: 1 molecule of X = 3C and 6H Subtraction (from 12C + 26H or  Compound X = C3H6 C12H26 ) to get C6H12 (award subtraction mark if candidate OR C12H26 – C4H10 arrived at C6H12 without explicitly – C2H4 showing the subtraction) C6H12 ÷ 2 = C3H6 Dividing by 2 Correct final answer (C3H6) 11.7.2 All hydrocarbons have induced dipole forces or London forces between their molecules. The larger kerosene molecules have a greater interactive surface and have a greater electron density. Bigger temporary dipoles are set up or easier dispersion of electrons therefore longer lasting dipoles. These factors result in STRONGER London forces. More energy is needed to overcome the stronger intermolecular forces (NOT bonds) or to separate the molecules therefore kerosene is a liquid at room temperature (higher b.pt.). Total: 200 marks IEB Copyright © 2017