Physical Sciences Grade 12 IEB Paper 2 2018 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2018 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 © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 2 of 13 QUESTION 1 MULTIPLE CHOICE 1.1 B 1.2 A 1.3 B 1.4 C 1.5 C 1.6 D 1.7 C 1.8 A 1.9 D 1.10 B QUESTION 2 2.1 Activated (Transition State) Complex Potential Energy Reactants (products lower (OR Cr + HCℓ) than reactants) Reaction Coordinate/ Course of Reaction IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 3 of 13 2.2 2.2.1 4,5 Volume of H2 gas (dm3) 3,0 1,5 Graph must have: Shallower gradient Same end volume 0 0 20 40 60 80 Time (s) 2.2.2 4,5 Volume of H2 gas (dm3) 3,0 1,5 Graph must have: Steeper gradient End volume at approximately 2,25 dm3 0 0 20 40 60 80 Time (s) 2.3 • The reaction could be done using a colorimeter/light meter. • which would measure the colour intensity of the green solution as a measure of concentration of chromium chloride (Cr3+) • at specific time intervals OR over a period of time. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 4 of 13 V 2.4 2.4.1 n= Vm (3) n= ( 22,4 ) n = 0,134 mol m ( 6,0 ) 2.4.2 • nCr = = = 0,1154 mol M ( 52 ) 3 • nH2 = (0,1154) × = 0,1731 mol 2 actual yield ( 0,134 ) • % Yield = × 100 = × 100 theoretical yield ( 0,1731) • % Yield = 77,41% Alternative: V(H2 ) = 0,1731 × 22,4 = 3,877 dm3 3 % Yield = × 100 3,877 = 77,37% = ∆V = (3 − 0) 2.4.3 Average Rate = 0,075 dm3 ⋅s−1 ∆t ( 40 ) 2.5 • Correct orientation of the colliding reactant particles • The sum of the kinetic energy (of the reacting particles) is greater than/equal to the activation energy. 2.6 • A higher concentration means that there is a greater number of particles per unit volume. • This increases the number of collisions that occur per unit time. • This therefore leads to an increase in the number of effective collisions per unit time, • leading to a higher reaction rate. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 5 of 13 2.7 2.7.1 The sharing of at least one pair of electrons by two atoms. 2.7.2 A measure of the tendency of an atom to attract a bonding pair of electrons. 2.7.3 • The difference in electronegativity between hydrogen and chlorine is greater than zero. • This results in unequal sharing of electrons, i.e. a polar covalent bond. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 6 of 13 QUESTION 3 3.1 5 minutes 3.2 2SO3  2SO2 + O2 3.3 When an external stress (change in pressure, temperature or concentration) is applied to a system in chemical equilibrium, the equilibrium point will change in such a way as to counteract the stress. 3.4 • Stress: increase in concentration of O2 • Le Châtelier's principle predicts the system will respond in order to decrease the concentration of O2 • Therefore, the forward reaction is (initially) favoured as it consumes O2 • Decreasing the amount of SO2 3.5 3.5.1 Forward 3.5.2 Exothermic n 3.6 3.6.1 All the concentrations decrease (since c = and V has increased). v 3.6.2 • From the rate equations we see that the rate of the forward reaction is proportional to cube of concentration (OR forward 1 rate 𝛼𝛼 3 ) whereas the rate of the reverse reaction is V proportional to the square of concentration (OR reverse 1 rate α 2 ) . V • Therefore the change in pressure has a greater effect on the forward reaction rate than the reverse. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 7 of 13 QUESTION 4 4.1 Increases 4.2 Reverse 4.3 Turns green CuC 24−  4.4 4.4.1 Kc = Cu (H2O )2+  C −  4  6   4.4.2 Concentrations: Cu (H2O )6 2+ 2− Reaction + 4C − ⇌ CuC 4 + 6H2O Initial 0 0 2 concentration Change in +1,1 +4,4 −1,1 concentration Equilibrium 1,1 4,4 0,9 concentration OR Moles: Cu (H2O )6 2+ 2− Reaction + 4C − ⇌ CuC 4 + 6H2O Initial moles 0 0 4 Change in +2,2 +8,8 −2,2 moles Equilibrium 2,2 8,8 1,8 moles Equilibrium 1,1 4,4 0,9 concentration Then: CuC 24−  Kc = Cu (H2O )2+  C −  4  6    Kc = ( 0,9 ) (1,1)( 4,4 ) 4 −3 Kc = 2,18 × 10 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 8 of 13 QUESTION 5 5.1 A proton acceptor. 5.2 It dissociates (almost) completely in water/solution. 5.3 KOH (OR LiOH/RbOH/CsOH) H2O 5.4 NaOH  Na+ + OH− 5.5 • The NaOH solution is more conductive. • NaOH is strong whilst butanoic acid is weak (OR NaOH dissociates completely, whilst butanoic acid ionises partially) • It forms a higher concentration of ions in solution (at equal concentrations). 5.6 H H H O H C C C C O H H H H 5.7 methyl propanoate OR ethyl ethanoate OR propyl methanoate IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 9 of 13 QUESTION 6 6.1 Increases Kw = �H3 O �[OH− ] + 6.2 −14 −5 − �10 � = �6,31 × 10 � [OH ] [OH− ] = 1,58 × 10−10 mol⋅dm−3 6.3 6.3.1 The point where an acid and base have reacted so neither is in excess. 6.3.2 nacid = 0,0165 mol n 0,0165 6.3.3 Vacid = = 0,21 c Vacid = 0,07857 dm3 Vacid = 78,57 cm3 6.4 6.4.1 A reaction with water where water itself is decomposed. 6.4.2 There is an increase in concentration of OH−. OR The production of OH− ions results in a decrease in the concentration of H3O+ (due to the upsetting of the equilibrium involving H3O+ and OH− ions in water) 6.5 • This statement is incorrect • pH depends on the concentration of hydronium ions (or hydroxide ions) • which is determined by both strength and concentration of the base IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 10 of 13 QUESTION 7 7.1 7.1.1 The substance that accepts electrons. 7.1.2 • Cℓ2 is a stronger oxidising agent (than Ni2+). • Therefore Cℓ2 is more likely to be reduced and thus accept electrons. 7.2 7.2.1 The electrode where oxidation takes place. 7.2.2 Ni OR nickel 7.3 Pt OR platinum 7.4 Ni | Ni2+(1 mol·dm−3) || Cℓ2(1 atm) | Cℓ−(1 mol·dm−3) | Pt at 25 °C 7.5 7.5.1 Ni  Ni2+ + 2e− 7.5.2 • The Ni electrode corrodes (loses mass). • The green colour of the Ni2+ electrolyte intensifies. 7.5.3 To complete the circuit. 7.5.4 • Ni2+ ions are being produced in this half-cell. • Anions enter the electrolyte from the salt bridge. • and Ni2+ cations exit the electrolyte into the salt bridge. 7.6 Increases IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 11 of 13 QUESTION 8 8.1 It is light. 8.2 It protects itself from corrosion. 8.3 Anode 8.4 Aℓ3+ + 3e−  Aℓ 8.5 8.5.1 • Less energy (OR electricity) is used to melt the electrolyte. • This saves money (cost is less). 8.5.2 • The cryolite is also electrolysed/decomposed. • This produces perfluorocarbons (PFCs) which are greenhouse gases 8.6 • O2 is produced at the anode. • This reacts with the carbon electrode (to produce CO2), causing the electrode to corrode (OR the carbon anode is continually oxidised) • C + O2  CO2 OR 2O2− + C  CO2 + 4e− 8.7 In order to melt aluminium oxide, • Many • Strong • ionic (electrostatic) forces/bonds need to be broken/overcome, • which requires much energy to separate the ions from the crystal lattice. 8.8 8.8.1 AℓCℓ3 + 3 Na → Aℓ + 3 NaCℓ 8.8.2 • nNa =3 × nA =3 × ( 7,56 ) =22,68 mol • m= Na = nM ( 22,68 )( 23 = ) 521,64 g IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 12 of 13 QUESTION 9 9.1 9.1.1 3−bromo−2,4−dimethylhexane 9.1.2 CH3CH2CH(CH3)CHBrCH(CH3)CH3 OR Br CH3 CH3 CH CH CH CH3 CH2 CH3 9.2 9.2.1 Alkenes 9.2.2 9.3 9.3.1 Hydroxyl group 9.3.2 Molecules with the same molecular formula but different positions of their same functional group OR same substituents. 9.3.3 Any one of the following: H H H H H H H H H H H C C C C C O H OR H C C C C C H H H H H H H H O H H H 9.4 9.4.1 A weak force of attraction between molecules or between atoms of noble gases. 9.4.2 • Compound B has London forces (only) • Compound C has hydrogen bonding • Hydrogen bonds are stronger than London forces (OR compound C has stronger intermolecular forces) • It is more difficult for the particles in compound C to flow past one another • resulting in compound C having a higher viscosity 9.5 C7H12 + 10O2  7CO2 + 6H2O IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 13 of 13 QUESTION 10 10.1 A compound/molecule containing only carbon and hydrogen atoms. 10.2 Pentane 10.3 10.3.1 Substitution 10.3.2 Addition 10.3.3 Condensation 10.4 Hydrohalogenation 10.5 Pent−1−ene 10.6 Unsaturated 10.7 Pentan−1−ol 10.8 H O H H H H H H C C O C C C C C H H H H H H H 10.9 2 Total: 200 marks IEB Copyright © 2018 PLEASE TURN OVER