Physical Sciences Grade 12 Paper 2 KZN Prelim 2024 Memo

Downloaded from hlayiso.com Physical Sciences/P2 1 September 2024 Preparatory Examination NSC FINAL GRADE 12 NATIONAL SENIOR CERTIFICATE PHYSICAL SCIENCES P2 (CHEMISTRY) PREPARATORY EXAMINATION SEPTEMBER 2024 MARKING GUIDELINES MARKS: 150 These Marking Guidelines consist of 14 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 2 September 2024 Preparatory Examination NSC QUESTION 1 1.1 A  (2) 1.2 A  (2) 1.3 B  (2) 1.4 D  (2) 1.5 A  (2) 1.6 B  (2) 1.7 C  Do not mark (2) 1.8 B  (2) 1.9 D  (2) 1.10 C  (2) [20] QUESTION 2 2.1.1 5 – ethyl – 2,6 - dimethylhept – 3 – yne Marking criteria:  correct stem i.e. hept – 3 – yne   substituents correctly identified i.e. ethyl, dimethyl  IUPAC name completely correct including numbering, sequence and hyphen  (3) 2.1.2 2,3 – dibromo – 5 – methylheptane Marking criteria:  correct stem and substituents i.e. dibromo, methyl and heptane  IUPAC name completely correct including numbering, sequence and hyphen  (2) 2.2.1 O ACCEPT O │ ║ │ ║ ― C ― C ― C ― R ― C ― R‫׳‬ │ │  (2) 2.2.2 alcohols accept alkanols (1) Copyright reserved

Downloaded from hlayiso.com Physical Sciences/P2 3 September 2024 Preparatory Examination NSC 2.2.3 A or E (2) Has multiple bonds between atoms of carbon.  2.3 H O H H │ ║ │ │ H ― C ― C ― C ― C ― H │ │ │ H H H H H H O │ │ │ ║ H ― C ― C ― C ― C ― H │ │ │ H H H Marking criteria:  functional group of first isomer correctly drawn  functional groupof second isomer correctly drawn  Whole structure of first isomer correctly drawn (4)  Whole structure of second isomer correctly drawn  2.4 Functional isomers (1) [15] QUESTION 3 3.1 The pressure exerted by a vapour at equilibrium with its liquid in a closed system.  Marking criteria: If any one of the underlined key words/phrases in the correct context (vapour pressure) is omitted, deduct 1 mark. (2) 3.2 100 kPa At the boiling point(149°C), the vapour pressure equals the atmospheric pressure.  (2) 3.3 P.  P has a lower boiling point than R.  (2) 3.4 Chain Length /branching/surface area  (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 4 September 2024 Preparatory Examination NSC 3.5 Marking criteria:  Relate boiling point with length of carbon chain/branching/number of side chains/surface area.   Compare the strength of the intermolecular forces.   Compare the energy required to overcome the intermolecular forces.  P has the lowest boiling point  and therefore has the shortest carbon chain/most number of branches/smallest surface area over which the intermolecular forces act.  Weakest/least intermolecular forces/Van der Waals forces/London forces Least energy needed to overcome the intermolecular forces OR Q has the highest boiling point  and therefore has the longest carbon chain/least number of branches/largest surface area over which the intermolecular forces act.  Strongest/most intermolecular forces/Van der Waals forces/London forces Most energy needed to overcome the intermolecular forces (4) 3.6 H │ H ― C ― H H H │ │ H ― C ― C ― C ― O ― H │ │ H H H ― C ― H │ H Marking criteria:  Functional group on first carbon  3 carbons in the longest chain   2 – methyl groups on the second carbon  (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 5 September 2024 Preparatory Examination NSC 3.7.1 149 (°C)  (1) 3.7.2 At 100 kPa, the vapour pressure corresponds to the atmospheric pressure T represents the highest boiling point.  (3) [18] QUESTION 4: DO NOT MARK 4.1.1 Hydrogenation (1) 4.1.2 Requires a catalyst/Pt/Pd/Ni (1) 4.2.1 H H H H H │ │ │ │ │ H ― C ═ C ― C ― C ― C ― H │ │ │ H H H Marking criteria:  double bond on first carbon  Whole structure correct  (2) 4.2.2 substitution  hydrolysis (2) 4.2.3 Dilute strong (2) 4.3.1 H2O/water (1) 4.3.2 Sulphuric acid/H2SO4/phosphoric acid/H3PO4 (1) 4.3.3 Hydration (1) 4.4.1 Elimination/dehydrohalogenation (1) 4.4.2 C5H11Br + NaOH → C5H10 + NaBr + H2O LHS RHS Bal (3) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 6 September 2024 Preparatory Examination NSC QUESTION 5 5.1  Change in concentration  of products/reactants per (unit) time.   Change in amount/number of moles/volume/mass  of products/reactants per (unit) time.   Amount/number of moles/volume/mass of products formed/reactants used per (unit) time.   Rate of change in concentration/amount/number of moles/volume/ mass.  (2 or 0) (2) 5.2  Reaction is exothermic/Temperature increases.   At a higher temperature particles move faster/have higher average kinetic energy. OR   More molecules have enough/sufficient kinetic energy for an effective collision. OR  More molecules have kinetic energy/Ek equal to or greater than the activation energy.  More effective collisions per unit time/second.  OR  Frequency of effective collisions increases. Reaction rate increases. (3) 5.3 Marking criteria  Equation 4  1,8  Substitute in equation  3 -1  Final answer: 1,1 dm3.min-1 change in volume of CO2(g) rate = Δt  4 - 1,8 =  3 - 1 = 1,1 dm3.min-1 (3) 5.4 LESS THAN. Less reactant particles per unit volume/lower concentration of HCl/Smaller surface of CaCO3  Less effective collisions per unit time/Lower frequency of effective collisions. (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 7 September 2024 Preparatory Examination NSC 5.5 Marking criteria: V  Formula: n   to calculate n(CO2)produced Vm 5,8  Correct substitution ( ) in the above formula  25,7  Ratio: n(CaCO3) used equals n(CO2) produced   Substitution of n(CaCO3) to get n m  Use n  to calculate m(CaCO3)initial  M  Final answer = reaction I  Greater mass of impure CaCO3 is required to produce the same volume of CO2/22,57 g of impure CaCO3 will produce less CO2 V n(CO2)produced   Vm 5,8   25,7 = 0,22568 mol We can also calculate number of moles n(CaCO3) used = n(CO2) produced of pure CaCO3 = 0,22568 mol  n(CaCO3) = 0,2257 mol m Use the ratio n(CaCO3)  M m 0,22568   100 m(CaCO3) = 22,57 g reaction I  All of the CaCO3 reacted.  OR Greater mass of impure CaCO3 is required to produce the same volume of CO2/22,57 g of impure CaCO3 will produce less CO2 (7) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 8 September 2024 Preparatory Examination NSC 5.6 Marking criteria:  Gradient of G is smaller/less steep  Curve G produces a smaller volume of CO2(g)  Volume of CO2(g) F   G 0 time (s) NB: If graphs are not labelled = 0/2 (2) [20] QUESTION 6 6.1 Double arrow in the equation.  (1) 6.2 Marking criteria: If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The underlined phrase must be in the correct context. When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that will oppose the disturbance.  (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 9 September 2024 Preparatory Examination NSC 6.3 INCREASE.  According to Le Chatelier’s Principle when the pressure increases, the reaction that leads to a decrease in the number of moles will be favoured.  In this case the forward reaction is favoured/equilibrium position shifts to the right. (3) 6.4 EXOTHERMIC. Number of moles/mass of reactant decreased. Forward reaction favoured. A decrease in temperature favours the EXOTHERMIC REACTION. (4) 6.5 Marking criteria:  Initial quantities of all substances  Quantity of SO2 at equilibrium  Using the correct mol ratio  Calculating the quantity(mol) at equilibrium of O2 and SO3 substances   Divide number of moles at equilibrium by 2 dm3  Kc expression  Correct substitution of equilibrium concentrations into Kc expression   Kc = 11,25 SO2 O2 SO3 Initial quantity (mol) 0,6 0,5 0,4  Change (mol) 0,2 0,1 0,2  Quantity at equilibrium (mol) 0,4 0,4 0,6  Equilibrium concentration (mol.dm-3) 0,2 0,2 0,3  [SO 3 ] 2 Kc   No Kc expression, correct substitution. 7 [SO 2 ] 2 [O 2 ] 8 [0,3] 2  =  [0,2] 2 [0,2] Wrong Kc expression 6 8 = 11,25  (8) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 10 September 2024 Preparatory Examination NSC QUESTION 7 7.1.1 A substance that can act as an acid and a base.  (2) 7.1.2 + H2O ⇌ + H3O+ LHS RHS Balancing OR + H2O ⇌ H2CO3 + OH- (3) 7.2.1 Marking criteria:  Formula: pH = - log[H3O+]  Substitute c(H3O+) in the formula pH = - log[H3O+]  Final answer pH = - log[H3O+]  = - log(2x0,15)  = 0,52 (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 11 September 2024 Preparatory Examination NSC 7.2.2 Marking criteria:  Calculate n(H+) from H2SO4  Substitute for c and V in n = cV to calculate n(OH-) from NaOH  Substitute 12,96 into pH = - log[H3O+] to calculate [H+]  Substitute [H+] in [H3O+][OH-] = 1 x 10-14 to calculate [OH-] in excess  Subtract n(H+) from n(OH-)TOTAL and relate to n(OH-)EXCESS. Marks are awarded for substitution. LHS RHS  Final answer n(H+) from H2SO4 = cV x 2 = (0,15)(0,02) x 2  = 0,006 mols n(OH-) from NaOH = cV = (0,25)(0,03)  = 0,0075 mols pH = - log[H3O+] 12,96 = - log[H3O+]  [H3O+] = 1,096 x 10-3 mol.dm-3 [H3O+][OH- ] = 1 x 10-14 (1,096 x 10-3)[ OH- ] = 1 x 10-14  [OH- ] = 0,091 mol.dm-3 - n(OH )excess = cV = (50 + X) (0,091) 1000 n(OH-)excess = - n(OH )TOTAL - n(H+) from H2SO4 (50 + X) (0,1)(X) (0,091)  = (0,0075) + - 0,006  1000 1000 X = 27, 98 cm3  (8) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 12 September 2024 Preparatory Examination NSC QUESTION 8 8.1 Notes  Accept any other correct formula from the data sheet. Any other formula using unconventional abbreviations, e.g. E°cell = E°OA - E°RA followed by correct substitutions Max: 3 4 Eᶿcell = Eᶿreduction - Eᶿoxidation  0,03  = 0,80  - Eᶿoxidation Eᶿoxidation = 0,77 V  (4) 8.2.1 Fe2+ → Fe3+ + e-  (2) Notes  Ignore phases  Fe2+ ← Fe3+ + e- ( 0 ) Fe3+ + e- ⇌ Fe2+ (0 ) 2 2 Fe2+ ⇌ Fe3+ + e- (1 ) Fe3+ + e- ← Fe2+ 2 ( ) 2 2  Ignore if charge on electron omitted. If a charge of an ion is omitted eg. Fe3+ + e- → Fe2+ Max: ( 0 ) 2 8.2.2 Pt(s)/Fe2+(aq)(1 mol.dm-3)/Fe3+(aq)(1 mol.dm-3)  //Ag+(aq)(1 mol.dm-3)/Ag(s)  (3) NB: Award full marks even if the phases and concentrations are not shown 8.2.3 Temperature 25°C or 298 K  Concentration of electrolytes = 1 mol.dm-3  (2) 8.2.4 From chemical to electrical (1) 8.4. INCREASES (1) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 13 September 2024 Preparatory Examination NSC QUESTION 9 9.1 A substance of which the aqueous solution contains ions/ A substance that dissolves in water to give a solution that conducts electricity/A solution/ dissolved substance that conducts an electric current through the movement of ions.  (2) 9.2 M Cu2+(aq) + 2e- → Cu(s)  Ignore phases Notes  Cu ← Cu2+ + 2e- ( ) Cu2+ + 2e- ⇌ Cu (1 ) 2 Cu2+ + 2e- ← Cu ( ) Cu2+ + 2e- ⇌ Cu ( )  Ignore if charge on electron omitted.  If a charge of an ion is omitted eg. Cu + 2 e- ← Cu Max: ( 1 ) 2 (3) 9.3 Cu/copper Copper is being plated on the metal M (3) 9.4 Marking criteria:  Calculate gain in mass of the bar i.e. 17,85 - 12,52 and substitute 63,5 in m the formula: n   M  Substitute 6,02 x 1023   Ratio of number of mols of e to number of moles of Cu: 2 : 1  Substitute in Q = IΔt, Calculate charge Final answer 3 hours m : n  M 17,85  12,52 : n   63,5 5,33 n(e) = 2( ) 63,5 = 0,168 mols n(e) = n = nNA Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 14 September 2024 Preparatory Examination NSC = 0,168 x 6,02 x 1023  = 1,011 x 1023 Q = ne = (1,011 x 1023)(1,6 x 10-19)  = 16 176 C Q = IΔt 16 176 = 1,5(Δt) Δt = 10 784 s (5) X = 3 hr 9.5 Zn2+ is a weaker oxidising agent than Cu2+ and will not be reduced.  (2) OR Cu2+ is a stronger oxidising agent than Zn2+, Cu2+ will be reduced to Cu. [15] TOTAL: 150 NOTE: The paper should be marked out 133 and then converted to 150. Copyright reserved Please turn over