Physical Sciences Grade 12 P2 NSC November Mock 2026 Memo

hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: CHEMISTRY (P2) MOCK EXAMINATION 2026 MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 10 pages. Copyright reserved Please turn over

Physical Sciences/P2 2 Hlayiso/Mock 2026 NSC – Marking Guidelines GENERAL MARKING GUIDELINES 1. Calculations: Marks are awarded for the correct formula, correct substitution and the correct final answer WITH the correct unit, as indicated by the ticks (✓). A formula is not awarded a mark if it is not followed by a substitution. If an incorrect formula is used, no further marks are awarded for that calculation. 2. Units: The final-answer mark is only awarded if the correct unit is given. Penalise an omitted or incorrect unit only ONCE per question. 3. Rounding: Accept final answers rounded to a minimum of TWO decimal places. Accept small differences caused by rounding of intermediate values. 4. Positive marking: Where an answer to a previous question is used in a later calculation, the later calculation is marked positively. 5. Negative marking: An incorrect answer that is based on a conceptual mistake cannot be correctly motivated. If the choice (e.g. INCREASES/DECREASES) is incorrect, no marks are awarded for the explanation. 6. Definitions: Where key words are underlined, deduct 1 mark for EACH key word or phrase that is omitted or used in an incorrect context. 7. Equations: Ignore phases. Ignore double arrows (⇌) unless stated otherwise. Unbalanced equation: maximum 2/3. Incorrect reactants or products: 0 for that part. 8. Organic names and formulae: Names must follow IUPAC conventions (hyphens, commas, alphabetical order of substituents). Condensed or semi-structural formulae are NOT accepted where STRUCTURAL formulae are asked: maximum 1/2 (or 2/3 in equations). Copyright reserved Please turn over

Physical Sciences/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 1 1.1 C ✓✓ (2) 1.2 B ✓✓ (2) 1.3 A ✓✓ (2) 1.4 D ✓✓ (2) 1.5 B ✓✓ (2) 1.6 A ✓✓ (2) 1.7 C ✓✓ (2) 1.8 D ✓✓ (2) 1.9 A ✓✓ (2) 1.10 B ✓✓ (2) [20] Copyright reserved Please turn over

Physical Sciences/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1.1 C✓ (1) 2.1.2 F✓ (1) 2.1.3 A and D ✓ (both required) (1) 2.2.1 3-chloro-2,4-dimethylhexane Marking criteria • Correct stem, i.e. hexane ✓ • All substituents (chloro and dimethyl) correctly identified ✓ • IUPAC name completely correct, including numbering, hyphens, commas and alphabetical order ✓ (3) 2.2.2 2,3-dimethylpent-1-ene Marking criteria • Correct stem and position of double bond, i.e. pent-1-ene ✓ • Substituents (dimethyl) correctly identified ✓ • IUPAC name completely correct, including numbering, hyphens and commas ✓ (3) 2.2.3 Pentan-3-one ✓✓ Marking: pentanone ✓; position 3 ✓. Accept: 3-pentanone (max 1/2). (2) 2.3.1 Esterification / condensation ✓ (1) 2.3.2 (Concentrated) sulphuric acid / H2 SO4 ✓ (1) 2.3.3 Ethanol ✓ (1) 2.3.4 H H H O H H H C C C C O C C H H H H H H Marking criteria • Functional group (ester group –COO–) correct ✓ • Whole structure correct ✓ Condensed formula (CH3 CH2 CH2 COOCH2 CH3 ): max 1/2. (2) 2.3.5 Ethyl butanoate Marking: ethyl ✓ butanoate ✓ (2) 2.4 Compounds with the same molecular formula, but different functional groups (belonging to different homologous series). ✓✓ (2) [20] Copyright reserved Please turn over

Physical Sciences/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 3.1 The temperature at which the vapour pressure of a substance equals atmospheric (external) pressure. ✓✓ (2) 3.2 The compounds have comparable (similar) molar masses. ✓ (1) 3.3 A✓ (1) 3.4 Both C and D have hydrogen bonds (and London forces) between their molecules. D (propanoic acid) has TWO sites for hydrogen bonding / forms dimers / has more extensive hydrogen bonding than C. ✓ The intermolecular forces in D are stronger than in C. ✓ More energy is needed to overcome the intermolecular forces in D than in C. ✓ (3) 3.5 Between pentane (A) molecules there are London forces only. ✓ Between butanal (B) molecules there are dipole-dipole forces (in addition to London forces). ✓ The intermolecular forces in B are stronger than in A. ✓ More energy is needed to overcome the intermolecular forces in B. ✓ OR: the reverse argument for A. (4) 3.6 LOWER THAN ✓ 2,2-dimethylpropane is more branched (more compact/spherical), so it has a smaller surface (contact) area and weaker London forces between molecules. ✓ (2) [13] Copyright reserved Please turn over

Physical Sciences/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 Addition / hydrohalogenation / hydrobromination ✓ (1) 4.1.2 Substitution / hydrolysis ✓ (1) 4.1.3 Elimination / dehydration ✓ (1) 4.2 H Br H H H C C C C H H H H H Marking: functional group (Br on C2) ✓; whole structure correct ✓. Condensed formula: max 1/2. (2) 4.3.1 Dehydrohalogenation / dehydrobromination ✓ (1) 4.3.2 Concentrated (strong) base, e.g. NaOH or KOH, dissolved in ethanol ✓ Strong heating (heat under reflux) ✓ (2) 4.3.3 But-2-ene ✓✓ Marking: butene ✓; position 2 ✓. (2) 4.4 H H H H H O H H C C C C H + H O H → H C C C C H H H H H H H H Marking criteria Reactants ✓ Product ✓ Balancing ✓ Accept H2 O as molecular formula. Ignore the catalyst (H3 PO4 /H2 SO4 ) if written above the arrow. Condensed formulae: max 2/3. (3) 4.5 C4 H8 + 6O2 → 4CO2 + 4H2 O Marking: reactants ✓ products ✓ balancing ✓ (3) 4.6 Propene ✓ (1) [17] Copyright reserved Please turn over

Physical Sciences/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 Change in concentration of reactants or products per unit time. ✓✓ OR Change in amount/number of moles/volume/mass of reactants or products per (unit) time. ✓✓ OR Rate of change in concentration/amount/volume/mass. ✓✓ (2) 5.2.1 Concentration of HCℓ (the acid) ✓ (1) 5.2.2 ANY ONE: Temperature / mass of CaCO3 / state of division (surface area) of CaCO3 / volume of HCℓ ✓ (1) 5.3 Powdered CaCO3 (experiment 2) has a larger surface area / more particles are exposed to the acid. ✓ More collisions take place (more particles collide with the correct orientation and sufficient energy). ✓ More effective collisions per unit time / higher frequency of effective collisions. ✓ (3) 5.4 At a higher temperature the (average) kinetic energy of the particles is higher. ✓ More particles have kinetic energy equal to or greater than the activation energy. ✓ More effective collisions per unit time / higher frequency of effective collisions. ✓ (3) 5.5 n(CO2 ) = rate × ∆t = (1,5 × 10–4 )(60) = 9 × 10–3 mol ✓ n(CaCO3 )reacted = n(CO2 ) = 9 × 10–3 mol (ratio 1 : 1) ✓ m = nM = (9 × 10–3 )(100) = 0,90 g ✓ mremaining = 2 – 0,90 = 1,10 g ✓ (4) 5.6 Volume of CO2 3 1 Time Marking criteria • Both graphs start at the origin and level off at the SAME final volume (CaCO3 is the limiting reactant) ✓ • Graph 3 has a steeper initial gradient than graph 1 ✓ • Graph 3 levels off before graph 1 ✓ (3) [17] Copyright reserved Please turn over

Physical Sciences/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 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) 6.2 Marking criteria • USING ratio CO : H2 : CH3 OH = 1 : 2 : 1 ✓ • Equilibrium amount of CO = initial – change (0,5 mol) ✓ • Equilibrium amount of H2 = initial – change (0,6 mol) ✓ • Equilibrium amount of CH3 OH = 0,3 mol ✓ • Divide by 2 dm3 ✓ • Correct Kc expression (formulae in square brackets) ✓ • Substitution of equilibrium concentrations into Kc expression ✓ • Final answer: 6,67 ✓ CO H2 CH3 OH Initial amount (mol) 0,8 1,2 0 Change in amount (mol) –0,3 –0,6 +0,3 ratio ✓ Equilibrium amount (mol) 0,5 ✓ 0,6 ✓ 0,3 ✓ Equilibrium concentration (mol·dm–3 ) 0,25 0,3 0,15 ÷2✓ [CH3 OH] 0,15 Kc = ✓ = ✓ = 6,67 ✓ [CO][H2 ]2 (0,25)(0,3)2 No Kc expression, correct substitution: max 7/8. Wrong Kc expression: max 5/8 (no marks for substitution and answer). (8) 6.3.1 DECREASES ✓ (1) 6.3.2 The forward reaction is exothermic (the reverse reaction is endothermic). ✓ An increase in temperature favours the endothermic (reverse) reaction. ✓ The concentration of CH3 OH decreases and the concentrations of CO and H2 increase (so Kc decreases). ✓ (3) 6.4.1 INCREASES ✓ (1) 6.4.2 An increase in pressure favours the reaction that produces the smaller number of moles (volume) of gas. ✓ The forward reaction produces fewer moles of gas (3 mol of reactant gas → 1 mol of product gas). ✓ The forward reaction is favoured. ✓ (3) 6.5 REMAINS THE SAME ✓ A catalyst increases the rates of the forward and reverse reactions equally (it does not change the position of the equilibrium). ✓ (2) [20] Copyright reserved Please turn over

Physical Sciences/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1 A base is a proton (H+ ion) acceptor. ✓✓ (2) 7.2.1 ACIDIC ✓ (1) 7.2.2 NH+4 (aq) + H2 O(ℓ) ⇌ NH3 (aq) + H3 O+ (aq) Marking: reactants ✓ products ✓ balancing ✓. Ignore phases; accept →. Accept: NH4 Cℓ + H2 O ⇌ NH3 + H3 O+ + Cℓ– (3/3). (3) 7.3.1 OPTION 1 m 3,2 n(NaOH) = = = 0,08 mol ✓ M 40 n 0,08 c= = = 0,32 mol·dm–3 ✓ V 0,25 [OH– ] = 0,32 mol·dm–3 [H3 O+ ][OH– ] = 1 × 10–14 ⇒ [H3 O+ ](0,32) = 1 × 10–14 ✓ [H3 O+ ] = 3,13 × 10–14 mol·dm–3 pH = – log(3,13 × 10–14 ) ✓ = 13,51 ✓ OPTION 2 m 3,2 c= = ✓✓ = 0,32 mol·dm–3 pOH = – log(0,32) = 0,49 ✓ MV (40)(0,25) pH = 14 – 0,49 ✓ = 13,51 ✓ (5) 7.3.2 Marking criteria • Formula n = cV / substitution for n(NaOH) ✓ • Substitution for n(H2 SO4 )initial ✓ • USING ratio n(H2 SO4 ) : n(NaOH) = 1 : 2 ✓ • n(H2 SO4 )excess = initial – reacted ✓ • n(H3 O+ ) = 2 × n(H2 SO4 )excess ✓ • Divide by total volume 0,055 dm3 ✓ • Formula pH = – log[H3 O+ ] with substitution ✓ • Final answer: 1,74 ✓ n(NaOH) = cV = (0,32)(0,025) = 8 × 10–3 mol ✓ n(H2 SO4 )initial = (0,15)(0,030) = 4,5 × 10–3 mol ✓ n(H2 SO4 )reacted = 12 (8 × 10–3 ) = 4 × 10–3 mol ✓ n(H2 SO4 )excess = 4,5 × 10–3 – 4 × 10–3 = 5 × 10–4 mol ✓ n(H3 O+ ) = 2(5 × 10–4 ) = 1 × 10–3 mol ✓ 1 × 10–3 [H3 O+ ] = ✓ = 0,018 mol·dm–3 0,055 pH = – log(0,0182) ✓ = 1,74 ✓ POSITIVE MARKING FROM 7.3.1 (concentration of NaOH). (8) [19] Copyright reserved Please turn over

Physical Sciences/P2 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 Temperature: 25 ◦ C (298 K) ✓ Concentration of the electrolytes: 1 mol·dm–3 ✓ (2) 8.2 Ensures electrical neutrality in the half-cells / completes the circuit (allows ions to move between the half-cells). ✓ (1) 8.3 Ni ✓ (1) 8.4 Ni(s) + 2Ag+ (aq) → Ni2+ (aq) + 2Ag(s) Marking: reactants ✓ products ✓ balancing ✓. Ni + 2Ag+ ⇌ Ni2+ + 2Ag: 2/3 Ni2+ + 2Ag → Ni + 2Ag+ : 0/3 Ignore phases. (3) 8.5 Eθcell = Eθcathode – Eθanode ✓ Eθcell = 0,80 – (–0,27) ✓✓ Eθcell = 1,07 V ✓ Any other formula using unconventional abbreviations, e.g. Ecell = EOA – ERA , followed by correct substitution: 3/4. (4) 8.6 INCREASES ✓ (1) [12] QUESTION 9 9.1 The chemical process in which electrical energy is converted to chemical energy. ✓✓ OR The use of electrical energy to produce a chemical change (a non-spontaneous reaction). ✓✓ (2) 9.2 Cu2+ (aq) + 2e– → Cu(s) ✓✓ Cu2+ + 2e– ⇌ Cu: 1/2 Cu → Cu2+ + 2e– : 0/2 Cu ← Cu2+ + 2e– : 2/2 (2) 9.3 Chlorine (gas) / Cℓ2 ✓ (1) m 0,635 9.4 n(Cu) = = = 0,01 mol ✓ M 63,5 n(Cℓ2 ) = n(Cu) = 0,01 mol (2 mol e– per mol of each) ✓ V = nVm = (0,01)(22,4) ✓ V = 0,22 dm3 (0,224 dm3 ) ✓ (4) 9.5 REMAINS THE SAME ✓ Cu is a stronger reducing agent than Cℓ– ions (and H2 O), so the copper anode is oxidised to Cu2+ ions. ✓ Cu2+ ions go into solution at the anode at the same rate as Cu2+ ions are reduced to Cu at the cathode. ✓ (3) [12] TOTAL: 150 Copyright reserved