Physical Sciences Grade 12 P2 2026 Memo (Eng)

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

Physical Sciences/P2 2 Hlayiso/Prediction 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 (consistent accuracy). 5. Negative marking: If the choice (e.g. INCREASES or DECREASES; EXOTHERMIC or ENDOTHERMIC) is incorrect, no marks are awarded for the explanation that follows. 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 as indicated). Copyright reserved Please turn over

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

Physical Sciences/P2 4 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 2 2.1.1 A✓ (1) 2.1.2 C✓ (1) 2.1.3 E and F ✓ (both required) (1) 2.2.1 2,4-dimethylhex-3-ene Marking criteria • Correct stem and position of the double bond, i.e. hex-3-ene ✓ • Substituents (dimethyl) correctly identified ✓ • IUPAC name completely correct, including numbering, hyphens and commas ✓ Accept: 2,4-dimethyl-3-hexene. (3) 2.2.2 3-bromo-2,5-dimethylhexane Marking criteria • Correct stem, i.e. hexane ✓ • All substituents (bromo and dimethyl) correctly identified ✓ • IUPAC name completely correct, including numbering, hyphens, commas and alphabetical order ✓ (3) 2.2.3 3-methylbutan-2-ol Marking: methylbutanol (correct stem and functional group) ✓; whole name completely correct (positions 3 and 2) ✓. Accept: 3-methyl-2-butanol. (2) 2.3 ANY ONE: H H H H H C C C C H H H O H C H H H 3-methylbutan-1-ol OR 2-methylbutan-2-ol OR 2-methylbutan-1-ol (drawn as structural formulae) Marking criteria • Functional group (–OH) on a DIFFERENT carbon atom of the same (branched) carbon chain ✓ • Whole structure correct ✓ Condensed or semi-structural formula: max 1/2. A straight-chain alcohol (e.g. pentan-2-ol) is a chain isomer: 0/2. (2) 2.4.1 Esterification / condensation ✓ (1) Copyright reserved Please turn over

Physical Sciences/P2 5 Hlayiso/Prediction 2026 NSC – Marking Guidelines 2.4.2 OPTION 1 Consider 100 g of Y: m 64,62 n(C) = = = 5,385 mol ✓ M 12 10,77 100 – 64,62 – 10,77 24,61 n(H) = = 10,77 mol; n(O) = = = 1,538 mol ✓ 1 16 16 C : H : O = 5,385 : 10,77 : 1,538 = 3,5 : 7 : 1 = 7 : 14 : 2 ✓ Empirical formula C7 H14 O2 : M = 7(12) + 14(1) + 2(16) = 130 g·mol–1 = M(Y) ∴ Molecular formula: C7 H14 O2 ✓ OPTION 2 m(C) = 0,6462 × 130 = 84 g ⇒ n(C) = 12 84 = 7 ✓ m(H) = 0,1077 × 130 = 14 g ⇒ n(H) = 14 ✓ m(O) = 130 – 84 – 14 = 32 g ⇒ n(O) = 32 16 = 2 ✓ Molecular formula: C7 H14 O2 ✓ (4) 2.4.3 H H H H O H H H C C C C C O C C H H H H H H H (Ethyl pentanoate: compound D is pentanoic acid, so alcohol X is ethanol.) Marking criteria • Functional group (ester group –COO–) correct ✓ • Whole structure correct ✓ Condensed formula (CH3 (CH2 )3 COOCH2 CH3 ): max 1/2. Positive marking from QUESTION 2.4.2 for an ester consistent with the molecular formula obtained and made from pentanoic acid. (2) [20] Copyright reserved Please turn over

Physical Sciences/P2 6 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 3 3.1 The pressure exerted by a vapour at equilibrium with its liquid in a closed system. ✓✓ (2) 3.2 The compounds have the same molar mass (molecular mass), i.e. 74 g·mol–1 . ✓ (1) 3.3 D / propanoic acid ✓ It has the lowest vapour pressure. ✓ (2) 3.4 Compound A (methyl ethanoate, an ester) has (London forces and) dipole-dipole forces between its molecules. ✓ Compound D (propanoic acid) has (London forces, dipole-dipole forces and) hydrogen bonds between its molecules. ✓ The intermolecular forces in A are weaker than those in D / hydrogen bonds are stronger than dipole-dipole forces. ✓ Less energy is needed to overcome the intermolecular forces in A, so more molecules of A evaporate / A has the higher vapour pressure. ✓ OR: the reverse argument for D. (4) 3.5.1 CHAIN ✓ (1) 3.5.2 Compound B is more branched / more compact / more spherical than C. ✓ B has a smaller surface (contact) area over which intermolecular forces act, so its intermolecular (London) forces are weaker. ✓ Less energy is needed to overcome the intermolecular forces in B, so more molecules evaporate / higher vapour pressure. ✓ OR: the reverse argument for C (longer chain, larger surface area, stronger intermolecular forces). (3) [13] Copyright reserved Please turn over

Physical Sciences/P2 7 Hlayiso/Prediction 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 Substitution ✓ (1) 4.1.2 Addition ✓ (1) 4.2.1 Hex-2-ene ✓✓ Marking: hexene ✓; position 2 of the double bond ✓. Accept: 2-hexene. (2) 4.2.2 H H H H H (hex-1-ene) H C C C C C C H H H H H H Marking criteria • Functional group (C=C double bond on carbon 1) ✓ • Whole structure correct ✓ Condensed formula (CH2 CHCH2 CH2 CH2 CH3 ): max 1/2. (2) 4.3 Secondary ✓ The carbon atom bonded to the halogen (Br) is bonded to TWO other carbon atoms. ✓ (2) 4.4 H Br H H H H H C C C C C C H + KOH H H H H H H → H H H H H H H C C C C C C H + KBr + H2 O H H H H Marking criteria • Organic reactant (2-bromohexane) as structural formula ✓ • Inorganic reactant (KOH) ✓ • Organic product (hex-2-ene) as structural formula ✓ • Inorganic products (KBr and H2 O) ✓ • Balancing ✓ Accept NaOH and NaBr. Ignore conditions written on the arrow. Accept 2-bromohexane → hex-2-ene + HBr (KOH/ethanol on arrow): organic reactant ✓, organic product ✓, balancing ✓, inorganic species ✓ (max 4/5). Hex-1-ene as organic product: max 4/5. Condensed or molecular formulae for the organic compounds: max 3/5. (5) 4.5.1 Dilute potassium hydroxide / KOH(aq) / NaOH(aq) / water / H2 O ✓ (1) 4.5.2 Mild heat / heat gently / dilute (aqueous) solution of the base ✓ Accept: heat. Do not accept: concentrated base, ethanol as solvent, or strong heat. (1) 4.6 2,3-dibromohexane ✓✓ (dibromohexane ✓; positions 2,3 ✓) (2) [17] Copyright reserved Please turn over

Physical Sciences/P2 8 Hlayiso/Prediction 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 the acid / H2 SO4 ✓ (1) 5.2.2 ANY ONE: Temperature / mass of Mg / state of division (surface area) of Mg / volume of acid ✓ (1) 5.3.1 The concentration (amount) of the reactants decreases / the amount (surface area) of Mg decreases (fewer effective collisions per unit time). ✓ (1) V 0,26 5.3.2 n(H2 ) = = ✓ = 1,08 × 10–2 mol Vm 24 n(H2 SO4 )used = n(H2 ) = 1,08 × 10–2 mol (ratio 1 : 1) ✓ ∆n 1,08 × 10–2 Rate = = ✓ ∆t 60 Rate = 1,81 × 10–4 mol·s–1 ✓ Accept answers from 1,80 × 10–4 to 1,81 × 10–4 mol·s–1 . Using 260 cm3 and 24 000 cm3 ·mol–1 is equivalent. (4) m 0,36 5.3.3 n(Mg) = = = 0,015 mol ✓ M 24 n(H2 ) = n(Mg) = 0,015 mol (Mg is the limiting reactant; ratio 1 : 1) ✓ V = nVm = (0,015)(24) = 0,36 dm3 (360 cm3 ) ✓ (3) 5.4 EQUAL TO ✓ (1) 5.5 Powdered Mg has a larger surface area / more Mg particles are exposed to the acid. ✓ More particles collide with the correct orientation and sufficient energy / more collisions occur. ✓ More effective collisions per unit time / higher frequency of effective collisions. ✓ (3) 5.6 Q✓ At the higher temperature (35 ◦ C) the average kinetic energy of the particles is higher, so the curve is broader/flatter and shifted to higher kinetic energies / more particles have kinetic energy equal to or greater than Ea . ✓ (2) [18] Copyright reserved Please turn over

Physical Sciences/P2 9 Hlayiso/Prediction 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 m and substitution for n(Cℓ ) (0,12 mol) ✓ • Formula n = M 2 • USING ratio PCℓ5 : PCℓ3 : Cℓ2 = 1 : 1 : 1 ✓ • Divide equilibrium amounts by 2 dm3 ✓ • Correct Kc expression (formulae in square brackets) ✓ • Substitution of 0,04 and equilibrium concentrations into the Kc expression ✓ • Equilibrium amount of PCℓ5 (0,18 mol) ✓ • Initial amount of PCℓ5 = equilibrium amount + change (0,30 mol) ✓ • Final answer: 62,55 g ✓ m 8,52 n(Cℓ2 ) = = = 0,12 mol M 71 PCℓ5 PCℓ3 Cℓ2 Initial amount (mol) x 0 0 Change in amount (mol) –0,12 +0,12 +0,12 ratio Equilibrium amount (mol) x – 0,12 0,12 0,12 x – 0,12 Equilibrium concentration (mol·dm–3 ) 0,06 0,06 ÷2 2 [PCℓ3 ][Cℓ2 ] (0,06)(0,06) Kc = ⇒ 0,04 = [PCℓ5 ] [PCℓ5 ] [PCℓ5 ] = 0,09 mol·dm–3 ⇒ n(PCℓ5 )eq = (0,09)(2) = 0,18 mol n(PCℓ5 )initial = 0,18 + 0,12 = 0,30 mol m = nM = (0,30)(208,5) = 62,55 g Accept 62,40 g – 62,55 g (rounding of M). No Kc expression, correct substitution: max 7/8. Wrong Kc expression: max 4/8 (no marks for substitution, equilibrium amount, initial amount and answer). (8) 6.3.1 PCℓ5 (g) was added / the concentration of PCℓ5 was increased. ✓ (1) 6.3.2 A catalyst was added. ✓ (1) 6.4.1 INCREASED ✓ (1) 6.4.2 The rate of the forward reaction increased more than the rate of the reverse reaction / the forward reaction is favoured. ✓ An increase in temperature favours the endothermic reaction. ✓ The forward reaction is endothermic (∆H > 0). ✓ (3) 6.4.3 GREATER THAN ✓ (1) Copyright reserved Please turn over

Physical Sciences/P2 10 Hlayiso/Prediction 2026 NSC – Marking Guidelines 6.5 INCREASES ✓ An increase in pressure (decrease in volume) favours the reaction that produces the smaller number of moles of gas, i.e. the reverse reaction (1 mol gas versus 2 mol gas). ✓ (2) [19] QUESTION 7 7.1.1 An acid that ionises incompletely (partially) in water. ✓✓ (2) 7.1.2 HOCℓ ✓ HOCℓ has a smaller Ka value, so it ionises to a smaller extent and its [H3 O+ ] is lower. ✓ (2) 7.1.3 BASIC ✓ (1) 7.1.4 OCℓ– (aq) + H2 O(ℓ) ⇌ HOCℓ(aq) + OH– (aq) Marking: reactants ✓ products ✓ balancing ✓. Ignore phases; accept →. Accept: NaOCℓ + H2 O ⇌ HOCℓ + Na+ + OH– (3/3). Also accept: NaOCℓ + H2 O ⇌ HOCℓ + NaOH (3/3). (3) 7.2.1 Phenolphthalein ✓ Oxalic acid is a weak acid and NaOH a strong base, so the pH at the equivalence point (endpoint) is greater than 7 (within the range of phenolphthalein). ✓ (2) 7.2.2 OPTION 1 n((COOH)2 ) = cV = (0,10)(0,025) = 2,5 × 10–3 mol ✓ n(NaOH) = 2(2,5 × 10–3 ) = 5,0 × 10–3 mol (ratio 1 : 2) ✓ n 5,0 × 10–3 c(NaOH) = = ✓ = 0,25 mol·dm–3 ✓ V 0,020 OPTION 2 ca Va na (0,10)(25) 1 = ✓ = ✓✓ cb = 0,25 mol·dm–3 ✓ cb Vb nb cb (20) 2 (4) Copyright reserved Please turn over

Physical Sciences/P2 11 Hlayiso/Prediction 2026 NSC – Marking Guidelines 7.2.3 Marking criteria • n(NaOH)initial = cV with substitution ✓ • n((COOH)2 ) = cV with substitution and USING ratio 1 : 2 ✓ • n(NaOH)excess = initial – reacted ✓ • Divide by total volume 0,025 dm3 ✓ • Use of Kw (or pOH) with substitution ✓ • Final answer: 12,85 ✓ n(NaOH)initial = (0,25)(0,015) = 3,75 × 10–3 mol n((COOH)2 ) = (0,10)(0,010) = 1,0 × 10–3 mol ⇒ n(NaOH)reacted = 2,0 × 10–3 mol n(NaOH)excess = 3,75 × 10–3 – 2,0 × 10–3 = 1,75 × 10–3 mol 1,75 × 10–3 [OH– ] = = 0,07 mol·dm–3 0,025 1 × 10–14 [H3 O+ ][OH– ] = 1 × 10–14 ⇒ [H3 O+ ] = = 1,43 × 10–13 mol·dm–3 0,07 –13 pH = – log(1,43 × 10 ) = 12,85 OR: pOH = – log(0,07) = 1,15; pH = 14 – 1,15 = 12,85. POSITIVE MARKING FROM QUESTION 7.2.2. (6) [20] QUESTION 8 8.1 Temperature: 25 ◦ C (298 K) ✓ Concentration of the electrolytes (solutions): 1 mol·dm–3 ✓ (2) 8.2 Ag ✓ The mass of the Ag electrode increases, so Ag+ ions are reduced (gain electrons) at this electrode / reduction takes place at Ag. ✓ (2) 8.3 Eθcell = Eθcathode – Eθanode ✓ 1,24 = 0,80 – Eθanode ✓✓ Eθanode = –0,44 V ✓ M is iron / Fe ✓ Any other formula using unconventional abbreviations, e.g. Ecell = EOA – ERA , followed by correct substitution: 4/5. (5) 8.4 Fe(s) + 2Ag+ (aq) → Fe2+ (aq) + 2Ag(s) Marking: reactants ✓ products ✓ balancing ✓. Ignore phases. Fe + 2Ag+ ⇌ Fe2+ + 2Ag: 2/3 Fe2+ + 2Ag → Fe + 2Ag+ : 0/3 POSITIVE MARKING FROM QUESTION 8.3 (metal identified there). (3) [12] Copyright reserved Please turn over

Physical Sciences/P2 12 Hlayiso/Prediction 2026 NSC – Marking Guidelines 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 2Br– (aq) → Br2 (ℓ) + 2e– ✓✓ Br2 + 2e– ← 2Br– : 2/2 2Br– ⇌ Br2 + 2e– : 1/2 Br2 + 2e– → 2Br– : 0/2 Ignore phases; accept Br2 (aq). (2) 9.3 Br– ions are a STRONGER reducing agent than H2 O. ✓ Br– ions are oxidised to Br2 (at the anode). ✓ H2 O is not oxidised to O2 . ✓ (3) m 1,27 9.4 n(Cu) = = = 0,02 mol ✓ M 63,5 n(e– ) = 2n(Cu) = 0,04 mol (Cu2+ + 2e– → Cu) ✓ N = nNA = (0,04)(6,02 × 1023 ) ✓ N = 2,41 × 1022 electrons ✓ (4) [11] TOTAL: 150 Copyright reserved