PHYSICAL SCIENCES: CHEMISTRY (PAPER 2)
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2019

QUESTION 1/VRAAG 1
1.1 C ✓✓ (2)
1.2 A ✓✓ (2)
1.3 C ✓✓ (2)
1.4 A ✓✓ (2)
1.5 D ✓✓ (2)
1.6 C ✓✓ (2)
1.7 D ✓✓ (2)
1.8 D ✓✓ (2)
1.9 C ✓✓ (2)
1.10 A ✓✓ (2)
[20]

QUESTION 2
2.1 Unsaturated ✓
ANY ONE

• C/It has a triple/multiple bond. ✓
• C/It has a triple/multiple bond between C atoms.
• C/It does NOT contain the maximum number of H atoms bonded to C atoms.
• Compound C is an alkyne. (2)

2.2
2.2.1 D ✓ (1)
2.2.2 B ✓ (1)
2.2.3 C ✓ (1)
2.2.4 E ✓ (1)
2.3
2.3.1 (1)
2.3.2 
 (2)
Marking criteria:

• Whole structure correct:²/₂
• Only functional group correct:½

IF

• More than one functional group: ⁰/₂
• If condensed or semi structural formula used: Max.½

2.3.3

Marking criteria:

• Three C atoms in longest chain. ✓
• One methyl subrstituent on C2

IF
Any error e.g. omission of H atoms, condensed or semi structural formula Max: ½ (2)
2.4
2.4.1
2,3-dibromo-5-methylheptane
Marking criteria:

• Correct stem i.e. heptane. ✓
• All substituents (bromo and methyl) correctly identified. ✓
• IUPAC name completely correct including numbering, sequence, hyphens and commas. ✓(3)

2.4.2 2C₄H₁₀ + 13O₂ ✓ →  8CO₂ + 10H₂O ✓ Bal ✓
Notes:

• Reactants ✓ Products ✓ Balancing ✓
• Ignore double arrows and phases.
• Marking rule 6.3.10.
• If condensed structural formulae used: Max.²/₃
• Accept coefficients that are multiples.(3)

[17]

QUESTION 3
3.1
3.1.1 Yes✓
ANY ONE:

• Compounds have the same molecular mass. ✓
• Only one independent variable. (2)

3.1.2 Functional group/Homologous series/Type of (organic) compound ✓ (1)
3.2 A/butane ✓
Lowest boiling point/weakest intermolecular forces. ✓(2)
3.3 Marking guidelines

• Type of IMF in A.
• BOTH B and C have hydrogen bonding.
• Compare number of sites for hydrogen bonding in B and C.
• Compare strength of IMFs.
• Compare energy required.
  
  
• Between molecules of butane/compound A are London forces/dispersion forces/induced dipole forces. ✓
• Molecules of compound B/propan-1-ol have one site for hydrogen bonding.✓
• Molecules of compound C/ethanoic acid have two/more sites for hydrogen bonding. ✓
• Strength of intermolecular forces increases from compound A/butane to compound B/propan-1-ol to compound C/ethanoic acid. ✓
  OR
  Intermolecular forces in compound A/butane are the weakest and intermolecular forces in compound C/ethanoic acid are the strongest.
• More energy is needed to overcome/break intermolecular forces in compound C than in the other two compounds. ✓(5)

3.4 Butan-1-ol ✓
Longer chain length./Larger molecule./Larger molecular mass./Larger molecular size./Stronger intermolecular forces./Larger surface area.✓(2)
[12]

QUESTION 4
4.1
4.1.1 Addition (polymerisation)✓(1)
4.1.2 Ethene ✓(1)
4.1.3 Polyethene/polythene ✓(1)
4.2
4.2.1 Dehydration/elimination ✓(1)
4.2.2 Catalyst/dehydrating agent/causes dehydration/removes water molecules ✓(1)
4.2.3 Prop-1-ene/propene (2 or 0)(2)
4.2.4

Marking criteria:

• Whole structure correct:
• Only functional group correct: Max: ½

IF:

• More than one functional group: ⁰/₂
• If condensed or semi structural formula used:Max.½ (2)

4.2.5 Addition/Hydration ✓(1)
4.2.6 Propan-2-ol✓✓
Marking criteria:

• Correct stem and functional group i.e propanol ✓
• Name completely correct ✓✓ (2)

[12]

QUESTION 5
5. 1 NOTE
Give the mark for per unit time only if in context of reaction rate.
ANY ONE

• Change in concentration ✓ of products/reactants per (unit) time. ✓
• Change in amount/number of moles/volume/mass of products or 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/of 0)(2)

5.2.2

Example:
Reaction rate increases with increase in concentration./Reaction rate is proportional to concentration.
IF
DIRECTLY proportional: Max: ½ (2)
5.2
5.2.1 Rate of the reaction ✓ (1)
5.3.2 B ✓ (1)
5.3.3

• At a higher temperature particles move faster/have a higher kinetic energy. ✓
• More molecules have enough/sufficient (kinetic) energy. ✓
• OR
  More molecules have (kinetic) energy equal to or greater than activation energy.
• More effective collisions per unit time/second./Increased frequency of effective collisions. 
• Reaction rate increases. ✓ (4)

5.3
5.3.1 Activation energy/(The boundary line for the) molecules with (adequate) kinetic energy to make effective collisions. ✓(1)
5.4 Curve Y/it was obtained for the reaction where a catalyst was added. ✓
OR
Curve X was obtained for the reaction in the absence of a catalyst. (1)
5.5 Marking guidelines

• Any formula: n= m or c= n ✓
                            M          V
• Substitute 0,1 dm³ in n = cV ✓
• Use mole ratio:
  n(S)expected = ½n(HCℓ)used ✓
• Substitution of 32 g∙mol-1 in n = m ✓
                                                      M
• SUBSTITUTE in:
  n(S) produced x 100/ m/(S) produced x 100
  n(S) expected             m(S) expected
• Final answer: 56,25% to 60% ✓
  

(6)
[18]

QUESTION 6
6.1 Reversible reaction/Both forward and reverse reactions can take place./Products can be converted back to reactants. ✓(1)
6.2 To favour the forward reaction/production of ammonia./To increase the yield of ammonia./Prevent the decomposition of NH₃. (1)
6.3 20(%) ✓ (1)
6.4
6.4.1 At 500 °C lower yield of ammonia:

• The (forward) reaction is exothermic./Reverse reaction is endothermic. ✓
• An increase in temperature favours the endothermic reaction. ✓
• The reverse reaction is favoured.✓
  OR
  At 350 °C higher yield of ammonia:
• The (forward) reaction is exothermic./Reverse reaction is endothermic. ✓
• A decrease in temperature favours the exothermic reaction. ✓
• The forward reaction is favoured. ✓(3)

6.4.2 At 350 atm higher yield of ammonia:

• An increase in pressure favours the reaction that produces the lower number of moles/number of molecules/volume of gas. ✓
• The forward reaction is favoured. ✓
  OR
  At 150 atm lower yield of ammonia:
• A decrease in pressure favours the reaction that produces the higher number of moles/number of molecules/volume of gas. ✓
• Reverse reaction is favoured. ✓(2)

6.5
6.5.1 1 mol N₂ reacts with 3 mol H₂ to produce 2 mol NH₃
∴ 2 mol N₂ reacts with 6 mol H₂ to produce 4 (mol) NH₃ ✓✓ (2 or 0)(2)
6.5.2 POSITIVE MARKING FROM QUESTION 6.5.1.
Marking criteria:

• Calculate 35% of 4 mol NH₃ (answer from Q6.5.1)✓
• Use mol ratio n(N₂) : n(H₂) : n(NH₃) = 1 : 3 : 2 ✓
• Equilibrium n(N₂) = initial n(N₂) - Δn(N₂)
  Equilibrium n(H₂) = initial n(H₂) - Δn(H₂)
• Divide by 0,5 dm³. ✓
• Correct Kc expression (formulae in square brackets). ✓
• Substitution of concentrations into correct Kc expression. ✓
• Final answer: 0,002 ✓
  Range: 0,00155 to 0,002 (1,55 x 10^-3 to 2 x 10^-3)

n(NH₃) = 35 × 4
              100
= 1,4 mol

ratio✓
Divide by 0,5 dm³ ✓

Kc =   [NH₃] ²   
        [H₂]³ [N₂] ✓
=     (2.8)²     
   (7.8)³(10.6) ✓
= 0,002 ✓ (7)

Wrong Kc expression: Max.⁴/₇
No Kc expression, correct substitution: Max.⁶/₇
[17]

QUESTION 7
7.1 A base forms hydroxide ions (OH-) in water/aqueous solution. ✓✓
IF:
A base ionises to form hydroxide ions (OH-).✓. Max.½ (2)
7.2 A strong base ionises/dissociates completely ✓ and a weak base ionises/dissociates incompletely. ✓ (2)
7.3 HCO^-₃(aq) + H₂O(ℓ) ✓ ⇌ H₂CO₃(aq) + OH^-(aq) ✓ Bal. ✓
Accept
NaHCO₃(aq) + H2O(ℓ) ⇌ H₂CO₃(aq) + NaOH(aq)
Notes:

• Reactants ✓ Products ✓ Balancing ✓
• Ignore single arrow.
• Marking rule 6.3.10.
• Ignore phases. (3)

7.4
7.4.1 pH = -log[H₃O+] ✓
= -log (0,2) ✓
= 0,70 ✓ (0,699) (3)
7.4.2 Titration of a weak base and a strong acid. ✓
OR
The endpoint will be at pH < 7. (1)
7.4.3 Marking guidelines:

• Any formulae: c= n  / n= m =/ C_a x V_a = na  /c = m 
                                V         M     c_b x V_b     nb       MV
• Substitute 0,2 mol∙dm^-3 & 20 x 10-3/0,02 dm³ or 20 cm³. ✓
• Use mol ratio n(XHCO³) : n(HCℓ) = 1 : 1 ✓
• Substitute n(XHCO³) or c(XHCO³) AND 0,4 g. ✓
• M(X) = 39 g∙mol^-1 ✓
• X = K/potassium. ✓

(6)
[17]

QUESTION 8
8.1 It is a conductor of electricity/a solid to connect wires to./Pt is inert or unreactive.✓OR
Cℓ−(aq) and chlorine gas are not solids and cannot be used as an electrode. (1)
8.2
8.2.1 Chemical (energy) to electrical (energy) ✓ (1)
8.2.2 Cℓ2 + 2e- → 2Cℓ− ✓✓
Marking guidelines

• Cℓ₂ + 2e- ⇌ 2Cℓ− ½            2Cℓ− ⇌ Cℓ₂ + 2e- ⁰/₂
  2Cℓ− ← Cℓ₂ + 2e- ²/₂          2Cℓ−- → Cℓ₂ + 2e- ⁰/₂
• Ignore if charge omitted on electron.
• If charge (-) omitted on Cℓ− (-):
  Max:½  Example: Cℓ₂ + 2e- → 2Cℓ ✓ (2)

8.2.3 Cr(s) | Cr₃+(aq) ✓|| Cℓ2(g) | Cℓ− (aq) | Pt(s) ✓
OR
Cr(s) | Cr³⁺(1 mol∙dm^-3) || Cℓ₂(g) | Cℓ− (1 mol∙dm^-3) | Pt(s)
Accept:
Cr | Cr³⁺ || Cℓ₂ | Cℓ− | Pt (3)
8.3

[13]

QUESTION 9
9.1 Electrolytic ✓ (1)
9.2 2H₂O + 2e- → H₂ + 2OH- ✓✓
Marking guidelines

• 2H₂O + 2e- ⇌ H₂ + 2OH- ½         H₂ + 2OH- ⇌ 2H₂O + 2e- ⁰/₂
  H₂ + 2OH- ← 2H₂O + 2e- ²/₂          H₂2 + 2OH- → 2H₂O + 2e- ⁰/₂
• Ignore if charge omitted on electron.
• If charge (-) omitted on OH-: Max: 21 Example: 2H₂O + 2e- → H₂ + 2OH ✓ (2)

9.3
9.3.1 Chlorine (gas) / Cℓ_2✓ (1)
9.3.2 P ✓ & Y ✓ (2)
9.4 Cathode✓
Reduction takes place here./Gains electrons.✓ (2)
9.5 CuCℓ₂(aq) ✓ → Cu(s) + Cℓ₂(g)✓ Bal✓
OR
Cu₂+(aq) + 2Cℓ- → Cu(s) + Cℓ₂(g)
Notes:

• Reactants ✓ Products ✓Balancing ✓
• Ignore double arrows.
• Marking rule 6.3.10.
• Ignore phases (3)

[11]

QUESTION 10
10.1
10.1.1 II – IV – III - I ✓ (1)
10.1.2 2NH₃ + H₂SO₄ ✓ → (NH₄)2SO₄ ✓ Bal ✓
Notes:

• Reactants ✓ Products ✓ Balancing ✓
• Ignore double arrows.
• Marking rule 6.3.10. (3)

10.1.3 Vanadium pentoxide (1)
10.1.4 SO₃(g) + H₂SO₄ ✓→ H₂S₂O₇ ✓ Bal ✓
Notes:

• Reactants ✓ Products ✓ Balancing ✓
• Ignore double arrows.
• Marking rule 6.3.10. (3)

10.1.5 Sulphuric acid will form (white) mists./The reaction is very exothermic/gives off too much heat./Corrosive reaction. ✓ (1)
10.2

[13]
TOTAL: 150