MEMORANDUM

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

QUESTION 2
2.1
2.1.1 Homologous series √ (1)
2.1.2 Unsaturated. √
Contains a triple bond √/multiple bonds  (2)
2.1.3 C₅H₉ √ (1)
2.1.4 5- ethlyl √ -2-methyl √ hex-3-yne √ OR 5-ethlyl-2-methyl-3-hexyne
Marking Criteria

• Hexyne √
• Side chains (ethyl and methyl) √
• Whole name correct √   (3)

2.2.1 
Marking criteria √

• Functional group correct (1/2)
• Whole structure correct (2/2) (2)

2.2.2 (Mild) heat √ in a water bath √ (2)
2.2.3 Esterification/ Condensation √ (1)
2.2.4 H₂SO₄ √ (1)
2.2.5 
Marking criteria √

• Functional group  √ correct (1/2)
• Whole structure correct (2/2) (2)

2.3
2.3.1 Compounds with the same molecular formula. √ but different structural formula. √ (2)
2.3.2 The C atom bonded to Br is bonded to one other C atom √√ OR The C atom bonded to Br is bonded to two hydrogen atoms (2)
2.3.3 1-bromo-2-methylpropane √√√
 Marking Criteria

• Propane √
• (bromo and methyl) √
• Whole name correct √  (3)

[22]

QUESTION 3
3.1 The temperature at which the vapour pressure of a liquid equals the atmospheric pressure. √√  (2)
3.2 A √ Compound A has the lowest boiling point. √ (2)
3.3.1 Molar mass/ Molecular size/ Surface area √  (1)
3.3.2 Hydrogen bond √ (1)
3.4

• The molar mass/Molecular size/ surface area of the compounds increases from top to bottom in the table. √
• The compounds have London forces and (Hydrogen bonds) √
• The strength of London forces increases with an increase in molar mass/molecular size/surface area. √
• More energy is needed to overcome the intermolecular forces √ (4)

3.5.1

Marking criteria

• 3 Carbon atoms in longest chain with correct functional group  (2)

3.5.2 B √ It has the same molecular formula √  (2)
3.5.3 Lower than √  (1)
3.5.4

• 2-methylpropan-1-ol has a smaller surface area than butan-1-ol. √
• Both compounds have London forces and (hydrogen bonds) √
• 2-methylpropan-1-ol have weaker London forces than butan-1-ol √
• Less energy is needed to overcome the intermolecular forces in 2-methylpropan-1-ol √

OR

• Butan-1-ol has a larger surface area than 2-methylpropan-1-ol √
• Both compounds have London forces and (hydrogen bonds) √
• Butan-1-ol have stronger Londonforces than 2-methylpropan-1-ol √
• More energy is needed to overcome the intermolecular forces in butan-1-ol √ (3)

3.6

• Methanoic acid and propan-1-ol have both Hydrogen bond and (London forces) √
• Methanoic acid have two sites for hydrogen bonds while propan-1-ol have only one site for hydrogen bonds. √
• Methanoic acid has stronger intermolecular forces than propan-1-ol √
• More energy is needed to overcome the intermolecular forces in methanoic acid. √

OR

• Methanoic acid and propan-1-ol have both Hydrogen bond and (London forces) √
• Methanoic acid have two sites for hydrogen bonds while propan-1-ol have only one site for hydrogen bonds. √
• Propan-1-ol has weaker intermolecular forces than methanoic acid √
• Less energy is needed to overcome the intermolecular forces in propan-1-ol. √ (4)

[22]

Related Items

QUESTION 4
4.1 Process of breaking down long chain hydrocarbons √/alkanes into smaller more useful chains √ (2)
4.2
4.2.1 C₄H₁₀ √√ (2)
4.2.2 HIGH TEMPERATURE √ OR HIGH PRESSURE  (1)
4.3
4.3.1 Substitution √ (1)
4.3.2 Elimination √ (1)
4.4 Hydrogenation √ (1)
4.5
4.5.1 Br₂/Bromine √  (1)
4.5.2 Pt/Ni/Pd/Platinum/Nickel/Palladium √ (1)
4.6 CH₃CH₂CH₂CHBr √√+ KOH →  CH₃CH = CHCH₃ √√ + KBr + H₂O
Marking criteria

• Organic reactant √√
• Organic product √√
• ALL Inorganic reagents correct √ (KOH, KBr and H₂O) (5)

[15]

QUESTION 5
5.1 Change in concentration (of reactant/products) per unit time √√
OR
Amount/Volume/Mass of reactant/product used/formed per unit time (2)
5.2 RELEASED √
ΔH < 0 √/Energy of products is less than the energy of reactants/ 
Reaction exothermic  (2)
5.3
5.3.1 INCREASES √ (1)
5.3.2 NO EFFECT √ (1)
5.4
5.4.1 After 120 s √ (1)
5.4.2 Concentration (of HCℓ) √ Surface area (of Zinc) √ OR temperature (2)
5.5
5.5.1 Rate / Tempo = Δv/Δt= 300 -0√ /120 – 0 √= 2,50 cm^3∙s^-1 √ (3)
5.5.2 (7)
5.6
5.6.1 Catalyst √ OR Increases reaction rate (1)
5.6.2

• Catalyst lowers activation energy/provides an alternative path of lower activation energy √
• More particles have sufficient Ek OR more particles have Ek>Ea √
• More effective collisions per unit time/Frequency of effective collisions increase √√ (4)

[24]

QUESTION 6
6.1 Reaction in which products can be converted back to reactants √√ (2 or 0) (2)
6.2
6.2.1 Decreases √ (1)
6.2.2 Increases √(1)
6.3

• (When the temperature is decreased) the exothermic reaction is favoured √
• Forward reaction is favoured √ (2)

6.4 CALCULATION USING MOLES
Marking guideline

• Formula  n=m/M
• Substitution of 28 g∙mol^-1
• Use of ratio N₂:H₂:NH₃
• Equilibrium amounts of N₂ and H₂ formed
• Division n equilibrium of N₂, H₂ and NH₃ by 2 
• Correct Kc expression
• Substitution into Kc expression
• Final answer

n = m/M √
= 7,84/28 √
= 0,28 mol
N₂ + 3 H₂ ⇋ 2 NH₃
ni (mol) 0,28 0,6 0
Δn (mol) 0,06 0,18 0,12 Ratio √
ne (mol) 0,22 0,42 √ 0,12
ce 0,22/2 0,42/2 0,12/2 Division by 2 √
(mol∙dm^-3) =0,11 0,21 0,06
Kc = [NH₃]²/[N₂].[H₂]³ √
= 0,06²/(0,11 x 0,21³) √
= 3,53 √

CALCULATION USING CONCENTRATION
Marking guidelines

• Formula c=m/MV or c=n/v
•  Substitution into c=m/MV or c=n/V
• Dividing by 2 for all concentration
• Use of ratio N₂:H₂:NH₃
• Equilibrium concentrations of N₂, H₂ and NH₃
• Correct Kc expression
• Substitution into Kc expression
• Final answer 
• (Any one)

(dividing by) √
N₂ + 3 H₂ → 2 NH₃
 (8)
6.5 As pressure increases percentage yield increases. √√  (2)
6.6 MOLE OPTION

0,84 mol van H₂ word benodig            0,2 mol N₂ word benodig 
∴ H2 is the limiting reagent √               ∴ H2 is the limiting reagent
Theoretical yield/Teoretiese opbrengs = 0,6 x 2/3 = 0,4 mol √
% Yield = actual yield /theoretical yield  × 100 %
= 0,12/0,4 x 100 √
= 30%
Pressure = 100 (atmospheres) √
CONCENTRATION OPTION

• V is constant
• Concentration ratio N₂ : H₂            Concentration ratio N₂ : H₂
                                    1: 3                                                1 : 3
  c(H₂) = 3 (0,14) √                              c(N₂) = 1/3 (0,3)
  c(H₂) = 0,42 mol∙dm^-3 of H₂ required c(N₂) = 0,1 mol∙dm^-3
  only 0,3 mol∙dm^-3 is available             0,14 mol∙dm^-3 is available

∴ H₂ is the limiting reagent √
Theoretical yield = 0,3×2/3 √= 0,2 mol∙dm^-3

• % yield = actual yield /theoretical yield ×100 %
• % yield = 0,06/0,2 ×100 % √
• % yield = 30 %
• Pressure 100 (atmosphere) √ (5)

[21]

QUESTION 7
7.1
7.1.1 Reaction of a salt with water √√ (2)
7.1.2 (Excess) Hydroxide/OH- ions are formed √
Hydroxide/ OH- is basic/alkaline √  (2)
7.1.3 H₂O √ and HCO₃^- √ (2)
7.1.4 HCO₃^- can accept a proton (H⁺) (to form H₂CO₃) √
HCO₃^- can donate a proton (H+) (to form CO₃^2-) √  (2)
7.2
7.2.1 Acid is a proton donor √√  (2)
7.2.2

• pH = - log [H₃O⁺] √
  1 = - log [H₃O⁺] √
  [H₃O⁺] = 0,1 mol∙dm ^-3 √
  [X] = 2 [H₃O⁺]
  [X] = 2(0,1)
  [X] = 0,2 mol∙dm ^-3 √  (4)

7.2.3 It is a solution of known concentration √√  (2)
7.2.4 Bromothymol blue √ Acid X is a strong acid. KOH is a strong base. √
There is no hydrolysis of the salt produced, therefore the equivalence point of the titration would be within the range of indicator. √  (3)
7.2.5  (7)
[26]
TOTAL: 150