PHYSICAL SCIENCES: CHEMISTRY (P2)
GRADE 12
NOVEMBER 2019
MEMORANDUM
NATIONAL SENIOR CERTIFICATE

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

QUESTION 2
2.1
2.1.1 C_nH_{2n - 2} ✓ (1)
2.1.2

Marking criteria

• Functional group correct. ✓
• 2 methyl substituents. ✓
• Whole structure correct:
  (3)

2.2.1 Compounds with the same molecular formula, ✓ but different positions of the side chain/substituents/functional groups ✓on the parent chain. (2)
2.2.3

Marking criteria

• Whole structure correct: ²/₃
• Only functional group correc Max: ½

OR: Any correct structure of an aldehyde with five carbon atoms. (2)

2.2
2.2.2 Pentan-3-one ✓✓
Marking criteria

• Functional group and correct position i.e. ✓
• Whole name correct ✓

2.3
2.3.1 Tertiary (alcohol) ✓
The C atom bonded to the functional group/hydroxyl (group)/-OH is bonded to three other C atoms. /The C-atom bonded to the hydroxyl (group) has no hydrogen atoms. ✓ (2)
2.3.2 2-methylbutan-2-ol
Marking criteria

• 2-methyl ✓
• Butan-2-ol ✓
• Any error e.g. hyphens omitted and/or incorrect sequence: (2)

2.3.3 2-methylbut-2- ene
Marking criteria

• 2-methyl
• But-2-ene/2-butene
• Any error e.g. hyphens omitted and/or incorrect sequence: (2)
  [16]

QUESTION 3
3.1 Marking guidelines
The underlined key phrases must be used in the CORRECT CONTEXT (pressure/boiling). 
The temperature ✓ at which the vapour pressure of a substance equals atmospheric/external pressure. ✓ (2)
3.2 (Q, R and S) have same molecular mass/formulae/number of carbon and hydrogen atoms/are (chain) isomers. ✓
OR
The compounds are all alkanes /same homologous series and have the same number of carbon atoms. (1)

Marking guidelines

• 55 (°C) ✓
• Compare all three compounds or Q and S in terms of branches/chain lengths / surface area. ✓
• Compare strengths of all three or Q and S’s IMF’
• Compare energy of all three ✓

3.3 55 (°C) ✓
Compare compound R with compounds Q and S:

• Compound R is less branched/compact/spherical/surface area than compound Q and more branched/compact/spherical/surface area than compound S. ✓ OR Q is the most branched/compact /spherical/surface area and S is least branced/compact/spherical/surface area.
• Intermolecular forces in compound R are stronger than in compound Q and weaker than in compound S. ✓
• More energy needed to overcome intermolecular forces in compound R than in compound Q and less energy needed to overcome (break) intermolecular forces in compound R than in compound S. ✓
  OR
• Compound R has a longer chain length than compound Q and a shorter chain length than compound S.✓ OR S has the longest chain length and Q the shortest.
• Intermolecular forces increase with increase in chain length. ✓
• More energy needed to overcome intermolecular forces as chain length increases. ✓(4)

3.4
3.4.1 P ✓✓ (2)
3.4.2 Marking guidelines

• Name type of IMFs in P/pentanal. ✓
• ame type of IMFs in
• Compare strength of IMFs. 
  OR Compare energy needed to overcome IMFs.
• In P/ pentanal/aldehydes: dipole-dipole forces ✓(in addition to London forces/dispersion forces/induced dipole forces).
• In T/pentan-1-ol: Hydrogen bonding.✓ (in addition to London forces/dispersion forces/induced dipole forces).
• Intermolecular forces in P/pentanal are weaker ✓than in T/pentan-1-ol OR dipole-dipole forces are weaker than hydrogen bonds OR intermolecular forces in T/pentan-1-ol are stronger than in P/pentanal.
  OR
  More energy needed to overcome/break intermolecular forces in T. (3)

[12]

QUESTION 4
4.1 Haloalkane/alkyl halide ✓ (1)
4.2
4.2.1 Elimination/dehydrohalogenation ✓ (1)
4.2.2 Substitution/hydrolysis ✓ (1)
4.2.3 Esterification/condensation ✓ (1)
4.3
4.3.1

• (Mild) heat/Heating
• Dilute (strong base)/(NaOH/KOH/LiOH) 
  OR
  Add water/H₂O/Voeg water/H₂O by (2)

4.3.2 Propan-1-ol
Marking criteria

• Correct stem and functional group i.e. propanol✓
• Whole name correct: ✓ (2)

4.4

Marking criteria

• Whole structure correct:
• Only functional group correct

Notes

• Accept –OH as condensed. 
• Condensed or semi-structural formula:
• Molecular formula:
  If functional group is incorrect
  If more than one functional group: (2)

4.5 POSITIVE MARKING FROM Q4.3.2 ONLY IF THE COMPOUND IN Q4.3.2 IS AN ALCOHOL.
4.5.1

Marking criteria

• Whole structure correct:
• Only functional group correct

Notes

• Condensed or semi-structural formula:
• Molecular formula: 20
• If functional group is incorrect (2)

4.5.2 (Concentrated) sulphuric acid/H₂SO₄ (1)
[13]

QUESTION 5
5.1 Exothermic
✓H < 0/Energy is released (2)
5.2 rate/tempo =- Δm 
                             Δt
= -0.25 - 2
        30
= 0,06 (g·s^-1) ✓
(0,0583 g·s^-1)

OR
rate/tempo =- Δm 
                       Δt
= -  - 1.75  
          30
= 0,06 (g·s^-1) ✓
(0,0583 g·s^-1) (3)

Notes
Accept negative answer i.e. - 0,06 g·s^-1.
5.3 Marking guidelines

• Calculate: m(CaCO₃) reacted of V(CO₂) produced. ✓
• Substitute: 100 g∙mol^-1. ✓
• USE mol ratio/GEBRUIK molverhouding: n(CO₂) : n(CaCO₃) = 1 : 1 ✓
• Use of 22,4 dm^-3∙mol^-1.✓
• Final answer: 0,18 dm³ (0,1792 dm³) ✓

OPTION 1
m(CaCO₃) = 40   x 2 ✓
                    100
= 0,8 g
n(CaCO₃)reacted = ^m/_M
= 0.8 
   100
= 8 x 10^-3 mol
n(CO₂) = n(CaCO₃) ✓
= 8 x 10^-3 mol
V(CO₂) = 8 x 10^-3 x 22,4 ✓
= 0,18 dm³ ✓

OPTION 2
For 2 g antacid
100 g ✓CaCO₃ ......22,4 dm³ ✓ CO₂
2 g CaCO₃ .....0,448 dm³ ✓
100% CO₂ .......... 0,448 dm³ ✓
40% CO₂ ............. 0,18 dm³ ✓ (5)

OPTION 3
100% CaCO₃ .........2 g
40% .......................0,8 g ✓
100 g ✓……. 1 mol
0,8 g .......... 8 x 10^-3 mol ✓
1 mol ...............22,4 dm³ ✓
8 x 10^-3 mol ......0,18 dm3 ✓
5.4 ANY ONE:

• Concentration (of acid)✓
• Size/mass of tablet/Identical tablet /Type of tablet.
• State of division / Surface area. (1)

5.5 Criteria for conclusion:

• Dependent [(reaction) rate/time] and independent (temperature) variables correctly identified.
• Relationship between the independent and dependent variables correctly stated.

Examples:

• Reaction rate (  1   ) increases with increase in temperature.
                         time
• Reaction rate (  1  ) decreases with decrease in temperature.
                         time
• Time taken for reaction decreases when temperature increases.
• Time taken for reaction increases when temperature decreases. (2)

IF
Reaction rate is DIRECTLY proportional to temperature:
5.6

• Increase in temperature increases the average kinetic energy/molecules move faster. 
• More molecules have enough/sufficient kinetic energy/More molecules have Ek > Ea. ✓
• More effective collisions per unit time/second. /Frequency of effective collisions increases. ✓ (3)

5.7 Marking guidelines

• For each value of temperature, the CURVE Y must be above the given CURVE
• CURVE Y must have an increasing rate with an increase in temperature.(2)

[18]

QUESTION 6
6.1 (The stage in a chemical reaction when the) rate of forward reaction equals the rate of reverse reaction. ✓✓
OR
(The stage in a chemical reaction when the) concentrations of reactants and products remain constant. (2)
6.2 CALCULATIONS USING NUMBER OF MOLES
6.2.1 Marking guidelines

• Substitute: 44 g∙mol^-1. ✓
• Equilibrium concentration of CO₂ multiply by 3 dm³
• Use mole ratio/Gebruik molverhouding: 1:2 / n(CO) = 2n(CO₂). ✓
• n(CO₂)change = n(CO₂)initial - n(CO₂)final
• Correct Kc expression (formulae in square brackets). ✓
• Substitution of concentrations into Kc expression. ✓
• Final answer: 12,24 (range: 11,85 – 12,66) ✓

OPTION 1
n(CO₂) = m 
               M
=60.8
   44
= 1,382 mol

Related Items

Use ratio
Divide/multiply by 3

K_C = [CO]²
         [CO]²
=(0.813)² 
   0.054
= 12.24

No Kc expression, correct substitution
Wrong Kc expression

OPTION 2
n(CO₂) = m 
               M
=60.8
   44
= 1,382 mol

n(CO₂)change = n(CO₂)initial- n(CO₂)final
= 1,382 – 0,162
= 1,22 mol

n(CO)change = 2(CO₂)
= 2(1,22) ✓
= 2,44 mol

n(CO)_eq = n(CO)change = 2,44 mol
c(CO) = ⁿ/_V
=2.44 
    3
= 0,813 mol∙dm^-3
K_C = [CO]²
         [CO]²
=(0.813)² 
   0.054
= 12.24 (Accept range: 11,85 – 12,66).

CALCULATIONS USING CONCENTRATION
Marking guidelines

• Substitute 44 g∙mol^-1. ✓
• Initial n(CO₂)divide by 3 dm³. ✓
• USE ratio/GEBRUIK verhouding: c(CO₂) : c(CO) = 1: 2 ✓✓
• Δc(CO₂) = c(CO₂)initial/begin - c(CO₂) eq/ewe
  c(CO)eq/ewe = c(CO)initial/begin + Δc(CO).
• Correct K_c expression (formulae in square brackets). ✓
• Substitution of concentrations into K_c expression. ✓
• Final answer: 12,15 (range/gebied: 11,85 – 12,66) ✓

OPTION 3
n(CO₂) = m 
               M
=60.8
   44
= 1,382 mol

K_C = [CO]²
         [CO]²
=(0.813)² 
   0.054
= 12.24

No Kc expression, correct substitution
Wrong Kc expression (7)

6.2.2 POSITIVE MARKING FROM Q6.2.1
n(C)reacted = n(CO₂)reacted
= 1,22 mol ✓
m(C) = nM
= 1,22(12)
= 14,64 g ✓

Marking guidelines

• USE mol ratio: n(C) = n(CO₂). ✓
• Substitute: 12 g∙mol^-1. ✓
• Final answer: 14,64 g. ✓ (3)

6.3
6.3.1 Remains the same ✓ (1)
6.3.2 Decreases✓

• (When pressure is increased) the reaction that leads to the smaller amount/number of moles/volume of gas is favoured. ✓
• The reverse reaction is favoured. / More CO₂ is formed. ✓ (3)

6.4
6.4.1 Endothermic ✓

• When the temperature increases the mol/percentage CO(g)/product increases/forward reaction is favoured.
• An increase in temperature favours the endothermic reaction (3)

6.4.2 POSITIVE MARKING FROM Q6.2.1.
Marking guidelines

• Calculate total volume/mol of gas at equilibrium: 0,162 + 2,44 = 2,606 dm³ /mol ✓ OR
  Calculate the total concentration at equilibrium: 0,054 + 0,813 = 0,867 mol∙dm^-3
• Calculate percentage of ANY one gas (CO₂ or CO). ✓
• Final answer: T = 827 °C ✓

OPTION 1
V_{total eq} = 0,162 + 2,44 ✓
= 2,606 dm³ (3)
% CO₂ =0.162 x 100 ✓
              2.606
= 6,225 %
OR
% CO =  2.44   x 100 ✓
              2.606
= 93,63 %

OPTION 2
c_{total eq} = 0,054 + 0,813
= 0,867 mol∙dm^-3
% CO₂ =0.054 x 100 ✓
              0.867
= 6,228 %
OR
% CO₂ =0.813 x 100 ✓
              0.867
= 93,77 %
∴ T = 827 °C ✓
[22]

QUESTION 7
7.1 Strong (acid)
Large Ka value > 1 / (HBr) ionises completely (2)
7.2 H₂O ✓
Br- ✓ (2)
7.3
7.3.1 Marking guidelines

• Formula: c = n  /n = cV / c_a × V_a =  n_a
                       V                c_b × V_b     n_b
• Substitution of: (0,5)(0,0165)/(0,5)(16,5) ✓
• Use mol ratio: 1:1/n(HBr) = n(NaOH) ✓
• Substitute: V = 0,09 dm³ /90 cm³ ✓
• Formula: pH = -log[H₃O+] ✓
• Substitute [H₃O+] in pH formula. ✓
• Final answer: pH = 1,04 (range: 1,036 – 1,05) ✓

OPTION 1
n(NaOH)reacted/reageer = cV ✓
= 0,5(0,0165) ✓
= 0,00825 mol
n(HBr)excess = n(NaOH) = 0,00825 mol ✓
c(H₃O+) = ⁿ/_V
=0.00825 
     0.09
= 0,092 mol·dm^-3
pH = -log[H₃O+] ✓
= -log(0,092) ✓
= 1,04 ✓

OPTION 2
c_a V_a =  n_a 
c_b V_b     n_b
   c_a(90)   = 1 
(0.5)(16.5)  1
c_a = 0,092 mol·dm^-3
pH = -log[H₃O+] ✓
= -log(0,092)✓
= 1,04 ✓ (7)
7.3.2 Marking guidelines

• Calculate n(HBr)initial: substitute (0,45)(0,09) in n = cV ✓
• Subtraction:
  n(HBr)reacted = n(HBr)initial – n(HBr)reacted with NaOH. ✓✓
  OR/OF: c(HBr)reacted = c(HBr)initial – c(H₃O+)excess
• Use mol ratio: n(Zn(OH)₂) : n(HBr) = 1 : 2 ✓
• Substitution of: 99 g∙mol^-1 ✓
• Final answer: 1,5964 g (range: 1,58 – 1,68) ✓

POSITIVE MARKING FROM Q7.3.1
OPTION 1
n(HBr)_initial = cV
= (0,45)(0,09) ✓
= 0,0405 mol
n(HBr reacted with Zn(OH)₂) = 0,0405 – 0,00825 ✓✓
= 0,03224 mol
n(Zn(OH)₂) = ½n(HBr) = ½(0,03224) = 0,016125 mol
m(Zn(OH)₂) = nM
= (0,016125)(99)✓
= 1,596 g ✓

OPTION 2
c(HBr) = 0,45 – 0,092 ✓✓
= 0,358 mol∙dm^-3
n(HBr reacted) = cV
= 0,358 x 0,09 ✓
= 0,0322 mol
n(Zn(OH)₂) = ½n(HBr) = ½(0,0322) ✓= 0,01611 mol
m(Zn(OH)₂) = nM
= 0,01611 x 99 ✓
= 1,595 g ✓ (1,60 g) (6)
[17]

QUESTION 8
8.1 Chemical to electrical✓ (1)
8.2 Provides path for movement of ions./ Completes the circuit./Ensures electrical neutrality in the cell./Restore charge balance. ✓ (1)
8.3 OPTION 1
E^θcell =E^θcathode - E^θanode
1.49 = 1.36 - E^θanode
E^θanode = 1.36 - 1.49
= -0.13(V)
X is Pb/Lead ✓

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:

OPTION 2
Cℓ₂ + 2e- → 2Cℓ-            E^θ = 1,36 V ✓
X → X²⁺ + 2e-                E^θ = 0,13 V ✓
Cℓ₂ + X → X²⁺ + 2Cℓ-    E^θ = 1,49 V ✓
X is Pb/Lead ✓ (5)

POSITIVE MARKING FROM Q8.3
8.4 X/Pb/Lead ✓ (1)
8.5
8.5.1 Reaction reached equilibrium./(In each half cell) the rate of oxidation is equal to rate of reduction./Rate of the forward reaction is equal to the rate of the reverse reaction. ✓ (1)
8.5.2 Increases ✓(1)
8.5.3

• [Cℓ-] decreases. ✓
• Forward reaction is favoured.✓ (2)

[12]

QUESTION 9
9.1 Marking guidelines
If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark.
The chemical process in which electrical energy is converted to chemical energy.✓✓
OR
The use of electrical energy to produce a chemical change.
OR
The process during which an electrical current passes through a solution/molten ionic compound. (2)
9.2
9.2.1 2H₂O(ℓ) + 2e- ✓ H₂(g) + 2OH-(aq) ✓✓
Ignore phases

Marking guidelines

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

9.2.2 Water/ H₂O ✓ (1)
9.3 H₂O is a stronger oxidising agent ✓ than Na+ ✓and will be reduced ✓ (to H₂).
OR
Na+ is a weaker oxidizing agent ✓ than H₂O ✓and therefore H₂O will be reduced✓ (toH₂)
OR
The half-reaction that produces H₂(g) has a more positive reduction potential (-0,83 V) ✓ than the half-reaction that produces Na (-2,71 V). ✓
Therefore water/H₂O will be reduced ✓ to H₂./Na+ will not be reduced to Na. (3)
[8]

QUESTION 10
10.1
10.1.1 Hydrogen/H₂ ✓ (1)
10.1.2 Nitrogen monoxide/NO ✓ (1)
10.1.3 Nitric acidr/HNO₃ ✓ (1)
10.2
10.2.1 (Catalytic) oxidation/Redox ✓ (1)
10.2.2 NH₃ + HNO₃ ✓ → NH₄NO₃ ✓ Bal ✓
Notes

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

10.3
10.3.1 (Total) percentage of nutrients/fertiliser/N,P,K. ✓ (1)
10.3.2 Marking guidelines

• Calculate mass fertiliser in A.
• Calculate mass fertiliser in B.✓
• Calculate mass P in A and B 
• Final answer
  B has more phosphorous than A. ✓

OPTION 1
Mass fertiliser in A:
Massa kunsmis in A:
m = 21 × 50 = 10,5 kg
      100
Mass fertiliser in B:
m = 20  × 40 = 10,8 kg
      100
Mass phosphorous in A
³/₈ × 10.5 = 3,94 kg
Mass phosphorous in B/
³/₈ × 10.8 = 4,05 kg
Fertiliser B has more phosphorous than fertiliser A. ✓

OPTION 2
Mass phosphorous in A
m = ³/₈ ×  21   × 50= 3,94 kg
               100
Mass(P) in B
m = ³/₈ ×  27  × 40= 4,05 kg
               100 
Fertiliser B has more phosphorous than fertiliser A.✓

OPTION 3
Mass phosphorous in A
%P = ³/₈ × 21= 7,88%
m(P) = 7.88  × 50 = 3,94 kg
            100
Mass(P) in B
%(P) = ³/₈ × 27 = 10,13%
m =10.13 × 40 = 4,05 kg
        100
Fertiliser B has more phosphorous than fertiliser A. (4)✓

OPTION 4
m = 21 × 50 = 10,5 kg
      100
Mass fertiliser in B:
m = 20  × 40 = 10,8 kg
      100
For the same NPK ratio ✓ the bag with more fertiliser will have more phosphorous ∴ bag B✓
[12]
TOTAL: 150