# Physical Sciences Grade 12 Paper 2 KZN Prelim 2024 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the KZN Prelim Exam 2024, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-p2-grade-12-september-2024-final-memo/195225d3-6778-4829-a154-15f437bba713
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2024
- Exam period: KZN Prelim Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 14

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### Page 1

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Physical Sciences/P2                         1           September 2024 Preparatory Examination
                                            NSC
                                               FINAL

                                       GRADE 12

                                     NATIONAL
                                SENIOR CERTIFICATE

                       PHYSICAL SCIENCES P2 (CHEMISTRY)

                          PREPARATORY EXAMINATION

                                  SEPTEMBER 2024

                             MARKING GUIDELINES

MARKS: 150

                        These Marking Guidelines consist of 14 pages.

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Physical Sciences/P2                              2         September 2024 Preparatory Examination
                                                 NSC

QUESTION 1

1.1      A                                                                                 (2)

1.2      A                                                                                 (2)

1.3      B                                                                                 (2)

1.4      D                                                                                 (2)

1.5      A                                                                                 (2)

1.6      B                                                                                 (2)

1.7      C  Do not mark                                                                    (2)

1.8      B                                                                                 (2)

1.9      D                                                                                 (2)

1.10     C                                                                                 (2)
                                                                                             [20]

QUESTION 2

2.1.1     5 – ethyl – 2,6 - dimethylhept – 3 – yne

          Marking criteria:
           correct stem i.e. hept – 3 – yne 
           substituents correctly identified i.e. ethyl, dimethyl
           IUPAC name completely correct including numbering, sequence and hyphen 

                                                                                             (3)

2.1.2     2,3 – dibromo – 5 – methylheptane
          Marking criteria:
           correct stem and substituents i.e. dibromo, methyl and heptane
           IUPAC name completely correct including numbering, sequence and hyphen 
                                                                                             (2)

2.2.1
                 O                        ACCEPT              O
             │   ║   │                                        ║
           ― C ― C ― C ―                                  R ― C ― R‫׳‬
             │       │                    
                                                                                             (2)

2.2.2     alcohols accept alkanols                                                          (1)

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Physical Sciences/P2                               3             September 2024 Preparatory Examination
                                                  NSC

2.2.3     A or E                                                                                    (2)
          Has multiple bonds between atoms of carbon. 

2.3

                         H   O   H   H
                         │   ║   │   │
                H      ― C ― C ― C ― C                  ― H
                         │       │   │
                         H       H   H
                         H   H   H   O
                         │   │   │   ║
               H       ― C ― C ― C ― C                  ― H
                         │   │   │
                         H   H   H

           Marking criteria:
            functional group of first isomer correctly drawn
            functional groupof second isomer correctly drawn
            Whole structure of first isomer correctly drawn                                        (4)
            Whole structure of second isomer correctly drawn 

2.4       Functional isomers                                                                        (1)
                                                                                                     [15]

QUESTION 3

3.1       The pressure exerted by a vapour at equilibrium with its liquid in a closed
          system. 

           Marking criteria:
           If any one of the underlined key words/phrases in the correct context (vapour
           pressure) is omitted, deduct 1 mark.

                                                                                                     (2)

3.2       100 kPa At the boiling point(149°C), the vapour pressure equals the
          atmospheric pressure.                                                                     (2)

3.3       P.  P has a lower boiling point than R.                                                  (2)

3.4       Chain Length /branching/surface area                                                      (1)

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Physical Sciences/P2                                 4             September 2024 Preparatory Examination
                                                    NSC

3.5
           Marking criteria:
                    Relate boiling point with length of carbon chain/branching/number of side
                     chains/surface area. 
                    Compare the strength of the intermolecular forces. 
                    Compare the energy required to overcome the intermolecular forces. 

          P has the lowest boiling point  and therefore has the shortest carbon
          chain/most number of branches/smallest surface area over which the
          intermolecular forces act. 
          Weakest/least intermolecular forces/Van der Waals forces/London forces
          Least energy needed to overcome the intermolecular forces

                                                      OR

          Q has the highest boiling point  and therefore has the longest carbon
          chain/least number of branches/largest surface area over which the
          intermolecular forces act. 
          Strongest/most intermolecular forces/Van der Waals forces/London forces
          Most energy needed to overcome the intermolecular forces
                                                                                                       (4)
3.6

                                            H
                                            │
                                        H ― C ― H

                                   H       H
                                   │       │
                               H ― C ― C ― C ― O ― H
                                   │       │
                                   H       H

                                        H ― C ― H
                                            │
                                            H
            Marking criteria:
             Functional group on first carbon
             3 carbons in the longest chain 
             2 – methyl groups on the second carbon 

                                                                                                       (3)
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Physical Sciences/P2                           5          September 2024 Preparatory Examination
                                              NSC

3.7.1     149 (°C)                                                                           (1)

3.7.2     At 100 kPa, the vapour pressure corresponds to the atmospheric pressure
          T represents the highest boiling point.                                           (3)
                                                                                              [18]

QUESTION 4: DO NOT MARK

4.1.1     Hydrogenation                                                                      (1)

4.1.2     Requires a catalyst/Pt/Pd/Ni                                                       (1)

4.2.1

                H           H      H   H   H
                │           │      │   │   │
            H ― C ═         C    ― C ― C ― C ― H
                                   │   │   │
                                   H   H   H

             Marking criteria:
              double bond on first carbon
              Whole structure correct 

                                                                                              (2)

4.2.2     substitution 
          hydrolysis                                                                         (2)

4.2.3     Dilute
          strong                                                                             (2)

4.3.1     H2O/water                                                                          (1)

4.3.2     Sulphuric acid/H2SO4/phosphoric acid/H3PO4                                         (1)

4.3.3     Hydration                                                                          (1)

4.4.1     Elimination/dehydrohalogenation                                                    (1)

4.4.2     C5H11Br + NaOH → C5H10 + NaBr + H2O LHS RHS Bal                                  (3)
                                                                                              [15]

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Physical Sciences/P2                               6           September 2024 Preparatory Examination
                                                  NSC

QUESTION 5

5.1         Change in concentration  of products/reactants per (unit) time. 
            Change in amount/number of moles/volume/mass  of products/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 0)                                                     (2)
5.2        Reaction is exothermic/Temperature increases. 
           At a higher temperature particles move faster/have higher average kinetic
             energy.
                                                   OR
          More molecules have enough/sufficient kinetic energy for an effective
             collision.
                                                   OR
           More molecules have kinetic energy/Ek equal to or greater than the
             activation energy.
           More effective collisions per unit time/second. 
                                                   OR
           Frequency of effective collisions increases.
          Reaction rate increases.
                                                                                      (3)

5.3
              Marking criteria
                      Equation
                                 4  1,8
                      Substitute        in equation 
                                  3 -1
                      Final answer: 1,1 dm3.min-1

                                       change in volume of CO2(g)
                       rate       =
                                                   Δt               
                                      4 - 1,8
                                 =               
                                      3 - 1
                              =       1,1 dm3.min-1
                                                                                                   (3)
5.4       LESS THAN.
          Less reactant particles per unit volume/lower concentration of HCl/Smaller
          surface of CaCO3 
          Less effective collisions per unit time/Lower frequency of effective
          collisions.                                                               (3)

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Physical Sciences/P2                                   7           September 2024 Preparatory Examination
                                                      NSC

5.5       Marking criteria:
                               V
           Formula: n            to calculate n(CO2)produced
                               Vm
                                           5,8
           Correct substitution (             ) in the above formula 
                                          25,7
           Ratio: n(CaCO3) used equals n(CO2) produced 

           Substitution of n(CaCO3) to get n

                        m
           Use n        to calculate m(CaCO3)initial 
                        M

           Final answer = reaction I 

          Greater mass of impure CaCO3 is required to produce the same volume of
          CO2/22,57 g of impure CaCO3 will produce less CO2

                                    V
         n(CO2)produced               
                                    Vm
                                     5,8
                                        
                                    25,7

                                = 0,22568 mol

                                                                We can also calculate number of moles
         n(CaCO3) used              = n(CO2) produced
                                                                of pure CaCO3
                                    = 0,22568 mol 
                                                                          n(CaCO3)
                                                                                             = 0,2257 mol
                                      m                         Use the ratio
                n(CaCO3)            
                                      M
                                          m
                     0,22568                
                                         100
                 m(CaCO3)            = 22,57 g

         reaction I 
         All of the CaCO3 reacted. 
                                                      OR
         Greater mass of impure CaCO3 is required to produce the same volume of
         CO2/22,57 g of impure CaCO3 will produce less CO2                                              (7)

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Physical Sciences/P2                                8             September 2024 Preparatory Examination
                                                   NSC

5.6
           Marking criteria:
            Gradient of G is smaller/less steep
            Curve G produces a smaller volume of CO2(g) 

                       Volume of CO2(g)
                                               F
                                                                                     
                                                         
                                                             G

                              0
                                                                 time (s)

         NB: If graphs are not labelled = 0/2
                                                                                                      (2)
                                                                                                      [20]

QUESTION 6

6.1      Double arrow in the equation.                                                                (1)

6.2
           Marking criteria:
           If any one of the underlined key words/phrases in the correct context is omitted,
           deduct 1 mark.
           The underlined phrase must be in the correct context.

          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)

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Physical Sciences/P2                             9             September 2024 Preparatory Examination
                                                NSC

6.3      INCREASE. 
         According to Le Chatelier’s Principle when the pressure increases, the
         reaction that leads to a decrease in the number of moles will be favoured. 
         In this case the forward reaction is favoured/equilibrium position shifts to the
         right.                                                                                      (3)

6.4      EXOTHERMIC.
         Number of moles/mass of reactant decreased.
         Forward reaction favoured.
         A decrease in temperature favours the EXOTHERMIC REACTION.                                  (4)

6.5        Marking criteria:
            Initial quantities of all substances
            Quantity of SO2 at equilibrium
            Using the correct mol ratio
            Calculating the quantity(mol) at equilibrium of O2 and SO3
           substances 
            Divide number of moles at equilibrium by 2 dm3
            Kc expression
            Correct substitution of equilibrium concentrations into Kc expression 
            Kc = 11,25
                                                    SO2      O2       SO3
           Initial quantity (mol)                     0,6    0,5      0,4      

           Change (mol)                               0,2    0,1      0,2      

           Quantity at equilibrium (mol)            0,4     0,4      0,6      

           Equilibrium concentration (mol.dm-3)       0,2    0,2      0,3      

                   [SO 3 ] 2
          Kc                         No Kc expression, correct substitution. 7
                 [SO 2 ] 2 [O 2 ]                                                  8
                   [0,3] 2
          =                   
                 [0,2] 2 [0,2]         Wrong Kc expression 6
                                                               8
              = 11,25 
                                                                                                       (8)

                                                                                                      [18]

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Physical Sciences/P2                          10           September 2024 Preparatory Examination
                                             NSC

 QUESTION 7

7.1.1    A substance that can act as an acid and a base.                                   (2)

7.1.2            + H2O ⇌            + H3O+ LHS RHS Balancing
         OR
                 + H2O ⇌ H2CO3 + OH-                                                          (3)

7.2.1
            Marking criteria:
             Formula: pH = - log[H3O+]
             Substitute c(H3O+) in the formula pH = - log[H3O+]
             Final answer

           pH        =   - log[H3O+]     
                     =   - log(2x0,15)   
                     =   0,52

                                                                                               (3)

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Physical Sciences/P2                                 11             September 2024 Preparatory Examination
                                                    NSC

7.2.2
           Marking criteria:
            Calculate n(H+) from H2SO4
            Substitute for c and V in n = cV to calculate n(OH-) from NaOH
            Substitute 12,96 into pH = - log[H3O+] to calculate [H+]
            Substitute [H+] in [H3O+][OH-] = 1 x 10-14 to calculate [OH-] in
              excess
            Subtract n(H+) from n(OH-)TOTAL and relate to n(OH-)EXCESS. Marks
              are awarded for substitution. LHS RHS
            Final answer

                 n(H+) from H2SO4         =   cV x 2
                                          =   (0,15)(0,02) x 2 
                                          =   0,006 mols

                n(OH-) from NaOH          =   cV
                                          =   (0,25)(0,03) 
                                          =   0,0075 mols

                            pH            =   - log[H3O+]
                            12,96         =   - log[H3O+] 
                            [H3O+]        =   1,096 x 10-3 mol.dm-3
                       [H3O+][OH- ]       =   1 x 10-14
               (1,096 x 10-3)[ OH- ]      =   1 x 10-14 
                            [OH- ]        =   0,091 mol.dm-3
                          -
                    n(OH )excess          =   cV
                                          =                 (50 + X)
                                                 (0,091)
                                                              1000
                       n(OH-)excess       =          -
                                              n(OH )TOTAL           -     n(H+) from H2SO4

                        (50 + X)                               (0,1)(X)
           (0,091)                       =   (0,0075) +                  -   0,006     
                          1000                                  1000
                             X            =   27, 98 cm3                                                (8)
                                                                                                         [16]

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Physical Sciences/P2                                     12        September 2024 Preparatory Examination
                                                        NSC

QUESTION 8

8.1
             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 Max: 3
                                                          4

          Eᶿcell           = Eᶿreduction     -    Eᶿoxidation 
          0,03            = 0,80           -    Eᶿoxidation
          Eᶿoxidation      = 0,77 V 
                                                                                                       (4)

8.2.1    Fe2+ → Fe3+ + e-                                                                            (2)

            Notes
                Ignore phases
                Fe2+ ← Fe3+ + e- ( 0 )                           Fe3+ + e- ⇌ Fe2+       (0 )
                                             2                                             2
                        Fe2+ ⇌ Fe3+ +   e-       (1 )             Fe3+ +   e-   ← Fe2+     2
                                                                                          ( )
                                                   2                                         2
                       Ignore if charge on electron omitted.
                        If a charge of an ion is omitted eg. Fe3+ + e- → Fe2+       Max: ( 0 )
                                                                                               2

8.2.2    Pt(s)/Fe2+(aq)(1 mol.dm-3)/Fe3+(aq)(1 mol.dm-3)  //Ag+(aq)(1 mol.dm-3)/Ag(s)  (3)

         NB: Award full marks even if the phases and concentrations are not shown

8.2.3    Temperature 25°C or 298 K 
         Concentration of electrolytes = 1 mol.dm-3                                                   (2)

8.2.4    From chemical to electrical                                                                  (1)

8.4.     INCREASES                                                                                    (1)
                                                                                                       [13]

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Physical Sciences/P2                                 13          September 2024 Preparatory Examination
                                                    NSC

QUESTION 9

9.1       A substance of which the aqueous solution contains ions/ A substance that
          dissolves in water to give a solution that conducts electricity/A solution/
          dissolved substance that conducts an electric current through the movement of
          ions.                                                                       (2)

9.2       M
          Cu2+(aq) + 2e- → Cu(s) 
          Ignore phases

             Notes
                 Cu ← Cu2+ + 2e-           (   )             Cu2+ + 2e- ⇌ Cu       (1 )
                                                                                         2
                       Cu2+ + 2e- ← Cu        ( )              Cu2+ + 2e- ⇌ Cu       (       )
                      Ignore if charge on electron omitted.
                      If a charge of an ion is omitted eg. Cu + 2 e- ← Cu    Max: ( 1 )
                                                                                         2

                                                                                                    (3)

9.3       Cu/copper
          Copper is being plated on the metal M
                                                                                                    (3)

9.4       Marking criteria:
           Calculate gain in mass of the bar i.e. 17,85 - 12,52 and substitute 63,5 in
                              m
             the formula: n     
                              M
           Substitute 6,02 x 1023 
           Ratio of number of mols of e to number of moles of Cu: 2 : 1
           Substitute in Q = IΔt, Calculate charge
          Final answer 3 hours

                m
          : n 
                M
                17,85  12,52
          : n                
                    63,5
                       5,33
          n(e) = 2(         )
                       63,5
                  = 0,168 mols

          n(e)    = n = nNA

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Physical Sciences/P2                           14            September 2024 Preparatory Examination
                                              NSC

                 = 0,168 x 6,02 x 1023 
                 = 1,011 x 1023
          Q      = ne
                 = (1,011 x 1023)(1,6 x 10-19) 
                 = 16 176 C

          Q            = IΔt
          16 176       = 1,5(Δt)
             Δt        = 10 784 s                                                              (5)
              X        = 3 hr

9.5       Zn2+ is a weaker oxidising agent than Cu2+ and will not be reduced.                (2)

                                               OR

          Cu2+ is a stronger oxidising agent than Zn2+, Cu2+ will be reduced to Cu.

                                                                                               [15]
                                                                                TOTAL:         150

NOTE:
The paper should be marked out 133 and then converted to 150.

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- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

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