Gr12 Physical Sciences P2 Solutions Learner Assistance Document 2020

Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA CURRICULUM DIRECTORATE LEARNER ASSISTANCE REVISION DOCUMENT SOLUTIONS GRADE 12 - 2020 PHYSICAL SCIENCES

CHEMISTRY- PAPER 2 ORGANIC CHEMISTRY Question 1 (2) 1.1 A  (2) 1.2 B  (2) 1.3 B  (2) 1.4 B  (2) 1.5 A  (2) 1.6 B  (2) 1.7 A  (2) 1.8 D  (2) 1.9 B  (2) 1.10 D  (2) 1.11 A  (2) 1.12 C  (2) 1.13 B  (2) 1.14 D  (2) [28] Question 2 2.1.1 Alkyne  (1) 2.1.2 Hydroxyl  (1) 2.1.3 C (1) 2.1.4 2-methylpentan-2-one  (2) 2.1.5 (2)    (3) 2.1.6 2C4H10 + 13O2 8CO2 + 16H2O bal (2) 2.2.1 Compounds with the same molecular formula but different positions of the side chain, substituent or functional group on the parent chain.  2.2.2 C and D  (1)

2.3 (2)  2.4 Propanoic acid  (2) [17] Question 3 3.1.1 B (1) 3.1.2 E (1) 3.1.3 F (1)    (3) 3.2.1 2-bromo-3-chloro-4-methylpentane 3.2.2 Ethene  (1) 3.3.1 (2)  3.3.2 (2)  3.4.1 Compounds that have the same molecular formula but different functional (2) groups.  3.4.2 B and F  (1) Question 4 4.1.1 Carboxyl group  (1) 4.1.2 Ketones  (1) 4.1.3 Addition  (1)

4.2 Ethene  (1) 4.3 CO2 / Carbon dioxide  (2) H2O / Water  4.4.1 (2)  4.4.2 (2)  [10] Question 5 (2) 5.1.1 B  (2) 5.1.2 (2)  5.1.3 CnH2n – 2  (1)    (3) 5.1.4 4-ethyl-5-methylhept-2-yne 5.1.5 Butan-2-one   (2) 5.2.1 Alkanes  (2) 5.2.2 (2)  2-methylpropane  5.3.1 Haloalkanes  (1) 5.3.2 Substitution  (1) Question 6 6.1 Temperature at which the vapour pressure of a substance is equal to the (2) atmospheric pressure. 

6.2.1 As the chain length increases, the boiling point increases  (1) 6.2.2 An increase in chain length leads to an increase in the strength of London (3) forces and molecular mass, resulting in higher boiling point and more energy required to overcome the London forces.  6.2.3 Alkene – London forces  (3) Alcohols – London forces, dipole-dipole forces and hydrogen bonds.  Since hydrogen bonds are stronger than London forces, the boiling point of alcohols will be higher.  [9] Question 7 7.1 Single bonds between carbon atoms  (1) 7.2.1 -O–H (1) 7.2.2 (2)  7.3.1 What is the relationship between the boiling point and chain length in (2) alkanes?  7.3.2 Alkanes have London forces. As chain length increases, so does London (2) forces.  7.4 Propane – London forces.  (2) Propanol – London forces, dipole-dipole forces and hydrogen bonds.  Since hydrogen bonding is stronger than london forces, the boiling point of propanol is greater than the boiling point of propaneas more energy is required to overcome the intermolecular forces. [10] Question 8 8.1.1 Alkene  (1) 8.2.1 Addition  (1) 8.2.2 Substitution  (1) 8.2.3 Dehydration  (1) 8.2.4 Addition  (1) 8.3 Propan-2-ol  (1) 8.4 Catalyst / Dehydration agent  (1) 8.5 Concentrated strong base / NaOH (2) Warm ethanolic , Mild heat any 2  [9]

Question 9 9.1 Haloalkane/alkyl halide  (1) 9.2 9.2.1 Elimination/dehydrohalogenation  (1) 9.2.2 Substitution/hydrolysis  (1) 9.2.3 Esterification/condensation  (1) 9.3 9.3.1  (Mild) heat/Heating/   Dilute (strong base) NaOH/KOH/LiOH)  OR Add water/H2OH2O by (2) 9.3.2 Propan-1-ol/1-propanol  Marking criteria  Correct stem and functional group i.e. propanol  Whole name correct propan-1-ol  (2) 9.4 9.5 POSITIVE MARKING FROM Q4.3.2 ONLY IF THE COMPOUND IN Q4.3.2 IS AN ALCOHOL. 9.5.1 H H  H H O H C C C O C C H  H H H 9.5.2 (Concentrated) sulphuric acidH2SO4  (1) [13]

CHEMICAL CHANGE Energy changes Class activity 1.1 A – Energy of reactants  (1) B – Activation energy for the forward reaction without the catalyst  (1) C - Activated complex  (1) (1) D – Activation energy for the forward reaction with the catalyst  (1) E – Energy of products  (1) (1) F – Enthalpy / heat of reaction  (1) 1.2 ∆H = E products – E reactants  = 60 – 40  = 20kJ  (3) 1.3 Exothermic  (1) 1.4 Non-spontaneous  (2) 1.5 Decreases  (1) [12] CHEMICAL EQUILIBRIUM Question 1 1.1.1 dynamic equilibrium (1) 1.1.2 NH3 was added to the system (1) 1.1.3 According to Le Chatelier an increase in the [NH3 ] will cause the equilibrium system to counteract the stress by favouring the side that decreases NH3. The reverse reaction is favoured (equilibrium system shifts to the left). The H2 and N2 increases.  (3)

1.1.4 (8) 1.1.5 (a) Decrease        (1) (b) According to Le Chatelier, an increase in temperature will cause the system to counteract by favouring the side that will decrease the temperature. The endothermic reaction will cause a decrease in temperature and thus the reverse reaction is favoured. When the reverse reaction is favoured [reactants] increases and [products] decreases. Kc is the ratio of [products] and [reactants]. The Kc value will therefore decrease. (5) 1.2.1 decrease (2) 1.2.2 remains the same          (2) [23] Question 2 2.1 2XA2 + A2 ⇌ 2XA3          (2) 2.2.1 When an external stress (change in pressure, temperature or concentration) is applied to a system in chemical equilibrium, the equilibrium point will change in such a way as to counteract the stress. (2) 2.2.2 Stress: Decrease pressure Response: Favour reverse reaction Graph shows rates of both reactions decreased but reverse decreased the least (favoured). Reason: the reverse reaction produces more moles of gasrelieving stress ofdecreased pressure. (3)

2.3 Kc = [XA]2[A2]  [XA3]2 (2) 2.4 Kc = (1,5)2(0,75) 12 = 1,69             (6) [15] Question 3 3.1 Forward reaction because the reaction rate increases / there were only products at the start of the reaction  (2) 3.2 Kc <1 The amount of reactants are greater than the amount of products      (2) 3.3 5 minutes        (1) 3.4 Equal to           (2) 3.5 • Both the forward and reverse reaction rates will initially decrease • The rate of the forward reaction will decrease more as the forward reaction is endothermic • Thus, the reverse reaction is initially favoured • Decreasing the yield of NO        (4) 3.6 (2)

3.7 Concentrations: (5) [18] Question 4 4.1.1 Greater than - steeper gradient at time ti /gradient decreases from t1 to t2 (3) 4.1.2 Equal to (1) 4.1.3 n(HCℓ) = 10,8 – 0,2 mol         (4)      4.1.4 n(H2O) = 0,4 mol n(Cℓ2) = 0,4 +  for method. (3) 4.1.5 (4)

4.1.6 (4) 4.2 Decrease.An increase in temperature will favour the reversereaction, as it isendothermic, to try to reduce the heat. This will decrease the yield low Kc. (4) 4.3.1 Decrease          (2) 4.3.2 Remain the same          (2) 4.3.3 Decreases (2) [29] Question 5 5.1 Small, thus low concentration of product. Equilibrium lies to the left{little NO (2) 5.2 0,2 mol (1) 5.3 1,8 mol (1) 5.4 (2) 5.5 (4) 5.6 Kc increased at higher temperature

Thus more products Thus equilibrium shifts to right Thus forward reaction is endothermic (4) [14] Question 6 6.1 2SO2 + O2 ⇌ 2SO2 (2) 6.2 concentration of SO3 decreases (2) 6.3 forward reaction rate increases more than the reverse thus the forward reaction is endothermic (2) 6.4 equilibrium (1) 6.5 both rates decrease immediately (2) Question 7 7.1 Reversible reaction         (1) 7.2 To favour the forward reaction/production of ammonia./ To increase the yield of ammonia./Prevent the decomposition of NH3.         (1) 7.3 20%            (1) 7.4.1 The (forward) reaction is exothermic     An increase in temperature favours the endothermic reaction The reverse reaction is favoured (resulting in a lower yield of ammonia) OR The (forward) reaction is exothermic A decrease in temperature favours the exothermic reaction The forward reaction is favoured (resulting in a higher yield of ammonia)        (3) 7.4.2 An increase in pressure favours the reaction that produces the lower number of moles/volume of gas The forward reaction is favoured (resulting in a higher yield of ammonia)OR A decrease in pressure favours the reaction that produces the higher number of moles/volume of gas Forward reaction is favoured (resulting in a higher yield of ammonia)  (2) [17]

ACIDS AND BASES Activity 1 1. B            (2)    2. B            (2) 3. A            (2) 4. C            (2) 5. B            (2) [10] Structured Activities Question 1 1.1.1 Each molecule can donate two hydrogen atoms  (2) 1.1.2 H2SO4  (1) 1.2.1 Strong acid  (1) 1.2.2 𝑝𝐻 = − log[𝐻3 𝑂+ ]  1 = − log[𝐻3 𝑂+ ]  [𝐻3 𝑂+ ] = 10−1 [𝐻3 𝑂+ ] = 0.1 𝑚𝑜𝑙. 𝑑𝑚-3  (3) [7] Question 2 (NORTH-WEST (SEPTEMBER) 2015) 2.1 An acid as a proton (H+) donor  (2) 2.2.1 HF / F- and H3O+ / H2O  (2) 2.2.2 Water (1) 2.3.1 Sulphuric acid ionises completely in water.  (1)

2.3.2 H2SO4 2H+ + SO42- [𝐻3 𝑂+ ] = 2 × 0,025 = 0.05 𝑚𝑜𝑙. 𝑑𝑚-3  𝑝𝐻 = − log[𝐻3 𝑂+ ]  𝑝𝐻 = − log[0.05]  𝑝𝐻 = 1.30  (4) [10] Question 3 (MPUMALANGA (SEPTEMBER) 2015) 3.1.1 A solution of precisely known concentration  (2) 𝑚 3.1.2 𝑐 = 𝑀𝑉  𝑚 0,2 =  (4) 56 × 0,3 𝑚 = 3,36 𝑔  3.1.3 𝑝𝑂𝐻 = − log[𝑂𝐻 − ]  𝑝𝑂𝐻 = − log[0,2]  𝑝𝑂𝐻 = 0,70  𝑝𝑂𝐻 + 𝑝𝐻 = 14 0,70 + 𝑝𝐻 = 14  𝑝𝐻 = 13,30  (5) 3.1.4 Bromothymol blue  (1) 3.1.5 𝐶𝑎 𝑉𝑎 𝑛 = 𝑛1  𝐶𝑏 𝑉𝑏 2 𝐶𝑎× 20 1  = 2 0,2×15 𝐶𝑎 = 0,075 𝑚𝑜𝑙. 𝑑𝑚-3  (4) [16] ELECTROCHEMICAL CELLS MULTIPLE CHOICE QUESTIONS 1.1 C (2) 1.2 D (2) 1.3 C (2) 1.4 A (2) 1.5A (2) [10]

Question 2 DBE FEB-MARCH 2017 2.1 2.1.1 Salt bridge  (1) 2.1.2 Voltaic / Galvanic cell  (1) 2.2 2.2.1 Decreases (1) 2.2.2 Increases  (1) 2.3 2.3.1 Y(s) → Y2+(aq) + 2e- Ignore phases OR Mg(s) → Mg2+(aq) + 2e- (2) 2.3.2 Y(s) |Y2+(aq) ||Aℓ3+(aq) | Aℓ(s) OR Mg(s) |Mg2+(aq) ||Aℓ3+(aq) | Aℓ(s OR Y(s) | Y2+ (1 mol∙dm-3)  || Aℓ3+(1 mol∙dm-3) | Aℓ(s)  Accept (3) Y | Y2+ || Aℓ3+ | Aℓ 2.4 OPTION 1 Notes E cell  E reduction  Eoxidation    Accept any other correct 0,7   1,66 - E  formula from the data sheet.  Any other formula using E  oxidation  2,36 (V)  unconventional abbreviations, e.g. Y is Mg   E°cell = E°OA - E°RA followed by correct substitutions OPTION 2 Aℓ3+(aq) + 3e- Aℓ(s) E ϴ = -1,66 V   2+ - Y(s) Y (aq) + 2e Eϴ = +2,36 V  Y(s) + Aℓ3+(aq) → Y2+(aq) + Aℓ(s) E ϴ = +0,7 V  ) Y is Mg [12]

Question 3 (DBE FEB-MARCH 2017) 3.1 Bauxite  (1) 3.2 Oxidation  (1) 3.3 Reduce melting point  OR To lower the temperature / energy needed to melt the Aℓ2O3. ACCEPT To dissolve the Aℓ2O3 so that it can electrolysed easier (1) 3.4 Aℓ3+(aq) + 3e- → Aℓ(s)  Ignore phases (2) 3.5 C + O2 → CO2 bal OR 2Aℓ2O3 + 3C → 4Aℓ + 3CO2 (3) [8]

Question 4 DBE FEB-MAR 2018 4.1 4.1.1 A substance that loses/donates electrons.  (1) 4.1.2 Platinum/Pt  (1) 4.1.3 Sn2+(aq)/tin(II) ions/tin(II)  (1) 4.1.4 Pt| Sn2+(1 mol∙dm-3) ,Sn 4+  (1 mol∙dm-3) || Ag+ (1 mol∙dm-3) | Ag(s)  (3) ACCEPT Pt| Sn2+ | Sn 4+ || Ag+ | Ag 4.1.5 E  E E   cell reduction oxidation = +0,80  - (+0,15)  = 0,65 V  (3) [9] Question 5 DBE MAY-JUNE 2017 5.1 Electrolytic (cell)  (1) 5.2 P (1) 5.3 5.3.1 Au(s) → Au3+(aq) + 3e-  (2) 5.3.2 (+)3  (1) 5.3.3 Electrical energy (is converted) to chemical energy.  (1) 5.3.4 Becomes smaller / thinner / eroded / decrease in mass.  (1) 5.4 ANY ONE  Increase in value. Protection against rust.  (1) 5.5 ANY ONE  Replace Au3+(aq) / electrolyte with Ag+(aq) / silver(I) solution / use a silver solution  Replace P / anode / gold with Ag(s) / silver [8]

CHEMICAL SYSTEMS Question 1 (DBE FEB-MARCH 2018 Q10) 1.1 1.1.1 N2(g) + 3H2(g) → 2NH3(g) bal      (1) 1.1.2 (NH4)2SO4          (1) 1.1.3 Ostwald process         (1) 1.1.4 Ammonium nitrate         (1) 1.2 1.2.1 The ratio of nitrogen (N), phosphorous (P) and potassium (K) in a certain fertiliser. (1) 1.2.2 Percentage fertiliser in the bag.       (1) 1.2.3 OPTION 1: % K =12 5 x 22%  = 9,17%  m(N) = 9,17 x 10 kg 100 = 0,92 kg         (4) OPTION 2: 100 22 x 10 = 2,2 kg  m(K) = 5  (2,2)  12 = 0,92 kg  [10] Question 2 2.1 2.1.1 II – IV – III - I  (1) 2.1.2 2NH3 + H2SO 4  → (NH 4)2SO4  Bal  2.1.3 Vanadium pentoxide  (1) 2.1.4 SO3(g) + H2SO 4 → H2 S 2 O 7  Bal  (3)

2.1.5 Sulphuric acid will form (white) mists./The reaction is very exothermic/gives off too much heat./Corrosive reaction.  (1) 2.2 Marking criteria  Calculate m(fertiliser).  2  Use ratio /m(P) = ½m(K)  X3  Use m(K) = 3,33 kg   Final answer: 3  OPTION 1 OPTION 3 20 20 m(fertiliser) =  50  (fertiliser) =  50  100 100 = 10 kg = 10 kg 2 m(P) = ½m(K)  m(K) = 10 X3 = ½(3,33) = 1,665 kg 2  3,33  = 10 m(X) = 10 – 3,33 – 1,665 X3 = 5,005  X = 3  OPTION 2 N : P : K = 5,005 : 1,665 : 3,33 2  20 m(K) =  x 50  = 3,33  =3:1:2 X  3 100  X = 3  X = 3  (4) [13] Question 3 3.1 3.1.1 Ammonia √ (1) 3.1.2 NO2 √ (1) 3.1.3 Catalytic oxidation of ammonia √ (1) 3.1.4 Platinum/Pt √ (1) 3.1.5 Ostwald (process) √ (1) 3.1.6 Haber (process) √ (1) 3.1.7 NH3 + HNO3 √ → NH4NO3 √ Bal. √ (3)

3.2 3.2.1 OPTION 1 OPTION 2 m(fertiliser = x 15 √ N : P : K 10 : 5 : 15 = 0,75 kg √ m(fertiliser) = x 15 = 4,5 kg m(P) = x 4,5 √ = 0,75 kg √ (2) 3.2.2 %fertiliser = 10 + 5 + 15 = 30% %filler = 100 – 30 = 70% m(filler) =  x 15  = 10,5 kg  (3) [13] The End