Greendale 6092/02 Chem Paper 1&2 ANS Prelim 2024

1 2024 Class Preliminary Examination Chemistry Secondary 4 Express Suggested answers Section A 1 2 3 4 5 6 7 8 9 10 D C C D B B C C A D 11 12 13 14 15 16 17 18 19 20 D D B A D A A A C C 21 22 23 24 25 26 27 28 29 30 B C D A D D B B B A 31 32 33 34 35 36 37 38 39 40 D B C A A C A D A C

2 Section B Qn Answer Mark 1(a) acid: hydrochloric acid other reactant: lead(II) nitrate [1] [1] 1(b) nitric acid [1] 1(c) Iron(II) carbonate is added in excess to ensure all the acid is completely reacted / used up. [1] TOTAL [4] 2(a) [1] 2(b) physical: melted, AND cooled AND cut into pellets chemical: cracking to break into smaller molecules [1] [1] 2(c)(i) [1] 2(c)(ii) similarity: The monomer 2 used to make both the polymers is the same dicarboxylic acid, difference: The other monomer used to make this polymer is a diol but the other monomer in (c)(i) is a diamine. [1] [1] 2(c)(iii) (c)(i): amide (linkage) AND Fig. 2.3: ester (linkage) [1] TOTAL [7] 3(a) Same empirical formula AND Mr AND different arrangement of atoms / different units present. [1]

3 Qn Answer Mark 3(b) isomer A: isomer B: [2] 3(c) alcohol: carboxylic acid: [1] [1] 3(d) empirical formula: C3H6O [2] TOTAL [7]

4 Qn Answer Mark 4(a) true false Atoms lose electrons more easily down group 1. ü Melting point decreases from fluorine to iodine. ü The strongest non-metal oxidising agent is at the top of the group. ü Metallic character increases across Period 3. ü [2] 4(b) Comparison of structure: lithium: giant metallic structure graphite: giant molecular structure (consisting of huge network of C atoms) oxygen: simple molecular structure consisting of discrete molecules [1] Comparison of bonding: lithium: strong electrostatic forces between lithium cations and sea of electrons graphite: strong covalent bonds between carbon atoms oxygen: weak intermolecular forces between discrete molecules [1] comparison between the melting points: Most energy needed to overcome the strong covalent bonds between carbon atoms; hence graphite has the highest melting point AND Least energy needed to overcome weak intermolecular forces in oxygen. [1] Electrical conductivity comparison: lithium: presence of delocalised / free moving / mobile electrons to conduct electricity AND oxygen: exist as molecules and no mobile charge carriers / no free moving electrons or ions to conduct electricity [1]

5 Qn Answer Mark graphite: each C atom is bonded to 3 other atoms and 1 free / non-bonded electron per C atom and there are free moving electrons to conduct electricity. [1] TOTAL [7] 5(a) NaH + H2O ® NaOH + H2 [1] 5(b) If the pH of the mixture is less than 10, it is a non-metal hydride; AND If the pH of the mixture is more than 10, it is a metal hydride; [1] 5(c) [2] 5(d) In solid state, ions held in fixed position / no free moving ions to conduct electricity. [1] In molten state, giant (crystal)/ (ionic) lattice structure breaks down AND free moving ions to conduct electricity. [1] 5(e) Student 1 is correct. AND SiH4 has the most number of H atoms; % by mass of hydrogen in SiH4 = 4/28 x 100% = 12.5% This is the highest compared to the rest: Eg: 3/30 = 10% for H in AlH3 Student 2 is wrong. AND Given the same number of H atoms, Eg: % of H in PH3 = 3/34 x 100% = 8.8% % of H in AlH3 = 3/30 x 100% = 10% [1] [1] TOTAL [8] 6(a)(i) 200 s [1] 6(a)(ii) average rate = 65/90 = 0.722 cm3 / s [1] 6(b) Gradient is larger than original / steeper AND Volume of gas produced is half – levels off at 45 cm3 [1]

6 Qn Answer Mark 6(c) Particles gain energy and move faster OR Greater fraction / more particles have energy greater than or equal to activation energy ; [1] Frequency of effective collisions increases, increasing rate of reaction ; [1] TOTAL [5] 7(a) 6.8 AND Comparing experiments 1 and 4, when concentration of S2O82– ions is constant, concentration of I– is doubled, rate of reaction is also doubled. AND comparing expt 4 and 5, concentration of iodide is doubled, rate from expt 4 to 5 should be 3.4 x 2 = 6.8 [1] 7(b) amount S2O82– ions in both experiments = 20/1000 x 0.008 = 0.00016 mol AND amount of I– ions in expt 4 = 10/1000 x 0.04 = 0.0004 mol AND amount of I– ions in expt 5 = 10/1000 x 0.08 = 0.0008 mol Mole ratio: S2O82– : I– 1 : 2 0.00016 : 0.00032 needed Since only 0.00032 mol needed to react with 0.00016 mol of S2O82–, I– ions in excess in both experiments, hence S2O82– ions is the limiting reactant. [1] [1] TOTAL [3] 8(a)(i) Reacts / dissolves in rain water to form acid rain AND Corrodes metal and limestone buildings [1] 8(a)(ii) equation in stage 1: SO2 + H2O ® H2SO3 equation in stage 2: H2SO3 + CaCO3 ® CaSO3 + H2O + CO2 [1] [1]

7 Qn Answer Mark 8(b)(i) As air to fuel ratio is higher, • higher concentration of O2 in air • occurrence of incomplete combustion of petrol is less AND • lead to less carbon monoxide formed. As the temperature of the internal combustion engine is lower, • O2 and N2 from air will less likely combine to form nitrogen monoxide. • lead to less nitrogen monoxide formed. [1] [1] [1] 8(b)(ii) 2CO + 2NO ® N2 + 2CO2 [1] 8(b)(iii) Oxidation state of carbon increases from +2 in CO to +4 in CO2; hence carbon undergoes oxidation. Oxidation state of nitrogen decreases from +2 in NO to 0 in N2; hence nitrogen undergoes reduction. [1] [1] TOTAL [9] 9(a) propyne [1] 9(b) CnH2n-2 [1] 9(c)(i) Energy absorbed to break 1 mole of C≡C bond, 2 moles of C – H bonds and 2.5 moles of O=O bonds is less than the energy released to make 2 moles of O – H bonds and 4 moles of C=O bonds. [2] 9(c)(ii) Amount of C2H2 = 1000/24 = 41.67 / 41.7 mol Energy released = 41.67 x 1410 = 58 750 kJ / 58 800 kJ [1] [1] 9(d)(i) C2H2Br2 / C2H2Br4 [1] 9(d)(ii) Reddish-brown aqueous bromine turns colourless. [1] TOTAL [8] 10(a) Forward reaction rate decreases over time, while backward reaction rate increases. Eventually, both forward and backward reaction rates are equal/same. [1] [1]

8 Qn Answer Mark 10(b) [3] 10(c) turned yellow/ orange [1] idea of hydroxide ions reacting with hydrogen ions AND increase / shift towards the forward reaction / more Meorꟷ is present in equilibrium [1] 10(d) 3 moles of gaseous reactant and 1 mole of gaseous product / counteract the decrease in pressure / to increase pressure; [1] shift towards the backward reaction AND less methanol produced as pressure decreases [1] 10(e)(i) percentage of PCl3 increases as temperature increases; [1] 10(e)(ii) increase temperature, more PCl3 formed hence shift towards the forward reaction to remove the heat “disturbance” [1] forward reaction must be endothermic, (as reaction mixture absorbs heat) [1] TOTAL [12]

9 Section C Qn Answer Mark 11(a) similarity at negative electrodes: At the negative electrode: H+ ions selectively discharged (over K+) in both electrolytes as hydrogen is below potassium in the reactivity series; AND 2H+(aq) + 2e– → H2(g) [1] Electrolysis of dilute potassium chloride: At the positive electrode: OH– ions selectively discharged (over Cl– ions), forming oxygen gas; AND 4OH–(aq) → 2H2O(l) + O2(g) + 4e– [1] electrolysis of concentrated potassium chloride: At the positive electrode: Cl– ions selectively discharged over OH– ions as higher concentration of chloride ions, forming chlorine gas. AND 2Cl–(aq) → Cl2(g) + 2e– [1] Electrolysis of dilute potassium chloride: Electrolyte: K+ and Cl– ions remain in the electrolyte, Universal Indicator remains green. [1] electrolysis of concentrated potassium chloride: Electrolyte: K+ and OH– ions remain in the electrolyte, increase in concentration of OH– ions over H+ ions. Universal Indicator changes from green to violet / purple / blue. [1] 11(b)(i) negative electrode: iron structure / iron / support AN