RVHS Prelim H1 Chemistry P2 Soln

River Valley High School 8873/02/PRELIMS/25 [Turn over 2025 JC 2 Preliminary Examination RIVER VALLEY HIGH SCHOOL JC2 H1 CHEMISTRY 8873 PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 2 1 (a) One mole of a substance contains exactly 6.02  1023 elementary entities. [1] (b) (i) Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) [1] (ii) Oxidising agent. The oxidation state of Mn in MnO4− decreases from +7 to +2 in Mn2+ and the oxidation state of Fe in Fe2+ increases from +2 to +3 in Fe3+. [2] (iii) The aliquot changes from yellow to orange. [1] (iv) MnO4 – + 8H+ + 5e → Mn2+ + 4H2O Fe2+ → Fe3+ + e ×5 MnO4 – + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O [1] (v) amount of MnO4− used = 32.00/1000  0.025 = 8.00  10−4 mol [1] (vi) amount of Fe2+ in 25.0 cm3 = 8.00  10−4  5 = 4.00  10−3 mol amount of Fe2+ in 250 cm3 = 4.00  10−3  250/25.0 = 4.00  10−2 mol [1] [1] (vii) Mass of Fe in wire = 4.00  10−2  55.8 = 2.232 g % by mass of Fe in wire = 2.232/3.35  100 % = 66.6 % The % of Fe in the ore is >60%, more ores can be purchased. [3] [Total: 11] 2 (a) (i) [2] (ii) There are 2 lone pairs of electrons and 2 bond pairs of electron s around O in HOCl whereas there are only 1 lone pair of electrons and 3 bond pairs of electrons in around N in NH2Cl. To minimise repulsion, HOCl has a bent shape while NH2Cl has a trigonal pyramidal shape. Since the lone pair – lone pair repulsion is greater in HOCl than the lone pair – bond pair repulsion in NH2Cl, the bond angle in HOCl is smaller at 104.5 than 107 in NH2Cl. [3]

2 River Valley High School 8873/02/PRELIMS/25 2025 JC 2 Preliminary Examination (iii) Yes, HOCl is a polar molecule. Cl is more electronegative than H, O−Cl is less polar than O−H. There is a net dipole moment in the HOCl molecule. [1] (b) CsCl has a giant ionic lattice structure while HOCl has a simple covalent structure. More energy is required to overcome the stronger electrostatic forces of attraction between Cs+ and C l− ions than the weaker hydrogen bonds between HOCl molecules. [2] [Total: 8] 3 (a) Intermediate 1: Intermediate 2: [2] (b) Step 1: HCl, heat or PCl3 or SOCl2 or PCl5 Step 2: ethanolic NH3, heat in a sealed tube Step 3: CH3COOH, DCC [3] (c) [1] [Total: 6]

3 River Valley High School 8873/02/PRELIMS/25 [Turn over 2025 JC 2 Preliminary Examination 4 (a) (i) Maximum temperature = 51.0 C 2 BFL drawn to extrapolate to 3 min [2] (ii) Highest temperature rise = 51.0 − 25.0 = 26.0 C [1] (iii) Heat evolved = 4.18  25.0  26.0 = 2717J = 2.72 kJ (3.s.f) [1] (iv) Amount of Mg in mass of FA 1 added into cup = 1.00 24.3 = 4.115  10−2 mol Amount of CuSO4 in 25.0 cm3 of FA 2 = 1.00 × 25.0 1000 = 2.50  10−2 mol CuSO4 is the limiting reagent. H = − − 2 2.717 2.50 10 = −108.68 kJ mol−1 = −109 kJ mol−1 [2] (b) Na2O(s) reacts vigorously with water to form NaOH(aq), which is a strongly alkaline, pH = 13. Na2O(s) + H₂O(l) → 2NaOH(aq) [2]

4 River Valley High School 8873/02/PRELIMS/25 2025 JC 2 Preliminary Examination (c) Down the Group, the atomic radius of rubidium is larger than sodium. The increase in nuclear charge is largely cancelled by the increase in shielding effect by the inner shells electrons. As a result, ionisation energy decreases in rubidium than in sodium, i.e. increasing ease of losing electrons by the metal. Thus, the tendency of the metal to be oxidised increases. [2] [Total: 10] 5 (a) (i) Comparing experiments 1 and 2, when [CO] is constant, [NO2] increases to 2.0 times as rate also increases to 2.0 times.  Order of reaction with respect to NO2 is 1. OR Using the substitution method, 11 2 2 21 22 (Rate) [NO ] [CO] (Rate) [NO ] [CO] mn mn k k = 11 2 2 =   2 2 (Rate) [NO ] (Rate) [NO ] m since [CO] is constant. =   -- -- 1.80 × 10 ³ 1.50 × 10 ³ 3.60 × 10 ³ 3.00 × 10 ³ m m=1 Comparing Experiments 2 and 3, when [NO2] is constant [CO] increases to 1.5 times as rate also increases to 1.5 times.  Order of reaction with respect to I– is 1. 22 3 3 22 23 (Rate) [NO ] [CO] (Rate) [NO ] [CO] mn mn k k = 22 3 3 (Rate) [CO] (Rate) [CO] n=   since [NO2] =   -- -- 3.60 × 10 ³ 2.50 × 10 ³ 5.40 × 10 ³ 3.75 × 10 ³ n n=1 [2] (ii) Rate = k[NO2][CO] [1]

5 River Valley High School 8873/02/PRELIMS/25 [Turn over 2025 JC 2 Preliminary Examination (iii) correct shape with t1/2 constant for 1st order [1] (iv) Adsorption: CO and NO ₂ molecules form weak bonds with the active sites on the platinum catalyst surface. This interaction weakens the internal bonds within the CO and NO ₂ molecules, making them more chemically reactive. Reaction: The adsorbed CO and NO ₂ molecules are held close together on the catalyst surface in the correct orientation. This allows the oxygen from NO₂ to react with CO, forming CO₂ and NO. Desorption: The weak bonds between the product molecules (CO ₂ and NO) and the platinum surface are broken . These products then diffuse away from the catalyst surface, freeing the active sites for more CO and NO₂ molecules to be adsorbed and reacted. [3] [Total: 7] 6 (a) Polymers are macromolecules with at least 100 repeat units of monomers and average molar mass of at least 1000 g mol–1. [1] (b) More items are made of PE and PP, so more will be thrown away as garbage. OR Low density (PE: 0.92, PP: 0.90) is less than seawater (1.03 g/ cm3), so they float easily and persist on the surface. [1] (c) As the number of electrons is much lesser / electron cloud size is smaller in PE and PP than PS, less energy is required to overcome the weaker instantaneous dipole - induced dipole interactions between PE and PP molecules than PS molecules. [1]

6 River Valley High School 8873/02/PRELIMS/25 2025 JC 2 Preliminary Examination (d) In the presence of alkali, PET, containing the ester bonds, can undergo hydrolysis. [1] (e) LDPE HDPE Plastic bags Cling wrap Crates Bottles Plastic bags and cling wrap are LDPE because they have lower tensile strength but greater ductility and flexibility while HDPE products have higher tensile strength and are more rigid. [2] (f) [2] (g) Environment Advantage: • Reduce oil usage and emissions of greenhouse gases associated with the production of plastics. • When PET is recycled, it does not go into landfill, which can run out over time. • PET does not become ocean waste and pollute the water. Either 1 Trade-off: • recycling process may generate more toxic materials . For example, when bleach is used during recycling, noxious Cl2 may be released to the atmosphere and this in turn may contribute to air pollution. • Leeching of toxic chemicals to the waterways. • Energy required to recycle plastics Either 1 [2] [Total: 10]

7 River Valley High School 8873/02/PRELIMS/25 [Turn over 2025 JC 2 Preliminary Examination 7 (a) (i) Nanoparticles are defined as particles with all dimensions between 1– 100 nm on the nanoscale. [1] (ii) As the surface area to volume ratio is high , a greater amount of sunscreen is in contact with the nanoparticles. The denser packing covers the skin more evenly/ does not leave bulky residue behind. [2] (iii) Due to the small size, nanoparticles can easily (either 1) • pass through the pores on the skin / absorb into the blood stream through the skin (AND cause cancer / become toxic to cells.) • Volatility is high for TiO2, increased the possibility of inhaling. [1] (b) Graphene is a single layer of carbon atoms arranged in hexagons as they are in graphite (or a single layer of