2025 H1 Chem JC2 Prelim P2 (for sharing) JPJC

© Jurong Pioneer Junior College [Turn over Name:____________________________________ Class:_____________ JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2025 CHEMISTRY 8873/02 Higher 1 29 August 2025 Paper 2 2 hours Additional materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class and exam index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. Section A (60 marks) Answer all the questions. Section B (20 marks) Answer one question. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 15 2 16 3 11 4 12 5 6 Section B 6 20 7 20 Penalty (delete accordingly) Missing/wrong units in final answer –1 / NA Total 80 This document consists of 24 printed pages and 4 blank pages.

2 © Jurong Pioneer Junior College 8873/02/J2 PRELIMINARY EXAMINATION /2024 Section A Answer all the questions in this section, in the spaces provided. 1 A unknown metal sulfate has the formula MSO4. The hydrated form of the metal sulfate has the formula MSO4.xH2O. For Examiner’s Use (a) Name the types of bonds that are present in the metal sulfate. Electrostatic forces of attraction between oppositely charged ions (ionic bond) and covalent bonds. [2] (b) When 6.24 g of the hydrated metal sulfate is strongly heated, thermal decomposition occurs. In addition to water, a solid oxide , MO, and a gaseous oxide are also produced. The solid oxide had a mass of 1.99 g. The volume of the gaseous oxide produced was 600 cm 3 at r.t.p. A strongly acidic solution is produced when the gaseous oxide is dissolved in water. (i) Write a balanced equation for the thermal decomposition of the hydrated metal sulfate. MSO4.xH2O → MO + SO3 + xH2O [1] (ii) Calculate the relative atomic mass of metal M. Hence, suggest the identity of metal M. Amount of gaseous oxide = (600 x 10–3) / 24 = 0.0250 mol 1 gaseous oxide ≡ 1 MO Amount of MO = 0.0250 mol Ar of M = (1.99 / 0.025) – 16 = 63.6 M is Cu [3] (iii) Using your answer in (b)(ii), calculate the integer x in MSO4.xH2O. Mr of hydrated metal sulfate = 63.6 + 32.1 + 64.0 + 18x = 159.7 + 18x Amount of MO = Amount of MSO4.xH2O 0.0250 = 6.24 159.7+18x 3.993 + 0.45x = 6.24 0.45x = 2.25 x = 4.99 x ≈ 5 [2] (c) The metal ion, M2+, can catalyse the hydrolysis of ester N into butanedioic acid and methanol as shown below. OH OH O O butanedioic acid methanol ester N + 2H2O → + 2CH3OH

3 © Jurong Pioneer Junior College 8873/02/J2 PRELIMINARY EXAMINATION /2024 [Turn over (i) State the number of σ bonds and π bonds in a molecule of butaedioic acid. 13 σ bonds, 2 π bonds [1] (ii) Draw the skeletal formula of ester N. O O O O [1] Butanedioic acid polymerises with 1,4–butanediamine to give nylon–4,4. (iii) State the type of reaction and the catalyst required for the polymerisation process to occur. condensation, dicyclohexylcarbodiimide DCC [2] (iv) Explain why clothing made from nylon–4,4 fibers are prone to creasing. The polymers in nylon–4,4 are linked by hydrogen bonds. Water molecules can penetrate the areas between the polymer chains and break the original hydrogen bonds . The amide groups will now form hydrogen bonds with water molecules. When the material is dried, the water molecules leave and new hydrogen bonds will be formed between the polymer chains (in different positions). As a result, this locks in place any creases that had formed. Hence polyamides are prone to creasing. [2] (v) Suggest why nylon–4,4 is less of an environmental concern than HDPE when disposed in a landfill. Landfills may contain acid (or base) waste or acids from natural processes and would degrade more rapidly than HDPE as the amide linkages undergo hydrolysis. Or other logical answers [1] [Total: 15] 2 The artificially produced isotope of cobalt, , is used as a source of x -rays in medical radiotherapy. For Examiner’s Use (a) Write the full electronic configuration of cobalt ion, . 1s2 2s2 2p6 3s2 3p6 3d7 [1] (b) In an experiment, a sample of cobalt was vaporised, ionised and passed through an electric field. It was observed that a beam of gives an angle of deflection of 8º.

4 © Jurong Pioneer Junior College 8873/02/J2 PRELIMINARY EXAMINATION /2024 (i) Calculate the angle of deflection of beam. + −           = 2 -22 Co SS 60 2 32 2 8 Angle Angle of deflection for = = 15.0  [1] (ii) Sketch and label on the diagram below, how the beams of would be deflected. [1] (c) Aqueous hydrogen peroxide, H 2O2, oxidises tartrate ion, C 4H4O62– in the presence of cobalt( II) chloride homogeneous catalyst in the equation given below. 5H2O2(aq) + C4H4O62–(aq) → 4CO2(g) + 2OH–(aq) + 6H2O(l) (i) State what is meant by the term homogeneous when it is applied to cobalt(II) chloride in the oxidation of tartrate ion by hydrogen peroxide. The Co2+ catalyst and H2O2 and C4H4O62– reactants are in the same/ aqueous phase. [1] (ii) By considering the structure of H2O2, suggest why the reaction between hydrogen peroxide and tartrate ion has a high activation energy. The lone pairs of electrons on the oxygen atoms of H2O2 repel the negatively charged tartrate ion. [1] source + - 8 source - + Deflection starts only after this boundary; else zero for all drawings  8 15

5 © Jurong Pioneer Junior College 8873/02/J2 PRELIMINARY EXAMINATION /2024 [Turn over (iii) The reaction between hydrogen peroxide and tartrate ion is very colourful as pink Co2+ is oxidised to green Co3+ by hydrogen peroxide in the first step. This is followed by the reduction of Co 3+ back to Co 2+ when it reacts with the tartrate ion . The colour changes continues until all the reactants are used up. Outline the mode of action of how cobalt( II) chloride catalyses the oxidation of tartrate ion by hydrogen peroxide. State the species involved in each step of the reaction. Step 1: Co2+ would first be oxidised by H 2O2 into Co3+. H2O2 is reduced to H2O/OH–. Step 2: Co3+ then oxides C4H4O62– into CO2 and Co2+ is regenerated. [2] (iv) Hence, construct an energy profile diagram to show changes, if any, to the enthalpy change or activation energy before and after the addition of cobalt(II) chloride. enthalpy / kJ mol-1 progress of reaction reactants products [2] (d) Hydrogen peroxide and iodide ions react in acidic solution as shown. H2O2(aq) + 2I–(aq) + 2H+(aq) → 2H2O(l) + I2(aq) A series of experiments were conducted to determine the rate equation of the reaction. (i) Use the following data to deduce the orders with respect to [H 2O2], [I–] and [H+], explaining your reasoning. Experiment [H2O2] / mol dm–3 [I–] / mol dm–3 [H+] / mol dm–3 Initial rate / mol dm–3 s–1 1 0.05 0.05 0.05 1.25 x 10–3 2 0.10 0.05 0.05 2.50 x 10–3 3 0.10 0.20 0.05 1.00 x 10–2 4 0.15 0.10 0.10 7.50 x 10–3 Activation energy without catalyst Activa