2025 DHS H1 Chem Prelim Paper 2 Suggested Solutions

© DHS 2025 8873/02 [Turn over Suggested Solutions DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H1 CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 8873/02 18 September 2025 2 hours READ THESE INSTRUCTIONS FIRST Write your centre number, index number, name and class at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. You may lose marks if you do not show your working or if you do not use appropriate units. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 11 2 9 3 11 4 10 5 14 6 5 Section B 7 / 8 20 Total 80 This document consists of 23 printed pages.

2 © DHS 2025 8873/02 Section A Answer all the questions in the spaces provided. 1 (a) Use of the Data Booklet is relevant to this question. Chlorine has 25 isotopes, ranging from 28Cl to 52Cl. (i) Complete Table 1.1 to show the number of each sub-atomic particle present in an 39Cl+ ion. Table 1.1 number of protons number of electrons number of neutrons 39Cl+ 17 16 22 [1] A sample of chlorine contains three isotopes. Data for two of the isotopes present in this sample is shown in Table 1.2. Table 1.2 mass number abundance 35 68.43% 37 21.70% The sample of chlorine has a relative atomic mass, Ar, of 35.53. (ii) Calculate the mass number of the third isotope. [2] Abundance of the third isotope = 100% – 68.43% – 21.70% = 9.87% Let the mass number of the third isotope be x. 35.53 = (35 × 68.43 100 ) + (37 × 21.70 100 ) + (𝑥 × 9.87 100) 𝑥 = 35.97 ≈ 𝟑𝟔 (𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑤ℎ𝑜𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟) A beam of 35Cl+ ions is deflected in an electric field. (iii) State one similarity and one difference in the expected behaviour of a beam of 35Cl– ions in the same electric field. Briefly explain your answer. [2] Similarity: The angle of deflection of both beams are of the same magnitude. This is because both ions have the same | 𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 | ratio.

3 © DHS 2025 8873/02 [Turn over Difference: The beams are deflected to opposite directions. The beam of 35Cl+ cations is attracted to the negative plate while the beam of 35Cl– anions is attracted towards the positive plate. (b) Beryllium chloride is a simple molecule with an electron–deficient central atom. (i) Draw a ‘dot–and–cross’ diagram for the BeCl2 molecule. [1] Bex xCl Cl (ii) Name the shape of the BeCl2 molecule. [1] Linear (iii) Deduce whether BeCl2 is a polar molecule. Explain your answer. [1] It is a non–polar molecule as the dipole moments cancel out. BeCl Cl BeCl2 reacts with S(CH 3)2 in a 1:2 ratio. Co -ordinate bonds are formed between Be and S atoms. (iv) Draw a diagram to show the structure of the product. You should also show the co-ordinate bonds clearly. [1] S(CH3)2 S(CH3)2 Be Cl Cl (v) Suggest the bond angle around beryllium in the product. Explain your answer using Valence Shell Electron Pair Repulsion theory. [2] The bond angle is 109.5°. The 4 bond pairs are arranged as far apart as possible to minimise repulsion. [Total: 11]

4 © DHS 2025 8873/02 2 (a) Hydrogen peroxide can undergo both oxidation and reduction. On its own, it disproportionates to form water and oxygen gas. 2H2O2(aq) → 2H2O(l) + O2(g) H = –196 kJ mol–1 (i) Sketch the energy profile diagram for the decomposition of H 2O2(aq) in Fig. 2.1. Label the activation energy Ea and the enthalpy change of reaction H. Fig. 2.1 [2] energy / kJ mol -1 2H2O2(aq) 2H2O(l) + O2(g) H = -196 Ea, uncatalysed = + 75.3 Ea, catalysed progress of reaction The uncatalysed reaction has an activation energy of +75.3 kJ mol –1 and occurs slowly at room temperature. Presence of iodide catalyst changes the activation energy of the decomposition reaction by 18.8 kJ mol–1. (ii) On Fig. 2.1, add the energy profile diagram for the catalysed reaction. [1] (iii) Determine the activation energy of the reverse reaction in the presence of iodide catalyst. [1] Ea of catalysed (forward) reaction = 75.3 – 18.8 = 56.5 kJ mol–1 Ea of reverse reaction with catalyst = 196 + 56.5 = 252.5 kJ mol–1

5 © DHS 2025 8873/02 [Turn over The iodide-catalysed decomposition of H2O2 occurs in two steps: step 1 H2O2 + I– → OI– + H2O step 2 H2O2 + OI– → H2O + O2 + I– (iv) Explain, in terms of oxidation state, what happens to H2O2 in step 1. [1] H2O2 is reduced as the oxidation state of oxygen is decreased from –1 in H2O2 to –2 in H2O or OI–. (b) Under acidic conditions, H2O2 oxidises I– to I2. (i) Write a balanced equation for the reaction between I– and H2O2 under acidic conditions. [1] [O] 2I– → I2 + 2e– [R] H2O2 + 2H+ + 2e– → 2H2O Overall: H2O2 + 2H+ + 2I– → 2H2O + I2 (ii) Describe the expected colour change as the reaction proceeds. [1] The mixture turns from colourless to brown. (c) In the presence of an acid, 0.100 mol of H2O2 reacts with 0.100 mol of hypobromous acid HOBr to form oxygen gas and a bromine–containing product. (i) State the change in oxidation state of bromine in this reaction. [1] Half-equation for oxidation of H2O2 H2O2 → O2 + 2H+ + 2e– Since H2O2 and HOBr reacts in a 1:1 ratio, each mole of HOBr gains 2 moles of electrons. The oxidation state of bromine is decreases by 2 OR from +1 to –1. (ii) Hence suggest a possible identity of the bromine–containing product. [1] Br– or HBr [Total: 9]

6 © DHS 2025 8873/02 3 (a) Fig. 3.1 shows the setup to carry out an experiment t o determine the standard enthalpy change of combustion of propan-2-ol, CH3CH(OH)CH3, an ingredient commonly used in alcohol wipes and hand sanitisers. It is a versatile reagent that can be used in the synthesis of many products. Fig. 3.1 Table 3.1 shows the experimental data obtained from the experiment. Table 3.1 initial mass of spirit burner and propan-2-ol 44.83 g final mass of spirit burner and propan-2-ol 41.98 g mass of water heated 500 g temperature rise 60.0 oC (i) Define the term standard enthalpy change of combustion. [1] Standard enthalpy change of combustion (Hc) of a substance is the energy released when one mole of the substance is completely burnt in excess oxygen under standard conditions. (ii) Write an equation which describes the standard enthalpy change of combustion, ∆Hc, of liquid propan-2-ol. [1] CH3CH(OH)CH3(l) + 9 2 O2(g) ⎯→ 3CO2(g) + 4H2O(l) (iii) Use the data in Table 3.1 to calculate the standard enthalpy change of combustion of liquid propan-2-ol. [Assume the specific heat capacity of water = 4.2 J g−1 K−1] [3] thermometer beaker containing 500 g of water spirit burner containing propan-2-ol

7 © DHS 2025 8873/02 [Turn over heat absorbed by water = mcT = (500) × 4.2 × 60.0 = 1.26 × 105 J mass of propan-2-ol burnt = 44.83 – 41.98 = 2.85 g n(propan-2-ol) burnt = 2.85  [3(12.0) + 8(1.0) + 16.0] = 4.75 x 10–2 mol Since temperature rises, the reaction is exothermic (i.e. H < 0). ∆Hc of liquid