2025 DHS H2 Chem Prelim Paper 2 Suggested Solutions

© DHS 2025 9729/02 [Turn over Suggested Solutions DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 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. 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 1 15 2 17 3 12 4 10 5 21 Total 75 This document consists of 16 printed pages.

2 © DHS 2025 9729/02 Answer all the questions in the spaces provided. 1 (a) Fig. 1.1 shows the relative first ionisation energies of six consecutive elements, A to F, in the Periodic Table with atomic number less than 20. The letters are not the symbols of the elements. Fig. 1.1 (i) Write an equation to represent the first ionisation energy of A. [1] A(g) → A+(g) + e− (ii) Explain why the first ionisation energy of F is more than that of E. [1] Both elements have similar shielding effect. However, F has a greater nuclear charge. Hence, F has a greater effective nuclear charge than E. The attraction between the nucleus and the outermost electron is stronger in F. (iii) Which element has the highest 4th ionisation energy? Explain your answer. [1] Element D. It is in Group 13 with outermost shell electronic configuration of ns2 np1. The 4th ionisation energy involves removal of an electron from an inner electron shell. Or 4th valence electron is removed from an inner electron shell. (iv) Element E is in Period 3. Identify element E and state the number of electron pairs in an atom of E. [1] element E is silicon number of electron pairs in an atom of E is 6 (b) 10.0 cm 3 of 0.10 mol dm −3 aqueous bromine was added to 50.00 cm 3 of a 0.10 mol dm−3 sodium hydroxide solution. The products formed were Br− and BrOx−. The excess sodium hydroxide required 15.00 cm3 of 0.20 mol dm−3 hydrochloric acid for complete neutralisation. Calculate the mole ratio between aqueous bromine and sodium hydroxide in the redox reaction.

3 © DHS 2025 9729/02 [Turn over Hence write a balanced equation for the reaction and deduce the value of x. [3] Initial amount of NaOH = 50.00 1000  0.10 = 5.00  10−3 mol amount of NaOH reacted with HCl(aq) = 15.00 1000  0.20 = 3.00  10−3 mol amount of NaOH reacted with aqueous Br2 = (5.00  10−3) – (3.00  10−3) = 2.00  10−3 mol amount of Br2 used = 10.00 1000  0.10 = 1.00  10−3 mol mole ratio of Br2 : NaOH = 1 : 2 Balanced ionic equation: Br2(aq) + 2OH−(aq) → Br−(aq) + BrO−(aq) + H2O(l) x =1 (c) The halogens Cl2 and I2 both react similarly with H2S. The reaction of Cl2 with H2S is shown in equation 1. equation 1 Cl2 + H2S → 2HCl + S (i) Predict which halogen, Cl2 or I2, has a greater reactivity when added to H2S. Explain your answer in terms of the role of the halogen in these reactions. [1] Cl2 has a greater reactivity than I2 as Cl2 is a stronger oxidising agent than I2. (ii) The white fuming gaseous products, HC l and HI, were collected in separate jars. A piece of red-hot wire was plunged into each jar and purple fumes were observed in one of them. Explain the observation. [2] Purple fumes observed is iodine vapour when HI thermally decomposed 2HI → H2(g) + I2(g) Bond energy of H–I < H–Cl the covalent bond length of H–I > H–Cl. A lower amount of energ y is required to overcome the weaker H –I covalent bond. Thermal stability of HI < HCl , HI decomposes readily in the presence of a red-hot wire to give purple iodine vapour. However, insufficient energy is provided to overcome H–Cl bond. HCl does not decompose. (iii) Both HI(g) and HCl(g) dissolve readily in water. Suggest a reagent, other than aqueous silver nitrate, that could be used to distinguish between the aqueous solutions of these two gases. Describe the expected observations. [2]

4 © DHS 2025 9729/02 Add Cl2(aq) or Cl2 gas into HCl(aq) and HI(aq) separately. With HCl (aq), solution remains colourless. With HI (aq), solution turns brown due to the formation of I2(aq). OR Add Br2(aq) into HCl(aq) and HI(aq) separately. With HCl (aq), solution turns orange. With HI (aq), solution turns brown due to the formation of I2(aq). Also accept: Add Pb(NO3)2(aq) into HCl(aq) and HI(aq) separately. With HCl (aq), white ppt of PbCl2. With HI (aq), yellow ppt of PbI2. OR Add Cu(NO3)2(aq) into HCl(aq) and HI(aq) separately. With HCl (aq), solution turns blue (due to Cu2+ present). With HI (aq), white ppt (due to CuI) in brown solution (due to I2) (d) Bromine and fluorine react together to give bromine trifluoride. Br2(l) + 3F2(g) → 2BrF3(l) Using the data in Table 1.1, together with data from the Data Booklet, construct a fully labelled energy cycle to calculate the average bond energy of the Br−F bond in BrF3. Table 1.1 process H / kJ mol−1 Standard enthalpy change of formation of BrF3(l) −301 Enthalpy change of vaporisation of Br2(l) +31 Enthalpy change of vaporisation of BrF3(l) +44 [3] Br2(l) + 3F2(g) 2BrF3(l) 2(−301) Br2(g) + 3F2(g) 2BrF3(g) 2Br(g) + 6F(g) 2(+44) 6 x BE(Br−F) +193 +3(+158) +31 By Hess’ Law,

5 © DHS 2025 9729/02 [Turn over 2(–301) = (+31) + (+193) + 3(+158) – 6BE(Br–F) – 2(+44) average BE(Br–F) = + 202 kJ mol–1 [Total: 15] 2 The kinetics of the Finkelstein reaction between bromobutane and sodium iodide in propanone forming solid sodium bromide was studied in a series of experiments. Br + NaI I + NaBr(s) (a) In experiment 1, 10.0 cm 3 of 0.10 mol dm –3 bromobutane and 15.0 cm 3 of 1.0 mol dm–3 sodium iodide were mixed. Fig. 2.1 shows the concentration of bromobutane against time, t, for this experiment. (i) Use the graph in Fig. 2.1 to determine the order of reaction with respect to bromobutane. Show your working clearly. [2] Fig. 2.1 Since t½ is constant at 420 s, the reaction is first order wrt bromobutane. (ii) By drawing a tangent at t = 0 s, determine the initial rate of reaction. Include its units. [2] rate of reaction = 0.04 610 = 6.56  10–5 mol dm–3 s–1 0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0 100 200 300 400 500 600 700 800 900 1000 time, t / s t½ t½ [bromobutane] / mol dm−3

6 © DHS 2025 9729/02 (b) In experiments 2 and 3, the time taken for a small and fixed amount of NaBr(s) to be formed was measured. The results obtained are found in Table 2.1. Table 2.1 experiment initial [bromobutane] / mol dm–3 initial [sodium iodide] / mol dm–3 time / s 2 0.60 0.60 21 3 0.40 0.40 47 (i) Use Table 2.1 to determine the order of reaction with respect to sodium iodide. Show your working clearly. [1] Since the extent of reaction is kept constant, relative rate  1 t . Let the order of reaction w.r.t. NaI be x. Comparing the rate of experiments 2 and 3, rate2 rate3 = k[bromobutane][NaI]x k[bromobutane][NaI]x 1/21 1/47 = k × 0.60 × (0.60)x k × 0.40 × (0.40)x 1.49 = ( 0.60 0.40) x x = 1 (to