2024 MI PU3 H2 Chem EOY P3 answers (final) w EC

Class Adm No Candidate Name: This question paper consists of 28 printed pages. 2024 Preliminary Examination Pre-University 3 H2 CHEMISTRY 9729/03 Paper 3 Free Response 13th Sep 2024 2 hours Candidates answer on separate paper. Additional materials: Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number on all the work you hand in. 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. If additional space is required, you should use the page at the end of this booklet. The question number must be clearly shown. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Question A B Total 1 2 3 4 / 5 Marks 19 23 18 20 80

2 Section A Answer all the questions in this section. 1 The halogens are elements in Group 17 of the Periodic Table . The name “halogen” translates to “salt producer”, due to their tendency to form salts readily in the presence of Group 1 metals such as sodium. (a) (i) Citing relevant data values, state and explain the trend of reactivity of the halogens as oxidising agents. [3] Electrode reaction Eo / V F2 + 2e– ⇌ 2F– +2.87 Cl2 + 2e– ⇌ 2Cl– +1.36 Br2 + 2e– ⇌ 2Br– +1.07 I2 + 2e– ⇌ 2I– +0.54 any 3, equation not required [1] Down Group 17, reactivity of the halogens decreases. [1] Down the group, Eo value becomes less positive and the halogens are less readily reduced. [1] Examiners’ Comments: - Many candidates can use the correct concept to explain. - Most common mistakes include the use bond energy to explain, not linking back to the question on reacti vity, and not giving at least 3 Eo values to state the trend. - (ii) Outline a procedure to demonstrate the relative reactivities of chlorine and bromine, given only the following reagents to choose from: Cl2(aq) Cl–(aq) Br2(aq) Br–(aq) [2] To a test tube containing 1 cm3 of Br–(aq), add 1 cm3 of Cl2(aq). [1] sufficient, but can also do the reaction between Cl– and Br2 The colourless solution will turn orange as the more reactive C l2 displaces Br – to produce Br2. [1] Examiners’ Comments: - Few candidates gave succinct procedure despite the question using “outline” as a command word. - A number of candidates also did not mention the initial colour of the solution for the observation.

3 [Turn over (b) Gaseous PCl3 and PCl5 exist in an equilibrium. PCl3(g) + Cl2(g) ⇌ PCl5(g) At 250 °C, the Kc for this equilibrium has a value of 26.0. (i) A sample of solid PC l5 is added into a sealed 2 dm 3 container at 250 °C, and the solid quickly sublimes. After equilibrium was established, there was found to be 0.20 mol of Cl2. Determine the initial mass of solid PCl5 added. [4] Let y be the initial [PCl5(g)]. [ ] / mol dm–3 PCl3(g) + Cl2(g) PCl5(g) initial 0 0 y change +x +x -x equilibrium x = 0.10 x = 0.10 y-x = y-0.10 ICE table with correct algebraic formulas [1] x = 0.20 / 2 = 0.10 mol dm-3 Kc = [PC𝑙5(g)] [PC𝑙3(g)][C𝑙2(g)] [1] 26.0 = 𝑦−0.1 0.12 y = 0.36 mol dm-3 [1] initial mass of PCl5(g) = (0.36 x 2) x [31.0 + 5(35.5)] = 150 g (3sf) [1] Examiners’ Comments: - A significant number of candidates did not have proper headers and units for their ICE table. - Some candidates calculated Kc using amount instead of concentration. (ii) In a modified experiment, some Na(s) was present in the container initially before addition of solid PCl5 (all other conditions were kept the same). Explain the effect of this on the position of equilibrium, and the value of Kc. [3] Na(s) reacts with Cl2(g) produced, decreasing [Cl2(g)]. [1] By Le Chatelier’s Principle, position of equilibrium shifts to the left to increase [Cl2(g)]. [1] No change to Kc as temperature remains constant. [1] Examiners’ Comments: - A number of candidates did not know the species that reacts with Na(s). - Majority of the candidates lost the second mark as they linked Kc to how POE shifted.

4 (iii) In the solid state, PCl5 exists as an ionic solid, with [PCl6]– as the anion. Identify a formula unit of the ionic solid. [1] [PCl4]+[PCl6]– [1] Examiners’ Comments: - Candidates are unfamiliar with the term “formula unit” and some gave just the cation or anion. (iv) Describe how PCl5 acts as a n acid in the formation of [PC l6]–, writing an equation to explain your answer. [1] PCl5 is a Lewis acid as the P atom accepts a lone-pair from Cl–. PCl5 + Cl– → [PCl6]– [1] Examiners’ Comments: - Majority of the candidates cannot give the correct equation. (v) Unlike PCl5, PI5 does not exist. Suggest an explanation for why this is the case. [1] The size / atomic radius of I is larger than C l, leading to very large and unfavourable steric hindrance around the P atom. [1] Examiners’ Comments: - A significant number of candidates did not use the correct correct to explain.

5 [Turn over (c) A 25.0 cm3 sample of water contains chlorine, Cl2, and monochloramine, NH2Cl. The amounts of chlorine and monochloramine in this sample can be analysed using the DPD-FAS titration method, where DPD serves as the indicator and Fe2+ the titrant. The concentration of chlorine is determined first: Step 1: Excess of indicator DPD (colourless) is added to the 25.0 cm 3 sample, turning the indicator magenta and reducing Cl2. DPD + Cl2 → DPD2+ + 2Cl– colourless magenta Step 2: The resultant mixture is then titrated against Fe2+. DPD2+ + 2Fe2+ → DPD + 2Fe3+ From the same mixture, the concentration of monochloramine is determined next: Step 3: A small amount of catalyst is added, catalysing the reduction of NH2Cl. DPD + NH2Cl + 2H+ → DPD2+ + NH4+ + Cl– Step 4: The resultant mixture is then titrated against Fe2+, same as Step 2. DPD2+ + 2Fe2+ → DPD + 2Fe3+ When 0.0010 mol dm -3 of Fe 2+ was used as the titrant, the titres obtained at the end of Steps 2 and 4 are 15.00 cm3 and 5.00 cm3 respectively. (i) Calculate the amount of Cl2 in the 25.0 cm3 water sample. [1] nFe2+ = 0.0010 x 15.00 1000 = 1.5 x 10-5 mol nCl2 = 1.5×10−5 2 = 7.5 x 10-6 mol [1] Examiners’ Comments: - Candidates did well for this question. (ii) Calculate the amount of NH2Cl in the 25.0 cm3 water sample. [1] nFe2+ = 0.0010 x 5.00 1000 = 5.0 x 10-6 mol nNH2Cl = 5.0×10−6 2 = 2.5 x 10-6 mol [1] Examiners’ Comments: - Candidates did well for this question.

6 (iii) Hence, determine the total concentration of Cl atoms (in g dm-3) in the water sample. [2] total nCl = 2(7.5x10-6) + (2.5x10-6) = 1.75 x 10-5 mol [1] [Cl] = (1.75×10−5) × 35.5 25/1000 = 0.0249 g dm-3 (3sf) [1] Examiners’ Comments: - A significant number of candidates did not multiply the amount of Cl atoms by 2 even though there are 2 Cl atoms in Cl2. [Total: 19]

7 [Turn over 2 (a) Compound A, C 5H8O2, is neutral . An orange precipitate is observed when 2,4-dinitrophenylhydrazine is added to it , but no precipitate is observed when Fehling’s reagent is added. 1 mol of A reacts with acidified KMnO 4 to produce CO2 gas and a single organic product X. A also turns orange acidified K2Cr2O7 green, forming another organic product Y.