H1 Chemistry Prelim P2 Answers ACJC

This document consists of 25 printed pages and 1 blank page. Anglo-Chinese Junior College JC2 Preliminary Examination Higher 1 CANDIDATE NAME FORM CLASS TUTORIAL CLASS 2CHX _ _ INDEX NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 8873/02 27 Aug 2025 2 hours READ THESE INSTRUCTIONS FIRST Write your name and index number in the spaces on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer one question. The use of 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 Examiners’ use only Section A 1 / 13 2 / 7 3 / 10 4 / 7 5 / 13 6 / 10 Section B 7 or 8 / 20 Total / 80

2 ACJC 2025 8873/02/2025 Section A Answer all questions in this section in the spaces provided. 1 In the presence of an acid catalyst, ethanal reacts with methanol to form 1,1-dimethoxyethane. CH3CHO(l) + 2CH3OH(l) ⇌ CH3CH(OCH3)2(l) + H2O(l) ethanal methanol 1,1-dimethoxyethane (a) 0.20 mol dm−3 of ethanal and 0.1 mol dm−3 of methanol were placed in a reaction vessel. Strong acid was then added as a catalyst. It was found that the equilibrium mixture contained 0.025 mol dm−3 of 1,1-dimethoxyethane. Calculate the equilibrium concentrations of the other reactants and products. CH3CHO CH3OH 1,1- dimethoxyethane H2O Initial/ mol dm−3 0.20 0.10 0.00 0.00 Change/ mol dm−3 – 0.025 – 0.05 + 0.025 + 0.025 At equilibrium/ mol dm−3 0.175 0.05 0.025 0.025 [3] (b) Write the expression for the equilibrium constant for this reaction, Kc, stating its units. Kc = [CH3CH(OCH3)2] [H2O]/ [CH3CHO] [CH3OH]2 mol-1dm3 [2] (c) Use your answers in (a) to calculate the value of Kc. Kc = [0.025] [0.025]/ [0.175] [0.05]2 = 1.43 mol-1dm3 [1] H+

3 ACJC 2025 8873/02/2025 [Turn over (d) 2-methylpropenoic acid can be obtained from 2-methylprop-2-en-1-ol. (i) To prevent the C=C bond from reacting in other reactions, reaction 1 is first carried out. Draw the structure of A. [1] (ii) Suggest the reagents used in reaction 1. HBr(g) ……………………………………………………………………………………...….[1] (iii) Name the type of reaction in reaction 2 and 3. reaction 2: …………………… reaction 3: …………………… [2] (iv) Draw the displayed formula of 2-methylprop-2-enoic acid. [1] (v) Suggest a chemical test to distinguish between 2-methylprop-2-en-1-ol and B. State the observations. ……………………………………………………………………………………………. ……………………………………………………………………………………….…[2] Either KMnO4 with dilute sulfuric acid, Heat 2-methylprop-2-enoic acid reaction 1 C4H9BrO reaction 2 reaction 3 2-methylprop-2-en-1-ol A A B Oxidation Elimination

4 ACJC 2025 8873/02/2025 B: Purple KMnO4 remains, 2-methylprop-2-en-1-ol: (purple) KMnO4 decolourises Or aqueous bromine B: orange Br2 remains, 2-methylprop-2-en-1-ol: (orange) Br2 decolourises [Total: 13] 2 (a) Sodium polyacrylate is a superabsorbent polymer that can absorb and retain large amounts of water relative to its own mass. It is widely used in diapers. (i) Draw the monomer used to make sodium polyacrylate. [1] (ii) Explain why this polymer is an addition polymer. The monomer has a C=C bond/pi bond which is broken during polymerisation. There is no elimination of small molecule. …………………………………………………………………………………….……… ……………………………………………………………………………………….…[2] (iii) Using the information provided above, explain, in terms of structure and bonding, the property of this hydrogel. The presence of -COO−Na+ in the polymer allows the formation of ion-dipole interactions water molecules hence making it hydrophilic. …………………………………………………………………………………………… …………………………………………………………………………………………[1] (iv) In an experiment, 4.0 x 10−6 mol of polymer absorbed 150 g of water. This mass of water is 300 times the mass of polymer used.

5 ACJC 2025 8873/02/2025 [Turn over Calculate the relative molecular mass of the polymer. Mass of polymer used = 150/300 = 0.5 g Mr = 0.50/4.0 x 10−6 = 125000 [1] (v) After polymerisation, sodium polyacrylate is further processed to form crosslinked sodium polyacrylate found in commercial products. The following diagram shows the structure of crosslinked sodium polyacrylate. Explain how crosslinking of sodium polyacrylate allows it to remain as a solid mass in wet diapers. ……………………………………………………………………………………………. ………………………………………………………………………………………….[2] Crosslinking of polymer chains forms covalent bonds between chains. Thus giving rise to the high tensile strength in this polymer/strengthening the structure of the polymer. [Total: 7]

6 ACJC 2025 8873/02/2025 3 (a) The noble gas xenon was once thought to be inert. It has since been discovered that xenon will react with strong oxidants. For example, xenon reacts with fluorine gas, forming a series of fluorides, XeF2, XeF4 and XeF6. (i) Draw dot-and-cross diagram of xenon difluoride and state its bond angle. 180° [2] (ii) Name the shape of xenon difluoride. ………………………………………………………………………………………….………………….[1] Linear (iii) Explain the three-dimensional arrangement using the principles of the VSEPR theory. ………………………………………………………………………………...………….. ……………………………………………………………………………………...….[2] To minimize repulsion, the electron pairs will spread themselves as far apart as possible. Since lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion and lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, hence the linear arrangement is adopted. OR The linear arrangement is adopted to maximise the separation between the two bond pairs by placing them on opposite sides of plane with the maximum angular separation of 180o so as to minimize the repulsion due to them resulting in a more stable arrangement.

7 ACJC 2025 8873/02/2025 [Turn over (b) The boiling points of some fluoride compounds are shown in Table 3.1. Table 3.1 Explain the difference in boiling points using structure and bonding. compound Mr boiling point/ oC CF4 88 −127.8 ClF3 92.5 11.8 PF3 88 −101.8 …………………………………………………...…………………………………...………….. …………………………………………………...…………………………………...………….. ……………………………………………………………………………………………...….[3] CF4, ClF3 and PF3 exist as simple molecular structures. Though C lF3 and PF 3 are polar, ClF3 has the highest bp due to largest and most polarisable electron cloud which implies that it has strongest id-id interactions. Largest amt of energy. is needed to overcome the interactions. Less energy is required to overcome the weaker instantaneous dipole -induced dipole between CF 4 molecule