ASRJC 2019 Prelim P4 Ans

ASRJC JC2 PRELIMS 2019 9729/04/H2 1 ANDERSON SERANGOON JUNIOR COLLEGE H2 Chemistry 9729 2019 JC2 Prelim Exam Paper 4 Solutions 1 (a) Results Chosen time/min Actual time t/min Initial burette reading/cm3 Final burette reading/cm3 Volume of FA 4 used/cm3 4 4 min 6s 4.1 0.00 15.90 15.90 8 8 min 3s 8.1 15.90 30.30 14.40 12 12 min 3s 12.1 0.00 12.90 12.90 16 15 min 56s 15.9 12.90 24.40 11.50 20 19 min 59s 20.0 24.40 34.50 10.10 (b) (i) (ii) Order of reaction is zero with respect to I2 because it is straight line with constant gradient / rate of reaction is independent of [ I2] / [I2] decreases linearly with time / [I2] does not affect rate / [I2] decreases at a constant rate. (c) (i) Rate = k [CH3COCH3][H+]

ASRJC JC2 PRELIMS 2019 9729/04/H2 2 (ii) Gradient = y2–y1 x2–x1 = 9 – 14.8 18.75 = –0.309 cm3 min–1 (3 s.f with units) Show coordinates on the graph or in the working (iii) rate of change = (c)(ii) x 0.01/1000 = –0.309 x 0.01/1000 = –3.09 x 10–6 = –3.09 x 10–6 mol min–1 (3 s.f.) (iv) I2 ≡ 2S2O32– rate of disappearance of I2 = 1 2 = 1 2 x |(c)(iii)| x 3.09 x 10–6 = 1.545 x 10–6 = 1.55 x 10–6 mol min–1 (3 s.f.) (v) rate of change of [I2] = –(c)(iv) ÷ 10/1000 = –1.545 x 10–6 ÷ 10/1000 ≈ –1.55 x 10–4 mol dm–3 min–1 (3 s.f with units) (vi) rate = k[CH3COCH3][H+] 1.545 x 10–4 = k [CH3COCH3][H+] 1.545 x 10–4 = k (1.00 x ¼)(1.00 x 2 x ¼) k = 1.24 x 10–3 mol–1 dm3 min–1 (3 s.f with units) (d) It is added to react with acid catalyst so that the reaction will stop / It is to quench the reaction by reacting with the acid catalyst. (e) The low concentration of iodine means that very little propanone and acid are reacted away from the reaction mixture and hence the concentration of propanone and acid remain effectively constant. Hence, the order of reaction with respect to iodine can be determined because any change in the rate is due to the change in concentration of iodine. (f) (i) nucleophilic substitution (ii) To prevent reaction of CH3CH2CH2Br with water/hydrolysis (if wet) (iii) Due to the low solubility of NaBr in propanone, NaBr formed gets precipitated out. Position of equilibrium shifts to the right. or Since sodium bromide is sparingly soluble in propanone, there will not be sufficient Br– ions to attack the electrophilic C attached to I, hence position of equilibrium favors right or Since NaI is much more soluble than NaBr, [NaI] is much greater than [NaBr]. Forward rate is much greater than the backward rate. Hence, the reaction favors r ight. (iv) Since 100% yield of pure 1–iodopropane is obtained and sodium iodide is used in excess, n(CH3CH2CH2Br) required = n(CH3CH2CH2I) obtained = 10 / 169.9 = 0.05886 mol mass of CH3CH2CH2Br = 0.05886 x 122.9 = 7.234 g minimum v olume of CH3CH2CH2Br = 7.234 / 1.35 = 5.36 cm3 rate of change is a negative value since [S2O32-] decreases over time Positive value since the direction of change, "disappearance", is specified rate of change is a negative value since [I2] decreases over time

ASRJC JC2 PRELIMS 2019 9729/04/H2 3 (v) Minimum mass of sodium iodide required = 0.05885 x 149.9 = 8.82 g Since sodium iodide is added in excess, mass of sodium iodide to be used > 8.82 g Preparation of crude 1–iodopropane 1. Weigh accurately about 10 g of sodium iodide in a 50 cm3 round–bottomed flask. 2. Using a 10 cm3 measuring cylinder, transfer 6 cm3 of 1–bromopropane to the flask. 3. Using a 50 cm3 measuring cylinder, transfer 25 cm3 of propanone to the flask. 4. Swirl the flask to ensure even mixing. 5. Add boiling chips (anti–bumping granules) to the mixture. 6. Using a water bath / heating mantle, gently heat the flask fitted with a reflux condenser for about 30 minutes. 7. Cool down the flask by removing the water bath / heating mantle. 8. Remove the flask and fractionally distil the mixture to obtain the pure 1–iodopropane at its boiling point of 102.6 oC. (Note: solid NaI and NaBr will remain in the flask) thermometer heating by heating mantle crude product mixture fractionating column product water in water out condenser

ASRJC JC2 PRELIMS 2019 9729/04/H2 4 2 (a) (i) Results mass of empty weighing bottle / g mass of weighing bottle + KI / g mass of weiging bottle + residual KI / g mass of KI used / g 2.018 initial temperature of water / °C minimum temperature reached / °C ∆TKI / °C 2.0 (ii) 4 2 6 0 1 2 3 4 5 6 7 8 8 10 12 ∆TKI / °C mass of KI / g 0 3.0

ASRJC JC2 PRELIMS 2019 9729/04/H2 5 (b) ∆Tm = 48.8 – 29.0 = 19.8 °C (read T max and T min correctly to ±½ small square, i.e. to 0.1 °C (1 d.p) t / min T / °C 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 28.0 30.0 32.0 34.0 36.0 38.0 40.0 42.0 44.0 46.0 48.0 50.0 52.0 29.0 48.8

ASRJC JC2 PRELIMS 2019 9729/04/H2 6 (c) (i) m = mass of FA 6 used 2 = 6.020 2 = 3.010 g Using the graph in (a)(ii), ∆TKI = 3.0 °C Or correctly calculates using ∆T per gram of KI by finding gradient Working on the graph, e.g. dotted lines or otherwise, must be shown. (ii) ∆Tm – ∆TKI = ∆TLiCl = 19.8 + 3.0 = 22.8 °C (1 d.p) (iii) ∆T per gram of LiCl =22.8 (6.020÷2) = 7.57 °C g−1 (3 s.f) (d) (i) Insoluble barium chromate is formed. (ii) The formation of insoluble barium sulfate will ensure that insoluble barium chromate is not formed during titration. (iii) (iv) 1. Weigh accurately about 3.00 g of BaCl2.xH2O in a pre–weighed weighing bottle, using a weighing balance (or electronic balance) 2. Dissolve this solid in a beaker with 30 cm3 of water. 3. Transfer the solution and washings into a 250 cm3 graduated flask and make up to the mark with deionised water 4. Stopper the graduated flask and shake the solution to obtain a homogeneous solution. 5. Reweigh the emptied weighing bottle. (v) First: sulfuric acid Second: potassium chromate(VI) Third: silver nitrate mass of AgCl volume of AgNO3 0

ASRJC JC2 PRELIMS 2019 9729/04/H2 7 3 (a) t est observations 1. Add 1 cm depth of FA 7 to a test–tube. Add aqueous ammonia slowly, with shaking, until no further change is seen. Off-white ppt. formed, insoluble in excess NH3(aq). Off-white ppt. rapidly turned brown on contact with air. White ppt, soluble in excess NH3 to form a colourless solution. Filter the mixture and add dilute nitric acid drop–wise to the filtrate, until no further change is seen. (You may continue with test 2, while waiting for the filtration process to complete.) Filtrate is colourless Residue is brown in colour White ppt refomed, soluble in excess acid 2. Add 1 cm depth of FA 7 to a test–tube. A dd aqueous barium nitrate to FA 7, White ppt formed followed by dilute nitric acid. White ppt insoluble in acid FA 7 contains the cations Mn2+ and Zn2+. identity evidence FA 7 contains cation: Mn2+ In test 1, when aq NH3 is added to FA 7, an off-white ppt of Mn(OH)2 is formed which is insoluble in excess aq. NH3. Mn(OH)2 is oxidised by oxygen in air to brown Mn(OH)3. FA 7 contains cation: Zn2+ In test 1, when aq NH3 is added to FA 7, a white ppt of Zn(OH)2 is formed, which is soluble in excess aq. NH3 to form a colourless solution of [Zn(NH3)4]2+ FA 7 contains the anion SO42–.

ASRJC JC2 PRELIMS 2019 9729/04/H2 8 (b) (i) test observations 1. Add 1 cm depth of FA 8 to a test–tube. Add FA 4 slowly, with shaking, until no further change is seen. Yellow solution turns brown/reddish brown/dark orange/violet Leave it to stand. Yellow solution reforms 2. Add 1 cm depth of FA 8 to a test–tube. Add 1 cm depth of FA 5 , with shaking, until no further change is seen. Orange/red–brown ppt. Effervescence observed. Colourless CO2 gas evolved gives a white ppt with Ca(OH)2/ limewater. 3. Add 1 cm depth of FA 8 to a test–tube. Add 1 cm depth of aqueous potassium iodide, with shaking, until no further change is seen. Yellow solution turns brown. (ii) Fe3+ has a high charge density. Hence, the cation can polarise the water molecules, giving rise to an acidic