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 s