YIJC 2020 Prelim P4 Ans (1)

1 [Turn over YIJC 2020 Prelim H2 Chemistry Prelim P4 Suggested Answers 1 (b) Volume of FA 3 added to mixture /cm3 6.00 11.00 15.00 20.00 Final burette reading /cm3 27.80 20.70 15.50 8.50 Initial burette reading /cm3 0.00 0.00 0.00 0.00 Volume of FA 2 added /cm3 27.80 20.70 15.50 8.50 (c) Vmax (FA 3) = 26.25 cm3 Vmax(FA 2) = 36.00 cm3 [6] (d) gradient = 36.0 − 0.0 0.0 − 26.25 = −1.37 (e) The graph has a negative gradient because the more M2CO3 added, the lesser the amount of H2SO4 left in the mixture and hence, lesser NaOH is needed to neutralise the H2SO4. (f) (i) Since Vmax(FA 2) = 36.00 cm3, nNaOH = 0.550 × 36.0 × 10-3 = 0.0198mol nH2SO4 = 1 2 × nNaOH = 1 2 × 0.0198 = 0.0099mol cH2SO4 = 0.0099 25.0 × 10-3 = 0.396 mol dm-3 (ii) nM2CO3 = nH2SO4 = 0.0099 mol Since Vmax(FA 3) = 26.25 cm3, cM2CO3 = 0.0099 26.25 × 10-3 = 0.377 mol dm-3

2 (iii) Mr of M2CO3 = 39.75 0.377= 105.4 Ar of M = 105.4-12.0-(16.0×3) 2 = 22.7 (g) Some NaOH will react with CO 2 in the air. Amount of NaOH left in FA 2 solution will decrease /[NaOH] decreases, resulting in a larger value of Vmax (FA 2) required to neutralise the H2SO4 solution. (h) % error for mixture 1 = 0.05 × 2 27.80 × 100% = 0.360% % error for mixture 2 = 0.05 × 2 8.50 × 100% = 1.18% The percentage error is smaller as the titre volume for mixture 1 is larger than that of mixture 2. (i) (i) Experiment Volume of FA 5 /cm3 Volume of FA 6 /cm3 Initial temperature of FA 5 /oC Maximum temperature /oC Maximum change in temperature/oC 1 20.0 40.0 30.4 37.4 7.0 2 40.0 20.0 30.8 38.1 7.3 (ii) Acid X is dibasic. Amount of water formed in experiment 2 is same as that in experiment 1. Heat released for experiment 2 is same as that of experiment 1. For the same volume of solution, change in temperature will be the same. 2 (a) Experiment Volume of FA 7 /cm3 Volume of FA 8 /cm3 Volume of FA 10/cm3 volume of deionised water/cm3 t/s 1/t/s-1 1 5.0 5.0 10.0 10.0 69 0.0145 2 10.0 5.0 10.0 5.0 35 0.0286 3 5.0 10.0 10.0 5.0 34 0.0294 4 5.0 5.0 20.0 0.0 18 0.0556 (b) Total volume should be constant so that the concentration of reactants is directly proportional to the volume used. (c) • [BrO3–] Comparing experiments 1 and 2, When [BrO3–] doubles, (keeping [Br–] and [H+] constant), relative rate doubles. Hence, order of reaction with respect to [BrO3–] is one. • [Br–] Comparing experiments 1 and 3, When [Br–] doubles, (keeping [BrO3–] and [H+] constant), relative rate doubles. Hence, order of reaction with respect to [Br–] is one. • [H+] Comparing experiments 1 and 4,

3 When [H+] doubles, (keeping [BrO3–] and [Br–] constant), relative rate quadruples. Hence, order of reaction with respect to [H+] is two. (d) Rate = k [BrO3–][Br–][H+]2 (e