MJC 2015 H1 CHEM P2 MS Prelims

1 ©meridian jc 2015 MJC Prelim H1 Paper 2 Suggested Answers 1 (a)(i) P Cl Cl Cl x x x x x (ii) (b) Both PCl3 and PCl5 have simple molecular structure. Size of electron cloud: PCl3 < PCl5 Extent of polarisation of electron cloud: PCl3 < PCl5 Extent of van der Waals’ forces between molecules: PCl3 < PCl5 Energy required to break forces between molecules: PCl3 < PCl5 Boiling point: PCl3 < PCl5 (c) (i). Rate of forward reaction = the rate of reverse reaction. (ii) Kc = 32 5 [PC ] [C ] [PC ] ll l mol dm–3 (iii) Kc = (0.1016) (0.1016) (0.0984) = 0.105 mol dm–3 (iv) When volume is de creased, pressure is increased . The equilibrium position shifts to the left to reduce number of mole of gas to decrease pressure . When pressure increased, concentration of reactant particles increases and particles are closer to each other. Frequency of effective collision of reactant particles increases so rate of reaction increases. (v) Reduction is due to oxidation number of P decrease from +5 in PC l5 to +3 in PCl3. Oxidation is due to oxidation number of C l increases from –1 in PCl5 to 0 in Cl2. 2(a)(i) Al (g)  Al+ (g) + e (ii) Across the period, increase in nuclear charge outweighs negligible increase in screening effect . Effective nuclear charge increases across the p eriod. Stronger

2 ©meridian jc electrostatic attraction between nucleus and first valence electrons. First ionisation energy generally increases. (iii) .The 3p electron in Al is further away from the nucleus than the 3s electron in Mg. (b)(i) (bii) SiCl4 + 2H2O  SiO2 + 4HCl PCl5 + 4H2O  H3PO4 + 5HCl (ci) (ii) 1. giant ionic lattice structure with strong electrostatic force between oppositely charged ions 2. amphoteric (iii) Al2O3 + 6HCl  2AlCl3 + 3 H2O Al2O3 + 2NaOH + 3H2O  2NaAl(OH)4 3(a)(i) ΔHrxn = [2(350) + 610 + 8(410) + 2(460)] – [3(350) + 360 + 460 + 9(410)] = –50 kJ mol–1 (ii) The bond energies from the Data Booklet are average values. (bi) No. of moles of butan-1-ol = 0.972 74.0 = 0.01314 mol Quantity of heat absorbed by water = 20482 J BeO Na2O Al2O3 SO2 1.87 2.52 1.83 0.86

3 ©meridian jc Quantity of heat released by reaction = 100 80 x 20482 J ΔHc(butan-1-ol) = – 25602.5 0.01314 = –1950 kJ mol–1 (ii) The process is not 100% efficient. Some of the heat released by the combustion of butan- 1-ol is lost to the surrounding. (c) Add aqueous iodine in sodium hydroxide to each alcohol respectively and heat. Yellow ppt of tri-iodomethane, CHI3, is observed for the secondary alcohol. No yellow ppt is observed for butan-1-ol. (d)

4 ©meridian jc Section B : Free Response Questions 4 (ai) Starting material to synthesize polypropene (used as plastic in food containers). (ii) Carbon monoxide causes blood poisoning which prevents transport of oxygen by haemoglobin. Unburnt hydrocarbon causes photochemical smog. (b) C H Percentage 82.8% 17.2% Divide by Ar 82.8/12 = 6.9 17.2 Simplest ratio 2 5 Empirical formula = C2H5 Molecular formula of T is C4H10. (c) CxHy + (x + 4 y ) O2  x CO2 + 2 y H2O Volume 10 cm3 100 – 25 50 cm3 = 75 cm3 Mole ratio 1 7.5 5 x = 5 x + 4 y = 7.5 => y = 10 Molecular formula of fragment W is C5H10. (d)(i) Ultraviolet light (ii) Trans: CH2Cl ClH2C and CH3 CH3 Cl Cl Cis: CH2ClClH2C and CH3CH3 Cl Cl

5 ©meridian jc (e)(i) H OH CH3 CH3CH3 (ii) CH3 Cl Cl CH3CH3 (iii) OH O CH3 and O CH3 CH3 (f) Angle of deflection for C2H4 – = –(4  3 52 ) x 1 28 = –2.5 (g) C25H52  C4H10 + C5H10 + 2C4H8 + C6H12 + C2H4 OR C25H52  C4H10 + C5H10 + C4H8 + C6H12 + 3C2H4 5(ai) To slow down / stop the reaction while preparing for titration. (iii) |Gradient| of reaction 1 = 20 44 = 0.455 |Gradient| of reaction 2 = 20 83 = 0.241 Since [chlorocyclopentane] doubles, rate of reaction doubles, order of reaction w.r.t chlorocyclopentane is 1 . Order of reaction w.r.t. OH – is 1 since reaction 1 has a constant half-life of 30 s. (iv) No change (v) Mechanism 1 , because the order of reaction for each of NaOH and chlorocyclopentane is 1.

6 ©meridian jc (b)(i) (ii) (c)(i) (ii) Step I – alcoholic NaOH, heat Step II – Br2 in CCl4 ,r.t.p. (iii) Heat each compound with aqueous NaOH followed by the addition of excess dilute HNO3 and lastly, add in AgNO 3(aq). If a white ppt of AgCl is formed, the compound is J. If a cream ppt of AgBr is formed, the compound is L. (iv) 1 BrCH2CH2OH 2 Substitution reaction 6(a)(i) Concentration of NaOH in g dm–3 = 0.70 10 = 0.070 g dm3 Concentration of NaOH in mol dm–3 = 0.070 23 + 16 + 1 = 1.75 x 10–3 mol dm3 reaction pathway potential energy +41 kJ mol–1 chlorocyclopentane + NaOH 184 kJ mol–1 T2 > T1 No. of particles with energy ≥ Ea at T1 No. of particles with energy ≥ Ea at T2 Number of reactant particles with energy E T1 T2 Ea Kinetics energy cyclopentanol + NaCl

7 ©meridian jc (ii) No. of moles of NaOH needed for neutralisation = 340.0 x 1.75 x 101000  No. of moles of tartaric acid = 3.50 x 10–5 mol Concentration of tartaric acid = 3.50 x 10–5  1000 0.25 = 1.40 x 10–3 mol dm3 (iii) The pH transition range of methyl orange does not lie within the range of rapid pH change over the equivalence point. (b)(i) Ka = ]OHC[ ]OHC][H[ 664 654  (ii) C4H6O6 (aq) C4H5O6 –(aq) + H+(aq) Initial conc 2 0 0 Change in conc -0.1 x 2 = -0.2 +0.2 +0.2 Equilibrium conc 1.8 0.2 0.2 Ka = [0.2]2 / 1.8 = 0.0222 mol dm-3 (c) A buffer solution is a solution which resists changes in pH when a small amount of acid and base is added to it. When a small amount of acid, H+ is added, A– + H+  HA When a small amount of base, OH– is added, HA + OH–  A– + H2O (d) Step I: Cl2 in CCl4 or Cl2(g), r.t.p in the dark Step II: ethanolic KCN, heat Step III: HCl(aq) or H2SO4(aq), heat (e) Acetoin undergoes condensation with 2,4-DNPH,  Acetoin has ketone or aldehyde group.

8 ©meridian jc Acetoin undergoes oxidation with alkaline iodine.  Acetoin has the –COCH3 and/or CH3CH(OH)– structure. After acetoin undergoes oxidation with KMnO4/H+, product of acetoin did not undergo neutralisation reaction with Na2CO3.  Product of oxidation did not have carboxylic acid group.  Acetoin has 2o alcohol group. E undergoes neutralisation reaction with Na2CO3.  E has carboxylic acid group. E undergoes substitution with bromine under UV light to give 2 monobromo compounds.  E contains 2 types of H atom for substitution. Acetoin: OH H3C CH C O CH3 Compound E: H3C CH C O OH CH3