PJC H2 CHEM P2 ANS

Pioneer Junior College H2 Chemistry Preliminary Examination 2009 Paper 2 Answer Scheme 1(a)(i) Relative atomic mass of an element is defined as the ratio of the average mass of one atom of the element to 121 the mass of an atom of 12C. (ii) Let the percentage abundance of Ga-69 be x. 69x + (1-x)71 = 69.7 x = 0.65 [1] Therefore percentage abundance of Ga-69 be 65% ; percentage abundance of Ga-71 be 35%. (iii) A lCl 3 – simple covalent/molecular structure – weak temporary induced dipole-induce dipole attraction between molecules. PC l5 – simple covalent/molecular structure – permanent dipole - permanent dipole attractions between molecules. A lCl 3 exist as dimers, with stronger id-id (due to greater no. of electrons) than pd-pd of PCl5. Hence m.p. of A lCl3 is higher than PCl5. (b) (i) pH (ii) A l2O3 insoluble in water ; high lattice energy(pH 7) P 4O10 + 6H2O → 4H3PO4 (pH about 2-3) (c)(i) MgCO 3 will decompose first because it is thermally less stable: ƒ Mg2+ has smaller ionic radius than Ca2+ ƒ Mg2+ (higher charge density than Ca2+) is more polarising than Ca2+. ƒ Mg2+ polarises C-O bond of CO 3 2- more easily/to a greater extent, weakens and breaks the C-O bon at a lower temperature Na2O MgO Al2O3 SiO2 P4O10 SO3 7 -

(ii) M(OH) 2.nH2O → M(OH)2 + n H2O M(OH) 2 → MO + H2O Mass of anhydrous M(OH) 2 = 1.575 – 0.720 = 0.855 g Mass of water lost = 0.090 g Amount of water lost from anhydrous M(OH) 2 = 5.00 x 10-3 No. of moles of M(OH) 2.n H2O = 5.00 x 10-3 No. of moles of water from M(OH) 2.nH2O = 0.720/18.0 = 0.040 5.00 x 10 -3 mol of M(OH)2.nH2O ≡ 0.040 mol of H2O 1 mol of M(OH) 2. n H2O ≡ 8 mol of H2O n = 8 M r of M(OH)2 = = 171 A r of M = 137 M is barium 2(a) (i) CH 4(g) + H2O (g) CO(g) + 3H 2(g) Initial/atm 1 1 0 0 Δ/atm - 0.43 -0.43 +0.43 +(0.43) x 3 Equilibrium/atm 0.57 0.57 0.43 1.29 Total Pressure = 0.57 + 0.57 + 0.43 + 1.29 = 2.86 atm (ii) Kp = ) p )( p ( ) p )( p ( O HCH 3 H CO 24 2 [1m] K p = = 2.84atm2 (iii) 1. By Le Chatelier’s Principle, the position of equilibrium will shift to the right to remove some of the added heat. The forward endothermic reaction is favoured; as a result, there will be higher p products ∝ [products] and lower p reactants ∝ [reactants]. Hence Kp increases with increasing temperature 2. K p remains constant as long as temperature is kept constant. Not affected by change in total pressure. (b) (i) Comparing expt A and B, when [H 2] is halved, keeping [NO] constant, rate is also halved. Order of reaction wrt H2 is 1. Substitute data from expt A and C into rate = k[H 2][NO]a where a = order of reaction wrt NO. 2.4 x 10-6 = k(0.0100)(0.025)a 1.62 x 10-4 k(0.0750)(0.075)a a = 2 Rate = k[H 2][NO]2

(ii) 1. NO N2O2 NO has an unpaired electron and is reactive; can dimerise readily to form N 2O2 2. Rate = k 2[N2O2][H2] = k2[H2] k1[NO]2 = k [H 2][NO]2 Therefore the rate equation is consistent with the proposed mechanism. 3(a)(i) 2Fe 3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(aq); E θ cell = +0.76 + (-0.54) V = +0.22 V The reaction is energetically feasible and produces I 2(aq) which forms I3 –(aq) – the brown solution. (ii) Fe 2+(aq) + 6F– (aq) → [FeF6]4– Ligand exchange takes place. F – ligands replace water ligands in the iron(II) complex ion. The [FeF 6]4–(aq) ion then reduces the brown iodine to colourless iodide. Thus the brown colour fades. I2(aq) + 2[FeF6]4–(aq) → 2I–(aq) + 2[FeF6]3–(aq); Eθ cell = +0.54 + (-0.40) V = +0.14 V (b) (i) Fe(IO 3)3(s) + aq Fe 3+(aq) + 3IO3 – (aq) [ F e 3+] = 3.6/(580.8) = 6.20 x 10-3 mol dm-3 [IO 3 –] = 3 x 6.20 x 10-3 = 0.0186 mol dm-3 K sp of Fe(IO3)3 = [Fe3+]eqm x [IO3 –]3 eqm = 3.99x 10-8 mol4 dm-12 at 298 K (ii) Due to the Common Ion Effect, the solubility decreases. Let the solubility be x mol dm -3 [IO3 –]total = 0.105 + 3x ≈ 0.105 mol dm-3; since 3x is small compared to 0.105 K sp of Ce(IO3)3 = [Ce3+]eqm x [IO3 –]3 eqm 3.99 x 10 -8 mol4 dm-12 = x(0.105)3 mol4 dm-12 x = 3.44 x 10 -5 (check: 3x = 1.03 x 10-4<< 0.105) Solubility of Fe(IO 3)3 in 0.100 mol dm-3 KIO3 = 3.44 x 10-5 mol dm-3.

4 ( a ) R = S = C O OH H C O O O H (b) Step 1: 1. Sn, concentrated HCl, heat under reflux 2 . N a O H ( a q ) Step 2: NaOH(aq), heat Step 3: H 2O(g), H3PO4(l), 300oC, 60 atm (or 1. conc H2SO4 2. H2O, warm) Step 4: KMnO 4, H2SO4(aq), heat (c) Nucleophilic substitution (d) (i) H4N + C H3 CH2 CH CH CH 2 OC O O (ii) H3N + HOOC COOH OHC O O + (iii) H2N C H3 CH2 CH Br CH CH 2 OC O O BrO H 5(a) X N H2 CH CH3 C O OH Y N H2 CH (CH2)4 C O OH NH2 Z N H2 CH (CH2)2 C O OH COOH CC l H H R C O H H H Rδ+ δ− HO - CH3COO-

(b) Location 1 H3N + CH (CH2)4 C O OH NH3 Location 2 H3N + CH CH3 C O O Location 3 N H2 CH (CH2)2 C O O COO