HCI H2 Chem Prelim P1 P2 Ans

HWA CHONG INSTITUTION 2008 C2 H2 CHEMISTRY 9746 PRELIMINARY EXAMINATION (SUGGESTED ANSWERS) PAPER 1 MCQ 1 C 6 C 11 B 16 C 21 D 26 B 31 B 36 B 2 B 7 B 12 D 17 C 22 D 27 B 32 D 37 B 3 B 8 D 13 B 18 C 23 C 28 C 33 B 38 C 4 C 9 A 14 A 19 C 24 A 29 A 34 D 39 A 5 C 10 B 15 D 20 A 25 D 30 D 35 D 40 A PAPER 2 STRUCTURED 1 (a) All 3 elements exist as simple covalent molecules with weak van der Waals’ forces (OR dispersion forces) between the molecules. S8 has the largest number of electrons, compared to P4 and Cl2; it has the strongest van der Waals’ forces, hence it needs the largest amount of energy to overcome these attractions during melting. (b) (i) disproportionation (ii) Reduction : S(s) + 2e− → S2−(aq) Oxidation : 2S(s) + 6OH −(aq) → S2O3 2−(aq) + 4e− + 3H2O(l) Overall : 4S(s) + 6OH −(aq) → S2O3 2−(aq) + 2S2−(aq) + 3H2O(l) (c) (i) AlCl3(s) dissolves in water to form [Al(H2O)6]3+(aq) OR Al3+(aq) ions. AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl–(aq) A hydrolysis reaction occurs for Al3+, making the solution acidic. [Al(H2O)6]3+(aq) + H2O(l) = [Al(H2O)5(OH)]2+(aq) + H3O+(aq) Or use “combined” equation: AlCl3(s) + 6H2O(l) = [Al(H2O)5(OH)]2+(aq) + H+(aq) + 3Cl–(aq) (ii) Al(OH)3 (iii) Al(OH)3(s) + OH–(aq) → [Al(OH)4]–(aq) OR A l(OH)3(s) + NaOH(aq) → Na[Al(OH)4](aq) (iv) starting pH at 7 for NaCl downward trend of pH HCI 2008 1

2 (a) Egs: Equation for reaction: N2(g) + 3H2(g) = 2NH3(g) Heterogeneous catalyst: Fe(s) Equation for reaction: CH2=CH2(g) + H2(g) → CH3CH3(g) Heterogeneous catalyst: Ni(s) Equation for reaction: 2H2O2(aq) → 2H2O(l) + O2(g) Heterogeneous catalyst: MnO2(s) (b) Adsorption of reactant molecules (N2 and H2) onto catalyst surface. State one effect of adsorption: process brings reactant molecules closer together, this increases their concentrations at the catalyst surface; weakens the bonds in them; and allows these molecules to be orientated in the right positions for reaction. The catalysed pathway thus involves lower activation energy. Product molecules desorb from catalyst surface, catalytic sites free to catalyse more reactions. (c) Impurities may bind (irreversibly or strongly) to catalytic sites on catalyst surface. This prevents adsorption of actual reactant molecules. HCI 2008 2

3 (a) (i) ∆H, r = –925 – 2(–286) = –353 kJ mol–1 (ii) Mg is the limiting reactant; so all 2.4 g of it reacts. 100 × 4.2 × (T – 25) = 2.4/24 × 353 × 1000 T = 109 oC So the heat is more than sufficient to raise the temperature. (b) (i) Mg2+ has a higher charge and smaller ionic size than Na+ (ii) ∆Hsoln = –L.E. + ∆Hhyd For NaOH: –44 = 896 + (–390) + ∆Hhyd (OH–) ∆Hhyd (OH–) = –550 kJ mol–1 For Mg(OH)2: x = 2995 – 1890 + 2(–550) = +5 kJ mol–1 (iii) There is a decrease in entropy (or ∆S is negative) as H2O molecules are attached to the metal ion and lose their degree of freedom. As more water molecules are attached to the Mg2+ ion on the average, the entropy (of the system) decreases more (or ∆S is more negative) than for the case of NaOH. (iv) Mg(OH)2 dissolves with the formation of more particles (aqueous ions); so the entropy (of the system) increases (or ∆S is positive). HCI 2008 3

4 (a) (i) C H Mass in 100 g 80 20 No. of moles 6.66 20 ratio 1 : 3 Empirical formula of B is CH3, B is likely to be C2H6 (ii) 2CH3CO2 –(aq) → 2CO2(g) + C2H6(g) + 2e– OR 2CH3CO2H(aq) → 2CO2(g) + C2H6(g) + 2H+(aq) + 2e– (iii) Anode: O2 evolved; Bulb brighter (b) M(OH)2.nH2O M(OH)2 + nH2O 1.295 g ∆ 0.590 g No. of moles of H2O = 0.590 / 18 = 0.0328 M(OH)2 MO + H2O ∆ 0.0740 g No. of moles of H2O = 0.0740 / 18 = 4.11 × 10–3 = No. of moles of M(OH)2 ∴ n = 0.0328 / 4.11 × 10–3 = 8 OR n = 0.590 / 0.0740 = 8 No. of moles of M(OH)2.8H2O = 4.11 × 10–3 Molar mass of M(OH)2.8H2O = 1.295 / 4.11 × 10–3 = 315 g mol–1 Ar of M = 315 – 2(16+1) – 8(18) = 137, ∴ M is barium (c) Let minimum [M(OH)2(aq)] be c mol dm–3 At the point of mixing, [ M 2+] = 5 × c / 105 mol dm–3 [SO4 2–] = 100 × 0.01 / 105 mol dm–3 For precipitation to just occur, ionic product = Ksp ∴ ( 5c / 105) (100 × 0.01 / 105 ) = 1.30 × 10–10 c = 2.87 × 10–7 mol dm–3 (d) (i) Step I: red P and Br2, heat OR PBr3, heat Step II: dilute H2SO4, heat (ii) CH3 C O C N (iii) Nucleophilic substitution (iv) H2, Ni, heat Reject: LiAlH4 in dry ether HCI 2008 4

5 (a) (i) ∆Hr = +610 – 350 + 436 – 2(410) = –124 kJ mol−1 (ii) The benzene ring is stabilised by resonance (or the six π electrons are delocalised around the ring) Hence, actual ∆Hhydrogenation is less exothermic than 3(–124) = –372 kJ mol–1 Enthalpy –205 resonance energy + 3H2(g) + 3H2(g) –372 (b) (i) Electrophilic substitution Equation showing nitronium ion reacting with benzene to form arenium ion Equation showing loss of H+ by arenium ion to form nitrobenzene *Must show: • slow step • correct electron movements in both steps • correct structure of all intermediates (ii) NO2 NO2 (c) I CH3 OR I CH3 HCI 2008 5

6 (a) (i) gly-pro-val-asp-gly-thr-phe-leu-ser-pro-glu (ii) position of his; relative positions of phe and glu glu his phe anode cathode (b) (i) The three-dimensional conformation of a protein where the secondary structure is held by interactions between the side chain groups (ii) phe contains a non-polar benzene group that has dispersion forces while thr has a OH group that participates in hydrogen bonding. As the strength of H bonding is stronger than VDW, the change in the R group can cause the folding of the enzyme to differ. (c) optical isomerism diagram of both isomers with mirror line (d) Step Error Step I HCN should be used for nucleophilic substitution of aldehyde so that cyanohydrin can be formed. Step III PCl5 will react with both acid and alcohol group. Step IV Concentrated NH3 should be used. Aqueous NH3 could only react with –COOH and not substitute out the halogen. HCI 2008 6