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 (i