2017 HCI H1 Chemistry Prelims P2 Answers

1 2017 H1 Chemistry Preliminary Examination Paper 2 Answers Section A 1 (a) (i) Since the forward reaction is exothermic, a low temperature will cause the position of equilibrium to shift to the right to produce more heat and thus producing more NH 3. However, the rate of reaction will be slow at low temperature, therefore, to increase the rate of reaction, a moderate temperature of 450oC is used. (ii) Iron catalyst provides an alternative reaction pathway with a lower activation energy. (iii) At a higher pressure, position of equilibrium shifts to the right so as to decrease the pressure by favouring the production of fewer number of moles of gases. The equilibrium is not affected as it is only affected by temperature. (iv) Kc = 3 22 2 3 ]][H[N ][NH (v) N2 (g) + 3H2 (g) 2NH3 (g) Let the number of moles of N2 reacted be x N2 3H2 2NH3 Initial no. of moles / mol 4 8 0 Change in no. of moles / mol - x - 3x + 2x Equilibrium no. of moles / mol 4 - x 8 - 3x 2x Given that the total number of moles of gases at equilibrium is 8, 4 - x + 8 – 3x + 2x = 8 Solving for x, x = 2 At equilibrium, there is 2 mol of N2, 2 mol of H2 and 4 mol of NH3. (vi) Kc =      3 2 22 24 22 = 1.00 mol-2 dm6

2 (b) (i) NH3(g) + 4 7 O2(g)  NO2(g) + 2 3 H2O(g) (ii) ΔH = ΔHf(NO2) + 2 3 ΔHf(H2O) - ΔHf(NH3) - 4 7 ΔHf(O2) ΔH = (-34) + ( 2 3 × -242) – (-46) – 0 = -351 kJ mol-1 (iii) ΔH = BE(bonds broken) - BE(bonds formed) Bonds broken = 3BE(N-H) + 4 7 BE(O=O) = 3 × 390 + 4 7 × 496 = 2038 kJ mol-1 Bonds formed = 3BE(O-H) + 2BE(N-O) = 3 × 460 + 2BE(N-O) = 1380 + 2BE(N-O) -351 = 2038 – 1380 – 2BE(N-O) BE(N-O) = +504 kJ mol-1 (iv) n(ammonia) = 17 44 = 2.59 mol = n(NO2) Theoretical mass of NO2 = 2.59 × 46 = 119.14 g Percentage efficiency = 119.14 89 × 100 = 74.7% (c) NH3 + CH3CH2Br  CH3CH2NH2 + HBr Ethanolic concentrated NH3, heat in a sealed tube The organic products belongs to the primary amine functional group. [Total: 20] 2 (a) (i) Al2O3 dissolves in both HCl and excess NaOH. P4O10 dissolves in NaOH. (ii) Al2O3(s) + 6HCl(aq)  2AlCl3(aq) + 3H2O(l) Al2O3(s) + 2NaOH(aq) + 3H2O(l)  2Na[Al(OH)4](aq) P4O10(s) + 12NaOH(aq)  4Na3PO4(aq) + 6H2O(l)

3 (b) (i) The aluminium atom in A lCl3 has only 6 electrons surrounding it hence it is electron-deficient. The lone pair of electrons from a chlorine atom in a neighbouring molecule would be donated via a dative bond, resulting in a dimer. (ii) Al2Cl6 has simple molecular structure. The dispersion forces between the A l2Cl6 molecules are weak and hence, only a small amount of energy is needed to overcome the weak interactions. [Total: 10] 3 (a) Cl2(g), AlCl3, (rt) (b) Step 2: oxidation Step 3: reduction (c) B dissociates to give H + and the conj