TJC Prelim P3 Solutions

1 O O •• Answers to 2008 TJC Prelim Paper 9746/3 Free Response 1. (a) (i) O N O N O N ClO [4] (ii) · Both have simple molecular structures and boiling involves breaking of intermolecular forces of attraction. · HNO3 has intermolecular hydrogen bonding which are stronger than the Van der Waals forces exist between NOCl molecules, hence more energy required to overcome, so higher boiling point. [2] (b) (i) PV = nRT n = RT PV = 40031.8 102010101 33 ´ ´´´ - · n = 0.608 mol [1] (ii) Let be degree of dissociation of NOCl. 2NOCl (g) 2NO (g) + Cl 2(g) Initial no of mol 0.5 0 0 Change in no of mol - (0.5) +(0.5) + 2 a (0.5) Equil no of mol 0.5(1-) +0.5 + 2 5.0 a · Total equilibrium number of mol = 0.5 – 0.5 + 2 5.0 a + 0.5 = 0.608 = 0.432 · % of NOCl dissociated = 10010 32.4 ´ % = 43.2 % [2] (iii) · Kp = 2 2 2 NOCl ClNO P PP NOClP = 2002.03.03.0 3.0 ´++ kPa = 75 kPa NOP = 8.0 3.0 x 200 kPa = 75 kPa 2ClP = 2008.0 2.0 ´ kPa = 50 kPa •• < 120o 120o <109.5o ·

2 10 20 25 30 0.5 0.3 0.2 time 0 NOCl NO Cl2 40 Number of moles · Kp = 2 2 75 5075 ´ = 50 kPa at 400 K [3] (iv) · When temp is increased, by Le Chatelier’s Principle, since reaction is endothermic, position of equilibrium shifts to the right to reduce temp. At the new equilibrium, lesser number of moles of NOCl will be present. of NOCl [3] (c) (i) · Increasing pKa: C < A < B · Alcohol B is least acidic as the negative charge of the alkoxide ion is localized on the O atom while for phenol A and C, the negative charge on O atom can be delocalized onto the benzene ring, giving a more stable phenoxide anion. · C is more acidic than A as C has an electron-withdrawing Cl atom which can further disperse the negative charge on O atom of the phenoxide ion, thus stabilizing the anion. For A, the – CH3 group is electron releasing and this will intensify the negative charge on the O atom, so destabilizing the phenoxide ion. [3] (ii) · Compound C is insoluble in water due to the hydrophobic benzene ring which forms Van der Waals forces with water, releasing insufficient energy to overcome intermolecular hydrogen bonds in water and in compound C. · Compound C is acidic and forms a soluble salt sodium phenoxide with aqueous NaOH. Sodium phenoxide is ionic and forms ion-dipole interactions with water molecules, releasing sufficient energy to overcome intermolecular hydrogen bonds in water and in compound C [2] [Total: 20] · 1 mark for correct shape of graph before 25 min (less steep initial gradient and longer time to reach equilibrium) · 1m for correct shape after 25 min (steeper initial gradient and shorter time to reach equilibrium) Time /min

3 2 (a) (i) Quantity of charge passed = 8.0 x 67 x 3600 = 1.93 x 106 C Number of moles of electrons passed = 1.93 x 106 / 96500 = 20.0 mol Mg 2e Number of moles of magnesium = 20.0 / 2 = 10.0 mol Mass of magnesium formed = 10.0 x 24.3 = 243 g Cl2 2e Number of moles of gas in one cylinder = PV / RT = 1240000 x 0.002 / (8.31 x 298) = 1.00 mol Number of moles of chlorine gas = 10.0 mol Number of cylinders filled = 10.0 / 1.00 = 10 [5] (ii) 1 mark for quoting the four relevant Eo values Cathode Anode Eo H2O/H2 = -0.83 V Eo O2/H2O = +1.23 V Eo Mg 2+ /Mg = -2.38 V Eo Cl2/Cl - = +1.36 V Since Eo H2O/H2 is more positive than Eo Mg 2+ /Mg, H2O will be preferentially reduced to H2 at the cathode, and effervescence of hydrogen gas would be observed.

Since Eo O2/H2O is less positive than Eo Cl2/Cl -, H2O will be preferentially oxidized to H2 at the anode, and effervescence of oxygen gas would be observed. [3] (b) (i) Magnesium oxide can be used to line furnaces because it has a giant ionic structure and has a high melting point, as Melting involves breaking of strong electrostatic forces of attraction between oppositely charged ions in the giant ionic lattice. [2] (ii) DSo is positive because in the reaction, there is an increase in the number of moles of gases (by 2.5), giving rise to a greater degree of randomness. [1] (iii) By Hess’ Law, DHo r = -602 + 2(+33.9) – (– 790) DHo r = +255.8 kJ mol-1 of Mg(NO3)2(s) Decomposition occurs when the reaction becomes spontaneous (i.e. DGo < 0) Hence, DHo - TDSo < 0 +255.8 – T (0.273) < 0 Decomposition temperature, T > 937 K [3] Mg(NO3)2(s) MgO(s) + 2NO2(g) + ½O2(g) DHr o -790 kJ -602 + 2(+33.9) kJ Mg(s) + N2(g) + 3O2(g)

4 (iv) Decomposition temperature of calcium nitrate is higher than magnesium nitrate because Cationic size of the Group II metal ions increases down the group as consecutive members have one more filled inner quantum shell, but charge the same (+2), hence charge density of cation decreases down the group Hence polarising power of the cation decreases so the electron cloud of the nitrate ion is less distorted for calcium nitrate. [2] (c) (i) Nucleophilic addition [1] (ii) CH3CH2MgBr CH2O (methanal) [2] (iii) [1] 3 (i) · X is a transition metal with a partially filled d-subshell. · In the presence of ligands, the d orbitals become non-degenerate. The d-subshell split into two energy levels due to the repulsion of the metal ion and the ligands. · When an electron from the d-orbital of lower energy is promoted to one of higher energy (d-d electronic transition), an amount of energy, E, in the visible region of the electromagnetic spectrum is absorbed. The light energy not absorbed will be seen as the colour of the complex. [3] (ii) · XO2+ + 2H+ + e X3+ + H2O · E = [-2.30-(-2.70)]/+4-(+3) = + 0.40 V [2] (iii) E Sn 4+/Sn 2+ = +0.15V E XO 2+/X 3+ = +0.40V · E XO 2+/X 3+ is more positive than E Sn 4+/Sn 2+, OR E = +0.40 – (+0.15) = +0.25V > 0, reaction is feasible. · XO2+ will be reduced to X3+ while Sn2+ is oxidised to Sn4+. [2] (iv) Select: +0.54V < E < +2.01V. E XO2 +/XO 2+ = +1.00V. Therefore either XO2 + or X