JJC_H2_CHEM_P3 ANS

2009 JJC Preliminary Examination Mark Scheme 1. (a) (i) The standard enthalpy change of combustion of a substance is the enthalpy change which occurs when one mole of the substance is completely burnt in oxygen/burnt in excess oxygen under standard conditions. [1] (ii) OH (s) + 7O2 (g)  6CO2 (g) + 3H2O (l) [1] (iii) –3054 = 6 (-394) + 3 (-286) - Hf (phenol) + 0 [1] Hf (phenol) = - 168 kJ mol-1 [1] (iv) - 53.6 kJ mol-1 [1] 1. (b) (i) Three electron withdrawing Cl atoms disperses the negative charge on the anion, stabilizing the anion. Hence 2,4,6-trichlorophenol is a stronger acid than phenol. [1] 2,4,6-trichlorophenol has a larger Ka value. [1] (ii) Test: Add Br2(aq) to both samples. [1] Observation: phenol will decolourise orange brown Br 2(aq) but 2,4,6-trichlorophenol will not. [1] (c) (i) [1] (ii) Step I: dil HNO3 / HNO3(aq) [1] Step II: Sn, conc HCl, heat [1] (iii) Bulky non-polar group in 1-naphthol hinders the formation of hydrogen bonding. [1] (d) (i) Transition element complexes are coloured because of electron transition between d orbitals. In a complex ion, the presence of ligands causes the 3d orbitals to split into 2 energy levels. The difference in energies between these two sets of 3d orbitals is small such that radiation from the visible region is absorbed when an electron moves from a d orbital of lower energy to an unfilled d orbital of higher energy. The blue colour seen is the complement of the colours absorbed. [3] (ii) The prescence of energetically accessible vacant orbitals to accept lone pair of electrons from ligands. [1] NO2 OH

1. (d) (iii) [2] (iv) Deep blue solution: [Cu(NH Cu H2O O: N: :O :N : H H H H H 2O : 3)4(H2O)2]2+ [1] 2. (a) (i) A weak acid is a proton donor that dissociates partially. [1] (ii) [H+] = 5.66 x 10–3 mol dm–3 [1] pH = 2.25 [1] (iii) Amount of acid used = 5 x 1–3 mol Amount of NaOH required = 5 x 10–3 mol Volume of NaOH required = 0.0500 dm3 = 50 cm3 [1] [OH–] = solution salt of conc acid cylicacetylsali of b K = 05 . 0 05 . 0 10 5 10 2 . 3 10 00 . 1 3 4 14        = 1.25 x 10 –6 mol dm–3 pOH = 5.90 pH = 8.10 [1] (b) (i) pH = pKa =  410 2 . 3 lg  = 3.49 [ 1 ] (ii) When H+ is added: C8H7O2COO– + H+ → C8H7O2COOH [1] When OH– is added: C8H7O2COOH + OH– → C8H7O2COO– + H2O [1] (iii) Amount of HCl added = 2.00 x 10–4 mol In the resultant solution: Amount of C8H7O2COO– left = 43 10 2 10 5 . 2    3 mol 10 30 . 2  Concentration of C8H7O2COO– = 10002 25 50 10 30 . 23     = 0.0299 mol dm –3 [1] [1] Page 2 of 7

2. (b) (iii) Amount of C8H7O2COOH = 43 10 2 10 5 . 2    3 mol 10 70 . 2  Concentration of C8H7O2COOH = 10002 25 50 10 70 . 23     = 0.0351