HCI H2 Chem 2013 Prelim P3 Soln

1 2013 C2 Chemistry Prelim Paper 3 – Answers 1 (a) NaX (s) + H2SO4 (l)  HX (g) + NaHSO4 (s) 2HI (g) + H2SO4 (l)  I2 (g) + SO2 (g) + 2H2O (l) 6HI (g) + H2SO4 (l)  3I2 (g) + S (s) + 4H2O (l) 8HI (g) + H2SO4 (l)  4I2 (g) + H2S (g) + 4H2O (l) Observations: white fumes of HCl Violet fumes of I2 Halides are reducing and the reducing power increases down the group. [5] (b) partition coefficient = (0.102  0.00120) = 85 Reciprocal of 85, i.e. 1.176 × 10−2 (to 4 sig. fig.) is also acceptable. I2 is more soluble in CCl4 than in H2O. [2] (c) (i) [I2]aq = 0.0850/85 = 0.00100 mol dm−3 [1] (ii) Since total iodine in the aqueous layer measures both I3 − and I2, Concentration of I3 − = 0.0170 – 0.001 = 0.0160 mol dm−3 Concentration of iodide ions = 0.100 – 0.0160 = 0.0840 mol dm−3 [2] (iii) Kc = [I3 −] / ([I−][I2]) I−(aq) + I2(aq) ⇌ I3 −(aq) eqm concentrations / mol dm−3 0.0840 0.001 0.0160 Substituting and solving, Kc = 190 mol−1 dm3 [3] (d) (i) A Lewis acid is not required because the partial positive charge on iodine is sufficiently positive. [1] 658

2 (ii) Mechanism: electrophilic substitution Correct arrows Correct intermediate slow/fast steps clearly indicated [3] (e) (i) Iodine undergoes disproportionation. Its oxidation number is intermediate and can be both oxidised and reduced simultaneously. OR The oxidation number of chlorine is the lowest possible so it cannot disproportionate. [2] (ii) 3ICl + 6OH− → IO3 − + 2I− + 3Cl− + 3H2O [1] HCl + – HCl 659

3 2 (a) (i) [1] (ii) 10.0s 1st order w.r.t CN–. Hence, when [CN –] doubles, rate doubles, time taken to consume the same amount of propanone will then be halved. OR CN– is a catalyst in this reaction. Hence, [CN–] stay constant throughout. Rate = k’[propanone] where k’ = k[CN–] t1/2 = ln 2 / k’ = ln 2 / k[CN–] Hence, when [CN–] doubles, the half-life will decrease by 2x. [1] (b) (i) [1] (ii) [3] [HCN] time 660

4 (c) Test 1: Reagents and Conditions: 1. dilute H2SO4, heat 2. neutral FeCl3 (aq) Observations: A: violet coloration B & C: No violet coloration Test 2: Reagents and Conditions: NaOH(aq), heat Observations: B: pungent NH3 gas evolved that turned moist red litmus blue A & C: No pungent NH3 evolved Test 3: Reagents and Conditions: Br2(aq) C: orange-yellow Br2(aq) decolourizes, (white ppt) A & B: No decolourization of orange-yellow Br2(aq) [6] (d) (i) CxHy  3CO2 + C4H6O3 + C3H6O S T Hence, x = 10 12(10) + y(1) = 136 y = 16 [1] (ii)  R undergoes reduction and has 3 C=C bond  Both S and T have  1 terminal C=C bond Or R:  S undergoes a nucleophilic substitution with thionyl chloride. Hence, S has –CO2H group.  The product then undergoes a condensation with ethylamine to give U. [5 max 3 marks] 661

5 S: or or [4] U: 662

6 3 (a) Mass of methyl butanoate = 18.42.0 100 = 0.368 g Mass of air = 0.368 610 10 = 36.8 x 104 g = 36.8 kg Volume of air = 36.8  1.20 = 30.7 m3 [2] (b) Any two of the following: Ethanoic acid / butanoic acid / methanoic acid Ethanoic acid is formed from the oxidation of ethanol Butanoic acid due to the acid-hydrolysis of methyl butanoate / ester. Methanoic acid due to acid -hydrolysis of ester and subsequent oxidation of methanol. [2] (c) Increase temperature will increase the rate. There is an increase in the fraction of molecules that have kinetic energy larger than or equal to the activation energy. [2] (d) The walls of the vial are thinned because the acids react ed Al2O3 in an acid -base reaction since Al2O3 is amphoteric , forming Al 3+ which is soluble . (also accept equation) [1] (e) (i) 2Al(OH)3  Al2O3 + 3H2O 2AlO2H  Al2O3 + H2O [2] (ii) ½Hf O = 102 – 3/2 (-242) – 1293 Hf O = -1656 kJ mol-1 Hr = ½ (-1656) + ½ (-242) – (-996) Hr = +47 kJ mol-1 [4] (f) β-damascone : Heart note Methyl butanoate : Head note β-damascone and β-ionone have similar electron cloud size. It needs similar amount of energy to break the dispersion forces / van der Waals forces of attraction between the molecules Methyl butanoate has smaller electron cloud size. It needs lesser amount of energy to break the dispersion forces / van der Waals forces of attraction between the molecules 663

7 [3] (g) (i) NH2OH [1] (ii) [1] (iii) [2] 664

8 4 (a) (i) Across the period, electronegativity increases. Down the group, electronegativity decreases. The elements along the diagonal have similar electronegativity, hence they have similar properties. [1] (ii) pH = 3 (accept any specific value from 1 to 4, but not a range) Due to the high charge-density of the Be2+, hydrolysis occurs as the O−H bonds in the H2O ligands are sufficiently polarised , weakened and break readily to donate a proton. [BeH2O)4]2+ (aq) + H2O(l) [Be(H2O)3OH] + (aq) + H3O+ (aq) (equation not required) [2] (b) (i) AlCl3 and NH3 react in a 1 : 1 mole ratio to form an adduct: (dative bond, correct bond angle 109.5 about Al and N) [2] (ii) (correct shape of graph with peak indicated at about mole fraction BeCl2 = 0.33) [1] (c) (i) Since the Mr is 290, there can only be 1 Pb in the compound. % Pb = 207/290 x 100% = 71.38% % F = 100 – 71.38 – 3.08 = 25.54% Pb : Be : F = 0.345 : 0.342 : 1.344 = 1 : 1 : 4 ; Empirical formula = PbBeF4 [2] Heat evolved / J mol–1 Mole fraction of BeCl2 0.33 0.50 0.25 0.75 665

9 (ii) 4 bond pairs,  shape is tetrahedral [2] (d) (i) The term ‘ground state’ is used to describe atoms whose electrons occupy their lowest possible energy levels. [1] (ii) Eu2+ has an electron configuration [Xe] 4f7, [1] (iii) Higher. Eu 2+ has a bigger ionic radius, hence lower charge-density and hence lower polarizing power than Be2+. The electron cloud of the NO 3 - is distorted to a smaller extent, the N – O bond is weakened to a smaller extent . More energy (high temperature) is needed to cause decomposition o