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