HCI Prelim P3 ANS

HWA CHONG INSTITUTION 2009 C2 Higher 2 CHEMISTRY 9746 PRELIMINARY EXAMINATION PAPER 3 Free Response Questions (ANSWERS) 1 (a) (i) -320 = 314 + 3/2(244) – 3BE(P−Cl) -32  BE(P−Cl) = +323 kJ mol –1 (ii) Shape with respect to each P: Trigonal pyramidal; Bond angle: 107 o (b) (i) Electrophilic addition Energy / kJ m ol–1 0 P (s) + 3/2 C l2 PCl3 -320 3/2 (+244) 3BE(PCl)P (g) + 3/2 Cl2 (g) +314 P (g) + 3Cl (g) PCl3 (g) +32 1

2 (ii) Rate = k[but-1-ene][HCl] (iii) BE / kJ mol–1 H−Cl 431 H−I 299 The rate of reaction will be faster. The rate determining step involves the breaking of the H-X bond. Since H-I bond is weaker than H−Cl, the rate of reaction should be faster if HI is used. (c) The product obtained will not rotate plane-polarised light, because the carbocation intermediate formed is planar. Hence in the second step, there is an equal chance of the C l– ion attacking the carbocation from either side of the plane to produce equal amounts of the two enantiomers, giving rise to a racemic mixture. The two enantiomers rotate plane-polarized light by equal amount and in opposite directions, hence cancelling each other’s optical activity. (d) (i) Both C l 2 and I 2 exist as diatomic molecules and have dispersion forces existing between molecules. I, being lower down in the group than C l, has a bigger electron cloud size and hence, stronger dispersion forces. Therefore I 2 is less volatile than Cl2. (ii) 4Cl2 (aq) + S2O3 2– (aq) + 5H2O (l)  8Cl – (aq) + 2SO4 2– (aq) + 10H+ (aq) I 2 (aq) + 2S2O3 2– (aq)  2I – (aq) + S4O6 2– (aq) E, / V Cl2 + 2e = 2Cl – +1.36 I2 + 2e = 2I – +0.54 E, Cl2/Cl– is more positive than E , I2/I–, indicating that C l2 is a stronger oxidizing agent than I2. Hence, Cl2 can oxidize S 2O3 2– to SO4 2– while I2 can only oxidize S 2O3 2– to S4O6 2–.

3 n. 2 (a) (i) When a re versible reaction reaches a state of dynamic equilibrium, reaction continues to occur but the rates of the forward and backward reactions are equal such that there is no net change in the concentrations of the reactants and products. (ii) K C CrO4 2–  2 H  2 initial [CrO 4 2–] = [CrO 4 2–] at eqm = [Cr2O7 2–] at eqm = Substituting into K C, [H +] = 1.793 x 10–6 mol dm–3 pH = -lg (1.793 x 10 –6) = 5.75 (iii) When chromate ions are precipitated, the chromate concentration will drop and the equilibrium will shift to the left. This increases the amount of H + present and lowers the pH value. (b)