HCI Prelim P1 P2 ANS

HWA CHONG INSTITUTION 2009 C2 Higher 2 CHEMISTRY 9746 PRELIMINARY EXAMINATION PAPER 1 Multiple Choice Questions (ANSWERS) Question Answer Question Answer 1 A 21 D 2 B 22 B 3 C 23 A 4 A 24 A 5 D 25 D 6 A 26 C 7 B 27 D 8 B 28 C 9 C 29 C 10 B 30 B 11 C 31 D 12 B 32 C 13 A 33 B 14 D 34 D 15 D 35 A 16 A 36 B 17 C 37 C 18 C 38 B 19 B 39 A 20 C 40 D 1

PAPER 2 Structured Questions (ANSWERS) 1 (a) The alkaline earth metals exhibit +2 oxidation state in their compounds because they have 2 valence electrons which are lost to achieve stable octet. They do not exhibit +1 oxidation state due to the low lattice energy of the compounds formed. +3 oxidation state is also not favourable due to the very high 3 rd ionization energy required to remove the third electron from the inner quantum shell. (b) (i) [Ba 2+] [F-]2 = 1.84 x 10–7 where [Ba2+] = 0.05 mol dm–3 [F-] =  (1.84 x 10–7 / 0.05) = 1.92 x 10–3 mol dm–3 (ii) [Ca2+]remaining = 3.45 x 10–11 / (1.92 x 10–3)2 = 9.38 x 10–6 mol dm–3 2 (a) (i) no. of moles of glycolic acid = 0.20/76.0 = 2.63 x 10–3 mol volume of NaOH required = 2.63 x 10–3/0.10 x 1000 = 26.3 cm3 (ii) no. of moles of CH2OHCOO- salt formed at equivalence = 2.63 x 10–3 mol conc. of CH2OHCOO- = 2.63 x 10–3 /(26.3 + 20.0) = 0.0568 mol dm–3 CH 2OHCOO- +H 2O = CH2OHCOOH + OH– Initial conc. / mol dm –3 0.0568 – – Eqm conc./ mol dm –3 0.0568 - x x x Kb of glycolate = Kw/Ka of glycolic acid = 1x 10–14/ 1.48 x 10–4 = 6.76 x 10–11 mol dm–3 Kb = x2/(0.0568 – x)  x2/0.0568 (assume x << 0.0568 mol dm–3) x = 1.959 x 10–6 mol dm–3 pOH = 5.708 pH = 14 – 5.708 = 8.29 (iii) Metacresol purple because its working range coincides with the sharp jump of the titration curve which lies in the alkaline pH region (equivalence pH = 8.29). (b) (i) Step 1: LiAlH 4 in dry ether Step 2: PCl5 (s), room temperature Step 3: NaOH in ethanol, heat 2

(ii) Increasing ease of hydrolysis: B, A, C C is an acid chloride which undergoes hydrolysis more easily than A (halogenoalkane), because the carbon in the –COCl group of C is bonded to two electronegative atoms, O and C l, thus the carbon is more electron deficient and more readily attacked by nucleophile (H2O) to undergo nucleophilic substitution. B undergoes hydrolysis less easily than A because in B, the C–C l bond has partial double bond character due to the overlap of the p-orbital of C l with the - electron cloud of the adjacent alkene carbons. Thus the C–C l bond in B is much stronger than in A, and is harder to break. OR The carbon of C–C l in B is less electron deficient than that in A due to the presence of the electron-rich  electron cloud of the double bond. (c) (i) K 2Cr2O7, dilute H2SO4, heat Orange dichromate turns green for glycolic acid. Solution remains orange for D. (ii) Iodine, aqueous NaOH, heat Yellow precipitate for E. No yellow precipitate for glycolic acid 3 (a) 123456789 1 0 1 1 1 2 1 3 1 4 pH rate of reaction (b) At very low and very high pH, the enzyme is denatured. This is due to the change in the shape of its tertiary structure / loss of the active site conformation / changes in the side-chain interactions of the peptide chain. Rate of decomposition of H2O2 3

(c) At higher temperature, average kinetic energ y of molecules increases and the proportion of molecules with KE  Ea increases. Hence frequency of effective collisions increases and rate of reaction increases. (d) At low [H 2O2], rate of reaction increases with increasing [H2O2] (or follow first order kinetics) as there are plenty of active sites available. At high [H 2O2], rate of reaction remains constant with increasing [H2O2] (or follow zero order kinetics) as the active sites are saturated. (e) (i) An ideal gas is a gas that obeys the ideal gas equation (pV=nRT), under all conditions of pressure and temperature. (ii) No. At high pressure, the gas molecules are close together such that the volume of the molecules is significant compared to the volume of the container (or such that intermolecular forces are stronger). 4 (a) (i) Q = mcT = 35.0 x 4.18 x 2.8 = 409.6 J H 2A + 2NaOH  2NaA + 2H2O Number of moles of mesoxalic acid = (15.0/1000) x 0.250 = 0.00375 mol Number of moles of NaOH = (20.0/1000) x 0.400 = 0.008 mol Therefore mesoxalic acid is the limiting reagent. Number of moles of water formed = 0.00375 x 2 = 0.0075 mol ∆H neut = –(409.6/1000) / 0.0075 = –54.6 kJ mol−1 Proportion of molecules Proportion of molecules with KEE o at 25 C a Propo rtion of molecules with KEEa at 35oC Ea Kinetic energy 25oC 35oC 4

(ii) Number of moles of NaOH = (10.0/1000) x 1.2 = 0.012 mol (still in excess) Temperature rise = (35/25) x 2.8 = 3.92 °C (b) Q: CH2=CHCH(Br)CH=CH2 R: CH2=CHCH(OH)CH=CH2 I: excess conc H 2SO4, 170 °C II: NaOH(aq), heat III: KMnO 4(aq), dilute H2SO4, heat 5 (a) (i)  dehydration of alcohol to form alkene: OR esterification:  nitration of benzene: OR esterification: OR hydration of alkene: (ii)  H2SO4 (l) + NaBr (s) → HBr (g) + NaHSO4 (aq)  H2SO4 (l) + 2HBr (g) → Br2 (g) + SO2 (g) + 2H2O (l) H2CC H 2 + H2OCH3CH2OH conc. H2SO4 RCO2H + R'OH RCO2R' + H2O conc. H2SO4 + conc. HNO3 conc. H2SO4 NO2 + H2O RCO2R' + H2O conc. H2SO4 RC O2H + R'OH CH3CH2OH conc. H2SO4 CH2=CH2 + H2O (b) (i) Tetrachloromethane / hexane (or any other non-polar organic solvent) (ii) The halide is iodide. 2I–(aq) + Cl2(g) → 2Cl–(aq) + I2(aq) (c) (i) HO HO CH2 C CH3 H NH3+ O2N 5

Note: nitro group can be anywhere on the ring and 1-3 substitutions. (ii) (iii) 6 (a) (i) A: [Cr(H 2O)6]3+ B: [Cr(H 2O)6]2+ C: Cr(OH) 3 or Cr(H2O)3(OH)3 O O CH2 C CH3 H NH C O CH3 CH3C O CH3C O H3CO H3CO CH2 C CH3 H NHCH3 D: CrO 4 2– or Na2CrO4 E: Cr N H2 H2 N NH2 NH2 H2N H2N 3+ (ii) Cr3+ has a high charge density. Hence [Cr(H2O)6]3+ can undergo hydrolysis in water to produce H+ ions, forming CO2 with carbonate ions. (iii) Ligand exchange 6

7 (b) Cr3+ has d3 electronic configuration. In an octahedral ligand field, the 6 NH 3 ligands will split the five degenerate 3d orbitals into 2 groups of different energy levels. The difference in the two energy levels, E, falls within the visible region of the electromagnetic spectrum. An electron in a lower d orbital energy level can absorb radiation in the visible spectrum and be promoted into the higher d orbital energy level. This d d electron transition gives rise to the colour as the complement of the absorbed colour. (c) (i) mass of chromium 1 3 3.04560 96500 52.01.45 g (ii) As the Cr 3+(aq) ions are discharged at the cathode, they are replenished by reduction of CrO4 2-(aq) ions at the cathode. CrO4 2-(aq) + 8H+(aq) + 3e  Cr3+(aq) + 4H2O(l)