HCI Prelim ANS

HCI (College) / 2009 C2 H1 Chemistry / Prelims  Answers  Paper 1 – MCQ 1 A 6 A 11 B 16 C 21 A 26 D 2 B 7 A 12 B 17 B 22 B 27 C 3 A 8 C 13 C 18 C 23 D 28 D 4 C 9 A 14 B 19 B 24 C 29 B 5 D 10 D 15 A 20 A 25 D 30 D Paper 2A – Structured 1 (a) (i) Relative atomic mass, Ar  average mass of one atom of an element 1 12 x the mass of one atom of 12C (ii) Ar of Cu: (30.8/100)(65) + (69.2/100)(63) = 63.6 (iii) Ion protons neutrons electrons 63Cu2+ 29 34 27 (b) (i) No of moles of excess HCl = 20 x 0.5 x 10-3 No of moles of NH3 neutralized by HCl = (40-20)x0.5x10-3 = 1.0x 10-2 No of moles of NH4 + = mole of NH3 neutralized = 1.0x10-2 Mass of NH 4 + in 2g of salt A = 18( 10-2) = 0.180 g (ii) Mass of SO4 2- ion in 2g of salt A = 2.33 x [( 32.1 +64)/ (137+96.1)] = 0.961 g (iii) No of moles of A: 2.00 / 399.5 = 0.00501 mol No of moles of NH4 + = 0.0100 mol No of moles of SO 4 2- = 0.960 / (32.1 + 16.0 x 4) = 0.0100 mol No of mole of A: NH 4 + : SO4 2- 1: 2 :2 [1] Cu(NH 4)2(SO4)2 . 6H2O 2 (a) XCln  XCln−2 + Cl2 5 3 ) 5 . 35 108 /( 19 . 1 ( ) 5 . 35 108 /( 714 . 0 ( 2       n n OR No. of moles of Cl− for 1st sample = mol 10 8.2935.5 108 1.19 3  (1) No. of moles of Cl− for 2nd sample = mol 10 4.9835.5 108 0.714 3  (2) (1) : (2) = 5 : 3 n = 5 8872/02/S/09 1

HCI (College) / 2009 C2 H1 Chemistry / Prelims  Answers  Ar of X = 124.2 5 ) 10 (8.29 0.294 0.5 3      X is Sb (b) (i) Both exist as giant ionic lattice held by strong electrostatic forces of attraction between oppositely charged ions.       rr q qLE q+, q- and r- are the same. r+ of KCl is larger, hence numerical value of LE is lower. Less energy is required to break the weaker lattice structure. (ii) NaCl exists as giant ionic lattice held by strong electrostatic forces of attraction between oppositely charged ions. A lCl3 is a simple covalent molecule with weak dispersion forces between the molecules. Less energy is required to separate the molecules than the ions.                             (c) (i) The position of equilibrium shifts to the right. By Le Chatelier’s principle, an increase in pressure favours the side of the reaction with less number of moles of gaseous molecules. This results in a decrease in pressure, thus re-establishing the equilibrium. (ii) When temperature is increased, the position of equilibrium shifts to the left. Forward reaction is exothermic as it involves bond formation. By Le Chatelier’s principle, the equilibrium will shift backward to remove the heat. (iii) As shown by in the diagram, at a lower temperature, the proportion of molecules with kinetic energy  activation energy Ea decreases. Frequency of effective collision decreases and thus reaction rate decreases. 3 (a) Average energy required when one mole of covalent bonds are broken in the gaseous state (b) (i) enthalpy change for bond breaking = 4(410) + 350 + 740 + 5/2 (496) = 3970 kJ mol -1 enthalpy change for bond forming = 4(460) + 4(740) = 4800 kJ mol -1 enthalpy change of combustion = 830 kJ mol-1 KE  EA at T2 K  Fraction of molecules with  KE  EA at T1 K  T1  T2  T1  T2 Fraction of molecules with  of  mole cules  proportion  8872/02/S/09 2

HCI (College) / 2009 C2 H1 Chemistry / Prelims  Answers  8872/02/S/09 3 (ii) Bond energies quoted in tables repres ent average (mean) bond energies derived from a full range of molecules containing that particular bond. Hence results from calculations using average bond energies will show discrepancies when compared with results from experiments with specific molecules. (c) (i) O C C O C CH OH CH2 C CH3 OH CN (ii) O C C O C CH OH CH2 C CH3 N H N NO2 NO2 (iii) O C C O C CH OH CH2 C CH3 HO HH H 4 Reaction I: concentrated HNO3, concentrated H2SO4, any temperature  55 °C was accepted Reaction II: limited amount of Cl2, presence of UV light (accept: heat) Reaction III: NaOH or KOH in ethanol, heat under reflux or heat (accept: warm) Reaction IV: KOH(aq) or NaOH(aq), heat under reflux or heat (accept: warm) A B C D BrOH CO2HBrCH2CH3 Cl NO2 CHCH3 CHCH2Br or CHCH2OH Reject dibromo-product

HCI (College) / 2009 C2 H1 Chemistry / Prelims  Answers  Paper 2B – Free Response 5(ai) ∆Hhyd becomes less exothermic down the group (or from Ca2+ to Sr2+ to Ba2+). Down the group, charge of the cations remains the same (+2) while the size of the cations increases;  charge density of the cations decreases. Thus, ion-dipole interactions formed between the cations and water molecules become weaker; less energy is released during hydration. (ii) BaSO4(s) + aq Ba 2+(aq) + SO4 2– (aq) +aq +2374 –1273 + ∆Hhyd(SO4 2–) Ba2+(g) + SO4 2–(g) OR “∆Hsoln = −LE + ∆Hhyd(Ba2+) + ∆Hhyd(SO4 2–)” By Hess’ law, +27 = +2374 – 1273 + ∆ H hyd(SO4 2–) ∆Hhyd(SO4 2–) = −1074 kJ mol−1 (bi) Ionic bonding between Ba2+ and C2O4 2− covalent bonding between C and O (ii) Trigonal planar around C, 120  (ci) pH of chlorides in water 0 7 14 NaCl MgCl2 AlCl3 SiCl4 PCl5 pH +27 kJ mol–1 NaCl  MgCl2  AlCl3 SiCl4 PCl5 8872/02/S/09 4

HCI (College) / 2009 C2 H1 Chemistry / Prelims  Answers  (ii) MgCl2 exists as a giant ionic lattice with strong electrostatic forces of attraction between oppositely charged Mg2+ and Cl ions. MgCl2(s) + 6H2O(l)  [Mg(H2O)6]2+(aq) + 2Cl(aq) [Mg(H2O)6]2+(aq) + H2O(l) [Mg(H2O)5OH]+ (aq) + H3O+(aq) SiCl4 exists as simple covalent molecules with weak dispersion forces between its molecules. SiCl4(l) + 2H2O(l)  SiO2(s) + 4HCl(aq) (di) ClCH2CH2Cl Cl2, CCl4 NCCH2CH2CNHO2CCH2CH2CO2H H2SO4(aq) heatbutanedioic acid CH2=CH2 ethanolic KCN heat (ii) Add aqueous bromine to both compounds. For ethene, orange solution decolourises, while for butanedioic acid, orange solution remains. OR Add Na2CO3(aq) to both compounds. For butanedioic acid, effervescence of CO2(g) observed, while for ethene, no gas observed. 6(ai) When a reversible reaction reaches a state of dynamic equilibrium, reaction continues to occur but the rates of the forward and backward reactions are equal. (ii) 2 22 4 2 7 2 ] [ ] [ ] [    H CrO O CrKc (iii) Initial [CrO4 2−] = 1000 100 194.2 8  = 0.4119 mol dm−3 [CrO4 2−] at eqm = (0.4119)5 1 = 0.08239 mol dm−3 (iv) [Cr2O7 2−] at eqm = 2(0.4119)5 4  = 0.1648 mol dm−3 [H+] = = 1.778 x 105.7510 −6 mol dm−3 Substituting into Kc,     2 62c ] 10 [1.778[0.08239] [0.1648]K 9 312  dm mol 10 7.67 (bi) 6I + Cr2O7 2 + 14H+  3I2 + 2Cr3+ + 7H2O (ii) 2S2O3 2− + I2  S4O6 2− + 2I 8872/02/S/09 5

HCI (College) / 2009 C2 H1 Chemistry / Prelims  Answers  n(S2O3 2−) used = 1000 21.50.20 = 4.30 x 10−3 mol n(I2) formed = 4.30 x 10−3  2 = 2.15 x 10−3 mol n(I−) reacted with Cr2O7 2− = 4.30 x 10−3 mol n(Cr2O7 2−) present in 25.0 cm3 = 4.30 x 10−3  6 = 7.167 x 10−4 mol n(Cr2O7 2−) in the 100 cm3 solution = 25.0 20010 7.1674 -  = 5.733 x 10−3 mol [Cr2O7 2−] in solution = 100/1000 10 5.733-3 = 0.0573 mol dm−3 (ci) A undergoes oxidation with K2Cr2O7 to give B and C  A contains an alcohol group B and C undergoes oxidative cleavage with hot conc. KMnO4  B and C contains an alkene group (double bond) B undergoes condensation reaction with 2,4-DNPH  B contains