NJC CHEM H1 Chemistry 2008 P2 answers

National Junior College 2008 Answer Keys to H1 Prelims P2 1 (a) (i) Protons Neutrons Electrons Relative Mass 1 1 1/2000 Charge +1 0 −1 Location in atom Nucleus Nucleus Orbitals (ii) Protons: 31; Neutrons: 38; Electrons: 31 (b) (i) Isotopes are atoms of the same element with the same no. of protons and electrons but different no. of neutrons. (ii) A r of Ga = ) ( ) (71 x5 269 x5 3 + = 69.8 (iii) Both have the same charge of +3. 69Ga3+ is deflected to a larger angle towards the negative plate as it is lighter in mass. 2 (a) (i) Bond energy of NO is the energy required to break one mole of covalent bonds between N and O in a gaseous NO molecule at standard conditions. (ii) ΔHr = 4[3(+390)] + 5(+496) + 4(−607) + 6[2(−460)] = − 788 kJ mol−1 (b) (i) ΔHf θ of O2(g) is zero as it is an element at standard state. (ii) ΔHr = 4(+90) + 6(−285.8) − 4(−46) = − 1170.8 kJ mol−1 (iii) The water in (aii) is in the gaseous phase while the water in (bii) is in the liquid phase. Bond energy value in (aii) is less exothermic as energy is required to vapourise the liquid water to gas. 3 (a) (i) C H H C CH3 Br C H CH3 H C H H C CH 3 H C Br CH3 H (iii) C H H C CH 3 OH C OH CH 3 H 1

National Junior College 2008 (ii) C H H C CH3 O CH3COOH+ (iv) C H H C CH3 Cl C Cl CH 3 H Cl C H H C CH3 Cl C Cl CH 3 H Cl (b) C H H C CH3 OH C OH CH 3 HC H H C CH3 C CH3 H + [O] + H2O Purple solution of KMnO4 decolourise, brown ppt of MnO2 formed. (c) Geometric isomerism CC CH2 CH3 CH3 H CC CH2 CH3 H CH3 trans cis 4 (a) (i) 4 HCl(aq) + MnO2(s) Æ MnCl2(s) + Cl2(g) + 2H2O(l) (ii) No. of moles of MnO2 = 9 86 5 . = 0.0575 mol No. of moles of hydrochloric acid = 6.0 x 25 x 10−3 = 0.15 mol HCl is limiting reagent. Volume of Cl 2 24 x4 15 0. = 0.9 dm3 (iii) HCl is the reducing agent. (b) (i) 50 ppm would endanger human health. In 1, 000, 000 dm 3 of air, there are 50 dm3 of Cl2 In 24 dm3 of air, volume of Cl2 = 24 x 10 50 6 = 1.2 x 10−3 dm3 No. of moles of Cl2 = 24 10 x 2 13−. = 5 x 10−5 mol 2

National Junior College 2008 (ii) 4 Cl2 (g) + S2O3 2− (aq) + 5 H2O (l) Æ 8 Cl− (aq) + 2 SO4 2− (aq) + 10 H+ (aq) (iii) No. of moles of Cl2 = 24 10 x 5003− = 0.0208 mol No. of moles of S2O3 2− = 4 0208 0. = 5.208 x 10−3 mol [S2O3 2−] = 1000 x700 10 x 208 53−. =7.44 x 10−3 mol dm−3 5 (a) (i) 11 12 13 14 15 16 17 Proton no. ionic radius Ionic radius decreases from Na+ to Si4+ as no. of protons increase but the number of electrons are the same, hence, p/e ratio increases and the valence electrons are drawn nearer to the nucleus due to increasing attraction. The same explanation applies for P3− to Cl−. The anions are bigger than the cations as they have one extra quantum shell hence the distance of the valence electrons from the nucleus will be greater. (ii) MgO, Al2O3 and SO2/SO3. MgO(s)+ 2 HCl (aq) Æ MgCl2(aq) + H2O(l) Al2O3 (s)+ 6 HCl (aq) Æ 2 AlCl3(aq) + 3 H2O(l) Al2O3 (s)+ NaOH (aq) Æ NaAl(OH)4(aq) SO2 (g)+ 2 NaOH (aq) Æ Na2SO3 (aq) + 2 H2O(l) SO3 (g)+ 2 NaOH (aq) Æ Na2SO4 (aq) + 2 H2O(l) 3

National Junior College 2008 (iii) SO2/SO3; MgO; Al2O3 SO2/SO3 exists as simple covalent molecules with weak Vander Waals forces holding the molecules together in fixed positions. These forces are weaker than the ionic bonds holding the ions in the MgO and Al 2O3 giant ionic lattice, thus requiring less energy to break. The ionic bonds between Mg 2+ and O 2− are weaker than the ionic bonds between Al3+ and O 2− as Al 3+ has higher charge density than Mg 2+, hence require more energy to break. (b) (i) For IO3 −, use expt 2 and 3, when increase 2x, initial rate increase 2x. Thus, 1st order wrt IO3 −. − 3IOV For I−, use expt 2 and 4, when increase 2x, initial rate increase 2x. Thus, 1 st order wrt I−. −IV For H+, use expt 1 and 2, but half all volumes in expt 1. When increase 2x, initial rate increase 4x. Thus, 2nd order wrt H+. +HV => R = k[H+]2[I−][ IO3 −] (ii) The total volume of the solutions is kept constant so the volume of reactants are proportional to the concentration of the reactants. Varying amounts of water was added to each solution until the final volume is 1000 cm 3 for all experiments. (iii) When temperature increases, kinetic energy of the molecules increases, thus there will be a greater fraction of molecules with energy colliding with energy more than activation energy, thus, more effective collisions. Fraction of molecules Energy 4

National Junior College 2008 6 (a) Reaction is still ongoing, but rate of forward equals rate of reverse reaction, thus, concentration of the reactants and products are constant. (b) Concentrated H2SO4 removes water, hence equilibrium shifts right to produce more ethyl propanoate. (c) (i) Kc = ] ][ [ ] ][[ C B waterpropanoateethyl (ii) acid B alcohol C ethyl propanoate water Initial conc x x - - Equilibrium conc x – 0.40 x – 0.40 0.40 0.40 Kc = 2 2 4 0 x 4 0 ) . ( ) . ( − = 3.5 2 2 4 0 x 4 0 ) . ( ) . ( − = 5 3. 4 0 x 4 0 . . − = 1.871 x = 0.614 Equilibrium concentrations of B and C = 0.214 mol dm−3 (iii) Forward reaction is endothermic. As temperature increases, equilibrium shift right to remove some of the added heat. Kc increase as product increase and reactant decrease. (d) (i) B is propanoic acid, C is ethanol. (ii) Compound D is chloroethane as ethanol undergo substitution reaction with PCl 5 to give chloroethane and white fumes of HCl. Compound D undergoes elimination with alcoholic KOH to give compound E which is ethene. Ethene undergoes addition reaction with bromine to give 1,2- dibromoethane hence decolourises Br 2. Compound D Compound E CCC l H H H H H CC H H H H 5

National Junior College 2008 (iii) CH3CH2COOH + 4 [H] Æ CH3CH2CH2OH + H2O LiAlH4, dry ether CH3CH2CH2OH + [O] Æ CH3CH2CHO + H2O K2Cr2O7, H2SO4, heat with immediate distillation (e) (i) Buffer (ii) CH3CH2COOH + OH− Æ CH3CH2COO− + H2O CH3CH2COO− + H+ Æ CH3CH2COOH 7 (a) 11 12 13 14 15 16 17 18 Proton no. 1st ionisation energies General increase in ionization energies as proton no. increases while screening effect is constant due to the same number of inner quantum shells. Hence, the valence electrons are more attracted to nucleus, requiring more energy to remove. First ionization energy of Al less than Mg as the most loosely held electron of Al is removed from the 3p orbital which is at a higher energy than the electrons in the 3s orbital of Mg. First ionization energy of S less than P as the paired electron of S in the same 3p orbital experience mutual electronic repulsion hence require more energy to remove than the unpaired electrons in the 3p orbitals of P. (b) (i) NaCl, AlCl3, PCl3 and PCl5 NaCl + aq Æ Na+ + Cl− NaCl dissolve in water to form a neutral solution pH = 7 AlCl3 + 6 H2O Æ [Al(H2O)6]3+ + 3 Cl− [Al(H2O)6]3+ + H2O [Al(H2O)5OH]2+ + H3O+ AlCl3 hydrolyses in water to form an acidic solution of pH = 3 6

National Junior College 2008 7 PCl3 + 3 H2O Æ 3 HCl + H3PO3 PCl5 + 4 H2O Æ 5 HCl + H3PO4 PCl3 / PCl5 hydrolyse in water to form an acidic solution of pH =2. (c) (i) Empirical formula is C4H9Cl n(48 + 9 + 35.5) = 92.5 n = 1 F is C4H9Cl (ii) F undergoes nucleophilic substitution with NaOH to give compound G. Since G H reacts with alkaline iodine, G is a methyl alcohol with