AJC H1 CHEM P2 MS

AJC JC2 PRELIM 2015 8872/02/H1 Anderson Junior College H1 Chemistry 8872 2015 JC2 Prelim Paper 2 Suggested Solutions Section A 1 (a) (i) Ionic radius of Si4+ is smaller than that of Na+. This is because both cations have the same number of electrons / electrons experience same screening effect, but Si4+ has more protons than Na+. Hence attraction of the electrons to the nucleus is stronger for Si4+. [3] (ii) P3‒ has a larger ionic radius than Si4+. This is because P3‒ has an additional quantum shell of electrons. [1]: stating difference in (i) and (ii) [1]: explanation in (i) [1]: explanation in (ii) (b) (i) Structure Bonding A Giant Ionic B Giant Covalent C Giant Ionic D Simple Covalent [1] for every 2 correct structures with corresponding bonding [2] (ii) Identity A MgO B SiO2 C Na2O D P4O10 [1] for every 2 correct identities (do not accept SO 2/SO3 for D, both are gases at rtp) [2] Explanation: A and C has good electrical conductivity when molten  giant ionic (i.e. can be Na2O, MgO or Al2O3) Since C reacts with water to form a strongly alkaline solution, C must be Na2O. Since A reacts very little with water to form a weakly alkaline solution, A must be MgO. Note: Al2O3 does not dissolve in water! B and D has poor electrical conductivity when molten  giant covalent (i.e. SiO2) or simple molecular (i.e. oxides of P or S) Since B does not dissolve in water, B must be SiO2. Since D reacts with water to form a strongly acidic solution, D must be P4O10, SO2 or SO3 (cannot be SO2/SO3 due to the high m.p. given). Note: P4O6 reacts with water to form an acidic solution of around pH 2. (c) NaCl has a giant ionic structure with strong electrostatic forces of attraction between Na+ and C l– ions. A lot of energy is required to break these strong ionic bonds in order to melt NaCl. Hence, it has a very high melting point. SiCl4 has a simple molecular structure with weak van der Waals’ forces between SiC l4 molecules. Hence, they have low melting points. [1] for one correct structure and bonding + comparison of energy and mp [2]

AJC JC2 PRELIM 2015 8872/02/H1 2 (a) (i) C14H30(l) + 2 43 O2(g) 14 CO2(g) + 15 H2O(l) [accept: 2 C14H30(l) + 43 O2(g) 28 CO2(g) + 30 H2O(l)] [1] (ii) C14H30(l) + 2 43 O2(g) 14 CO2(g) + 15 H2O(l) Bonds broken Energy Bonds formed Energy 13 C‒C 13 × 350 14×2 C=O 14 × 2 × 740 30 C‒H 30 × 410 15×2 O ‒H 15 × 2 × 460 2 43 O=O 2 43 × 496 34520 kJ mol‒1 (given out) 27514 kJ mol‒1 (taken in) Hf o (kerosene) = +27514 ‒ 34520 = ‒7006 kJ mol‒1 [1] for the 5 bonds and their correct bond energies [1] for correct no of bonds broken and formed [1] for final answer [3] (iii) 1 mol of kerosene produces 7006 kJ of energy  198 g of kerosene produces 7006 kJ of energy  1 g of kerosene produces 7006 ÷ 198 = 35.4 kJ of energy Diesel fuel is more efficient, as it produces more energy per unit mass/gram. [1] for calculation ‒ allow ecf from (ii) [1] for conclusion (correct choice of fuel + explanation) [2] (b) (i) H 1 is 2Hf o (SO2) [1] (ii) H1 = 2 × Hc o (SO2) = 2(‒99) = ‒198 kJ mol‒1 H2 = 2 × Hf o (SO2) = 2(‒297) = ‒594 kJ mol‒1 H3 = H2 + H1 = ‒792 kJ mol‒1 Hf o (SO3) = 2 1H3 = ‒396 kJ mol‒1 [1] for application of Hess’ Law (allow ecf from (i)) [1] for Hf o (SO3) [2]

AJC JC2 PRELIM 2015 8872/02/H1 3 (a) (i) CH3CH(OH)CH3 [1] (ii) C O OC H H H CH H H C H H C H H [1] (iii) CH3CO2CH(CH3)2 + OH– CH3COO– + (CH3)2CHOH [1] (iv) CH3CH=CH2 / propene [1] (v) Add aq. Br2 to a sample of the reaction mixture. Orange aq. Br2 solution will be decolourised in the presence of an alkene. [1] [1] (b) CH3 CH2NH2CH2Cl limited Cl2(g) uv light excess concentrated NH3 in ethanol heat in sealed tube [1]: correct intermediate [1]: correct reagents and conditions for each step [3]

AJC JC2 PRELIM 2015 8872/02/H1 4 (a) n(ethanol) = 9.2 ÷ 46.0 = 0.200 mol [1] (b) (i) To keep the temperature constant so that the position of equilibrium will not shift due to changes in temperature. [1] (ii) To quench the reaction mixture by reacting away the H2SO4 catalyst [1] (c) (i) redox / oxidation [1] (ii) MnO4 ‒ + 2H2O + 3e MnO2 + 4OH‒ reduction CH3CH2CHO + 3OH‒ CH3CH2CO2 ‒ + 2H2O + 2e oxidation overall: 2MnO 4 ‒ + 3CH3CH2CHO + OH‒ 3CH3CH2CO2 ‒ + 2MnO2 + 2H2O [1] (iii) propanal(l) ethanol(l) acetal F(l) water(l) initial amount / mol 0.1 0.2 0 0 change in amount / mol ‒ x ‒ 2x + x + x equilibrium amount / mol 0.1 ‒ x 0.2 ‒ 2x x x [1] (iv) n(KMnO4) used in the titration = 0.0253 × 1.00 = 0.0253 mol [1] (v) 1 mol of propanal requires 3 2 mol of MnO4 ‒ for reaction. Hence (0.1 ‒ x) mol of propanal requires 3 2x-0.2 mol of MnO4 ‒ for reaction. 1 mol of alcohol requires 3 4 mol of MnO4 ‒ for reaction. Hence (0.2 ‒ 2x) mol of ethanol requires 3 8x-0.8 mol of MnO4 ‒ for reaction. Total n(MnO4 ‒) used in the titration = 3 2x-0.2 + 3 8x-0.8 = 0.0253  x = 0.09241 equilibrium amount of: propanal = 0.1 ‒ 0.09241 = 0.00759 mol ethanol = 0.2 ‒ 2(0.09241) = 0.0152 mol acetal = 0.0924 mol water = 0.0924 mol [1] for using mole ratio and total n(KMnO 4) in working [1] for x [1] for propanal and ethanol [3]

AJC JC2 PRELIM 2015 8872/02/H1 (d) Kc = 2 2 [ethanol] [propanal] O][acetal][H = 2)0.0250 0.0152)(0.0250 0.00759( )0.0250 0.0924)(0.0250 0.0924( = )0.0250 0.0152(0.00759)( .0924)(0.0924)(0 2 = 122 mol‒1 dm3 [1] for expression [1] for Kc value [1] for units [3]

AJC JC2 PRELIM 2015 8872/02/H1 5 (b) (i) Reaction I: dilute H2SO4, heat (under reflux) Reaction II: excess concentrated H2SO4, 170 / 180 oC [4] (ii) C C O O C C O O H H HH G [1] (iii) The presence of π bond in the C=C double bond prevent free rotation about the double bond. [1] (iv)  C C O O C C O OH H H H    H C C O HO C C O H H O [1] correct O–H bond [1] one lone pair in correct place in H bond [1] correct dipoles on at least one O–H bond [3] (v) G can form intramolecular hydr ogen bonding leading to less extensive intermolecular hydrogen bonding than fumaric acid, resulting in lower melting point. [1] (b) (i) n(NaOH) = 0.05 x 32 ÷ 1000 = 1.6 x 10–3 mol n(fumaric acid) = 1.6 x 10–3 ÷ 2 = 8.0 x 10–4 mol [fumaric acid] = 8.0 x 10–4 ÷ 20/1000 = 0.0400 mol dm–3 [2] (ii) Phenolphthalein. The pH at the end (equivalence) point for this titration is around 8.5. Phenolphthalein has a working pH range that lies within the range of rapid pH change for the titration. [3] (c) (i) H]CO[CH ]][HCOO[CHK 23 - 3 a   [1] (ii) 1.84 x 10–5 = 0.30 ](0.20)[H [H+] = 1.84 x 10–5 x 0.20 0.30 = 2.76 x 10–5 mol dm–3 pH = –lg (2.76 x 10–5) = 4.56 [2] (iii) H+ + CH3CO2   CH3CO2H OH– + CH3CO2H  CH3CO2 – + H2O (no mark for using “HA”) [2]

AJC JC2 PRELIM 2015 8872/02/H1 6 (a) (i) Al2Cl6 Al Cl Cl Cl Cl Al x x x x x x Cl Cl [1]: correct dot and cross diagram [1]: with two dative bonds (circled) between C l and A l (each with 2 dots or 2 crosses) clearly indicated [2] (ii) Increasing the temperature favours the forward endothermic reaction. Position of equilibrium shifts right to absorb the increase of heat energy. This results in an increase in amount of AlCl3 and decrease in amount of Al2Cl6. [1] [1] (b) An orange solution with pH 3 is observed. AlCl3 dissolves with slight hydrolysis A lCl3 + 6H2O Al(H2O)6 3+ + 3Cl- Al(H2O)6 3+ + H2O Al(OH)(H2O)5 2+ + H3O+ [1]: correct colour (accept yellow) [1]: correct pH [1]: correct equations Explanation not required AlCl3 dissolves in water where Al3+ ions form complex ions [Al(H2O)6]3+. A lCl 3(s) + 6H2O(l) [Al(H2O)6]3+(aq) + 3Cl–(aq