RI 2015 H1 CHEM P2 ANS Prelims

Suggested Solutions to RI Y6 2015 H1 Prelims Paper 2 Section A (40 marks) 1 (a) (i) (ii) The order of reaction is the sum of all the powers of all the reactants’ concentrations in the rate equation. Or If rate = k[A]m[B]n where k is a rate constant and [A], [B] are concentrations of A and B respectively, Order of reaction = m + n (iii) From the graph, time taken, t1, for [product] to increase from 0 to ½Co = 1.45 min time taken, t2, for [product] to increase from ½Co to ¾ Co = 1.35 min If t1 ≈ t2≈1.40 min, then t1/2 is constant. Since t1/2 is constant, the reaction is first-order with respect to the reactant. Rate = k[N2O5] (iv) t1/2 = ln 2/ k k = ln 2/ 1.40 mins = 0.495 min–1

3 (b) (i) Pressure used is high as a high pressure favours the forward reaction. By Le Chatelier’s Principle, when pressure increases, the system will favour the side with fewer number of gaseous molecules. Hence, position of equilibrium shifts to the right and this increases yield. Since the forward reaction is exothermic, a lower temperature would result in a higher yield of ammonia. However, the rate of production is too slow at low temperature. On the other hand, a high temperature increases the rate of production but results in lower yield and higher production cost. Thus, a compromise is needed and a moderately high temperature (of 450 oC) is used to ensure a reasonable rate of production and yield. (b) (ii) This reduces the surface area of the catalyst, hence reducing the activity of the catalyst. (b) (iii) There is a general increasing trend in the ionisation energies, from left to right, as we remove one additional electron. This is because once the first electron is removed from the neutral atom, each successive electron is removed from an ion of increasing positive charge which attracts the electrons more strongly. There is a significant difference between the third and the fourth ionisation energies because the fourth electron is removed from an inner principal quantum shell. Therefore there are 3 electrons in the outermost shell. (iv) 1s2 2s2 2p6 2 (a) In graphite, the carbons are arranged in six -membered hexagonal rings with 3 σ bonds for each carbon, the single electron not used for bonding forms delocalised pi bonds /electron cloud above and below the layers of graphite rings, allowing for the movement of electrons for electr ical conductivity in a direction parallel to the layers. In diamond, the carbons are arranged in a network of tetrahedral carbons with four σ bonds each, this network of σ bonds do not allow for electrical conductivity. (b) (i) K2Cr2O7, H2SO4 heat with immediate distillation NaOH(aq) heat (ii) NaOH, I2(aq), heat or KMnO4/K2Cr2O7, H2SO4, heat or Tollen’s/Fehling’s, warm (iii) Butanal will react faster than butanone as the H atom provides less steric hindrance than the methyl group on butanone. (c) Chlorobutane will have a white precipitate upon addition of AgNO 3 while no change will be seen for chlorobenzene. The p orbital on the chlorine atom on chlorobenzene is able to overlap with the pi electron cloud of the benzene ring, creating a partial double bond between the carbon and the chlorine. Hence there is no hydrolysis reaction with NaOH, resulting in no free Cl- ion to form AgCl(s) when AgNO3 is added.

4 3 (a) If the concentration of H + ions in the blood increases, the HCO 3 – present reacts with the additional H+ ions and thus the pH remains virtually unchanged. or HCO3 –(aq) + H+(aq)  H2CO3(aq) (b) In 100 cm3 sample, mass of Mg2+ = 127 mg = 0.127 g mass of Ca2+ = 40 mg = 0.04000g Amount of Mg2+ = 0.127 ÷ 24.3 = 0.005226 mol Amount of Ca2+ = 0.04000 ÷ 40.1 = 0.0009975 mol Mg2+(aq) + Na2CO3(aq)  MgCO3(s) + 2Na+(aq) Ca2+(aq) + Na2CO3(aq)  CaCO3(s) + 2Na+(aq) Amount of MgCO3 = 0.005226 mol Amount of CaCO3 = 0.0009975 mol Maximum combined mass of CaCO3 and MgCO3 precipitated = (0.005226 × 84.3) + (0.0009975 × 100.1) = 0.540 g (c) Aqueous BaCl2/ Ba(NO3)2. Accept aqueous Ba2+. White ppt. (d) (i) 2Mn2+(aq) + O2(aq) + 4OH–(aq)  2 MnO2(s) + 2H2O(l) (ii) Amount of S2O3 2– reacted = 5.5 1000 × 0.020 = 0.0001100 mol. Amount of dissolved O2 = Amount of S2O3 2– reacted ÷ 4 = 0.0001100 ÷ 4 = 2.750 × 10–5 mol. Mass of dissolved oxygen = (2.750 × 10–5) × 32.0 = 8.800 × 10–4 g = 8.800 × 10–1 mg Concentration of dissolved oxygen = (8.800 × 10–1) ÷ ( 50 1000) = 1.76 × 10 mg dm–3 (e) Ice is a molecular solid in which the lattice forces are hydrogen bonds. The hydrogen bonds and covalent bonds are arranged tetrahedrally around each H2O molecule. This results in a very open structure/ empty spaces and prevents the molecules from getting too close to one another. As a result, there are more molecules per unit volume in liquid water than in ice. (f) Sparingly soluble, reacts to form a weak base . MgO(s) + H2O(l) ⇌ Mg(OH)2(aq)

5 Section B (40 marks) 4 (a) (i) A – ketone B – secondary alcohol (ii) Phenol dissociates to give the phenoxide ion, C6H5O . The p-orbital of O overlaps with the -electron cloud of the benzene ring so that the negative charge on O delocalises into the benzene ring. The dispersal of negative charge stabilises the phenoxide ion so that it is more stable than the alkoxide ion. An alcohol dissociates to give the alkoxide ion, RO–. ROH ⇌ RO + H+ The electron-donating alkyl group intensifies the negative charge on O atom. Charge on RO ion also remains localised on a single electronegative O. The alkoxide ion is, therefore, the least stable and most likely to accept a proton. (iii) Concentrated H2SO4 , 170C or (iv) Geometrical isomerism arises due to the restricted rotation in the presence of  bonds. (v) Reagent Product Observation 2,4-DNPH Orange ppt K2Cr2O7/ H2SO4(aq), heat Orange K2Cr2O7 turned green, white ppt formed (b) (i) Ethyl methanoate (ii) OH O- + H+

6 (iii) A strong acid is fully dissociated whereas a weak acid is partially dissociated. (iv) pH = -log (0.2) = 0.70 (c) 5 (a) (i) Atomic radius decreases from aluminium to silicon to phosphorus. Across a period, electrons are being added to the same outermost principal quantum shell thus shielding effect remains approximately constant and nuclear charge increases, leading to increasing effective nuclear charge. (ii) AlCl3, which have cations of very high charge density, dissolve in water to form acidic solutions. The small, highly polarising cation weakens the O –H bonds of the water molecules in its surrounding sphere of coordination and results in the release of hydrogen ions in solution. AlCl3(s) + 6H2O(l)  [Al(H2O)6]3+(aq) + 3Cl–(aq) [Al(H2O)6]3+(aq) + H2O(l) ⇌ [Al(H2O)5(OH)]2+(aq) + H3O+(aq) Acidic chlorides of silicon and phosphorus dissolve readily in water to form white fumes of hydrogen chloride. The resulting solution is strongly acidic due to the hydrogen chloride produced. SiCl4(l) + 2H2O(l)  SiO2(s) + 4HCl(g) PCl3(l) + 3H2O(l)  H3PO3(aq)+ 3HCl(g) PCl5(s) +