RVHS Prelim P3 Student ANS

H2 Chem Prelim 2 Paper 3 Suggested Answers 1 RVHS 2015 1 (a) (i) Across the period, the atomic number increases, increasing nuclear charge. Hence, ionic radii of isoelectronic ions decrease from Na+ to Si4+, and from P3 to Cl. The ions of Na, Mg, Al and Si loses electrons to form cations of one isoelectronic group containing 10 electrons, while the ions of P, S and Cl gains electrons in the outermost valence shell to form anions of another isoelectronic group containing 18 electrons. Between the two isoelectronic groups, the cations have smaller ionic radii than the anions as there is one less quantum shell of electrons. [4] (ii) 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) PCl3(l) + 3H2O(l)  H3PO3(aq) + 3HCl(aq) Or PCl5(s) + 4H2O(l)  H3PO4(aq) + 5HCl(aq) [2] (b) (i) Ca(OH)2(aq) + 2HClO(aq)  Ca(ClO)2(aq) + 2H2O(l) [1] Na+ Mg2+ Al3+ Si4+ P3 S2 Cl  11 12 13 14 15 16 17 Radius/ nm Atomic number

H2 Chem Prelim 2 Paper 3 Suggested Answers 2 RVHS 2015 (ii) pH = pKa + log [ClO] [HClO] 7.6 = log(3.5  108) + log [ClO] [HClO] [ClO] [HClO] = 1.39 Amount of Ca(OH)2 added = 56.0 74.1 = 0.7557 mol Amount of OH = 0.7557  2 = 1.511 mol Let the amount of HClO present = x when freshly treated. Ca(OH)2 + 2HClO  Ca(ClO) 2 + 2H2O Initial amt / mol 0.7557 x 0  Change in amt / mol 0.7557 1.511 +0.7557  Eqm amt / mol 0 x1.511 0.7557  Ca(ClO)2  2ClO [ClO] [HClO] = 2×0.7557 V x  1.511 V = 1.39 x = 2.598 mol [HClO] = 2.598 2500×1000 = 1.04  106 mol dm3 [2]

H2 Chem Prelim 2 Paper 3 Suggested Answers 3 RVHS 2015 (c) (i) PbCl2(s) ⇌ Pb2+ + 2Cl Eqm conc / mol dm3 2.45  102 2  2.45  102 Ksp = [Pb2+][Cl]2 = (2.45  102)(2  2.45  102)2 = 5.88  105 mol3 dm9 (ii) Maximum [Cl] in water = 250×103 35.5 = 7.04  103 mol dm3 Precipitation will occur when I.P > Ksp I.P. = [Pb2+][Cl]2 = [Pb2+] (7.04  103)2 = 5.88  105 mol dm3 [Pb2+] = 1.19 mol dm3 = 2.45  102 g dm3 Minimum mass of Pb required for a 50 cm3 sample = 2.45  102  50 1000 = 12.3 g

H2 Chem Prelim 2 Paper 3 Suggested Answers 4 RVHS 2015 (d) C:H is approximately 1:1  M contains a benzene ring M reacts with ferric chloride to give purple solution  M contains a phenol group M undergoes oxidation with hot acidified K2Cr2O7 with no CO2 evolved to give N  M contains an alcohol N undergoes nucleophilic substitution with 1 mol of phosphorus chloride to give P  N contains a carboxylic acid , P is an acyl chloride , M is a primary alcohol . P undergoes acid-metal reaction with sodium, and (intramolecular) condensation  Q contains an ester (OR Q is neutral with 2 O atoms  Q contains an ester ) M: N: P: Q: [8] [Total: 20] Cl

H2 Chem Prelim 2 Paper 3 Suggested Answers 5 RVHS 2015 2. (a) 𝑝𝑉 = 𝑛𝑅𝑇 𝑛 = (20 × 101000)(2.20 × 10−3) (8.31)(273 + 450) 𝑛 = 0.740 𝑚𝑜𝑙 [1] (b) At 450 C, NH 3 has higher kinetic energy and is able to overcome intermolecular forces of attraction/ Hydrogen bonds between NH 3 molecules. Hydrogen bonding / IMF between molecules becomes negligible and NH3 behaves more like an ideal gas. [2] (c) (i) 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) [2] (ii) ∆G = ∆H - T∆S Process (2) has negative entropy change since number of moles of gas decrease from 3 to 1 . ( -T∆S) is positive and a lower temperature will result in a more negative ∆G. [2] (iii) x = (80.8) + (207) + (52.2) x = 340 kJ mol1 [2] (d) Step 1: HNO3 + 2H2SO4 ⇌ NO2 + + 2HSO4  + H3O+ [1] Step 2: [3] NH3(aq) + HNO3(aq) → NH4NO3(aq) N2(g) + 2H2(g) + 3/2O2(g) (-80.8) + (207) x 52.2

H2 Chem Prelim 2 Paper 3 Suggested Answers 6 RVHS 2015 (e) Ammonia is a stronger base compared to phenylamine[*]. The lone pair of electron on N atom in phenylamine is delocalized into the benzene ring[*], and is less available for protonation[*] compared to the lone pair of electron on ammonia. [2] (f) (i) It is the regular coils and foles on a localised segment of the polypeptide strand that is stabilised by hydrogen bonds between – NH and –CO group on the polypeptide backbone. [1] (ii) [3] (ii) pH 3 pH 7 [2] [Total: 20] CH C O O - NH3 + CH2 CH2 C O OH CH C O O - NH3 + CH2 CH2 C O O -

H2 Chem Prelim 2 Paper 3 Suggested Answers 7 RVHS 2015 3. (a) (i) The order of reaction with respect to a reactant is defined as the power to which the concentration of a reactant is raised to in the experimentally-determined rate equation. OR The overall order of reaction is defined as the sum of the powers to which the concentrations of reactants are raised to in the experimentally-determined rate equation. [1] (ii) Expt No. Time / min Relative rate / mol dm3 min1 1 1.00 1.00 2 1.20 0.833 3 1.20 0.833 4 0.96 1.04 Comparing experiments 1 and 2, When [CN] is increased to 1.2 times while [(CH 3)2C=O] and [H +] are kept constant, rate increases to 1.2 times . Hence, order of reaction with respect to CN is 1. Comparing experiments 2 and 3, When [H+] is increased to 1.2 times while [(CH 3)2C=O] and [CN ] are kept constant, rate does not change . Hence, order of reaction with respect to H+ is 0. Comparing experiments 2 and 4, When [(CH3)2C=O] is increased to 1.25 times while and [H +] and [CN] are kept constant, rate increases to 1.25 times . Hence, order of reaction with respect to (CH3)2C=O is 1. [3] (iii) Rate = k [(CH3)2C=O] [CN] [1] (iv) [2]

H2 Chem Prelim 2 Paper 3 Suggested Answers 8 RVHS 2015 (v) At a higher temperature, there is a larger number of reactant particles with energy greater than or equal to the activation energy , resulting in an increase in the frequency of effective collisions between the reactant particles and hence an increased rate of reaction. [3] (b) (i) C Cl O OH [1] (ii) [3]

H2 Chem Prelim 2 Paper 3 Suggested Answers 9 RVHS 2015 (c) (i) C6H5COOH ⇌ C6H5COO– + H+ Initial conc/ mol dm–3 0.200 0 0 Change in conc/ mol dm–3 x +x +x Eqm conc/ mol dm–3 0.200  x x x Ka = 104.19 Ka = 6.46  10–5 Ka = x2 / 0.200  x Assume that 0.200  x  0.200, x2 / 0.200 = 6.46  10–5 x = [H+] = 3.59  10–3 mol dm–3 pH = lg (3.59  10–3) = 2.44 [3] (ii) C6H5COO– + H2O ⇌ C6H5COOH + OH– [1] (iii) [2] [Total: 20]

H2 Chem Prelim 2 Paper 3 Suggested Answers 10 RVHS 2015 4 (a) Option B [1] Amount of LiNO3 = 3.00 × 0.95 × 1/ Mr = 0.04136 mol Amount of gas = 5/4 × 0.0414 = 0.0517 mol Volume of gas = 0.0517 × 24000 = 1240 cm3 [3] (b) Be2+ is smaller and more positively-charged than Li+. Hence, Be2+ has a higher charge density than Li+. Therefore, Be 2+ has a greater polarising power and weakens the N O bonds in the NO3 – anion to a greater extent. Therefore, Be(NO3)2 has a lower thermal decomposition temperature. [3] (c)  In the presence of ligands, the set of degenerate 3d orbitals of transition metal is split into 2 groups with different energy with a small energy gap. __ __ __ __ __ __ __ E d-orbitals __ __ __ in free ion octahedral complex  d electrons from the lower energy level can be promoted to the higher ener