HCI Prelim ANSWERS P2

2015 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 1 HCI 2015 C2 CHEMISTRY PRELIM PAPER 2 ANSWERS 1 (a) Ammonia is more soluble in water than it is in trichloromethane as the molecules are able to form hydrogen bonds with water molecules. Hence, [NH 3(aq)] is higher than [NH3(trichloromethane)]. [1] (b) M2+(aq) + nNH3(aq) ⇌ [M(NH3)n]2+(aq) [1] or [M(H2O)6]2+(aq) + nNH3(aq) ⇌ [M(NH3)n]2+(aq) + 6H2O(l) Addition of acid (to the aqueous layer) removes the ‘free’ ammonia. Position of equilibrium shifts to the left. The titration results will give the total amount of ammonia (‘free’ plus complexed) in the aqueous layer. [1] (c) 1) Using separate burettes, place 50.0 cm 3 of aqueous M 2+, 50.0 cm 3 of aqueous ammonia and 100.0 cm3 of trichloromethane into a stoppered bottle. 2) Stopper the bottle and shake vigorously for about 5 minutes before leaving to equilibrate to room conditions. 3) Use a dry pipette to transfer 25.0 cm3 of the bottom organic layer into a conical flask. (Place a finger over the top of pipette or fit the pipette filler before lowering into the bottom layer. This will minimise the amount of top aqueous layer that enters the pipette.) 4) Add an equal volume of water and 2 drops of methyl orange indicator. 5) Fill a burette with hydrochloric acid. Record the initial burette reading. 6) Titrate the ammonia with hydrochloric acid (shaking between each addition) until the indicator changes colour from yellow to orange. Record the final burette reading. 7) Repeat the titration until titres are within 0.10 cm3 of one another. M1  apparatus to measure M2+(aq), NH3(aq) and CHCl3 (burettes / pipettes)  for mixing without spillage (stoppered bottle / stoppered conical flask / separating funnel). M2  volume of M2+(aq) = 50 cm3, volume of NH3(aq) > 30 cm3 M3  volume of CHCl3 allows at least three titrations (e.g. > 325cm3 or 310cm3)  allows mixture to reach equilibrium (e.g. shake for 5 minutes / shake and allow to stand) M4  apparatus for titration (dry pipette, burette & conical flask)  suitable indicator (e.g. methyl orange) and correct colour change at end -point (e.g. yellow to orange). M5  uses dry pipette to remove bottom organic layer; for separating funnel, collects bottom layer into dry beaker or dry conical flask before withdrawing aliquot with a dry pipette. M6  adds equal volume of water to organic layer before titration M7  records initial and final burette readings  repeats titrations until titre values are 0.10 cm3 (d) (i) [NH3(aq)] = x / 0.4 = 2.5x mol dm–3 [1] (NH3) added = 1.0 mol Total ‘free’ ammonia = x + (2.5x)(2) = 6x mol (volume of aqueous layer is 2 dm3) (M2+) = 0.1 mol (NH3) complexed = (1 – 6x) mol n = h(NH3)complexed h(M2+) = 10 (1 – 6x) [1]

2015 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 2 2 (a) (i) [H+] = 10–9.2 = 6.31  10–10 mol dm–3 [1] [S2–] = 1.0 ´10-23 (6.31´10-10)2 = 2.51  10–5 mol dm–3 [1] (ii) Ksp = [Fe2+][S2–] [Fe2+] = 4.9 ´10-18 2.51´10-5 = 1.95  10–13 mol dm–3 [1] (b) –102 = 415 + 279 + 337 + 762 + 1560 – 1981 – 1372 + [1] = –102 kJ mol–1 [1] (c) (i) = 2.303  8.31  298  lg(4.9  10–18) = –98.7 kJ mol–1 [1] (ii) –98.7 = –102 – 298 = -102 + 98.7 298 [1] = –11.1 J mol–1 K–1 [1] (iii) entropy decreases [1] as the ppt forms because ions in the aqueous form are converted to ions held in the solid lattice (decrease in disorder) and there are fewer ways of distributing the energy / arranging the particles. [1] 3 (a) (compound R) = (compound P) = (1.00 / 327.1) = 0.003057 mol Mass of compound P = 0.003057 × 137.0 × (100/50)4 = 6.70 g [1] (b) LiAlH4 in dry ether [1] (c) (i) +4 [1] (ii) 1s22s22p63s23p63d5 [1] (iii) 4s and 3d subshells are close in energy level, once 4s electrons are removed, some or all 3d electrons may be removed without requiring significantly more energy. [1]

2015 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 3 (iv) Eϴ of MnO4 −/Mn2+ (i.e. +1.52 V) is more positive than Eϴ of MnO2/Mn2+ (i.e. +1.23 V), or Eϴ of MnO 4 −/MnO2 (i.e. +1.67 V) is more positive than Eϴ of MnO 2/Mn2+ (i.e. +1.23 V ), hence KMnO 4 is a stronger oxidising agent than MnO 2. [1] If KMnO 4 is used, product would be a carboxylic acid instead of an aldehyde. [1] (v) K2Cr2O7, dilute H2SO4, heat with immediate distillation [1] (d) (i) [2] (ii) Diagram [1] At room temperatures, t he proportion of molecules with kinetic energy greater than or equal to the activation energy is low hence there is no significant reaction. [1] At higher temperatures, t he proportion of molecules with kinetic energy greater than or equal to the activation energy increases. [1] Thus, the frequency of effective collisions increases, and reaction rate increases. [1] (iii) Less energy is required to break weaker C‒Br bond than C‒C l bond, rate of reaction using BrCH2CH2CCH will be faster than that of using ClCH2CH2CCH. [1]

2015 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 4 (e) (i) [1] (ii) K2CO3 acts as base to deprotonate H on amine group in compound Q, making the deprotonated amine a stronger nucleophile than compound Q. [1] 4 (a) (i) Anode: Zn + 4OH–  [Zn(OH)4]2– + 2e– [1] Cathode: O2 + 2H2O + 4e–  4OH– [1] (ii) = 0.40 – 1.65 = –1.25 V [1] (iii) Q = It = nF ne = (0.0384)(50) / 96500 = 1.99  10–5 mol [1] nO2 = 1.99  10–5 / 4 = 4.97  10–6 mol VolO2 = 4.97  10–6  24000 = 0.119 cm3 [1] VolO2 = cross sectional area of glass tube  (a2 – a1) 0.119 = 0.0314  (a2 – 1.00) a2 = 4.80 cm [1] (b)  Zn2+ ionic radius (0.074 nm) is smaller than Ba 2+ (0.135nm) while the charge remains the same.  Charge density of Zn2+ is higher than Ba2+.  Zn2+ has a higher polarizing power than Ba2+ and can thus better distort the electron cloud of CO3 2–, weakening the covalent bond in CO3 2– to a bigger extent.  This result in a greater ease of decomposition for ZnCO 3. Hence, decom position temperature of ZnCO3 is lower than BaCO3. [3] 5 (a) (i) Proline [1] (ii) 4 structures: [3] (b) (i) Hydrogen bond between oxygen of C=O in a peptide bond with the –CONH– hydrogen from another peptide bond across the interior of the molecule with the following indicated: lone pair on oxygen, dipoles on polar N–H bond, label. [1]

2015 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 5 (ii) The N at the aze residues will not have a H bonded to it after the peptide linkage is formed. [1] Therefore there will be fewer stabilising internal hydrogen bonds possible. Thus the stability of the folded structure will be weakened. [1] (iii) Circle phenyl or benzyl group of phe, isopropyl group of val, isobutyl group of leu , alkyl group of pro (any one) [1] Type of interaction: Dispersion forces / van der Waals forces / instantaneous dipole - induced dipole [1] 6 (a) As magnesium is less electronegative than carbon, it will take on a partial negative charge, making it an electron rich site. [1] (b) Step II: Nucleophilic addition [1] Step III: Acidic hydrolysis [1] (c) PCl5 or SOCl2 [1] White fumes of HC l observed, confirming the presence of the alcohol functional group present in Z (which is absent in X and 3-bromoprop-1-ene). [1] or Na [1]. Effervescence of colourless gas (H2) observed. [1] (d) (i) Top box: water out Bottom box: water in [1] (ii) Not a wise decision. Replacing it with a stopper cap will result in a closed system; as reflux is involved, pressure might build up, leading to unwanted consequences. [1] (e) (f) (i) 1st IE of Zn (908 kJ mol –1) is