HCI Prelim P2 Mark Scheme

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 Paper 2 1 (a) 1. Thermal stability decreases from H–F to H– I [1] 2. Quote bond energy data:H–F +562 > H–C l +431 > H–Br +366 > H–I +299 kJ mol–1 and state that this H–X bond is broken during thermal decomposition. [1] (b) (i) It is the heat released when 1 mol of gaseous F – is dissolved in an infinite volume of water (or completely dissolved in water) at 298 K and 1 bar. [1] F–(g) → F–(aq) [1] (ii) H+(g) + e– + F(g) –328 energy /kJ mol–1 +1310 H +(g) + F–(g) H(g) + F(g) –1091 –515 +562 = –1606 HF(g) +48 HF(aq) ΔH dissociation H +(aq) + F–(aq) The steps to be drawn are: 1. +562 linking HF(g) → H(g) + F(g) 2. +1310 linking H(g) + F(g) → H +(g) + e– + F(g) 3. –328 linking H +(g) + e– + F(g) → H+(g) + F–(g) 4. –1091 –515 linking H +(g) + F–(g) → H+(aq) + F–(aq) 1 link correct  1 mark 2-3 links correct  2 marks 4 links correct  3 marks (iii) ΔHdissociation = +48 + 562 + 1310 – 328 – 1606 = –14 kJ mol–1 [1] allow ecf from (a)(ii) HWA CHONG INSTITUTION 2018 C2 H2 CHEMISTR Y PRELIMINARY EXAM SUGGESTED SOLUTIONS

1 (c) (i) The hydrogen bond is formed because: 1. there is a H atom bonded to the highly electronegat ive F atom of one HF molecule [1] 2. and there is a lone pair on the F atom of another HF molecule [1] (ii) 1. label δ+ δ– for the HF molecule that provides the protonic H 2. hydrogen bond link H δ+ atom of one HF molecule to lone pair on F atom of the second HF molecule 3. label H–F coval ent bond 0.092 nm 4. label hydrogen bond 0.163 nm [½] each (d) [1] (e) (i) 1. Reaction between N 2 and H2 gives fewer number of moles of gases (from 4 mol of gases to 2 mol) or fewer gas molecules [1] 2. Number of ways to distribute particl es and/or energy decreases, disorder decreases. Hence entropy decreases [1] (ii) [1] (iii) N 2(g) + 3H 2(g) = 2NH3(g) initial mol 1 3 0 change –0.2 –0.2×3 +0.2×2 eqm mol 0.8 2.4 0.4 Sum=3.6 eqm partial pressure 0.8/3.6 ×20 = 4.44 2.4/3.6 ×20 = 13.33 0.4/3.6 ×20 = 2.22 K p = (2.22)2 4.44(13.33)3 = 4.69 × 10–4 MPa–2 [1] for all three equilibrium partial pressures [1] for Kp value (allow ecf from partial pressures) [1] for Kp units

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 Alternative working: N 2(g) + 3H 2(g) = 2NH3(g) initial MPa x 3x 0 change –0.2x –0.6 x +0.4x eqm MPa 0.8x 2.4 x 0.4x Sum=3.6 x 3.6x = 20 MPa  x = 5.556 Kp = (଴.ସ௫)మ (଴.଼௫)(ଶ.ସ௫)య = 4.69 × 10–4 MPa–2 2 (a) Nucleophilic substitution NaOH(aq), heat OR ethanolic KCN, heat OR ethanolic concentrated NH3, heat (in sealed tube) [1] Accept Elimination, ethanolic NaOH, heat Accept (alkaline) Hydrolysis, NaOH(aq), heat (b) (i) alcohol, phenol, carboxylic acid (any 2) [1] (ii) The melting point of sodium, 98°C, is relatively low so it can easily be melted to react with the organic compound in the molten form. [1] The standard reduction potential E o(Na+/Na) = −2.71 V is very negative, which shows that sodium is a strong reducing agent / Na can easily be oxidised to Na+, so sodium can react with / reduce the organic compound in the fusion reaction. [1] (c) (i) In the presence of ligands, the partially filled degenerate 3d orbitals of Fe2+ split into two different energy levels. The difference in the two energy levels, ΔE, is small, and falls within the visible region of the electromagnetic spectrum. There are vacancies in the higher energy d orbitals. An electron in a lower energy d orbital can absorb radiation in the visi ble spectrum and be promoted to the higher energy d orbital. This d-d electron transition gives rise to the colour. The violet colour seen is the complement of the absorbed colour which is yellow. [3] (ii) Element Na Fe C N O S Percentage by mass 27. 1 16.4 17.7 24.7 4.7 9.4 Ar 23.0 55.8 12.0 14.0 16.0 32.1 Molar ratio [1] 1.178 0.294 1.475 1.764 0.294 0.293 Simplest ratio 4 1 5 6 1 1 Compound A: Na4FeC5N6OS [1] (iii) NOS− [1]

2 (d) [1] 3 (a) (i) Hydrolysis [1] (ii) No. of moles of N2 formed = 2.50 × 10–3 mol Max vol. of N2 gas formed = 2.50 × 10–3 × 24.0 × 103 = 60 cm3 [1] [1] for finding and annotating the two half-lives in graph [1] explanation: from the graph, t 1/2 is constant to about 14.5 min hence reaction is first order with respect to D. (iii) k = ln 2 / t1/2 = ln 2 / 14.5 = 0.0478 min–1 [1m ans ; 1m unit] (b) (i) [1]: correct order of H + with justification, e.g. comparing expt 1 and 2, when [H+] increases by 0.01/0.005 = 2 ti mes, rate increased by 2.69 × 10–2/1.35 × 10–2 = 2 times. Reaction is first order with respect to H+. [1] Rate = k [H+][D] ecf from order of H+ expt pH Rate / mol dm –3 s–1 [H+] / mol dm–3 1 2.0 2.69 × 10–2 0.0100 2 2.3 1.35 × 10–2 0.00500 3 2.4 1.07 × 10–2 0.00398 (ii) [3] total 4 (a) (i) pH = 1.27 [H+] = 0.0537 mol dm−3 Let a = [H2C2O4] used Ka = (0.0537)2 / a – 0.0537 a = 0.1027 mol dm–3 [1] ݔ 100 × 1.50 90.0 × 1000 = 0.1027 x = 0.616% [1] with ecf (ii) C2O42– + H2O = HC2O4– + OH– [1]

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 [1] Explanation: C2O42– undergoes hydrolysis to form OH– [OH–] > [H+] indicates pH is more than 7 (b) (i) ligand exchange reaction [1] [Fe(H2O)6]3+ + 3 C2O42– → [Fe(C2O4)3]3– + 6 H2O [1] (ii) H2C2O4 formed / neutralization / acid-base reaction when H2SO4 is added. The decrease in [C 2O42–] causes the position of equilibrium in equilibrium [Fe(H2O)6]3+ + 3 C2O42– = [Fe(C2O4)3]3– + 6 H2O to shift to the left .Hence, the presence of [Fe(H2O)6]3+ gives rise to a pale yellow solution . [1] mention H 2C2O4 formed/ neutralization / acid-base reaction occur and decrease in [C2O42–] [1] equilibrium shift left, forming yellow [Fe(H2O)6]3+ (iii) Fe3+ + e Fe2+ Eo = +0.77 V 2CO2 + 2H+ + 2e H2C2O4 Eo = −0.49 V 2Fe3+ + H2C2O4 → 2Fe2+ + 2CO2 + 2H+ [1] E ocell= +1.26V > 0 (spontaneous) [1] The Fe 3+(aq) formed oxidises H 2C2O4 to form CO 2 while itself is reduced to green Fe2+(aq) [1]. 5 (a) Electrophilic substitution [1] [2] mechanism (b) (i) [1] Idea along the lines of “2- or 4- chloro/bromobenzaldehydes can be formed despite –CHO being 3-directing” (ii) • restricted rotation about C=N bond [1] • two different groups on C and on N on each end of the C=N double bond [1] (c) (i) [1] (ii) G , AlCl3 (or FeCl3), heat [1]

3 (iii) H: [0.5] Step 3: I2(aq), NaOH(aq), heat/warm [0.5] followed by dilute H2SO4 [0.5] Step 4: NaBH 4 in methanol (or H2, Ni, high pressure) [0.5] OR H: [0.5] Step 3: NaBH 4 in methanol or LiAlH4 in dry ether (or H2, Ni, high pressure) [0.5] Step 4: 1) I2(aq), NaOH(aq), heat/warm [0.5] followed by dilute H2SO4 [0.5] (d) Oxidation half-equation: C6H5CHO + 3OH− → C6H5CO2− + 2e− + 2H2O [0.5] Reduction half-equation: [Ag(NH3)2]+ + e− → Ag + 2NH3 [0.5] Overall equation: C6H5CHO + 3OH− + 2[Ag(NH3)2]+ → C6H5CO2− + 2H2O + 2Ag + 4NH3 [1] (e) (i) 22 yx d − [1] yzd [1] (ii) Highest energy d-orbital: 2z d [0.5] Lowest energy d-orbitals: xyd and 22 yx d − [0.5] 6 (a) (i) 24 [1] x y z z y x

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 (ii) ΔGo = –nF Eocell = –(24)(96500)(0.85) = –1 968 600 J mol –1 ≈ –1970 kJ mol –1 [1] allow ecf from (a)(i) (iii) Eocell = +0.40 – Eo(COଷ ଶି/C6H12O6) = +0.85 Eo(COଷ ଶି/C6H12O6) = 0.40 – 0.85 = –0.45 V [