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]