RI H2 Chem 2013 Prelim P3 Soln

-1- Raffles Institution 2 Oct 2013 H2 Chemistry Prelim Exam 2013 – Paper 3 (Suggested Answers) Question 1 (a) The reaction involves autocatalysis where CH 3CH2COOH (or H + from the dissociation of CH3CH2COOH) is the autocatalyst. The rate of reaction gradually increases at first as the autocatalyst pr oduced catalyses the reaction. However, the rate decreases subsequently as the [CH 3CH2COOCH2CH3] decreases. (b) (i) Plotted the graph for Experiment 2.  smooth curve with 6 correctly plotted points (ii) To find the order of reaction with respect to the ester. Consider the graph for experiment 1. 1st t1/2 = 31.0 min (with construction lines shown on the graph) 2nd t1/2 = 32.0 min (with construction lines shown on the graph) Since t1/2 is constant, order of reaction with respect to ethyl propanoate = 1 Alternative answer Can also use the graph for experiment 2 to find two t1/2 values. 1st t1/2 = 63.0 min (with construction lines shown on the graph) 2nd t1/2 = 63.0 min (with construction lines shown on the graph) Since t1/2 is constant, order of reaction with respect to ethyl propanoate = 1 To find the order of reaction with respect to NaOH. For experiment 1, initial rate = 0.010/24.0 = 4.17 x 10–4 mol dm–3 min–1 For experiment 2, initial rate = 0.010/48.0 = 2.08 x 10–4 mol dm–3 min–1 When [NaOH] is doubled, initial rate is also doubled (4.17x10-4/2.08x10-4 = 2). Hence order of reaction with respect to NaOH = 1 Alternative answer Consider the graph for experiment 2. t1/2 = 63.0 min  t1/2 for experiment 2 = (2)(t1/2 for experiment 1)  reaction rate is halved when [NaOH] is halved. Hence order of reaction with respect to NaOH = 1 (iii) Rate = k [CH3CH2CO2CH2CH3] [NaOH] (iv) For experiment 1, initial rate = 0.010/24.0 = 4.17 x 10–4 mol dm–3 min–1 initial [NaOH] = 2.0 mol dm–3 initial [CH3CH2CO2CH2CH3] = 0.0200 mol dm–3 Hence k = 4.17 x 10–4/[(2.0)(0.0200)] = 0.0104 mol–1 dm3 min–1 (v) Refer to the graph. Sketch with correct co-ordinates and 2 half-lives. Half-life = 31 to 32 min 226

-2- Question 1 (b) (vi) (c) (i) The conjugate base of the ester is stabilised as the negative charge can be dispersed by delocalisation into the adjacent C=O group. Note: Accept all sensible answers (ii) (1) The conjugate base of ethane is unstable as the electron-donating methyl group will intensify the negative charge. Note: Accept all other sensible answers. (ii) (2) The –COOH group of ethanoic acid is more acidic than the -hydrogen atom. Note: Accept all other sensible answers. (iii) O OCH2CH3 O O OCH2CH3 O O OCH2CH3 O O OCH2CH3 O (iv) OCH3CH3O O O 227

-3- Question 2 (a) Nickel is regarded as a transition element because it is able to form at least one stable ion (e.g. Ni2+) with a partially filled d-subshell. Ni2+ 1s2 2s2 2p6 3s2 3p6 3d8 (b) (i) (ii) Under standard conditions, [M2+] = 1.0 mol dm-3  lg [M2+] = lg 1.0 = 0 From the graph, when lg [M2+] = 0, e.m.f. = Eo cell = +0.12 V Ni(s) + M2+(aq)  Ni2+(aq) + M(s) Eo cell = Eo M2+/M – Eo Ni2+/Ni = Eo M2+/M – (– 0.25) = +0.12 V Eo M2+/M = –0.13 V (iii) At equilibrium, Ecell = 0 V From the graph, when Ecell = 0 V, lg [M2+] = –4.10  [M2+] = 10-4.10 = 7.943 x 10-5 mol dm-3 [Ni2+] = 1.0 mol dm-3 Kc = = = 1.26 x 104 (iv) Reaction: Cu2+(aq) + M(s)  Cu(s) + M2+(aq) Eo cell = 0.34 – (–0.13) = +0.47 V Since Eo cell > 0, the reaction is feasible under standard conditions. Any one of the following observations:  blue solution of CuSO4 fades  pink deposit of Cu forms  the rod of metal M dissolves Assumption: M2+(aq) is colourless. 228

-4- Question 2 (c) (i) In both cases, there are 4 electron pairs around the central O atom giving rise to a tetrahedral electron-pair geometry to minimize repulsion. In the [Ni(H 2O)6]2+ ion, there are 3 bond pairs and 1 lone pair around the O atom. In the isolated H2O molecule, there are 2 bond pairs and 2 lone pairs around the O atom. Since the isolated H2O molecule has one more lone pair around the O atom and the strength of electron-pair repulsion decreases in the following order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, the H-O-H bond angle is smaller in the isolated H2O molecule. (ii) The two solutions have different colours despite containing the same nickel(II) ion because of the presence of different ligands. Both the H2O and NH3 ligands split the five 3d-orbitals of the Ni2+ ion into two sets of slightly different energy levels but to different extents. Hence [Ni(H2O)6]2+(aq) and [Ni(NH3)6]2+(aq) ions absorb different wavelengths of light from the visible spectrum for d-d transitions (i.e. the promotion of electrons from the lower energy d orbitals to the higher energy d orbitals). Consequently different colours, corresponding to the complements of the different colours absorbed, are observed for the two different complex ions. (d) (i) (ii) The bond energy of a X –Y bond is the average amount of energy required to break 1 mole of the X–Y bonds in the gaseous state to form X(g) and Y(g) atoms. (iii) Ni(s) + 4C(s) + 2O2(g) Ni(g) + 4C(s) + 2O2(g) Ni(CO)4(l) Ni(CO)4(g) Ni(g) + 4CO(g)0 Energy / kJ mol-1 H4 H1 4 BE(C Ni) H24 H3 or Ni(s) + 4C(s) + 2O2(g) Ni(g) + 4CO(g) Ni(CO)4(l) Ni(CO)4(g) Ni(s) + 4CO(g) 0 Energy / kJ mol-1 H4 H1 4 BE(C Ni)H24 H3 (d) (iii) By Hess’ law, H1 = H4 + (4)(H2) – 4 BE(C–Ni) – H3 –636 = +431 + (4)(–111) – 4 BE(C–Ni) – 29 4 BE(C–Ni) = 594 BE(C–Ni) = 594/4 = 148.5 = 149 kJ mol-1 229

-5- Question 3 (a) Si > S8 > P4 or Si, S8, P4 Si has a giant molecular structure . It has the highest melting point because a lot of heat energy is required to break the extensive network of relatively strong Si–Si covalent bonds. Both P and S have simple molecular structures, with P e xisting as non-polar P4 molecules and S existing as non-polar S8 molecules. S has a lower melting point than Si because less heat energy is needed to overcome the weaker instantaneous dipole-induced dipole attractive forces among the S 8 molecules than breaking the covalent bonds in Si. P has the lowest melting point among the three elements because the instantaneous dipole- induced dipole forces present are even weaker than those in S since the P 4 molecule has fewer electrons and hence a smaller electron cloud size than that of the S8 molecule. (b) (i) A: Na2O pH = 13.70 pOH = 14 – 13.70 = 0.30 [OH-] = 10-0.30 = 0.501 mol dm-3 Amount of NaOH = (0.100)(0.501) = 0.0501 mol Na2O(s) + H2O(l)  2NaOH(aq) Amount of Na2O = (½)(0.0501) = 0.0251 mol (ii) B: SiCl4 SiCl4(l) + 2H2O(l)  SiO2(s) + 4HCl(aq) Or SiCl4(l) + 4H2O(l)  SiO2 .2H2O(s) + 4HCl(aq) Or SiCl4(l) + 4H2O(l)  Si(OH)4(s) + 4HCl(aq) (c) (i) 2 IF5 ⇌ IF4 + + IF6 –  x = 4 and y = 6 Consider the IF4 + ion. Around the I atom, there are 5 electron pairs. To minimise repulsion, the electron-pair geometry is trigonal bipyramidal. Since there are 4 bond pairs and 1 lone pair, the shape of the IF4 + ion is distorted tetrahedral (or seesaw or sawhorse). (ii) The mechanism involved in the reaction is electrophilic addition. 230

-6- Question 3 (d) Initially, the solution of Br2 in hexane is red-brown in colour. When K I(aq) is added and the mixture shaken, Br 2 goes to th