AJC H2 Chem 2012 Prelim P3 Soln

2 2012 H2 Chemistry Preliminary Examination Paper 3 Solutions 1 (a) (i) 4FeCr2O4 + 8Na2CO3 + 7O2 2Fe2O3 + 8Na2CrO4 + 8CO2 a = 8, b = 7, c = 2, d = 8, e = 8 1 Any dilute acid e.g. dilute H2SO4 or dilute HCl (b) voltmeter electron flow V salt H2 at 25 °C & 1 atm Pt electrode bridge platinised Pt electrode 1.00 mol dm–3 Cr2O7 2– (aq), 1.00 mol dm–3 H+ (aq) 1.00 mol dm–3 Cr3+ (aq), 1.00 mol dm–3 H+ (aq) correct components of each half–cell direction of electron flow (c) Zn2+ + 2e Zn –0.76 V Cr3+ + e Cr2+ –0.41 V Zn reduces Cr3+ to blue Cr2+(aq); E ocell = +0.35 V > 0, 2H+ + 2e H2 0.00 V The acid present in the solution oxidises blue Cr2+ back to green Cr3+; Eocell = + 0.41 V, producing H2 gas. (i) Mole ratio of CrOx n– : Cr2O7 2– = 7.5 x 10–3 : 2.5 x 10–3 = 3 : 1 Thus, 3 mol of CrOxn– disproportionate to 1 mol of Cr2O72– and 1 mol of Cr3+ Let O.N. of Cr be y. y – 3 = 2(6) – 2y y = 5 O.N. of Cr in the ion CrOxn– = +5 (e) (i) Fe3+ + e Fe2+ +0.77 V Cr3++ e Cr2+ –0.41 V Eocell = +1.18 V Fe3+ + Cr2+ Fe2+ + Cr3+

2 (ii) When cell discharges, iron half –cell becomes less positive and the chromium half–cell becomes more positive. Passage of current would stop. Membrane a llows anions to pass through to maintain electrical neutrality in the two half– cells. (not allowing the positive ions to pass through, otherwise self–discharge or short circuit occurs) [2] (iii) Eocell = Eored – Eoox 1.59 = +0.40 – Eoanode Eoanode = –1.19 V Anode: Zn + 4OH– Zn(OH)42– + 2e (or Zn + 2OH– Zn(OH)2 + 2e) (iv) Q = 0.8 x 1.9/65.4 x 2 x 96500 = 4486 C Current = 4486 / (30 x 24 x 60 x 60) = 1.73 x 10–3 A (v) ICB is expected to discharge faster as it involves simple electron transfer reactions, while energy is needed to break covalent bonds in O2 (and H2O) in the zinc–air battery. (vi) Zn–air battery has higher energy density as it uses air as the oxidising agent, unlike the heavier oxidising agent, MnO 2, used in the alkaline battery. Thus, more zinc can be packed within a cell of similar weight. 2 (a) (i) It is the amount of energy evolved when 1 mo le of solid ionic compound is formed from its constituent gaseous ions at 298 K and 1 atm. (ii) I: As the cationic radius increases down the group, inter–ionic distance increases OR state that L.E. a and cationic radius (r+) increases down the group. Thus, the strength of electrostatic forces of attraction between the M 2+ and SO42– ions decreases. Hence, the magnitude of L.E. decreases. II: DHhydration of the cation a + + r q As the cationic radius increases down the group, OR the charge density of the cation decreases. Thus, the strength of ion– dipole interactions formed between M 2+ and water molecules decreases and hence DHhydration becomes less exothermic. (iii) DHsolution = – L.E. + DHhydration of M2+ + DHhydration of anion For Group II sulfates, the decrease in |L.E| is less than that of | DHhydration of M2+|. Thus, DHsolution becomes more positive down the group and solubility decreases For Group II hydr oxides, size of the OH – anion is much smaller than that of the SO 42– anion. The decrease in |L.E| is more than that of | DHhyd of M2+|. Thus, DHsolution becomes more negative down the group and solubility increases. OR The size of the cation is much smaller than that of the anion (SO 42–), thus the decrease in |L.E| is less significant than the decrease in the |DHhydration|.

3 Conversely, the size of the OH – anion is much smaller than that of the SO 42– anion, thus the decrease in |L.E| is more significant than the decrease in the |DHhydration|. (b) (i) For Hg2SO4, min. [SO42–] = 7.4 x 10–7 / (0.1/2)2 = 2.96 x 10–4 mol dm–3 For CaSO4, min. [SO42–] = 2.4 x 10–5 / (0.1/2) = 4.80 x 10–4 mol dm–3 (ii) Since min. [SO 42–] required to precipitate fi rst trace of Hg2SO4 is lower, Hg2SO4 will be precipitated first. When max. Hg 2SO4 is precipitated, i.e. when first trace of CaSO4 appears, [SO42–]in solution = 4.80 x 10–4 mol dm–3 \ [Hg+]in solution = [7.4 x 10–7 / (4.80 x 10–4)]1/2 = 0.0393 mol dm–3 n(Hg +) in 50 cm3 solution = 0.0393 x 50/1000 = 1.97 x 10–3 mol n(Hg+) precipitated = 0.1 x 25.0/1000 – 1.97 x 10–3 = 5.30 x 10–4 mol 2Hg + + SO42– Hg2SO4 n(Hg 2SO4) precipitated = ½ x n(Hg+) precipitated = ½ x 5.30 x 10–4 = 2.65 x 10– 4 mol Max. mass of Hg2SO4 precipitated = 2.65 x 10– 4 x (2 x 201 + 32.1 + 4 x 16.0) = 0.132 g (c) (i) L.E of CaO = –(– 147 + 753 + 590 + 1150 + ½ (496) + 178) + (–635) = – 3410 kJ mol–1 (3 s.f.) CaO(s) Ca(s) + ½O2(g) –635 +178 ½ (+496) 0 Ca(g) + O(g) Ca(g) + ½O2(g) Ca2+(g) + 2 e– + O(g) +590 + 1150 Ca2+(g) + O2–(g) –147 + 753 L.E Energy / kJ mol–1

4 (ii) magnitude of the L.E. of Ca3N2 should be larger than that of CaO due to a larger charge on N3– OR due to more ionic attractions between 5 mol of ions in 1 mol of Ca3N2 than between 2 mol of ions in 1 mol of CaO. 3 (a) (i) sp2 hybrid orbitals (ii) H2SO4 + SO3 HSO4– + HSO3+ (iii) Electrophilic substitution [1] CH3 SO3H+ SO3H H CH3 slow SO3H H CH3 + HSO4 - SO3H CH3 fast + H2SO4 correct arrow movement correct carbocation intermediate and regeneration of catalyst (iv) The lone pa ir of electrons on N atom is delocalised into the benzene ring. This further intensifies the negative charge on the sulfonate anion, making it less stable. (v) Add aqueous bromine to each compound. Observation: orange solution of bromine turns colourless and a white precipitate is formed for 4– aminobenzenesulfonic acid. Bromine solution remains orange and no precipitate formed for 4–methylbenzenesulfonic acid OR Heat each compound with KMnO4 in dilute H2SO4. Observation: purple KMnO 4 turns colourless for 4–methylbenzenesulfonic acid. KMnO 4 remains purple for 4–aminobenzenesulfonic acid. OR Add cold HCl / H 2SO4 Observation: 4– aminobenzenesulfonic acid dissolves in HC l but 4– methylbenzenesulfonic gives two immiscible layers.

5 (c) (i) OH NO2 COOH NO2 NO2 O OH NH2 OH O NH O K L M N O Reaction Type of reaction Deduction Treating compound K, C8H9NO3, with hot acidified potassium manganate(VII) gives compound L, C7H5NO4. oxidation loss of 1 C Þ side–chain oxidation occurred. L produces effervescence with sodium hydrogen carbonate. acid–base reaction Carboxylic acid group in L reacted with sodium hydrogen carbonate. L contains carboxylic acid group, –COOH. Heating compound K under reflux with acidified potassium dichromate(VI) gives compound M, C 8H7NO4. oxidation gain in 1 O, loss of 2 H Þ 1o alcohol in K is oxidised to –COOH in M. Compound M is treated with tin and concentrated hydrochloric acid, followed careful neutralisation, compound N, C 8H9NO2 is obtained. reduction –NO2 group in M is reduced to –NH2 group in N.

6 1 mole of compound N reacts with 2 moles of aqueous bromine. electrophilic substitution N is a phenylamine. -Br is substituted at positions 2, 4 or 2, 6 w.r.t the –NH 2 group Treatment of compound N with anhydrous phosphorus pentachloride produces compound O, C8H7NO. N is first converted to acyl chloride, by PCl5 via nucleophilic substitution. The acyl chloride formed then undergoes intra– molecular condensation with the –NH 2 group to form a cyclic amide O. The –CH2COOH group must be adjacent to the –NH2 group in N to enable ring formation/formation of a cyclic amide (ii) Compound N will dissolve in the aqueous layer while compound O will remain in the organic layer. The basic – NH2 group in compound N will react with cold dilute HCl to give an ionic product which can form strong ion–dipole interactions with water molecules. Compou