RI 2008 A-Level H2 Chemistry Solutions

This document is copyrighted, please do not reproduce it without permission © Raffles Institution H2 Chemistry 9746 Nov 2008 N08-Paper 1 – MCQ BACDC CADAC DCDDA CBCAB CAACC BCCCD BDBBD CABCA N08-Paper 2 – Structured Questions 1(a)(i) HO H O H H H H C C 1(a)(ii) C C H H H H O H O H HO H O H H H H C C hydrogen bond 1(b)(i) Ethene 1b(ii) Decolourisation of purple KMnO4(aq). 1(c) The three functional groups are:  secondary alcohol,  phenol and  primary amine. 1(d)(i) 1(d)(ii) CH(OH)CH2NH2 OH OHBr BrBr CH(OH)CH2NH2 O Na O Na + + 1(d)(iii) 1(d)(iv) + OH OH C O CH2NH3 CH(OH)CH2NH3 OH OH + Cl 2(a)(i) Ksp of AgBr = [Ag+] [Br–] 2(a)(ii) Ksp of AgBr = [Ag+] [Br–] = (7.1 x 10–7)(7.1 x 10–7) = 5.0 x 10–13 mol2 dm–6 2(b)(i) For AgCl, Gppt = 2.303RT log Ksp = (2.303)(8.31)(298)(log 2.0 x 10–10) = 5.53 x 104 J mol–1 = 55.3 kJ mol-1 2(b)(ii) Gppt = Hppt  TSppt TSppt = Hppt  Gppt Sppt = [–66.0  (55.3)]/298 = 0.0359 kJ mol–1 K–1 = 35.9 J mol–1 K–1 2(b)(iii) The negative sign indicates that entropy of the system decreases. The entropy decreases because of the formation of an ordered solid lattice. 2(c) For AgF, Gppt = 2.303RT log Ksp = (2.303)(8.31)(298)(log 1.006) = +14.8 J mol–1 Since Gppt > 0, precipitation of AgF does not occur i.e. AgF is soluble in water at 298 K. 3(a)(i) CH3 CH COOH CH3 3(a)(ii) CO2H CO2H 3(b)(i) Ecell = 1.52 – 0.70 = +0.82 V + –

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 3(b)(ii) 5C6H4O2 + 2MnO4– + 6H+  5C6H6O2 + 2Mn2+ + 8H2O or + 8 H2O+ 2 Mn2+5 + 2 MnO4 5 O O + 6H + OH OH 3(b)(iii) H2(g) would reduce quinone to quinol. This is because the E for 2H+ + 2e– ⇌ H2 is less positive than +0.70 V. Alternative answer: SO2(g) 3(c) LiAlH4 in dry ether Alternative answer: NaBH4 3(d)(i) Reagent: 2,4-dinitrophenylhydrazine Quinone will react to produce an orange precipitate. 3(d)(ii) Reagent: Na Quinol will react to produce effervescence of H2 that gives a ‘pop’ sound with a lighted splint. Or Reagent: neutral FeCl3(aq) Quinol will react to produce a violet/purple solution. 3(d)(iii) Reagent: Br2(aq) For quinone, decolourisation of brown Br 2(aq) will occur. For quinol, decolourisation of brown Br 2(aq) and formation of a white precipitate will be observed. 4(a) Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9 4(b)(i) [Ag(NH3)2]Cl 4b(ii) Cation in CuCl2(aq): [Cu(H2O)6]2+ Cation in C: [Cu(NH3)4(H2O)2]2+ or [Cu(NH3)4]2+ 4(c)(i) CuCl2 + 2HCl  H2[CuCl4] D CuCl2 + Cu + 2HCl  2H[CuCl2] E 4(c)(ii) Complex ion in D: [CuCl4]2– Complex ion in E: [CuCl2]– 4(c)(iii) ligand exchange 4(c)(iv) D is tetrahedral in shape. Also accepted: square planar 4(c)(v) 4(d)(i) 4(d)(ii) +1 reduction CuCl 4(e)(i) Molar ratio of Cu to F to K = 5 .63 5 .21 : 0 .19 7 .38 : 1 .39 8 .39 = 0.338 : 2.036 : 1.018 = 1 : 6 : 3 Empirical formula of G: K3CuF6 4(e)(ii) +3 4(f)(i) Both E and F contain Cu(I). Cu(I) does not have any empty or partially-filled d orbitals. Consequently d-d transitions are not possible and hence both E and F are colourless. 4(f)(ii) G contains Cu( III) which has electronic configuration 1s2 2s2 2p6 3s2 3p6 3d8. There are partially filled 3d orbitals. Consequently d-d transitions are possible giving rise to colour observed for G.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution N08-Paper 3 – Free Response Questions 1(a) The bond energy of a X –Y bond is the average amount of energy required to break one mole of the X–Y bonds in the gas phase. 1(b)(i) The reaction can be carried out by inserting a hot wire (or glass rod) into a test tube containing the gas or by heating. (Note: mention of a catalyst here is incorrect.) 1(b)(ii) The extent of the given decomposition reaction increases down the group. 1(b)(iii) H = 2BE(H–X )  BE(X–X)  BE(H–H) For X = Cl, H = (2)(431) – 436 – 244 = +182 kJ mol–1 For X = I, H = (2)(299) – 436 – 151 = +11 kJ mol–1 1(b)(iv) The decomposition reaction becomes less endothermic down the group. Hence the extent of the reaction increases down the group. 1(c) Type of reaction: electrophilic addition Mechanism: CH2 CH2 + X X slow Step 1: Formation of carbocation CH2 CH2 X + + X a carbocation Step 2: Formation of addition product + X + X CH2 CH2 fast X CH2 CH2 X 1(d)(i) The boiling point increases from CH 3CH2Cl to CH3CH2I. This is due to increasing strength of instantaneous dipole-induced dipole interactions arising from increasing number of electrons from CH3CH2Cl to CH3CH2I. 1(d)(ii) The C –X bond polarity decreases from CH3CH2Cl to CH 3CH2I. This is because of decreasing electronegativity value from Cl to Br to I. 1(d)(iii) The reactivity towards nucleophilic reagents increases from CH 3CH2Cl to CH 3CH2I. This is due to decreasing strength of the C –X bond from X = Cl to X = I. 1(e) P is a tertiary alkyl bromoalkane with the following structure: Note: Q is the corresponding tertiary alcohol (i.e. –Br being replaced by –OH), which is resistant to oxidation by acidified K2Cr2O7. The four alkenes are as shown below. Cis-trans isomers Note: P cannot be since such a bromoalkane produces five alkenes, shown below, upon undergoing elimination. + –   denotes chiral carbon

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2(a) Products at the cathode: NaOH(aq) and H2(g) Anode: 2R–CO2–  R–R + 2CO2 + 2e– Cathode: 2H2O + 2e-  2OH– + H2 Overall equation: 2R–CO2– + 2H2O  R–R + 2CO2 + 2OH– + H2 2(b) 2CH3CO2–  CH3CH3 + 2CO2 + 2e– Quantity of charge passed = current x time = (2.0)(40 x 60) = 4800 C Amount of electrons = 4800/96500 = 0.04974 mol Amount of ethane = (½)(0.04974) = 0.02487 mol Molar mass of ethane = 30.0 g mol-1 Mass of ethane produced = (0.02487)(30.0) = 0.746 g 2(c) Let A be RCO2H. RCO2H + NaOH  RCO2Na + H2O Amount of NaOH = (11.4/1000)(0.100) = 1.14 x 10–3 mol Amount of RCO2H = 1.14 x 10 –3 mol Molar mass of A = 0.100/(1.14 x 10 –3) = 87.7 g mol–1 Hence A is either CH3CH2CH2CO2H or (CH3)2CHCO2H. (Note: you are asked for a structure hence the molecular formula C3H7CO2H is not accepted.) 2(d) Assume C is an ideal gas and apply the ideal gas equation. pV = nRT n = RT pV = ) (8.31)(380 106) x 105)(87 x (1.01 = 2.783 x 10–3 mol Mass of 2.783 x 10-3 mol of C = 0.20 g Mr of C = 3 -10 x 2.783 20. 0 = 71.8 Since C is an alkane, C has molecular formula, C5H12. 2(e)(i) P4O10 + 6H2O  4H3PO4 or P2O5 + 3H2O  2H3PO4 2NaOH + CO2  Na2CO3 + H2O or NaOH + CO2  NaHCO3 2(e)(ii) Mass of H2O = 1.55 g Amount of H2O = 0 .18 55. 1 = 0.0861 mol Mass of CO2 = 3.03 g Amount of CO2 = 0 .44 03. 3 = 0.0689 mol 2(e)(iii) Molar ratio of C to H = Amount of C : Amount of H = 0.0689 : (2)(0.0861) = 2 : 5 Hence empirical for