DHS Prelim P3 Answer scheme

1 © Dunman High School 2015 H2 Chemistry Prelim P3 2015 Year 6 H2 Chem Preliminary Examination Paper 3 (Suggested Solutions) Answer any four questions on writing paper. 1 (a) (i)  The difference in ionic radius/ size between the cation and anion is large. This leads to poor packing of the ions in the solid state. The cationic radius is large which results in a low magnitude of the lattice energy/ less exothermic lattice energy. OR the dispersion of charge of the cation due to resonance weakens the ionic bonds /interaction. (ii) A large amount of energy is required to overcome the strong ionic bonds/ interaction in [emim]BF4 hence they have high boiling point. (iii) It allows volatile organic compounds to be easily removed from the ionic liquids by simple distillation. OR They can be used as a medium for reactions involving ionic reagents since ionic reagents can dissolve readily in it. (b) Let the equilibrium partial pressure of NO be 2x. N2 (g) + O2 (g) 2NO (g) Initial partial pressure/atm 0.80 0.80 0 Change in partial pressure/atm –x –x +2x Equilibrium partial pressure/atm 0.8 – x 0.8 – x 2x 𝐾𝑝 = (𝑃𝑁𝑂)2 (𝑃𝑁2)(𝑃𝑂2) 0.40 = 4𝑥2 (0.8−𝑥)2 x = 0.1922 PO2 = PN2 = 0.80 – 0.1922 = 0.608 atm PNO = 2 x 0.1922 = 0.384 atm Note: Since the moles of reactants : products is 1 : 1, total pressure remains unchanged as volume of flask is fixed at 1 dm3. PO2 + PN2 + PNO = 0.80 x 2 = 1.60 atm.

2 © Dunman High School 2015 H2 Chemistry Prelim P3 (c) (d) The catalyst will increase the rate of both the forward and the reverse reactions hence equilibrium will be established more quickly. This means that the same equilibrium partial pressure of N 2 and NO will be established before 30 minutes. The rate of change of partial pressure for N 2 and NO will also be faster OR the graphs have steeper gradient s. The equilibrium partial pressure of all species will stay unchanged. (e) ΔH = B.E. (bonds broken) – B.E. (bonds formed) = 994 + 496 – 2(607) = +276 kJ mol–1 > 0 Since ΔH of the fwd reaction is endothermic, equilibrium will shift forward when temperature increases. 0.608 0.384 NO N2

3 © Dunman High School 2015 H2 Chemistry Prelim P3 (f) (g) L: 0.2 mol of L undergoes redox with Na to form 0.2 mol of H2 gas  There are 2 alcohol groups in L L turns moist red litmus paper blue  It contains a basic group/ it is basic/ it contains an amine group. L has a higher pKb than (CH3CH2)2NH but a lower pKb than C6H5NH2  L is less basic than (CH3CH2)2NH but more basic than C6H5NH2.  Lone pair of N atom in L less available than that of (CH 3CH2)2NH due to the electron withdrawing alcohol groups.  Lone pair of N atom in L more available than that of C 6H5NH2 as there is no delocalisation of the lone pair on N for L.  L has pKb of 4.0 (accept any value between 3.1 – 9.4) 2 (a) (b) (i) 2 Ca(IO3)2 (s)  2 CaO (s) + 2 I2 (g) + 5 O2 (g) (ii) cation Cationic size/nm Ca2+ 0.099 Ba2+ 0.135 The charge density ( α radiusionic charge ) of Ba2+ is smaller than Ca2+. Hence, polarizing power of Ba 2+ is smaller and the electron cloud of the iodate ion is polarized to a smaller extent by Ba2+. The iodine–oxygen bond in the iodate ion becomes less weakened by Ba 2+ and more energy is required to break the bonds. Thus, barium iodate(V) is thermally more stable. Ca Ix x x x x xx OO Ox  2+ 2

4 © Dunman High School 2015 H2 Chemistry Prelim P3 (c) When iodine reacts with the alkene, electrophilic addition occurs. M r of the original alkene is 310 – 2(127) = 56 => the alkene has 4 carbon atoms. Ie C4H8. For the alkene to have geometric isomerism, it is but—2–ene. The alkene is Liquid cylcohexane is non –polar, iodine is also non –polar. Iodine vapour simply dissolves in liquid cyclohexane as it is able to form favourable van der Waals forces of attractions with cyclohexane as the weak van der Waals attractions between iodine molecules is similar in strength with the weak van der Wa als in cyclohexane. (d) (i) Gθ = ΔHθ – TΔSθ = +80000 – (273+25)(152) = 34 704 J mol–1 Gθ = –2.303 RT lg Ksp 34 704 = –2.303 (8.31)(298) lg Ksp Ksp = 8.22 x 10–7 mol3 dm–9 (ii) 2IO3 – + 12H+ + 10e– I2 + 6H2O E = +1.20 V O2 + 2H+ + 2e– H2O2 E = +0.68 V E cell = 1.20 – 0.68 = +0.52 V > 0. Hence reaction is feasible. Equation: 5H2O2 + 2IO3 – + 2H+  I2 + 6H2O + 5O2 Observation: colourless solution turns brown (with black iodine solids) and effervescence of oxygen gas seen. (e) (i) CH3 OH NHCH3 CH3 D is OHCH3 Cl CH3 C is OH (ii) E is CH3CHICOCH2CH2OH O O O CH3 F C5H8O3 is C C CH3 H CH3 H C C CH3 H H CH3

5 © Dunman High School 2015 H2 Chemistry Prelim P3 3(a) E / V Fe3+ + e Fe2+ +0.77 I2 + 2e 2I +0.54 Quote 3 E ; 2 E Fe2+ + e Fe 0.44  E cell = E(Fe2+/Fe)  E(I2/ I) = 0.44  (+0.54) = 0.98 V < 0  Oxidation of I by Fe2+ is a non–spontaneous reaction.  Thus, there is no reaction between I and Fe 2+ when solutions of iron( II) sulfate and potassium iodide are mixed.  E cell = E(Fe3+/Fe2+)  E(I2/ I) = +0.77  (+0.54) = +0.23 V > 0  This is a spontaneous reaction.  I is oxidised by Fe 3+ to form I2 while the Fe 3+ is reduced to Fe 2+ if solutions containing Fe3+ and I are mixed. (b) (i)  Ag+(aq) + SCN(aq) AgSCN(s) (ii)  The red colouration is due to the formation of [Fe(H2O)5SCN]2+ complex. (iii) Ag+(aq) + SCN(aq) AgSCN(s) K1 [Fe(H2O)6]3+ (aq) + SCN(aq) [Fe(H2O)5SCN]2+ (aq) K2  Equilibrium constant for the formation of AgSCN, K1, is much larger than equilibrium constant for the formation of [Fe(H2O)5SCN]2+, K2.  [Fe(H2O)5SCN]2+ is formed only after the precipitation of AgSCN is completed.  Any excess SCN subsequently added after the end –point reacts with Fe 2+ in the analyte solution to form the red complex. (iv)  The solubility and hence the solubility product of CH 3CO2Ag varies with temperature.  The silver nitrate  sodium ethanoate mixture is kept at 25 C, so that the [Ag+] in the filtrate corresponds to the amount of solid CH 3CO2Ag dissolved at 25 C.  Titration involves the precipitation of AgSCN , whose solubility varies with temperature.  Thus, the titration mixture should also be kept at 25 C to ensure that the end– point is accurately determined. (v)  The titration must be carried out in acidic solution to prevent the precipitation of Fe3+ as Fe(OH)3. (c)(i)  Graph I is a titration curve of pAg against volume of KSCN.  At the beginning of the titration, Ag+ is in excess.  As KCN (titrant) is added, [Ag +] decreases. This corresponds to increasing pAg values.

6 © Dunman High School 2015 H2 Chemistry Prelim P3 (ii) Let s be the equilibrium concentration of Ag+. AgSCN(s) Ag+(aq) + SCN(aq) eqm [ ] / mol dm3  s s Ksp(AgSCN) = [Ag+][SCN] 1.03  10–12 = s2  s = 1.01  106 mol dm-3  pAg = pSCN = log10(1.01  106) = 5.99 (iii)  Equivalence point occurs at the point at which Graphs I and II intersect.  Volume of KSCN needed = 8.8 cm3 (iv) In the saturated solution of CH3CO2Ag, [Ag+] = [CH3CO2 ] = 0.20 )10.0)(8.8( = 0.0440 mol dm3 Thus, Ksp(CH3CO2Ag) = [Ag+][CH3CO2 ] = (0.0440)2 = 1.94  103 mol2 dm6 – (d)  Solid residue obtained by filtration of titration mixture is AgSCN. Ag C N Mass / g per 100 g sample 80.6 8.96 10.5 No. of moles 108 6.80 = 0.746 0.12 96.8 = 0.747 0.14 5.10 = 0.750 Ratio of atoms 1 1 1  Empirical form