SAJC 2008 H2 Prelim Paper3 Solutions

Suggested solutions for SAJC 2008 Prelim Paper 3 (H2 Chemistry) 1(a)(i) CrO4 3- + 8H+ + 2e Î Cr3+ + 4H2O (ii) 2CrO4 3- + 2H+ Î Cr2O7 2- + H2O + 2e (iii) 3CrO4 3- + 10H+ Î Cr3+ + Cr2O7 2- + 5H2O (b) No. of mole of S2O3 2- = 0.30 x 32.00/1000 = 0.0096 mol No. of mole of I2 = ½ x 0.0096 = 0.0048 mol Since Cr2O7 2- ≡ 3I2 From equation (iii), 3CrO4 3- ≡ Cr2O7 2- Thus, Cr2O7 2- ≡ 3 I2 ≡ 3CrO4 3- No. of mole of CrO4 3- = No. of mole of K3CrO4 = No. of mole of I2 = 0.0048 mol Mass of K3CrO4 used = 0.0048 x 233.3 = 1.12 g 1(c) Mass = density x volume = 7.3 x 0.5 x 1.0 x 10-5 x 106 = 36.5 g No. of mole of Cr = 52 5 . 36 = 0.702 mol Cr2O7 2- (or Cr+6) + 6e Î Cr(s) 6 F ≡ 1 mole Cr 6 x 0.702 mol x 96500 C ≡ 40 x time Time = 10161.45 sec = 2.82 hr 1(d) H 2O is preferentially reduced rather than Al3+ ions. 1(e) Cr3+(aq) ion is coloured due to the splitting of the d-orbitals. The electronic repulsion between the water ligands and Cr 3+ ion causes the degenerate 3d orbitals to undergo splitting, resulting in 2 groups of non-degenerate orbitals, with a small energy gap between them. By absorption of light energy, electrons can be promoted from a lower energy d- orbital to a higher energy d-orbital. This energy is related to the wavelength of the light absorbed, and the light not absorbed is thus seen as the colour of the complex. 1fi) CH 3CHC lCHC lCH 3 CH 3CHCNCH(CH 3)CN CH3CHCOOHCH(CH 3)COOH KCN alcoholic, heat C CH3C H3C O O O O HCl(aq), heat conc H 2SO4, heat CH 2(OH) 2 (ii) Geometric isomerism 1

H3C C=C CH3 H H H3C C=C H CH3H (iii) CH3 NCH3CH2 CH3 CH3 Cl- CH3 NCH3CH2 CH3 CH3 Cl- CH3 NCH3CH2 CH3 + CH3Cl 2(a)(i) 2 assumptions for Ideal gas are : - molecules in gaseous state do not exert any force or negligible forces of attraction; - volume of molecules is negligibly small compared with that of the container. ii) Set II because carbon dioxide will deviate most from ideal gas behaviour at low temperature and high pressure due to presence of attractive forces and significant molecular volume. iii) Attractive forces operate among molecules at relatively short distances. At high pressure, density of gas increases, the molecules are much closer to one another and hence, gases condense into liquid. 2(b)(i) Let the solubility of Ca(H2PO4)2 = s mol dm-3 Ca(H2PO4)2(s) Ca 2+(aq) + 2 H2PO4 -(aq) s s 2s KSP = 1.0 x 10-5 = (s) x (2s)2 = 4s3 s = 3√ [(1.0 x 10-5 ) /4] = 1.36 x 10-2 mol dm-3 No. of moles of Ca(H2PO4)2 that can be dissolved in 100 cm3 of water = 1.36 x 10-2 x 100/1000 = 1.36 x 10-3 mol 2(b)(ii) I. P. = [Ca2+] [H2PO4 -]2 = [ 30/90 x 3.0 x 10-3] x [ 60/90 x 3.0 x 10-2]2 = (1 x 10-3) x (2 x 10-2)2 = 4 x 10-7 mol3 dm-9 Since I. P. < KSP, no ppt. occurs. 2

2(c)(i) BeO and Al2O3 are amphoteric as they show both acidic and basic properties. BeO(s) + 2H+(aq) Î Be2+(aq) + H2O(l) BeO(s) + 2OH- (aq) + H2O(l) Î [Be(OH)4]2-(aq) OR: Al2O3(s) + 6H+(aq) Î 2Al3+(aq) + 3H2O(l) Al2O3(s) + 2OH-(aq) + 3H2O(l) Î 2[Al(OH)4] - (aq) (ii) Be(NO3)2(s) → BeO(s) + 2NO2(g) + ½ O2(g) (iii) Be(NO3)2 > Sr(NO3)2 easier to decompose harder to decompose Down the group, the cationic size increases and hence the charge density decreases. The polarizing power of the cations decreases down the group. Be 2+ being smaller than Sr2+, has the greater charge density and higher ability to polarise/distort the large electron cloud of the nitrate ion compared to Sr 2+. As a result, the N-O bond is weakened and the nitrate ion is more easily decomposed. Thus, Be(NO3)2 is easier to decompose than Sr(NO3)2. 2(d)(i) MnO2(s) + 4H+(aq) + 2e Î Mn2+(aq) + 2H2O(l) +1.23 V Fe2+(aq) Î Fe3+(aq) + e -0.77 V MnO2(s) + 4H+(aq) + 2Fe2+(aq) Î Mn2+(aq) + 2H2O(l) + 2Fe3+(aq) +0.46 V Black solid pale green pale pink yellow Since E0 overall = positive, the reaction is feasible. (ii) Cl2(g) + 2e Î 2Cl-(aq) +1.36 V 2Br-(aq) Î Br2(aq) + 2e - 1.07 V Cl2(g) + 2Br-(aq) Î 2Cl-(aq) + Br2(aq) +0.29 V Yellow-green red-brown Since E0 overall = positive, the reaction is feasible. 3ai) By measuring the changes in pressure/volume, or changes in colour with colorimeter. ii) Graph of Time vs [N 2O5] gives constant half-life Î 1st order reaction w.r.t. N2O5. Graph with at least two constant half-lifes, 22 min. iii) k = ln 2/t 1/2 = 0.693/22 = 0.0315 min-1 3

3aiv) bii) (ii) O N 1050 N O O 1200 O O 3c ; Since D reacts with NaOH (aq) to form E & F, D contains either an ester or an amide linkage. ; Since C 2H4O2 gives effervescence with Na2CO3, C2H4O2 contains –COOH group. C2H4O2 is CH3COOH. ; n F = 1.52/152 = 0.01 mol; n H2SO4 = 0.40 / 1000 x 25 = 0.01 mol Since n F / n H2SO4 = 1; hence F has two amine groups. ; Since F gives a white precipitate with Br 2, F must have a phenylamine group. ; Since F reacts with K 2Cr2O7, F contains a primary or secondary alcohol group. ; Since F reacts with alkaline aqueous iodine, F must contain the structure – CH3(OH)H and F must be secondary alcohol. (no marks given if students give - CH3C=O as F can be oxidised.) ; Since H reacts with alkaline aqueous iodine, H must contain the structure – CH3C=O. 4

3c E: CH3COO-Na+ F: NH2 NH2C H HO CH3 NH2 NH2C H HO CH 3 Br Br Br G: NH2 NH2C O H3C NHCOCH3 NHCOCH3C H HO CH3 H: D: ΔHatm(nitroglycerin) 4ai) C 3H5(NO3)3(l) Î 3C(g) + 5H(g) + 3N(g) + 9O(g) ΔHformation= -364 ΔHatm= 3(+715) B.E.= 5/2(+436) B.E.= 3/2(+994) B.E.= 9/2(+496) 3C(s) + 5/2 H2(g) + 3/2 N2(g) + 9/2 O2(g) ΔHformation + ΔHatm = 3(+715) + 5/2(+436) + 3/2(+994) + 9/2(+496) ΔHatm(nitroglycerin) = +364 + 6958 = + 7322 kJ mol-1 ΔHdecomposition aii) C3H5(NO3)3(l) Î 3CO2(g) + 5/2H2O(g) + 3/2N2(g) + 1/4O2(g) aiii) ΔHdecomposition = Σ(ΔHformation Products) – Σ(ΔHformation Reactants) = 3(-394) + 5/2(-242) – (-364) = - 1423 kJ mol-1 aiv) ΔGθ decomposition = ΔHdecomposition - T ΔSθ = - 1423 – [298 x (+208/1000)] = - 1485 kJ mol-1 Since ΔGθ is negative, the reaction is spontaneous at 250C. 4av) The spontaneity is not affected by changes in temperature because ΔHdecomposition is negative, T is always positive and ΔSθ is positive, i.e. ΔG = ΔH - TΔS = negative – [(+positive) x (+positive)] = negative Thus, the decomposition reaction of nitroglycerin is spontaneous at all temperatures. 5

4(bi) In the presence of water, AlCl3 undergoes hydrolysis to form an acidic solution. [Al(H2O)6]3+ (aq) + H2O (l) [Al(H2O)5(OH)]2+ (aq) + H3O+ (aq) T