RVHS H2 CHEM P2 ANS Prelim

RIVER VALLEY HIGH SCHOOL YEAR 6 PRELIMINARY EXAMINATION H2 CHEMISTRY 9647/02 Paper 2 Structured Questions Suggested Answers RIVER VALLEY HIGH SCHOOL YEAR 6 PRELIMINARY EXAMINATION H2 CHEMISTRY 9647/02 Paper 2 Structured Questions Suggested Answers 1. (a) (b) Amount of CaCO3 = 0.200 / Mr = 0.200 / 100.1 = 1.998 × 10-3 mol Amount of HCl required = 1.998 × 10-3 × 2 = 3.996 × 10-3 mol If limestone sample contains 100 % CaCO3, Volume of 0.100 M HCl needed = (3.996 × 10-3 / 0.100) × 1000 = 40.0 cm3 Volume used should be at least 40.0 cm3. (c) Amount of CaCO3 = 0.200 / Mr = 0.200 / 100.1 = 1.998 × 10-3 mol

River Valley High School 9647/02/PRELIM/11 Preliminary Examination 2011 2 Amount of CO2 gas evolved = 1.998 × 10-3 mol Volume of CO 2 gas = 1.998 × 10-3 × 24000 = 48 cm3 Volume of apparatus should be at least 48 cm3. (d) Measure the initial volume of gas, Vi. Lower the vial containing the rock sample into the acid to start the reaction. Measure the final volume of gas, V f, when bubbling has stopped / plunger of gas syringe has stopped moving. Let the volume of CO2 gas collected be V cm3, V = Vf - Vi Amount of CO2 gas = V / 24000 mol Amount of CaCO3 in rock sample = (V / 24000) × 1 = V / 24000 mol Mass of CaCO3 in rock sample = (V / 24000) × 100.1 = 100.1V / 24000 % mass of CaCO3 in rock sample = 100.1V 1 100%24000 0.200 = 2.09V % (e) CO2 gas may escape through the joints in the above experimental setup. Grease the joints to prevent the escape of gas. Or any reasonable answers. 2. (a) Stage 1: CH3COOH + SOCl2  CH3COCl + SO2 + HCl Stage 2: CH3COCl + C6H5NH2  N H O + HCl (b) Thionyl chloride (and/or Ethanoyl chloride formed) will react with water [1], (eventually leading to a lower yield of product). (c) (i) Ethanol is very soluble in water and separation of the two layers would be very difficult. (ii) To allow for better separation of the aqueous and organic layer.

River Valley High School 9647/02/PRELIM/11 Preliminary Examination 2011 3 (iii) (d) (i) HCl + NaHCO3  NaCl + CO2 + H2O (ii) Release the pressure in the separating funnel at intervals. (e) (i) Anhydrous calcium chloride/magnesium sulfate/sodium sulfate. (ii) nCH3COOH = 61.05 60 = 0.105 mol nSOCl2 = 20 0.94 119.1 = 0.158 mol  CH3COOH is the limiting reagent. Theoretical yield of N-phenylethanamide = 0.105  135 = 14.18 g % yield = 8.94 14.18 100%= 63.0% (iii) Recrystallisation 3. (a) (i) Nitric acid is an oxidising agent. (ii) 2Al + 3H2O  Al2O3 + 6H+ + 6e 6NO3 − + 12H+ + 6e  6NO2 + 6H2O 2Al + 6NO3 − + 6H+  Al2O3 + 6NO2 + 3H2O (iii) 2Al(s) + 2NaOH(aq) + 6H2O(l)  2NaAl(OH)4 (aq) + 3H2(g) (b) Aluminium oxide has a giant ionic structure while aluminium chloride has a simple molecular structure . A larger amount of energy is needed to overcome the strong electrostatic force of attraction between A l3+ ions and O2− ions than the weaker van der Waals’ forces between aluminium chloride molecules. Organic layer containing N-phenylethanamide Aqueous layer containing phenylammonium salt

River Valley High School 9647/02/PRELIM/11 Preliminary Examination 2011 4 (c) (i) Mg(NO3)2(s)  MgO(s) + 2NO2(g) + ½O2(g) (ii) Mg2+ has a smaller ionic radius compared to Ba 2+. Thus Mg 2+ has a higher charge density and is able to polarise the large NO3  anion to a larger extent and the N O bonds are weakened more. Hence magnesium nitrate decomposes at a lower temperature. (iii) Aluminium nitrate has a lower decomposition temperature. Al3+ has a smaller ionic radius (0 .050 nm) and a larger charge. Thus Al3+ has a higher charge density and greater polarising power. (c) (i) Reaction I: HCN, NaCN / trace amounts of base Reaction II: dilute H2SO4, heat (ii) H3CCC OH OH CH3 HH A H3CC C CH2 CH2 CH3 OH OH NH2 NH2 B 4. (a) Amount of Cr deposited on metal object = 3.7/52 = 7.115  102 mol Cr3+(aq) + 3e  Cr(s) Amount of e = 7.115 x 102  3 = 0.2135 mol Q = I  t 0.2135  96500 = I  1.5  60  60 I = 3.82 A (b) The reduction potential for the reduction of H 2O is more positive/less negative than that for A l3+. Therefore, H2O will be preferentially reduced at the cathode to form H 2 and aluminium will not be plated on the metal object. Anode: Al(s)  Al3+(aq) + 3e Cathode: 2H2O(l) + 2e  H2(g) + 2OH(aq) (c) Electronic configuration of Cr 3+: 1s22s22p63s23p63d3 In the isolated chromium(III) ion, all five 3d-orbitals are degenerate. In the metal complex, the different extent of inter-electronic repulsion between the central chromium(III) ion and water ligands causes the five 3d- orbitals to split into two groups with an energy gap between them. Electrons can move from a 3d-orbital of lower energy to one of higher energy if they absorb sufficient energy (E). The energy absorbed E in these d-d transitions is in the visible region of light spectrum and the colour of the ion is the complement of the colour absorbed.

River Valley High School 9647/02/PRELIM/11 Preliminary Examination 2011 5 (d) Formula of P: [Cr(H2O)4Cl2]Cl.2H2O Formula of Q: [Cr(H2O)5Cl]Cl2.H2O P contains 1 free Cl– ion which can be precipitated as AgCl by Ag+. Q contains 2 free Cl– ion which can be precipitated as AgCl by Ag+. [Cr(H2O)4Cl2]Cl.2H2O(s) + aq  [Cr(H2O)4Cl2]+(aq) + Cl(aq) + 2H2O(l) [Cr(H2O)5Cl]Cl2.H2O(s) + aq  [Cr(H2O)5Cl]2+(aq) + 2Cl(aq) + H2O(l) (e) Energy LE = ∆H hyd  ∆Hsoln = ( 4563  942)  (35) =  5470 kJ mol1 5. (a) (i) K 2 butanal p propene CO H p pp p (ii) CH3CH=CH2(g) + CO(g) + H2(g) CH3CH2CH2CHO(g) Initial p 40 40 40 0 Eqm p 40  x 40  x 40  x x propenep = COp = 2Hp = 40 – 39.6 = 0.400 atm Cr3+(g) + 3NO3 -(g) Cr3+(aq) + 3NO3 -(g) Cr3+(aq) + 3NO3 -(aq) 4563 3(314) = 942 Cr(NO3)3(s) LE 35

River Valley High School 9647/02/PRELIM/11 Preliminary Examination 2011 6 K p 3 39.6 619(0.400) 2atm (iii) At high pressure, the equilibrium position lies more to the right to reduce the number of gaseous molecules, thus increases the yield of butanal. (b) (i) Step I : nucleophilic substitution (ii) (iii) Intermediate product from step 1: Intermediate product from step 2: OH OH O O Structure of Wittig reagent: (c) (i) Rate = k( butadienep) m (ii) Clear header(s) that shows how data is processed such that order of reaction can be determine, e.g. 2 butadiene rate (p ) , OR –lg rate and –lg( butadienep) O R ( butadienep) 2. Time /s pbutadiene /atm Rate /atm s1 butadiene rate p /s1 2 butadiene rate (p ) /atm1 s1 0 0.917 9.48 x 105 1.03 x 104 1.13 x 10 3 1000 0.827 7.75 x 105 9.37 x 10 5 1.13 x 10 3 2000 0.753 6.45 x 105 8.57 x 10 5 1.14 x 10 3

River Valley High School 9647/02/PRELIM/11 Preliminary Examination 2011 7 3000 0.691 5.45 x 105 7.90 x 10 5 1.14 x 10 3 4000 0.638 4.66 x 105 7.30 x 10 5 1.14 x 10 3 (iii) The reaction is second order with respect to butadiene because 2 butadiene rate (p ) is a constant throughout the reaction. Or other suitable deduction based on graph plotted. Suitable axes and units; units not required for log or ln values Best fit line, with y-intercept, showing the correct trend. (iv) 2 butadiene rate (p ) lg rate butadienelg(p ) rate Or sketch of other suitable graph showing the correct trend. 300 K 500 K k time 300 K 500 K Grad = k 2 butadiene(p ) 300 K 500 K - lg