SAJC H2 CHEM P1 P2 P3 ANS Prelim

1 2011 A LEVEL PRELIM P1 SOLUTIONS 1 A 11 D 21 B 31 D 2 D 12 A 22 C 32 B 3 B 13 C 23 C 33 B 4 C 14 C 24 C 34 A 5 D 15 C 25 D 35 A 6 B 16 C 26 D 36 B 7 A 17 A 27 A 37 D 8 C 18 C 28 C 38 A 9 A 19 B 29 C 39 B 10 C 20 D 30 B 40 C 2011 Prelims (P2): Suggested Answer 1 (a) No of moles of H2SO4 = 10/1000 x 1 = 0.01 mol No of moles of NaOH = 40/1000 x 1.50 = 0.06 mol (excess) No of moles of H2O formed = 0.02 mol Amount of heat produced = 0.02 x 57 400 = 1148 J Temperature rise = 1148/(50 x 4.18) = 5.5 oC (b) Volume of FA 1 / cm3 10 20 30 40 Volume of FA 2 / cm3 40 30 20 10 Temperature rise, T / oC 5.5 11.0 8.2 / 8.3 4.1 / 4.2 (c) 1. Measure 10.0 cm3 of FA 1 using a 20 cm3 measuring cylinder. 2. Pour the FA 1 into a styrofoam cup. 3. Measure the temperature of FA 1 in the cup using a thermometer. 4. Measure 40.0 cm3 of FA 2 using another 50 cm3 measuring cylinder. Measure the temperature of FA2. 5. Take the average of the temperatures of FA1 and FA2 as the initial temperature. 6. Pour the FA 2 into the Styrofoam cup containing FA 1 and stir using the thermometer. 7. Record the maximum temperature of the mixture. 8. Repeat the experiment using the various volumes of FA 1 and FA 2 shown in the table in (b). OR 1. Measure 10.0 cm3 of FA 1 using a 20 cm3 measuring cylinder. 2. Pour the FA 1 into a styrofoam cup. 3. Measure 40.0 cm3 of FA 2 using another 50 cm3 measuring cylinder. 4. Pour the FA 2 into the Styrofoam cup containing FA 1. 5. Stir the mixture gently and record the initial temperature of the mixture using a

2 thermometer. 6. Record the maximum temperature of the mixture. 7. Repeat the experiment using the various volumes of FA 1 and FA 2 shown in the table in (b). (d) Source of error and suggestion to minimise error Heat absorbed by the calorimeter Calibration of calorimeter to account for heat absorbed (see calibration of calorimeter) Heat loss to surrounding Provide lagging on the body of the calorimeter OR Use a lid to cover the calorimeter OR Consider cooling correction (see next section) Measuring cylinders not precise in measuring volumes of FA 1 and FA 2. Use a burette to measure accurately the volumes of FA 1 and FA 2. (e) From the graph, find the point of intersection of the two lines to obtain the vol of FA 2 with maximum temperature rise. Assume vol of H2SO4 in FA 1 is V1 cm3 at point of intersection V1/1000 x [H2SO4] = (50 – V1) /1000 x 1.5/2 [H2SO4] = (50 – V1) x 1.5 2 V1 2 (a) (i) Powdered iodine has larger surface area for effective collision. Thus the rate of reaction will increase. (ii) The reaction is exothermic and has to be kept in control OR iodine may oxidise ethanol. (iii) To ensure the reaction is complete. (b) (i) CH3CH2OH + CH3CH2I  CH3CH2OCH2CH3 + HI Iodoethane undergoes nucleophilic substitution with ethanol acting as the nucleophile. (ii) I2 + 2e 2I- E = +0.54 O2 + 4H+ + 4e 2H2O E = +1.23 Overall: 4I- + O2 + 4H+  2I2 + 2H2O E overall = +1.23 - +0.54 = +0.69 V (c) (i) To remove / react H3PO3. (ii) To remove water soluble impurities / ethanol / ether / last traces of sodium carbonate. (iii) K2CO3 / CuSO4 / CoCl2 / MgSO4 / Na2SO4 / CaSO4 (iv) The boiling temperature of fixed at 71°C to ensure that the distillate is pure.

3 (d) (i) Cl is more electronegative than I will pull bond pairs of electrons towards itself. Hence, there is lesser repulsion between the bond pairs and the bond angle of PCl3 is smaller. (ii) PI3 + 3H2O  H3PO3 + 3HI (iii) White fumes were seen for the reaction of PCl3 and water due to the formation of HCl gas. The HI formed for the reaction of PI3 and water undergoes thermal decomposition to form purple I2 fumes. The size of I is bigger than Cl. Thus the atomic overlap between H and I atoms is less effective. The bond strength of H-I is weaker, hence it is more easily broken to form I2. 3 (a) (i) The formation of B is faster than C OR Step 1 is faster than Step 2 as it has a steeper gradient OR the maximum concentration is attained faster in the formation of B than C. (ii) Step 2 (iii) t1/2 = 0.40 min 2 half-lives constant (iv)

4 (b) 4 (a) (i) OR NiOOH + H2O + e  Ni(OH)2 + OH- Iron: Fe + 2OH- Fe(OH)2 + 2e (ii) Iron to Nickel oxide-hydroxide (iii) 1.2 = E°cell (NiOOH) – (+0.55) Thus, E°cell (NiOOH) = +1.75V (iv) NiOOH + H2O + e Ni(OH)2 + OH- When the acid is added, it removes the OH- ions and hence reduces its concentration leading to the equilibrium above to shift to the right to replenish the OH- ions removed. Hence the E°cell at the cathode becomes greater than +1.75V, resulting in the e.m.f becomes greater than +1.2V (v) When the volume of the hydroxide increase, there is no change in concentration of the OH- hence no shift in equilibrium Hence, there is no change in e.m.f (b) (i) 2Ni(OH)2 + Fe(OH)2 2NiOOH + Fe + 2H2O (ii) Mass of Fe required = 3 x 0.95 = 2.85g No. of moles of Fe = 2.85/55.8 = 0.05107 No of moles of electrons required = 0.05107 x 2 = 0.1022 Time / min Concentration / mol dm-3 1.2 - - 1.0 - - 0.8 - - 0.6 - - 0.4 - - 0.2 - - 0.0 | | | | | | | | | | | | | | | | | | | | | | | | 3 1 2 4 5 6 A B C

5 Thus, Total amount of electricity required = 0.1022 x 96500 = 9857.5C Thus minimum current = 9857.5 / (2 x 60 x 60) = 1.37A (iii) Total amount of electricity = 1.37 x (2 x 60 x 60) = 9864 C Total number of electrons = total number of nickel oxide-hydroxide = 9864 / 96500 = 0.102 [1] Total mass of nickel oxide-hydroxide = 0.102 x (58.7 + 32 +1) = 9.35g Actual mass of nickel oxide-hydroxide = 9.35/0.95 = 9.85g (iv) O2 + 2H2O + 4e 4OH- E°red = +0.40V Having a more negative than +1.75V ( E°red of the reaction at the nickel (III) oxide-hydroxide), the hydroxide ions are also discharged to produce oxygen gas. 5 (a) (i) Fe(s) + 3/2 Cl2(g) FeCl3(s) (ii) 1s2 2s2 2p6 3s2 3p6 3d5 (iii) Fe(H2O)6]3+ + H2O [Fe(H2O)5(OH)]2+ + H3O+ 2H+ + CO3 2- CO2 + H2O (iv) A: [Fe(H2O)6]3+.3Cl- OR [Fe(H2O)6]Cl3 OR [Fe(H2O)6]3+ B: Fe(OH)3 (s) C: Fe(OH)2 (s) D: neutral FeCl3(aq) (v) Ligand exchange [Fe(EDTA)]- (b) (i) I2 + 2Cu2+ 2I+ + 2Cu+ Redox (ii) Electrophilic substitution

6 Suggested Solution P3 2011 Prelim H2 Chemistry 1. a) i. Unique sequence of amino acids held by peptide bonds. Any permutation of Ala, Gln and Lys. To take note of Isoelectric point with respect to pH = 7.3. Diagram involving 3 amino acids. Either diagram below is accepted. 1. a) ii. disrupting the tertiary (R-groups of glutamine would form hydrogen bond OR lysine will form ion-dipole interactions with the –OH functional group in ethanol) disrupting the secondary structure (peptide linkages would now form hydrogen bonds with ethanol instead of other peptide linkages) Diagrams drawn are as represented below to illustrate the same points above: causing the tertiary and secondary structures to unwind and become a random coil thus denaturing / loses the unique 3D conformation of the protein. 1. b) i. 1. acidified potassium dichromate, heat and distill 2. HCN, trace amounts of NaOH (aq), @ 10 – 20 oC 3. H2SO4(aq), heat ii. 1. alkaline iodine solution, heat 2. acidify with mineral acid / H+ (aq) 1. c) i. CH3CH2`OH + H2O  CH3COOH + 4H+ + 4e-