CJC H1 CHEM P1 P2 ANS Prelim

Paper 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C A C C D B A B B D B A C A A 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 C C C A C A C D B B B B D C A Paper 2 1(a) (i) (ii) H = 180 – 282 = -102 kJ mol-1 (iii) The activation energy will decrease. (iv) H = 150 – ½ (496) = -98 kJ mol-1 (v) Bond energy is an estimated value / average value only. (b) (i) Amount of unreacted iodine = 2 102 . 01000 60 . 23   = 2.36 x 10 -4 mol (ii) Amount of iodine reacted with sulfur dioxide = 410 36 . 2 01 . 01000 40    = 1.64 x 10-4 mol (iii) I2 +2e  2I- SO2 + 2H2O  SO4 2- + 4H+ + 2e Balanced equation: SO2 + I2 + 2H2O  2I- + SO4 2- + 4H+ Amount of SO2 present in wine = 1.64 x 10-4 mol (iv) Concentration of SO 2 = 1000 50 10 64 . 14 = 3.28 x 10-3 mol dm-3 [Total: 12] 8872 / CJC JC 2 CHEMISTRY Promotional Exam 09 ANSWERS

2(a) (i) The first ionisation energy of an element is defined as the amount of energy required to remove one electron from each atom in a mole of gaseous atoms producing one mole of gaseous ions with one positive charge. (ii) Be(g)  Be+( g ) + e (b) N 1s 2 2s2 2px 12py 12pz 1 O 1s 2 2s2 2px 22py 12pz 1 Less energy is required to remove a paired 2p electron in O since repulsion is experienced between the paired electrons. (c) Ionisation energy / arbitrary units Ionisatio n number 1 2 3 4 5 6 (d) (i) SiO2 is giant covalent that requires a lot of energy to break the strong covalent bonds between atoms during melting. SiCl4 is simple covalent that requires little energy to break the weak Van der Waals’ forces between molecules during melting. (ii) SiCl 4 + 2H2O  SiO2 + 4 H C l pH = 2 (3 is also accepted) [Total: 10] 8872 / CJC JC 2 CHEMISTRY Promotional Exam 09 ANSWERS

3(a) CH2Cl O CH2OHO H (ecf from B) E: HCN, trace amount of NaOH, (10 – 20 oC ) F: H C l ( a q ) , r e f l u x (b) Test: Add Tollen’s reagent to J, K and L and warm. Observation: Only Y gives silver precipitate, the others will not. OR Add 2,4-DNPH and warm, only Y gives orange precipitate, the others will not. Test: Add Br 2 in CCl4 or aqueous to J and L. Observation: Only J decolourises the reddish-brown Br2. L will not. Test: Add I2, NaOH (aq) and warm L. Observation: L gives yellow precipitate OR Add PCl 5 to L, white fumes (HCl) is evolved. Answers of the correct logic or sequence will be given marks. (c) [Total: 12] 4(a) The sailors ate fruits / vegetables. (b) OR (c) S t a r c h (d) Amt of triiodide solution = Amt of vit C in 25.0 cm 3 sample = 4 10 006 . 7 031 . 01000 60 . 22     Amt of vit C in 100 cm3 sample = 34 10 80 . 2 10 006 . 70 . 25 100      Mass of vit C in tablet = mg g493493 . 0 176 10 80 . 2 3   Common dose of vitamin C is likely to be 500 mg. [Total: 6] CH3 CO2H CO2H Cl Acidified KMnO4, Reflux [1] Cl2, AlCl3 [1] [1] A CH2Cl B CH2NH2 C CH2Cl D C O C CC O OH O H H CH O H CH2 O H C O C CC O OH O H H CH O H CH2 O H 8872 / CJC JC 2 CHEMISTRY Promotional Exam 09 ANSWERS

5(a) (i) It is the enthalpy change when one mole of water is formed in the neutralisation of an acid and alkali, the reaction being carried out in aqueous solution under standard conditions, at 25 oC, 1 atm. (ii) Amt of CH 3CO2H = 31000 40  = 0.012 mol Amt of KOH = 4 . 11000 60  = 0.084 mol (limiting) = Amt of H2O formed = 0.084 mol Heat absorbed by water = (40 + 60) x 4.2 x 10.5 = 4410 J Hn = - 084 . 0 4410 = - 52500 J mol-1 = -52.5 kJ mol-1 (iii) The value in a(ii) is less exothermic/ less negative/ differs from that of the reaction between a strong acid and alkali because the enthalpy change due to the conversion of CH3CO2H into ions has to be considered. (b) (i) Le Chatelier's Principle states that if a change is made to a system in equilibrium, the system reacts in such a way as to tend to oppose the change , and a new equilibrium i s f o r m e d . (ii) Low temperature , so that equilibrium will lie to the left which is the backward exothermic reaction. Good air flow, so that equilibrium will lie to the left to decrease the amount of oxygen gas. (iii) There will be an increase in the production of SO3. Equilibrium will lie to the right to decrease the amount of oxygen gas when there is good air flow and the low temperature favours the forward exothermic reaction. (iv) K c =     2 2 2 2 3 O SO SO (v) 2SO 2(g) + O2(g) 2SO3(g) I 2 1 0 C -1.9 -0.95 +1.9 E 0.1 0.05 1.9 Equilibrium concentration of SO 2 = 2 1 . 0 = 0.05 mol dm-3 (Recall vol is 2 dm3.) Equilibrium concentration of O2 = 2 05 . 0 = 0.025 mol dm-3 Kc =      14440025 . 0 05 . 0 95 . 0 2 2 2 2 2 2 3  O SO SO mol-1 dm3 (c) (i) Substitution. (ii) Compound Y has a high boiling point because it is giant ionic and requires a lot of energy to break the strong electrostatic forces of attraction between the oppositely charged ion. Compound Y is soluble in water because it can form ion-dipole interaction with the m o l e c u l e s o f w a t e r . [Total: 20] 8872 / CJC JC 2 CHEMISTRY Promotional Exam 09 ANSWERS

6(a) H H Equation:  A has a benzene ring as C:H ratio is almost 1:1.  A neutral liquid A, C10H12O2 when heated with aqueous hydrochloric acid gave two products, B and C.  A undergoes acid hydrolysis to form B and C.  B gave a yellow precipitate when warmed with aqueous alkaline iodine.  B has –CH3CH(OH) structure.  When heated with acidified sodium dichoromate (VI), B gave D, C8H8O.  B undergoes oxidation to form a ketone, D.  D reacts with 2,4-dinitrophenylhydrazine to give orange crystals, E.  D is a ketone. (b) (i) S 2O8 2- and I- will repel each other/ No catalyst is added. (ii) Using experiments 1 and 3, When the concentration of I- increased by 1.5 times, the rate increased by 1.5 times. 1st order with respect to I-. Using experiments 2 and 3, When concentration of S2O8 2- is halved, the rate is halved. 1st order with respect to S2O8 2-. (iii) Rate k[I -][S2O8 2-]   C  O  C  CH3 A C  OH B    CH O CH3 3 C = O C DCH CO H  3 2 CH3 C = N  N    CH3 H NO2 E NO2 H H   O  + H C  OH + CH CO H C  O  C  CH3 2 3 2    CH3 CH O 3 8872 / CJC JC 2 CHEMISTRY Promotional Exam 09 ANSWERS

(c) Al3+ has a high charge density, hence when AlCl3 dissolves in water, the surrounding water molecules in the hydrated Al 3+ ion is polarised. The O-H bond in water is polarised and H+ is easily liberated. AlCl3(s) + 6H2O  [Al(H2O)5(OH)]2+(aq)+ H+(aq) + 3Cl–(aq) (Equation without Cl- is accepted.) PCl3 is covalent and it dissolves vigorously in water to give an acidic solutio