NJC H2 Chem 2013 Prelim P2 Soln

NJC SH2 Prelim 2013 H2 Chemistry Paper 2 1 SH2 H2 Prelim H2 2013 Answers Paper 2 1(a) (i) Ca(IO3)2 (s) Ca2+ (aq) + 2IO3 ‒ (aq) ΔHsol > 0 As temperature is increased, solubility of calcium iodate(V) increases. When temperature is increased, equilibrium shifts right to favour the forward endothermic reaction to partially absorb the extra heat. Hence more Ca(IO3)2 dissolved in water. (ii) [1] 1(b) (i) 2S2O3 2- + I2  2I‒ + S4O6 2- Titrate the iodine produced with thiosulfate solution. (ii) Preliminary calculation Volume of Ca(IO 3)2 used for reaction = 25.0 cm3 (pipette capacity) No. of moles of IO 3 ‒ in the 25.0 cm3 solution = 1 .390 6 . 4 x 2 x 1000 25 = 5.896 x 10-4 mol IO3 ‒ : I 2 : S 2O3 2- 1 : 3 : 6 124

NJC SH2 Prelim 2013 H2 Chemistry Paper 2 2 No. of moles of S 2O3 2- required for titration = 6 x 5.896 x 10 -4 = 3.538 x 10 -3 mol Vol. of S 2O3 2- required for titration = 3.538 cm 3 Volume is too small and will results in high experimental error. Therefore, concentration of FA3 is too high and dilution is required. Volume of H + required for titration = 3.538 cm 3 (no. of moles of H + required is same as S 2O3 2- and their concentrations are the same) Volume of I - required for titration will be less than volume of H + required. Since the I- and H+ can be in excess, 5 cm3 each of FA2 and FA4 are used. Dilution Desired amount of S 2O3 2- from the burette = 25 cm3 Dilution factor = 538. 3 25 Volume of FA 3 to be diluted in a 250 cm 3 volumetric flask = 250 ÷ 538. 3 25 = 35.38 cm 3 1. Using a burette, transfer 35.40cm 3 of FA3 into a clean 250 cm 3 volumetric flask. 2. Top up to the mark with deionized water, stopper, invert and shake to ensure homogeneous solution. 3. Label this solution FA5. Procedure 1. Filter the suspension, FA1, into a clean and dry conical flask using a dry filter paper and funnel. 2. Pipette 25.0 cm 3 of the filtrate into a 250 cm3 conical flask. 3. Using 2 separate 10 cm 3 measuring cylinders, add 5 cm 3 each of FA2 and FA4 into the same conical flask and swirl to ensure mixing. 4. Titrate the contents of the conical flask against FA5 from a burette till a colour change from brown to pale yellow is observed. 5. Add 3 drops of starch indicator and continue titration till the blue-black colour is discharged. 6. Record the volume of FA5 required for titration. 7. Repeat steps 2 to 6 till consistent titration results of within 0.10 cm 3 are obtained. 125

NJC SH2 Prelim 2013 H2 Chemistry Paper 2 3 1(c) experimental data, x = volume of S2O3 2- required for titration (dm3) n(S 2O3 2-) required for titration = (x) × concentration of S2O3 2- = a [IO3 -] = )10 25( 6 3 a = b mol dm-3 [Ca 2+] = ½ b mol dm -3 Ksp = [Ca 2+] [IO3 -]2 = (½ b) (b 2) = ½ b 3 2(a) Rxn A : Nucleophilic addition Rxn B: Hydrolysis (b)(i) No naked flame around / heating should be done with electric isomantle as chemicals are flammable. Perform experiment in fume cupboard as chemicals are toxic. (ii) To exclude any trace of water/moisture or to ensure chemicals are dry. (iii) CH3CH2I + Mg  CH3CH2MgI I = 2(b)(iv) nGrignard reagent produced in step II = 0.06173 mol npentan-2-one = = = 0.0568 mol < nGrignard reagent Since pentan-2-one reacts with Grignard reagent in the mole ratio of 1:1, hence Grignard reagent is in excess. (v) C OH CH2CH3 CH3 CH2CH2CH3 126

NJC SH2 Prelim 2013 H2 Chemistry Paper 2 4 (vi) CH3CH3 (ethane) (vii) Ethoxyethane forms an immiscilble layer above the aqueous layer because of :  unfavourable interaction between ethoxyethane and water or forces of attraction between water molecules are stronger than the forces of attraction between ethoxyethane and water  ethoxyethane is less dense than water (0.713 g cm 3 as against 1 g cm3) 2(b)(viii) Chemicals added Impurity removed 1 water Mg salts like MgI2 2 saturated sodium hydrogencarbonate HCl 3 sodium thiosulfate iodine 4 saturated sodium chloride H2O 2 (c) 3 (a)(i) Free Radical Substitution Initiation: (*must use half arrows to show homolytic bond cleavage) 127

NJC SH2 Prelim 2013 H2 Chemistry Paper 2 5 Propagation: (*must show formation of C5H8Cl2) Termination: (*any two possible termination steps) 3(a)(ii) Type of reaction: Elimination Reagents and conditions: Ethanolic KOH, heat 3(a)(iii) The negative charge on the carbon atom is delocalised across the 5 carbon atoms in the ring due to continuous overlap of p orbitals, thus the anion is stabilised by resonance. As the conjugate base is relatively stable, the dissociation cyclopentadiene to give H+ is possible. 128

NJC SH2 Prelim 2013 H2 Chemistry Paper 2 6 3(b)(i)Heat evolved, Q = mcT = (1)(4.18)(100  27) = 305.14 kJ (heat gained by water) Since the process is only 65% efficient, Total amount of heat released from combustion = 305.14  65 100 = 469.4 kJ Amount of C5H10 = 70 10 = 0.1429 mol Hc =  0.1429 469.4 =  3286 =  3290 kJ mol1 (3 s.f) 3(b)(ii) By Hess’s Law, Hrxn = 29 + (2800) + 2(286) + 3286 = 57.0 kJ mol1 (3 s.f.) 4(i) HO O HO N CH3 morphine * * * * * 5 chiral carbon atoms 129

NJC SH2 Prelim 2013 H2 Chemistry Paper 2 7 (ii) Reaction: HO O O N CH3 H3C CH3 + Cl HO O O N CH3 H3C C Cl H H H 4(iii) (iv) Neutral FeCl3(aq) or freshly prepared FeCl3(aq)/ FeCl3 solution Morphine: Violet complex observed Codeine: No violet complex observed or remained yellow HO O O N CH3 H3C CH3 + Cl +  HO O NaO N CH3 C O H3C ONa 130

NJC SH2 Prelim 2013 H2 Chemistry Paper 2 8 5a(i) K C Fe O % by mass 26.8 16.5 12.8 43.9 Ar 39.1 12 55.8 16 Divided by Ar 0.6854 1.375 0.2293 2.743 Mole ratio 3 6 1 12 Formula: K3[Fe(C2O4)3] or [Fe(C2O4)3]3– 5a(ii) [Fe(H2O)6]3+ + 3 C2O4 2– [Fe(C2O4)3]3– + 6 H2O 5a(iii) Shape of complex ion: Octahedral 5(b) [Fe(H2O)6]3+ + 3 C2O4 2– [Fe(C2O4)3]3– + 6 H2O ---- (1) (from 5(a)(ii)) Colour change from emerald green to yellow: H2C2O4 formed when H 2SO4 is added. The decrease in [C 2O4 2– ] causes the position of equilibrium in equilibrium (1) to shift to the left. Hence, the presence of [Fe(H2O)6]3+ gives rise to a yellow solution. Colour change from yellow to pale green with the liberation of CO2 gas: Fe3+ + e Fe2+ Eo red = +0.77 V H2C2O4 2CO2 + 2H+ + 2e Eo ox = +0.49 V 2Fe3+ + H2C2O4  2Fe2+ + 2CO2 + 2H+ Eo cell=+1.26V (Feasible) The Fe 3+(aq) formed oxidises H 2C2O4 to form CO 2 while itself is reduced to Fe 2+(aq), giving rise to a pale green solution. Fe O C C O O O O C C O O O O CCO O O 3 131

NJC SH2 Prelim 2013 H2 Chemistry Paper 2 9 6 (a)(i) Anode: 2Cl –  Cl2 + 2e Cathode: 2H 2O + 2e  2OH– + H2 6(a)(ii) Cl2 + 2 NaOH → NaClO + NaCl + H2O ( cold or rm temp) OR 3Cl2 + 6 NaOH → NaClO3 + 5NaCl + 3H2O (high temp) Ionic equations accepted (iii) To ensure there is no back flow of solution from cathode department to anode so that hydroxide ions formed at cathode do not flow to anode and be oxidised instead of chloride ions. or