SAJC H1 CHEM P2 ANS

SAJC 2011 H1 Chemistry Prelims Paper 2 Solutions Section A (40 marks) Answer all the questions in this section in the spaces provided. 1 (a) (i) BeC l2 is a covalent compound while BeF 2 is an ionic compound. The electron cloud of C l- anion is larger than that of F - anion. Hence, C l- anion is more readily polarised by the Be 2+ cation which has a high charge density, giving rise to its covalent character. (ii) (iii) Let x be the mole fraction of BeC l2 and mole fraction of Be 2Cl4 = 1 – x. x(80) + (1-x)(160) = 126 80x = 34 x = 0.425, 1 – x = 0.575 (b) (i) (ii) % Mass of Be in Beryl = 100 x 3) x 16 (28.1 62 x 27 9 x 3 9 x 3 +++ = 5.022 % 5.022% /barb2right 30.6 kg 100% /barb2right 100 x 5.022 30.6 = 609.3 kg

2 % Mass of Beryl in ore = 11.4% 100 x 5330 609.3 = 1 (c) (i) ∆∆ ∆∆ H1: Enthalpy change of formation of Be 3N2 ∆∆ ∆∆ H3: Bond dissociation energy of N 2 ∆∆ ∆∆ H4: Sum of 1 st and 2 nd IE of Be ∆∆ ∆∆ H6: Lattice Energy (ii) Lattice energy = -588 – (324 + 994 + 3(900 + 1760) + 1404) Lattice energy = - 11290 kJ mol -1 (iii) /uni2502/uni2206HLE /uni2502α/uni2502 q+q -/r+ + r-/uni2502 Product of q+ and q- of Be 3N2 > Product of q+ and q- of Li 2O Since product of charges outweighs the sum of ionic radii, the lattice energy of Li 2O is less exothermic than Be 3N2. 2 (a) First ionisation energy is the energy required to r emove 1 mole of electrons from 1 mole of gaseous atoms to form a mole of singly charged gaseous ions. (b) (i) P: Group VI Q: Group II (ii) Physical properties: Electrical conductor at aqueous/ molten state. High melting and boiling point. Soluble in water/insoluble in organic solvents. 3 (a) SiO 2 is insoluble in water due to the high bond dissociation energy needed to break the strong Si—O covalent bonds. Na 2O(s) + H 2O( l) → 2NaOH(aq) Na 2O is readily soluble in water to form a strongly alkaline solution. SiO 2(s) + 2NaOH(aq) → Na 2SiO 3(aq) + H 2O( l) SiO 2 is an acidic oxide which reacts readily with alkalis to form salt and water.

3 (b) (i) No. of moles of S 2O3 2- = 30/1000 x 0.100 = 0.00300 mol No. of moles of Br 2 = 4 x 0.00300 = 0.01200 mol S2O3 2- ≡ 4Br 2 ≡ 6 SiF 4 No. of moles of SiF 4 = 3 x 0.006 = 0.018 mol Mass of SiF 4 = 0.018 x (28.1 +19.0 x 4) = 1.87 g (ii) Disproportionation 3Br 2 + 6OH - /barb2right 5Br - + BrO 3 - + 3H2O 4 (a) Kc = )2 18 . 0)( 2 24 . 0( )2 80 . 0()2 40 . 2( ]][ [ ][ ][ 3 24 3 2 =OHCH CO H = 64 mol 2 dm -6 (b) When temperature increases (from 550 oC to 850 oC), Percentage of H 2 increases, thus forward reaction is favoured, equi librium has shifted right. By L.C.P, endothermic reaction is fa voured as it helps to absorb the excess heat. Hence, production of H 2 is endothermic. When volume changes from Z to X, Percentage of H 2 increases thus forward reaction is favoured. When equilibrium shifts to the right, the number of moles of gas inc reases. By L.C.P. pressure has decreased / volume has increased. Hence, increasing order: Z < Y < X 5 (a) Add Tollen’s reagent and warm. Silver mirror is observed for cinnamaldehyde, but no silver mirror for ethyl cinnamate. OR

4 Fehlings solution, warm/ 2,4-dinitrophenylhydrazine, warm (b) CH 3CH 2OH, concentrated H 2SO 4, heat (c) Ethyl cinnamate is more volatile. Both compounds are simple molecular. Less energy is required to overcome the weaker van der Waals’ forces/pd-pd interactions between ethyl cinnamate molecules than the stronger hydrogen bonding between cinnamic acid molecules.

5 Section B (40 marks) Answer two out of three questions in this section on the writing papers provided. 6 (a) (i) According to VSEPR theory, lone pair-bond pair repu lsion is greater than bond pair-bond pair repulsion. Thus NO 2 - is non-linear but NO 2 + is linear. (ii) Hydrazine is able to form intermolecular hydrogen bonds with water. CO 2 is a non polar molecular that can only act as hydr ogen bond acceptors to water. Hence the interaction of CO 2 with water involves less extensive hydrogen bonding. (iii) Aluminium oxide is a giant ionic lattice structure with strong electrostatic forces of attraction between its oppositely charged ions whilst aluminium iodide is simple covalent molecule with weak induce d dipole- induced dipole intermolecular forces of attraction. Induced dipole- induced dipole is weaker than ionic bonding hence more energy is needed to overcome the ionic bonds.

6 (b) (i) (ii) (c) (i) or [H +][ A-]/[H A] (ii) HA is a stronger acid than 2-methylpropanoic acid. Cl, an electron-withdrawing group in HA will help to disperse the negative charge on the oxygen, stabilising the carboxylate anion and hence causing it to be a stronger acid. (iii) [H +] = 0.001 mol dm -3 Since [H +] << [H A] Thus, H A is a weak acid. (iv) A buffer solution is one whose pH remains almost un changed when a small amount of acid and base is added.

7 (I) When a small amount of acid is added: H + + A- /barb2right HA (II) When a small amount of base is added: HA + OH - /barb2right A- + H 2O

8 7 (a) (i) 1s 22s 22p 63s 23p 6 (ii) CH 3CH 2OH + H 2O /barb2right CH 3COOH + 4H + + 4e - 3CH 3CH 2OH + 2Cr 2O7 2- + 16H + → 3CH 3COOH + 4Cr 3+ + 11H 2O (iii) moles of potassium dichromate = 0.015 x 0.017 = 2.55 x 10 -4 mol moles of ethanol needed = 2.55 x 10 -4 x 3/2 = 3.825 x 10 -4 mol minimum vol of ethanol = (3.825 x 10 -4) x 24000 = 9.18 cm 3 (iv) Angle of deflection is proportional to charge / mas s ratio of the ion. Chromate(VI ) and dichromate both have the same charge. However , mass of dichromate(VI ) is higher than chromate( VI ), hence chromate( VI ) will be deflected to a larger extent when passed through an electric field. (b) i. Excess concentrated sulfuric acid, 170° C ii. P I3 , heat or H 3PO 4 with Na I iii. KMnO 4 / H 2SO 4, heat OR K 2Cr 2O7 / H 2SO 4, heat (iv) 2-propanol upon oxidation (reaction (iii)) gives a ketone that will give an orange precipitate with 2,4-dinitrophenylhydrazine. 1-propanol upon oxidation gives a carboxylic acid that will not give an orange precipitate with 2,4-dinitrophenylhydrazine. (c) /uni2206Ho oxidation = + (-1371) – (-876) = -495 kJ mol -1 (d) Information Deduction Geraniol oxidised to give R + S and effervescence. Geraniol has alkene group. Effervescence of CO 2 due to terminal alkene or ethanedioic acid.

9 R and S give an orange ppt with 2,4-DNPH R and S are ketones as they are products of oxidation. R and S form a yellow ppt in alkaline aq. iodine. R and S have RCOCH 3. S forms effervescence in sodium hydrogen carbonate. S has RCOOH. Or (CH 3)2C=CHCH=C(CH 3)CH 2CH 2OH R S

10 8 (a) (i) Using experiments 2 and 3, [FA4] is kept constant, when [FA3 ] increases by 1.2 times, rate increases by 1.2 times. Hence, order of reaction with respect to [ FA3 ] is 1. Using experiments 1 and 2, 6 7 1n 1n 1.53x10 8.17x10 (15) k(25) (20) k(10) − − = 1n 0.4005 )25 10 ( n = = Hence, order of reaction with respect to [ FA4 ] is 1. (ii) Rate = k[2-bromo-3-methylpentane][sodium hydroxide] (iii) Catalyst would lower the activation energy by provi ding an alternat