PJC H1 Chemistry 2008 JC2 H1 Chemistry Prelim Paper 2 Suggested Answers

1 Pioneer Junior College JC2 Preliminary Examination 2008 H1 Chemistry Paper 2 Suggested Answers Section A 1 (a) Number of Particle Electric charge Mass number Protons Electrons Neutrons X 0 32 16 16 16 Y 1- 81 35 36 46 Z 3+ 70 31 28 39 (b) Source beams of particle Y and particle Z - + Particle Y Particle Z 2 (a) Electrons repel each other and the electron pairs would arrange themselves as far apart as possible. Or lone pair-lone pair repulsion > bond pair - lone pair> bond pair - bond pair (b) . . O C O x . x . x x . . . . x .x . . . . . O ... . 2- . . O S O x . x . x x . . . . x .x . . . . . O . .. . 2- x x

2 For CO 3 2-, there are 3 bond pairs and 0 lone pair. By VSEPR, the shape is trigonal planar. For SO 3 2-, there are 3 bond pairs and 1 lone pair. By VESPR, the shape is trigonal pyramidal. 3 (a) Phosphorus burns with a blue flame. P4 + 5O2 → P4O10 Magnesium burns readily in oxygen wi th a bright white flame, leaving a white residue of MgO. 2Mg + O 2 → 2MgO (b) Calcium has one more quantum shell of electrons than magnesium. Due to more effective shielding by inner she ll electrons, the valence electrons are less attracted to the nucleus than m agnesium and thus it would lose the electrons more readily to form ions. Or As calcium has a larger atomic radius than magnesium, the valence electrons of calcium are further away from the nucleus. Thus, calcium will lose electrons more readily to form ions than magnesium. 4 (a) (i) Standard enthalpy change of formation of a compound is the heat change when one mole of compound is formed from its elements in their standard states under standard conditions of 298 K and 1 atm. (ii) Enthalpy / kJ mol-1 0 BF3(g) ΔHo at(B) B(s) + 3/2F2(g) B(g) + 3/2F2(g) B(g) + 3F(g) ΔHo f(B) 3/2 B.E. (F-F) 3 B.E. (B-F)

3 (iii) By Hess’ Law, 3B.E.(B-F) = -(-1137) + (573) + 3/2(+158) B.E. (B-F) = +649 kJ mol -1 4 (b) (i) Number of moles of propan-1-ol burnt = 0.60 / [3(12.0) + 8(1.0) + 16.0)] = 0.010 mol (ii) Let the final temperature be T°C Amount of heat liberated from combustion = 0.010 x 2021 = 20.21 kJ 1000 ) 0 . 21 T ( 18 . 4 200− × × = 20.21 T = 45.1°C (iii) Heat loss to surroundings or incomp lete combustion of propan-1-ol resulting in less heat produced. 5 (a) CH2CN OH CH2COOCH3 O A B (b) Step 1: KCN, ethanol, heat Step 2: HCl, heat Step 3:KMnO 4, H2SO4 (aq), heat (c) Step 1: substitution Step 2: acid hydrolysis Step 3: oxidation

4 6 (a) Molar mass of C15H15NO2 = (12.0 x 15) + (1.0 x 15) + 14.0 + (16.0 x 2) = 241 g mol-1 Amount of C 15H15NO2 in 2 capsules = (50 x 10 -3 x 2) / 241 = 0.000415 mol (b) 2 hours Amount of mefenamic acid (mole) Time (hours) Half-life ≈ 2 hours (c) (i) Solubility of mefenamic acid = (20 x 10-3) / 241 = 8.30 x 10-5 mol dm-3 (ii) Although mefenamic acid contains carboxylic acid group and amine group that are capable of form ing hydrogen bonds with water molecules, the presence of the tw o large hydrophobic aromatic rings interfere with the formation of hydrogen bonds results in its low solubility in water. (d) NaOH(aq) / Na2CO3(aq) at room temperature

5 (e) Overdose = 740 mg kg-1 Mass of mefenamic acid that resu lt in an overdose in a 65 kg patient = 740 x 65 = 48100 mg = 48.1 g Since the bioavailability of mefenamic acid is given to be 90%, maximum mass of mefenamic acid that can be ingested before an “overdose” occurs = 48.1 / 0.9 = 53.4 g Section B 1 (a) (i) [H+] = 10-2.49 = 3.24 x 10-3 mol dm-3 (ii) 3-chloropropanoic acid is a weak acid because [H +] << [3-chloropropanoic acid], indicating it undergoes partial dissociation. (iii) Ka = ] COOH CH C CH [ ] H ][ COO CH C CH [ 2 2 2 2 l l + − = ) 10 24 . 3 100 . 0 ( ) 10 24 . 3 ( 3 2 3 − − × − × = 1.08 x 10-4 mol dm-3 (b) (i) Concentration of NaOH = ⎟ ⎠ ⎞⎜ ⎝ ⎛÷⎟ ⎠ ⎞⎜ ⎝ ⎛ × 1000 50 . 22100 . 0 1000 0 . 25 = 0.111 mol dm-3 (ii) Phenolphthalein is a suitable indi cator because it works for a weak acid-strong base titration (i.e. t he working range of the indicator coincides with the region of sharp pH changes at the end point). There is a distinct colour change from colourless to pink at the end point. (c) (i) A buffer solution is one which resists changes in pH on dilution or on addition of small amount of acid or alkali. (ii) On addition of small amount of OH-, CH 2ClCH2COOH + OH- → CH2ClCH2COO- + H2O C H 2ClCH2COOH will react with the OH - added. With no significant increase in [OH-], pH is maintained at a near constant. On addition of small amount of H+, CH 2ClCH2COO- + H+ → CH2ClCH2COOH

6 CH 2ClCH2COO- will react with the H + added. With no significant increase in [H+], pH is maintained at a near constant. (d) CH2ClCH2COOH ⇌ CH2ClCH2COO- + H+ less stable CH 3CHClCOOH ⇌ CH3CHClCOO- + H+ more stable 2-chloropropanoic acid will be a stronger acid because the electron- withdrawing Cl is nearer to the carboxylate gr oup. It helps to disperse the negative charge on the carboxylate anion mo re, thus stabilising the anion to a greater extent. As a result, 2- chloropropanoic acid tends to undergo dissociation to a greater extent, making it a stronger acid. (e) (i) CH3CH2CH2Cl CH3CH=CH2 CH3CH(OH)CH2OH CH3COCHO 1 2 3 Step 1: KOH in ethanol, heat Step 2: KMnO 4 in NaOH(aq), room temperature Step 3: K 2Cr2O7 in dilute H2SO4, distil (ii) 6H5CH2CH2OH C6H5CH=CH2 C6H5CH(Br)CH3 C6H5CH(NH2)CH3 1 2 3 C Step 1: concentrated H2SO4, 170ºC Step 2: HBr(g) Step 3: excess NH3 in ethanol, heat in sealed tube 2 (a) A is sodium. Sodium chloride has a giant ionic lattice structur e with strong ionic bonds / electrostatic attractions between the sodium and chloride ions . A large amount of energy is required to overcome the ionic bonds, thus it has a high melting point. Sodium chloride dissolves in water, without undergoing hydrolysis, thus giving a solution of pH 7. NaC l(s) + aq → Na+(aq) + Cl-(aq)

7 2.0 x 10 -3 mol of chloride of B ≡ 5 . 35 108 435 . 1 + mol of AgCl 2.0 x 10 -3 mol of chloride of B ≡ 0.0100 mol of AgCl ≡ 0.0100 mol of Cl- 1 mol of chloride B ≡ 5 mol of Cl- B is phosphorus. PC l5 + 4H2O → H3PO4 + 5HCl Phosphorus pentachloride has a simple molecular structure with weak van der Waals’ forces of attraction (induced dipole – induced dipole attractions) between its molecules. As a relatively small amount of energy is required to overcome the intermolecular forces of attractions, it has a relatively low melting point of 180ºC. Phosphorus pentachlorid e hydrolyses in water , giving hydrochloric acid which reacts with silver nitrate to form a white precipitate of silver chloride. (b) (i) Kc = ] C ][ SO [ ] C SO [ 2 2 2 2 l l At 3 minutes, K c = (0.02) / (0.05 x 0.08) = 5.00 mol-1 dm3 At 9 minutes, K c = (0.04) / (0.08 x 0.06)