2019 JPJC Prelim H2 Chem P2 ANS

2019 JPJC JC2 H2 Chemistry (9729) 1 2019 Prelim Paper 2 Answers JURONG PIONEER JUNIOR COLLEGE 2019 JC2 H2 CHEMISTRY (9729) Preliminary Examination Paper 2 (Suggested Answers) 1 (a) (i) Particles: Cu2+ cation Interactions: metallic bonding (ii) Particles: I2 molecule Interactions: instantaneous dipoleinduced dipole interactions (b) The layers of Cu2+ cations can slide past each other without breaking the metallic bond (or with the mobile valence electrons holding them together). (c) Electrical conductivity: Mobile valence electrons available to conduct electricity in copper but no mobile charge carriers available in iodine to conduct electricity. or Solubility in organic solvent: Iodine is able to form favourable instantaneous dipoleinduced dipole interactions with the organic solvent, but copper is unable to form favourable interactions with the organic solvent. (d) (i) Atomic mass is the weighted average of the mass of isotopes. (ii) Percentage abundance of the isotopes in this sample differs from what is normally obtained. or There are more than 2 types of isotopes of copper present in the sample. (e) Since angle of deflection  charge mass , angle of deflection of I− = 1127 263    (7.0 ) = 1.7  (1dp) (f) X Cl Br I E(X2/X) / V +1.36 +1.07 +0.54 Since E(X2/X–) becomes less positive down the group , it implies that the tendency of X2 to be reduced to X  decreases and hence, the oxidising p ower of X 2 decreases down the group. 127I 63Cu2+

2019 JPJC JC2 H2 Chemistry (9729) 2 2019 Prelim Paper 2 Answers (g) (i) [O]: 2IO3 + 12H+ + 10e  I2 + 6H2O x V [R]: H2O2 + 2H+ + 2e  2H2O +1.77 V E cell = E red  E ox +0.57 = (+1.77)  E(IO3/ I2) E(IO3/ I2) = +1.20 V (ii) Note the E cell value on the voltmeter at first instance of cell being connected. 2 (a) soln hyd ve ve ve LEHH         Down the group, both LE of MCO 3 and Hhyd of M2+ becomes less exothermic (or more endothermic or magnitude/value of LE of MCO 3 and Hhyd of M2+ decreases) since the radius of M2+ increases down the group. However, the decrease in LE is less than the decrease in Hhyd since M 2+ is smaller than CO32. Hence, Hsoln becomes less exothermic and the solubility of MCO3 decreases down the group. (b) (i) Ksp of SrCO3 = [Sr2+][CO32] = (1.05  105)2 = 1.10  1010 mol2 dm6 (ii) For precipitation to occur, ionic product  Ksp. ionic product of SrCO3 = [Sr2+]min[CO32]  Ksp. [Sr2+]min(0.02) = 1.1 × 1010 minimum [Sr2+] to precipitate SrCO3 = 5.50 × 109 mol dm3 ionic of SrF2 = [Sr2+]min[F]2  Ksp. [Sr2+] min(0.1)2 = 2.5 × 109 minimum [Sr2+] to precipitate SrF2 = 2.50 × 107 mol dm3 SrCO3 will precipitate first because a lower [Sr 2+] is requir ed to form the precipitate. 1 mol dm3 H+(aq), I2(aq), IO3(aq), 298 K salt bridge V e e 1 mol dm3 H2O2(aq), H+(aq), 298 K Pt(s) anode (negative) cathode (positive) Pt(s)

2019 JPJC JC2 H2 Chemistry (9729) 3 2019 Prelim Paper 2 Answers (c) The student’s response is incorrect as SrCO 3 has a higher decomposition temperature than CaCO3. Thermal stability of Group 2 carbonate depends on the charge density of M 2+, not lattice energy. Since Sr2+ has a larger radius, Sr 2+ has a lower charge density than Ca 2+. Hence, Sr 2+ polarises large CO32 less and weaken the C –O bond to a smaller extent than that in BaCO3. Hence, SrCO 3 has a higher thermal stability and has a higher decompos ition temperature. (d) (i) Electrode A: Pb(s)  Pb2+(aq) + 2e Electrode B: Pb2+(aq) + 2e  Pb(s) Electrode C: 4OH(l)  O2(g) + 2H2O(g/l) + 4e (ii) Xn+ + ne  X n(X) deposited = 1  119 = 0.008403 mol ne passed = 3240  96500 = 0.03358 mol mole ratio X : e 0.008403 : 0.03358 1 : 4 Hence, the value of n is 4. (e) 3 (a) (i) Compound B Compound C (ii) step 2 : acidified K2Cr2O7(aq), heat under reflux step 4 : Al2O3, heat (or excess conc. H2SO4(l), heat) By Hess’ Law, 148 + 736 + 1450 + 2∆Hf(OH–(g)) + (–2993) = –925 ∆Hf(OH–(g)) = –133 kJ mol–1

2019 JPJC JC2 H2 Chemistry (9729) 4 2019 Prelim Paper 2 Answers step 5 : LiAlH4, dry ether (iii) Type of mechanism: nucleophilic addition HCN  H+ + CN (or NaCN  Na+ + CN) (b) (i) average t½ = ½ (48 + 48) = 48 min Since the half–lives are constant at about 48 min, the order of reaction with respect to [C6H5CH(CH3)Cl] is 1. 48 96 t½ = 48 min t½ = 48 min 0.025 0.0125 [C6H5CH(CH3)Cl] / mol dm–3 time / min Experiment 1 [CH3O] = 0.5 mol dm3 Experiment 2 [CH3O] = 1.0 mol dm3

2019 JPJC JC2 H2 Chemistry (9729) 5 2019 Prelim Paper 2 Answers (ii) From the graph of [OH–] = 1.0 mol dm–3 Initial rate = 0.02 0.05 40 0   = 7.50  10–4 mol dm–3 min–1 From the graph of [OH–] = 1.0 mol dm–3 Initial rate = 0.02 0.05 80 0   = 3.75 x 10–4 mol dm–3 min–1 When [CH3O] doubles, the rate is doubled. Hence, the order of reaction with respect to [CH3O–] is 1. (iii) rate = k [C6H5CH(CH3)Cl] [CH3O] units of k =   31 23 mol dm min mol dm   = mol–1 dm3 min–1 (iv) Experiment 3: t½ of C6H5CH(CH3)Cl =   ln2 2.0k = ½  48 = 24 min (c) [C6H5CH(CH3)Cl] / mol dm–3 time / min Experiment 1 [CH3O] = 0.5 mol dm3 Experiment 2 [CH3O] = 1.0 mol dm3 (40, 0.02) (80, 0.02)

2019 JPJC JC2 H2 Chemistry (9729) 6 2019 Prelim Paper 2 Answers 4 (a) (i) Volume of pure methanoic acid in one ant = (7.5  103)  50 100  100 80 = 4.688  103 cm3 No. of ants required to produce 1 cm3 of pure methanoic acid = 1 1000  (4.688  103) = 214 (ii) volume of HCOOH injected by one ant = (7.5  103)  50 100 = 3.75  103 dm3 mass of HCOOH injected by one ant = (3.75  103)  1.2 = 4.50 x 10-3 g no. of ants needed = (1.8  0.2)  (4.50 x 10-3) = 80 (iii) 2HCOOH + Na2CO3  2HCOONa + H2O + CO2 (iv) n(HCOOH) injected by one bee = (5.4  103)  30.0 = 1.80  104 mol Since 1Na2CO3  2HCOOH, n(Na2CO3) required = ½  (1.80  104) = 9.00  105 mol Mass of Na2CO3 needed to neutralise one bee sting = (9.00  105)  106.0 = 0.00954 g (b) A B C (c) (i) Let compound J be CxHyOz. CxHyOz + (x + y 4 – z 2 )O2  xCO2 + ( y 2 )H2O n(H2O) collected = 0.038  18.0 = 0.002111 mol After cooling: volume of gases = V(unreacted O2) + V(CO2) =140 cm3 After reaction with NaOH: V(CO2) = 90 cm3 V(unreacted O2) = 50 cm3 V(reacted O2) = 150 – 50 = 100 cm3 n(CO2) evolved = 90 1000  24.0 = 0.00380 mol n(reacted O2) = 100 1000  24.0 = 0.00417 mol

2019 JPJC JC2 H2 Chemistry (9729) 7 2019 Prelim Paper 2 Answers CxHyOz + (x + y 4 – z 2 )O2  xCO2 + ( y 2 )H2O 0.00417 0.00380 0.00211 2 10 1.8 9 1 5 x = 9, y = 10 and z = Hence, the molecular formula of J is C9H10O3. (ii) reaction 1 phenol reaction 2 carboxylic acid (iii) C7H6O3 (iv) CH(CH3)OH is present. (v) (vi) To form K ( i.e. benzoic acid and phenol), the acid used must be a stronger acid than benzoic acid and phenol (e.g. H2SO4 which is a strong acid). Since the Ka of CF 3CO2H is higher than that of benzoic acid and phenol, CF3CO2H is a stronger acid and hence, it can be used as a replacement for sulfuric acid. (vii)  Ka of ethanoic acid is higher than that of phenol, p–p orbital overlap results in the delocalisation of lone pair on O  into C=O of CH3COO, dispersing the negative charge over two electronegative O atoms and stabilises CH3COO more than phenoxide ion. Hence, ethanoic acid is a stronger acid and has a higher Ka than phenol.  Ka of trifluoroethanoic acid is higher than that of ethanoic acid. The three electron –withdrawing (or electronegative) F