JPJC 2023 Prelim P2 Answers

© Jurong Pioneer Junior College [Turn Over Answers to 2023 JC2 Preliminary Examination H2 Chemistry (9729) Paper 2 1 (a) (i) Since F is more electronegative than Cl, C−F bond is more polar. [1m] (ii) Both are polar and thus have permanent dipole -permanent dipole (pd -pd) attractions and instantaneous dipole -induced dipole attractions between molecules. Since CH3Cl has more electrons per molecule than CH3F, more energy is required to overcome the stronger instantaneous dipole-induced dipole (id- id) attractions between molecules in CH3Cl than id-id attractions between molecules in CH3F. Thus, CH3Cl has higher boiling point. [1m] (iii) CH3Cl and CH3CH2F are isoelectronic so id-id attractions are similar. The additional electron-donating CH3 group attached to C of C−F bond makes C−F bond less polarised than C−Cl bond. Hence, less energy is required to overcome the weaker permanent dipole -permanent dipole attractions between molecules in CH3CH2F than pd-pd attractions between molecules in CH3Cl. Thus, CH3CH2F has lower boiling point. [1m] (b) Lone pairs in axial positions. Lone pairs in equatorial positions. Lone pairs in axial and equatorial positions. number of lone pairs at 90 relative to one another 0 0 1 number of bond pair and lone pair at 90 relative to one another 6 4 3 relative stability most stable [1m] least stable [1m] [1m]

2 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2023 (c) (i) [3m] (ii) E cell = (+2.01) – (+0.54) = +1.47 V [1m] G = −nFE cell = −(2)(96500)(+1.47) = −284 kJ mol−1 [1m] (iii) From the equation, S  0 as the reaction proceeds with no change in number of moles of aqueous particles and hence the reaction is enthalpy driven. [1m] −   −  ? 0 0 ve =TG H S Since G < 0, H must be negative and thus the reaction is exothermic. [1m] 2 (a) (i) Since  (q q ) (r r )LE +− +− + and (ionic) radius increases from Mg2+ to Ba2+ [1m], the magnitude of HLE of Group 2 sulfates decreases from MgSO 4 to BaSO4. [1m] (ii) Since (ionic)) radius increases and thus charge density decreases from Mg2+ to Ba2+ [1m], strength of ion-dipole interactions formed between M2+ with water molecules decreases. Thus Hhyd of M 2+ becomes less negative (i.e. less exothermic) from Mg2+ to Ba2+. (iii) Hsol = − HLE + Hhyd (M2+) + Hhyd (SO42−) Both Hhyd(M2+) and HLE become less exothermic/ less negative. HLE becomes less exothermic/ less negative by a smaller extent than that of Hhyd(M2+). [1m] Thus Hsol becomes more endothermic / less exothermic and h ence solubility decreases from MgSO4 to BaSO4 as shown in Table 2.1. [1m] (iv) Solubility of CaSO4 = − 34.7 10 1001000 = 4.7 × 10−2 mol dm−3 Ksp = [Ca2+][SO42−] = (4.7 × 10−2)2 = 2.209 × 10−3 mol2 dm−6 [1m] Ionic Product =         0.0100 100 0.0200 100 200 200 = 5.00 × 10−5 mol2 dm−6 Since ionic product < Ksp, no precipitation occurs. [1m] [I2(aq)] = [I−(aq)] = 1 mol dm−3 at 298 K salt bridge V e− [S2O82−(aq)] = [SO42−(aq)] = 1 mol dm−3 at 298 K Pt(s) (−) (+) Pt(s) > e−

3 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2023 (b) [2m] for correct diagram Hf  of SO42−(g) = − [(1473 + 180 + 502 + 966) – (2469)] = −652 kJ mol−1 [1m] 3 (a) (i) Reactant molecules must possess energy  Ea AND collide with correct orientation. [1m] (ii) rate: At higher total pressure, molecules are closer together so frequency of effective/successful collision increases so rate increases. [1m] yield: Position of equilibrium shifts to the righ t to favour the side with less number of moles of gaseous molecules so as to decrease some of the increased pressure. Thus, yield of SO3 increases. [1m] (b) (i) Kc = 2 3 2 22 [SO ] [SO ] [O ] AND units: mol−1 dm3 (ii) 2SO2(g) + O2(g)  2SO3(g) Initial amount / mol 2.00 1.00 0 Change −2x −x +2x Eqm amount / mol 2 − 2x 1 − x 2x Total amount of gases at eqm = (2 − 2x) + (1 − x) + 2x = 3 − x Mole fraction of SO3 = − 2x (3 x) = 0.82  x = 0.872 energy /kJ mol−1 0 BaSO4(s) Ba(s) + S(s) + 2O2(g) Ba(g) + S(s) + 2O2(g) +180 Ba2+(g) + 2e− + S(s) + 2O2(g) +(502 + 966) −1473 Ba2+(g) + SO42−(g) −2469 Hf  of SO42−(g)

4 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2023 n(SO2) at eqm = 2 − 2(0.872) = 0.256 mol n(O2) at eqm = 1 − 0.872 = 0.128 mol n(SO3) at eqm = 2(0.872) = 1.744 mol Kc = 2 2 () ( ) ( ) 1.744 40.0 0.256 0.128 40.0 40.0 = 1.45  104 mol−1 dm3 [1m] (c) (i) Heterogeneous catalysis since the solid catalyst and gaseous reactants are in different phases. [1m] (ii) Vanadium in V2O5 can vary its oxidation state because variable number of 3d and 4s electrons can be involved in bonding since 3d and 4s electrons have similar energies. [1m] OR Vanadium in V2O5 has low-lying partially filled 3d orbitals for adsorption of reactant molecules onto its surface. [1m] (d) 4 (a) Relative reducing power of halides: chloride < bromide < iodide HI/Iodide has the strongest reducing power as it is able to reduce H2SO4 to H 2S such that the oxidation number of S decreases from +6 in H2SO4 to ‒2 in H2S. [1m] HBr/Bromide has a weaker reducing power than HI/iodide as it is only able to reduce H2SO4 to SO2 such that the oxidation number of S decreases from +6 in H2SO4 to +4 in SO2. [1m] HCl/Chloride has the weakest reducing power as it is unable to reduce H2SO4. [1m] [1m] E [2m]

5 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2023 (b) (i) This is to keep [Fe3+] constant so that rate is only affected by the changes in [I–] (OR zero order wrt [Fe3+]) [1m] (ii) (ii) At t = 0, rate = 0.012 0.004 0 24 − − = 0.000333 mol dm−3 s−1 [1m] + tangent At t = 20 s, rate = 0.0101 0.004 0 49 − − = 0.000124 mol dm−3 s−1 [1m] + tangent 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0.012 0 10 20 30 40 50 60 [I−]/mol dm−3 time / s (0, 0.0101) (24, 0.004) (49, 0.004) (0, 0.012)

6 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2023 (iii) At t = 0, [I−] = 0.012 mol dm−3 At t = 20 s, [I−] = 0.0076 mol dm−3 [1m] Since Fe3+ is used in large excess, rate  [I−]a a0.000333 0.012= ( ) 0.000124 0.0076 2.69 = 1.58a  a = 2 (shown) [1m] (iv) Let rate = k [Fe3+]a [–]2 mol dm−3 s−1 = (mol–2 dm6 s–1) (mol dm−3)a+2  a = 1 OR At t = 0 s, 0.000333 = (15.4)(0.150)a(0.012)2  a = 1 OR At t = 20 s, 0.000124 = (15.4)(0.150)a(0.0076)2  a = 1 rate = k [Fe3+] [–]2 [1m] (v) 5.38 × 10−3 = k (0.040) (0.080)2  k = 21.0 [1m] k is a larger value as compared to 15.4 for the first experiment in (a).  the second experiment was carried out at a higher temperature. [1m] (c) Step 2 is the rate-determining step. [1m] Rate equation for step 2: rate = k2 [FeI2+] [I–] ---(1) From step 1, at equilibrium, rate of forward reaction = rate of reverse reaction kf [Fe2+] [I–] = kr [FeI2+] [FeI2+] = ( f r k k ) [Fe2+] [I–]---(2) Substitute (2) into (1), rate = k2 ( f r k k ) [Fe2+] [I–]2  overall rate equation is rate = k [Fe2+] [I–]2 (shown) 5 (a) (i) (ii) reaction 1: (CH3)2CHCl, anhydrous AlCl3, heat [1m] reaction 2: conc. HNO3, conc. H2SO4, T < 50 C [1m] (iii) [1m]

7 © Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2023 (iv) In acidic solution, is soluble in water and cannot be separated out . [1m] Excess NaOH deprotonates to give via acid-base reaction, which is sparingly soluble/ insoluble in water and will precipitate out, allowing it to be separated out. [1m] (v) (vi) (b) (i) C14H18N2O5 [1m] (ii) (iii) N in amide is less electronegative than O in ester so the carbonyl C (of amide) is less electron -deficient and thus less susceptible to nucleophilic attack. [1m] (iv) [1m