2019 JPJC Prelim H2 Chem P3 ANS

 Jurong Pioneer Junior College [Turn Over JURONG PIONEER JUNIOR COLLEGE 2019 JC2 H2 CHEMISTRY (9729) Preliminary Examination Paper 3 (Suggested Answers) 1 (a) For HI, violet fumes is observed. E(HCl) = +431 kJ mol1; E(HI) = +299 kJ mol1 Since E(HI) < E(HCl), H‒ I bond is weaker and is more easily broken than H‒C l bond. Hence, H I is less thermally stable and decomposes more readily ( i.e. HI decomposes upon contact with the hot wire). OR radius of Cl = 0.099 nm; radius of I = 0.133 nm Since I has a larger radius than C l, H‒I bond is longer and weaker so it is more easily broken than H‒C l bond. Hence, H I is less thermally stable and decomposes more readily (i.e. HI decomposes upon contact with the hot wire). (b) (i) MnO2(s) + 4H+(aq) + 2e  Mn2+(aq) + 2H2O(l) (ii) E cell = E red  E ox (+1.23) – (+1.36) = 0.13 V < 0 (not energetically feasible) Overall eqn: MnO2 + 4H+ + 2Cl  Mn2+ + 2H2O G = zFE cell where z = total number of electrons transferred per overall eqn = (2)(96500)(0.13) = +25090 J ol1 = +25.1 kJ mol1 Since G > 0, the reaction is not energetically feasible (or not spontaneous) and thus unlikely to occur. (iii) When concentrated HC l is used, [H +] is high and this causes the position of equilibrium of MnO2 + 4H+ + 2e  Mn2+ + 2H2O to shift right so as to react away some H + ions. This makes E(MnO2/Mn2+) becomes more positive than +1.23 V such that the Ecell becomes positive and  energetically feasible. When concentrated HC l is used, [C l] is high and this causes the position of equilibrium of Cl2 + 2e  2Cl to shift left so as to react away some C l ions. This makes E(Cl2/Cl) becomes more negative (or less positive) than +1.36 V such that the Ecell becomes positive and  energetically feasible. The continuous removal of Cl2 gas from the reaction mixture as it is evolved shifts the position of equilibrium of C l2 + 2e  2Clto the left to form back some C l2. This makes E(Cl2/Cl) becomes more negative (or less positive) than +1.36 V such that the Ecell becomes positive and  energetically feasible.

2  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2019 (c) (i) Type of mechanism: electrophilic addition (ii) Propene can use its two  electrons to form a dative bond with H+ (or to accept a H+). (iii) Nucleophilic substitution (iv) p–p orbital overlap results in the delocalisation of lone pair on O of CO into the  electron cloud of C=O, forming a partial double bond character in the C–O bond. This strengthens the C –O bond in –CO2H, making it difficult to break. Hence, carboxylic acid is unreactive towards HCl. (d) (e) (i) valalavalserglyargasnleuglyval or leuglyval valalavalserglyargasn or serglyargasnleuglyval alavalser (ii) The lone pair on N a is in the hybridised sp 2 orbital which is perpendicular to the p–orbitals of adjacent C=C and =C –N bonds. Hence, the lone pair on N a is not delocalised and is available for protonation. or The lone pair on N b is in the unhybridised p orbital which is parallel to the p – orbitals of adjacent C=N and C=C. Hence, the lone pair on N b is delocalised and less/not available for protonation.

3  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2019 [Turn Over (iii) When enzyme is used in the decarboxylation reaction, it provides an alternative reaction pathway with lower activation energy, Ea’. This increases the number of reacting particles with energy  Ea’. Hence, the frequency of effective collisions increases and the rate increases. (iv) 2 (a) For ideal gas, the volume of gas particles is negligible compared to the volume of gas/container. For ideal gas, there is no/negligible forces of attractions between the gas particles. For ideal gas, the collision between the gas particle is perfectly elastic. (b) (i) pK  3 2 2 O 3 O p p (ii) Using pV = nRT where V is in m3, c = n V mol m3 = RT p mol m3 = 1000RT p mol dm3 Kc = 2 2 3 3 [O ] [O ] = ( ) ( )  2 3 O 23 O 1000RT 1000RT p p = p 1000RTK = (6.1  1062)(1000)(8.31)(273+25) = 1.5  1055 Pa1 (shown) (iii) Since Kp << 1, it implies that the position of equilibrium 3O 2  2O3 lies on the far left and hence, the forward reaction hardly occurs and there are much less products than reactants at equilibrium. rate [histidine] 0 linear  1st order wrt [histidine] horizontal  zero order wrt [histidine] number of particles 0 energy/ kJ mol1 Ea Ea’ no. of particles with energy ≥ Ea’ for catalysed reaction no. of particles with energy ≥ Ea for uncatalysed reaction

4  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2019 (iv) 3O2(g)  2O3(g) initial conc. / mol dm3 10.0  4 = 2.50 0 change 1.5x +x eqm conc. / mol dm3 2.50 – 1.5x x     2 3 c 3 2 O O K  =   2 3 2.50 1.5 x x Since Kc << 1, we can assume that x << 2.50 such that 1.5  1055  2 32.50 x equilibrium [O2], x = 1.53  1027 mol dm3 (c) (i) 3SO2 + O3 + 3H2O  3H2SO4 (ii) More energy is required to remove H + from negatively charged ion HSO 4 than from electrically neutral H2SO4 due to greater electrostatic force of attraction. Hence, HSO4 is a weaker acid than H2SO4. (iii) SO42 + H+  HSO4 initial amt 0.200  10 1000 = 0.002 0.100  5 1000 = 0.0005 0 change 0.0005 0.0005 +0.0005 final amt / mol 0.0015 0 0.0005 System: acidic buffer (mixture contain weak acid HSO4 and its salt SO42) pH = pKa + lg     salt acid = [lg(0.012)] + lg 0.0015 0.0005   = 2.40 (d) p–p orbital overlap results in the delo calisation of lone pair of electrons on O of phenol into benzene ring, making the lone pair less available for donation. Hence, phenol is a weaker nucleophile than alcohol (or less reactive towards nucleophilic reaction). Reagent and condition: (1) NaOH(aq), room temperature (2), CH3COCl(l), room temperature (e) (i)

5  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2019 [Turn Over (ii) (iii) excess (CH3)3NH2, ethanol, heat in sealed tube (iv) If step 2 is carried ou t first, the phenolic group can undergo electrophilic substitution. (v) The N of salbutamol still possesses a lone pair and can function as a nucleophile for further substitution to occur. 3 (a) (i) 1s2 2s2 2p6 3s2 3p6 3d5 4s1 (ii) The maximum oxidation state is determined by the total number of 4s electrons and unpaired 3d electrons. Since the total number of 4s and unpaired 3d electrons increases from Sc to Mn and decreases from Mn to Zn, the maximum oxidation state increases from Sc to Mn then decreases from Mn to Zn. (iii) Sc3+: 1s2 2s2 2p6 3s2 3p6 Fe3+: 1s2 2s2 2p6 3s2 3p6 3d5 Iron(III) salt are usually coloured due to dd transition. The presence of ligands causes the d orbitals to split into 2 different energy levels with relatively small energy difference, E. Radiation from the visible light region of the electromagnetic spectrum, corresponding to E, is absorbed when an electron transits from a d orbital of lower energy to partially filled d orbital of higher energy. Hence, iron(III) salts are coloured and the colour observed is the complement of the colours absorbed. On the other hand, Sc3+ has no 3d electron and hence electron transition between d orbitals is not possible and radiation from visible light region is not absorbed.

6  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2019 (iv) (b) (i) Cu2+ + 4Cl + Cu  2[CuCl2] ---(1) Cu2+ undergoes redox reaction with Cu foil to form a stable colourless [CuC l2] complex in the presence of Cl due to ligand exc