2024 Prelims RI H2 Chem P3 (Ans)

© Raffles Institution 2024 9729/03/S/24 2024 Y6 H2 Chemistry Preliminary Exams Paper 3 – Suggested Solutions Section A 1(a)(i) Kc = [SO3]2 [SO2]2[O2] mol−1 dm3 1a(ii) Kc = [SO3]2 [SO2]2[O2] = (4.60 1.00)2 (0.500 1.00 )2(0.100 1.00 ) = 846.4 mol−1 dm3 = 846 mol−1 dm3 (3 s.f.) 1(a)(iii) pV = nRT p = (0.5+0.1+4.6)(8.31)(273+450) 1 1000 = 31.24 x 106 Pa p = 31.2 MPa 1(a)(iv) Partial pressure of SO3 = 4.60 0.5+0.1+4.6 × (31.24 ×106) = 27.6 MPa 1(a)(v) Let y be the amount of O2 added into the system 2 SO2 + O2 ⇌ 2 SO3 Initial amt / mol 0.500 0.1 + y 4.6 Change in amt / mol −0.1 −0.05 +0.1 Equilibrium / mol 0.400 0.05 + y 4.7 Kc is a constant as temperature remained constant At equilibrium, (4.70 1.00)2 (0.400 1.00 )2(0.05+y 1.00 ) = 846.4 mol−1 dm3 4.72 = 846.4(0.4)2(0.05 + y) 0.05 + y = 0.1631 y = 0.113 mol 1(b)(i) When temperature increases, equilibrium position shifts left to favour the backward endothermic reaction to counteract the increase in temperature. Hence, the amount of SO2 will increase.

© Raffles Institution 2024 9729/03/S/24 1(b)(ii) 1(c)(i) 800 K 1(c)(ii) As temperature increases, ∆Gr of reaction 1 becomes more positive. Thus, position of equilibrium of reaction 1 lies more to the left, causing the positions of equilibrium for both reactions 2 and 3 to shift to the left, which result in a lower proportion of H2SO4. 1(d)(i) - An ideal gas consists of particles of negligible volume . The size of the gas particles is negligible compared to the volume of the container. - The gas particles exert negligible attractive forces on one another. - Collisions between gas particles are perfectly elastic. 1(d)(ii) At moderately high pressure, the gas particles come closer together and intermolecular attractive forces between the gas particles become significant. OR At very high pressure, the gas particles are much closer together and the gas occupies a smaller volume. As such, the volume of the gas particles is not negligible as compared to the volume of the container. 1(e)(i) Above Tc, kinetic energy of the gas particles is so high such that the intermolecular forces of attraction are overcome at all pressures. 1(e)(ii) The intermolecular forces of attraction present in steam are stronger hydrogen bonds compared to the weaker instantaneous dipole-induced dipole interactions (id-id) between carbon dioxide molecules.

© Raffles Institution 2024 9729/03/S/24 More energy is required to overcome the stronger hydrogen bonds as compared to weaker id-id, hence steam has a higher critical temperature. 2(a)(i) Standard enthalpy change of formation is the energy change when 1 mole of the pure substance in a specified state is formed from its constituent elements in their standard states under standard conditions of 1 bar and 298 K. 2(a)(ii) By Hess’ Law, (+107.5) + (+494) + (−896.3)+ (−656.0) = ∆Hf(NaHCO3(s)) ∆Hf(NaHCO3(s)) = −951 kJ mol−1 (3 s.f.) 2(a)(iii) amount of sodium hydrogencarbonate = 11.2 84.0 = 0.1333 mol q = −(+18.7 x 0.13333) = −2.493 kJ ∆T = −2.493 × 1000 (100)(4.18) = −5.96 C or K 2(a)(iv) ∆G = ∆H − T∆S Since ∆Gsolution < 0, and ∆Hsolution > 0, −T∆Ssolution must be less than 0. Hence, the sign for ∆Ssolution is also positive. 2(a)(v) The hydroxide ion is smaller than hydrogencarbonate ion, while having the same charge. Therefore, the hydroxide ion has a higher charge density. Hence, the standard enthalpy change of hydration of hydroxide is more exothermic. 2(b)(i) Ksp = [Cu2+][CO32−] units = mol2 dm−6 energy / kJ mol−1 0 Na(g) + 1 2 H2(g) + C(s) + 3 2 O2(g) +107.5 Na(s) + 1 2 H2(g) + C(s) + 3 2 O2(g) NaHCO3(s) ∆Hf(NaHCO3(s)) +494 Na+(g) + 1 2 H2(g) + C(s) + 3 2 O2(g) + e− Na+(g) + HCO3−(g) −656.0 −896.3

© Raffles Institution 2024 9729/03/S/24 2(b)(ii) [CO32−] to precipitate PbCO3 = 7.40 x 10-14 1 2 (0.1) = 1.48 x 10-12 mol dm−3 [CO32−] to precipitate CuCO3 = 1.40 x 10-10 1 2 (0.2) = 1.40 x 10-9 mol dm−3 Since [CO 32−] needed to precipitate PbCO 3 < CuCO 3, PbCO 3 will be precipitated first. 2(b)(iii) NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq) [Cu(H2O)6]2+(aq) + 2OH−(aq) ⇌ [Cu(OH)2(H2O)4](s) + 2H2O(l) – (1) Upon addition of aqueous ammonia, a weak base, [OH −] increases and shift the position of equilibrium of (1) to the right, resulting in the formation of a light blue precipitate of Cu(OH)2 since ionic product of Cu(OH)2 > Ksp(Cu(OH)2). [Cu(H2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4]2+(aq) + 6H2O(l) – (2) Upon addition of excess aqueous ammonia, [NH3] increases and shifts the position of equilibrium of (2) shifts to the right and decreases the concentration of [Cu(H2O)6]2+. Hence, the position of equilibrium of (1) shifts to the left, causing the blue precipitate of [Cu(OH) 2(H2O)4](s) to dissolve since ionic product Cu(OH)2 < Ksp(Cu(OH)2), giving a dark blue solution. 2(c)(i) When mole fraction of CO32− = 0.05, pH is about 9 – 9.25. [H+] = 1.00  10−9 mol dm−3

© Raffles Institution 2024 9729/03/S/24 2(c)(ii) 2(c)(iii) As carbon dioxide emission increases, concentration of H 2CO3 in seawater increases. pH of seawater decreases. From Fig. 2.1, the mole fraction of carbonate ions decreases with decreasing pH, Hence, it is more difficult for the marine creatures to build their shells. 3(a) Since Cl is more electronegative than I, the Cl atom is more electron-withdrawing than the I atom and the negative charge is dispersed to a greater extent in CH3CHClCOO− ion than in CH3CHICOO− ion. Hence CH3CHClCOO− ion is more stable. The greater stability of the conjugate base anion explains the higher acidity of CH3CHClCOOH. 3(b)(i) 3(b)(ii) Step 1: H2SO4(aq), K2Cr2O7(aq) and heat Step 2: HCN with trace KCN Step 3: H2SO4(aq), heat 3(c) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 pH mole fraction X H2CO3 HCO3− CO32−

© Raffles Institution 2024 9729/03/S/24 trans isomer acceptable 3(d) The C≡C group in alkynes do not have a partial positive charge to attract nucleophiles. OR The alkyne is electron rich and repels electron rich nucleophiles. 3(e)(i) 3(e)(ii) Due to ring strain/ bond strain , making cyclooctyne less stable and lowering the activation energy required for reaction. Note: The alkyne −C≡C− carbon atoms are sp hybridized and the ideal VESPR bond angle to minimise repulsion is 180 . In the ring structure of cyclooctyne, the bond angles from the −C≡C− are less than 180. 3(f)(i) The azide group has a net positive charge OR lost electron density coordinating to Cu Thus, it is more electron deficient and a better electrophile. The alkyne group is next to the Cu and sideways overlap between the Cu orbital with the lone-pair electrons is possible with the alkyne  electron cloud. Hence the Cu lone pair electrons can be delocalised with the alkyne group, making it more electron rich and susceptible to electrophilic attack. 3(f)(ii) and 3(f)(iii)

© Raffles Institution 2024 9729/03/S/24 Section B 4(a) Aqueous Cu 2+ exists as [Cu(H 2O)6]2+ complex and Cu 2+ has an electronic configuration of [Ar]3d9. The presence of H 2O ligands split the 3d orbitals into two sets of slightly different energy levels. Since the 3d subshell in Cu 2+ is partially filled , electrons in the lower energy d orbitals can absorb energy corresponding to certain wavelengths from the visible spectrum and get promoted to the higher -energy d orbitals (d-d transition). The colour observed is the complement of the colour absorbed. Even though aqueous Zn2+ exists as [Zn(H 2O)6]2+ complex, Zn2+ has an electronic configuration of [Ar]3d 10. Hence,