2022 RVHS JC2 H2 CM Prelim P2 (Solutions)

River Valley High School 9729/02/PRELIM/22 [Turn over 2022 Preliminary Examination Suggestion Solutions for 2022 H2 Chemistry Prelim Paper 2 1 (a) The enthalpy change of solution of a substance is the enthalpy change when one mole of the substance is completely dissolved to give an infinitely dilute solution, so that no further enthalpy change takes place on adding more solvent. [1] (b) (i) Assuming no heat loss to surroundings, amount of NH4NO3  26000 = 50  4.18  5 amount of NH4NO3 = 0.04019 mol minimum mass of NH4NO3 = 0.04019  (2(14.0) + 4(1.0) + 3(16.0)) = 3.22 g [1] (ii) [2] (c) By Hess’ Law, Hsoln(NH4Cl) = – (–705) + [(–307) + (–381)] = +17.0 kJ mol–1 [3] (d) Since Cl− has a smaller ionic radius and higher charge density than Br−, it forms stronger ion-dipole interactions with water molecules. Thus, Cl− has a larger magnitude of Hhyd than Br−. [2] [Total: 9]

2 River Valley High School 9729/02/PRELIM/22 2022 Preliminary Examination 2 (a) It is a d-block element that is able to form one or more stable ions with a partially filled d subshell. [1] (b) Both elements have giant metallic lattice structure and exhibit metallic bonding. In Co, both the 4s and 3d electrons can be contributed to form the sea of delocalised electrons as they are very close in energy. The resulting cobalt ion has a higher positive charge and a smaller ionic radius/ higher charge density . This results in stronger electrostatic forces of attraction between the metal cations and the sea of delocalised electrons in Co as compared to Ca, which only contributes 2 valence electrons per Ca atom to form Ca2+. [2] (c) (i) 1s22s22p63s23p63d7 [1] (ii) O2 + 2H2O + 4e– ⇌ 4OH– E = +0.40 V Overall equation: 4[Co(NH3)6]2+ + O2 + 2H2O → 4[Co(NH3)6]3+ + 4OH– Ecell = (+0.40) – (+0.11) = +0.29 V Since E cell > 0, the reaction is feasible. Hence, yellow -brown [Co(NH3)6]2+ is oxidised by oxygen in air to produce red-brown [Co(NH3)6]3+ [2] (iii) (iv) [3] [1] (d) (i) Amount of D reacted with EDTA = 18.75 1000 × 0.0400 = 7.500 × 10–4 mol Amount of D in 3.501 g sample = 500 25.0 × 7.500 × 10−4 = 0.01500 mol Molar mass of D = 3.501 0.01500 = 233.4 g mol–1 [2]

3 River Valley High School 9729/02/PRELIM/22 [Turn over 2022 Preliminary Examination (ii) Amount of D reacted with AgNO3 = 23.34 233.4 = 0.1000 mol Amount of AgCl2 formed = 14.340 (107)+(35.5) = 0.1000 mol Mole ratio of D: AgCl is 1 : 1 [1] (iii) 17a + 35.5(3) = 233.4 – 58.9 Solving: a = 4 Cation: [Co(NH3)4Cl2]+ [2] (iv) [2] [Total: 17] 3 (a) (i) Enantiomerism [2] (ii) An electrophile is an electron pair acceptor and is electron deficient. [1]

4 River Valley High School 9729/02/PRELIM/22 2022 Preliminary Examination (iii) [2] (iv) [1] (b) [2] (c) All three substances have simple molecular/ covalent structures . More energy is needed to overcome the stronger hydrogen bonds between CH3OH molecules than the weaker instantaneous dipole-induced dipole (id-id) interactions between CH 3SH or CH 3SeH molecules. Thus CH 3OH has the highest boiling point. As CH3SeH has a larger number of electrons than CH3SH, more energy is needed to over the stronger id-id interactions between CH3SeH molecules than the weaker id -id interactions between CH 3SH molecules. Thus CH3SeH has a higher boiling point than CH3SH. [3] (d) (i) S is positive as there is an increase in disorder as the amount of gas molecules increases from 0 mol to 3 mol. [1] (ii) ΔG⦵ = ΔH⦵ – TΔS⦵ = +129 – (130 + 273)(0.332) energy Reaction coordinate CH2CH2 + Br2 CH2BrCH2Br CH2BrCH2+ + Br– Ea(2) Ea(1) H Ea(1) : activation energy for first step Ea(2) : activation energy for second step

5 River Valley High School 9729/02/PRELIM/22 [Turn over 2022 Preliminary Examination = – 4.80 kJ mol−1 Since G < 0, reaction is spontaneous at 130oC. [2] [Total: 14] 4 (a) (i) 𝑝𝑉 = 𝑛𝑅𝑇 𝑝𝑉 = 𝑚 𝑀𝑟 𝑅𝑇 𝑀𝑟 = 𝑚𝑅𝑇 𝑝𝑉 𝑀𝑟 = (1.50)(8.31)(327 + 273) (1.60 × 105)(250 × 10−6) = 187.0 = 187 (𝑡𝑜 3 𝑠. 𝑓. ) [1] (ii) Let the mole fraction of AlCl3 be x. 133.5x + (1 − x)(267) = 187 133.5x + 267 − 267x = 187 133.5x = 80 x = 0.59925 x = 0.6 (to 1 d.p) Mole fraction of AlCl3 = 0.6 Mole fraction of Al2Cl6 = 1 – 0.59925 = 0.40075 = 0.4 (to 1 d.p) [1] (iii) 𝑃𝐴𝑙𝐶𝑙3 = (0.6)(1.60 × 105) = 96000 𝑃𝑎 𝑃𝐴𝑙2𝐶𝑙6 = (0.4)(1.60 × 105) = 64000 𝑃𝑎 𝐾𝑝 = (𝑃𝐴𝑙𝐶𝑙3) 2 𝑃𝐴𝑙2𝐶𝑙6 𝐾𝑝 = (96000)2 64000 = 144000 𝑃𝑎 [3]

6 River Valley High School 9729/02/PRELIM/22 2022 Preliminary Examination (b) Since the reaction is endothermic, increasing the temperature will favour the forward reaction/ shift the position of equilibrium to the right to absorb some of the extra heat.Thus Kp will increase. [2] (c) (i) [1] (ii) Elimination [1] (e) (i) [1] (ii) [3] (iii) Nucleophilic substitution [1] [Total: 14]

7 River Valley High School 9729/02/PRELIM/22 [Turn over 2022 Preliminary Examination 5 (a) (i) [2] (ii) Stage I HNO3 + 2H2SO4  NO2+ + 2HSO4− + H3O+ Stage II [2] (ii) [1] (b) (i) Hydrolysis or acid-base [1] (ii) Amount of Mg = 1.5 ÷ 24.3 = 6.17 × 10−2 mol Amount of bromopropane = (5 × 1.35) ÷ 123 = 5.49 × 10−2 mol Magnesium is in excess [2] (iii) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) [1] (iv) Upper [2]

8 River Valley High School 9729/02/PRELIM/22 2022 Preliminary Examination Butanoic acid is more soluble in diethyl ether than water and will dissolve in the organic layer [1]. As diethyl ether is less dense than water, butanoic acid will be found in the upper layer. (v) 1-bromobutane Butanoic acid will undergo an acid -base reaction with NaOH to form the soluble salt, sodium butanoate, which will dissolve in the aqueous layer due to the formation of favourable ion -dipole interactions. 1-bromobutane will stay in the organic layer to be removed. [3] (vii) Water [1] (viii) Accept any range within 154-174 C. [1] (c) (i) Ka = [H ][X ] [HX] +− Since [H +] = [X –] and assuming that the degree of dissociation is small, 10–4.82  2[H ] 0.20 + [H+] = 1.74  10–3 mol dm–3 pH = −lg (1.74  10–3) = 2.76 [2] (d) (i) pH = 14.95 ÷ 2 = 7.48 [1] (ii) H2O(l)  H+(aq) + OH−(aq) Kw = 1.00 × 10−14 mol2 dm−6 D2O(l)  D+(aq) + OD−(aq) Kw = 1.12 × 10−15 mol2 dm−6 The Kw value of D2O is lower than the Kw value of H2O. The position of equilibrium for D 2O lies more on the left/extent of ionisation of D2O is lower than H 2O. Therefore the O-D bond is likely to be a stronger bond that is more difficult to break and dissociate. [2] [Total: 21]