2025 RI H2Chem Prelims P3 Answers

© Raffles Institution 2025 9729/03/S/25 2025 Y6 H2 Chemistry Preliminary Exams Paper 3 – Suggested Solutions Section A 1(a)(i) • The gas particles have negligible volume or size of the gas particles is negligible compared to the volume of the container. • The gas particles exert negligible attractive forces on one another. • The collisions between the gas particles are perfectly elastic. 1(a)(ii) 1(a)(iii) At moderately high pressure , NH3 molecules come closer together and the intermolecular attractive forces (hydrogen bonds) between NH3 molecules become significant. This causes the gas to occupy a volume smaller than that of an ideal gas. 1(a)(iv) 27 C = 300 K and 227 C = 500 K When compressed at constant T to half V, Apply p1V1 = p2V2 4.00  5 = p2  2.5 p2 = 8.00 atm When heated from 300 K to 500 K at constant V, Apply p2/T2 = p3/T3 8.00 / 300 = p3 / 500 p3 = 13.3 atm Alternative method using pV/T: p1V1/T1 = p2V2/T2 4.00  5 / 300 = p2  2.5 / 500 p2 = 13.3 atm 1(b)(i) It is the energy released when 1 mol of ammonia gas is completely burnt in excess oxygen under standard conditions of 298 K and 1 bar. p pV/T (ii) ideal gas R = 8.31 (iii) NH3

© Raffles Institution 2025 9729/03/S/25 1(b)(ii) By Hess’ law, ∆Hr = −4(−46.1) + 6(436) + 3(496) – 12(460) – 6(44.0) = −1495.6 kJ mol−1 Therefore, ∆Hc of NH3(g) = −1495.6 / 4 = −373.9 = −374 kJ mol−1 1(b)(iii) The reaction is accompanied by a decrease in number of gaseous particles, resulting in less disorder. 1(b)(iv) ∆Gr = ∆Hr − T∆Sr ∆Gr = −1495.6 – 298(−583/1000) = −1322 = −1320 kJ mol−1 Since ∆Hr is negative and ∆Sr is also negative, this means ∆Gr is more negative as lower temperature where negative ∆ Hr term outweighs the positive −T∆ Sr term. The reaction is spontaneous at low temperatures. 1(b)(v) At lower temperatures, particles have insufficient energy to overcome the high activation energy needed to break the strong NN and O=O bonds to form NOx. Or The NN bond is very strong, forming NOx from N2 requires a lot of energy to break this bond, which is not favoured at low temperatures. 1(c)(i) Acid. NH3 donates H+ to H− in NaH to form NH2− and H2. Or Oxidising agent. NH3 is reduced as the oxidation number of H decreases from +1 in NH3 to 0 in H2. 1(c)(ii) Reducing agent. NH3 is oxidised as oxidation number of N increases from −3 in NH3 to −2 in N2H4. ∆Hr

© Raffles Institution 2025 9729/03/S/25 1(c)(iii) Nucleophile. Electron pair on N is donated to the electron deficient carbon in C−Br. 2(a)(i) Kp = PNH3 2 PH2 3 PN2 unit: atm−2 Since the initial molar ratio and change in molar ratio of H2 and N2 are in the ratio of 3:1, the equilibrium amount molar ratio will also be 3:1. Thus, PH2 = 3 4 (200 - 35) = 123.75 atm PN2 = 1 4  (200 - 35) = 41.25 atm Kp = PNH3 2 PH2 3 PN2 = (35)2 (123.75)3 (41.25) = 1.57  10−5 atm−2 2(a)(ii) As temperature increases, Kp decreases. This shows that the equilibrium position shifts left with increasing temperature to absorb heat energy. Hence, the backward reaction is endothermic and the forward reaction has a negative enthalpy change (i.e. exothermic). 2(a)(iii) When small amount of inert gas is added into the reaction vessel under constant temperature and pressure, total volume of the gaseous system is increased (as the system must expand to keep its total pressure constant). Concentrations (or partial pressures) of the reactants and products are decreased. The system counteract the change shifting the position of equilibrium so as to re -establish the equilibrium, hence, the equilibrium position will shift to the left , i.e. the side involving greater number of moles of gas. 2(a)(iv) Iron has partially filled 3d subshell for the ready exchange of electrons to and from reactant molecules, facilitating the formation of weak bonds with the reactant molecules. 2(a)(v) Since the forward reaction is exothermic , a lower temperature would result in a higher yield of ammonia . However, the rate of production is too slow at low temperature, hence a moderately high temperature of 450 C is used to ensure a reasonable rate of production and yield. The forward reaction takes place with a reduction in the number of gaseous particles and a high pressure will favour the desired reaction (increase yield) . However, too high a pressure increases cost of production / increases safety concerns. Thus, a moderate pressure of 200 atm is used. Iron catalyst is added to increase the rate of reaction and reduce the time taken to reach equilibrium.

© Raffles Institution 2025 9729/03/S/25 2(b)(i) A ligand is a chemical species which can form a dative/co-ordinate bond simultaneously with the central metal ion (or atom) through lone pair of electrons. 2(b)(ii) Isomers A and B: Isomer C that does not rotate plane polarised light: 2(c)(i) 1s22s22p63s23p63d7 2(c)(ii) In the presence of ligands causes the splitting of the five 3d orbitals in Co2+ into two sets of slightly different energy levels. Since the 3d subshell in Co2+ is partially filled, electrons from 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 transitions). The colour observed is the complement of the colour absorbed. 2(c)(iii) Due to the large energy gap between the d -orbitals, it is energetically more favourable for electrons to pair up in d -orbitals in the lower energy level despite inter-electronic repulsion. 2(c)(iv) dx2-y2

© Raffles Institution 2025 9729/03/S/25 2(c)(v) Co2+ ion in [Co(CN)4]2– 3(a)(i) LiAlH4 in dry ether 3(a)(ii) 3(b)(i) nucleophilic addition 3(b)(ii) cis-trans isomerism 3(b)(iii) +3 3(c)(i) 3(c)(ii) SO2 and Cl− 3(d)(i) The orbital containing the lone pair of electrons on the nitrogen atom overlaps with the  electron cloud of the adjacent C=O bond and the lone pair of electrons is delocalised. Hence, this lone pair of electrons on the nitrogen is not available for donation/coordination to an electron-deficient species. 3(d)(ii) energy

© Raffles Institution 2025 9729/03/S/25 3(e)(i) Evidence Deduction T decolourises Br2(aq). T undergoes electrophilic addition reaction T contains C=C bond(s) T, C8H8, is reacts with hot, acidified KMnO4, to form U, C4H4O6 as the only carbon-containing product. T undergoes strong oxidation/oxidation cleavage U contains either ketone and/or carboxylic acids functional groups. Since no CO2 is formed and T has 8 C while U has 4 C, 1 mole of T formed 2 mole U U forms an orange ppt with 2,4-DNPH. U undergoes condensation reaction. U contains ketone functional group(s). 1 mol of U reacts with excess Na2CO3(aq) to form 1 mole of CO2 gas. U undergoes acid-base reaction. U contains 2 carboxylic acid functional group(s). 3(e)(ii) Section B 4(a)(i) Down group 17, the electron clouds of the halogens become larger and more polarisable. Hence, more energy is required to overcome the increasing strength of the instantaneous dipole -induced dipole interactions between the halogen molecules down the group, leading to decreasing volatility. 4(a)(ii) Cl2(aq) + NaBr(aq) Cl2(aq) + 2Br–(aq) → Br2(aq) + 2Cl–(aq) • Orange colour due to production of Br2(aq) is observed. I2(aq) + NaCl(aq) No reaction occurs. • Brown colour due to unreacted I2(aq) is observed. Br2(aq) + NaI(aq) Br2(aq) + 2I–(aq) → I 2(aq) + 2Br–(aq) • Brown colour due to production of I2(aq) is observed.

© Raffles Institution 2025 9729/03/S/25 4(b)(i) Hydrogen chloride is thermally stable (does not decompose). Hydrogen bromide and hydrogen iodide thermally decom