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−