2025 RI Prelims H2 Chemistry P3 suggested solutions

© 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. Comments:  Most common error is the omission of “particles” when discussing negligible volume and attractive forces. Please note that volume of the gas (without “particles”) refers to the volume of the container.  The third basic assumption should be about perfectly elastic collisions between gas particles. 1(a)(ii) Comments:  A small group of students did not label the axes correctly as required by the question.  Note that the y-intercept has a value of R = 8.31 (not 1). 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. Comments:  Students should state clearly it was the “significant attractive forces” that resulted in a smaller volume, not simply because the molecules are “closer together”. p pV/T (ii) ideal gas R = 8.31 (iii) NH3

© Raffles Institution 2025 9729/03/S/25 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 Comments:  Some students approached this question by calculating the amount of gas using the ideal gas equation, pV=nRT in which case will require the conversion of values to S.I. units for correct computation. I.e. 1 atm = 101325 Pa, 1 dm3 = 110 m−3, T in Kelvin 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. Comments:  Standard enthalpy of combustion is always exothermic, hence energy is released.  Question required specific reference to ammonia gas and not a generic compound or substance.  Student should include the standard conditions explicitly as 1 bar and 298 K.

© 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 Comments:  When drawing energy cycles, each distinct enthalpy change process should be shown clearly and separately, not combined.  Some students did not take into account the correct stoichiometry required for the formation of ammonia gas and the vapourisation of liquid water and did not multiply the respective enthalpy changes with the corresponding coefficients.  State symbols must be shown for all species.  Majority of students did not realise that the enthalpy change for reaction 1 is 4 x ∆Hc  Cultivate the good habit of doing a quick check on the answers. A large number of students calculated ∆Hc as a positive value which cannot be correct as combustion processes are always exothermic. 1(b)(iii) The reaction is accompanied by a decrease in number of gaseous particles, resulting in less disorder. Comments:  Generally well done. ∆Hr

© Raffles Institution 2025 9729/03/S/25 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. Comments:  In ∆ Gr = ∆Hr− T∆Sr , please note that ∆Hr = 4 x ∆Hc  A handful of students missed out on the conversion of ∆Sr to the same units as ∆Hr.  The temperature under standard conditions is 298 K, not 273 K. 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. Comments:  Some students simply stated “high activation energy” or “endothermic reaction” and were not given credit. It is important to highlight the key reason being the presence of strong NN bond. 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. Comments:  As an acid, ammonia donates a proton (H+), not H atom.  If using oxidation number to support the answer, the reference to specific atom (not a substance or compound) must be clearly stated. E.g. oxidation number of H, N, Cl etc. 1(c)(ii) Reducing agent. NH3 is oxidised as oxidation number of N increases from −3 in NH3 to −2 in N2H4. Comments:  A significant number of students erroneously thought that the oxidation number of Na decreased from +3 to +1. Note that the oxidation number of Na in NaOCl and NaCl is unchanged at +1. Na is a group 1 metal and the maximum oxidation number is +1. 1(c)(iii) Nucleophile. Electron pair on N is donated to the electron deficient carbon in C−Br. Comments:  Clear understanding of nucleophile donating an electron pair is expected. A nucleophile is not defined as being “electron rich”.

© Raffles Institution 2025 9729/03/S/25 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 Comments:  Students should not use square brackets for partial pressure write Kp expression, as they are used to represent concentration.  Some students did not read the instructions carefully and forgot to perform calculation for Kp.  There is no need to convert the pressure into Pa as it will be more tedious in the calculation for Kp subsequently. 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). Comments:  Students should use the change in position of equilibrium (POE) explanation based on the data given instead of using ∆Gr = ∆Hr − T∆Sr or using bond energy to calculate ∆Hr for the explanation of this question.  Students should always explain why the endothermic reaction is favoured (i.e. to absorb heat) by using Le Chateliers’ Principle.  Students should explicitly answer to the question by indicating that the sign of ∆H is negative in their answer. 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.

© Raffles Institution 2025 9729/03/S/25 Comments:  Most students incorrectly wrote that the pressure is increased althou