2024 RI Prelim H2 Chem Paper 3 Suggested solutions

© 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 Comments:  Generally well done.  Students are reminded to use square brackets ‘[ ]’ to represent concentration as the question is asking for Kc expression.  Students are reminded to read the question carefully, and to remember to include the correct units. 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.) Comments:  Generally well done.  Students are reminded to present their final answers to 3 significant figures. 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 Comments:  A common mistake involves wrong conversion of volume (from dm3 to m3).  A more efficient way to arrive at the total pressure is to sum up the total amount of gases, to calculate the total pressure (instead of repeating the calculation three times to calculate each gas’s partial pressure). 1(a)(iv) Partial pressure of SO3 = 4.60 0.5+0.1+4.6 × (31.24 ×106) = 27.6 MPa Comments:  Generally well done.  Students should read the question carefully as the question is asking for the equilibrium partial pressure of SO3 and not SO2, hence the equilibrium amount used should be 4.60 mol.

© Raffles Institution 2024 9729/03/S/24 1(a)(v) Let y be the amount of O2 added into the system 2 SO 2 + O 2 ⇌ 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 Comments:  This question was poorly answered.  Common mistakes involve students forgetting to include the change in amount of O2 and hence their new equilibrium amount of O2 was wrongly calculated as 0.10 + y. 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. Comments:  Generally well done. 1(b)(ii)

© Raffles Institution 2024 9729/03/S/24 Comments:  This question was poorly answered.  From t0 to t1, forward rate = backward rate (the system is at dynamic equilibrium).  At t 1 when the temperature increases sharply, o Both forward and backward rate will be higher as compared to before t1. o The increase in temperature will increase the backward rate to a larger extent than the forward rate (system will favour the backward endothermic reaction when temperature increases). o Between t1 and t2, backward rate decreases as the concentrations of the products decreases while the forward rate increases as the concentrations of the reactants increases.  At t 2 and beyond, forward rate = backward rate.  Both forward rate and backward rate at t2 will be higher than the initial rate from t0 to t1 due to the higher reaction temperature. 1(c)(i) 800 K Comments:  Generally well done 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. Comments:  Students should focus on reaction 1 and use either ∆Gr (more positive) or ∆H (negative) and deduce where the position of equilibrium for reaction 1 lies. 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. Comments:  Generally well done.  Students should be clear in their explanation, and the assumption should be referring to the gaseous particles. (which includes both molecules and atoms) 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.

© Raffles Institution 2024 9729/03/S/24 Comments:  Generally well done 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. Comments:  The ∆H and ∆S is not constant at extreme temperature or pressure.  Students should explain why gaseous particles cannot be liquefied despite the intermolecular forces of attraction being significant at high pressure. 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. More energy is required to overcome the stronger hydrogen bonds as compared to weaker id-id, hence steam has a higher critical temperature. Comments:  Students are reminded to use clear phrasing when describing intermolecular hydrogen bonding. Unclear phrasing can lead to ambiguity related to intramolecular hydrogen bonding or the O−H covalent bond.  CO 2 is a linear molecule where there is no net dipole moment. Hence, it is a non-polar molecule, with only instantaneous dipole-induced dipole interactions between molecules. 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. Comments:  Standard enthalpy change of formation can be either endothermic or exothermic and therefore should be define as energy change.  A common mistake is to miss out ‘in a specified state’. Students should take note that the energy change is dependent on the state of the substance, hence it is essential to specify the states in the definition. Take for example, H2(g) + 1 2 O2(g)  H2O(l) and H2(g) + 1 2 O2(g)  H2O(g) represent the standard enthalpy change of formation of water and steam respectively. The latter would have a more positive / less negative standard enthalpy change of formation as energy is needed to further vaporise the water to steam.

© Raffles Institution 2024 9729/03/S/24 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.) Comments:  Some common mistakes in the energy level diagrams are missing labelled vertical axis and incorrect direction of arrows. The direction of the arrows should to correspond to the energy change (upwards for endothermic and downwards for exothermic processes).  The most common mistake is to form the HCO3−(g) first. As this process involves electron as a reactant, the electron should first be produced from the ionisation of the Na(g).  Another common mistake is to combine the atomisation and 1st IE of sodium. For energy cycle, it is more appropriate to do these two processes separately. This is because atomisation must be carried out before ionisation. By combining these two processes, it poses the ambiguity on which process happens first.  Other common mistakes include missing state symbols or missing electron as a product of the ionisation of the Na(g). energy / kJ mol−1 0 Na(g) + 12 H2(g) + C(s) + 32 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