2023 RI Prelim P3 suggested solutions with examiner comments

© Raffles Institution 2023 9729/03/S/23 2023 Y6 H2 Chemistry Preliminary Exam Paper 3 Suggested Solutions Section A 1(a)(i) Both diamond and silicon have giant covalent structures. During melting, strong Si– Si and C –C covalent bonds between atoms are broken in silicon and diamond respectively. Si–Si covalent bonds are weaker than C –C as the valence orbitals of Si are more diffuse than that of C, thus overlapping of the orbitals of Si is less effective than that of C. Hence less energy is required to break the weaker Si–Si bonds during melting compared to C–C bonds in diamond. Examiner’s Comment • This question was not well answered. Many students misread the question and tried to compare SiO2 instead of Si with diamond. • Note that when melting a giant covalent structure, the covalent bonds between Si atoms are broken in silicon, and between C atoms in diamond. To compare the strengths of covalent bonds between atoms, we need to compare the effectiveness of overlap between Si orbitals, with overlap between C orbitals. • It is not acceptable to use bond energy values to compare covalent bond strengths, without further explanation. • Some students were unsure of the structure of Si, mistaking it for being simple covalent, or similar to that of graphite. • Please be clear and specific with terms used e.g. the orbital overlap is between the atoms, not molecules. 1(a)(ii) Al2O3 is amphoteric/reacts with both acids and bases, while SiO2 is acidic/reacts with bases only. Al2O3(s) + 6H+(aq) → 2Al3+(aq) + 3H2O(l) Al2O3(s) + 2OH–(aq) + 3H2O(l) → 2[Al(OH)4]–(aq) SiO2(s) + 2OH–(conc) → SiO32–(aq) + H2O(l) or Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l) Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na+[Al(OH)4]–(aq) SiO2(s) + 2NaOH(conc) → Na2SiO3(aq) + H2O(l) Examiner’s Comment • Students generally recognised that A l2O3 is amphoteric and SiO 2 is acidic, but had difficulty balancing the equations for their reactions. 1(b)(i) For reaction 2, ∆H = 2(−110.5) − (−910.9) = +689.9 kJ mol−1 Assuming that ∆H  and ∆S are independent of temperature, ∆G2500K = +689.9 − (2500)(+361 x 10–3) = −213 kJ mol−1 (3s.f.) Reaction 2 has negative ∆ G2500K and is spontaneous , whereas reaction 1 has positive ∆G2500K and is non−spontaneous. Hence reaction 2 is the preferred method for the extraction of Si from SiO2.

© Raffles Institution 2023 9729/03/S/23 Examiner’s Comment • To calculate ∆G for reaction 2 at 2500 K, ∆H for reaction 2 needs to be calculated first, using ∆Hro = Σn∆Hfo (products) − Σm∆Hfo (reactants). Many students also calculated ∆H correctly by using an energy cycle. • Students are reminded to present their final calculated answers for each part of a question to 3 significant figures, and to include the correct units in their answers. • A number of students used “−110” instead of “−110.5” as given in the question in their calculations. Please always use exactly what is given in a question (or in the Data Booklet). 1(b)(ii) ∆S is positive as there is an increase in the number /moles/amount of gas eous particles (from 0 to 2) in the equation. Examiner’s Comment • This question was generally well answered. Note that it is important to mention “gaseous” particles in the answer. • A number of candidates reported the increase incorrectly e.g. forming 1 mol of gas when the equation clearly shows 2 moles of CO molecules formed. Some students also cited gaseous ions / atoms, which were incorrect here. 1(c) Or 2Ca(s) + Si(s) + 2O2(g) CaSiO4(s) Hatom [Ca] x 2 2Ca(g) + Si(s) + 2O 2(g) LE [CaSiO4] 2 × (1st IE [Ca] + 2nd IE [Ca]) 2Ca2+(g) + 4e− + Si(s) + 2O2(g) +13872 kJ mol−1 2Ca 2+(g) + SiO44−(g)  ∆Hf [Ca2SiO4]

© Raffles Institution 2023 9729/03/S/23 Using Hess’ Law, −2306 = 2(+121) + 2(+590 + 1150) + 13872 + LE [Ca2SiO4] LE [Ca2SiO4] = −19900 kJ mol−1 (3s.f.) Examiner’s Comment • Some students did not show a good understanding of the definitions of enthalpy changes, or were careless in writing equations. Many students neglected to multiply ∆Hatom or IE by 2 times. • Note that the lattice energy of an ionic compound is the energy released when 1 mole of the solid ionic compound is formed from its constituent gaseous ions under standard conditions, hence it is an exothermic reaction. Some students wrote the equation for LE as the breaking of the ionic lattice instead and obtained a positive answer for LE. • Students are reminded to read the question carefully and to make use of the equation given for the formation of SiO 44−. Since electrons are required for this reaction, Ca should be ionised first to obtain 4 electrons. • Equations were often not balanced in terms of charges due to missing/additional electrons. • Where energy level diagranms are drawn as the answer, the vertical axis should be drawn and labelled with “Energy/kJ mol−1”. 1(d)(i) The standard electrode potential, E, of a half –cell is the electromotive force / potential difference , measured at 298 K, between the half –cell and the standard hydrogen electrode, in which the concentration of any reacting species in solution is 1 mol dm–3 and any gaseous species is at a pressure of 1 bar. Examiner’s Comment • This question was poorly answered. Students are reminded to learn their definitions well. • The standard conditions need to be specified clearly. 1(d)(ii) Cathode: O2 + 2H2O + 4e− → 4OH− Anode: Fe → Fe2+ + 2e− (x2) Overall: O2 + 2H2O + 2Fe → 4OH− + 2Fe2+ Ecell = +0.40 − (−0.44) = +0.84 V Examiner’s Comment • This question was generally well -answered, although some students calculated Ecell wrongly. • An irreversible arrow should be used for the overall equation. 1(d)(iii) ∆G = –nFEcell = −4(96500)(+0.84) = −324240 J mol−1 = −324 kJ mol−1 Examiner’s Comment • This question was generally well-answered, although some students missed out the negative sign for the formula for ∆G. • Note that the units for ∆G using this formula is J mol−1.

© Raffles Institution 2023 9729/03/S/23 1(d)(iv) (I) Oiling the bicycle chain minimises contact between H2O / O2 and Fe, thus rusting stops/slows down. (II) Zn2+ + 2e− ⇌ Zn E = −0.76 V Fe2+ + 2e− ⇌ Fe E = −0.44 V Zn is more easily oxidised than Fe, since E(Zn2+/Zn) is more negative than E(Fe2+/Fe). Hence Zn rusts in place of Fe. Examiner’s Comment • This question was generally well-answered. • Students are reminded that standard electrode potentials, E , represent reduction processes. Hence if E(Zn2+/Zn) is more negative than E(Fe2+/Fe), it means that the reduction of Zn 2+ to Zn is less spontaneous than that of Fe 2+ to Fe, so the oxidation of Zn is more spontaneous than that of Fe. • Students also need to compare the standard electrode potentials of E(Zn2+/Zn) and E(Fe2+/Fe), rather than just quoting the values. 1(e) Examiner’s Comment • This question was not well attempted. Students are reminded that the industrial process of anodising of aluminium is part of the learning outcomes of the chapter on electrolysis. • Many students incorrectly drew electrochemical cells or setups with a salt bridge, when an electrolysis cell was required. • Many students omitted labelling the electrolyte or labelled it wrongly as containing Al3+. Note that the Al object is coated with Al2O3 and not Al. • At the anode (attached to the positive terminal of the battery), water is discharged forming O 2, hence the A l object to be anodised should be placed at the anode, so that