2023 RI H2 Chem Prelims P3 Answers

© 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. 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) 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. 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.

© Raffles Institution 2023 9729/03/S/23 1(c) Or 2Ca(s) + Si(s) + 2O2(g) CaSiO4(s) Hatom [Ca] x 2 2Ca(g) + Si(s) + 2O2(g) LE [CaSiO4] 2  (1st IE [Ca] + 2nd IE [Ca]) 2Ca2+(g) + 4e− + Si(s) + 2O2(g) +13872 kJ mol−1 2Ca2+(g) + SiO44−(g) Using Hess’ Law, −2306 = 2(+121) + 2(+590 + 1150) + 13872 + LE [Ca2SiO4] LE [Ca2SiO4] = −19900 kJ mol−1 (3s.f.) 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. 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 1(d)(iii) G = –nFEcell = −4(96500)(+0.84) = −324240 J mol−1 = −324 kJ mol−1 Hf [Ca2SiO4]

© 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+ + 2