JPJC 2023 Prelim P3 Answers

© Jurong Pioneer Junior College Answers to 2023 JC2 Preliminary Examination H2 Chemistry (9729) Paper 3 1 (a) 2CO + 2NO → 2CO2 + N2 [1m] (b) The permanent dipole -permanent dipole attractions between CO molecules are stronger than the instantaneous dipole-induced dipole attractions between N 2 molecules and hence, more significant / is not negligible. [2m] (c) (i) Amount of CO gas = 364 (12.0 +16.0) = 13.0 mol Amount of H2 gas = 5 2(1.0) = 2.50 mol Amount of CH3OH gas = 171 (12.0 + 4.0 +16.0) = 5.34 mol Total amount of gas = 13.0 + 2.50 + 5.34 = 20.84 mol Mole fraction of CO = 13.0 20.84 = 0.624 Mole fraction of H2 = 2.50 20.84 = 0.120 Mole fraction of CH3OH = 5.34 20.84 = 0.256 [1m] for all three mole fractions Kp = 3 2 CH OH 2 CO H P P .(P ) = 2 0.256×60 (0.624×60).(0.120×60) = 7.91  10−3 atm−2 [1m] Partial pressure of all three gases calculated [1m] value of Kp (ii) Increasing the temperature shifts the position of equilibrium to the left to favour the backward endothermic reaction so as to absorb some heat. The equilibrium yield of methanol decreases. [1m] (d) (i) [2m] cycle Hc (CH3OH(l)) = -(-129) – (-111) + (-394) + 2(-286) = −726 kJ mol−1 [1m] (d) (ii) Methanol is a liquid and is easier to transport / Synthesis gas is gaseous state, need to be stored under pressure. [1m] CH3OH(l) + O2(g) CO2(g) + 2H2O(l) CO(g) + 2H2(g) + O2(g) C(s) + 2O2(g) + 2H2(g) Hc (CH3OH(l)) −129 −394 + 2(−286) −111

2 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 1 (e) (i) HCOCl + AlCl3 → HCO+ + AlCl4- [1m] (ii) [2m] mechanism • full arrow from −electrons of benzene ring to C of HCO+ electrophile • correct arenium ion + delocalisation of +ve charge over the 5 sp2 C • full arrow from the –bond of C–H bond to the +ve charge of arenium ion. • correct product formed + balanced eqn with the regeneration of AlCl3 catalyst. (iii) To retain the extra stability due to the delocalised  electron cloud of benzene ring, benzene favours substitution reaction over addition reaction which would otherwise, destroy the delocalised 6 electron cloud and its associated stability. [1m] (i) A: B: C: [1m] for each structure (ii) Step 1: HCN, trace amount of NaCN / KCN / NaOH / KOH Step 3: HCl(aq) / H2SO4(aq), heat [1m] each 2 (a) They have giant covalent structure. Decreasing amount of energy is required to overcome the decreasing strength of covalent bonds between the respective atoms due to increase in radius from C to Ge, resulting in an increase in bond length between their respective atoms. Hence, the melting points decreases from carbon to germanium. [2m] (b) SiCl4 hydrolyses completely in water to give an acidic solution of pH 1. A solid forms. Hydrolysis: SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl (aq) AlCl3, dissolves to give [Al(H2O)6]3+ which hydrolyses partially in water to give an acidic solution of pH 3