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. Hydration : AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl−(aq) Hydrolysis : [Al(H2O)6]3+(aq)  [Al(H2O)5OH]2+(aq) + H+(aq) [3m]

3 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 [Turn Over 2 (c) The larger the Ka value, the stronger the acid. HCO2H is a stronger acid than CH3CO2H as the electron-donating -CH3 group intensifies the negative charge on CH3COO−, making it less stable than HCOO−. Methanesulfonic acid is a stronger acid than ethanoic acid as the negati ve charge of CH3SO2O− can be delocalised over more oxygen atoms, dispersing the negative charge and making CH3SO2O− more stable. OR Methanesulfonic acid is a stronger acid than ethanoic acid as it has an additional electron-withdrawing oxygen atom attached to S, which helps to disperse the negative charge and making CH3SO2O− more stable. [2m] (d) (i) [3m] mechanism • + on C and − on O of C=O bond • full arrow from lone pair from O of OH− to the +C of C=O group • full arrow from −bond of C=O to the -O of C=O group • correct negatively charged intermediate • full arrow from lone pair on O− of intermediate to form C=O bond • full arrow from C−O bond to O of −OCH3 group • correct carboxylic acid and alkoxide products formed (ii) The three methyl groups bonded to the carbon adjacent to the ester group results in steric hindrance, which makes it more difficult for the OH − nucleophile to approach the C=O of the ester group. OR The (CH 3)3C- group is more electr on-donating than -CH3, making the carbon of C=O of the ester group less electron-deficient and less easily attacked by the OH- nucleophile. [1m]

4 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 2 (e) Information Type of reaction Deduction Ester D is hydrolysed by aqueous alkali, followed by acidification to give E and F. alkaline hydrolysis The organic products are alcohol and carboxylic acid. F reacts with Na2CO3(aq) but E does not. acid-carbonate / acid- base reaction F has carboxylic acid / -COOH group. E has alcohol group. F gives orange ppt with 2,4-DNPH but E does not. condensation F contains ketone or aldehyde group but not E F does not react with Fehling’s solution. - F is not an aliphatic aldehyde. F contains ketone group. E reacts with acidified K2Cr2O7 on prolonged heating to give F. oxidation E has a secondary alcohol group and a primary alcohol group. Both E and F have four carbon atoms each. D, E and F reacts with alkaline I2(aq) to give yellow ppt. oxidation with alkaline I2(aq) / positive iodoform test F is a ketone with - COCH3 group E has -CH(OH)CH3 group D also have the above groups. Reasoning [4m] [1m] for each correct structure

5 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 [Turn Over 2 (f) (i) Amount of Sn2+ in 25.0 cm3 of H = 5 18.50× ×0.0202 1000 = 9.25  10−4 mol [1m] Amount of Sn2+ in 25.0 cm3 of the solution treated with H2 = 5 27.70× ×0.0202 1000 = 1.39  10−3 mol [1m] (ii) Amount of Sn4+ originally present in H = 1.385  10−3 - 9.25  10−4 = 4.60  10−4 mol Ratio of Sn4+ : Sn2+ = 4.60  10−4 : 9.25  10−4 = 0.50 [1m] Average oxidation number of Sn in the oxide = 1 3 {1(+4) + 2(+2)} = + 8 3 Sn : O ratio = 8/3 : 2 = ¾ : 1 = 3: 4 Alternatively, 1 Sn4+: 2 O2− and 2 Sn2+ : 2 O2−  3 Sn : 4 O The formula of oxide C is Sn3O4. [1m] 3 (a) (i) 10NaN3 + 2KNO3 → 16N2 + 5Na2O + K2O Amount of NaN3 : amount of N2 = 1 : 1.6 Using idea gas equation, amount of N2 gas to give the pressure required is n. PV = nRT n = 5 -31.50×10 ×75.0×10 8.31×(35.0 + 273) = 4.40 mol [1m] Amount of NaN3 required = 4.40  1 1.6  (23.0 + 42) = 179 g [2m] (ii) SiO2 + K2O → K2SiO3 OR SiO2 + Na2O → Na2SiO3 [1m] (iii) (b) (i) [1m] [1m]

6 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 3 (b) (ii) Step 1: LiAlH4, dry ether [1m] Step 3: cold, alkaline KMnO4 [1m] [1m] for each correct structure (c) (i) Ka1 = 10−pKa1 = -4.810 = 1.58  10−5 mol dm−3 [H+] = -51.58×10 ×0.10 = 1.26  10−3 mol dm−3 pH = -lg(1.25  10−3) = 2.9 or 2.90 [1m] (ii) [3m] (d) (i) 3Ba(NH2)2 → Ba3N2 + 4NH3 [1m] (ii) Ca2+ has a smaller radius and hence, a higher charge density than Ba2+. The ability of Ca2+ to polarise the electron cloud of the large NH2− anion and weaken the N-H bonds in NH2− is greater than that of Ba2+. Hence, Ca(NH2)2 will decompose more readily. [2m] Volume of NaOH added /cm3 10 20 5 15 4.8 10.7 2.90 30 13.0 pH 0

7 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 [Turn Over 4 (a) (i) The energy required / enthalpy change that occurs when 1 mole of electrons is removed from a mole of gaseous atoms to form a mole of singly charged gaseous ions. [1m] (ii) From Sc to Cu, nuclear charge increases. However, there is addition of electrons to an i