JPJC Prelim P2 Ans

¤ Jurong Pioneer Junior College [Turn Over Mark Scheme for JC2 H2 Chemistry (9729) Preliminary Examination Paper 2 1 (a) (i) Li+: 1s2 H–: 1s2 [1m] for both (ii) Both have same number of quantum shells and no shielding by inner shell electrons. Li+ has a larger nuclear charge/more protons and hence smaller radius. [1m] (iii) LiAlH4 o LiAl + 2H2 [1m] (iv) LiAl has giant metallic structure. [1m] Large amount of heat energy is required to overcome the strong electrostatic attraction between a lattice of cations and delocalised electrons. [1m] (v) The effect of a smaller Ar is more significant than that of a smaller volume of Li atom due to its smaller (atomic) radius. [1m] (b) (i) (ii) H is 2-hydroxybutanoic acid. [1m] NaBH4 reduces ketone only while LiAlH4 reduce both ketone and carboxylic acid, forming back butane-1,2-diol [1m] (c) (d) J CO2 H2O Amount/mol 4.32 192 = 0.0225 5.94 44.0 = 0.135 1.62 18.0 = 0.0900 Mol ratio 0.0225 0.0225 = 1 0.135 0.0225 = 6 0.0900 0.0225 = 4 Mole ratio of J : C : H = 1 : 6 : 8 [1m] DFFHSW´µ Let molecular formula of J be C6H8Ox. 6(12.0) + 8(1.0) + 16.0x = 192.0 Ÿ x = 7 ? molecular formula is C6H8O7. [1m] Since J : NaOH = 1 : 3, J has 3 ‒COOH groups. [1m] for any achiral C6H8O7 e.g. [1m] dotted lines between O and H for two hydrogen bonds [1m] dipoles on 2 O-H, lone pair on O + label ´K\GURJHQERQGµIRU one hydrogen bond hydrogen bond [1m]

2 ¤ Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2021 2 (a) (i) x reaction of solid NaCl with conc. H2SO4 role of conc. H2SO4: Bronsted acid (9) explanation: H2SO4 donates a H+ to form HSO4‒ OR Cl‒ accepts a H+ to form HCl. [1m] x reaction of solid NaI with conc. H2SO4 role of conc. H2SO4: Oxidising agent (9) explanation: O.S. of I increases from ‒1 in NaI to 0 in I2 OR O.S. of S decreases from +6 in H2SO4 to ‒2 in H2S. [1m] 2(9): [1m] (ii) HI/I‒/NaI is a stronger reducing agent than HCl/Cl‒/NaCl. [1m] (iii) 2NaBr(s) + 3H2SO4(l) o 2NaHSO4(s) + Br2(g) + SO2(g) + 2H2O(l) [1m] for Br2 + SO2; [1m] for balanced eqn (b) (i) There is p-p orbital overlap between C of C=O bond and the adjacent C of C=C. [1m] (ii) (iii) Electron-withdrawing C=O in the secondary carbocation that forms Q intensifies the positive charge, making it less stable and less readily formed than the primary carbocation that forms P. [1m] (c) (i) step 1: hydrolysis/ nucleophilic substitution [1m] step 2: condensation [1m] (91) (92) (93) (94) (95) 5(9): [2m] 2-4(9): [1m]

3 ¤ Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2021 (ii) T U V (iii) When the reactants are adsorbed on the catalyst surface (9), the bonds in the reactants are weakened (9) which lowers Ea. Surface concentration of reactants also increases. Thus, the rate of the reaction increases (9). The products are desorbed (9) from the catalyst surface, making it available for adsorption of new reactant molecules. 4(9): [2m]; 2-3(9): [1m] (iv) Cis-trans isomerism arises due to restricted rotation about the C N bond which has two different groups attached to each C and N [1m] [1m] [1m] [1m] [1m] [1m]

4 ¤ Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2021 3 Fig. 3.1 Vfinal = 252 ½(252) = 126 ¾(252) = 189 1st t½ = 720 s 2nd t½ = 700 s

5 ¤ Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2021 (a) (i) [1m] for 2 correctly read t½ values clearly indicated on graph Half-lives are approximately constant so the reaction is first order wrt [C6H5N2Cl] [1m] (ii) rate = k [C6H5N2Cl] Average t½ = ½ (720 + 700) = 710 s k = ln 2 710 = 9.76 u 104 s1 [1m] 3sf; ecf t½ from (a)(i); [1m] units (iii) Water is a solvent so it is in large excess. Thus only a small proportion of water is reacted and hence concentration of water remains effectively constant throughout the progress of the reaction. [1m] (b) (i) n(N2) = uu36(101 10 252 10 (8.31)(45+273) )( ) = 9.63 u 103 mol [1m] Since HCl { N2, [H+] = [HCl] = u39.63 10 1001000

= 0.0963 mol dm3 [1m] ecf from n(N2) pH = ‒lg 0.0963 = 1.02 [1m] ecf from [H+] (ii) [1m] for any of the following methods x Use an electronic weighing balance to measure the mass of the solution at various time intervals x Use a conductivity meter to measure the conductivity of the solution due to production of H+ and Cl‒ at various time intervals x Titrate quenched samples taken from the main reaction mixture at various time intervals with NaOH(aq) of known concentration (c) a R=A e E Tk §·¨¸©¹ a1ln = ln A R Ek T gradient = ‒a R E= ( 6.60) ( 13.40) 0.00310 0.00360 = ‒13600 K‒1 [1m] Ea = gradient u (R) = (‒13600) u (‒8.31) = +113 000 J mol1 or +113 kJ mol1 [1m] 3sf + units; ecf from gradient

6 ¤ Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2021 4 (a) (i) [1m] correct displayed structure with two dative bonds (represented by Cl o Al; lone pair not required) (ii) (iii) 8E(Al௅Cl) = ௅ (+116) ௅ (²1401) + 2(+326) + 3(+244) [1m] ecf from multiplier E(Al௅Cl) = +334 kJ mol‒1 [1m] 3sf (iv) Al2Cl6 dissolves in water to form [Al(H2O)6]3+ in water. Al2Cl6 + 12H2O o 2[Al(H2O)6]3+ + 6Cl Since Al3+ has a high charge density (9), Al2Cl6/[Al(H2O)6]3+ undergoes partial hydrolysis (9) in water. Al3+ polarises the coordinated H2O molecule and weakens the O௅H bond (9) which breaks to release H+, giving rise to a weakly acidic solution of pH 3 (9). 4(9): [2m]; 2-3(9): [1m] [Al(H2O)6]3+ [Al(H2O)5(OH) ]2+ + H+ OR Al2Cl6 + 12H2O 2[Al(H2O)5(OH) ]2+ + 2H+ + 6Cl OR Al2Cl6 + 12H2O 2[Al(H2O)5(OH)]Cl2 + 2HCl Al2Cl6(s) 0 8 u E(Al‒Cl) energy / kJ mol1 Al2Cl6(g) +116 2Al(g) + 6Cl(g) 2Al(s) + 3Cl2(g) 2Al(g) + 3Cl2(g) 2(+326) 3(+244) ‒1401 (9) (9) (9) (9) 4(9): [2m] 2-3(9): [1m] [1m]

7 ¤ Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2021 (b) (i) Cl௅ anion and H2O molecules are attracted to the anode. E/V O2 + 4H+ + 4e 2H2O +1.23 Cl2 + 2e 2Cl +1.36 --- (1) High [Cl] shifts the position of equilibrium (1) to the left (9), making E(Cl2/Cl) less positive (or more negative) than E(O2/H2O)/+1.23V (9). So Cl is more easily oxidised than H2O, producing Cl2 gas instead of O2 gas. (9) award only with mention of shift in position of equilibrium 4(9): [2m]; 2-3(9): [1m] (ii) I u t = ne u F 0.250 u (2 u 60 u 60) = ne u 96500 ne, amount of e = 0.0187 mol [1m] Amount of Fe = 0.521 55.8 = 0.00934 mol Since n(Fe) : ne = 0.00934 : 0.0187 = 1 : 2 x = 2 (c) ClF3 [ClF4]– shape: ………………………………… shape: …………………………………. 5 (a) CFCs are non-toxic / non-flammable. [1m] (b) R-11 / CCl3F is a better refrigerant. [1m] From Table 5.1, CCl3F has a higher boiling point than NH3 so CCl3F has stronger IMF. Both CCl3F and NH3 have simple covalent structures. Due to greater number of electrons per (CCl3F) molecule (9), more energy (9*) is needed to overcome the stronger (9*) instantaneous dipole-induced dipole interaction between CCl3F molecules (9) than the weaker hydrogen bonds between NH3 molecules (9). Thus, CCl3F is less easily vaporised and has lower vapour pressure than NH3. 4(9): [2m]; 2-3(9): [1m] (9*) stronger + more energy Cl F F F Cl F F F F ‒ (9) [1m] T-shape (9) square planar (9) [1m] [1m]

8 ¤ Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2021 (c) (i) A radical is a species with unpaired electron(s) [1m] (ii) Name the type of bond breaking: homolytic fission [1m] (iii) First propagation step: Clx + O3 o ClOx + O2 [1m] Second propagation step: ClOx + O3 o 2O2 + Clx [1m] (iv) 2Clx o Cl2 OR ClOx + Clx o ClO‒Cl (or Cl2O) OR 2ClOx o