2019 DHS Prelim H2 Chem P1 ANS

2019 Y6 Prelims H2 Chemistry 9729 Paper 1 Suggested Solutions  DHS Chemistry Unit Page 1 of 6 Answer Key 1 2 3 4 5 6 7 8 9 10 C C B C C D C D A D 11 12 13 14 15 16 17 18 19 20 C C B A D C A B A D 21 22 23 24 25 26 27 28 29 30 D C C C B D B A A B 1 C From Reaction (1) 2NaN3 ≡ 2Na ≡ 3N2 From Reaction (2) 10Na(s) + 2KNO3(s) → K2O(s) + 5Na2O(s) + N2(g) 10Na ≡ 1 N2 Combining reactions (1) and (2), 2NaN3 gives 3N2 directly, while 2Na gives 1/5 N2 Hence, 2NaN3 gives a total of 16/5 N2 1.25NaN3 ≡ 2N2 2 C Moles of sodium percarbonate = 10/1000 x 0.1 = 0.001 mol Moles of CO2 = (48/1000) / 24 = 0.002 mol Since (Na2CO3)x•y(H2O2) ≡ xCO2 x = 2 Moles of KMnO4 = 24/1000 x 0.05 = 0.0012 mol 2KMnO4 ≡ 5 H2O2, Moles of H2O2 = 5/2 x 0.00120 = 0.003 mol Since (Na2CO3)x•y(H2O2) ≡ y H2O2 y = 3 y/x = 3/2 3 B Note that elements in the same group (e.g. O and S) will have the same number of unpaired electrons in their ground state. Group 17 elements have only 1 unpaired electron. Hence, option 1 is incorrect.  1 Electronic configuration of Ti: [Ar]3d24s2 (2 unpaired electrons) Cl: [Ne] 3s23p5 (1 unpaired electron)  2 Electronic configuration of Si: [Ne] 3s23p2 (2 unpaired electrons) O: 1s22s22p4 (2 unpaired electron)  3 Electronic configuration of Ni: [Ar]3d84s2 (2 unpaired electrons) S: [Ne] 3s23p4 (2 unpaired electron) 4 C pV = nRT For Graphs A and B, For a fixed mass of gas at constant T, pV = constant. Since we are plotting a p against pV graph, the graph should be a vertical line instead (i.e. x = constant graph) For Graphs C and D, For a fixed mass of gas at constant p, V = (nR/p)T The graph should be y=mx graph (upward sloping straight line that passes through origin), with gradient = (nR/p) Since I has a higher Mr than J, number of moles of I will be smaller than J and hence graph of I should have a smaller gradient. 5 C Carbonate Ethanoate C O O O 2- CH3 C O O - Nitrate Phenoxide N O O O - In ethanoate ion, the shape with respect to the methyl carbon is tetrahedral. The common feature is the delocalization of electrons. 6 D  1 At constant temperature, Kc remains unchanged. Hence G remains unchanged.  2 Reaction is spontaneous when G < 0. For 0 < Kc < 1, ln Kc is negative and hence G > 0 (non-spontaneous)  3 Adding a catalyst increases rate of both forward and backward reaction to the same extent, hence the position of the equilibrium remains unchanged. Kc remains unchanged and hence G remains unchanged.

Dunman High School 2019 Prelims – H2 Chemistry 9729/01 Solutions  DHS Chemistry Unit Page 2 of 6 7 C N2H5+ + NH3 NH4+ + N2H4 Acid1 Acid2 NH3 + HBr NH4+ + Br– Acid1 Acid2 N2H4 + HBr N2H5+ + Br– Acid1 Acid2 For all cases, since Kc > 1, strength of acid1 > acid2. From all 3 equations, we can deduce that Strength of acid: HBr > N2H5+ > NH4+ Hence order in decreasing pH: NH4+ > N2H5+ > HBr Note: HBr is a strong acid. So the options can be narrowed down to options B and C. 8 D  A KOH (strong base) is in excess and resultant solution will have pH >7  B This happens at equivalence point where KOH exactly neutralises benzoic acid. Salt hydrolysis occurs and basic salt will be formed. Resultant solution will have pH > 7.  C Benzoic acid is in excess and acidic buffer will be formed. Since volume of KOH added is half the equivalence volume, this is at maximum buffer capacity. Resultant solution will have pH =pKa.  D Benzoic acid is in excess and acidic buffer will be formed. Resultant solution will have pH < 7. Since volume of KOH added is less than half the equivalence volume, pH will be less than that at maximum buffer capacity. Hence pH < pKa and option D will have the lowest pH. 9 A  1 Precipitate will start to occur when mass of solute dissolved exceed its solubility. For 10 g of KCl, first trace of ppt will start to appear at 30 C. For 50 g of KClO3, first trace of ppt will start to appear at about 76 C.  2 At 10 C, both solutes have exceeded its solubility and hence a saturated solution of KCl and KClO3 is formed.  3 At temperatures between 30 C and 76 C, KClO3 ppt will be formed but not KCl. Mass of KClO3 ppt = 50 – 37 = 13g 10 D 2NH3(g)+ H2O2(l)  N2H4(l) + 2H2O(l) Hreaction = 241 kJ mol1 Hreaction = [ Hf(N2H4(l)) + 2 Hf(H2O(l))]  [2Hf(NH3(g)) + Hf(H2O2(l))] 241 = [Hf(N2H4(l)) + 2( 286)]  [2(46) + (188)] Hf(N2H4(l)) = +51 kJ mol1 Decomposition of N2H4(l) is N2H4(l)  N2(g) + 2H2(g) The reverse of formation of hydrazine, therefore, H for decomposition of hydrazine is 51 kJ mol1 11 C CO(g) + 2H2(g) CH3OH(g) For the forward reaction, S is negative as there is a decrease in the number of moles of gaseous molecules (resulting in a less disordered system). H is negative as “h igher yield of methanol can be achieved at a lower temperature.” That is, at lower temperature, position of equilibrium is shifted to the right, favouring an exothermic reaction. G = H  TS At T=0 K, G = H G is negative. Gradient is S. Since S is negative, gradient is positive.

Dunman High School 2019 Prelims – H2 Chemistry 9729/01 Solutions  DHS Chemistry Unit Page 3 of 6 12 C From the first table, comparing expt 2&3, [P] x2, rate also x2, order of reaction with respect to [P] is 1. Comparing expt 1&2, [P] x2, [Q] x2, rate x2. Since rate  [P], two times increase in [Q] has no effect on rate. Order of reaction with respect to [Q] is 0. Therefore, rate equation is rate =k[P] From the second table, if we disregard the temperatures, rate4 = k(0.10) rate5 = k(0.20) rate6 = k(0.30) Taking temperatures into consideration , relative rates are as follows . (Using 20 C of expt 6 as the reference) rate4 = k(0.10) x 2 x 2 = 0.40k rate5 = k(0.20) x 2 = 0.40k rate6 = k(0.30) x 1 (reference expt) = 0.30k Hence, only statements 1 and 3 are correct. 13 B Element X (=Al) Y (=Na) Z (=P) Oxide of the element Al2O3 Na2O P4O10 Chloride of the element AlCl3 NaCl PCl5 Oxides with HCl(aq) Al2O3 + 6HCl → 2AlCl3 + 3H2O Na2O +2 HCl → 2NaCl + H2O PCl5 merely hydrolyses in aqueous solution of hydrochloric acid to give H3PO4+5HCl SiO2 is acidic and does not react with HCl. Oxides with water: Al2O3 has very exothermic lattice energy. Does not dissolve in water. Na2O forms NaOH in water. Na2O + H2O → 2NaOH P4O10 + 6H2O → 4H3PO4 Chlorides with water: AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl−(aq) Al3+(aq) has high charge density, it undergoes partial hydrolysis in water. [Al(H2O)6]3+(aq) ⇌ [Al(H2O)5(OH)]2+(aq) + H+(aq) NaCl simply dissolves in water. PCl5 undergoes complete hydrolysis in water. PCl5 + 4H2O→H3PO4+5HCl 14 A From the table, since M2 displaces K2 and L2, M2 is the strongest oxidising agent. If we consider the more common halogens (for convenience) – Cl2, Br2 and l2, then M2 is Cl2.  1 Since L2 cannot displace other halogens, L2 is the weakest oxidising agent. L2 is I2. So, when L2 (ie. I2) , the weakest oxidizing agent reacts with thiosulfate ions, S 2O32, oxidation state of sulfur changes from +2 in S2O32 to +2.5 in S4O62 On the other hand, when M2 (ie. Cl2 reacts with S2O32, oxidation state of sulfur changes from +2 in S2O32 to +6 in SO42. Statement 1 is not correct.  2 L is I which forms AgI with silve