JPJC Prelim H2 Chemistry P1 (Worked Solution)

2021 JPJC JC1 H2 Chemistry (9729) 1 2021 Preliminary Exam Paper 1 (Worked Solution) JURONG PIONEER JUNIOR COLLEGE 2021 JC2 H2 Chemistry (9729) Preliminary Exam Paper 1 (Worked Solutions) Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans 1 B 6 A 11 C 16 B 21 A 26 D 2 C 7 A 12 B 17 D 22 B 27 C 3 C 8 A 13 C 18 D 23 C 28 B 4 D 9 A 14 B 19 C 24 D 29 A 5 B 10 D 15 C 20 D 25 A 30 B 1 Answer: B The reaction causes the original particle to lose 1 proton, 1 electron and gain 1 neutron. Options A and C both have the same number of electrons and protons before and after the reaction. Option D shows a decrease in number of electrons and protons (36 in Kr to 35 in Br) but the number of neutrons remain unchanged (both particles have 40 neutrons). 2 Answer: C From Figure 1, large increase between the 2nd and 3rd electron removed indicates that there are two valence electrons in element P. Hence element P is in group 2. P+: ns1 From the electronic configuration of P+, the point corresponding to the second IE of element P is C. 3 Answer: C 8A SiCl4 has 4 bond pairs and no lone pairs around Si Î bond angle is 109.5 °. 8B SO2 has 2 bond pairs and 1 lone pair around S Î bond angle is < 120 °. 9C IF2– has 2 bond pairs and 3 lone pairs around I Î bond angle is 180 °. 8D CH3+ has 3 bond pairs and no lone pairs around C Î bond angle is 120 °. 4 Answer: D 9A and B The structures of N2O4 and (CH3CO2H)2 dimers are shown below. 9C The CH3 groups in (CH3CO2H)2 is still tetrahedral around the C, thus the dimer is non- planar. 8D Due to p-p orbital overlap, the electrons are delocalised across the O=N-O bond, making all the nitrogen-oxygen bonds to be of the same strength and thus of the same length.

2021 JPJC JC1 H2 Chemistry (9729) 2 2021 Preliminary Exam Paper 1 (Worked Solution) 5 Answer: B §· § · § · u u u ¨¸ ¨ ¸ ¨ ¸©¹ © ¹ © ¹ r 92.2 4.7 3.1 of Si 28 29 30 28.109100 100 100A 8A Ar of Si = 28.099 8C Ar of Si = 28.668 8D Ar of Si = 28.854 6 Answer: A { 42 33 42 2 2 Since 1 CH 2 O 50 cm of CH requires 100 cm of O for complete reaction. Since CO will be absorbed by the alkaline KOH, it will not be collected. Volume of gas collected volume of O left 150 100 350 cm 7 Answer: A u uu ' u u ' uu ' u uu 'u? y u u uu kJ kJ Total amount of heat evolved 1371 46.0 300Total amount of heat transferred 300 J 1000 300 1371 300 46.0efficiency 100% 100%1000 46.0 1000 1371 m cTcT cT m cT m 8 Answer: A Using ΔG = ΔH – TΔS, the negative gradient of the graph in the Ellingham diagram corresponds to ΔS of the reaction. Reaction II has a gradient of zero, that means ΔS = 0 (reject options B and C) Reaction I has a positive gradient, that means ΔS < 0 Î decrease in disorderedness Reaction III has a negative gradient, that means ΔS > 0 Î increase in disorderedness 9 Answer: A The graph shows that when pressure increases, % products at equilibrium decreases. Î POE shifts left to form less gas molecules (reject options B and C) The graph also shows that when temperature increases, % products at equilibrium increases Î POE shifts right to favour endothermic reaction (reject option D)

2021 JPJC JC1 H2 Chemistry (9729) 3 2021 Preliminary Exam Paper 1 (Worked Solution) 10 Answer: D (2 and 4 only) At time t, the change caused POE to shift right to form more SO3. 81 Adding a catalyst will not cause a shift in POE. 92 When temperature decreases, POE shift right to favour exothermic reaction. 83 The addition of inert gas at constant volume results in the partial pressures of all gases to remain unchanged. POE will not shift. 94 At equilibrium before time t: >@ >@>@

? 23 2 SO SO 2 2 3 13 c 22 22 From graph: 0.5 mol and 0.3 mol 0.80 0.500.4 =0.25 mol2 0.3SO 10 14.4 mol dm 0.5 0.25SO O 10 10 O nn n K 11 Answer: C u ªº? u¬¼ 14 w 6.22 7 OH H lg 2.4 10 7.4 6.22 OH 10 6.03 10 p pK p 12 Answer: B Since equal volumes of the solutions are used:

§·u§· u¨¸¨¸©¹©¹ 22 63 9 2 0.100 1.0 10ionic product of PbX 1.25 10 mol dm22 Since ionic product is less than the Ksp of PbCl2 and PbBr2 but larger than the Ksp of PbI2, only PbI2 will precipitate out. 13 Answer: C When temperature increases, POE shifts left to favour endothermic reaction Î Kp will decrease (reject options A and B) When temperature increases, both forward and backward rate increases. Since POE is shifting left, the backword rate increases more. (reject option D) 14 Answer: B (1 and 2 only) Using the Boltzmann Distribution: 91 increasing temperature increases the number of molecules with energy ≥ Ea 92 When more gas is added at the same temperature and volume, the number of molecules with a particular energy all increases. 83 Compressing the gas increases pressure, but will not change the energy distribution of the molecules. 15 Answer: C (2 and 3 only) 81 A catalyst reduces the activation energy of the reaction by providing an alternative pathway/mechanism of lower activation energy. The KE of the reacting particles are not affected. 92 A catalyst increases both the forward and backward rate of a reversible reaction by the same extent. 93 See definition of the catalyst in option 1.

2021 JPJC JC1 H2 Chemistry (9729) 4 2021 Preliminary Exam Paper 1 (Worked Solution) 16 Answer: B From Data Booklet: 2H+ + 2e H2 0.00V For the metal not to dissolve, E cell for the reaction must be negative. 8A Cr2+ + 2e Cr –0.91 V Î E cell = 0.00 – (–0.91) = +0.91 V 9B Cu2+ + 2e Cu +0.34 V Î E cell = 0.00 – (+0.91) = –0.34 V 8C Fe2+ + 2e Fe –0.44 V Î E cell = 0.00 – (–0.44) = +0.44 V 8D Pb2+ + 2e Pb –0.13 V Î E cell = 0.00 – (–0.13) = +0.13 V 17 Answer: D 8A To obtain pure copper, the pure copper electrode should be the anode (negative electrode) which is electrode Q. 8B [CuSO4] remains unchanged as the amount of Cu2+ that is oxidised at the anode is replenished by the amount of Cu that is reduced at the cathode. u { u u option C is wrong e 40.0 26.8 60 96500 0.666 mol Since 1 Cu 2e 0.666Mass of Cu 63.5 21.2 g ( )2 26.47 21.2% by mass of Ag 100 20 %26.47 Qn n 18 Answer: D From graph, the sharp drop in ΔH (and hence the boiling point) from B to C signifies the change from giant structure to simple covalent molecules (group 14 to group 15 element). 8A Element E is in group 17 and Element F is in group 18. Both are non-polar molecules thus will both be soluble in warm benzene (non-polar solvent). 8B Elements A, B and C are in groups 13, 14 and 15 respectively. The chlorides of these elements are acidic since the structure of the chlorides are becoming increasingly covalent in nature. 8C Element G is in group 1 and Element D is in group 16. The oxide of G (basic oxide) and the oxide of D (acidic oxide) will form a neutral salt. 9D Oxide of A could be Al2O3 which will react with excess NaOH to from Al(OH)4– complex.

2021 JPJC JC1 H2 Chemistry (9729) 5 2021 Preliminary Exam Paper 1 (Worked Solution) 19 Answer: C 20 Answer: D Reagent and conditions to convert alcohol to alkene: conc. H2SO4, heat Î reagent L is conc. H2SO4 (reject options A and B) Reagent M cannot be ethanolic as the alkene will dissolve in the organic solvent, making collection difficult. (reject option C) 21 Answer: A (1, 2 and 3) 91 92 93 22 Answer: B 9A (CH3)3COH acts as an base to accept H+ to from the conjugate acid (CH3)3CO+H2 8B HCl acts as an acid to protonate the –OH group in (CH3)3COH to make it a better leaving group so as to form the carbocation in step 2. 9C 9D Steps 2 and 3 are representative of the SN1 mechanism.

2021 JPJC JC1 H2 Chemistry (9729) 6 2021 Preliminary Exam Paper 1 (Worked Solution) 23 Answer: C If the reaction is an elimination reaction or an SN1 reaction, the resulting mixture will be a racemic mixture with no optical activity. (reject options A and D) If the reaction is an SN2 reaction