2024 NYJC H1 Chem 8873 P1 Answers

J2 Prelim Exam 2024 H1 Chemistry Paper 1 Answers and Comments Page 1 of 7 Nanyang JC J2 Prelim Exam 2024 H1 Chemistry 8873/01 Paper 1 MCQ Answers and Comments Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans 1 B 6 C 11 A 16 B 21 A 26 C 2 C 7 A 12 D 17 B 22 A 27 D 3 C 8 A 13 D 18 C 23 D 28 C 4 C 9 D 14 B 19 A 24 B 29 A 5 D 10 B 15 D 20 B 25 A 30 B 1 B n(ethanol) = 2046.0 = 0.4347 mol n(propanoic acid) = 3074.0 = 0.4054 mol propanoic acid is the limiting reagent m(ethyl propanoate) = 0.4054102.0 = 41.35g yield of ester = (2241.35)100% = 53% 2 C AlxCy + H2O → gas gas + O2 → CO2 + H2O n(CO2) = 0.07224 = 0.003 mol n(C) in every mol of gas = n(C) in every mol of compound A = 0.003 mol Option A: n(Al2C3) = 0.14490.0 = 0.0016 mol n(C) in Al2C3 = 3  0.0016 = 0.00480 mol Option B: n(Al3C4) = 0.144129.0 = 0.00112 mol n(C) in Al3C4 = 4  0.00112 = 0.00447 mol Option C: n(Al4C3) = 0.144144.0 = 0.0010 mol n(C) in Al4C3 = 3  0.0010 = 0.00300 mol Option D: n(Al5C3) = 0.144171.0 = 0.000842 mol n(C) in Al5C3 = 3  0.000842 = 0.00253 mol 3 C Zn → Zn2+ + 2e− Mole ratio of Zn : VO2+ : e−  1 : 2 : 2  ½ : 1 : 1  1 mol of VO2+ gained 1 mol of e− Since ON of V in VO2+ = +5, final ON of V in Y is +4 V5+ + e− → V4+ Ion Y is VO2+ (ON of V is +4) 4 C + X → + p+ The number of subatomic particles (protons & neutrons) should be balanced. no. of protons no. of neutrons 32 15P and p+ 15 + 1 = 16 (32 – 15) + 0 = 17 32 16S 16 32 – 16 = 16 X 0 1

J2 Prelim Exam 2024 H1 Chemistry Paper 1 Answers and Comments Page 2 of 7 5 D Al: 1s2 2s2 2p6 3s2 3p1 Option A is incorrect as e lectrons are present in five different energy levels. Option B is incorrect as t here are 7 electrons in p orbitals and 6 electrons in s orbitals. Option C is incorrect as the occupied orbital of lowest energy (1s) is spherical. 6 C 1: bond angle wrt O decreases from 109o in ice (0lp, 4bp – 2 covalent bonds and 2 H- bonds: see below) to 105o in water (2bp, 2lp) 2: bond angle wrt S increases from <120o in SO2 (2bp, 1lp) to 120o in SO3 (3bp, 0lp) 3: bond angle wrt C increases from 109o in RCH(OH)R’ (4bp, 0lp) to 120o in ketone RCOR’ (3bp, 0lp) C OH H R R' 109o R C O R'120o 4: bond angle wrt Al decreases from 120o in AlCl3 (3bp, 0lp) to 109o Al2Cl6 (4bp, 0lp) 7 A CCl4 and C 6H14 are non -polar molecules that forms only id-id interactions. For A, CH2Cl2 and (CH 3)2CO are polar molecules that can form pd-pd interactions. For B and C, stronger pd -pd interactions are broken and weaker id -id interactions are formed between non-polar molecule mixed with a polar molecule hence heat is taken in. Similarly for D, strong hydrogen bonds between CH 3CH2OH molecules are broken and only weak id -id interactions are formed in the mixture. Hence, A is the best answer where comparable pd -pd interaction is formed between two polar molecules. 8 A Since m.p. of Y is very low and it also has poor electrical conductivity, it should have a simple molecular structure  based on options, Y is SiCl4 Since m.p. of X is very high and it also has good electrical conductivity, it could have a giant metallic structu re or giant molecular structure (graphite). It could also be giant ionic structure if the electrical conductivity is in molten or aqueous state.  based on options, X is C(graphite) Since m.p. of of W is very high and it also has slight electrical conductivity, it should be silicon (metalloid). Diamond is unable to conduct electricity as all 4 valence electrons are used to form covalent bonds and none are delocalised to be able to act as charge carriers. Hydrogen bonds

J2 Prelim Exam 2024 H1 Chemistry Paper 1 Answers and Comments Page 3 of 7 9 D LE   qq rr +− +−+  q+q− for CaCl = (+1)(−1); q+q− for CaCl2 and MgCl2 = (+2)(−2). Since lattice energy depends more on product of ionic charges ( q+q−) than on the sum of ionic radius ( r+ + r−); hence magnitude of lattice energy of CaCl 2 and MgCl2 is greater than that of CaCl. Between CaCl 2 and MgCl 2: r + Mg2+ < r + Ca2+. Therefore, r + + r − for MgCl 2 < CaCl 2. Hence, magnitude of lattice energy of MgCl2 is greater than that of CaCl2. 10 B Hf (CO2): It is energy change when one mole of CO 2(g) is formed from its constituent elements, carbon and oxygen in their standard states under standard conditions. C(s) + ½ O2(g) → CO2(g) Option B is correct as the standard enthalpy change of combustion of carbon is the energy change when one mole of carbon is completely burnt in excess oxygen under standard conditions. The chemical equation is the same as Hf (CO2). C(s) + ½ O2(g) → CO2(g) Option A is incorrect as it only considers the bonds formed without considering bonds broken in carbon and oxygen. (See formula below). Hrʅ=  BE(bonds broken)–  BE(bonds formed) Option C is incorrect as the standard state of carbon is graphite, not diamond. Graphite is the elemental form with the lowest energy. Option D is incorrect as the standard enthalpy change is measured under standard conditions of 298 K and 1 bar. 11 A mol ½ H2SO4 NaOH ½Na2SO4 H2O initial 0.1 0.1 0 0 change −0.05 −0.1 +0.1 +0.1 final 0.05 0 0.1 0.1 NaOH is the limiting reagent. Hence, the number of moles of H2O formed = 0.1 mol q = mcT = (50.0 + 100)(1)(4.18)(29.0 − 20.0) = 5643 J H = – 2 q n(H O) = – 35643 10 0.1 − = –56.4 kJ mol−1 12 D If the temperature of the gas is increased, average kinetic energy of the particles increases. T he maximum of the curve becomes lower and moved to the right. The fraction of molecules that react in the presence of a catalyst is shown by + . 13 D Rate of a chemical reaction can be measured by change in concentration of reactants and products. 14 B

J2 Prelim Exam 2024 H1 Chemistry Paper 1 Answers and Comments Page 4 of 7 Decomposition of NH4Cl(g) NH4Cl(g) ⇌ NH3(g) + HCl(g) H = +314 kJ mol–1 EA refers to the activation energy required for the reaction to overcome before the reaction can take place. Reactant, NH 4Cl will have lower energy level than products, NH 3 + HCl since H = +314 kJ mol–1 15 D A is wrong as catalyst only increases the rate of reaction and not the yield of the reaction. B is wrong as the reaction is exothermic. An increase in temperature will favour the backward endothermic reaction instead to reduce the added heat. Hence, decreasing the yield of reaction. C is wrong as the higher pressure will favour the reaction that produced a lesser amount of gas eous particles which is the forward reaction. Hence the position of equilibrium will shift to the right to reduce the total amount of gaseous particles. D is correct as the addition of a catalyst and/or a high temperature will increase the rate of reaction. 16 B 2NO(g) + Cl2(g) Ý 2NOCl(g) Initial 2.32 X 0 Change −1.77 −0.885 +1.77 Final 0.55 X − 0.885 1.77 Kc = [NOCl]2 / [NO]2[Cl2] 2807 = (1.77/1)2 / (0.55/1)2((X − 0.885)/1) X − 0.885 = 0.003689 X = 0.08886 = 0.889 17 B A lower temperature at T2 will favour the forward exothermic reaction to compensate for the lost of heat. Hence, I2 will decrease more. Hence, the equilibrium amount of I2 should be lower than 0.23. At T2, the rate of reaction will also be slower. Hence, the gradient of the graph should be gentler. 18 C Bronsted Lowry acid: donates hydrogen ions Lewis acid: accepts a lone pair of electrons Arrhenius acid: produces hydrogen ions in aqueous solution 19 A Option 1 is correct. [H]+ dissociated from acid 1 = 10−4 = 0.00