RI H2 Chemistry 2020 to 2022 TYS Solutions

© Raffles Institution 2020 A-Level H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C D B D A B B A D B C C D C B 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 B C D D B C B A D C D A C A D Q1 (C) There are a total of five 3d orbitals. However, only the 3 dx2-y2 , 3d xy, 3d xz and 3d yz orbitals have four lobes. Q2 (D) Since there is a large jump from the 8 th to 9 th ionisation energy, the element has eight valence electrons and is a Group 18 element, i.e. Ar. Q3 (B) When nucleon number = 217, atomic number no. of neutrons no. of electrons Po2+ 84 217 – 84 = 133 84 – 2 = 82 At3+ 85 217 – 85 = 132 85 – 3 = 82 Rn4+ 86 217 – 86 = 131 86 – 4 = 82 Fr5+ 87 217 – 87 = 130 87 – 5 = 82 From the above table, only At3+ has 50 more neutrons than electrons. Q4 (D) The three alkanes are constitutional isomers with the same Mr and number of electrons. All three are simple, non-polar molecules with only instantaneous dipole-induced dipole (id-id) interactions. Since boiling involves overcoming the intermolecular forces of attraction (NOT covalent bonds), the differences in boiling points among the three alkanes are due to differences in the strength of their id-id interactions. Pentane, a s traight–chained hydrocarbon, has greater surface area for intermolecular interactions compared to its branched isomers, 2-methylbutane and 2,2-dimethylpropane. Thus, intermolecular id-id interaction is the strongest in pentane. As branching increases from 2 -methylbutane to 2,2- dimethylpropane, the surface area for intermolecular interaction decreases. Hence the boiling point decreases from pentane to 2–methylbutane to 2,2–dimethylpropane. Q5 (A) Both liquids are initially at 20 oC. Since stronger intermolecular forces between CHC l3 and CH3COCH3 are formed (compared to their original intermolecular forces), energy is released upon mixing and initial temperature of the mixture will be above 20 oC. Since the intermolecular forces between CHCl 3 and CH3COCH3 are stronger (than their original intermolecular f orces) and require more energy to overcome, the boiling point of the mixture will be above 61 oC. Q6 (B) Since all four gases behave as ideal gases and temperature is kept constant, pV = nRT = constant Graph of pV against V is a horizontal straight line. Hence, the gas with a larger n will have a larger pV. Since Mr of CH4 < Ne < N2 < Cl2, for equal masses of the four gases, amount, n, of CH4 > Ne > N2 > Cl2. Hence, graph B corresponds to Ne.

© Raffles Institution Q7 (B) H2(g) + I2(g) ⇌ 2HI(g) initial amt / mol 0 0 0.040 change in amt / mol +x +x –2x eqm amt / mol x x 0.040 – 2x At equilibrium, ntotal = x + x + (0.040 – 2x) = 0.040 mol Ptotal = 1.0 atm Kp = PHI 2 PH2PI2 = ( 0.040 – 2x 0.040 × 1.0) 2 ( x 0.040 × 1.0)2 = 54 x = 0.004279 mol PHI = 0.040 – 2(0.004279) 0.040 × 1.0 = 0.79 atm Q8 (A) Cationic radius of Mg2+ < Ca2+, resulting in Mg 2+ having a higher charge density and stronger polarising power than Ca2+. Consequently, there is greater extent of distortion of the electron cloud of the CO 32– anion and hence greater extent of weakening of covalent bonds within the CO 32– anion for Mg CO3. Less heat energy is required to break the covalent bonds within the CO 32– anion, causing the decomposition temperature of MgCO3 to be lower. Q9 (D) n(B2O3) = 2.50 2(10.8) + 3(16.0) = 0.03592 mol n(CO2) = 0.80 12.0 +2(16.0) = 0.01818 mol ratio of B : C in boron carbide = 2(0.03592) : 0.01818 = 4 : 1 Hence, the empirical formula of boron carbide is B4C. Q10 (B) Since the reaction is zero order with respect to I2, the rate of reaction is constant and independent of [I2]. Hence, the graph of [I2] against time is a downward sloping straight line with a constant gradient (since rate = – gradient). Q11 (C) ∆G = ∆H – T∆S A reaction is spontaneous when ∆G < 0. statement ∆H ∆S ∆G 1 > 0 < 0 > 0 at all temperatures 2 < 0 > 0 < 0 at all temperatures 3 < 0 < 0 < 0 at low temperatures (when the negative ∆H outweighs the positive – T∆S) Q12 (C) (This is an autocatalytic reaction where Mn2+ acts as the autocatalyst.) For the graph of volume of CO 2 against time, the gradient of the graph at a particular time gives the instantaneous rate of the reaction. Since the gradient of the graph at t2 is greater than that at t1, the reaction is occurring at a faster rate at t2. n(MnO 4−) added = n(C2O42−) added = 25.0 / 1000 × 0.01 = 0.000250 mol Since mole ratio of MnO4− : C2O42− = 2 : 5, C2O42− is the limiting reagent. mole ratio of C2O42− : CO2 = 5 : 10 n(CO2) = 0.000250 / 5 × 10 = 0.0005 mol At s.t.p., z = 0.0005 × 22.7 = 0.0114 dm3 = 11.4 cm3 Q13 (D) Since t½ = 40 min, 120 min = 3 t½ At r.t.p., n(O2) formed at 120 min = 6.00 / 24 = 0.250 mol mole ratio of H2O2 : O2 = 2 : 1 Let the initial amount of H2O2 be x mol. amt of H2O2 / mol amt of O2 / mol t = 0 min x 0 t = 40 min 1 2 x 1 2 (x – 1 2 x) t = 80 min 1 4 x 1 2 (x – 1 4 x) t = 120 min 1 8 x 1 2 (x – 1 8 x) = 0.250 x = 0.5714 mol Initial concentration of H2O2 = 0.5714 / 200 × 1000 = 2.9 mol dm−3

© Raffles Institution Q14 (C) Kc = [R][S]2 [P][Q] mol dm−3 Kc for experiment 1 = 0.0375 mol dm−3 Kc for experiment 2 = 0.0510 mol dm−3 Comparing experiments 1 and 2, as temperature is increased from 300 K to 400 K, Kc increases. This shows that the equilibrium position shifted right with increasing temperature. Hence, the forward reaction must be endothermic. Q15 (B) A suitable indicator is one where its pH range coincides with the region of rapid pH change in the titration curve (i.e. the pH range of the indicator must fall on the vertical portion of the titration curve). equivalence point rapid change in pH occurs around Suitable indicator first approx. 3 – 6 naphthyl red second approx. 8.5 – 10.5 thymol blue Q16 (B) A Incorrect. Point X is an intermediate. A transition state cannot be isolated (unstable) and exists at a potential energy maximum. B Correct. See above diagram. C Incorrect. Although step 1 is an endothermic process, the reaction involves both the breaking of C=O π bond and formation of C-C bond. Besides, bond formation is always an exothermic process. D Incorrect. The reaction pathway diagram does not give any conclusion about the reversibility of the reaction. Q17 (C) A Incorrect. There are only two c onstitutional isomers of C4H10: B Incorrect. Constitutional (Structural) isomers have the same molecular formula but different structural formula, i.e. different arrangement of atoms. But-1-ene (C4H8) and pen-1-ene (C5H10) are not constitutional isomers as they have different molecular formula. C Correct. Constitutional isomers with different functional groups can differ in their chemical properties. E.g. cyclobutane (C 4H8) does not undergo electrophilic addition but but-1- ene (C 4H8) does. D Incorrect. Refer to definition of constitutional isomers in option B. Q18 (D) O HO Mr = 288 cold KMnO4 O HO Mr = 322 OHOH * ***** * * Z Q19 (D) statement 1 Incorrect. •Cl and •CClF2 should be the major free radical products formed in the initiation step as the C-Cl bond is weaker and can be broken more easily than the C -F bond. statement 2 Correct. X• is consumed in the first step and regenerated in the second step of the chain reaction. statement 3 Incorrect. The termination step should involve the reaction between radicals instead. E.g. 2X• → X2 Ea(step 2) Ea(step 1)

© Raffles Institution Q20 (B) The reactions occurring in the catalytic converter are: 2CO(g) + O2(g) → 2CO2(g) Carbon monoxide is o