RI 2023 Y5 Promo Solutions

© Raffles Institution 2023 9729/S/23 Page 1 2023 Y5 H2 Chemistry Promotion Examination – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer B C C D A B B D B C A C D C A MCQ worked solutions A1 (B) After cooling back to room temperature and pressure, H2O condenses back to liquid state. 4CH3NH2(g) + 9O2(g) → 4CO2(g) + 2N2(g) + 10H2O(l) Initial V/ cm3 20 60 0 0 0 Change in V/cm3 −20 −45 +20 +10 − Final V/ cm3 0 15 20 10 − Final total volume = 15 + 20 + 10 = 45 cm3 A2 (C) Mol ratio of MnO42− to MnO4− = 3 : 2 2 mol of MnO42− is oxidised to 2 mol of MnO4−. Since 3 mol of MnO 42− underwent disproportionation, 3−2 = 1 mol of MnO 42− is reduced to the manganese- containing product. Oxidation no. increased from +6 to +7 from MnO42− to MnO4−. Hence, amt of e − lost by 2 mol of MnO 42− = 2 mol Total amt of e− lost = total amt of e− gained, Amt of e− gained by 1 mol of MnO42− = 2 mol Hence, the oxidation number of Mn decreases by 2 to form the manganese- containing product. Oxidation number of the product = +6 − 2 = +4 The only manganese- containing product in the options with a oxidation number of +4 is MnO2. A3 (C) The data for gold should be ignored as the question is about the Ar value for copper. A r = (63×0.62 + 65×0.32) ÷ 0.94 = 63.7 A4 (D) Option 1 is incorrect as 36S has a nucleon number of 36 while 37Cl has a nucleon number of 37 . The charge of the ion indicates the number of electrons lost or accepted and does not affect the nucleon number. Option 2 is correct as the S atom accepts 2 electrons to form an ion with an outer electronic configuration of 3s 2 3p6, while the Cl atom accepts 1 electron to form an ion with an outer electronic configuration of 3s2 3p6 Option 3 is in correct. 36S2− has 16+2 = 18 electrons and 36−16= 20 neutrons . 37Cl− has 17+1 = 18 electrons and 37−17= 20 neutrons. A5 (A) As Ba is in group 2, it loses 2 valence electrons to form Ba2+. Hence, the anion has a charge of 2−. In the O22− anion, each of the 2 electrons from Ba is accepted by each of the 2 oxygen atoms. A6 (B) Recall that a gas deviates from ideal gas behavior at high pressure and low temperature. The higher the pressure and the lower the temperature, the greater the deviation from ideality.

© Raffles Institution 2023 9729/S/23 Page 2 A7 (B) Assuming ideal gas behaviour, pV = nRT n = pV RT Since pressure is kept constant at 1 atm, no. of moles of air in the heated balloon initial no. of moles of air in the balloon = 101325 × 5500 R × (65 + 273) × R × (25 + 273) 101325 × 5000 = 0.97 Note: Since p and R are constants, students can also make use of n ∝ V T and thus n2 n1 = V2 T2 × T1 V1 . A8 (D) COCl2(g) ⇌ CO(g) + Cl2(g) Kc = a = [CO][Cl2] [COCl2] Let Kc’ be the equilibrium constant of the reaction below. 2CO(g) + 2Cl2(g) ⇌ 2COCl2(g) Kc’ Kc’ = [COCl2]2 [CO]2[Cl2]2 = ( [COCl2] [CO][Cl2])2 = ( [CO][Cl2] [COCl2] )–2 = (a)–2 = 1 a2 A9 (B) Statement 1 is correct. As the equilibrium system is homogeneous (all species are in the same phase), H 2O is not a solvent, [H 2O] is not constant and should appear in the equilibrium constant expression. Statement 2 is incorrect. Equilibrium constant, K c (or Kp), is only affected by temperature changes. Increasing amount of CH3OH, and hence [CH 3OH] will not have any effect on equilibrium constant. Statement 3 is correct. Adding HCO2C2H5 to the equilibrium mixture will shift the equilibrium position of  to the left, leading to an increase in [ HCO2H]. The increase in [HCO2H] in turn result in the equilibrium position of  to shift to the right, leading to an increase in concentration and hence amount of HCO 2CH3.  HCO2H(l) + CH3OH(l) ⇌ HCO2CH3(l) + H2O(l) K1  HCO2H(l) + C2H5OH(l) ⇌ HCO2C2H5(l) + H2O(l) K2 A10 (C) ∆Hro = ∆Hfo[PCl5(l)] – [∆Hfo[PCl3(l)] + ∆Hfo[Cl2(g)]] Since ∆Hfo of an element at its standard state = 0, ∆Hfo[Cl2(g)] = 0, and the question states that ∆Hfo[PCl5(l)] < 0 and ∆Hfo[PCl3(l)] < 0, ∆Hfo[PCl5(l)] – ∆Hfo[PCl3(l)] = –124 ∆Hro is less negative than ∆Hfo[PCl5(l)]. A11 (A) Reaction 1 H2SO4 + 2NaOH → Na2SO4 + 2H2O n(H2SO4) = 2.00 X 10–3 mol n(NaOH) = 3.00 X 10–3 mol NaOH is the limiting reagent. n(H2O) = 3.00 X 10–3 mol q1 = 50 X 4.18 X ∆T1 ∆H1 = 1 3 50 4.18 3.00 10 x xT x − ∆− J mol–1 Reaction 2 HCl + NaOH → NaCl + H2O n(HCl) = 2.00 X 10–3 mol HCl is the limiting reagent. n(H2O) = 2.00 X 10–3 mol q2 = 50 X 4.18 X ∆T2 ∆H2 = 2 3 50 4.18 2.00 10 x xT x − ∆− J mol–1 Since reactions 1 and 2 are neutralisation reactions between strong acids and a strong base, ∆H1 = ∆H2 1 3 50 4.18 3.00 10 x xT x − ∆− = 2 3 50 4.18 2.00 10 x xT x − ∆− 1 2 3 1.52 T T ∆ = =∆ From Pg 11 of Chemical Energetics 1 lecture notes: The neutralisation reaction between a weak acid – strong base is less exothermic than that of the strong acid – strong base neutralisation reaction. This is because during the neutralisation reaction, the CH 3COOH molecules undergo further dissociation to produce H+ ions for reaction with the OH- ions. Some energy is consumed to bring about further dissociation of the CH 3COOH molecules during the neutralisation reaction. So, the energy released from the reaction will be less. In this case, ∆ 𝐻𝐻neut o between CH3COOH and NaOH is less negative than that between HC l (or H 2SO4) and NaOH . Hence, options B and D are incorrect.

© Raffles Institution 2023 9729/S/23 Page 3 A12 (C) 200 → 100 → 50 → 25 24 h = 3 x t1/2 ⇒ t1/2 = 8 h A13 (D) Option 1 is incorrect. The shapes of the two graphs reflect how [H2O2] changes with time for an autocatalytic reaction. However, the catalytic decomposition of H2O2 is not an autocatalytic reaction. Option 2 is incorrect. The amount of product obtained is independent of the amount of catalyst used. Using double the mass and hence amount of catalyst does not give a higher yield of product. Option 3 is correct. Refer to Pg 40 of Reaction Kinetics lecture notes. At low [H 2O2], the reaction is first order with respect to H2O2 as not all the active sites of the catalyst are occupied. At high [H 2O2], the reaction is zero order with respect to H2O2 as all the active sites of the catalyst are saturated with H2O2. The rate of reaction increases with the amount of catalyst used. A14 (C) Option A is incorrect as a meso compound has a plane symmetry, cannot rotate plane- polarised light and thus is not optically active. Option B is incorrect as enantiomers interact differently with chiral molecules. Option C is correct as a racemic mixture is made of equal proportion of two enantiomers and enantiomers do not have a plane of symmetry. Option D is incorrect as a molecule with only one chiral centre must be chiral, and cannot be a meso compound, and hence will be able to rotate plane-polarised light. A15 (A) Cis-hex-3-ene has a higher boiling point than trans-hex-3-ene as cis-hex-3-ene is slightly polar while trans-hex-3-ene is non-polar. Cis-hex-3-ene has a more negative ∆Hc trans-hex-3-ene as cis-hex-3-ene is less stable due to steric strain arising from crowding between two alkyl groups on the same side of the C=C. energy + 9O2(g) + 9O2(g) 6CO2(g) + 6H2O(l) more exothermic ∆Hc o less stable higher energy more stable lower energy less exothermic ∆Hc o 1st t1/2 2nd t1/2 3rd t1/2

© Raffles Institution 2023 9729/S/23 Page 4 Section B B1(a) Hybridisation of carbon atom in ethene: sp2 The two atoms form a σ–b ond via the head–on overlapping of two sp2 hybrid orbitals. They also form a π–bond via the side–on overlapping of the two unhybridised p orbitals.