HCI 2021 Prelim Paper 1 Solutions

2021 HCI C2 H2 Chemistry Preliminary Exam / Paper 1 ANSWERS: Comments 1 B In order to find the number of valence electrons in the G+ ion, we take the difference between all the successive ionisation energies to find the one with the greatest difference. No of electrons removed 1 2 3 4 5 6 Ionisation energy/ kJ mol –1 1907 2914 4964 6274 21 267 25 431 Difference in IE 1007 2050 1310 14993 4164 After analysing the table, we can see that the greatest jump in ionisation energy is when the 5th electron is removed. This means that the G+ ion has 4 electrons in its valence shell. Therefore, G would have 5 electrons in the valence shell, and belong to group 15. 2 A 1 HCHO, CH3OH HCHO CH3OH Intermolecular force: Permanent dipole-permanent dipole interactions Intermolecular Force: Hydrogen bonding CH3OH has the higher boiling point. (strength of dispersion forces similar since number of electrons for each molecule is similar.) HWA CHONG INSTITUTION 2021 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS 1 2 3 4 5 6 7 8 9 10 B A A C C D B C D D 11 12 13 14 15 16 17 18 19 20 A D C A A D B B C D 21 22 23 24 25 26 27 28 29 30 C C D C B B A D B B

2 BF3: Simple covalent compound. Dispersion forces present between the non-polar molecules of BF3. AlF3: Ionic compound. Ionic bonds present between oppositely charged ions. AlF3 has the higher boiling point. 3 CH3CH2CH2CH2CH3, C(CH3)4 Both compounds are simple non-polar covalent molecules, with same number of electrons. However, C(CH3)4 is a branched chain hydrocarbon while CH3CH2CH2CH2CH3 is a straight chain hydrocarbon. CH3CH2CH2CH2CH3 has linear shape and greater surface area of contact between molecules, hence more extensive dispersion forces. CH3CH2CH2CH2CH3 has the higher boiling point. 3 A Option A: Both NH3 and HF can form only one hydrogen bond on average per molecule. Hence while it is true that HF has a higher boiling point than NH 3, it is because the hydrogen bonds between HF molecules are stronger than the hydrogen bonds between NH3 molecules. F more electronegative than N, so there is higher partial positive charge on the H of HF, hence the hydrogen bonds between HF molecules are stronger. Option B: In the solid state, H2O molecules are held at fixed positions and arranged in an orderly manner to form a regular lattice such that hydrogen bonding is maximised to four per molecule. The hydrogen bonding between water molecules in ice are positioned in a roughly tetrahedral shape around each O atom. This produces an open lattice, with empty spaces between the H2O molecules. The more random arrangement of hydrogen bonding in liquid water results in H2O molecules packing much more closely and, together, take up less space. So the lattice structure of ice occupies a larger volume for the same mass of liquid water, hence ice has a lower density than liquid water. Option C: CH3CO2H dimerises in non-aqueous solvents as shown in the following diagram. Option D: In 2-nitrophenol, the –OH and –NO2 groups can form intramolecular hydrogen bonding, so there is less extensive intermolecular hydrogen bonding between 2-nitrophenol molecules compared to between 4-nitrophenol molecules. Therefore, less energy is needed to overcome the intermolecular hydrogen bonding for 2-nitrophenol to boil, hence its lower boiling point. (The words 'less extensive'

2021 HCI C2 H2 Chemistry Preliminary Exam / Paper 1 refers to lower average number of hydrogen bonds present in the same mass of 2- nitrophenol molecules as compared to 4-nitrophenol molecules.) Refer to Section 8.3 on hydrogen bonding in Chemical Bonding Lecture Notes. 4 C For options A, B and D, the graph is a straight line passing through the origin. For option C, the graph is a horizontal line parallel to the x axis. In order to derive the shape of the graph, start from the ideal gas equation: pV = nRT For each of the options, find the relationship the different terms have based on the ideal gas equation. E.g. for option B pV= nRT Since density = ௠௔௦௦ ௏ pV= ௠௔௦௦ ெೝ RT p= ௠௔௦௦ ௏ ோ் ெೝ Since ோ் ெೝ is constant, p is directly proportional to density.

5 C CxHy (l) + (x+௬ ସ) O2 (g) → xCO2 (g) + ௬ ଶH2O (l) Initial gas volume/ cm3 - 100 - - Final gas volume/ cm3 - 25 60 Reacting volume/ cm3 75 60 Since the gaseous mixture contracted by 15 cm3 after the reaction, the total volume of excess O2 and CO2 = 85cm3 When the gaseous mixture was passed through NaOH (aq), the volume of gas contracted another 60cm3. This implies that the volume of CO2 is 60 cm3. The volume of excess oxygen is therefore 25cm3. Hence 100 – 25 = 75 cm3 of O2 reacted with the hydrocarbon. The ratio of vol of O2 reacted: vol of CO2 produced = x+௬ ସ : x = 75 : 60 Hence the ratio of x: y = 1:1. Any hydrocarbon with C:H of 1:1 could be the answer. 6 D 1 Relative atomic mass = 0.9499 × 32 + 0.0075 × 33 + 0.0425 × 34 + 0.0001 × 36 2 This is a correct definition. 3 The angle of deflection in an electric field is directly proportional to the charge/ mass ratio. Since the charges of the isotopes are the same (all +1), the lightest isotope will have the largest angle of deflection. 7 B No. of moles of zinc = 13.1/ 65.4 = 0.200 mol No. of moles of Cu2+ = ଵହ଴ ଵ଴଴଴×1 = 0.150 mol Cu2+ is the limiting reagent. q=msolutionc' T = 150 × 4.20 × 15 = 9450 J 'Hrxn = − ଽସହ଴ ଴.ଵହ = − 63 000 J mol–1 = – 63.0 kJ mol–1

2021 HCI C2 H2 Chemistry Preliminary Exam / Paper 1 8 C 1 The enthalpy change of neutralisation is defined as the energy evolved when one mole of water is formed from a reaction between an acid and a base. This equation shows the formation of 2 moles of water. Hence this equation is twice that of the enthalpy change of neutralisation. 2 As ethanoic acid is a weak acid, its conjugate base is a weak base that is able to hydrolyse partially in water to give OH– ions. CH3CO2 – + H2O ⇌ CH3CO2H + OH– As such the pH of a solution of CH3CO2Na is greater than 7. 3 When a weak acid is neutralised by a base, the enthalpy change of neutralisation will be slightly less exothermic as energy is absorbed to ionise the un-ionised weak acid. 9 D Given the rate equation, rate = k[BrO3–][Br–][H+]2 To determine x: Since the order with respect to [H+] is 2, when the [H+] is doubled from 0.40 to 0.80 mol dm–3 in experiment 1 to 2, 1/t will be four times its original value. Hence x = 0.10 × 4 = 0.40 s–1. To determine y: Since the order with respect to [Br–] is 1, when the [Br–] is doubled from 0.04 to 0.08 mol dm–3 in experiment 1 to 3, 1/t will be double its original value. Hence y = 0.10 × 2 = 0.20 s–1. To determine z: When [H+] and [BrO3–] are both doubled from experiment 1 to 4, if there is no change to [Br–], then 1/t would be 0.10 x 2 x 22 = 0.80. Since actual 1/t is 1.60 s–1, the [Br–] must also be higher resulting in a faster rate of reaction. Compare the initial rate of 0.10 s–1 and 1.60 s–1 : Let the multiplication factor of [Br–] be a a × 2 × 22 = (1.60 ÷ 0.10) a = 2 z = 2 × 0.04 = 0.08 mol dm–3

10 D The overall equation is the sum of all individual steps: 2A o C C + B o D D + B o E 2A + 2B + C + D o C + D + E Overall: 2A + 2B o E Statement 1 is correct. The rate equation of the slow step is: rate = k2PBPC Since the concentration of intermediates do not appear in the rate equation, the PC is replaced by the PA (assuming step is a fast equilibrium) i.e PC = K1PA2 Hence, rate = kPA2PB where k = k2K1 This suggests that the initial rate of formation of E is proportional to the initial partial pressure of A, raised to the power of 2. Thus, statement 3 is correct. Analysis of the units of the rate constant: atm s–1 = (Units of k) × atm2 ×atm Units of k =