2019 ACJC Prelim H2 Chem P1 ANS

1 © ACJC 2019 9729/01/Prelim/2019 [Turn over ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY Preliminary Examination CHEMISTRY 9729/01 Higher 2 Paper 1 Multiple Choice 18 September 2019 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, index number and tutorial class on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 18 printed pages. 9729/01/Prelim/19 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2019 Department of Chemistry [Turn over

2 © ACJC 2019 9729/01/Prelim/2019 [Turn over 1 An aqueous mixture of sodium carbonate and sodium hydrogencarbonate was titrated with hydrochloric acid and the pH was recorded. What is a suitable indicator to use for detecting the first end point and the ratio of sodium carbonate to sodium hydrogencarbonate in the mixture? indicator ratio A methyl orange 1:1 B methyl orange 1:2 ✓ C thymol blue 1:1 D thymol blue 1:2 Solution The working range of methyl orange is between pH 3.1 to 4.4 and thus is not suitable for detecting the first end point between carbonate and H + at a pH that is above 7. Thus thymol blue is a more suitable choice. CO32– + H+ ⟶ HCO3– HCO3– + H+ ⟶ H2CO3 If 10 cm3 of H+ is needed to react with the carbonate ions present in solution, then the same volume of H+ is needed to react with the hydrogencarbonate produced after the first end point, i.e. 10 cm 3. Thus the other 10 cm 3 portion of H + is u sed to react with the initial hydrogencarbonate present. Hence ratio of carbonate to hydrogencarbonate is 1:1. Comment: Students who were not careful erroneously chose D because of the convenient 1:2 ratio that they see from the graph. 0 5 10 15 20 25 30 35 40 pH volume of HCl added / cm3

3 © ACJC 2019 9729/01/Prelim/2019 [Turn over 2 The most common oxidation state of americium, Am, in aqueous solution is +3. Recently, Cu3+ has been shown to quantitatively oxidise Am3+(aq) in dilute HNO3, while itself is reduced to Cu2+. In an experiment, 20.0 cm 3 of 0.0120 mol dm –3 Am3+(aq) was found to require 24.00 cm3 of 0.0300 mol dm–3 Cu3+ for complete oxidation. What is the formula of the americium-containing species formed? A AmO+ B AmO2+ C AmO22+✓ D Am2O22+ Solution n(electrons) = n(Cu3+) = 7.20 X 10-4 change in oxidation state of Am = (7.20 X 10-4) / (20.0 X 0.0120 / 1000) = +3 Initial oxidation state of Am = +3 +3 = +6 3 The graph shows the second ionisation energies for ten consecutive elements. Which of the following could be X? A oxygen B fluorine✓ C neon D sodium Solution 2nd IE: X+(g) → X2+(g) + e 1st Dip is for this case: 1s2 2s2 to 1s2 2s2 2p1 (2p electron further away) 2nd Dip is for this case: 1s2 2s2 2p3 to 1s2 2s2 2p4 (inter-electron repulsion) X+(g) has this electronic configuration: 1s2 2s2 2p4 Hence X is 1s2 2s2 2p5. X is Fluorine. 2nd IE / kJ mol Proton Number −1 X

4 © ACJC 2019 9729/01/Prelim/2019 [Turn over 4 Which of the following graphs are correct about a fixed amount of an ideal gas? 1 2 3 A 1 and 2 only B 1 and 3 only C 2 and 3 only ✓ D all of the above Solution Using the ideal gas equation, pV = nRT, p= nRT V . Since n, R and T are constant for (1), then p has an inverse relationship with V producing a "y=1/x" graph in (1). pV has a direct positive relationship T as seen in the ideal gas equation and thus produces a "y=x" graph in (2). pV T =nR, where nR is a constant. Thus we expect to see a "x = k" graph in (3). 5 Two single -neck round -bottomed flasks were evacuated and insulated from the surroundings. They were filled separately with gaseous ammonia and gaseous hydrogen chloride at room temperature and connected with a gas tap joint. When the gas tap joint is opened, the two gases are allowed to mix. What is the final pressure of the resultant gas mixture? A more than 3.5 atm but less than 7.0 atm B exactly 3.5 atm ✓ C more than 0.5 atm but less than 1.0 atm D exactly 0.5 atm ρ pV T 0 0 p V 0 0 pV T/K 0 0 NH3 3.0 atm 250 cm3 HCl 4.0 atm 250 cm3 constant T

5 © ACJC 2019 9729/01/Prelim/2019 [Turn over Solution NH3(g) + HCl(g) ⟶ NH4Cl(s) All the NH3(g) will react with HCl(g) to produce NH4Cl(s) which will deposit on the walls/bottom of the flasks. The solids’ volume is negligible compared to the volume of the flasks. That supposedly leaves us with 0.5 atm of HC l(g) in a combined volume of 500 cm3. But the reaction is exothermic and heat is given off. Since the flasks are insulated, the system will heat up and we expect the pressure of the system to be higher than 0.5 atm. Comment: This question was poorly done with approximately half the cohort having chosen B, thinking that it was a simple mixing of the two gases. 6 When 1.00 g of ethanol was burned under a beaker of water, it was found that 100 cm 3 of water was heated from 15 ºC to 65 ºC. The process was known to be only 70% efficient. Use these data and values from the Data Booklet to calculate the enthalpy change of combustion of ethanol. A −209 kJ mol–1 B −673 kJ mol–1 ✓ C −1373 kJ mol–1 D +1373 kJ mol–1 Solution Heat absorbed by water = mc∆T = (100)(4.18)(65 – 15) = 20,900 Joule Heat given out by combustion = Heat absorbed by water ÷ 0.70 = 29,857 Joule ∆Hc(ethanol) = − Heat / Amount of Limiting Reagent = − 29,857 / (1 ÷ 46.0) = −1,373,000 J mol−1 7 Some standard enthalpy changes are given below. ∆Ho / kJ mol–1 Ca2+(g) + aq ⟶ Ca2+(aq) –1650 Cl–(g) + aq ⟶ Cl–(aq) –364 Ca2+(g) + 2Cl–(g) ⟶ CaCl2(s) –2258 What is the standard enthalpy change of solution of calcium chloride? A +244 kJ mol–1 ✓ B –120 kJ mol–1 C –2378 kJ mol–1 D –4636 kJ mol–1

6 © ACJC 2019 9729/01/Prelim/2019 [Turn over Solution ∆Hsoln = ∑∆Hhyd(Ca2+ + 2Cl–) – LE = –1650 + 2(–364) – (–2258) = –120 kJ mol–1 Students would have chosen A if they had only taken the ∆Hhyd for 1 mole of Cl–. Students would have chosen C if they misinterpret ∆Hsoln as the summation of ∆Hhyd values. Option D is just a summation of all three given enthalpy changes, factoring in ∆H hyd for 2 moles of Cl–. 8 Hydrogen peroxide reacts with acidified iodide ions, liberating iodine. H2O2 + 2I⎺ + 2H+ → I2 + 2H2O In the investigation of this reaction, the following results were obtained. initial concentrations of reactants / mol dm−3 initial rate of formation of iodine / mol dm−3 s−1 [H2O2] [I ⎺ ] [H+] 0.01 0.01 0.10 2.0 x 10−6 0.03 0.01 0.10 6.0 x 10−6 0.03 0.02 0.10 1.2 x 10−5 0.03 0.02 0.20 1.2 x 10−5 Which of the following statements are correct? 1 The rate equation can be written as: rate = k [H2O2][I⎺ ]. 2 The reaction is second order with respect to H+