2021 SAJC Prelim P2 Answer

ST ANDREW’S JUNIOR COLLEGE JC2 Preliminary Examinations Higher 2 CANDIDATE NAME CLASS 2 0 S CHEMISTRY Paper 2 Structured Questions Candidate answer on the Question Paper. Additional Materials: Data Booklet 9729/02 1 September 2021 2 hours READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of __ printed pages. For Examiner’s Use Q1 9 Q2 9 Q3 24 Q4 20 Q5 13 Total 75

Answer all questions in the spaces provided. 1 (a) Fig. 1.1 shows a sketch of the logarithm of the first ten ionisation energies, log (IE), of element A against the number of electrons removed. Fig. 1.1 (i) Write an equation for the second ionisation energy of element A. [1] A+(g)  A2+(g) + e– Must have state symbols (ii) Explain the following features of Fig. 1.1.  the general trend in ionisation energy shown in the graph  the significant jump in values from the 5th to the 6th ionisation energy [4] Ionisation energies increased as more electrons were removed. The nuclear charge remained the same , but it attracted a decreasing number of electrons . This led to an increase in the attractive forces on the remaining electrons and hence more energy was required to remove the next electron. OR Successive IE increases as remaining electrons will be attracted more strongly to increasingly positively charged ion and hence more energy was required to remove the next electron The 6th electron is from the inner quantum shell. The decrease in distance of this electron (6th electron) from the nucleus , and less shielding resulted in a 0 1 2 3 4 5 6 7 8 9 10 11 log (IE) Number of electrons removed

stronger attraction to nucleus , significant increase in the energy required to remove this electron /more difficult to remove this inner shell electron. (iii) Element A is in Period 3. Identify element A using information from Fig. 1.1 and complete it s electronic configuration. [1] Element A is phosphorus / P. 1s2 2s2 2p6 3s2 3p3 (b) Sulfur is another element in Period 3. It forms covalent bonds with other elements in covalent compounds such as SO2, SO3, SCl2. (i) Describe a covalent bond. [1] A covalent bond is the electrostatic attraction between a shared / bonding pair of electrons and two positively charged nuclei. (ii) Predict the intermolecular force which exists between SO3 molecules. Explain how this force arises. You may use diagrams where appropriate to illustrate your answer. [2] The SO 3 molecules are held together by instantaneous dipole-induced dipole (id-id) attractions. In each electron cloud, electrons are constantly moving. The electron density of a cloud can be unsymmetrical at any moment , resulting in an instantaneous dipole. This dipole can induce a dipole in the neighbouring electron cloud, causing an attraction between them. OR Diagram marks depend on labels used Label for id-id Uneven electron cloud Labels for instantaneous dipole and induced dipole

Two electron clouds with different labels [Total: 9 marks] 2 Esters are commonly used as artificial flavourings in food products such as cakes and sweets. However, the shelf life of acidic foods containing esters are relatively short as esters are prone to acid-catalysed hydrolysis. The kinetics of the acid -catalysed hydrolysis of the ester , methyl propanoate, CH3CH2CO2CH3, was investigated in an experiment. (a) Write a balanced equation for the hydrolysis of methyl propanoate. [1] CH3CH2CO2CH3 + H2O  CH3CH2CO2H + CH3OH (b) 0.200 mol of ester was hydrolysed by heating with water and hydrochloric acid catalyst in a 1 dm 3 mixture. In a second experiment, hydrochloric acid was replaced with sulfuric acid of the same concentration. When the reaction is complete, 0.200 mol of propanoic acid is obtained. The following results were obtained. 0.000 0.020 0.040 0.060 0.080 0.100 0.120 0.140 0.160 0.180 0.200 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 time / s [CH3CH2COOH] / mol dm–3 Experiment 1 HCl catalyst Experiment 2 H2SO4 catalyst t1/2 = 300s t1/2 = 300s

(i) Based on the results obtained, show that the reaction is first order with respect to the ester and first order with respect to H+ ions. [4] [CH3CH2COOH]max = 0.200 mol dm–3 Show 2 half life on the graph (either experiment 1 or 2) correctly e.g.  from 0  0.100  from 0.100  0.150  does not have to be successive (one after the other e.g. 0.02 -> 0.11) Since half life is constant at 300 s (experiment 1) / 150 s (experiment 2), reaction is first order with respect to [ester]. Show 2 tangent to curve at t=0 clearly and calculate gradient of both tangent to curve When [H+] x 2, initial rate x 2, reaction is 1st order with respect to [H+] OR Show 2 half life of the other experiment on the graph clearly rate = k[ester][H+] 1/2 ln 2 ln 2 ' [ ]t k k H  When [H+] x 2, half life x ½ Reaction is 1st order with respect to H+ (ii) State the units of the rate constant, k. [1] Units of rate constant k = mol–1 dm3 s–1 (c) With the aid of a sketch of the Boltzmann distribution, explain how an increase in temperature changes the rate of ester hydrolysis. [3]

When temperature is increased, the molecules gain kinetic energy and move about faster. This increases the number of molecules having energy greater than or equal to activation energy . As a result, the frequency of effective collisions increases. Reaction rate thus increases. [Total: 9 marks] 3 Scuba diving is an extreme sport where divers carry their own source of breathing gas to explore the open sea. Table 3.1 lists the compositions by volume of two breathing gases used by scuba divers. Table 3.1 Mixture Composition by volume Nitrox 32% oxygen, 68% nitrogen Heliox 21% oxygen, 79% helium (a) (i) State one assumption of the kinetic theory as applied to an ideal gas. Hence, deduce, with reasons, whether Nitrox or Heliox shows greater devi ation from ideal gas behaviour. [2] The intermolecular forces between the gas particles are negligible/insignificant. Nitrox shows greater deviation as nitrogen has a larger electron cloud size than helium, so its intermolecular forces/ id-id are less negligible /more significant. OR

The particle size /volume is negligible as compared to the size /volume of the container. Nitrox shows greater deviation as nitrogen molecule is larger than helium atom, so its particle size/volume is less negligible/ more significant. (ii) Use of the Data Booklet is relevant to this question. A canister of breathing gas has a density of 435 g m–3 at standard temperature and pressure (s.t.p). Calculate the average molar mass of the mixture. Hence, with reference to Table 3.1, deduce the identity of the breathing gas. [3] pV = nRT 5 (435)(8.31)(273) 10 mpV RTM mRT RTM pV p      = 9.87 g mol–1 The breathing gas is Heliox. (b) At sea level, the atmospheric pressure is 1 atm. In the sea, there is an increase in pressure of 1 atm for every 10 m of depth below the surface. For example, at 10 m, the pressure on the diver's body is 2 atm and at 20 m, the corresponding pressure is 3 atm. When a diver inhales from the gas canister, the air that enters the diver's