2020 YIJC Prelim Exam Paper 3 Solutions

2020 YIJC Preliminary Examination H2 Chemistry Paper 3 (Suggested Answers) 1 Chlorine is an element in Group 17 of the Periodic Table. It exists as a diatomic gas at room temperature. Chlorine can also be found in many inorganic and organic compounds. (a) Predict the colour of the final solutions when  Cl2(aq) is added to KBr(aq)  I2(aq) is added to KCl(aq) Write equations for any reactions that occur. State clearly if no reaction occurs. [2]  Cl2 + 2Br  2Cl+ Br2 colour of final solution: orange  No reaction colour of final solution: brown (b) Describe and explain the trend in the thermal stabilities of the hydrogen halides HCl, HBr and HI. [2] The thermal stabilities of the hydrogen halides decrease down the Group from HCl to HBr to HI. This is because  the size of the halogens increases from Cl to I and the valence orbital used for bonding is larger and more diffuse;  the effectiveness of the orbitals overlap decreases;  this result in the weaker covalent bond formed between the hydrogen and halogen atoms or quoting H-X bond energies;  lesser amount of energy is required to overcome the covalent bond between the hydrogen and halogen atoms.

(c) Chlorine dioxide, C lO2 was discovered in 1811 and has been widely used for bleaching purposes in the paper industry and for treatment of drinking water. (i) ClO2 can be synthesised from the reaction between NaC lO3 and H2O2 in an acidic medium. Construct a balanced equation for the reaction between ClO3− and H2O2. [2] reduction: ClO3− + 2H+ + e−  ClO2 + H2O oxidation: H2O2  O2 + 2H+ + 2e− 2ClO3− + H2O2 + 2H+  2ClO2 + 2H2O + O2 (ii) In the process of handling C lO2, the concentration of the C lO2 in aqueous solution must not exceed 45 parts per million by mass, as it will cause irritation to the eyes and nose. 1 part per million can be expressed as 1 mg of the substance in 1 kg of water. An aqueous solution of C lO2 was prepared by adding 150 cm 3 of 1.0  103 mol dm−3 NaClO3 to 100 cm3 of 2.0  103 mol dm3 H2O2. Using your equation in (c)(i), c alculate the concentration of the C lO2(aq) in parts per million, State whether it has exceeded the safety limit. [Assume the density of the final solution is 1 g cm3.] [3] Amount of NaClO3 = (1.0  103) × 0.15 = 1.50 × 104 mol Amount of H2O2 = (2.0  103) × 0.10 = 2.00 × 104 mol Limiting reagent: NaClO3 Amount of ClO2 produced = 1.50 × 104 mol Mass of ClO2 produced = (1.50 × 104) × 67.5 = 0.010125 g = 10.125 mg Total mass of solution = 250 g = 0.250 kg [ClO2] in ppm = 10.125 ÷ 0.250 = 40.5 ppm It has not exceeded the safety limit.

The electrode potential for the reduction of chlorine dioxide is shown. ClO2(aq) + 4H+(aq) + 5e− ⇌ Cl−(aq) + 2H2O(l) Eo = +1.50V (iii) State what is meant by the term standard electrode potential. [1] The standard electrode potential of an element is the potential difference between the element and the aqueous solution of its ion at 1 mol dm3 relative to that of the standard hydrogen electrode at 298 K and 1 bar. (iv) Draw a fully labelled diagram of the electrochemical cell you would set up in order to measure the standard reduction potential of ClO2/Cl− in the laboratory. Indicate clearly the positive and negative electrodes. [3] high resistance voltmeter (−) (+) Another cell composing of a standard ClO2/Cl half-cell and a standard Fe3+/Fe2+ half- cell was set up. (v) Calculate ΔGo, in kJ mol–1, for the reaction that occurs [2] ClO2 + 4H+ + 5e−  Cl− + 2H2O Fe2+  Fe3+ + e− 5Fe2+ + ClO2 + 4H+  5Fe3+ + Cl− + 2H2O Eocell = +1.50 – (+0.77) = +0.73 V ΔGo = nFEocell =  5 × 96500 × 0.73 = 352225 J mol1 = 352 kJ mol1 Pt(s) [H+] = 1 mol dm–3 H2(g) at 298K & 1 bar salt bridge Pt(s) [ClO2] = 1 mol dm–3 [Cl‒] = 1 mol dm–3 [H+] = 1 mol dm–3 V

(vi) State and explain what happens to the standard cell potential, Eocell, when chlorine gas is bubbled into the Fe3+/Fe2+ half-cell. [2] When chlorine is bubbled into into the Fe3+/Fe2+ half-cell, Fe2+ will be oxidised to Fe3+ causing [Fe2+] to decrease (or [Fe3+] to increase). As such the Fe3+ + e ⇌ Fe2+ equilibrium shifts to the right and Eoreduction(Fe3+/Fe2+) to become more positive and Eocell less positive. (d) Describe and explain the relative ease of hydrolysis of chloroethane, chlorobenzene and ethanoyl chloride. [3] reactivities: ethanoyl chloride > chloroethane > chlorobenzene Ethanoyl chloride is more reactive than chloroethane towards nucleophilic substitution reactions because the + charge on the carbonyl carbon is higher due to the additional electron withdrawing O atom bonded to it or the reactive carbon of ethanoyl chlori de is sp 2 hybridised and is trigonal planar which has less steric hindrance than the tetrahedral carbon in chloroethane. Chlorobenzene is less reactive than chloroethane because the CCl bond in chlorobenzene is stronger than that in chloroethane as it has a partial double bond character due to the delocalisation of the lone pair of electrons on the C l atom into the benzene or the rear side of the carbon–halogen bond is blocked by the benzene ring and the  electron cloud of the benzene ring will repel the lone pair of electrons of an incoming nucleophile , rendering approach of the nucleophile difficult. [Total: 20]

2 (a) Propanone reacts with iodine in acidic conditions as shown below. CH3COCH3(aq) + I2(aq)  CH3COCH2I(aq) + HI(aq) The rate of the reaction may be followed by measuring the concentration of the iodine at regular time intervals. Four separate sets of experiments were carried out in which the initial concentrations of the reactants were varied as shown in Table 2.1. Table 2.1 experiment initial concentration × 102 / mol dm3 propanone hydrochloric acid iodine 1 5 5 5 2 5 5 10 3 2 5 5 4 5 10 5 The results of experiment 1 and 2 are shown in Figure 2.1. time / min Fig. 2.1 0 1 2 3 4 5 6 7 8 9 10 11 0 10 20 30 40 50 60 70 80 [I2] × 102 / mol dm3 experiment 2 experiment 1

The results shown in Table 2.2 were obtained for experiment 3 and 4. Table 2.2 experiment initial rate / mol dm3 min1 3 2.70  10–4 4 1.25  10–3 (i) Using Fig. 2.1, determine the order of reaction with respect to [ I2]. Draw clearly any construction lines on Fig. 2.1 and show all your working. [1]  To determine the order of reaction with respect to iodine, initial rate of experiment 1 = – (5  2) × 102 0  48 = 6.25  10–4 mol dm–3 min–1 initial rate of experiment 2 = – (10  7) × 102 0  49 = 6.1224  10–4 mol dm–3 min–1 When the concentration of iodine increases by 2 times, the rate of the reaction remains constant. Hence the order of reaction with respect to iodine is 0 order. (ii) Determine the orders with respect to [H