SAJC Prelim P3 MS

1 [Turn over NAME Class ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION Chemistry (9729) Paper 3 Free Response 15 September 2017 2 hours Additional Materials: Data Booklet, Writing Paper READ THESE INSTRUCTIONS: Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Marks [60] Section B Answer one question. Marks [20] 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 XX printed pages (including this page).

2 [Turn over Section A Answer all the questions in this section. 1 (a) Iodine is a lustrous purple-black solid at standard conditions that sublimes readily to form a violet gas. (i) Give the equation that represents the second ionisation energy of iodine. [1] I+(g) I 2+ (g) + e [1] H>0 is not required since this is not a definition question. (ii) The graph below shows the second ionisation energies of eight elements with consecutive atomic number. Which of the above elements, A to H, is iodine? Explain your answer. [2] Element E. [1] The sharp drop from G to H indicates that G is from Group 1 where the second ionisation energy involves the removal of an electron from the inner principal quantum shell, which is nearer to the nucleus, requiring more energy. [1] Hence Element E is in Group 17 and is iodine. (iii) Explain the trend in second ioni sation energies from elements A to G, including the irregularity for element B. [4] A B C D E F G H Second Ionisation energy/ kJ mol-1 Proton number

3 [Turn over The second IE generally increases from A to G because from A to G, there is an increase in proton number and hence nuclear charge while shielding effect is almost constant as electrons are added to the same quantum shell. [1] Hence effective nuclear charge increases and the attraction for the outermost electron becomes increasingly stronger. [1] More and more energy is required to remove the strongly attracted valence electron as we move across the period. Second IE of B is lower than that of A because the electron is removed from the valence 5p subshell which is further away from the nucleus compared to the 5s subshell in A. [1] It also experiences additional screening effect by the two 5s. These factors outweigh the effect of increase in nuclear charge from A to B, resulting in a weaker attraction by the nucleus.[1] Less energy is required to remove the outermost electron in B than that in A. (iv) Suggest, with reason, which of the above elements, A to H, can form an amphoteric oxide. Hence, write an equation to show the reaction of this amphoteric oxide with hydrochloric acid. (You are not required to provide the identity of the element in the equation.) [2] Element A has amphoteric oxide since it is in Group 13 [1] and its oxide would have both ionic and covalent character. A2O3 + 6HCl  2ACl3 + 3H2O [1] or In2O3 + 6HCl  2InCl3 + 3H2O [1] No ecf from part (ii) (b) Hydrogen peroxide reacts with acidified iodide ions to liberate iodine, according to the following equation: H2O2(aq) + 2H+(aq) + 2I(aq)  2H2O(l) + I2(aq)

4 [Turn over The rate of reaction can be measured by the increase in the concentration of iodine formed over time. The reaction was determined to be zero order with respect to [H+]. The following results were obtained by varying the concentrations of hydrogen peroxide and iodide ions. Expt Initial [H2O2(aq)] / mol dm-3 Initial [I(aq)] / mol dm-3 Initial rate / mol dm3 min1 1 0.020 0.040 1.2 x 10-4 2 0.020 0.050 1.5 x 10-4 3 0.050 0.040 3.0 x 10-4 4 0.020 0.500 1.5 x 10-3 5 0.050 1.000 7.5 x 10-3 (i) What is understood by the terms order of reaction and half-life. [2] The order of reaction with respect to a reactant is the power to which the concentration of that reactant is raised in the experimentally determined rate equation. [1] Or Let the Rate = k[A]m[B]n, where m and n are the order of reaction wrt [A] and [B] respectively. [1] Half-life, t½, is the time taken for the reactant concentration to decrease to half of its original value.[1] (ii) Determine the order of the reaction with respect to [H2O2] and [ I] and hence suggest the units of the rate constant of this reaction. [3] Let rate= k[H2O2(aq)]m[I(aq)]n Compare experiments 1 & 2, keeping [H2O2(aq)] constant nm nm 4 4 )050.0()020.0(k )040.0()020.0(k 10x5.1 10x2.1   (or use inspection method)

5 [Turn over n = 1 Rate of reaction is 1st order with respect to [I(aq)] [1] Compare experiments 1 & 3, keeping [I-(aq)] constant nm nm 4 4 )040.0()050.0(k )040.0()020.0(k 10x0.3 10x2.1   (or use comparing method) m = 1 Rate of reaction is 1st order with respect to [H2O2(aq)] [1] Hence, rate= k[H2O2(aq)][I-(aq)] units of k = mol–1 dm3 min–1 [1] (iii) The half-life of hydrogen peroxide in experiment 4 was 9.24 min. Predict the half-life of hydrogen peroxide in experiment 5. [1] For Expt 4 and 5, since [ I(aq)] >> [H 2O2(aq)], [ I -(aq)] is approximately constant. Thus, rate = k’[H2O2(aq)] (a pseudo first order reaction) where k’ = k[I-(aq)] t1/2 = ][k 2ln 'k 2ln  I t1/2 of H2O2 in experiment 4 = 9.24 min (for [I(aq)] = 0.500 mol dm-3) t1/2 of H2O2 in experiment 5 = 4.62 min (for [I(aq)] = 1.00 mol dm-3) [1] To investigate the rate of the above reaction, a teacher suggested titrating the iodine formed with sodium thiosulfate solution at specified time intervals. (iv) Write an equation for the reaction between iodine and thiosulfate. [1]

6 [Turn over I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) [1] State symbols not required but penalized when wrong ones are provided. (v) Suggest how the reaction can be quenched at specified time intervals. [1] The reaction can be quenched by:  adding NaOH/NaHCO3/Na2CO3 to remove the H+(aq)  sudden cooling of the reaction mixture  sudden dilution through the addition of large volume of water any of the above method [1] (vi) With reference to the Data Booklet, explain why hydrochloric acid is not a suitable acid used for the reaction between hydrogen peroxide and iodide. [2] Cl2 + 2e− 2Cl− Eo = + 1.36 V H2O2 +2H+ +2e− 2H2O Eo = + 1.77 V Eocell = +1.77 – 1.36 = +0.41 V [1 for quoting and calculating] H2O2 can oxidise chloride to chlorine [1] while it itself is reduced to H 2O. Hence, the oxidation of iodide may not be complete. [Total:19] 2 Metals have been used widely since ancient times. (a) An electrochemical cell is constructed using solutions of NaHSO4, H2SO3, and MnSO4 with suitable electrodes. The relevant half reactions are:

7 [Turn over HSO4– (aq) + 3H+(aq) + 2e– H2SO3(aq) + H2O Eo = +0.17 V Mn2+(aq) + 2e– Mn(s) Eo = –1.18 V (i) Draw a fully labelled diagram of the above electrochemical cell to measure the cell potential under standard conditions, indicating clearly the direction of the electron flow in the external circuit. [3] [1] – e- flow [1] – voltmeter, salt bridge (can BOD from diagram) + Mn and Pt electrodes [1] – label all the solutions (HSO4- , H2SO3, H+) & 1 mol dm-3 and temperature (ii) Write a balanced equation for the reaction that would take place if the electrodes