2021 SAJC Prelim P3 Answer

ST ANDREW’S JUNIOR COLLEGE JC2 Preliminary Examinations Higher 2 CANDIDATE NAME CLASS 2 0 S CHEMISTRY Paper 3 Free Response Candidate answer on the Question Paper. Additional Materials: Data Booklet 9729/03 15 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. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all the questions. Section B Answer one question. 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 39 printed pages. For Examiner’s Use Q1 24 Q2 17 Q3 19 Q4/5 20 Total 80

Section A Answer all the questions in this section. 1 (a) Iodates are compounds that contain the IO3 anion. (i) Draw the dot-and-cross diagram of IO3. [1] (ii) Use your knowledge of VSEPR theory to name the shape of and state the bond angle for IO3. Explain your reasoning. [3] Electron pairs repel each other and arrange themselves as far apart as possible to maximise stability and minimize (electrostatic) repulsion Lone pair-lone pair repu lsion are strong er than lone pair -bond pair rep ulsion which are stronger than bond pair-bond-pair repulsion. IO3 has trigonal pyramidal[√] shape and 107o since there are 3 bond pairs and 1 lone pair of electrons around the central atom I. (iii) Explain why BrO3 has a larger bond angle than IO3. [1] Br is more electronegative than I, hence it draws the bond pair of electrons closer to itself, resulting in greater bond pair-bond pair repulsion, and a larger bond angle. (b) The decomposition of hydrogen peroxide, H2O2, can be catalysed by acidified IO3. 2H2O2 → 2 H2O + O2 With the aid of relevant data from the Data Booklet and the information below, show that IO3 is a suitable catalyst for the decomposition of H2O2 under standard conditions. IO3 + 6H+ + 5e⇌  1 2I2 + 3H2O Eo = +1.19 In your answer, give relevant equations for the reactions that occur. [3] Step 1 O2 + 2H+ + 2e– → H2O2 Eo = +0.68 5H2O2 + 2H+ + 2IO3– → I2 + 5O2 + 6H2O [1/2 for either half-eqn or full eqn] Eocell = +1.19 – (+0.68) = +0.51 V IO O O x x x x x x x x xx x x x x x xx xx

Step 2 H2O2 + 2H+ + 2e– → 2H2O Eo = +1.77 5H2O2 + I2 → 4H2O + 2IO3 + 2H+ [1/2 for either half-eqn or full eqn] Eocell = +1.77 – (+1.19) = +0.58 V (c) The kinetics of the catalytic decomposition of H2O2 by IO3 can be investigated. 2H2O2(aq)   2H2O(l) + O2(g)  The concentration of H 2O2 remaining can be determined by titrating with standard acidified KMnO4. Briefly outline how you would determine rate of the catalysed decomposition of H2O2 using acidified KMnO4. [3] Add (1 cm3 or any small amount < 10 cm3) of IO3 catalyst to a solution of H2O2 and start the stopwatch. At regular time interval (of 5 min), pipette 25.0 (or 10) cm3 aliquot of the reaction mixture and quench using a large volume of cold water. Titrate the quenched mixture with the standard acidified KMnO4 Since volume of KMnO4 used  amount of H2O2 remaining, plot the graph of volume of KMnO4 used against time. (Instantaneous) rate is found by drawing a tangent to the curve and finding its gradient g1, where rate = g1.

(d) (i) A student collects some data for the reaction of H2O2 with acidified IO3, as shown in Table 1.2. Experiment [H2O2]/ mol dm3 [IO3]/ mol dm3 [H+]/ mol dm3 Initial rate/ mol dm3 s1 1 0.050 0.070 0.025 1.47 x 105 2 0.100 0.070 0.050 2.94 x 105 3 0.100 0.140 0.025 5.88 x 105 4 0.150 0.140 0.025 8.82 x 105 Table 1.2 Determine the order of reaction with respect to [H2O2], [IO3] and [H+]. Show your reasoning. [3] Let Rate = k[H2O2]x[IO3-]y[H+]z Comparing expt 3 and 4, when [H2O2] is 1.5 times, initial rate is 1.5 times. First order wrt [H2O2] Comparing expt 1 and 3, Rate3 Rate1 = k[0.100]1[0.140]𝑦[0.025]z k[0.0500]1[0.070]𝑦[0.025]z y = 1 First order wrt [IO3-] Comparing expt 1 and 2, Rate2 Rate1 = k[0.100]1[0.070]1[0.050]z k[0.050]1[0.070]1[0.025]z z = 0 Zero order wrt [H+] (ii) Hence, write the rate equation for the reaction, and calculate a value for the rate constant using experiment 1. Include units in your answer. [2] Rate = k[H2O2][IO3-] k = 1.47 x 10-5/(0.05)(0.07) = 0.0042 mol-1dm3s-1 (e) (i) NH4IO3 is an unstable compound that readily decomposes when warmed as shown: NH4IO3(s) → 1 2N2(g) + 1 2O2(g) + 1 2I2(s) + 2H2O(l) ΔHr o Using the information given below in Table 1.1, calculate the standard enthalpy change of ΔHr o for the above reaction. [1]

Substance Standard enthalpy change of formation / kJ mol1 NH4IO3  417.4 H2O  286 Table 1.1 ΔHo= ΔHf(products) – ΔHf(reactants) = 2(286)  (417.4) = 154.6 kJ mol1 (ii) Explain how the value and sign of ΔGr o would compare to the value and sign of ΔHo for the decomposition of NH4IO3. [2] Since there is an increase in the number of gaseous molecules (from 0 to 1 mol) hence increase in disorderliness of system, ∆So > 0. Given that ∆Ho < 0 and ∆Go = ∆Ho - T∆So, ∆Go will have a negative sign and be of a bigger value than ∆Ho. (f) (i) When 4.00 g a Group 2 metal iodate was heated strongly, 0.947 g of a metal oxide, a purple gas and a colourless gas which rekindles a glowing splint were produced. Determine the identity of the Group 2 metal. [1] Let x be the Ar of the Group 2 metal 4.00 x+2(126.9+16.0x3) = 0.947 x+16 x = 87.5 which is close to Ar of Strontium (ii) Using your answer in (f)(i), write a balanced equation, with state symbols, for the decomposition of the Group 2 iodate. [1] Sr(IO3)2(s)  SrO(s) + I2(g) + 5 2O2(g)

(iii) To deduce which compound, calcium iodate or barium iodate, has a higher decomposition temperature. The following explanation was provided by a student: 'Calcium iodate has a higher decomposition temperature than barium iodate. The Ca2+ ion is a smaller ion than Ba 2+, hence the lattice energy of calcium iodate is more exothermic than that of barium iodate.' Comment on why the student’s answer was wrong and hence suggest a more appropriate answer. [3] Lattice energy should not be used to determine relative thermal stability of Group 2 iodates and barium iodate should have a higher decomposition temperature. Barium iodate has a higher decomposition temperature because Ba2+ has larger ionic radius and hence has a lower charge density and lower polarising power. Hence, Ba2+ polarises the electron cloud of the iodate ion to a lesser extent, weakening the I–O bonds to a lesser extent in barium iodate. [Total:24 marks]

2 Lithium is one of the most abundant elements on the Earth, and has gained widespread use in a variety of applications, from chemical synthesis to battery technology. (a) Lithium aluminium hydride is often in reactions to reduce organic compounds. However, it is not the only reducing agent available, and different reducing agents work on different functional groups. Compound A Draw the structures of the organic products formed when Compound A reacts with the following reducing agents: (i) LiAlH4 [1] (ii) NaBH4 [1] (iii) H2 with Ni catalyst, heat [1]

(b) Fig 2.1 shows how the standard cell potential between the Li +/Li half -cell and the Cl2/Cl– hal