PJC H2 CHEM P3 (Answers)

PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION HIGHER 2 CANDIDATE NAME CT INDEX GROUP NUMBER CHEMISTRY 9647/03 Paper 3 Free Response 24 September 2014 2 hours Candidates answer on separate paper. Additional Materials: Answer Paper Graph Paper Data Booklet Cover Page READ THESE INSTRUCTIONS FIRST Write your name, CT group and index number on all work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough workings. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer any four questions. A Data Booklet is provided. You are reminded of the need for good English and clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At then end of the examination, fasten all your work securely together. This document consists of 20 printed pages. 1 3

© PJC 2014 9647/03/JC2 Prelim/2014 2 Answer any four questions. 1 (a) The graph below shows the second and th ird ionisation energies for the first d-block elements scandium to zinc. Ionisation energy / kJ mol -1 By writing relevant electronic configuration, explain why (i) the second ionisation energy of ch romium is higher than that of manganese. Cr + :1s2 2s2 2p6 3s2 3p6 3d5 Mn+ :1s2 2s2 2p6 3s2 3p6 3d5 4s1 The second electron removed from chro mium is from an inner 3d subshell which is closer to the nucleus and required a higher energy to remove it. (ii) the third ionisation energy of iron is lower than that of manganese. Fe 2+ : 1s2 2s2 2p6 3s2 3p6 3d6 Mn2+ : 1s2 2s2 2p6 3s2 3p6 3d5 The third electron from Fe experiences inter-electronic repulsion as it is removed from one 3d orbital containing a pair of electrons and required a lower energy to remove it. [4] (b) An aqueous solution contains a mixture of iron(III) and zinc ions. (i) Draw a diagram to show the bonding in the hexaaquairon(III) complex ion. 2nd IE 3rd IE

© PJC 2014 9647/03/JC2 Prelim/2014 3 . Fe H2O H2O H2O H2O OH2 OH2 3+ (ii) Describe, in a sequence of steps, how you would separate the two cations so that they are obtained as Fe 3+(aq) and Zn2+(aq). Write equations for all the reactions that iron(III) ion and its compound have undergone. You are only provided with HNO3(aq) and NaOH(aq). 1. To 2 cm 3 of solution in a test tube, add NaOH(aq) until in excess 2. Filter the mixture 3. Add excess HNO3(aq) to the residue to obtain Fe3+(aq) 4. Add excess HNO3(aq) to the filtrate to obtain Zn2+(aq) Precipitation reaction: Fe3+(aq) + 3OH(aq) → Fe(OH)3(s) Acid Base reaction: Fe(OH)3(s) + 3HNO3(aq) → Fe(NO3)3(aq) + 3H2O(l) [5] (c) An aqueous iron( III) solution can be used as a homogeneous catalyst for the reaction between iodide ions and peroxodisulfate ions, S2O8 2. By considering relevant Eo values from the Data Booklet, describe and explain the role of iron( III) ions in the reaction between I and S 2O8 2. Write equations and calculate the Eo cell for the reactions that occur. [4] S2O8 2–(aq) + 2e– 2SO4 2–(aq) E = +2.01V Fe3+(aq) + e– Fe2+(aq) E = +0.77V I2(aq) + 2e– 2I –(aq) E = +0.54V Step 1: Formation of intermediate 2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(aq) E cell = +0.77 + (-0.54) = +0.23 V > 0 Step 2: Regeneration of the catalyst 2Fe2+(aq) + S2O8 2–(aq) → 2Fe3+(aq) + 2SO4 2–(aq) E cell = 0.77 + (+2.01) = +1.24 V > 0 Reaction is energetically feasible.

© PJC 2014 9647/03/JC2 Prelim/2014 4 The ease of interconversion between the +2 and + 3 oxidation states of iron and the reaction of oppositely charged ions (Fe3+ and I as well as Fe2+ and S2O8 2) in the same physical state provide an alternative reaction pathway of lower activation energy. (d) The kinetics of the uncatalysed reaction between peroxodisulfate ions and iodide ions can be investigated experimentally. S 2O8 2(aq) + 2I(aq) → 2SO4 2(aq) + I2(aq) To find the rate equation: rate = k[S 2O8 2(aq)]a[I(aq)]b for this reaction, a continuous method with sampling is used. In an experiment, 50.0 cm 3 of 0.200 mol dm 3 of aqueous sodium iodide was mixed with 50.0 cm 3 of 2.00 mol dm 3 aqueous sodium pero xodisulfate. At various time intervals, 10.0 cm 3 of the reaction mixt ure was withdrawn and quenched with 50 cm 3 of ice-cold water. The re sultant mixture was titrated against 0.0250 mol dm3 aqueous potassium thiosulfate, K2S2O3, using starch as an indicator. The reaction between thiosulfate and iodine is as follows: 2S 2O3 2 + I2 → S4O6 2 + 2I The results are shown below: Time / min 0 2 4 6 12 16 Volume of K2S2O3 (aq) / cm3 0 9.50 17.00 22.50 32.25 35.50 (i) Show that 40.00 cm3 of aqueous potassium thiosulfate is required to react with 10.0 cm 3 of the reaction mixture when the reaction between peroxodisulfate and iodide ions is complete. n(I ) = (50.0/1000)(0.200) = 0.010 mol n(S2O8 2) = (50.0/1000)(2.00) = 0.10 mol n(I2) produced in 100 cm3 of reaction mixture = 0.010 / 2 = 0.00500 mol n(I2) in 10.0 cm3 = 0.00500 / (100/10.0) = 0.000500 mol n(S2O3 2) = 0.000500 x 2 = 0.00100 mol volume of S2O3 2 needed = 0.00100 / 0.0250 = 0.040 dm3 = 40.00 cm3

© PJC 2014 9647/03/JC2 Prelim/2014 5 (ii) By drawing a suitable graph, use it to show that the reaction is first order with respect to I-. (iii) The order of reaction with respect to S 2O8 2 is reported to be one. You are required to conduct a second experim ent using the same experimental procedures to confirm the order of reaction. Suggest suitable concentrations of aqueous sodium iodide and sodium peroxodisulfate to be used and explain how the data obtained could be used to confirm the order of reaction. Mix 50.0 cm 3 of 0.200 mol dm-3 of aqueous sodium iodide with 50.0 cm 3 of 4.00 mol dm 3 (or any other appropriate concentration) aqueous sodium peroxodisulfate and plot a volume of S 2O3 2 needed against time, the gradient of the graph at time = 0 should be doubled compared to the first experiment if it is first order wrt S2O8 2. Or Mix 50.0 cm 3 of 0.200 mol dm 3 of aqueous sodium peroxodisulfate with 50.0 cm3 of 2.00 mol dm 3 aqueous sodium iodide and plot a volume of S2O3 2 needed against time, the half life shou ld be constant if it is first order wrt S2O8 2-. [7] [Total: 20]

© PJC 2014 9647/03/JC2 Prelim/2014 6 2 The halogens are an important class of inorganic elements that forms a large variety of halogen-containing products, many of which are useful to us. Three members of the series, namely chlorine, bromine and iodine, were discovered in the 19 th century by Humphry Davy, Antoine-Jérôme Balard and Bernard Courtois respectively. (a) A student carried out a series of redox reactions on three unknown halogens, R2, S2 and T2, and their respective halides. Aqueous solutions of R2 and S2 are brown in colour, while an aqueous solution of T2 is colourless. For each experiment, an unknown halogen was added to a solution containing an unknown halide. This was fo llowed by the addition of tetrachloromethane to the resultant solution. The following table shows the result s an