VJC 2020 H2 Chem Prelim P2 Ans

 VJC 2020 9729/02/PRELIM/20 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 ……………………………………………….………….. …………………………….. CHEMISTRY 9729/02 Paper 2 Structured Candidates answer on the Question Paper. 15 September 2020 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the 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 working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 20 2 / 11 3 / 8 4 / 19 5 / 17 Total / 75 This document consists of 16 printed pages.

 VJC 2020 9729/02/PRELIM/20 2 Answer all the questions in the spaces provided. 1 (a) (i) Complete the diagram to show the relative energies of all the electrons in an iron atom. [1] (ii) Write an equation for the second ionisation energy of iron.  Fe+(g)  Fe2+(g) + e [1] (iii) Explain why the second ionisation energy of iron is lower than that of chromium.  For the second ionisation energy, 4s electron is removed from iron while 3d electron is removed from chromium . As 4s electron is further away from the nucleus than 3d electron, less energy is needed to remove it, resulting in a lower second ionisation energy. [1] (b) A piece of rusted iron was analysed to find out how much of the iron had been oxidised to rust (also known as hydrated iron(III) oxide). A small sample of the rusted iron was dissolved in excess dilute sulfuric acid to give 250 cm3 of solution. The resultant solution contains Fe2+ and Fe3+ from iron and rust respectively. (i) 25.0 cm 3 of this solution required 16.90 cm 3 of 0.0200 mol dm 3 KMnO4 for complete oxidation of Fe2+. Calculate the amount of Fe2+ in 25.0 cm3 of the solution. 5Fe2+ + MnO4 + 8H+  5Fe3+ + Mn2+ + 4H2O Amount of Fe2+ in 25.0 cm3 of the solution = 5  0.0200  16.90  103  = 0.00169 mol [1] (ii) To another 25.0 cm3 of the solution, an oxidising agent was added to convert all the Fe2+ ions present to Fe3+ ions. The Fe3+ ions were then titrated with 0.100 mol dm 3 EDTA4 solution and 17.60 cm3 was required. Assuming 1 mol of EDTA4 reacts with 1 mol of Fe3+, calculate the amount of Fe3+ in 25.0 cm3 of the solution. Amount of Fe3+ in 25.0 cm3 of the solution = 0.100  17.60  103  = 0.00176 mol [1] energy 1s  2s  2p    3s  3p    3d      4s 

3  VJC 2020 9729/02/PRELIM/20 [Turn over (iii) From your answers in (b)(i) and (b)(ii), calculate the amount of original Fe3+ in 250 cm3 of the rusted iron solution. Amount of Fe3+ in 250 cm3 solution = (0.00176 – 0.00169)  250/25.0  = 7.00  104 mol [1] (iv) Determine the percentage of iron that had been oxidised to rust in the sample. Percentage of iron that had rusted = 0.0007 0.0176  100%  = 3.98% [1] (c) (i) Explain if a real gas, such as N2O3, behaves more or less ideally at:  high pressures  At high pressures, gas particles are close r together. The gas particles occupy significant volume compared to the volume of the container. (OR F orces of attraction between the gas particles are significant / no longer negligible .) A real gas behaves less ideally at high pressures.  high temperatures  At high temperatures, forces of attraction between the gas particles are insignificant / negligible as the gas particles have sufficient (high kinetic) energy to overcome them. A real gas behaves more ideally at high temperatures. [2] (ii) At room temperature, N2O3 dissociates as shown. N2O3(g) ⇌ NO(g) + NO2(g) This dissociation involves homolytic fission of N–N bond. Draw a ‘dot-and-cross’ diagram for the N2O3 molecule.  [1] (iii) Hence, state the shape around each N atom in the N2O3 molecule.  Bent and trigonal planar [1] (iv) Suggest the common feature of NO and NO 2 that allows each of them to undergo dimerisation to form N2O2 and N2O4 respectively.  Both molecules have a single unpaired electron on N atom . ( OR Both are free radicals.) [1]

 VJC 2020 9729/02/PRELIM/20 4 (d) (i) With an appropriate sketch of Boltzmann distribution, explain how a catalyst is able to increase the rate of a chemical reaction for a given temperature T.   A catalyst speeds up the rate of reaction by providing a different/alternative reaction path which has a lower activation energy. As shown on the diagram , the number of molecules with energy greater or equal to the lowered activation energy Ea(cat) will increase. This results in an increase in the frequency of effective collisions. Hence, the rate of reaction increases. [2] (ii) On the sketch in d(i), draw a new distribution curve at a higher temperature T’, clearly labelled T’. [1] (e) X is an organic liquid that is found as a contaminant in crude oil. It contains only C, H and S. When 0.841 g of X was subjected to complete combustion, 1.76 g of CO 2, 0.360 g of H 2O and a gas which can decolourise acidified potassium manganate (VII) were formed. (i) What is the empirical formula of X? mC + mH + mS = 0.841 g mC = 12.0 44.0  1.76 = 0.480 g mH = 2 x 1.0 18.0  0.360 = 0.040 g mS = 0.841 – 0.480 – 0.040 = 0.321 g Element C H S  Mass 0.480 0.040 0.321 Mole ratio 0.480  12.0 = 0.04 0.040  1.0 = 0.04 0.321  32.1 = 0.01 Simplest mole ratio 4 4 1  Empirical formula is C4H4S. [2] (ii) X is unsaturated and has a structure in which all the carbon and sulfur atoms are connected together in a five-membered ring. Draw the structure of X.  [1] (iii) Hence, construct the equation for the complete combustion of X.  C4H4S + 6O2  4CO2 + 2H2O + SO2 [1] Ea Ea(cat) number of molecules energy No. of molecules with energy greater or equal to Ea No. of molecules with energy greater or equal to Ea(cat) T’ T

5  VJC 2020 9729/02/PRELIM/20 [Turn over (iv) X behaves similarly like benzene in its reaction s with electrophile. Suggest a poss