JJC 2015 H1 CHEM P2 Ans Prelims

[Turn over JURONG JUNIOR COLLEGE 2015 JC 2 PRELIMINARY EXAMINATION Higher 1 CANDIDATE NAME Suggested Answers CLASS 15S CHEMISTRY 8872/02 Paper 2 Structured Questions 28 August 2015 2 hours Candidates answer Section A on the Question Paper. Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class and exam index number 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. Section A Answer all the questions. Section B Answer two questions on separate answer paper. 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. For Examiner’s Use 1 2 3 4 B5 B6 B7 Total This document consists of 14 printed pages and no blank page.

2 © Jurong Junior College 8872/02/ J2 PRELIMINARY EXAMINATION/2015 [Turn over Section A Answer all the questions. 1 Ethanoic acid, also known as acetic acid, is a colourless liquid that has a strong and distinct pungent and sour smell. Aside fr om culinary uses, as flavouring and as a preservative, ethanoic acid is used to make many polymers and fibres. It can be made by the following method: For Examiner’s Use (a) Draw the structure of compound A. [1] (b) (i) Ethanoic acid can be formed from the oxidation of ethanol. Suggest suitable reagent and conditions to perform this reaction. acidified KMnO4/ K2Cr2O7, heat (ii) Write a balanced equation to represent the reaction in (b)(i). CH3CH2OH + 2[O]  CH3CO2H + H2O [3] (c) 2-chloroethanoic acid can be formed from ethanoic acid using suitable reagents and conditions via substitution reaction. Explain why 2-chloroethanoic acid and ethanoic acid has different acidity. 2-chloroethanoic acid is a stronger acid Cl is an electron withdrawing atom which disperses the negative charge on the O− and stabilises the ClCH2CO2 − ion. [2]

3 © Jurong Junior College 8872/02/ J2 PRELIMINARY EXAMINATION/2015 [Turn over 1 (d) A gas is said to exhibit ideal gas behavior if it obeys the ideal gas equation pV = nRT where p = pressure of gas in Pa, V = volume of gas in m3, n = amount of gas (mol), R = 8.31 J K−1 mol−1 (a constant), and T = temperature in K. For Examiner’s Use Three experiments were performed where different masses of ethanoic acid were expanded into a vessel of volume 0.0150 m 3 at different pressures and at a temperature of 100 oC. Experiment Mass of ethanoic acid/ g Pressure/ Pa Relative molecular mass 1 0.002 6.346 65.1 2 0.015 31.011 100 3 0.050 86.128 120 Accept exact answer 100.0, 120.0 (i) Assuming ideal behaviour, calculate the relative molecular mass of ethanoic acid for each experiment 1 to 3 and enter the values into the above table. (ii) In gaseous state just above the boiling point, the monomer and dimer forms of ethanoic acid exist together in equilibrium. 2CH3COOH  (CH3COOH)2 Draw a fully labelled diagram to illustrate the bond formed when ethanoic acid dimerises. 1 (d) (iii) Use Le Chatelier’s Principle to predict and explain how an increase in pressure will affect the position of equilibrium in d(ii) and the trend in relative molecular mass of ethanoic acid. For Examiner’s Use When pressure increases, equilibrium position will shift to right to reduce the number of gas particles, so as to reduce the increased [4] +- - -+ Hydrogen bonds H3C O: C O  H CH3 H  O C :O −

4 © Jurong Junior College 8872/02/ J2 PRELIMINARY EXAMINATION/2015 [Turn over pressure. The equilibrium mixture will contain a higher proportion of dimers, giving a higher Mr. (e) The salt of ethanoic acid, sodium ethanoate, may be added to food as a seasoning. It is often used to give potato chips a salt and vinegar flavor. The enthalpy change for the solution of sodium ethanoate in water, ∆H1 can be determined by using the energy cycle given below. CH 3COO–(g) + Na+(g) CH 3COONa(s) CH 3COO–(aq) + Na+(aq) ∆H2 = –780 kJ mol–1 ∆H3 = –763 kJ mol–1 (i) Name the enthalpy change that is represented by ∆H3. Lattice energy of CH3COONa (ii) With reference to the energy cycle and the data given above, calculate ∆H1. –763 = –780 – ∆H1 ∆H1 = –17 kJ mol−1 [2] [Total: 12] 2 (a) The element iron is a transition metal that is very common in our planet. It has many uses in different industries and even in the human body. It has four naturally occurring isotopes. isotope Relative abundance / % 54Fe 5.845 56Fe 91.754 57Fe 2.119 58Fe 0.282 For Examiner’s Use (i) Write the full electronic configuration of iron. ∆H2 ∆H3 ∆H1

5 © Jurong Junior College 8872/02/ J2 PRELIMINARY EXAMINATION/2015 [Turn over 1s2 2s2 2p6 3s2 3s2 3p6 3d6 4s2 (ii) Use the relative abundance data to calculate the relative atomic mass of iron to 3 decimal places. Show your working. Ar = 54(0.05845) + 56(0.91754) + 57(0.02119) + 58(0.00282) = 55.910 [2] (b) Iron is used as a catalyst in the manufacture of ammonia in the Haber process. N2(g) + 3H2(g)  2NH3(g) (i) State the optimum industrial conditions of temperature and pressure used in the Haber process. 450oC, 250 atm (ii) Explain why these particular conditions are chosen. At low temperature, even though equilibrium yield of NH 3 is high as equilibrium position in N2(g) + 3H 2(g)  2NH3(g) shifts to the right to favour the exothermic reaction. However, equilibrium is reached at a slower rate. Hence a moderately high temperature is chosen . A high pressure is used as equilibrium position in N2(g) + 3H2(g)  2NH3(g) shifts to the right to reduce number of gas molecules, resulting in high equilibrium yield of NH 3 which is achieved at a faster rate. Pressure which is too high is not used because it would require a higher cost. 2 (b) (iii) With the aid of the Boltzmann dist ribution curve, explain how the presence of a catalyst affects the rate of this process. For Examiner’s Use A catalyst provides an alternative reaction path of lower activation energy. [7]

6 © Jurong Junior College 8872/02/ J2 PRELIMINARY EXAMINATION/2015 [Turn over (A.E. for catalysed rxn) (A.E.for uncatalysed rxn) Thus, the number of molecules with energy  lowered activation energy/Ea’, increases. Therefore, the frequency of effective collisions between molecules increases and hence, the reaction rate increases. (c) Nitrogen is one of the Period 2 elements and it forms the nitride ion, N3−. State and explain how you would expect the radius of the following ions to compare with the radius of nitride ion.  fluoride ion, F −  lithium ion, Li + Both N3− and F− are isoelectronic and have similar shielding effect by inner shell electrons, but F− has a greater nuclear charge than N3−. Hence F− has a smaller radius than N3−. Li+ has one less quantum shell of electrons compared to N3−. Hence shielding effect by inner shell electrons is lesser for Li+. Hence Li+ has a smaller radius than N3−. [2] [Total: 11] 3 Halogenoalkanes are used widely in the industry to produce organic compounds commer