MI 2008 PU2 Promotional H1Chem P2 ans

Class Adm No Candidate Name: _______________________ Promotional Examinations 2008 Pre-university 2 CHEMISTRY HIGHER 1 8872/2 Monday 8th Sept 2008 2 hours Additional materials: Answer paper Data Booklet 5 pieces of foolscap paper Cover page READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, index number and class 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. Section A Answer all questions. Section B Answer any two questions on the writing papers. At the end of the examination, hand in your answers to Section A and Section B separately. The number of marks is given in brackets[ ] at the end of each question or part question. For Examiner's Use Section A B5 B6 B7 Total This question paper consists of 16 printed pages.

Section A: Structured Questions (40m) Answer ALL questions in the spaces provided. 1. Acidified KMnO 4, which oxidizes iron(II) compounds, is used to estimate the amount of iron present in substances such as flour. 1.00g of flour was shaken vigorously with dilute H2SO4 and the volume of the solution was made up to 100cm3 in a volumetric flask. A 10.0cm3 portion of this solution was titrated against 5.00x10 -6 moldm-3 of KMnO4 solution. 11.00cm3 of the KMnO 4 solution was required to completely oxidized the iron(II) salt to iron(III) salt. (i) Write a balanced equation to show the above reaction. [2] Fe2+ Æ Fe3+ +e MnO4 - + 8H+ +5e Æ Mn2+ + 4H2O Overall equation : 5Fe2+ + MnO4 - + 8H+ Æ 5Fe3+ + Mn2+ + 4H2O ; ecf given (ii) Calculate the concentration of Fe2+ in the flour solution. [2] No. of mol MnO4 - = 5.00x10-6 x 0.011 = 5.50x10-8 No. of mol Fe2+ in 10.0cm3 = 5.50x10-8 x 5 = 2.75x10 -7 ; ecf given No. of mol of Fe2+ in 100cm3 = 2.75x10-7 x 10 = 2.75x10-6 Concentration of Fe2+ = 2.75x10-6 x (1000/100) = 2.75x10-5 moldm-3 ; ecf given [Turn over 2 (iii) Hence, determine with calculation, if this flour meets the requirement of containing 16.5mg of iron per 100g of flour by law. [2] 16.5mg = 1.65 x 10-2g Mass of Fe in 1.00g of flour = 2.75x10-6 x 55.8 = 1.535 x 10-4g Mass of Fe in 100g of flour = 1.535 x 10-4 x 100 = 1.535 x 10-2 g ; The flour does not meet the requirement. ; 2 half-equation 1m marks given only with calculation [Total marks: 6m]

2. Sulphur and its isotopes are mainly used in medical applications. Sulphur also known as brimstone has been used for thousands of years to cure skin ailments. Ancient cultures relied on sulphur to treat a variety of illnesses as well as to make dyes and other useful products. (a) The relative isotopic mass and percentage abundances of the isotopes of a sample of sulphur are shown in the table. relative isotopic mass % abundance 32S 94.93 34S 5.07 Calculate the relative atomic mass for sulphur in the sample. [1] Relative atomic mass = 32 x 0.9493 + 34 x 0.0507 = 32.1 ; (b) Oxides of sulphur contribute towards air pollution. Draw a dot and cross diagram to show the bonding in a molecule of sulphur trioxide, SO 3. Hence state the shape and bond angle of SO3 [ 3 ] Shape of molecule : Trigonal planar, 120o ; ; S as the central atom. Each O atom forms a double bond with sulphur. 12 valence electrons surrounding the central atom. ; (c) The melting points of silicon dioxide and sulphur trioxide are shown as follows: compound melting point / oC SiO2 2830 SO3 17 Briefly relate these melting points to the structure and bonding in each of these oxides. [3] SiO 2 has a giant covalent structure with strong covalent bonds between silicon and oxygen atoms. ; SO 3 has a simple covalent structure with weak Van Der Waals forces of attraction between its molecules. ; Hence, silicon dioxide requires more energy to overcome the strong covalent bonds between atoms as compared to the weak Van Der Waals forces of attraction between sulphur trioxide molecule. Melting point of silicon dioxide will be higher. ; [Turn over 3

(d) The lattice energy of sodium oxide is – 2000 kJ mol -1. Explain how you would expect the numerical magnitude of the lattice energy of sodium sulphide, Na2S, to compare with that of sodium oxide, Na2O. Hence, predict the relative melting points of Na2O and Na2S . [ 2 ] Lattice energy is proportional to - - r r q q α L.E. ++ + Since the product of the charges of the cation and anion is the same for both compound but the ionic radius of sulphide ion is larger than that of oxide ion due to an increased in the number of shells of electrons, hence, the lattice energy of sodium sulphide will be less exothermic since the electrostatic forces of attraction is weaker . ; (Writing this explanation, students have to write formula for lattice energy). Hence sodium sulphide will have have a lower melting point since its ions are held together by a slightly weaker electrostatic force as compared to sodium oxide. ; Alt explanation Comparing the lattice energy of Na2O and Na2S, the lattice energy of Na2S will be less exothermic because the interionic distance of Na2S will be larger than that of Na2O due to the increase in size of the anion. Hence, the forces of attraction is weaker ; Hence sodium sulphide will have a lower melting point since its ions are held together by a slightly weaker electrostatic force as compared to sodium oxide. ; [Total marks: 9m] [Turn over 4

3. (a) A dilute aqueous solution of hydrogen peroxide, used to bleach hair, decomposes as shown by the equation. 2H2O2(aq) Æ 2H2O(g) + O2(g) The decomposition of hydrogen peroxide is known to be a first order reaction. (i) Explain the meaning of the phrase in Italics. [1] A first order reaction is a reaction whereby the rate of reaction is proportional to the concentration of the reactant, in this case H2O2. ; (ii) Hence, write a rate equation to show the decomposition of hydrogen peroxide. [1] Rate = k[H2O2] ; (b) Graphical method can also be used to determine the order of reaction for the decomposition of hydrogen peroxide. Hence, sketch (i) a graph of concentration of hydrogen peroxide against time (ii) a graph of rate of reaction against concentration of hydrogen peroxide to show that the decomposition of hydrogen peroxide is a first order reaction. [3] (i) (ii) Show constant half-life ; Shape of graph ; ; [H2O2] [Rate] [H2O2] [Turn over 5

(c) (i) Calculate the standard enthalpy change of reaction for the decomposition of hydrogen peroxide. [ Given : ΔHf o (H2O2) = -187.83 kJ mol-1, ΔHf o(H2O) = -285.83kJ mol-1 ] [2] ΔHrxn = [2(-285.83) + 0] – [2(-187.83)] = -196 kJ mol-1 (ii) Hence, sketch a clearly labelled energy profile diagram to illustrate the decomposition of H2O2. [ 1 ] ; Ecf given [Total marks: 8m] [Turn over 6

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4. Trifluroethyl vithyl ether was the first fluorinated volatile compound to be used in anaesthetia. A series of fluorinated compounds was investigated by Robbins(1946) who observed that the anaesthetic potency of a fluorinated compound was greatly increased