SAJC Prelim P2 MS

Turn Over NAME Class ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION Chemistry (9729) Paper 2 Structured Questions 11 September 2017 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS: Write in dark blue or black pen. You may use an 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. 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. For Examiner’s use: Question 1 2 3 4 5 6 Total Marks 17 5 10 16 6 21 75 This document consists of XX printed pages (including this page).

2 Answer all the questions 1 Sulfur is a common element on Earth that forms many important chemical compounds. One of these compounds is sodium thiosulfate, Na2S2O3, an ionic compound used to treat several medical conditions, such as cyanide poisoning and fungal growths. (a) (i) Draw a dot -and-cross diagram for sodium thiosulfate, Na 2S2O3. State the shape and bond angle in the thiosulfate ion. [2] shape around sulfur – tetrahedral bond angle - 109° 1 mark for correct dot-cross diagram 1 mark for both shape and bond angle Double bond between S-S atoms is accepted Additional electrons can be shown as either dot or cross. Triangle/square/circle/any other shape cannot accept. (ii) Below are the melting points of sodium thiosulfate and sulfur. Compound Melting point / °C Sodium thiosulfate 49 Sulfur, S8 115 Explain why sulfur has a higher melting point than sodium thiosulfate. [3] Sodium thiosulfate has a giant ionic lattice structure with electrostatic forces of attraction between Na+ and S2O32- ions. [1] Sulfur/S8 is a simple non-polar covalent molecules with instantaneous-dipole- induced-dipole (id-id) interactions between its molecules.[1]

3 Turn Over The large number of electrons in each sulfur molecule leads to strong id-id interactions, which require more energy to overcome compared to the ionic bonds in sodium thiosulfate, hence it has a higher melting point than sodium thiosulfate[1] (b) Another important sulfur compound is sulfuric acid, H 2SO4. Before the Contact Process was discovered, concentrated sulfuric acid for industrial purposes was produced by the following method. The mineral pyrite, FeS2, was first heated in air and oxidised to solid Fe 2(SO4)3 and sulfur dioxide gas. Fe2(SO4)3 decomposes at 480 °C to form iron(III) oxide and sulfur trioxide gas. The sulfur trioxide gas could be mixed with any volume of water to produce sulfuric acid of the desired concentration. However, this process was expensive and not efficient. (i) Write a balanced equation, with state symbols for the reaction between pyrite, FeS2, and oxygen to form Fe2(SO4)3. [1] 2FeS2(s) + 7O2(g)  Fe2(SO4)3(s) + SO2(g) [1] (ii) With the aid of an equation, define the term standard enthalpy change of formation of gaseous sulfur trioxide, SO3. [2] Standard enthalpy of formation of gaseous SO3 is the energy released or required when 1 mole of gaseous SO3 is formed from its constituent elements under standard conditions of 298 K and 1 bar. [1] 1/8 S8(s) + 3/2 O2(g)  SO3(g) [1] (iii) Given the following information, determine the enthalpy change of formation of gaseous sulfur trioxide SO3. Fe2(SO4)3(s)  Fe2O3(s) + 3SO3(g) ∆Hrxn = +95 kJ mol–1

4 Substance ∆Hf / kJ mol–1 Fe2(SO4)3(s) –2107 Fe2O3(s) –824 [3] 2 marks for energy cycle 3 x ∆Hf (SO3(g) ) + (–824) = –2107 + 95 3 x ∆Hf (SO3(g) ) = –1188 ∆Hf (SO3(g)) = –396 kJ mol–1 [1] OR ∆Hrxn = ∆Hf(products) – ∆Hf(reactants) = +95 [1] –824 – (–2107 + 3∆Hf(SO3(g) ) = +95 [1 mark for correct substitution of values] 3∆Hf(SO3(g) ) = –1188 ∆Hf (SO3(g)) = –396 kJ mol–1 [1] Fe2(SO4)3(s) Fe2O3(s) + 3SO3(g) 2Fe(s) + 6O2(g) + 3/8S8(s) Fe2O3(s) + 9/2O2(g) + 3/8S8(s) –2107 +95 3 x ∆Hf (SO3(g)) –824

5 Turn Over (iv) Use the appropriate bond energies given in the Data Booklet and the data below to calculate another value for the standard enthalpy change of formation of gaseous sulfur trioxide SO3. ⅛S8(s) + O2(g)  SO2(g) ∆Hf = –297 kJ mol–1 [3] ∆Hf (SO3(g)) = –549 kJ mol–1 2 marks for energy cycle, ½ mark for each correct arrow with reactants and products 1 mark for calculation (v) Suggest a reason for the discrepancy between the values in (b)(iii) and (b)(iv). [1] The bond energy data from the data booklet are only average values and would not apply exactly to particular compounds.[1] (c) The value of pV is plotted against p for 1 mol of oxygen O2, where p is the pressure and V is the volume of the gas at 300 K. 1/8S8(s) + 3/2 O2(g) SO2(g) + ½O2(g) SO3(g) SO2(g) + O(g) ∆Hf (SO3(g)) –297 –BE (S=O) = –500 ½ x BE (O2) = ½ x 496 = +248

6 (i) On the diagram above, draw and label the graph of pV against p for SO 3 at 300 K. [1] Can cut the ideal gas line at any point, but negative and positive deviation must be more than O2. (ii) Explain the difference between the graph of SO3 and the graph of O2. [1] SO3 has more electrons than O2, hence it has stronger instantaneous-dipole- induced-dipole interactions and has greater deviation from ideality. [1] OR SO3 has a larger size, hence the volume of SO3 compared to the total volume occupied by the gas is more significant. [1] [Total: 17] 2 Transition elements show typical properties that distinguish them from s -block elements, such as calcium. These include variable oxidation states in their compounds, and the formation of coloured complex ions. (a) An ion of manganese has 3 electrons in its 3d subshell. Suggest the oxidation state of this manganese ion. [1] +4 [1] pV O2 0 p Ideal gas SO3

7 Turn Over (b) The following t able gives data about some physical properties of the elements calcium, chromium, and manganese. Property Ca Cr Mn Atomic radius (metallic) / nm 0.197 0.129 0.132 Ionic radius (2+) / nm 0.099 0.073 0.083 Melting point / K 1112 1907 1246 Density / g cm–3 1.54 7.19 7.43 Electrical conductivity / x 106 S cm–1 0.298 0.0774 0.00695 (i) Explain why the atomic radii of chromium and manganese are similar to each other. [2] Mn has more proton and hence greater nuclear charge. However, the two paired electrons in the 4s subshell of Mn also experience interelectronic repulsion.[1] The effect on the radius due to nuclear charge is counteracted by the interelectronic repulsion between the two paired electrons in the 4s subshell.[1] Accept Proton number increases and hence nuclear charge increases from Cr to Mn. Shielding effect increases because the electrons are added to the inner 3d subshell. [1] The effect on the radius due to nuclear charge is counteracted by the effect on the radius due to the shielding effect / effective nuclear charge is approximately constant/similar. [1] (ii) Explain why the density of manganese is significantly greater than that of calcium, usi ng relevant data from the table and the Data Booklet. (No calculations are required.) [2] Density = mass / volume Manganese has atomic mass of 54.9, which is greater than the atomic mass of calcium, 40.1. [1] Manganese has atomic radius of 0.132 n m, which is less than the atomic radius of calcium, 0.197 nm. [1] Or

8 Manganese has ionic radius of 0.08