MI H1 CHEM P2 Answer

Class Adm No Candidate Name: This question paper consists of X printed pages and X blank page. 2015 Promotional Examination II Pre-university 2 H1 CHEMISTRY 8872 / 02 Paper 2 Structured Questions 16 Sept 2015 2 hours Candidates answer Section A on the Question Paper Additional Materials: Cover page Data Booklet Writing paper READ THESE INSTRUCTIONS FIRST 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 writing paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluids. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all the questions. Section B Answer two questions on the separate writing papers. Start each new question on a fresh sheet of writing 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 Section A Section B Total Question No 1 (12m) 2 (11m) 3 (7m) 4 (10m) 5 (20m) 6 (20m) 7 (20m) 80 Marks Obtained

2 Section A (40 marks) Answer ALL questions in this section in the spaces provided. 1 Phosphorus is an essential part of life for both human beings and plants. Phosphorus compounds can be found in the minerals in bones and teeth. Phosphorus is also a vital element for plants which is present in the form of phosphates in the fertiliser to maximise the growth of the plant. For Examiner’s Use (a) Phosphoric acid reacts with potassium hydroxide as shown in the following reaction. H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(l) Some thermochemical data are shown below. ΔHθ/ kJ mol-1 Standard enthalpy change of formation of H3PO4(aq) -1277 Standard enthalpy change of formation of KOH(aq) -482 Standard enthalpy change of formation of K3PO4(aq) -1950 Standard enthalpy change of formation of H2O(l) -286 (i) Define standard enthalpy change of formation of H3PO4 (aq). [1] The standard enthalpy change of formation of aqueous H3PO4 is the enthalpy change when one mole of aqueous H3PO4 is formed from its constituent elements in their standard states (H2, P 4 and O2) under standard conditions. (ii) Calculate the standard enthalpy change for the reaction between phosphoric acid and potassium hydroxide based on the information provided. [2] ΔHθ rxn = -1950 + (-286 x 3) – [(-1277) +(-482 x 3)] ; = -85.0 kJ mol-1 ; (b) Phosphorus can react with chlorine to form phosphorus pentachloride and phosphorus trichloride. (i) The boiling point s of phosphorus and chlorine are 280 °C and –34 °C respectively. Explain, in terms of structure and bonding, the difference in boiling points observed in phosphorus and chlorine. [2] Both phosphorus and chlorine have simple molecular structures and are held

3 [Turn over by temporary dipole-induced dipole forces of attraction between molecules.; More energy is required to overcome the stronger temporary dipole-induced dipole forces of attraction between P 4 molecules due to the larger electron cloud size of P4 compared to Cl2. ; For Examiner’s Use (ii) Draw a molecule of phosphorus pentachloride, PC l5, showing clearly its shape. Indicate the bond angles on the molecule. [1] P Cl Cl Cl Cl Cl 120 o 90 o

4 (iii) PCl3 and PC l5, both important in the production of other phosphorous compounds, coexist in equilibrium through: PCl3(g) + Cl2(g) ⇌ PCl5(g) At 250 C, 0.289 g of PC l3 and 2.16 g of C l2 were mixed in a 2.50 dm 3 flask. The equilibrium mixture was found to contain 0.105 g PCl5. Determine the value of Kc. [3] Initial amount of PCl3 = 0.289 31.0+35.5×3=0.002102 mol Initial concentration of PCl3 = 0.002102 2.5 = 0.000841 mol dm-3 Initial amount of Cl2 = 2.16 35.5×2 = 0.03042 mol Initial concentration of Cl2 = 0.03042 2.5 = 0.01217 mol dm-3 ; Amount of PCl5 at equilibrium= 0.105 31.0+35.5×5 = 0.0005036 mol Equilibrium concentration of PCl5 = 0.0005036 2.5 = 0.0002014 mol dm-3 Equilibrium concentration of Cl2 = 0.01217 - 0.0002014 = 0.0119 mol dm-3 Equilibrium concentration of PCl3 = 0.000841 - 0.0002014 = 0.000640 mol dm-3 ; Kc= [PCl5] [Cl2][PCl3] = 0.0002014 0.0119×0.000640 = 26.4 mol-1 dm3 ; (final answer with units)

5 [Turn over (c) Radioactivity, or radioactive decay, follows first -order kinetics and it involves the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. An example of radioactive isotope of phosphorus is 32P which is used as biochemical tracer in the identification of malignant tumours. 32P has a half-life of 14 days. For Examiner’s Use (i) Define the term half-life of a reaction. [1] Half-life is the time taken for the concentration of reactant to fall to half of its initial concentration. (ii) If the initial mass of 32P present was 0.0168g, calculate the mass of 32P at the end of 42 days. [2] No of half-lives= 42/14 = 3 ; Mass of 32P at the end of 42 days = 0.0168 / (23) = 0.00210g ; [Total:12]

6 2 (a) Bauxite is an aluminium ore containing aluminium oxide and other impurities such as silica, iron oxides and titanium dioxide. The Bayer process is an important industrial process to obtain pure aluminium oxide from bauxite. In the Bayer process, bauxite is first reacted with hot sodium hydroxide. This converts the aluminium oxide in bauxite into a soluble salt, S, and water. The solid impurities are then filtered off. The second step in the process involves bubbling carbon dioxide gas into the products of the earlier step, producing aluminium hydroxide precipitate and sodium carbonate. In the third step, the precipitate obtained is heated to its decomposition temperature, producing pure aluminium oxide and water vapour. (i) The soluble salt, S, has the following composition by mass: Na, 28.0%; Al, 33.0%; O, 39.0%. Determine the molecular formula of S given that it has a molecular mass of 82.0. [3] Na Al O % 28.0 33.0 39.0 amount 28.0/23.0 = 1.217 33.0/27.0 = 1.222 39.0/16.0 = 2.438 ratio 1 1 2 (NaAlO2)n = (23.0 + 27.0 + 32.0)n = 82.0 n = 1 ; molecular formula = NaAlO2 ; working in table ; For Examiner’s Use

7 [Turn over (ii) Hence using your answer in (a)(i), write balanced equations for the first two steps of the process. [2] Al2O3 + 2NaOH  2NaAlO2 + H2O ; 2NaAlO2 + CO2 + 3H2O  2Al(OH)3 + Na2CO3 ; (iii) During an industrial production of aluminium oxide via the Bayer process, a 1 tonne sample of bauxite was used. The aluminium oxide produced reacted completely with 4060 dm3 solution of 5