VJC Prelim P3 ANS

1  VJC 2015 9647/03/PRELIM/15 [Turn over Victoria Junior College 2015 H2 Chemistry Prelim Exam 9647/3 Suggested Answers 1 Magnesium ethanoate is commonly used as a source of magnesium or ethanoate ions in chemistry experiments. One of the more prevalent uses of magnesium ethanoate is in the mix ture called calcium magnesium ethanoate (CMA). It is a mixture of calcium ethanoate and magnesium ethanoate. CMA acts as a powerful SO 2, NO x, and toxic particulate emission control agent in coal combustion processes to reduce acid rain, and as an effective catalyst for the facilitation of coal combustion. (a) Samples of anhydrous magnesium ethanoate, Mg(CH 3COO)2 and barium ethanoa te, Ba(CH3COO)2, were heated to a temperature of 600 oC causing them to thermally decompose. White residues were formed in both cases and a common gaseous product, X, with molecular formula of C3H6O was also given off. X gives a yellow precipitate upon warming with aqueous alkaline iodine. In addition, the decomposition of magnesium ethanoate also produced a second gaseous product which formed a white precipitate with calcium hydroxide solution. (i) Identify the gaseous product, X. Propanone, CH3COCH3 [1] (ii) Write a balanced equation with state symbols, for each decomposition reaction. Mg(CH3COO)2 (s) MgO(s) + CO2(g) + CH3COCH3(g) Ba(CH3COO)2 (s) BaCO3(s) + CH3COCH3(g) [2] (iii) Account for the difference in the decomposition products. The Mg2+ ion is much smaller than Ba2+ ion, hence its higher charge density enables it to polarise the CH3COO- ion and distort the C-O bond to a greater extent resulting in complete decomposition. [2]

2  VJC 2015 9647/03/PRELIM/15 [Turn over (b) Magnesium, is an extremely important light wei ght structural metal which can be produced by the electrolysis of magnesium chloride . Magnesium chloride can be prepared from magnesium oxide w hich is obtained from sea -water containing a significant amount of Mg2+ and Ca2+. The steps involved are shown below: The numerical values of the relevant solubility products are given below. Magnesium carbonate 1.0 x 10-5 Calcium carbonate 8.7 x 10-9 Magnesium hydroxide 1.1 x 10-11 Calcium hydroxide 5.5 x 10-6 (i) Calculate and compare the solubility between magnesium carbonate and magnesium hydroxide. Ksp (Magnesium carbonate) = [Mg2+][CO3 2-] 1.0 x 10-5 = (x)(x) x = 3.16 x 10-3 mol dm-3 Ksp (Magnesium hydroxide) = [Mg2+][OH-]2 1.1 x 10-11 = (y) (2y)2 y = 1.40 x 10-4 mol dm-3 Solubility of MgCO3 is higher than that of Mg(OH)2 [3] (ii) Assuming the concentration of magnesium ions in the filtrate in step 2 is 3.0 x 10 -5 mol dm-3, calculate the concentration of hydroxide ions present in the filtrate when trace of solid magnesium hydroxide first appears. First trace of ppt appears when Ionic product = Ksp [Mg2+][OH-]2 = 1.1 x 10-11 (3.0 x 10-5)(z)2 = 1.1 x 10-11 z = 6.06 x 10-4 mol dm-3 [1] Sea water (containing Mg2+ and Ca2+) filtrate Step 1 Controlled addition of CO3 2- filter Step 2 addition of OH-(aq) Mg(OH)2 Step 3 heat MgO Step 4 MgCl2 Step 5 Mg

3  VJC 2015 9647/03/PRELIM/15 [Turn over (iii) Explain why the addition of carbonate ions in step 1 has to be controlled. To ensure maximum precipitation of Ca 2+ in the form of CaCO 3 which is then removed while preventing the precipitation of Mg2+ in Step 1. [1] (iv) Give a reason why the electrolysis of magnesium chloride is preferred to that of magnesium oxide. The melting point of MgCl2 is lower than MgO and hence less energy will be required to melt MgCl2. [1] (c) Some organic and inorganic compounds are class ified under a category called ‘non - existent compounds’. They are called non -existent because so far chemists had been unable to synthesise them. Some of the reasons why these compounds are unstable are:  unfavourable bond energy terms  a redox incompatibility of the ions making up the compound  reaction occurring between the ions For the following cases, suggest an explanation for each observation. You may use data from the Data Booklet to assist you in your answers. Give relevant equations to support your answers. (i) When aqueous sodium carbonate is a dded to aqueous aluminium chloride, a colourless gas is liberated and the precipitate formed is not aluminium carbonate, Al2(CO3)3. Aluminium carbonate is not precipitated because A l3+(aq) is acidic and will react with the carbonate, liberating carbon dioxide gas. When aluminum chloride dissolves in water, it undergoes hydrolysis to produce an acidic solution as follows: Al(H2O)6 3+ + H2O Al(H2O)5(OH)2+ + H3O+ This acidic solution will react with sodium carbonate to produce carbon dioxide: 2H3O+ + CO3 2-  CO2 + 3H2O [OR 2H+ + CO3 2-  CO2 + H2O] The precipitate is aluminium hydroxide formed due to hydrolysis of sodium carbonate. CO3 2- + H2O HCO3 - + OH- Al3+(aq) + 3OH-(aq)  Al(OH)3(s) [4]

4  VJC 2015 9647/03/PRELIM/15 [Turn over (ii) Caprolactum, a monomer of the polymer nylon 6, exists as an alicyclic ring structure and not as 6 -aminohexanoic acid, a straight chain aliphatic structure as most monomers do. O H C N H2NCH2CH2CH2CH2CH2CO2H Caprolactum 6-aminohexanoic acid 6-aminohexanoic acid can undergo internal nucleophilic substitution [OR cyclisation] to form caprolactum. H O O H NH HOC C N CH2 CH  + H2O CH2CH2CH2 Energy requirements: To break NH and CO bonds = + 390 + 360 = +750 kJ mol¯1 To form OH and CN bonds = – 460 – 305 = –765 kJ mol¯1 Overall H = –15 kJ mol-1. Reaction is exothermic and feasible. [Note: Overall there is also an increase in entropy of the system.] [3] (iii) It is possible to find MnCl2 in the laboratory but not MnCl3. Cl2 + 2e- 2Cl- Eo = +1.36 V Mn3+ + e- Mn2+ Eo = +1.49 V 2Mn3+ + 2Cl-  2Mn2+ + Cl2 = 1.49 – (1.36) = +0.13 V Since > 0, a redox reaction will occur in which Mn 3+ is reduced to Mn 2+ and Cl- is oxidized to chlorine gas when Mn3+ is combined with Cl-. [2] [Total: 20]

5  VJC 2015 9647/03/PRELIM/15 [Turn over 2 (a) The Contact Process is an industrial process for manufacturing sulfuric acid. The key stage in this process is the reaction between sulfur dioxide and oxygen. 2SO2(g) + O2(g) ⇌ 2SO3(g) H = –197 kJ mol–1 Vanadium(V) oxide , V 2O5, is used as a heterogeneous catalyst for this process. Describe the mode of action by which V2O5 fulfils this role. Availability of partially filled 3d orbitals in vanadium allow reactant molecules to be adsorbed onto metal surface. Reactant molecules are brought closer together and bonds within them are weakened. Activation energy is lowered. Subsequently, products formed desorb from the catalyst surface. [3] (b) In an experiment to determine the Kp of the above equilibrium, a mixture containing 0.200 mol of SO 2 and 0.100 mol of O 2 was heated in a closed flask and allowed to reach equilibrium at 550oC and 3.5 atm. The flask was then rapidly cooled to liquefy the SO 3 so that it can be separated from the gaseous SO 2 and O 2. Excess water was carefully added to the liquid SO 3, causi