VJC 2015 H1 CHEM P2 ANS Prelims

©VJC 2015 8872/02/PRELIM/15 1 1 (a) The dissolved oxygen concentration (DOC) in rivers and lakes is important for aquatic life. If the DOC falls below 5 mg dm–3, most species of fish cannot survive. [1 g = 1000 mg] Environmental chemists can determine the DOC in water using the procedure below: Step 1 : A sample of river water is shaken with aqueous Mn 2+ and aqueous alkali. The dissolved oxygen oxidises the Mn 2+ to Mn3+, forming a pale brown precipitate of Mn(OH)3. O2(aq) + 4Mn2+(aq) + 8OH(aq) + 2H2O(l)  4Mn(OH)3(s) Step 2 : The Mn(OH) 3 precipitate is then reacted with an excess of aqueous potassium iodide, which is oxidised to iodine. 2Mn(OH)3(s) + 2I–(aq) → I2(aq) + 2Mn(OH)2(s) + 2OH–(aq) Step 3 : The iodine formed is then determined by titration with aqueous sodium thiosulfate, Na2S2O3(aq). 2S2O3 2–(aq) + I2(aq)  S4O6 2–(aq) + 2I–(aq) 25.0 cm3 of a sample of river water was analysed using the above procedure. The titration requires 24.60 cm3 of 0.00100 mol dm–3 Na2S2O3(aq). (i) Calculate the amount of oxygen present in the 25.0 cm3 sample of river water. Amount of S2O3 2– reacted in titration = 0.00100  24.60 1000 = 2.46  10–5 mol 4S2O3 2–  2I2  4Mn(OH)3  O2 Amount of O2 present = ¼  2.46  10–5 = 6.15  10–6 mol [2] (ii) Hence, calculate the dissolved oxygen concentration (DOC), in mg dm–3, in the river water. DOC = 6.15  10–6  32.0  1000 0.025 = 7.87 mg dm–3 [1] (iii) Comment on whether there is enough oxygen for fish to survive in that river. Since the actual DOC exceeds 5 mg dm –3, there is enough oxygen for the fish to survive. [1] (b) The presence of nitrite ions, NO 2 –, in the river water interferes with this method because NO 2 – ions can also react with iodide ions. During the reaction, NO 2 – is 2015 VJC H1 Chemistry Prelim Exam 8872/2 Suggested Answers Section A Answer all the questions in this section in the spaces provided.

©VJC 2015 8872/02/PRELIM/15 2 reduced to NO while iodide is oxidised to iodine. (i) Construct the half –equation for the reduction of NO 2 – to NO in an alkaline medium. NO2 – + H2O + e–  NO + 2OH– [1] (ii) Hence, give the overall equation for the reaction between NO 2 – and iodide ions. 2I– + 2NO2 – + 2H2O  I2 + 2NO + 4OH– [1] (c) (i) An oxide of manganese contains 72.0% by mass of manganese. Determine the empirical formula of this oxide. Mn O Mass % 72.0 100 – 72.0 = 28.0 Number of moles of atoms 72.0 ÷ 54.9 = 1.311 28.0 ÷ 16.0 = 1.75 ÷ by smaller number 1 1.335 Simplest ratio 1 × 3 = 3 1.335 × 3 = 4 Empirical formula of the oxide is Mn3O4. [2] (ii) The oxide in (i) is actually an equimolar mixture of two oxides. The oxidation state of manganese in one of the oxides is +3. Deduce the formulae of the two oxides. Mn2O3 and MnO [1] (iii) Hence state the full electronic configuration of manganese existing in the lower oxidation state. Lower oxidation state of Mn is +2. Electronic configuration of Mn2+: 1s22s22p63s23p63d5 [1] [Total: 10]

©VJC 2015 8872/02/PRELIM/15 3 2 (a) (i) Explain what is meant by the term standard enthalpy change of combustion , using ethene, C 2H4 as an example. Illustrate your answer with a balanced equation, including state symbols. It is the enthalpy change when 1 mole of ethene is completely burnt in an excess of oxygen under standard conditions at 298K and 1 atm. C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) [2] (ii) When 0.65 g of ethene was burnt under a container with 100 g of water, the temperature of the water rose from 28 oC to 86.5 oC. The process is known to be 75% efficient. Use these data and those relevant in the Data Booklet to calculate the enthalpy change of combustion of ethene. Heat evolved, Q = mct = 100  4.18  (86.5 – 28) = 24453 J = 24.45 kJ (75%) Actual heat produced = 24.45  100 75 = 32.6 kJ Amount of ethene used = 0.65 28.0 = 0.0232 mol Hc of ethene = – 32.6 0.0232 = – 1405 kJ mol–1 [2] (iii) The heat released when 1 g of a substance is combusted is known as its fuel value (in kJ g-1). Calculate the fuel value of ethene. Fuel value of ethene = 1405 26.0 = 54.0 kJ g–1 [1] (b) Sorbic acid, a preservative used in cheese has the following structure: CH3-CH=CH-CH=CH-CO2H z y x It is unsaturated like ethene. (i) State the type of isomerism exhibited by sorbic acid and explain how it arises. Draw the structural formulae of all the possible isomers. Geometric or cis -trans isomerism arises due to the presence of a C=C bond which prevents free rotation and 2 different groups are across the C=C bond. Total of 4 isomers are possible since both C=C bonds are able to exhibit cis-trans isomerism.

©VJC 2015 8872/02/PRELIM/15 4 [4] (ii) State the type of hybridisation found on the carbon atoms labelled x, y and z. Both Cx and Cy are sp2 hybridsed Cz is sp3 hybridised [All 3 correct – 1m ; only 2 correct – 0.5 m] [1] [Total: 10]

©VJC 2015 8872/02/PRELIM/15 5 3 Thionyl chloride, SO2Cl2, consists of two very important elements from Period 3. It is a liquid at room temperature. (a) When a sample containing 2 moles of gaseous SO2Cl2 is placed in a 2.0 dm3 vessel, it decomposes to SO2 and Cl2 as shown in the equation below. SO2Cl2(g) ⇌ SO2(g) + Cl2(g) H negative At equilibrium, the total concentration of the mixture is determined to be 1.56 mol dm–3. (i) Calculate the value of Kc, giving its units. Initial [SO2Cl2] = 2 2.0 = 1.0 mol dm–3 SO2Cl2(g) ⇌ SO2(g) + Cl2(g) initial conc. 2 / 2.0 = 1 0 0 change conc. –x +x +x eqm. conc. 1 – x x x Total concentration at equilibrium = (1 – x) + x + x = 1.56  x = 0.56 mol dm–3 [SO2Cl2]eqm = 1 – 0.56 = 0.44 mol dm–3 [SO2]eqm = [Cl2]eqm = 0.56 mol dm3 Kc = [SO2][Cl2] [SO2Cl2] = 0.56 × 0.56 0.44 = 0.713 mol dm3 [3] (ii) Draw the dot -and-cross diagram for SO 2Cl2 and state the shape of the molecule. 4 bond pairs, 0 lone pairs therefore tetrahedral. [2] (iii) Explain how an increase in temperature would change the value of Kc. Kc value would decrease. The reaction is exothermic hence according to Le Chatelier’s Principle , an increase in temperature would cause the position of equilibrium to shift left to remove excess heat . This would result in less product form. [2] (b) Phosphorus, another Period 3 element, forms a wide range of chlorides. Most famous are PCl3 and PCl5. (i) Explain briefly why PC l5 is a solid whereas SO2Cl2 is a liquid at room temperature. PCl5 is a non -polar molecule with instantaneous dipole -induced dipole interactions (dispersion forces) whereas SO2Cl2 is a polar molecule with permanent dipole -permanent dipole interactions . Since the number of

©VJC 2015 8872/02/PRELIM/15 6 electrons in PC l5 is larger, id-id interactions will be stronger than the pd - pd interactions, thus making PCl5 a solid but SO2Cl2 a liquid. [2] (ii) PCl5 is an acidic chloride that hydrolyses in water to produce HC l and H3PO4. Similarly, SO2Cl2 reacts with water to form two strong acids. Write a chemical equation, with state symbols, to depict the hydrolysis of SO2Cl2 in water. SO2Cl2(l) + 2H2O(l)  H2SO4(aq) + 2HCl(aq) [1] [Total: 10]

©VJC 2015 8872/02/PRELIM/15 7 4 This question is about the chemistry of some organic compounds. (a) Pyruvic acid, CH3COCO2H, is an important component in living cells as it is involved in the aerobic process of supplying energy. The fl