CJC H2 Chem 2012 Prelim P2 Soln

3 9647/02/CJC JC2 Preliminary Exam 2012 (c) The temperature of the Bunsen flame varies depending on the ratio of the fuel to oxygen burnt. Besides keeping to the same fuel to oxygen ratio, s uggest how you would control another factor in the heating to ensure a fair comparison of the rate of decomposition of different carbonates. ……………………………………………………………………………………………………. …………………………………………………………………………………………………. [1] (d) Other than the use of safety goggles, state one hazard that must be considered when planning the experiment and suggest how you would keep this risk to a minimum. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. …………………………………………………………………………………………………. [2] (e) With reference to the apparatus in (b), show how you would calculate the mass of each carbonate used in the experiment. [Mr: MgCO3 = 84.3; CaCO3 = 100.1; SrCO3 = 147.6; BaCO3 = 197.0] [2] · Bunsen burner at same distance from the reaction vessel. 1. “hot” apparatus – use heat-proof gloves or let apparatus cool before handling 2. potential suck back (if water allowed to suck back, hot boiling tube would crack and shatter) – remove delivery tube from water when heating is stopped. Note: 1. Vol of CO2(g) collected must not exceed capacity of collection device. 2. Mass of each carbonate used must contain the same number of moles. Let volume of CO2(g) collected = 40 cm3 Since molar gas volume at r.t.p. = 24 dm3, mol of CO2 = = 1.67 × 10–3 mol MCO3 ® MO + CO2 \ minimum mol of MCO3 = mol of CO2 = 1.67 × 10–3 mol Mr of MCO3 = (Ar of M ) + [12.0 + 3(16.0)] = (Ar of M ) + 60.0 Let mol of carbonate = 2.00 × 10–3 mol mass of MCO3 = nMr = 2.00 × 10–3 × [(Ar of M ) + 60.0] \ mass of MgCO3 = (2.00 × 10–3) × 84.3 = 0.169 g mass of CaCO3 = (2.00 × 10–3) × 100.1 = 0.200 g mass of SrCO3 = (2.00 × 10–3) × 147.6 = 0.295 g mass of BaCO3 = (2.00 × 10–3) × 197.0 = 0.394 g [vol and mol of CO2(g)] [same mol of each carbonate]

4 9647/02/CJC JC2 Preliminary Exam 2012 (f) Draw a table with appropriate headings (and units) to show the data you would record and the values you would calculate in order to plot a suitable graph to show the variation in the rates of decomposition of the carbonates. Sketch, and explain, the shape of the graph you would expect from your results. Label clearly the axes. Explanation: ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. [4] [Total: 12] 0 Mr of MCO3 / s–1 Mr time / s time 1 / s–1 MgCO3 84.3 CaCO3 100.0 SrCO3 147.6 BaCO3 197.0 · Down the group, as the cation increases in size (while the charge remains unchanged), the charge density decreases and the polarising power of the cation also decreases. · Hence, the larger cations polarise (distort ) the carbonate anion less and the compound is thus relatively stable to heat.] [decrease in charge density of cation; less polarisation of anion]

5 9647/02/CJC JC2 Preliminary Exam 2012 2 (a) Carbon dioxide is useful in beverage carbonation. Cylinders of pressurised carbon dioxide are used to produce carbonated drinks. One such cylinder has an internal volume of 3.0 dm3 and contains 4.6 kg of carbon dioxide. (i) Calculate the pressure (in Pascals) the carbon dioxide gas would exert inside the cylinder at 28 °C. (ii) To find the pressure of a fixed amount of carbon dioxide gas under certain conditions, the van der Waals’ equation should be used. T R n b) n (VV n ap 2 2 = -÷ø öçè æ + Without further calculation, explain how the pressure obtained using the above equation would differ from that in (a)(i). ………………….…………………………………………………………………………….. ………………….…………………………………………………………………………….. [3] (b) Real gases like carbon dioxide can be liquefied at low temperatures just by applying pressure. Gases can be liquefied by pressure alone if their temperature is below their critical temperature, T c. The critical temperature of carbon dioxide is 31.1 °C. (i) Explain why real gases like carbon dioxide can be liquefied just by applying pressure. ………………….…………………………………………………………………………….. ………………….…………………………………………………………………………….. (ii) By considering structure and bonding, suggest a value for the critical temperature of methane and give a reason for your choice. ………………….…………………………………………………………………………….. ………………….…………………………………………………………………………….. ………………….…………………………………………………………………………….. [2] pV = nRT p (3 x 10-3) = 4.6×103 44 (8.31)(28+273) p = 8.72 x 107 Pa The pressure obtained would be lower since intermolecular forces of attraction exist between CO2 molecules. At high pressure, the molecules are very close together, and the intermolecular forces of attraction become significant. Any value less than that of carbon dioxide will be accepted as the answer. The van der Waals’ forces of attraction between methane molecules is weaker compared to that between carbon dioxide molecules because CH4 has a smaller electron cloud.

6 9647/02/CJC JC2 Preliminary Exam 2012 (c) Beyond the critical temperature and pressure, carbon dioxide exists as a supercritical fluid, a state that resembles a gas but has density closer to that in the liquid phase. Carbon dioxide is now well established as a solvent for use in extraction. (ii) Suggest why supercritical carbon dioxide is preferred as a solvent to extract caffeine from solid coffee over organic solvents like benzene. ……………..………………………………………………………………………………….. ……………..…………………………………………………………………………………. ………………….……………………………………………………………………………. (iii) Suggest why small amounts of ethanol need to be added to supercritical carbon dioxide in the extraction of polyphenols. An example of a polyphenol is shown below. OCH3 HO OH ……………..……………………………………………………………………………………. ………………….……………………………………………………………………………….. ……………..……………………………………………………………………………………. [2] (d) Ethanedioate ions, C 2O42–, can be oxidised by hot acidified aqueous potassium manganate(VII) to form carbon dioxide. (i) Draw the structure of ethanedioate ions, C2O42–, and give the bond angle around the central carbon atom. C C O O -O O- 120° Carbon dioxide is non-toxic while benzene is toxic and should be kept away from food and beverages. OR The carbon dioxide can be easily removed as a gas by depressurizing. The ethanol molecules added can form hydrogen bonds with the phenol groups present and this increase the solubility of polyphenols.

7 9647/02/CJC JC2 Preliminary Exam 2012 (ii) Construct a balanced equation for the reaction between ethanedioate ions and hot acidified potassium manganate(VII). (iii) 1.63 g of a salt, KHC 2O4∙H2C2O4, was dissolved in distilled water and made up to 250 cm 3 solution. Calculate the volume of 0.020 mol dm –3 of KMnO 4 required to react with 20.0 cm3 of the KHC2O4∙H2C2O4 solution. [Mr of KHC2O4∙H2C2O4 = 218.1] [4] The graph of rate against time for the reaction between acidified potassium manganate(VII) and ethanedioate ions is shown below. (e) (i) The reaction between acidified potassium manganate(VII) and ethanedioate ions is usually carried out at a higher temperature of 60 °C . Suggest why the rate of this reaction is slow at room temperature. ………………….………………………………………………………………………… ……………..……………………………………………………………………………… C2O42-→ 2CO2 + 2e- MnO4- + 8H