VJC 2022 JC1 Promo H2 Chem P2 Solutions

1 © VJC 2022 9729/02/PROMO/2022 [Turn over 2022 VJC H2 Chemistry Promo Exam Paper 2 [Suggested Answers] Section A: Structured Questions [30 marks] Answer all the questions in the spaces provided. S1 (a) The engine capacity of a car refers to th e total volume of the combustion cylinders in the engine when the pistons are pushed out of the cylinders. The following experiment was conducted using a test engine where the combustion cylinders have a total volume of 200 cm3. • Ethane gas, CH3CH3, was mixed with air and injected into the cylinders. • The resultant fuel-air mixture (0.0280 mol) was then compressed and ignited. • The pistons are pushed out during the combustion process to bring the volume of the cylinders to a maximum. The final pressure and temperature was recorded to be 1.23 × 106 Pa and 753 oC respectively. (i) Using ideal gas equation, calculate the total number of moles of gaseous species in the cylinder after combustion of the fuel–air mixture. pV = nRT (1.23 x 106)(200 x 10–6) = n(8.31)(753 + 273) n = 0.0289 mol Examiner’s Comments: Generally well done. Students need to be careful to substitute the correct value (1.23 not 1.20, or 8.31 not 8.13) for calculation. [2] (ii) Using your answer in (a)(i), calculate the change in the total number of moles of gaseous species for the combustion process. Change in number of moles of gaseous species = 0.0289 – 0.0280 = +8.53 x 10–4 mol Examiner’s Comments: Generally well done too. [1] (iii) The complete combustion of 1 mol of ethane gas at 753 oC is represented by the following equation: CH3CH3(g) + 7 2 O2(g) → 2CO2(g) + 3H2O(g) Using your answer in (a)(ii), calculate the number of moles of CH 3CH3 in the fuel-air mixture. Change in number of moles of gaseous species / 1 mol CH3CH3 combusted = (2 + 3) – (1 + 3.5) = +0.5 mol 0.5 mol change in gaseous species / 1 mol CH3CH3 combusted = 8.53 x 10–4 mol change in gaseous species / x mol CH3CH3 combusted No. of moles of CH3CH3 in the fuel-air mixture = x

2 © VJC 2022 9729/02/PROMO/2022 [Turn over = 8.53 x 10–4 ÷ 0.5 = 1.71 x 10–3 mol Examiner’s Comments: Many students failed to appreciate that the answers for (a)(ii) is the change of volume during combustion. Hence many went to try to calculate using ICE box method with 2 unknown parameters (that was not required at all) and in the end couldn’t solve the question. [2] (b) In another experiment, liquid pentane, CH3(CH2)3CH3, was used as the fuel. The energy delivered by the test engine was found to be 2.80 × 104 kJ h–1. Additional data: • Enthalpy change of combustion of liquid pentane = –3510 kJ mol–1 • Density of liquid pentane = 0.626 g cm–3 • Molar mass of pentane = 72.0 g mol–1 (i) Calculate the total amount of energy delivered by the combustion of pentane when the test engine was kept running for two hours. Energy delivered = 2.80 x 104 × 2 = 5.60 x 104 kJ [OR 5.60 x 107 J] Examiner’s Comments: Overall, the question was well done. [1] (ii) Calculate the volume of pentane (in cm3) required needed to deliver the energy calculated in (b)(i). ΔHc of pentane = – Energy delivered ÷ Amount of pentane –3510 = – 5.60 x 104 ÷ Amount of pentane Amount of pentane = 5.60 x 104 ÷ 3510 = 16.0 mol Mass of pentane = 16.0 × 72.0 = 1150 g Volume of pentane = 1150 ÷ 0.626 = 1840 cm3 Examiner’s Comments: Generally quite well done. Some students made careless mistakes in (i) calculating amount of pentane from the ΔHc data, and (ii)calculating volume of pentane from amount of pentane. [2] (c) The fuel energy value (in kJ g –1) of a substance is the heat energy released when 1 g of that substance is combusted. It is known that ethanol , C 2H5OH, has a fuel energy value of 23.3 kJ g–1. Liquid ethano l is commonly found in canisters used for portable gas stoves during camping trips. A camper found a left -over canister of ethanol an d carried out an experiment to determine the mass of ethanol left in the canister. This was done using a bomb calorimeter as shown below:

3 © VJC 2022 9729/02/PROMO/2022 [Turn over In the bomb calorimeter, the bomb was filled with oxygen gas at high pressure and was immersed into a container filled with water. The liquid ethanol was then ignited using ignition wires . T he maximum temperature of the water inside the calorimeter was measured. Additional data: Mass of water used 200 g Change of temperature 24.0 oC Specific heat capacity of water 4.18 J g–1 K–1 Heat capacity of calorimeter (excluding water) 19.3 J K–1 (i) Write an equation for the standard enthalpy change of combustion of liquid ethanol, C2H5OH(l). C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) Balanced equation showing complete combustion of 1 mol of ethanol Correct state symbols Examiner’s Comments: Common mistakes: -H2O liberated should exist as liquid state and not gaseous state as ΔHc of liquid ethanol is measured at standard condition of 298K and 1 bar. -It should be 3 mol H2O and not 7/2 mol of H2O reacted. [1] (ii) Using relevant data provided, calculate the mass of ethanol left in the canister. Heat released from combustion of ethanol = Heat absorbed by water + calorimeter = mcwater∆T + Ccalorimeter∆T = (200 × 4.18 × 24.0) + (19.3 × 24.0) = 20500 J = 20.5 kJ The mass of ethanol left = 20.5 / 23.3 = 0.880 g

4 © VJC 2022 9729/02/PROMO/2022 [Turn over OR The mass of ethanol left = 20500 / 23300 = 0.880 g Examiner’s Comments: Common mistakes: -Heat absorbed by calorimeter was left out in the calculation or calculated wrongly. - ΔT term was wrongly expressed in (24 + 273) K. Do not add 273Kto 24 oC, the unit of ΔT in oC is the same as the one in K since ΔT refers to temperature difference. -Misinterpreted f uel energy in terms of kJ mol –1 instead of kJ g –1 (as given in the question). As such, some students assumed 0.880 represents the amount of ethanol burnt and calculated the mass of ethanol burnt wrongly by multiplying 0.880 by the Mr of ethanol. -Mass of ethanol burnt was calculated wrongly as (1 –0.880)g. However, 0.880 g is indeed the mass of ethanol left as this is the mass of ethanol burnt which heats up water and calorimeter by 24 oC. Anyway, initial mass of ethanol was also not given to be 1 g. [2] (d) Tetracarbonylnickel, Ni(CO) 4, was first synthesised in 1890 by Ludwig Mond by the direct reaction of nickel metal with carbon monoxide. This pioneering work foreshadowed the existence of many other metal carbonyl compounds, including those of vanadium and chromium. Some data for the synthesis of tetracarbonylnickel, Ni(CO)4, are shown below: Equation Ho / kJ mol–1 So / kJ K–1 mol–1 Ni(s) + 4CO(g) → Ni(CO)4(g) –161 –0.420 (i) Explain the significance of the sign of So. The sign of So is negative. This is due to a decrease in number of moles of gas particles in the reaction, leadin