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 us