RI 2021 Prelim P2 Answers

9 © Raffles Institution 2021 9729/02/S/21 2021 H2 Chemistry Prelim Paper 2 – Suggested Solutions General comments • There is a common theme in the examiner comments – it is apparent that many students do not read the question carefully to answer the question. Many students do not give the necessary details dictated by the question. Please read carefully! • The handwriting of some scripts left much to be desired. No marks are awarded for answers which cannot be read. Please write clearly for your own sake. • Please space out the answers. Do not squeeze the drawing of structures into the space of one line. Structures which cannot be clearly seen were not awarded marks. There is sufficient space given for each part. • In the drawing of mechanisms, please show the lone pair of electrons clearly – make them bigger and darker. • Please use ink throughout. Do not use pencil as pencil markings are often unclear and may be deemed as rough work. 1(a)(i) Heat gained by water = 4.18 x 10.0 x 300 = 12540 J Heat gained by copper can = 0.384 x 10 x 250 = 960 J Total heat gained = 12540 + 960 = 13500 J = 13.5 kJ Examiner Comments • This was generally well done. • Some students forgot to account for the heat gained by copper can even though the question clearly stated that consideration. 1(a)(ii) Amt of ester = 0.980 / 74.0 = 0.0132 mol Theoretical heat energy released = 0.0132 x 1592 = 21.1 kJ Examiner Comments • This was generally well done. • Note that there is no need for put negative sign here as the heat energy calculated is for the heat released. • A sign would be necessary for heat change as it could be a positive or negative change. 1(a)(iii) Method 1 - Find % heat loss to surroundings 13.5 kJ (expt) / 21.1 kJ (theoretical) = 64% of heat energy transmitted (36% heat loss) Amt of ethyl ethanoate = 0.948 / 88.0 = 0.010773 mol Heat transferred = 4.18 x 11.5 x 300 + 0.384 x 11.5 x 250 = 15525 J = 15.5 kJ Theoretical heat released = 15525 / 0.64 = 24257 J = 24.3 kJ H = − 24.3 kJ / 0.01077 = −2252 = −2250 kJ mol−1 Method 2 - Find thermal capacity of apparatus OR by proportion Thermal capacity = theoretical energy released / observed temperature change Thermal capacity = 21.1 kJ / 10 K = 2.11 kJ K–1 Theoretical heat produced from combustion = 2.11 kJ K–1 x 11.5 K = 24.3 kJ Amt of ethyl ethanoate = 0.948 / 88.0 = 0.010773 mol H = − 24.3 kJ / 0.01077 = −2252 = −2250 kJ mol−1 Examiner Comments • Similar to part (i), students must account for the heat transfer to both water and the copper can. • It is stated clearly in the question that the percentage heat loss (not absolute heat loss) is the same ac ross both experiments. A number of students failed to consider this important information resulting in loss of marks. • Many students struggled with the calculation of percentage heat loss to determine the theoretical heat transferred during reaction. Firstly, student