2024 MI PU3 H2 Chem EOY P4 answers (final) w EC

Class Adm No Candidate Name: This question paper consists of 19 printed pages and 1 blank page. 2024 Preliminary Examination Pre-University 3 H2 CHEMISTRY 9729/04 Paper 4 Practical 27 Sep 2024 2 hours 30 minutes Candidates answer on the Question paper. READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number on all the work you hand in. Give details of the practical shift and laboratory where appropriate, in the boxes provided. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. You may lose marks if you do not show your working or if you do not use appropriate units. Qualitative Analysis Notes are printed at the back of the Question Paper. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Question 1 2 3 4 Total Marks 14 18 14 9 55 Shift Laboratory

2 1 Determination of enthalpy change of reaction FA 1 is solid sodium hydrogencarbonate, NaHCO3 FA 2 is 1.50 mol dm-3 sulfuric acid, H2SO4 (also required in both question 2 and 3) Sodium hydrogencarbonate is commonly known as baking soda and is used as a reagent in various reactions. It is soluble in water according to equation 1. equation 1 NaHCO3(s) + aq ⟶ Na+(aq) + HCO3–(aq) H1 It can also react with acids in both solid and aqueous state. equation 2 2NaHCO3(s) + H2SO4(aq) ⟶ Na2SO4(aq) + 2H2O(l) + 2CO2(g) H2 equation 3 2NaHCO3(aq) + H2SO4(aq) ⟶ Na2SO4(aq) + 2H2O(l) + 2CO2(g) H3 In this question, you will carry out experiments to determine H1 and H2, then use Hess’s Law to determine H3. For Examiners’ Use (a) In this experiment, you will determine the maximum temperature change when a known mass of solid sodium hydrogencarbonate, FA 1, reacts with sulfuric acid, FA 2. Then, you will determine H2. In an appropriate format in the space provided below, record  all weighings to an appropriate level of precision,  all values of temperature to an appropriate level of precision. Procedure 1. Weigh the capped bottle containing FA 1. 2. Place one polystyrene cup inside a second polystyrene cup. Place these in a glass beaker to prevent them from tipping over. 3. Use a measuring cylinder to transfer 25 cm3 of FA 2 into the first polystyrene cup. Cover the cup with the lid provided. 4. Stir FA 2 in the cup gently with the thermometer. Read and record its temperature. 5. Transfer all the FA 1 to the polystyrene cup. Stir the mixture. 6. Continue to stir the mixture. Observe the temperature and record the value that shows the maximum change from the initial temperature. 7. Reweigh the empty bottle and its cap. 8. Record the maximum temperature change and the mass of FA 1 used.

3 [Turn over (i) Results Mass of capped bottle with FA 1 / g 9.05 Mass of capped bottle with residual FA 1 / g 5.07 Mass of FA 1 used / g 3.98 Initial temperature of FA 2 / °C 30.8 Final temperature of mixture / °C 22.6 Maximum temperature change / °C 8.2 [5] [1] correct headers and units [1] mass readings to 2 d.p. and temperature readings to 1 d.p. [1] correctly determined maximum temperature change and mass of FA 1 used [2] T/m accuracy Examiners’ Comments: - Common mistakes include omitting the word “capped”, or using the phrasing “maximum temperature” instead of “final temperature of mixture”. - It was immediately obvious that some students faked their values when their reaction mixture increased in temperature, and it is reasonable to expect to be penalised for it. Additionally, depending on the nature of the question, having the wrong sign could potentially have implications on subsequent questions. (ii) Calculate the heat change, q, using the values you obtained in (a)(i). You should assume that the specific heat capacity of the solution is 4.18 J g1 K1, and that the density of the solution is 1.00 g cm3. q = mcT = 25.0 × 4.18 × 8.2 = 856.9 ≈ 857 J (3sf) heat change = 857 J [1] Examiners’ Comments: - Students performed unexpectedly poorly for this question, showing fundamental misconceptions. - Common mistakes include using the mass of the solid only, or summing up the mass of the solid with that of the solution. The specific heat capacity, c, value of 4.18 is specifically for water. Since the solid is going to fully react / dissolve, the volume of the solution (which we use to determine the mass of water) does not change much – so we always omit the mass of solid added. - Some students included a negative sign for their answers, but energy is a scalar quantity (only magnitude). It will also result in fewer mistakes if the sign is only included in ΔH calculations later based on whether the reaction is exo or endo.

4 (iii) Hence, determine the enthalpy change of reaction, H2. [Ar: Na, 23.0; H, 1.0; C, 12.0; O, 16.0] Amount of FA 1 = 3.98 ÷ 84.0 = 0.04738 mol Amount of FA 2 = 0.025 × 1.50 = 0.0375 mol FA 1 is limiting reagent [1] calculation of both amounts required H2 = 𝑞 𝑎𝑚𝑡 𝑜𝑓 𝐿𝑅 × CLR = + 856.9 0.04738 × 2 = +36170 J mol-1 = +36.2 kJ mol−1 (3sf) [1] sign required H2 = +36.2 kJ mol−1 [2] Examiners’ Comments: - Students performed poorly for this question, with many not being familiar with the equation for enthalpy change of reaction , as well as lacking conceptual understanding of the coefficient of LR (CLR). - A significant number of students also did not show calculation to determine the LR, which is required before the calculation of the enthalpy change of reaction. (iv) Calculate the percentage error of the temperature change when using the thermometer. Percentage error = 2(0.1) 𝑇 × 100% = 0.2 8.2 × 100%= 2.44 % (3sf) percentage error = 2.44 % [1] Examiners’ Comments: - Students performed poorly for this question. - All students had to do was to relate this question to the usual one for titre value. The thermometer given had a smallest interval of 0.2 °C, thus the uncertainty of a single reading is ±0.1 °C. Two readings had to be taken to determine for ΔT. (v) Suggest the effect on the value for T when 50 cm3 of FA 2 is used instead of 25 cm 3 in the experiment. [1] FA 1 is the limiting reagent and thus q will not change. Since mass of the solution is doubled, T will be halved. Examiners’ Comments: - Students performed poorly for this question. - For stronger students who understood what the question was testing for, many did not recognise that the specific volumes given required “halved” as part of the answer requirement. (b) A second experiment was conducted to find out H1 and the results of the experiment are presented in Table 1.1. Table 1.1 mass of FA 1 used / g 4.00 volume of water used / cm3 50.0 initial temperature of water / °C 29.4 minimum temperature reached / °C 21.0

5 [Turn over (i) Use the information in Table 1.1 to determine a value of H1. q = mcT = 50 × 4.18 × (29.4 - 21.0) = 1756 J [1] no double penalty based on (a)(ii) amount of FA 1 = 4 / 84.0 = 0.04762 mol H1 = +1756 / 0.04762 = +36.9 kJ mol-1 [1] sign required H1 = +36.9 kJ mol-1 [2] Examiners’ Comments: - Refer to comments in (a)(iii). (ii) Use Hess’s Law and your answers from (a)(iii) and (b)(i) to determine a value for H3 for t