2021 HCI Prelim Paper 4 Solutions

2021 HCI C2 H2 Chemistry Prelim / Paper 4 HWA CHONG INSTITUTION 2021 C2 H2 CHEMISTRY PRELIM PAPER 4 SUGGESTED SOLUTIONS 1(a) Sample answer: experiment 1 experiment 2 TFA1 / oC 29.4 29.2 TFA2 / oC 29.6 - Tmax / oC 38.2 35.0 Tave / oC 29.5 - 'Tmax / oC 8.7 5.8 mass of capped bottle and FA3 = 7.56 g mass of capped bottle and residual FA3 = 5.50 g mass of FA3 added = 2.06 g Records temperature / oC and mass / g Correct headers and units (included in the header or with each entry in the table) [1] All temperatures recorded to 0.1 oC, and mass to 0.01 g [1] Correctly calculates Tave and 'Tmax for experiment 1 and 'Tmax for experiment 2 and mass of FA3 added is about 2.1-2.2 g. [1] In general, the recording of data is well done – the requisite data was captured, and the precision used was appropriate. There was a wide variety of formats but they were generally acceptable. It is highly recommended that recordings should be grouped in a systematic manner, so that data can be retrieved quickly for calculations. The most common omission was the absence of 'Tmax for experiment 1 and experiment 2, though they may be determined in the calculation of q1 and q2. Candidates were not penalized for this, but it is a good experimental practice to record and process all the necessary data coherently, especially when the question asks for it. The instructions requested for candidates to weigh the bottle, the contents and the cap, but some candidates either failed to include the mass of the cap before, or after emptying the bottle. This caused the measurement of the mass of solid used to be inaccurate. This is a reminder to be systematic in your experimental procedures. 1(b)(i) q1 = mc'T = (50.0 + 20.0) × 4.18 × '𝑇௠௔௫ J (accept kJ) [1] nH2SO4 = 1.00 × 20/1000 = 0.02 mol 'H1 = −௤భ ଴.଴ଶ J mol1 (accept kJ mol1) [1]

The quantity, q1, refers to the amount of energy transferred to the mixture by the (exothermic) chemical reaction. This quantity is measured in joules or kilojoules, and does not have a sign. The mass, m, refers to the total mass of mixture present, i.e. 70.0 g, which is calculated using the total volume of mixture multiplied by density of mixture. '𝑇௠௔௫ refers to the temperature change of the mixture, i.e. the difference between Tmax and Tavg. The enthalpy change of reaction, 'H1, is the amount of heat given out per mole of the reaction in equation 1, i.e. when 2 mol of NaOH reacts with 1 mol of H2SO4. To find 'H1, in this case, we divide q1 by the amount of H2SO4 since it is the limiting reagent. 'H1, is negative because this is an exothermic reaction, i.e. the temperature of the reaction mixture increases. The appropriate sign has to be incorporated into the calculation at this point. 1(b)(ii) q2 = mc'T = 50.0 × 4.18 × '𝑇௠௔௫ J (accept kJ) [1] n(citric acid) = mass added ÷ 210 = c mol 'H2 = −௤మ ௖ J mol1 (accept kJ mol1) [1] All comments about q1 are applicable to q2. In addition, the mass, m, refers to the mass of mixture, i.e. 50.0 g, which is calculated using the volume of mixture multiplied by density of mixture. It does not include the mass of the solid as we assumed that the volume of mixture remains constant at 50.0 cm3 after adding the solid. An extension question to think about, before turning to the last page to check your answer. Q: Another student mixed 50.0 cm3 of 1.00 mol dm–3 sodium hydroxide with an excess amount of citric acid solid using the procedure for experiment 2. Explain why using excess citric acid solid would make his value for 'H2 incorrect. 1(b)(iii) 'H3 = 'H2 – ଷ ଶ× 'H1

2021 HCI C2 H2 Chemistry Prelim / Paper 4 Correctly constructs energy cycle or uses equations in working [1] Correctly applies Hess’ Law [1] Constructing this Hess cycle requires an addition of stoichiometric amounts of NaOH on both sides of equation 3. This allows 'H1 and 'H2 to be incorporated into the complete cycle. x 'H1 depicts the reaction between 2 mol of NaOH and 1 mol of H2SO4 while the Hess cycle depicts the reaction of 3 mol of NaOH. Hence, we need to multiply 'H1 by 3/2 in the cycle. x In the statement of Hess’s law, the sign of 'H1 was changed because the direction of the arrow was reversed. 1(c)(i) Draws two best-fit straight lines, extend both lines until they cross and VFA1 at point of intersection is 34.5 ± 0.5 cm3. [1] The best-fit lines are generally well-drawn. Do note that the graphs were obtained by plotting the temperature change of seven separate solutions containing different proportions of NaOH and H2SO4. This is not a thermometric titration. As such, best-fit lines, rather than curves, were expected. Candidates should follow the instructions in the question which says ‘draw two best-fit straight lines’. An extension question to think about, before turning to the last page to check your answer. Q: Explain why in these experiments the change in temperature, 'T, is proportional to the amount of water formed. 1(c)(ii) At the point of intersection, volumes/ amount of NaOH and H2SO4 reacted are exact / neither NaOH and H2SO4 are in excess. [1] The maximum amount of water is formed so the maximum amount of heat is evolved and 'T is maximum. [1] Or words to the effect Most students mistook the experiment for a thermometric titration, and failed to recognize that at the point of intersection, neither NaOH nor H2SO4 are in excess. It was not sufficient to say that the NaOH or H2SO4 is fully neutralized, because there is always a complete reaction of either NaOH or H2SO4 at each point of the graph. Most students were also unable or did not explain why 'T is maximum at the point of intersection. A clear reference to the concept of the enthalpy change of neutralization was required to earn credit. The maximum amount of heat was evolved because the most amount of water was produced by the given proportion of NaOH and H2SO4.

1(c)(iii) At the point of intersection, VFA1 = VNaOH = v cm3 VH2SO4 = 50 – v cm3 n(H2SO4) = 1.00 × (50 – v)/1000 n(NaOH) = 2 × n(H2SO4) [1] [NaOH] in FA1 = ௡(ே௔ைு) ௩ × 1000 mol dm3 [1] This question was generally well done. 2 General comments: Many candidates found this question on chemical equilibrium challenging. A large number of plans were poorly described and unlikely to yield any results if carried out. On the other hand, candidates who read the information in the question carefully and wrote their answers according to the given outline, bullet points and requirements were usually able to come up with coherent plans. Candidates are reminded to write legibly and present their answers clearly. For example, number the experimental steps in part (a) and write mathematical expressions in part (b). Candidates should not give answers that are not required. For example, the question did not require a justification of the quantities used (since the quantities are given) and treatment of results is not considered a part of the experimental procedure in (a). (a) The marks in the six boxes are awarded as follows: Appropriate apparatus: electronic balance, burette/micropipette, thermostatically controlled water bath 3 × [½] Leave mixture in sealed bottle (or sealed conical flask) for at least 48 hr 2 × [½] Do not accept beaker and test-tube Dilute equilibrium mixture (in a conical flask) with rinsing / Add water (directly) to mixture in conical flask and titrate whole mixture 2 × [½] Mark is lost if water is added to initial mixture or repeats titration or eqm mixture is placed in burette instead of conical flask Titrate quickly with NaOH [½] Use suitable indicator (phenolphthalein/ thymol blue/ thymolphthalein) * Do not accept methyl orange and correct colour change (colourless to pink/ yellow to blue/ colourless to blue) 2 × [½] Record masses (allow TARE) and initial and final burette readings 2 × [½]