ASRJC H2 Chem 2022 Prelim P3 Soln

ASRJC JC2 PRELIM 2022 9729/03/H2 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2022 JC 2 PRELIMINARY EXAMINATION SOLUTIONS WITH MARKERS COMMENTS NAME:______________________________ ( ) CLASS: 22 / _____ CHEMISTRY 9729/03 Paper 3 Free Response Questions 16 September 2022 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Please look through the marks scheme and pay attention to the comments from markers. As you read, take note of:  What are your misconceptions?  What are common errors to avoid?  How should you improve next time? Do pay attention as your tutor go through in class and be proactive in asking questions. If you do not ask now, then when? Remember the power of tiny gains and be a little better every day!

1 (a) (i) N OO [1] (ii) bent, sp2 [1] Comments: (i) Many students did not know that in some molecules (eg NO2), the total number of valence electrons is an odd number. You can easily check by summing the total number of valence electrons before drawing the dot -and-cross diagram. In this case NO2 has 5+6+6=17 valence electrons. This means that there will b e at least one unpaired electron in the molecule! You can see this in your Chemical Bonding lecture notes, Pg 20 for more example and details. NO. .. x O.. ... x. x .x.x. is incorrect as there are 9 electrons around N. N, in period 2, CANNOT expand octet. It will only have a maximum of 8 electrons. (ii) In counting the number of bond pairs, each double bond or dative bond is counted as one bond pair. One lone electron is also counted as one lone pair as it also exerts repulsion, though less than a full 2-electron lone pair. This concept has been covered in our JC1 Chemical Bonding tutorial question 8! When writing the hybridisation state such as sp2, ‘sp’ is written in small letters and ‘2’ is written in superscript (raised). (iii) Bond length is the distance between the nuclei of the two atoms in the bond. [1] (iv) The 2p orbital of N overlaps with the 2p orbital of the singly bonded O atom allows the lone pair of electrons on the O atom to be delocalised into the N=O bond. [1] As a result, both the N –O bonds have partial double bond character . And the observed bond length is intermediate between a single and double bond. [1]  Mention delocalization of electron pair of O into the N=O bond  Leading to partial double bond character Comments: (iii) The words ‘nuclei’ and ‘bond’ are often missing. (iv) Many students attempted to explain the difference in bond length/bond strength of N=O and N -O instead of answering to the question on why the N -O in NO 2 is intermediate between a single and double bond. (FYI) Delocalisation of lone pair on O in N=O bond:

ASRJC JC2 PRELIM 2022 9729/03/H2 [Turn over (b) (i) Energy is released as a N-N covalent bond is formed using the single unpaired electron of each of the N atoms of two NO2 radicals, thus forming N2O4. O N O N O O O N O N O O + [1] mention energy released due to bond formation between 2 N atoms or it is N -N bond. Comments: Common misconceptions observed:  N of one NO2 radical forms a bond with O of another NO2 radical  The bond formed between N of one NO 2 radical and N of another NO 2 radical is dative  Energy is absorbed during bond formation (ii) 2 NO2(g) N2O4(g) H = –58.0 kJ mol–1 reddish brown colourless As temperature increases from 273K, POE shifts to the left to decrease the temperature by absorbing the excess heat, favouring the endothermic reaction. This POE shift is accompanied by increase in number of gaseous particles (2 vs 1), which contributes to increase in pressure. Hence the gradual increase in pressure as observed. At temperature T, N2O4(g) has completely converted to NO 2 molecules. As temperature increases from T, the number of moles of NO 2 molecules does not change further. The pressure of the gaseous sample then increases proportionally with increase in temperature. Comments: Some students did not apply the concept of Le Chatelier’s Principle to the reversible reaction to explain the increase in pressure from 273 K to T K. It is insufficient to just write that position of equilibrium shifts left. The full explanation as above should be given in your answer. Students did not realise that at temperature T, all the N 2O4 molecules has been converted to NO2 molecules. The position of equilibrium lies fully to the right then and there is no further increase in the amount of gas particles (iii) Observation: light brown/yellow gaseous mixture becomes darker brown, OR [1] the colour of the gaseous mixture becomes darker brown Comments: This part is well-done. As the question is about noting visible changes to the system, answers which discussed about the change to the gradient (or shape) of the graph or the pressure of the container were not accepted. (c) (i) 2NO2 + H2O 2H+ + NO2– + NO3– [1] Disproportionation [1] [1] [1] [1]

Comments While students showed confidence in cancelling out common terms to arrive at the correct equation, many incorrectly stated the equation as simply redox. Disproportionation is a more specific answer to this context as the same element (N) is both reduced and oxidised. (ii) TEA: catalyst as it is reacted/consumed in Step II and regenerated in Step IV. [1] A: intermediate as it is produced/formed in in Step III and reacted in Step IV. [1] State role + reason Comments: Good effort there in including the details of both species observed in the specific steps of the mechanism in relation to the roles. Answers which explained that they have identified catalyst as it is chemically unchanged after the reaction or intermediate as it does not appear in the overall equation are not given credit as these answers do not clearly explain how the answer was derived based on the mechanism. Ambiguous answers such as Stage II or Step 2 were accepted this time even though they differ from t he labelling used by the question. P lease be more mindful in A levels. (iii) (HOCH2CH2)3NH+(aq) + H2O(l) (HOCH2CH2)3N(aq) + H3O+(aq) pKa = 14 – 6.23 = 7.77  Ka = 10−7.77 = 1.70×10–8 mol dm–3 [H3O+] = √(1.70 × 10−8)(6.00 × 10−2) = 3.19× 10–5 mol dm–3 [1] pH = –log (3.19× 10–5) = 4.50 [1] Comments: This question was badly done. Many students tried to solve in terms of K b which will not work as F is an (conjugate) acid. Students should have solved in terms of K a for this weak acid. Many students fumbled in this question evidently from their weird assumptions such as [OH-] = [F] or [F] = [TEA]. (iv) [NO2]𝑎𝑣 = (4.13×10-3)(7.1×10-2) (0.91×10−4)(1.56×10-5)(10×24×60×60) = 0.239 µg m-3 [1] for correct answer Since the [NO2]𝑎𝑣 is below / less than / does not exceed 25 µg m–3, the quality of air in terms of NO2 level is good / satisfactory / acceptable. [1] allow ecf (accept words to the same effect) Comments: A handful of students struggled with the unit conversion from cm 2 to m2. Others who did not obtain full credit for this part often left out the basis on their comment on air quality. [Total: 18]

ASRJC JC2 PRELIM 2022 9729/03/H2 [Turn over 2 (a) (i) Category Following must be correctly drawn and/or labelled. Any item missing/incorrect in each category will have the mark deducted. Marks 1 Complete set of apparatus including:  two separate half -cells (of which one is the S tandard Hydrogen Electrode)  salt bridge  voltmeter  electrodes 1 2  half-cells reactants  conditions (concentration of solutions, temperature