MCQ Answers for 2019 H2 Chemistry Paper 1

MCQ answers

1. B
2. C
3. B
4. C
5. C
6. A
7. D
8. B
9. D
10. D
11. A
12. B
13. D
14. D
15. D
16. A
17. B
18. A
19. A
20. D
21. D
22. C
23. A
24. C
25. B
26. B
27. C
28. C
29. B
30. B

The standard for this MCQ paper is pretty easy, there wasn’t really any trick questions or questions which are ambiguous. IMO.
Only the last question will take a bit longer to calculate, but you should have plenty of buffer from the easy organic questions earlier.

Weightage of Paper 1 is 15% so each question is worth 0.5%. Accounting for careless mistakes, most students should score  more than 24 to
have a decent chance to score distinction.

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Suggest type question

Suggest Type Questions

Firstly, suggest type questions took up 12/75 (16%) marks in 2018 paper 2 and ~24/80 (30%) marks in paper 3. So we cannot ignore its importance.

 Secondly, most “suggest” questions aren’t totally novel questions, they are just contextual. Meaning we can look at the question and derive the answer from the information provided.

 In this article, I wish to highlight some ways to think about how to approach such questions. Please refer to the 2018 paper when reading through this document, it will give you some context.

Some simple general questions

• What is the difference between A & B?
• Is this linked to the earlier parts of the qn? What is the direction/flow of the qn?
• Why do we need to do ______? What if we DON’T do _____? What will happen?
• To explain given observations, think about which topic and then narrow down.
• For organic mechanisms, look for similarities with the 6 mechanisms that we have in our syllabus.
• Suggest questions are usually 1-2 marks, try not to vomit out a long answer, it wastes time and doesn’t answer the qn.
• How does the change affect the eqm? Are reactants/pdts removed or added?
• To answer qns on suggesting alternative reagents, think abt the properties of the orginal reactant that is important, that defines its role. Then look for the same in the replacement.
• To answer qns about the advantages or disadvantages, think abt what would be an ideal substance/reactant. Usually, it has to do with ease of storage, transportation, non-reactive, not pollutive, not corrosive/dangerous/poisonous or costs. Really depends on the context. E.g. what means a gd fuel (easy to be fully combusted), battery (storage, current), solvent (not reactive with reactants)
• To answer qns how to improve the expt, think abt your aim and assumptions. What would be the ideal expt and what would cause errors?

Suggest Type Questions in 2018 Paper 2 and 3

2018 P2 Q1c(iii) Suggest why, at room temperature, methyl isocyanate is a gas but A is a solid.

CH₃NCO           ->       C₆H₉N₃O₃

Methyl isocyanate                                   A

Thought process: Given physical properties, we will need to explain this using our knowledge of chemical bonding. A is simply a trimer of methyl isocyanate, so what’s the difference between the two molecules?

Take note of the earlier parts of the question, usually they will “lead” you to a certain direction when answering the question. In part (ii), we are told A has no dipole moment, so the higher IMFs is probably not due to pd-pd interactions.

Q2f(ii) Ammonia can be formed from its elements.

N₂ (g) +  H₂ (g)  <-> NH₃ (g)           ∆G_f⁰ = -16.6 kJ mol^-1

Suggest whether the ratio of [products]/[reactants] at equilibrium for the formation of ammonia at 298K will be less than, equal to or greater than 1. Give a reason for your answer.

Question to ask yourself:

What did part (i) ask about? Is it a clue to how we can answer part (ii)?

Given the negative Gibbs free value, what can you tell about the equilibrium position?

What does the equilibrium position tell you about the ratio of the amount of products vs the amount of reactants?

Standard questions commonly posed as suggest type questions

P2 Q4 (b)(ii) Methylbenzene reacts with fuming sulfuric acid, H₂S₂O₇, to form a mixture of three isomers, G, H and J, in equilibrium.

Suggest why there is a lower concentration of G than of J in the equilibrium mixture.

Qn to ask: What’s the difference between G and H/J?

How might this difference hinder the electrophilic attack? (since it is formed in lower conc)

c(iii) K also undergoes a reaction with fuming sulfuric acid.

Suggest the order of reactivity of benzene, methylbenzene and K with fuming sulfuric acid. Explain your answer.

Qn to ask: Order of reactivity depends on whether the substituent group is activating or deactivating. Think about how the substituents increase/decrease the electron density on the benzene ring.

Is –COCH₃ activating or deactivating? In all fairness, this is really a very standard qn.

The importance of reading and underlining key ideas in the question stem…

“When the external voltage is increased beyond E_cell, current will flow and a chemical reaction takes place.”

AgCl + e^– <-> Cl^– + Ag           E⁰ = -0.22V

O₂ + 4H⁺ + 4e^– <-> 2H₂O     E⁰ = +1.23V

Qn 5a(i) Suggest why there is a flow of electrons when the external voltage is +1.45V.

Qn to ask: E⁰_cell is exactly 1.45V and since the question stem stated that the current will only flow when external voltage is greater than 1.45V. What does this tell us about the E_cell?

Under what kind of conditions will the E­_cell drop?

(vi)  Suggest why the water sample is stirred constantly during the measurement.

Qn: What do we want to measure in this electrolysis?

What would happen if we do NOT stir the water sample constantly? Can this idea be applied to other electrolysis/galvanic cell?

(c)(iii) Suggest why the actual value for the pressure recorded in vessel Y was slightly different from the value calculated by this method. Assume oxygen gas behaves as an ideal gas under these conditions.

Qn:  What’s the difference between vessel X and Y?

The question did not specify whether the pressure is more or less in vessel Y so we can either explain it is lower or higher.

We cannot say due to IMF (which is what we do normally) as qn already assumes oxygen behaves ideally.

If explaining it is lower: think in terms how there could be less gaseous oxygen molecules

If explaining it is higher: think maybe how other dissolved gases in water might have escaped.

Can this be applied to other scenarios?

Paper 3

Q1e(ii) In some circumstances, double bonds will undergo a nucleophilic addition reaction. Suggest reasons to explain Fig. 1.2.

Suggest reasons to explain Fig 1.2. Use the concepts of electronegativity, electronic and steric effects, and delocalization in your answer.

Qn to ask: why doesn’t 1,3-butadiene undergo nucleophilic addition with HCN? (same reason why LiAlH₄ cannot reduce alkenes) Think about the difference between alkene and carbonyl compounds. Hint: Electronegativity, polarity.

As for why B cannot be formed, but C can be formed: Think in terms of steric and electronic effects, the difference between the attack position.

What is the difference between the C=C in 1,3-butadiene and the C=C in 4-methyl-1-penten-3-one? (Only delocalization hasn’t been used)

Q2b(i) HCN can be oxidized to cyanogen, C₂N₂.

C₂N₂ (g) + 2H⁺ (aq) + 2e^– <-> 2 HCN (aq)                 E⁰ = + 0.37 V

Suggest a suitable oxidant to oxidise HCN to C₂N₂ in acidic solution, using the Data Booklet. Write an equation for the reaction and calculate the E⁰_cell.

Note: Use of Data Booklet means you need to quote data from it.

Qn: If the above half equation is the oxidation half cell, what must be the E⁰ of the reduction half cell in order to have a spontaneous reaction?

(c) (ii) Suggest why the units of k₂ are s^-1, despite its rate-determining step being a bimolecular reaction.

Qn: What is the difference between reaction 2 and 3? In order to have units of rate constant as s^-1, what must be the total units of the reactants? (hint: think about H₂O)

(e) Cyanogen bromide, BrCN, undergoes addition reactions with alkenes. With propene, isomer D is produced rather than isomer E.

Suggest a mechanism for this reaction and use it to explain the preferential production of isomer D.

Qn: Think about which portion of BrCN will get the partial positive charge. Which intermediate carbocation formed will be more stable?

Important exam skill alert!

3c(ii) The addition of NH₃(aq) to a solution containing Cu²⁺(aq) produces a deep blue solution. This solution changes from deep blue to purple when ethane-1,2-diamine, H₂NCH₂CH₂NH₂, is added.

Suggest an explanation for this observation and write an equation for the reaction occurring. [The formula of ethane-1,2-diamine can be shortened to en in the formulae and equations of your answer.]

Qn: Change of colour is due to change of ligands, so you need to state en is the stronger ligand. This qn requires an eqn and is only 2 marks hence there’s no need to explain using ∆E and wavelength etc.

Usually when the qn wants you to account for the colour, it is about 3 marks. (eg. Qn 5a(i))

Thus an important skill is also to know how much to write and is all the details required. Time is precious, don’t regurgitate everything you know about the topic unless required.

Is this a novel qn? Not really, but it’s a “suggest” qn as you are not required to remember the ligands’ strength.

Section B

Q4d(v) The entropy change for the exothermic oxidation of powdered charcoal (carbon) and tetranitratoethane is large and positive. Suggest two reasons why the entropy change for the reaction is large and positive. Explain your answer.

Thought process: We are again presented with observation in this suggest type questions and asked to explain it using our content knowledge.

This is very common and forms the bulk of “suggest” type of questions. The context is novel but the concept is not. Usually an analogous example can be found in our syllabus or in the question context.

So we will need to explain two parts, why entropy is large and why it’s positive. Part iv requires us to write the equation so that we can see how the entropy has changed. How many ways are there to increase entropy? This is an easy qn.

Q5b(i) When a solution of sodium vanadate (V) is gradually acidified, the colour changes from colourless to orange to red, and finally to a yellow solution which contains the ion VO₂⁺.

The standard reduction potentials, E⁰, for the electrode reactions VO₃^–/VO²⁺ and VO₂⁺/VO²⁺ are both +1.00 V.

Suggest how the oxidising abilities of VO₃^– and VO₂⁺ will change as pH increases, and hence suggest which will be the more powerful oxidant at pH 7. Explain your answer.

Qn: [H⁺] DECREASES as pH increases. How will that affect the two equilibriums? For oxidising power, is it better to have a lower or higher E⁰?

Write out the two eqms to see clearly, since the E⁰ for both eqms is the same, you will need to differentiate which eqm is more affected by a change in [H⁺].

Parting words…

Hopefully, you now have a good understanding of suggest type questions and how to approach them.

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Dilution of Buffer Solutions

Dilution Effects on pH changes – Buffer

For avoidance of doubts, in this post, we are referring to acidic buffer.

In Acid-base Equilibria, there is a portion of the syllabus on buffers. Normal questions will ask for an outline of how the buffer works. They might ask you what is the mass of strong acid or base to a weak acid/base to add to create a buffer at a certain pH.

A more advanced question will ask you what effects will dilution have on the pH. Generally, dilution has NO effect on the pH. Why is this so? If you look at the buffer formula, pH = pKa + lg [salt]/[acid], dilution does not affect the [salt]/[acid] ratio. i.e. they are diluted to the same extent or volume would cancel out.

But the issue comes when the question further asks you will the pH change be larger if you dilute the buffer? We have to be careful here to watch out for the conditions. It is important to note whether the total volume of the buffer changed. Why is this important?

3 Scenarios

Firstly, if the volume changed drastically, i.e. large volumes of water was added to the buffer, the pH will tend to move towards 7. Which is the pH of water. This is because concentration of H+ ions tends to be closer to the amount from auto-ionization of water.

Second, if the total volume remained constant, it means that dilution AND removal of some of the solution had happened. (since we have to add water which would increase the total volume) In that case, although the [salt]/[acid] ratio will remained constant, there will be a larger change in ratio when a small amount of strong acid is added. Note that since some of the solution was removed, the number of moles of salt and acid decreased.

e.g.
Before dilution: [salt]/[acid] ratio is 5/5. So when a small amount of strong acid is added, say 0.1 mol, the ratio becomes (5-0.1)/(5+0.1),
4.9/5.1

After dilution: ratio becomes 0.5/0.5. So when a small amount of acid is added, the ratio becomes 0.4/0.6

It should be apparent that the salt to acid ratio is significantly different after dilution. Thus the pH change will also be larger.

Third, if the total volume DID NOT remain constant, it means that total volume has increased. If so, then the number of moles of salt and acid remained the same after dilution. So when a small amount of strong acid is added, the ratio is 4.9/5.1, as if there was no dilution! Why?

e.g. for a buffer with 5 mol of both salt and acid in 1 dm3 of water, we dilute it with another 1 dm3 of water. The ratio will remain the same since the volume will cancel out.

Hence you see that what’s crucial is actually the number of moles of salt and acid, as these will determine how large is the pH change.

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