Chemistry SEE marking guide

Chemistry SEE marking guide External assessment

Queensland Curriculum & Assessment Authority

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SEE 1 — Combination response (50 marks)

SEE 2 — Combination response (120 marks)

Assessment objectives This assessment instrument is used to determine student achievement in the following objectives:

1. describe and explain chemical equilibrium systems, oxidation and reduction, properties andstructure of organic materials, and chemical synthesis and design

2. apply understanding of chemical equilibrium systems, oxidation and reduction, propertiesand structure of organic materials, and chemical synthesis and design

3. analyse evidence about chemical equilibrium systems, oxidation and reduction, propertiesand structure of organic materials, and chemical synthesis and design

4. interpret evidence about chemical equilibrium systems, oxidation and reduction, propertiesand structure of organic materials, and chemical synthesis and design

5. investigate phenomena associated with chemical equilibrium systems, oxidation andreduction, properties and structure of organic materials, and chemical synthesis and design

6. evaluate processes, claims and conclusions about chemical equilibrium systems, oxidationand reduction, properties and structure of organic materials, and chemical synthesis anddesign

7. communicate understandings, findings, arguments and conclusions about chemicalequilibrium systems, oxidation and reduction, properties and structure of organic materials,and chemical synthesis and design.

Note: Objective 1 is not assessed in SEE 1.

Objectives 5, 6 and 7 are not assessed in SEE 2



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Purpose This document is an External assessment marking guide (EAMG).

The EAMG: • Provides a tool for calibrating external assessment markers to ensure reliability of results

• Indicates the correlation, for each question, between mark allocation and qualities at each level of the mark range

• Informs schools and students about how marks are matched to qualities in student responses

Mark allocation Where a response does not meet any of the descriptors for a question or a criterion, a mark of ‘0’ will be recorded. Where no response to a question has been made, a mark of ‘N’ will be recorded. Allow FT mark(s) – refers to ‘follow through’, where an error in the prior section of working is used later in the response, a mark (or marks) for the rest of the response can still be awarded so long as it still demonstrates the correct conceptual understanding or skill in the rest of the response.



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SEE1 External assessment marking guide Paper 1 — Short response

Q Sample response The response:

1 a) 0.060 mol L–1 • states 0.060 [1 mark]

b) K = [B] [C] / [A] = (0.09 × 0.15) / 0.06 = 0.23

• uses the correct formula [1 mark] • demonstrates correct substitution

[1 mark] • correctly calculates K [1 mark]

c)

• states 4, 2, 1, 3 [1 mark]

d) The change caused the equilibrium to shift to the right. Initially, the concentration of all gases decreased. Immediately after, the concentration of gas A (reactant) gradually decreased, while the concentration of gases B and C (products) increased. This suggests the equilibrium shifted towards products to favour the production of more gas molecules.

• states the equilibrium has shifted to the right [1 mark]

• provides appropriate reasoning, using evidence from the graph [1 mark]



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e) The temperature was increased. The change caused the concentration of gases B and C (products) to gradually increase and the concentration of gas A (reactant) to gradually decrease. This is characteristic of a temperature change that added more stress to the reactant side of the equation. As the reaction is endothermic, this suggests the temperature was increased.

• identifies that temperature increased [1 mark]

• provides appropriate reasoning [1 mark]

2 a) Gradient = 224 –143130 –74

= 1.45 As the molecular mass increases, the boiling point increases by approximately 1.45 °C for every 1 g mol–1 increase in molar mass.

• identifies a relationship between molecular mass and boiling point [1 mark]

• uses data from the graph to support the response [1 mark]

b) The boiling point increases more rapidly for alkanes than alcohols as molecular mass increases.

• determines that boiling point increases more rapidly for alkanes than alcohols as molecular mass increases [1 mark]

3 a)

There is a peak between 1050 and 1410, indicating a C–O bond. There is a peak between 3200 and 3600, indicating an O–H bond.

• identifies C–O peak [1 mark] • identifies O–H peak [1 mark] • provides reasoning for both peaks

[1 mark]

b) The mass spectrum suggests the molecular mass of the compound is 46 g mol–1. It is known that the substance is an alcohol, so referring to the boiling point data, we can extrapolate the graph to predict that the boiling point of the substance is approximately 76 °C.

• infers the molecular mass of compound is 46 g mol–1 [1 mark]

• determines the boiling point is 76 °C [1 mark]



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3 c) Structure:

The peak at 31 m/z could represent the CH2OH fragment: 12.01 + 16 + (3 × 1.01) = ~31. The peak at 15 m/z could represent the CH3 fragment.

• determines the structure of the alcohol [1 mark]

• provides two pieces of evidence from the graph [2 marks]

Paper 2 — Extended response

Q Sample response The response: M

4 a) Calculates concentration of NaOH for the standardisation calculation

nstandard = cV = 0.10 mol L–1 x 0.0250 L = 0.0025 mol KHC8H4O4 nNaOH = 1

1 x 0.0025 mol

= 0.0025 mol NaOH nNaOH = cV 0.0025 mol = c x 0.0225 L 0.00250.0225

= c 0.11 mol L–1 = c

• provides pertinent mathematical operation/s correctly performed

• correctly determines [NaOH]

3

• provides pertinent mathematical operation(s) • determines a consequentially correct [NaOH]

2

• correctly determines a value associated with standardisation

1

• does not satisfy any of the descriptors above. 0



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4 a) Calculates total acidity for the titratable acidity calculation

nNaOH = 0.11 mol L–1 x 0.00725 L = 0.00080 mol NaOH nH2T = ½ x 0.00080 mol NaOH = 0.00040 mol H2T 0.00040 mol H2T: 0.010 L 0.040 mol H2T: 1 L m = n x M = 0.040 mol x 150 g mol–1 = 6.0 g L–1 tartaric acid

• provides pertinent mathematical operation(s) correctly performed

• correctly determines total acidity

4

• provides pertinent mathematical operation(s) • determines a consequentially correct total acidity

3

• correctly determines 2 values associated with total acidity

2

• correctly determines a value associated with total acidity

1


Plots findings on graph for communication

• accurately plots fifth datapoint 1




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4 b) Analyses data and draws a conclusion for identifying trends, patterns and relationships

ΔTA / °C = Δ titratable acidity / Δ temperature = (10.5 – 6) g L–1 / (15 – 20)°C = –0.9 g L–1 °C–1 Results show that as the mean daily temperature increases, the titratable acidity decreases by approximately 0.9 g L–1 °C–1 for wines grown in regions with temperatures between 15 °C and 20 °C. The data was sufficient to establish a trend and the correlation is quite strong; however, more wine samples would decrease the likelihood that this trend was due to chance. Additionally, the assumption that all acids in the wine are tartaric acid could affect the accuracy of results, but the extent depends on the acid composition of the wine. It indicates in the stimulus that the majority of acids in wine are tartaric or malic acid. As both of these acids are diprotic, they would be indistinguishable in an acid–base titration, so this assumption would not have a great effect on the calculated TA. However, the presence of monoprotic or triprotic acids such as citric acid could make the total acidity inaccurate. There was less than 0.5% uncertainty for all titre means, indicating the student was precise when conducting the titration. Overall, there is reasonable evidence to suggest that as the mean daily temperature increases, the titratable acidity decreases by approximately 0.9 g L–1 °C–1. However, this only applies to white wines grown in regions of Australia with mean daily temperatures between 15 °C and 20 °C, and the trend is based on only 5 data points, so may not be reliable.

• identifies that titratable acidity decreases by 0.9 g L–1 °C–1

2

• identifies that as the mean daily temperature increases, titratable acidity decreases

1


for identifying uncertainty and limitations

• considers uncertainty and limitations in response

2

• considers uncertainty or limitations in response 1


for providing a justified conclusion

• provides a justified conclusion linked to the research question

2

• provides a reasonable conclusion relevant to the research question

1




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4) c) Results indicate that titratable acidity decreases by approximately 0.9 g L–1 for each 1 °C increase in mean daily temperature. The correlation is quite strong, but reliability is weak because only 5 samples were tested. Potential effect based on experimental findings: Min. potential decrease in TA = 0.9 g L–1 °C–1 x 0.3 °C = 0.27 g L–1

Max. potential decrease in TA = 0.9 g L–1 °C–1 x 4.8 °C = 4.32 g L–1

The Intergovernmental Panel on Climate Change (IPCC) indicated that global surface temperature could rise 0.3 to 4.8 °C during the 21st century. Based on the experimental findings, this would lead to a 0.27–4.32 g L–1 decrease in acid content.

The stimulus states that wine grapevines are currently grown in regions of the world where temperatures during the growing season average 10–20 °C. Based on the most extreme prediction, grapes grown in the warmest regions of the world would produce wines with a total acidity of approximately 1.68 g L–1 (6 g L–1 – 4.32 g L–1). This would mean that the total acidity of wine grown in these regions would be well below the acceptable total acidity range of 5–9 g L–1 and could mean that the pH would not be low enough to ensure chemical stability.

In saying this, there is no evidence that the observed trend would continue at higher temperatures, and a decrease in total acidity would not necessarily affect the pH, as buffers in the wine could mitigate the effect of changes in acid content. Additionally, these predictions are based on limited experimental data and do not account for other factors that may be influenced by increased global temperatures, such as climate and soil type. Furthermore, only one variety of wine (white wine) was tested. While this is an important element of fair testing, it should not be assumed that the observed trend applies to all varieties of wine.

for the evaluation of the claim

• integrates experimental findings and information from the stimulus to quantitatively predict the effect of global warming on the wine-making industry

3

• considers the experimental findings to draw a reasonable conclusion regarding the effect of global warming on the wine-making industry

2

• applies insufficient or inappropriate findings of the experiment to the claim

1


for quality of evidence and limitations

• shows understanding of the features of the evidence that affect its ability to substantiate the claim

3

• shows understanding of some of the features of evidence that affect its ability to substantiate the claim

2

• provides simplistic statements about the quality of evidence

1


for communication

• uses scientific language fluently and concisely to communicate ideas

2

• uses scientific language competently to communicate ideas

1


https://en.wikipedia.org/wiki/Intergovernmental_Panel_on_Climate_Change



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To conclude, the experiment provides some evidence to support the claim that global warming will have a significant effect on the wine-making industry by decreasing the total acidity of wines; however, further data is needed to ensure the reliability and validity of experimental findings. Additionally, total acidity is only one aspect of wine quality. To fully evaluate the claim, data relating to the effect of temperature on other aspects of wine quality (such as alcohol content, sweetness and tannin content) would be needed.


4 d) • The experiment could be repeated using a different variety of wine, e.g. a red wine. Obtaining data for different varieties could provide broader support for/against the claim, because an increase in global temperature could affect some types of grapes (and therefore grape growers) more than others.

• Wines from different parts of the world could be tested to give greater variation in mean daily temperature, as growing temperatures in some countries are as low as 10 °C. This would increase the domain of the data and provide more holistic evidence about the effect of increased temperature on the global wine-making industry.

• The pH of each wine could be recorded as part of the experiment to determine the degree of correlation between total acidity and pH. This could also be used to assess the potential effect of increased temperature on the appearance and chemical stability of wine.

New research question: How is the pH of red wine affected by the mean daily temperature that grapes are exposed to during the ripening process, for mean daily temperatures between 10 °C and 20 °C?

for suggestions

• identifies 3 appropriate modifications 3

• identifies 2 appropriate modifications 2

• identifies 1 appropriate modification 1


for modifications

• justifies 3 modifications 3

• justifies 2 modifications 2

• justifies 1 modification 1


for the new research question

• states a research question that − is relevant to the modifications − identifies the dependent and independent

variables

2



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• states a research question that is relevant tothe modifications

OR • states a research question that identifies the

dependent and independent variables

1



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SEE 2 External assessment marking guide Paper 1: Multiple choice

Question Response

1 D

2 D

3 A

4 A

5 B

6 D

7 B

8 C

9 C

10 D

11 B

12 A

13 D

14 D

15 A

16 B

17 C

18 D

19 C

20 D



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Paper 1: Short response Q Sample response The response:

21 a) Monosaccharide • provides monosaccharide [1 mark]

b) Both starch and cellulose form 1-4 glycosidic links. However, starch is a polymer of α–glucose and cellulose is a polymer of β–glucose. Starch forms a linear polymer due to 1-4 α–glycosidic links (amylose) and a branched polymer due to 1-4 and 1-6 α–glycosidic links (amylopectin). Cellulose only exists as a linear polymer with 1-4, β–glycosidic links.

• identifies that both contain 1-4 links[1 mark]

• identifies that starch is a polymer ofα–glucose and cellolose is a polymerof β–glucose [1 mark]

• indicates that starch can be linear dueto 1-4 α–glycosidic links (amylose)and branched due to 1-4 and 1-6α–glycosidic links [1 mark]

• indicates that cellulose only exists asa linear polymer with 1-4, β–glycosidiclinks [1 mark]

22 a) • circles the red species [1 mark]



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b) 𝐾𝐾a =

[2.0 × 10−4][2.0 × 10−4]0.034

= 1.18 × 10−6

p𝐾𝐾a = –log [1.18 × 10−6] = 5.9 p𝐾𝐾a = 5.9 (to two significant figures)

• demonstrates substitution correctlyperformed [1 mark]

• determines 𝐾𝐾a = 1.2 × 10−6 [1 mark]• determine p𝐾𝐾a = 5.9 [1 mark]

c) Phenol red changes colour over a pH range because the molecular form (HIn(aq)) and ionic form (ln−(aq)) are different colours.

When [HIn(aq)] = [ln−(aq)], the pH = p𝐾𝐾a and phenol red changes colour.

When pH < p𝐾𝐾a, the [HIn(aq)] > [ln−(aq)] and phenol red turns yellow. When pH > p𝐾𝐾a, the [HIn(aq)] < [ln−(aq)] and phenol red turns red.

• indicates pH colour range is due tomolecular form and ionic form beingdifferent colours [1 mark]

• identifies phenol red changes colourwhen pH = p𝐾𝐾a [1 mark]

• indicates when pH < p𝐾𝐾a equilbriumfavours the molecular form (HIn), thesolution is yellow. When pH > p𝐾𝐾aequilbrium favours the ionic form (ln−),the solution is red [1 mark]

23 a) i) Na(l) and Cl2(g) • provides Na(l) and Cl2(g) [1 mark]

ii) H2(g) and O2(g) • provides H2(g) and O2(g) [1 mark]

b) In a dilute solution of aqueous sodium chloride, sodium ions, chloride ions, hydrogen ions, hydroxide ions and water molecules are present. The concentration and the Eo value of the species create competition at the electrodes and affect the products formed. Na+ and H+ compete to be reduced at the cathode. The E0 value for reducing H+ is more positive; therefore, H+ is preferentially reduced and H2 gas is formed rather than Na metal. Cl− and OH− compete to be oxidised at the anode. As the concentration of Cl− is low in a dilute NaCl solution, OH− is preferentially oxidised and O2 gas is produced rather than Cl2 gas.

• identifes that Na+, Cl−, OH−, H+, andH2O are present [1 mark]

• identifies that concentration and Eo

values of the species affects products[1 mark]

• identifies that H+ is preferentiallyreduced, producing H2 gas due to amore positive Eo value [1 mark]

• identifies that OH− is preferentiallyoxidised, producing O2 gas due to ahigher concentration of ions [1 mark]



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24 a) As the temperature increases, the [H3O+] increases. 2H2O(l) ⇌ H3O+(aq) + OH−(aq) Therefore, the equilibrium shifts towards the products. Increasing temperature shifts equilibrium in the endothermic direction, therefore the self-ionisation of water is endothermic.

• identifies [H3O+] increases astemperature increases [1 mark]

• identifies equilibrium shifts towardsthe products and the endothermicdirection [1 mark]

• determines self-ionisation of water isendothermic [1 mark]

b) pH = –log [H+] = 6.63 [H+] = 10−6.63

= 2.34 x 10−7 𝐾𝐾w = [H+] [OH−]

= (2.34 x 10−7)2 𝐾𝐾w = 5.48 x 10−14 (to 3 significant figures)

• determines[H+] = 2.34 x 10−7 [1 mark]

• determines consequentially correct Kw

[1 mark]

25 a) N2O4(g) ⇌ 2NO2(g) • provides N2O4(g) ⇌ 2NO2(g) [1 mark]

b) 80 seconds • identifies time as 80 [1 mark]



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c) • indicates that at 100 s concentrationof N2O4 and NO2 would halve [1 mark]

• indicates that after 100 sconcentration of N2O4 would decreaseto a new equilibrium at a lowerconcentration[1 mark]

• indicates that after 100 sconcentration of NO2 would increaseto a new equilibrium at a higherconcentration[1 mark]

26 The peak in the IR spectrum at 1700–1750 corresponds to a C=O bond in either an aldehyde or a ketone. There is no stretch in the IR peak at 2720–3100, therefore, the molecule is not an aldehyde and must be a ketone Fragment at m/z = 43 is CH3CH2CH2

+ and COCH3 +

Fragment at m/z = 71 is CH3CH2CH2CO+ Empirical formula = 86. This corresponds to the molecular mass shown on the mass spectrum. Therefore, the molecular formula is C5H10O.

• identifies IR peak at 1700–1750corresponds to a C=O bond inaldehyde or ketone [1 mark]

• indicates that X is a ketone [1 mark]• identifies mass fragment at- 43 m/z as CH3CH2CH2

+ and COCH3 +

OR - 71 m/z as CH3CH2CH2CO+ [1 mark]

• uses mass spectrum data to showthat the molecular formula for X isC5H10O [1 mark]

• provides correct structural formula forpentan-2-one [1 mark]



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27 a) Z • provides Z [1 mark]

b) Q2+ • provides Q2+ [1 mark]

c) Z(NO3)2(aq) and A(s) • provides Z(NO3)2(aq) and A(s)[1 mark]

d) Z2+ + 2e− ⇌ Z A2+ + 2e− ⇌ A X2+ + 2e− ⇌ X Q2+ + 2e− ⇌ Q Z can reduce X2+, Q2+ and A2+, therefore it is the strongest reducing agent. X can only reduce Q2+ therefore it is the second weakest reducing agent. Q cannot reduce any metal ions, therefore it is the weakest reducing agent.

• provides half-equations in correctorder [1 mark]

• provides reasoning for- Z as strongest reducing agent

[1 mark] - Q as weakest reducing agent

[1 mark] - A as stronger reducing agent than X

[1 mark]

Paper 2: Short response Question Sample response The response:

1 a) Zinc is oxidised when Zn changes to Zn2+. Copper is reduced when Cu2+ changes to Cu. Therefore, the reaction can be classified as redox as Zn is oxidised and Cu2+ is reduced.

• identifies that- zinc is oxidised [1

mark] - copper is reduced [1

mark] - reaction is redox [1

mark]

b) i) 4H+(aq) + 2NO3 −(aq) + 2e− → 2NO2(g) + 2H2O(l) • provides 2H+(aq) +

e− → H2O(l)[1 mark]

• provides NO3 −(aq) +

e− → NO2(g)[1 mark]



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Question Sample response The response:

ii) Eored = 0.46 + 0.34 = +0.80 V • determines Eored =+0.80 V [1 mark]

c) i) For hydrochloric acid:Eocell = Eored – Eoox = +0.00 – (+0.34) = –0.34 Reaction is non-spontaneous, therefore HCl cannot dissolve Cu. For nitric acid: Eocell = +0.46 (positive), therefore the reaction is spontaneous. HNO3 can dissolve Cu.

• determines Eocell forHCl equals –0.34 andEocell for HNO3 equals+0.46 V[1 mark]

• determines reactionbetween Cu and HCl isnot spontaneous andtherefore Cu will notdissolve [1 mark]

• indicates reactionbetween Cu and HNO3is spontaneous andtherefore Cu willdissolve [1 mark]

ii) Reduction:4H+(aq) + 2NO3

−(aq) + 2e− → 2NO2(g) + 2H2O(l), Eo = +0.80V or H+(aq) + 2e− → H2(g), Eo = 0.00 V The Eo value for NO3

− is more positive than H+(aq), therefore NO3 − is a

stronger oxidising agent.

Therefore NO3 − reduced in preference to H+ and NO2(g) formed

• identifies that, inHNO3, H+(aq) andNO3

– compete to bereduced [1 mark]

• indicates that NO3 − is

stronger oxiding agent[1 mark]

• determines NO3 − is

preferentially reducedtherefore NO2(g)formed[1 mark]

2 a) Esterification • identifies the reactionas esterification [1mark]



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b) 𝐾𝐾c = [𝐶𝐶9𝐻𝐻8𝑂𝑂4][𝐶𝐶2𝐻𝐻4𝑂𝑂2][𝐶𝐶7𝐻𝐻6𝑂𝑂3][𝐶𝐶4𝐻𝐻6𝑂𝑂3]

• provides[𝐶𝐶9𝐻𝐻8𝑂𝑂4][𝐶𝐶2𝐻𝐻4𝑂𝑂2][𝐶𝐶7𝐻𝐻6𝑂𝑂3][𝐶𝐶4𝐻𝐻6𝑂𝑂3]

[1 mark]

c) 𝐾𝐾c < 1 The equilibrium lies towards the reactants, therefore, the concentration of the reactants is greater than the concentration of the products.

• identifies thatequilbrium lies towardsthe reactants [1 mark]

• identifies thatreactants > products[1 mark]

d) Molar mass of aspirin = (12.01 x 9) + (16.00 x 4) + 8.08 = 180.17 g • determines molarmass of aspirin is 180g [1 mark]

• determines naspirin =2.78 × 10−3[1 mark]

• determines nacid [1mark]

• determines mass [1mark]

Moles of aspirin produced = 0.5 g / 180.17 g = 2.78 × 10−3 mol

Ratio 1:1 45.0% efficient

Moles of salicylic acid = 2.78 × 10 −3

0.45= 6.17 × 10−3 mol

Mass of salicylic acid = 6.17 × 10−3 × 138.13 = 0.852 g = Mass = 852 mg (to three significant figures)



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e) An increase in 𝐾𝐾c means that equilibrium has shifted towards the products. An increase in temperature shifts equilibrium in the endothermic direction. Le Châtelier’s principle means that when a system at equilibrium experiences an increase in temperature, the equilibrium shifts in the endothermic direction to decrease the temperature. As the forward reaction increases, the system as written must be endothermic.

• identifies thatincreased temperaturemeans increasedproducts [1 mark]

• identifies thatequilibrium has shiftedin the endothermicdirection [1 mark]

• uses Le Châtelier’sprinciple to explain ashift in equilibrium foran increase intemperature [1 mark]

• identifies the forwardreaction asendothermic [1 mark]

3 a) C2H12O6(aq)yeast �⎯⎯� 2CH3CH2OH(aq) + 2CO2(g) • provides correct

reactants and products[1 mark]

• correctly balances theequation[1 mark]

• indicates that yeast isrequired as a catalyst[1 mark]

b) Molar mass (ethanol) = 46.08 g Molar mass (glucose) = 180.18 g

atom economy = 2×46.08180.18

× 100

• shows substitutioncorrectly performed [1mark]

• determines atomeconomy [1 mark]

atom economy = 2×46.08180.18

× 100 = 51.148 % ≈ 51%

Atom economy = 51%



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c) Hydration atom economy = 100% Fermentation atom economy = 51% Therefore, production of ethene by hydration is greener.

• determines atomeconomy for hydrationreaction is 100% [1mark]

• identifies thathydration reaction isgreener [1 mark]

d) Use of renewable feedstocks Design for energy efficiency

• provides use ofrenewable feedstocks[1 mark]

• provides design forenergy efficiency [1mark]

4 a) unsaturated • provides unsaturated[1 mark]

b) 2–methylpropene • provides 2-methylpropene [1mark]

c) 2(CH3)2C = CH2 + 2H2O 𝐻𝐻+ �⎯� (CH3)3C(OH) + (CH3)2CHCH2OH • identifies (CH3)3C(OH)

as a product[1 mark]

• identifies(CH3)2CHCH2OH as aproduct[1 mark]

d) Major product (Markovnikov’s rule) (CH3)3C(OH)

• identifies tertiaryalcohol as the majorproduct produced [1mark]



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e) (CH3)2CHCH2OH is a primary alcohol and (CH3)3C(OH) is a tertiary alcohol.

Therefore, they have different boiling points.

Experimental technique: distillation

The hydroxyl group of a primary alcohol is more exposed than it is in a tertiary alcohol, therefore is easier for the primary alcohol to form more hydrogen bonds.

Therefore, the intermolecular forces are stronger and the boiling point higher for (CH3)2CHCH2OH.

• identifies the productsare primary andtertiary alcohols [1mark]

• identifies boiling pointas physical propertythat can be used toseparate the alcohols[1 mark]

• identifies distillation asa suitableexperimentaltechnique [1 mark]

• links the position of thehydroxyl group in theprimary alcohol toincreased hydrogenbonding [1 mark]

• indicates that strongerintermolecular forcesresult in a higherboiling point for theprimary alcohol [1mark]

5 a) ethanamine • provides ethanamine[1 mark]

b) H2O(l) C2H5NH3

+(aq) • provides H2O(l) [1

mark]• provides C2H5NH3

+(aq)[1 mark]

c) The hydrochloric acid (H+) reacts with the OH− ions and decreases their concentration.

• indicates that [OH−]will decrease[1 mark]



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According to Le Châtelier’s principle, this will make the equilibrium position shift to the right to counteract decrease in OH− (products). Therefore, the concentration of A will decrease.

• indicates that theequilibrium will shiftright [1 mark]

• indicates that A willdecrease [1 mark]



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d) [CH3CH2NH2] = 2.0 − x ≈ 2.0 as 2.0 ≫ 𝐾𝐾b • indicates assumptionto support[CH3CH2NH2] ≈ 2.0[1 mark]

• indicates [C2H5NH3 +] =

[OH−] [1 mark]• shows substitution

correctly performed [1mark]

• determines[OH−] = 3.35 × 10−2[1 mark]

• determines pOH[1 mark]

• determines pH[1 mark]

Let x = [C2H5NH3 +] = [OH−]

𝐾𝐾b = [C2H5NH3

+][OH−][CH3CH2NH2]

5.6 × 10−4 = [x][x][2.0]

= x2

2.0x2 = 1.12 × 10−3

x = 0.033466 = [OH−]

pOH = –log [OH−] = 1.475 ≈ 1.5

pH = 14 − 1.5 = 12.5 pH = 12.5 (to 1 decimal place)



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e) Heat CH3Br with KCN under reflux to produce CH3CN and KBr

CH3CN(aq) + 2H2(g)heat with Ni catalyst�⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯� CH3CH2NH2(aq)

• indicates CH3Br reactswith KCN to produceKH3CN and KBr[1 mark]

• identifies reaction isheated under reflux inethanol [1 mark]

• indicates CH3CN reactswith H2(g) to produceCH3CH2NH2 [1 mark]

• indicates heat andNi/Pt/Pd catalystrequired [1 mark]

• represents one of thereactions as abalanced chemicalequation [1 mark]


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Chemistry SEE marking guide