Deduce the ratio of Fe²⁺:Fe³⁺ in Fe₃O₄.

State the type of spectroscopy that could be used to determine their relative abundances.

State the number of protons, neutrons and electrons in each species.

Iron has a relatively small specific heat capacity; the temperature of a 50 g sample rises by 44.4°C when it absorbs 1 kJ of heat energy.

Determine the specific heat capacity of iron, in J g⁻¹K⁻¹. Use section 1 of the data booklet.

Write the half-equation for the reduction of hydrogen peroxide to water in acidic solution.

Deduce a balanced equation for the oxidation of Fe2+ by acidified hydrogen peroxide.

1:2 ✔

Accept 2 Fe³⁺: 1 Fe²⁺
Do not accept 2:1 only

mass «spectroscopy»/MS ✔

Award [1 max] for 4 correct values.

specific heat capacity « = $\frac{q}{m\times ∆T}/\frac{1000 \mathrm{J}}{50 \mathrm{g}\times 44 \mathrm{K}}$» = 0.45 «J g⁻¹K⁻¹» ✔

H₂O₂(aq) + 2H⁺(aq) + 2e⁻→ 2H₂O(l) ✔

H₂O₂(aq) + 2H⁺(aq) + 2Fe²⁺(aq) → 2H₂O(l) + 2Fe³⁺(aq) ✔

Estimate the H−N−H bond angle in methanamine using VSEPR theory.

Ammonia reacts reversibly with water.

NH₃(g) + H₂O(l) $\rightleftharpoons$ NH₄⁺(aq) + OH⁻(aq)

Explain the effect of adding H⁺(aq) ions on the position of the equilibrium.

Hydrazine reacts with water in a similar way to ammonia. Deduce an equation for the reaction of hydrazine with water.

Outline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride.

Hydrazine has been used as a rocket fuel. The propulsion reaction occurs in several stages but the overall reaction is:

N₂H₄(l) → N₂(g) + 2H₂(g)

Suggest why this fuel is suitable for use at high altitudes.

Determine the enthalpy change of reaction, ΔH, in kJ, when 1.00 mol of gaseous hydrazine decomposes to its elements. Use bond enthalpy values in section 11 of the data booklet.

N₂H₄(g) → N₂(g) + 2H₂(g)

The standard enthalpy of formation of N₂H₄(l) is +50.6 kJ$$mol⁻¹. Calculate the enthalpy of vaporization, ΔH_vap, of hydrazine in kJ$$mol⁻¹.

N₂H₄(l) → N₂H₄(g)

(If you did not get an answer to (f), use −85 kJ but this is not the correct answer.)

Calculate, showing your working, the mass of hydrazine needed to remove all the dissolved oxygen from 1000 dm³ of the sample.

Calculate the volume, in dm³, of nitrogen formed under SATP conditions. (The volume of 1 mol of gas = 24.8 dm³ at SATP.)

107^°

Accept 100° to < 109.5°.

Literature value = 105.8°

[1 mark]

removes/reacts with OH⁻

moves to the right/products «to replace OH⁻ ions»

Accept ionic equation for M1.

[2 marks]

N₂H₄(aq) + H₂O(l) $\rightleftharpoons$ N₂H₅⁺(aq) + OH^–(aq)

Accept N₂H₄(aq) + 2H₂O(l) $\rightleftharpoons$ N₂H₆²⁺(aq) + 2OH^–(aq).

Equilibrium sign must be present.

[1 mark]

bubbles
OR
gas
OR
magnesium disappears

2NH₄⁺(aq) + Mg(s) → Mg²⁺(aq) + 2NH₃(aq) + H₂(g)

Do not accept “hydrogen” without reference to observed changes.

Accept "smell of ammonia".

Accept 2H⁺(aq) + Mg(s) → Mg²⁺(aq) + H₂(g)

Equation must be ionic.

[2 mark]

no oxygen required

[1 mark]

bonds broken:
E(N–N) + 4E(N–H)
OR
158 «kJ$$mol^–1» + 4 x 391 «kJ$$mol^–1» / 1722 «kJ»

bonds formed:
E(N≡N) + 2E(H–H)
OR
945 «kJ$$mol^–1» + 2 x 436 «kJ$$mol^–1» / 1817 «kJ»

«ΔH = bonds broken – bonds formed = 1722 – 1817 =» –95 «kJ»

Award [3] for correct final answer.

Award [2 max] for +95 «kJ».

[3 marks]

OR
ΔH_vap= −50.6 kJ$$mol⁻¹ − (−95 kJ$$mol⁻¹)

«ΔH_vap =» +44 «kJ$$mol⁻¹»

Award [2] for correct final answer.

Award [1 max] for −44 «kJ$$mol⁻¹».

Award [2] for:
ΔH_vap − = 50.6 kJ$$mol⁻¹ − (−85 kJ$$mol⁻¹) + = 34 «kJ$$mol⁻¹».

Award [1 max] for −34 «kJ$$mol⁻¹».

[2 marks]

total mass of oxygen «= 8.0 x 10^–3 g$$dm^–3 x 1000 dm³» = 8.0 «g»

n(O₂) «$=\frac{8.0\text{ g}}{32.00\text{ g}\text{mo}{\text{l}}^{-1}}=$» 0.25 «mol»

OR
n(N₂H₄) = n(O₂)
«mass of hydrazine = 0.25 mol x 32.06 g$$mol^–1 =» 8.0 «g»

Award [3] for correct final answer.

[3 marks]

«n(N₂H₄) = n(O₂) $=\frac{8.0\text{ g}}{32.00\text{ g}\text{mo}{\text{l}}^{-1}}=$» 0.25 «mol»

«volume of nitrogen = 0.25 mol x 24.8 dm³$$mol^–1» = 6.2 «dm³»

Award [1] for correct final answer.

[1 mark]

Determine the limiting reactant showing your working.

The mass of copper obtained experimentally was 0.872 g. Calculate the percentage yield of copper.

The reaction was carried out in a calorimeter. The maximum temperature rise of the solution was 7.5 °C.

Calculate the enthalpy change, ΔH, of the reaction, in kJ, assuming that all the heat released was absorbed by the solution. Use sections 1 and 2 of the data booklet.

State another assumption you made in (b)(i).

The only significant uncertainty is in the temperature measurement.

Determine the absolute uncertainty in the calculated value of ΔH if the uncertainty in the temperature rise was ±0.2 °C.

Sketch a graph of the concentration of iron(II) sulfate, FeSO₄, against time as the reaction proceeds.

Outline how the initial rate of reaction can be determined from the graph in part (c)(i).

Explain, using the collision theory, why replacing the iron powder with a piece of iron of the same mass slows down the rate of the reaction.

n_CuSO4 «= 0.0800 dm³ × 0.200 mol dm^–3» = 0.0160 mol AND

n_Fe «$\frac{3.26\text{g}}{55.85\text{g}\text{mo}{\text{l}}^{-1}}$» = 0.0584 mol ✔

CuSO₄ is the limiting reactant ✔

Do not award M2 if mole calculation is not shown.

ALTERNATIVE 1:
«0.0160 mol × 63.55 g mol^–1 =» 1.02 «g»  ✔

«$\frac{0.872\text{g}}{1.02\text{g}}\times 100=$» 85.5 «%»  ✔

ALTERNATIVE 2:
«$\frac{0.872\text{g}}{63.55\text{g}\text{mo}{\text{l}}^{--1}}=$» 0.0137 «mol»  ✔

«$\frac{0.0137\text{mol}}{0.0160\text{mol}}\times 100=$» 85.6 «%»  ✔

Accept answers in the range 85–86 %.

Award [2] for correct final answer.

ALTERNATIVE 1:

q = «80.0 g × 4.18 J g^–1 K^–1 × 7.5 K =» 2.5 × 10³ «J»/2.5 «kJ» ✔

«per mol of CuSO₄ = $\frac{-2.5\text{kJ}}{0.0160\text{mol}}=-1.6\times {10}^2$ kJ mol^–1»

«for the reaction» ΔH = –1.6 × 10² «kJ» ✔

ALTERNATIVE 2:

q = «80.0 g × 4.18 J g^–1 K^–1 × 7.5 K =» 2.5 × 10³ «J»/2.5 «kJ» ✔

«n_Cu = $\frac{0.872}{63.55}$ = 0.0137 mol»

«per mol of CuSO₄ = $\frac{-2.5\text{kJ}}{0.0137\text{mol}}=-1.8\times {10}^2$ kJ mol^–1»

«for the reaction» ΔH = –1.8 × 10² «kJ» ✔

Award [2] for correct final answer.

density «of solution» is 1.00 g cm⁻³

OR

specific heat capacity «of solution» is 4.18 J g⁻¹ K⁻¹/that of «pure» water

OR

reaction goes to completion

OR

iron/CuSO₄ does not react with other substances ✔

The mark for “reaction goes to completion” can only be awarded if 0.0160 mol was used in part (b)(i).

Do not accept “heat loss”.

ALTERNATIVE 1:

«$0.2{}^{\circ }\text{C}\times \frac{100}{7.5{}^{\circ }\text{C}}=$» 3 %/0.03 ✔

«0.03 × 160 kJ» = «±» 5 «kJ» ✔

ALTERNATIVE 2:

«$0.2{}^{\circ }\text{C}\times \frac{100}{7.5{}^{\circ }\text{C}}=$» 3 %/0.03 ✔

«0.03 × 180 kJ» = «±» 5 «kJ» ✔

Accept values in the range 4.1–5.5 «kJ».

Award [2] for correct final answer.

initial concentration is zero AND concentration increases with time ✔

decreasing gradient as reaction proceeds ✔

«draw a» tangent to the curve at time = 0 ✔

«rate equals» gradient/slope «of the tangent» ✔

Accept suitable diagram.

piece has smaller surface area ✔

lower frequency of collisions

OR

fewer collisions per second/unit time ✔

Accept “chance/probability” instead of “frequency”.

Do not accept just “fewer collisions”.

Ethane-1,2-diol can be formed according to the following reaction.

2CO (g) + 3H₂ (g) $\rightleftharpoons$ HOCH₂CH₂OH (g)

(i) Deduce the equilibrium constant expression, K_c, for this reaction.

(ii) State how increasing the pressure of the reaction mixture at constant temperature will affect the position of equilibrium and the value of K_c.

Position of equilibrium:

K_c:

(iii) Calculate the enthalpy change, ΔH^θ, in kJ, for this reaction using section 11 of the data booklet. The bond enthalpy of the carbon–oxygen bond in CO (g) is 1077kJmol^-1.

(iv) The enthalpy change, ΔH^θ, for the following similar reaction is –233.8 kJ.

2CO(g) + 3H₂(g) $\rightleftharpoons$ HOCH₂CH₂OH (l)

Deduce why this value differs from your answer to (a)(iii).

Determine the average oxidation state of carbon in ethene and in ethane-1,2-diol.

Ethene:

Ethane-1,2-diol:

Explain why the boiling point of ethane-1,2-diol is significantly greater than that of ethene.

Ethane-1,2-diol can be oxidized first to ethanedioic acid, (COOH)₂, and then to carbon dioxide and water. Suggest the reagents to oxidize ethane-1,2-diol.

(i)
$\ll K_{\text{C}}=\gg \frac{\left[\text{HOC}{\text{H}}_{\text{2}}\text{C}{\text{H}}_{\text{2}}\text{OH}\right]}{{\left[\text{CO}\right]}^{\text{2}}\times {\left[{\text{H}}_{\text{2}}\right]}^{\text{3}}}$ 

(ii)
Position of equilibrium: moves to right OR favours product
K_c: no change OR is a constant at constant temperature

(iii)
Bonds broken: 2C≡O + 3(H-H) / 2(1077kJmol^-1) + 3(436kJmol^-1) / 3462 «kJ»

Bonds formed: 2(C-O) + 2(O-H) + 4(C-H) + (C-C) / 2(358kJmol^-1) + 2(463kJmol^-1) + 4(414kJmol^-1) + 346kJmol^-1 / 3644 «kJ»

«Enthalpy change = bonds broken - bonds formed = 3462 kJ - 3644 kJ =» -182 «kJ»

Award [3] for correct final answer.
Award [2 max] for «+»182 «kJ».

(iv)
in (a)(iii) gas is formed and in (a)(iv) liquid is formed
OR
products are in different states
OR
conversion of gas to liquid is exothermic
OR
conversion of liquid to gas is endothermic
OR
enthalpy of vapourisation needs to be taken into account

Accept product is «now» a liquid.
Accept answers referring to bond enthalpies being means/averages.

Ethene: –2

Ethane-1,2-diol: –1

Do not accept 2–, 1– respectively.

ethane-1,2-diol can hydrogen bond to other molecules «and ethene cannot»

OR

ethane-1,2-diol has «significantly» greater van der Waals forces

Accept converse arguments.
Award [0] if answer implies covalent bonds are broken

hydrogen bonding is «significantly» stronger than other intermolecular forces

acidified «potassium» dichromate«(VI)»/H⁺ AND K₂Cr₂O₇/H⁺ AND Cr₂O₇^2-

OR

«acidified potassium» manganate(VII)/ «H⁺» KMnO₄ /«H⁺» MnO₄^-

Accept Accept H₂SO₄ or H₃PO₄ for H⁺.
Accept “permanganate” for “manganate(VII)”.

Write an equation for the complete combustion of ethyne.

Deduce the Lewis (electron dot) structure of ethyne.

Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C₂H₆.

Identify the type of interaction that must be overcome when liquid ethyne vaporizes.

Product A contains a carbon–carbon double bond. State the type of reactions that compounds containing this bond are likely to undergo.

State the name of product B, applying IUPAC rules.

Determine the enthalpy change for the reaction, in kJ, to produce A using section 11 of the data booklet.

The enthalpy change for the reaction to produce B is −213 kJ. Predict, giving a reason, which product is the most stable.

The IR spectrum and low resolution ¹H NMR spectrum of the actual product formed are shown.

Deduce whether the product is A or B, using evidence from these spectra together with sections 26 and 27 of the data booklet.

Identity of product:

One piece of evidence from IR:

One piece of evidence from ¹H NMR:

Suggest the reagents and conditions required to ensure a good yield of product B.

Reagents:

Conditions:

Deduce the average oxidation state of carbon in product B.

Explain why product B is water soluble.

C₂H₂ (g) + 2.5O₂ (g) → 2CO₂ (g) + H₂O (l)
OR
2C₂H₂ (g) + 5O₂ (g) → 4CO₂ (g) + 2H₂O (l)  [✔]

  [✔]

Note: Accept any valid combination of lines, dots and crosses.

«ethyne» shorter AND a greater number of shared/bonding electrons
OR
«ethyne» shorter AND stronger bond  [✔]

London/dispersion/instantaneous dipole-induced dipole forces  [✔]

Note: Do not accept just “intermolecular forces” or “van der Waals’ forces”.

«electrophilic» addition/A_«E»  [✔]

Note: Accept “polymerization”.

ethanal  [✔]

«sum of bond enthalpies of reactants =» 2(C–H) + C≡C + 2(O–H)
OR
2 × 414 «kJ mol^–1» + 839 «kJ mol^–1» + 2 × 463 «kJ mol^–1»
OR
2593 «kJ»  [✔]

«sum of bond enthalpies of A =» 3(C–H) + C=C + C–O + O–H
OR
3 × 414 «kJ mol^–1» + 614 «kJ mol^–1» + 358 «kJ mol^–1» + 463 «kJ mol^–1»
OR
2677 «kJ»  [✔]

«enthalpy of reaction = 2593 kJ – 2677 kJ» = –84 «kJ»  [✔]

Note: Award [3] for correct final answer.

B AND it has a more negative/lower enthalpy/«potential» energy
OR
B AND more exothermic «enthalpy of reaction from same starting point»  [✔]

Identity of product: «B»

IR spectrum:
1700–1750 «cm^–1 band» AND carbonyl/CO group present
OR
no «band at» 1620–1680 «cm^–1» AND absence of double bond/C=C
OR
no «broad band at» 3200–3600 «cm^–1» AND absence of hydroxyl/OH group  [✔]

Note: Accept a specific value or range of wavenumbers and chemical shifts.

¹H NMR spectrum:
«only» two signals AND A would have three
OR
«signal at» 9.4–10.0 «ppm» AND «H atom/proton of» aldehyde/–CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of alkyl/CH next to» aldehyde/CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of» RCOCH_2- present
OR
no «signal at» 4.5–6.0 «ppm» AND absence of «H-atom/proton next to» double bond/C=C   [✔]

Note: Accept “two signals with areas 1:3”.

Reagents:
acidified/H⁺ AND «potassium» dichromate«(VI)»/K₂Cr₂O₇/Cr₂O₇^2–  [✔]

Conditions:
distil «the product before further oxidation»  [✔]

Note: Accept “«acidified potassium» manganate(VII)/KMnO₄/MnO₄^–/permanganate”.

Accept “H₂SO₄” or “H₃PO₄” for “H⁺”.

Accept “more dilute dichromate(VI)/manganate(VII)” or “excess ethanol”.

Award M1 if correct reagents given under “Conditions”.

–1  [✔]

Any three of:
has an oxygen/O atom with a lone pair  [✔]

that can form hydrogen bonds/H-bonds «with water molecules» [✔]

hydrocarbon chain is short «so does not disrupt many H-bonds with water molecules» [✔]

«large permanent» dipole-dipole interactions with water [✔]

Almost all candidates recognized the products of the complete combustion of ethyne, and over two thirds managed to balance the equation. It was good to see candidates using integers for the balancing.

The majority of candidates drew the Lewis structure of ethyne. A few teachers commented that they did not cover alkynes assuming they are not included in the syllabus. Please check the current syllabus carefully when preparing students.

A very well answered question. The vast majority of candidates understood that triple bonds are stronger than single bonds and result in a shorter bond length. It was disappointing, however, to see a considerable number of candidates stating that ethane has a double bond.

Some candidates could not relate evaporation of a liquid to the breaking of its intermolecular forces and gave irrelevant answers such as “evaporation”. Other candidates gave general answers such as “the intermolecular forces” or used the term “van der Waals’ forces” which did not gain credit as too vague. The current guide is clear that “London/dispersion forces” is the appropriate term to use for instantaneous dipole-induced dipole forces. Less than 40 % of the candidates scored the mark. It was disappointing to see some candidates state “covalent bonding” as the type of interaction that must be overcome when liquid ethyne vaporizes. Some teachers thought the wording of the question may have been vague and candidates may have been confused about what was meant by the “type of interaction”.

About 60 % of the candidates stated “addition” as the type of reactions that compounds containing carbon-carbon double bonds underwent. It was disappointing to see a variety of answers including substitution, condensation and combustion showing a total lack of understanding. Some candidates gave specific types such as "bromination" or “hydration” which did not receive the mark.

60 % of the candidates were able to name compound B as ethanal. Some candidates did not recognize it as an aldehyde and gave names related to carboxylic acids or other homologous series. Other candidates called it methanal.

Candidates were confident in using average bond enthalpies for calculating the enthalpy change for the reaction. Mistakes included forgetting to include the breaking of the O-H bonds in water and reversing the signs.

Reasonably well answered. About half of the candidates showed understanding of the relation between stability and the enthalpy change from the same starting materials. ECF was applied in this question based on the answer in part (iii).

The majority of candidates handled this question competently and nearly half of the candidates obtained both marks. They obtained the value of the absorption from the spectra provided and compared it to the values in the data booklet to deduce the identity of the product. Common mistakes included not identifying the peaks and signals precisely (for example C=O instead of CHO for ¹H NMR signal at 9.4-10.0 ppm). Some teachers commented that the TMS signal should not have been included as the SL do not know about it. Other teachers commented that using the 'actual' rather than an ‘idealized’ IR spectrum may have caused confusion due to the peak at around 3400 cm^-1 which could be confused for O-H in alcohols. Thankfully both of these answers were hardly seen in the scripts. The peak at 3400 cm^-1 was not at all broad and did not confuse the majority of students. Please note that real spectra are usually used in examination papers, and it is worth encouraging students to check more than one peak to confirm their deductions.

Surprisingly, this question was not answered well by the majority of the candidates. However, it did discriminate well between high-scoring and low-scoring candidates. Common mistakes included incorrect formulas (such as K₂CrO₇), missing the acidic conditions and stating “reflux” instead of “distillation”. Many candidates gave completely irrelevant reagents and conditions such as “oxygen, pressure and a nickel catalyst”. It is possible that some candidates did not think of “distillation” as a “condition”.

About 60 % of the candidates determined the average oxidation state of carbon in ethanal. A couple of teachers commented that asking SL students to determine an “average oxidation state” seems a little difficult. Please note that this term has been used in recent papers whenever there are two or more atoms of the element in different parts of the compound. There was no evidence of confusion on the part of the candidates and most answered the question well.

This was a challenging question with a demanding markscheme. Most students missed the fact that ethanal can form hydrogen bonds with water. And students who did state this often achieved only 1 out of the 3 marks because they did not offer a full explanation. Some candidates stating "hydrogen bonding" showed confusion by mentioning the hydrogen of the aldehyde group. Few identified the lone pairs on oxygen as the reason for the ability to hydrogen bond. Most candidates just stated that ethanal is polar and dissolves in polar water achieving no marks. However, one mark was awarded for “dipole-dipole interactions with water”.

Outline why the rate of the reaction decreases with time.

Sketch, on the same graph, the expected results if the experiment were repeated using powdered magnesium, keeping its mass and all other variables unchanged.

Nitrogen dioxide and carbon monoxide react according to the following equation:

NO₂(g) + CO(g) $\rightleftharpoons$ NO(g) + CO₂(g)               ΔH = –226 kJ

Calculate the activation energy for the reverse reaction.

State the equation for the reaction of NO₂ in the atmosphere to produce acid deposition.

concentration of acid decreases
OR
surface area of magnesium decreases

Accept “less frequency/chance/rate/probability/likelihood of collisions”.

Do not accept just “less acid” or “less magnesium”.

Do not accept “concentrations of reagents decrease”.

[1 mark]

curve starting from origin with steeper gradient AND reaching same maximum volume

[1 mark]

«E_a(rev) = 226 + 132 =» 358 «kJ»

Do not accept –358.

[1 mark]

2NO₂(g) + H₂O(l) → HNO₃(aq) + HNO₂(aq)
OR
2NO₂(g) + 2H₂O(l) + O₂(g) → 4HNO₃(aq)

Accept ionised forms of the acids.

[1 mark]

Using the graph, estimate the initial temperature of the solution.

Determine the maximum temperature reached in the experiment by analysing the graph.

Calculate the concentration of ethanoic acid, CH₃COOH, in mol dm^–3.

Determine the heat change, q, in kJ, for the neutralization reaction between ethanoic acid and sodium hydroxide.

Assume the specific heat capacities of the solutions and their densities are those of water.

Calculate the enthalpy change, ΔH, in kJ mol^–1, for the reaction between ethanoic acid and sodium hydroxide.

Explain the shape of curve X in terms of the collision theory.

Suggest one possible reason for the differences between curves X and Y.

21.4 °C

Accept values in the range of 21.2 to 21.6 °C.

29.0 «°C»

Accept range 28.8 to 29.2 °C.

ALTERNATIVE 1

«volume CH₃COOH =» 26.0 «cm³»

«[CH₃COOH] = 0.995 mol dm^–3 \( \times \frac{{50.0\,{\text{cm³}}}}{{26.0\,{\text{cm³}}}} = \)» 1.91 «mol dm⁻³»

ALTERNATIVE 2

«n(NaOH) =0.995 mol dm^–3 x 0.0500 dm³ =» 0.04975 «mol»

«[CH₃COOH] = $\frac{0.04975}{0.0260}$ dm³ =» 1.91 «mol dm^–3»

Accept values of volume in range 25.5 to 26.5 cm³.

Award [2] for correct final answer.

«total volume = 50.0 + 26.0 =» 76.0 cm³ AND «temperature change 29.0 – 21.4 =» 7.6 «°C»

«q = 0.0760 kg x 4.18 kJ kg^–1 K^–1 x 7.6 K =» 2.4 «kJ»

Award [2] for correct final answer.

«n(NaOH) = 0.995 mol dm^–3 x 0.0500 dm³ =» 0.04975 «mol»

OR

«n(CH₃COOH) = 1.91 mol dm^–3 x 0.0260 dm³ =» 0.04966 «mol»

«ΔH = $-\frac{2.4\text{kJ}}{0.04975\text{mol}}=$» –48 / –49 «kJ mol^–1»

Award [2] for correct final answer.

Negative sign is required for M2.

«initially steep because» greatest concentration/number of particles at start

OR

«slope decreases because» concentration/number of particles decreases

volume produced per unit of time depends on frequency of collisions

OR

rate depends on frequency of collisions

mass/amount/concentration of metal carbonate more in X

OR

concentration/amount of CH₃COOH more in X

Identify the type of reaction in step 1.

Calculate the standard enthalpy change, ΔH^Θ, of step 2 using section 13 of the data booklet.

Determine the standard enthalpy change, ΔH^Θ, of step 1.

Suggest one reason why the calculated value of ΔH^Θ using Hess’s Law in part (c) can be considered accurate and one reason why it can be considered approximate.

«electrophilic» addition/A_E
OR
reduction ✔

Accept “hydrogenation”.

«(−286 kJ) + (−1411 kJ) =» −1697 «kJ» ✔

«−1697 kJ + 1561 kJ =» −136 «kJ»

OR

«ΔH^Θ = Δ$H_{\text{f}}^{\theta }$ (products) − Δ$H_{\text{f}}^{\theta }$ (reactants) = −84 kJ − 52 kJ =» −136 «kJ» ✔

Accurate:
no approximations were made in the cycle
OR
values are specific to the compounds
OR
Hess’s law is a statement of conservation of energy
OR
method is based on a law
OR
data in table has small uncertainties ✔

Approximate:
values were experimentally determined/had uncertainties
OR
each value has been determined to only three/four significant figures
OR
different sources have «slightly» different values for enthalpy of combustion
OR
law is valid until disproved
OR
law of conservation of energy is now conservation of mass–energy
OR
small difference between two quite large terms «leads to high percentage uncertainty» ✔

Hydrogen gas can be formed industrially by the reaction of natural gas with steam.

                                          CH₄(g) + H₂O(g) → 3H₂(g) + CO(g)

Determine the enthalpy change, ΔH, for the reaction, in kJ, using section 11 of the data booklet.

Bond enthalpy for C≡O: 1077 kJ mol⁻¹

Outline why no value is listed for H₂(g).

Determine the value of ΔH^Θ, in kJ, for the reaction using the values in the table.

Outline why the value of enthalpy of reaction calculated from bond enthalpies is less accurate.

bonds broken: 4(C–H) + 2(H–O)/4(414) + 2(463)/2582 «kJ»

bonds made: 3(H–H) + C≡O/3(436) + 1077/2385 «kJ»

ΔH «= ΣBE_{(bonds broken)} – ΣBE_{(bonds made)} = 2582 – 2385» = «+» 197 «kJ»

Award [3] for correct final answer.

Award [2 max] for –197 «kJ».

[3 marks]

$\mathrm{\Delta }{H}_{\text{f}}^{\mathrm{\Theta }}$ for any element = 0 «by definition»

OR

no energy required to form an element «in its stable form» from itself

[1 mark]

ΔH^Θ « = $\sum \mathrm{\Delta }{H}_{\text{f}}^{\mathrm{\Theta }}$_(products) – $\sum \mathrm{\Delta }{H}_{\text{f}}^{\mathrm{\Theta }}$_(reactants) = –111 + 0 – [–74.0 + (–242)]»

= «+» 205 «kJ»

[1 mark]

«bond enthalpies» averaged values «over similar compounds»

OR

«bond enthalpies» are not specific to these compounds

[1 mark]

Calcium carbonate is heated to produce calcium oxide, CaO.

CaCO₃(s) → CaO (s) + CO₂(g)

Calculate the volume of carbon dioxide produced at STP when 555 g of calcium carbonate decomposes. Use sections 2 and 6 of the data booklet.

Thermodynamic data for the decomposition of calcium carbonate is given.

Calculate the enthalpy change of reaction, ΔH, in kJ, for the decomposition of calcium carbonate.

The potential energy profile for a reaction is shown. Sketch a dotted line labelled “Catalysed” to indicate the effect of a catalyst.

Outline why a catalyst has such an effect.

Write the equation for the reaction of Ca(OH)₂ (aq) with hydrochloric acid, HCl (aq).

Determine the volume, in dm³, of 0.015 mol dm⁻³ calcium hydroxide solution needed to neutralize 35.0 cm³ of 0.025 mol dm⁻³ HCl (aq).

Saturated calcium hydroxide solution is used to test for carbon dioxide. Calculate the pH of a 2.33 × 10⁻²mol dm⁻³ solution of calcium hydroxide, a strong base.

Determine the mass, in g, of CaCO₃(s) produced by reacting 2.41 dm³ of 2.33 × 10⁻² mol dm⁻³ of Ca(OH)₂ (aq) with 0.750 dm³ of CO₂ (g) at STP.

2.85 g of CaCO₃ was collected in the experiment in e(i). Calculate the percentage yield of CaCO₃.

(If you did not obtain an answer to e(i), use 4.00 g, but this is not the correct value.)

Outline how one calcium compound in the lime cycle can reduce a problem caused by acid deposition.

«n_CaCO3 = $\frac{555 \mathrm{g}}{11.09 \mathrm{g} {\mathrm{mol}}^{-1}}$=» 5.55 «mol» ✓

«V = 5.55 mol × 22.7 dm³ mol⁻¹ =» 126 «dm³» ✓

Award [2] for correct final answer.

Accept method using pV = nRT to obtain the volume with p as either 100 kPa (126 dm³) or 101.3 kPa (125 dm³).

Do not penalize use of 22.4 dm³ mol^–1 to obtain the volume (124 dm³).

«ΔH =» (−635 «kJ» – 393.5 «kJ») – (−1207 «kJ») ✓

«ΔH = + » 179 «kJ» ✓

Award [2] for correct final answer.

Award [1 max] for −179 kJ.

Ignore an extra step to determine total enthalpy change in kJ: 179 kJ mol⁻¹ x 5.55 mol = 993 kJ.

Award [2] for an answer in the range 990 - 993« kJ».

lower activation energy curve between same reactant and product levels ✓

Accept curve with or without an intermediate.

Accept a horizontal straight line below current line with the activation energy with catalyst/E_cat clearly labelled.

provides an alternative «reaction» pathway/mechanism ✓

Do not accept “lower activation energy” only.

Ca(OH)₂ (aq) + 2HCl (aq) → 2H₂O (l) + CaCl₂ (aq) ✓

«n_HCl = 0.0350 dm³ × 0.025 mol dm⁻³ =» 0.00088 «mol»
OR
n_Ca(OH)2 = $\frac{1}{2}$ n_HCl/0.00044 «mol» ✓

«V = $\frac{{\displaystyle \frac{1}{2}\times 0.00088 \mathrm{mol}}}{0.015 \mathrm{mol} {\mathrm{dm}}^{-3}}$ =» 0.029 «dm³» ✓

Award [2] for correct final answer.

Award [1 max] for 0.058 «dm³».

Alternative 1:

[OH⁻] = « 2 × 2.33 × 10⁻² mol dm⁻³ =» 0.0466 «mol dm⁻³» ✓

«[H⁺] = $\frac{1.00\times {10}^{-14}}{0.0466}$ = 2.15 × 10⁻¹³ mol dm⁻³»
pH = « −log(2.15 × 10⁻¹³) =» 12.668 ✓

Alternative 2:

[OH⁻] =« 2 × 2.33 × 10⁻² mol dm⁻³ =» 0.0466 «mol dm⁻³» ✓

«pOH = −log (0.0466) = 1.332»

pH = «14.000 – pOH = 14.000 – 1.332 =» 12.668 ✓

Award [2] for correct final answer.

Award [1 max] for pH =12.367.

«n_Ca(OH)2 = 2.41 dm³ × 2.33 × 10⁻²mol dm⁻³ =» 0.0562 «mol» AND
«n_CO2 =$\frac{0.750 {\mathrm{dm}}^3}{22.7 \mathrm{mol} {\mathrm{dm}}^{-3}}$=» 0.0330 «mol» ✓

«CO₂ is the limiting reactant»

«m_CaCO3 = 0.0330 mol × 100.09 g mol⁻¹ =» 3.30 «g» ✓

Only award ECF for M2 if limiting reagent is used.

Accept answers in the range 3.30 - 3.35 «g».

«$\frac{2.85}{3.30}$ × 100 =» 86.4 «%» ✓

Accept answers in the range 86.1-86.4 «%».

Accept “71.3 %” for using the incorrect given value of 4.00 g.

«add» Ca(OH)₂/CaCO₃/CaO AND to «acidic» water/river/lake/soil
OR
«use» Ca(OH)₂/CaCO₃/CaO in scrubbers «to prevent release of acidic pollution» ✓

Accept any correct name for any of the calcium compounds listed.

Explain the general increasing trend in the first ionization energies of the period 3 elements, Na to Ar.

Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.

State an equation for the reaction of phosphorus (V) oxide, P₄O₁₀ (s), with water.

Describe the emission spectrum of hydrogen.

Identify the strongest reducing agent in the given list.

A voltaic cell is made up of a Mn²⁺/Mn half-cell and a Ni²⁺/Ni half-cell.

Deduce the equation for the cell reaction.

The voltaic cell stated in part (ii) is partially shown below.

Draw and label the connections needed to show the direction of electron movement and ion flow between the two half-cells.

increasing number of protons

OR

increasing nuclear charge

«atomic» radius/size decreases

OR

same number of shells

OR

similar shielding «by inner electrons»

«greater energy needed to overcome increased attraction between nucleus and electrons»

atomic/ionic radius increases

smaller charge density

OR

force of attraction between metal ions and delocalised electrons decreases

Do not accept discussion of attraction between valence electrons and
nucleus for M2.

Accept “weaker metallic bonds” for M2.

P₄O₁₀ (s) + 6H₂O (l) → 4H₃PO₄ (aq)

Accept “P₄O₁₀ (s) + 2H₂O (l) → 4HPO₃ (aq)” (initial reaction).

«series of» lines

OR

only certain frequencies/wavelengths

convergence at high«er» frequency/energy/short«er» wavelength

M1 and/or M2 may be shown on a diagram.

Mn

Mn (s) + Ni²⁺ (aq) → Ni (s) + Mn²⁺ (aq)

wire between electrodes AND labelled salt bridge in contact with both electrolytes

anions to right (salt bridge)
OR
cations to left (salt bridge)
OR
arrow from Mn to Ni (on wire or next to it)

Electrodes can be connected directly or through voltmeter/ammeter/light bulb, but not a battery/power supply.

Accept ions or a specific salt as the label of the salt bridge.

Outline why ozone in the stratosphere is important.

State one analytical technique that could be used to determine the ratio of ¹⁴N:¹⁵N.

A sample of gas was enriched to contain 2 % by mass of ¹⁵N with the remainder being ¹⁴N.

Calculate the relative molecular mass of the resulting N₂O.

Predict, giving two reasons, how the first ionization energy of ¹⁵N compares with that of ¹⁴N.

Suggest why it is surprising that dinitrogen monoxide dissolves in water to give a neutral solution.

absorbs UV/ultraviolet light «of longer wavelength than absorbed by O₂»  [✔]

mass spectrometry/MS  [✔]

« $\frac{(98\times 14)+(2\times 15)}{100}=$» 14.02  [✔]

«M_r = (14.02 × 2) + 16.00 =» 44.04  [✔]

Any two:
same AND have same nuclear charge/number of protons/Z_eff  [✔]

same AND neutrons do not affect attraction/ionization energy/Z_eff
OR
same AND neutrons have no charge [✔]

same AND same attraction for «outer» electrons [✔]

same AND have same electronic configuration/shielding [✔]

Note: Accept “almost the same”.
“same” only needs to be stated once.

oxides of nitrogen/non-metals are «usually» acidic  [✔]

60 % of the candidates were aware that ozone in the atmosphere absorbs UV light. Some candidates did not gain the mark for not specifying the type of radiation absorbed.

Well answered. More than half of the candidates stated mass spectrometry is used to determine the ratio of the isotopes.

Many candidates successfully calculated the relative atomic mass of nitrogen in the sample. M2 was awarded independently of M1, so candidates who calculated the relative molecular mass using the A_r of nitrogen in the data booklet (14.01) were awarded M2. Many candidates scored both marks.

This was a challenging question for many candidates, while stronger candidates often showed clarity of thinking and were able to conclude that the ionization energies of the two isotopes must be the same and to provide two different reasons for this. Some candidates did realize that the ionization energies are similar but did not give the best reasons to support their answer. Many candidates thought the ionization energies would be different because the size of the nucleus was different. Some teachers commented that the question was difficult while others liked it because it made students apply their knowledge in an unfamiliar situation. The question had a good discrimination index.

Only a quarter of the candidates answered correctly. Some simply stated that N₂O forms HNO₃ with water which did not gain the mark.

Determine the coefficients that balance the equation for the reaction of lithium with water.

Calculate the molar concentration of the resulting solution of lithium hydroxide.

Calculate the volume of hydrogen gas produced, in cm³, if the temperature was 22.5 °C and the pressure was 103 kPa. Use sections 1 and 2 of the data booklet.

Suggest a reason why the volume of hydrogen gas collected was smaller than predicted.

The reaction of lithium with water is a redox reaction. Identify the oxidizing agent in the reaction giving a reason.

Describe two observations that indicate the reaction of lithium with water is exothermic.

2 Li (s) + 2 H₂O (l) → 2 LiOH (aq) + H₂(g) ✔

$n_{Li}«\frac{0200 g}{6.94 g}=»0.0288«mol»$✔

«n_LiOH = n_Li»

$\left[\mathrm{LiOH}\right] «=\frac{0.0288 mol}{0.5000 d{m}^3}=»0.0576 «mol d{m}^{-3}»$ ✔

Award [2] for correct final answer.

$«n_{H_2}=\frac{1}{2}\times 0.0288 mol=0.0144 mol»$

$«V=\frac{nRT}{P}=»\left(\frac{0.0144 mol\times 8.31 J{K}^{-1}mo{l}^{-1}\times \left(22.5+273\right)K}{103 kPa}\right) «\times {10}^3»$✔

$V=343 «{\mathrm{cm}}^3»$✔

Award [2] for correct final answer.

Accept answers in the range 334 – 344 cm³.

Award [1 max] for 0.343 «cm³/dm³/m³».

Award [1 max] for 26.1 cm³ obtained by using 22.5 K.

Award [1 max] for 687 cm³ obtained by using 0.0288 mol.

lithium was impure/«partially» oxidized

OR

gas leaked/ignited ✔

Accept “gas dissolved”.

H₂O AND hydrogen gains electrons «to form H₂»

OR

H₂O AND H oxidation state changed from +1 to 0 ✔

Accept “H₂O AND H/H₂O is reduced”.

Any two:

temperature of the water increases ✔

lithium melts ✔

pop sound is heard ✔

Accept “lithium/hydrogen catches fire”.

Do not accept “smoke is observed”.

This part-question was better answered than part (ii). 50% of the candidates drew a correct arrow between n=2 and n=3. Both absorption and emission transitions were accepted since the question did not specify which type of spectrum was required. Some teachers commented on this in their feedback. Mistakes often included transitions between higher energy levels.

Determine the standard enthalpy change, $∆H^{⦵}$, for this reaction, using section 11 of the data booklet.

Calculate the standard enthalpy change, $∆H^{⦵}$, for this reaction using section 12 of the data booklet.

Bonds broken: $8(\mathrm{C}–\mathrm{H})+2(\mathrm{C}–\mathrm{C})+5(\mathrm{O}=\mathrm{O})$ $/8\times 414 «\mathrm{kJ} {\mathrm{mol}}^{-1}»+2\times 346 «\mathrm{kJ} {\mathrm{mol}}^{-1}»+5\times 498 «\mathrm{kJ} {\mathrm{mol}}^{-1}»/6494 «\mathrm{kJ}»$ ✔

Bonds formed: $6(\mathrm{C}=\mathrm{O})+8(\mathrm{O}–\mathrm{H})/6\times 804 «\mathrm{kJ} {\mathrm{mol}}^{-1}»+8\times 463 «\mathrm{kJ} {\mathrm{mol}}^{-1}»/8528 «\mathrm{kJ}»$ ✔

«Enthalpy change$=$bonds broken$-$bonds formed $=6494 \mathrm{kJ}-8528 \mathrm{kJ}=»-2034 «\mathrm{kJ}»$ ✔

Award [3] for correct final answer.

$4(-241.8 «\mathrm{kJ}»)$ AND $3(-393.5 «\mathrm{kJ}»)$ AND $«1»(-105 «\mathrm{kJ}»)$ ✔

$«{\mathrm{\Delta H}}^{⦵}=4(-241.8 «\mathrm{kJ}»)+3(-393.5 «\mathrm{kJ}»)–«1»(-105 «\mathrm{kJ}»)=»-2043 «\mathrm{kJ}»$ ✔

Award [2] for correct final answer.

Award [1 max] for $\mathit{-}\mathit{2219}\mathit{ }«kJ»$.

This question was generally well answered. The mean mark on the question was 1.8 out of 3 marks. Mistakes included three C–C (instead of two), missing C–C bonds completely, subtracting bonds formed bonds broken, 3 and 4 instead of 6 and 8 for bonds formed coefficients, and bond enthalpies of C=O and O=O double bonds treated as single bonds.

About half of the candidates obtained both marks. In the incorrect answers coefficients were sometimes ignored, some candidates used the wrong state for water, and some candidates did not realize the heat of formation of O₂ was zero and inserted the bond enthalpy for O₂ in the calculation.

State the type of reaction which converts ethene into C₂H₅Cl.

Write an equation for the reaction of C₂H₅Cl with aqueous sodium hydroxide to produce a C₂H₆O compound, showing structural formulas.

Write an equation for the complete combustion of the organic product in (b).

Determine the enthalpy of combustion of the organic product in (b), in kJ mol⁻¹, using data from section 11 of the data booklet.

State the reagents and conditions for the conversion of the compound C₂H₆O, produced in (b), into C₂H₄O.

Explain why the compound C₂H₆O, produced in (b), has a higher boiling point than compound C₂H₄O, produced in d(i).

Ethene is often polymerized. Draw a section of the resulting polymer, showing two repeating units.

«electrophilic» addition ✔

NOTE: Do not accept “nucleophilic addition” or “free radical addition”.
Do not accept “halogenation”.

CH₃CH₂Cl (g) + OH⁻ (aq) → CH₃CH₂OH (aq) + Cl⁻ (aq)
OR
CH₃CH₂Cl (g) + NaOH (aq) → CH₃CH₂OH (aq) + NaCl (aq) ✔

C₂H₆O (g) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)
OR
CH₃CH₂OH (g) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g) ✔

bonds broken:
5(C–H) + C–C + C–O + O–H + 3(O=O)
OR
5(414«kJ mol⁻¹») + 346«kJ mol⁻¹» + 358«kJ mol⁻¹» + 463«kJ mol⁻¹» + 3(498«kJ mol⁻¹») / 4731 «kJ» ✔

bonds formed:
4(C=O) + 6(O–H)
OR
4(804«kJ mol⁻¹») + 6(463«kJ mol⁻¹») / 5994 «kJ» ✔
«ΔH = bonds broken − bonds formed = 4731 − 5994 =» −1263 «kJ mol⁻¹» ✔

NOTE: Award [3] for correct final answer.

K₂Cr₂O₇/Cr₂O₇²⁻/«potassium» dichromate «(VI)» AND acidified/H⁺
OR
«acidified potassium» manganate(VII) / «H⁺» KMnO₄ / «H⁺» MnO₄⁻ ✔

NOTE: Accept “H₂SO₄” or “H₃PO₄” for “H⁺”.
Do not accept “HCl”.
Accept “permanganate” for “manganate(VII)”.

distil ✔

C₂H₆O/ethanol: hydrogen-bonding AND C₂H₄O/ethanal: no hydrogen-bonding/«only» dipole–dipole forces ✔

hydrogen bonding stronger «than dipole–dipole» ✔

NOTE: Continuation bonds must be shown.
Ignore square brackets and “n”.

The diagram shows the Maxwell-Boltzmann curve for the uncatalyzed reaction.

Draw a distribution curve at a lower temperature (T₂) and show on the diagram how the addition of a catalyst enables the reaction to take place more rapidly than at T₁.

The hydrogen peroxide could cause further oxidation of the methanol. Suggest a possible oxidation product.

Determine the overall equation for the production of methanol.

8.00 g of methane is completely converted to methanol. Calculate, to three significant figures, the final volume of hydrogen at STP, in dm³. Use sections 2 and 6 of the data booklet.

Determine the enthalpy change, ΔH, in kJ. Use section 11 of the data booklet.

Bond enthalpy of CO = 1077 kJ mol⁻¹.

State the expression for K_c for this stage of the reaction.

State and explain the effect of increasing temperature on the value of K_c.

curve higher AND to left of T₁ ✔

new/catalysed E_a marked AND to the left of E_a of curve T₁ ✔

Do not penalize curve missing a label, not passing exactly through the origin, or crossing x-axis after E_a.
Do not award M1 if curve drawn shows significantly more/less molecules/greater/smaller area under curve than curve 1.
Accept E_a drawn to T₁ instead of curve drawn as long as to left of marked E_a.

methanoic acid/HCOOH/CHOOH
OR
methanal/HCHO ✔

Accept “carbon dioxide/CO₂”.

CH₄(g) + H₂O(g) $\rightleftharpoons$ CH₃OH(l) + H₂(g) ✔

Accept arrow instead of equilibrium sign.

amount of methane = « $\frac{8.00 \mathrm{g}}{16.05 \mathrm{g} {\mathrm{mol}}^{-1}}$ = » 0.498 «mol» ✔

amount of hydrogen = amount of methane / 0.498 «mol» ✔

volume of hydrogen = «0.498 mol × 22.7 dm³mol⁻¹ = » 11.3 «dm³» ✔

Award [3] for final correct answer.
Award [2 max] for 11.4 «dm³ due to rounding of mass to 16/moles to 0.5. »

Σbonds broken = 4 × 414 «kJ» + 2 × 463 «kJ» / 2582 «kJ» ✔

Σbonds formed = 1077 «kJ» + 3 × 436 «kJ» / 2385 «kJ» ✔

ΔH «= Σbonds broken − Σbonds formed =( 2582 kJ − 2385 kJ)» = «+»197«kJ» ✔

Award [3] for final correct answer.
Award [2 Max] for final answer of −197 «kJ»

$K_{\mathit{c}}=\frac{\left[\mathrm{CO}\right]{\left[{\mathrm{H}}_2\right]}^3}{\left[{\mathrm{CH}}_4\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}$ ✔

K_c increases AND «forward» reaction endothermic ✔

Identify the type of reaction.

Outline the role of the hydroxide ion in this reaction.

Suggest, with a reason, why 1-iodopentane reacts faster than 1-chloropentane under the same conditions. Use section 11 of the data booklet for consistency.

Sketch labelled Maxwell–Boltzmann energy distribution curves at the original temperature (T₁) and the new lower temperature (T₂).

Explain the effect of lowering the temperature on the rate of the reaction.

«nucleophilic» substitution/S_N2 ✔

Do not accept if “electrophilic” or “free radical” substitution is stated.

«acts as a» nucleophile/Lewis base
OR
donates/provides lone pair «of electrons»
OR
attacks the «partially» positive carbon ✔

bond enthalpy C–I lower than C–Cl
OR
C–I bond weaker than C–Cl ✔

«weaker bond» broken more easily/with less energy
OR
lower Ea «for weaker bonds» ✔

Accept the bond enthalpy values for C–I and C–Cl for M1.

peak at T₁ to right of AND lower than T₂ ✔

lines begin at origin AND T₁ must finish above T₂ ✔

«rate is» lower AND «average» kinetic energy of molecules is lower
OR
«rate is» lower AND less frequent collisions
OR
«rate is» lower AND fewer collisions per unit time ✔

«rate is» lower AND fewer/smaller fraction of molecules/collisions have the E ≥ E_a ✔

Lower «rate» needs to be mentioned once only.

Do not accept “fewer collisions” without reference to time/frequency/probability for M1.

Ethyne, like ethene, undergoes hydrogenation to form ethane. State the conditions required.

Outline the formation of polyethene from ethene by drawing three repeating units of the polymer.

Under certain conditions, ethyne can be converted to benzene.

Determine the standard enthalpy change, ΔH^ϴ, for the reaction stated, using section 11 of the data booklet.

                                        3C₂H₂(g) → C₆H₆(g)

Determine the standard enthalpy change, ΔH^Θ, for the following similar reaction, using ΔH_f values in section 12 of the data booklet.

3C₂H₂(g) → C₆H₆(l)

Explain, giving two reasons, the difference in the values for (b)(i) and (ii). If you did not obtain answers, use −475 kJ for (i) and −600 kJ for (ii).

One possible Lewis structure for benzene is shown.

State one piece of physical evidence that this structure is incorrect.

State the characteristic reaction mechanism of benzene.

nickel/Ni «catalyst»

high pressure

OR

heat

Accept these other catalysts: Pt, Pd, Ir, Rh, Co, Ti.

Accept “high temperature” or a stated temperature such as “150 °C”.

[2 marks]

Ignore square brackets and “n”.

Connecting line at end of carbons must be shown.

[1 mark]

ΔH^ϴ = bonds broken – bonds formed

«ΔH^ϴ = 3(C≡C) – 6(C=C)_benzene/3 × 839 – 6 × 507 / 2517 – 3042 =»

–525 «kJ»

Award [2] for correct final answer.

Award [1 max] for +525 «kJ»

Award [1 max] for:

«ΔH^ϴ = 3(C≡C) – 3(C–C) – 3(C=C) / 3 × 839 – 3 × 346 – 3 × 614 / 2517 – 2880 =» –363 «kJ».

[2 marks]

ΔH^Θ = ΣΔH_f(products) – ΣΔH_f(reactants)

«ΔH^Θ = 49 kJ – 3 × 228 kJ =» –635 «kJ»

Award [2] for correct final answer.

Award [1 max] for “+635 «kJ»”.

[2 marks]

ΔH_f values are specific to the compound

OR

bond enthalpy values are averages «from many different compounds»

condensation from gas to liquid is exothermic

Accept “benzene is in two different states «one liquid the other gas»“ for M2.

[2 marks]

equal C–C bond «lengths/strengths»

OR

regular hexagon

OR

«all» C–C have» bond order of 1.5

OR

«all» C–C intermediate between single and double bonds

Accept “all C–C–C bond angles are equal”.

[1 mark]

electrophilic substitution

OR

S_E

[1 mark]

Determine the molar enthalpy of combustion of an alkane if 8.75 × 10⁻⁴ moles are burned, raising the temperature of 20.0 g of water by 57.3 °C.

Formulate equations for the two propagation steps and one termination step in the formation of chloroethane from ethane.

«q = mcΔT = 20.0 g × 4.18 J g⁻¹°C⁻¹ × 57.3 °C =» 4790 «J» ✔

«$∆H_{\text{c}}\frac{4790 \mathrm{J}}{{\displaystyle \frac{1000}{8.75\times {10}^{-4} \mathrm{mol}}}}=$» –5470 «kJ mol^–1» ✔

Award [2] for correct final answer.

Accept answers in the range –5470 to –5480 «kJ mol⁻¹».

Accept correct answer in any units, e.g. –5.47 «MJ mol⁻¹» or 5.47 x 10⁶ «J mol⁻¹».

Cl· + C₂H₆ → ·C₂H₅ + HCl ✔

·C₂H₅ + Cl₂ → Cl· + C₂H₅Cl ✔

·C₂H₅ + Cl· → C₂H₅Cl
OR
Cl· + Cl· → Cl₂
OR
·C₂H₅ + ·C₂H₅ → C₄H₁₀ ✔

Do not penalize incorrectly placed radical sign, eg C₂H₅·.

Calculate the enthalpy change, in kJ, for the spray reaction, using the data below.

The energy released by the reaction of one mole of hydrogen peroxide with hydroquinone is used to heat 850 cm³ of water initially at 21.8°C. Determine the highest temperature reached by the water.

Specific heat capacity of water = 4.18 kJ$$kg⁻¹$$K⁻¹.

(If you did not obtain an answer to part (i), use a value of 200.0 kJ for the energy released, although this is not the correct answer.)

Identify the species responsible for the peak at m/z = 110 in the mass spectrum of hydroquinone.

Identify the highest m/z value in the mass spectrum of quinone.

ΔH = 177.0 – $\frac{189.2}{2}$ –285.5 «kJ»

«ΔH =» –203.1 «kJ»

Accept other methods for correct manipulation of the three equations.

Award [2] for correct final answer.

[2 marks]

203.1 «kJ» = 0.850 «kg» x 4.18 «kJ$$kg^–1$$K^–1» x ΔT «K»
OR
«ΔT =» 57.2 «K»
«T_final = (57.2 + 21.8) °C =» 79.0 «°C» / 352.0 «K»

If 200.0 kJ was used:
200.0 «kJ» = 0.850 «kg» x 4.18 «kJ$$kg^–1$$K^–1» x ΔT «K»
OR
«ΔT =» 56.3 «K»
«T_final = (56.3 + 21.8) °C =» 78.1 «°C» / 351.1 «K»

Award [2] for correct final answer.

Units, if specified, must be consistent with the value stated.

[2 marks]

C₆H₄(OH)₂⁺

Accept “molecular ion”.

Do not accept “C₆H₄(OH)₂” (positive charge missing).

[1 mark]

«highest m/z» 108

Only accept exactly 108, not values close to this.

[1 mark]

Draw arrows in the boxes to represent the electron configuration of a nitrogen atom.

Draw the Lewis (electron dot) structure of the ammonia molecule.

Deduce the expression for the equilibrium constant, K_c, for this equation.

Explain why an increase in pressure shifts the position of equilibrium towards the products and how this affects the value of the equilibrium constant, K_c.

State how the use of a catalyst affects the position of the equilibrium.

Determine the enthalpy change, ΔH, for the Haber–Bosch process, in kJ. Use Section 11 of the data booklet.

Calculate the enthalpy change, ΔH^⦵, for the Haber–Bosch process, in kJ, using the following data.

$∆H_{ \mathrm{f}}^{⦵}\left({\mathrm{NH}}_3\right)=-46.2 \mathrm{kJ} {\mathrm{mol}}^{-1}$.

Suggest why the values obtained in (d)(i) and (d)(ii) differ.

State the relationship between NH₄⁺ and NH₃ in terms of the Brønsted–Lowry theory.

Determine the concentration, in mol dm^–3, of the solution formed when 900.0 dm³ of NH₃(g) at 300.0 K and 100.0 kPa, is dissolved in water to form 2.00 dm³ of solution. Use sections 1 and 2 of the data booklet.

Calculate the concentration of hydroxide ions in an ammonia solution with pH = 9.3. Use sections 1 and 2 of the data booklet.

Accept all 2p electrons pointing downwards.

Accept half arrows instead of full arrows.

Accept lines or dots or crosses for electrons, or a mixture of these

$K_c=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$ ✔

shifts to the side with fewer moles «of gas»
OR
shifts to right as there is a reduction in volume✔

«value of » K_c unchanged ✔

Accept “K_c only affected by changes in temperature”.

same/unaffected/unchanged ✔

bonds broken: N≡N + 3(H–H) / «1 mol×»945 «kJ mol^–1» + 3«mol»×436 «kJ mol^–1» / 945 «kJ» + 1308 «kJ» / 2253 «kJ» ✔

bonds formed: 6(N–H) / 6«mol»×391 «kJ mol^–1» / 2346 «kJ» ✔

ΔH = «2253 kJ – 2346 kJ = » –93 «kJ» ✔

Award [2 max] for (+)93 «kJ»

–92.4 «kJ» ✔

«N-H» bond enthalpy is an average «and may not be the precise value in NH₃» ✔

Accept it relies on average values not specific to NH₃

conjugate «acid and base» ✔

amount of ammonia $⟨⟨=\frac{P.V}{R.T}=\frac{100.0 kPa\times 900.0 d{m}^3}{8.31 J K^{-1}mo{l}^{-1}\times 300.0 K}⟩⟩ = 36.1 «\mathrm{mol}»$ ✔

concentration $⟨⟨=\frac{n}{V}=\frac{36.1}{2.00}⟩⟩=18.1 «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

Award [2] for correct final answer.

[OH⁻] $⟨⟨=\frac{{\mathrm{K}}_{\mathrm{W}}}{\left[{\mathrm{H}}^{+}\right]}=\frac{{10}^{-14}}{{10}^{-9.3}}={10}^{-4.7}⟩⟩=2.0 \times {10}^{-5} ⟨⟨\mathrm{mol} {\mathrm{dm}}^{-3}⟩⟩$ ✔

Most students realised that the three p-orbitals were all singly filled.

Even more candidates could draw the correct Lewis structure of ammonia, with omission of the lone pair being the most common error.

Most students could deduce the equilibrium constant expression from the equilibrium equation.

Many students realised that increasing pressure shifts an equilibrium to the side with the most moles of gas (though the "of gas" was frequently omitted!) but probably less than half realised that, even though the equilibrium position changes, the value of the equilibrium constant remains constant.

It was pleasing to see that about a third of students gaining full marks and an equal number only lost a single mark because they failed to locate the correct bond enthalpy for molecular nitrogen.

Very few students could determine the enthalpy change from enthalpy of formation data, with many being baffled by the absence of values for the elemental reactants and more than half who overcame this obstacle failed to note that 2 moles of ammonia are produced.

About half the candidates recognised the species as a conjugate acid-base pair, though some lost the mark by confusing the acid and base, even though this information was not asked for.

About 40% of candidates gained full marks for the calculation and a significant number of others gained the second mark to calculate the concentration as an ECF.

This question was very poorly answered with many candidates calculating the [H⁺] instead of [OH^-].

Describe the structure and bonding in solid sodium oxide.

Write equations for the separate reactions of solid sodium oxide and solid phosphorus(V) oxide with excess water and differentiate between the solutions formed.

Sodium oxide, Na₂O:

Phosphorus(V) oxide, P₄O₁₀:

Differentiation:

Sodium peroxide, Na₂O₂, is formed by the reaction of sodium oxide with oxygen.

2Na₂O (s) + O₂ (g) → 2Na₂O₂ (s)

Calculate the percentage yield of sodium peroxide if 5.00 g of sodium oxide produces 5.50 g of sodium peroxide.

Determine the enthalpy change, ΔH, in kJ, for this reaction using data from the table and section 12 of the data booklet.

Outline why bond enthalpy values are not valid in calculations such as that in (c)(i).

The reaction of sodium peroxide with excess water produces hydrogen peroxide and one other sodium compound. Suggest the formula of this compound.

State the oxidation number of carbon in sodium carbonate, Na₂CO₃.

«3-D/giant» regularly repeating arrangement «of ions»
OR
lattice «of ions»  [✔]

electrostatic attraction between oppositely charged ions
OR
electrostatic attraction between Na⁺ and O²⁻ ions  [✔]

Note: Do not accept “ionic” without description.

Sodium oxide:
Na₂O(s) + H₂O(l) → 2NaOH (aq)  [✔]

Phosphorus(V) oxide:
P₄O₁₀ (s) + 6H₂O(l) → 4H₃PO₄ (aq)  [✔]

Differentiation:
NaOH / product of Na₂O is alkaline/basic/pH > 7 AND H₃PO₄ / product of P₄O₁₀ is acidic/pH < 7  [✔]

n(Na₂O₂) theoretical yield «= $\frac{5.00\text{ g}}{61.98\text{ mo}{\text{l}}^{-1}}$» = 0.0807/8.07 × 10⁻² «mol»
OR
mass Na₂O₂ theoretical yield «= $\frac{5.00\text{ g}}{61.98\text{ mo}{\text{l}}^{-1}}$ × 77.98 gmol⁻¹» = 6.291 «g»  [✔]

% yield «= $\frac{5.50\text{ g}}{6.291\text{ g}}$ × 100» OR «$\frac{0.0705}{0.0807}$ x 100» = 87.4 «%»  [✔]

Note: Award [2] for correct final answer.

ΣΔH_f products = 2 × (−1130.7) / −2261.4 «kJ»  [✔]

ΣΔH_f reactants = 2 × (−510.9) + 2 × (−393.5) / −1808.8 «kJ»  [✔]

ΔH = «ΣΔH_f products − ΣΔH_f reactants = −2261.4 −(−1808.8) =» −452.6 «kJ»  [✔]

Note: Award [3] for correct final answer.

Award [2 max] for “+452.6 «kJ»”.

only valid for covalent bonds
OR
only valid in gaseous state  [✔]

NaOH  [✔]

Note: Accept correct equation showing NaOH as a product.

IV  [✔]

Disappointingly many students did not realise that sodium oxide was held by ionic bonds, many said it was covalent or metallic bonding. The ones that knew it was ionic failed to describe it adequately to earn the 2 marks.

Very few students could correctly write out the two equations and so often were unable to realise it was acid/base behaviour that would differentiate the oxides.

Many candidates were able to correctly calculate the % yield but some weaker candidates just used 5.0/5.5 to find %.

The calculation of the enthalpy change using enthalpies of formation was generally answered well but common mistakes were students forgetting to multiply by 2 or adding extra terms for oxygen.

Most students didn’t gain a mark and “values are average” was the most common incorrect answer. The fact this was an ionic compound did not register with them. Some students did gain a mark for stating that the substances were not in a gaseous state.

Some students correctly identified sodium hydroxide as the correct product, but hydrogen, oxygen and sodium oxide were common answers.

Oxidation number of +4 was often correctly identified.

State the electron configuration of the Cu⁺ ion.

Copper(II) chloride is used as a catalyst in the production of chlorine from hydrogen chloride.

4HCl (g) + O₂ (g) → 2Cl₂ (g) + 2H₂O (g)

Calculate the standard enthalpy change, ΔH^θ, in kJ, for this reaction, using section 12 of the data booklet.

The diagram shows the Maxwell–Boltzmann distribution and potential energy profile for the reaction without a catalyst.

Annotate both charts to show the activation energy for the catalysed reaction, using the label E_{a (cat)}.

Explain how the catalyst increases the rate of the reaction.

Solid copper(II) chloride absorbs moisture from the atmosphere to form a hydrate of formula CuCl₂•$x$H₂O.

A student heated a sample of hydrated copper(II) chloride, in order to determine the value of $x$. The following results were obtained:

Mass of crucible = 16.221 g
Initial mass of crucible and hydrated copper(II) chloride = 18.360 g
Final mass of crucible and anhydrous copper(II) chloride = 17.917 g

Determine the value of $x$.

State how current is conducted through the wires and through the electrolyte.

Wires: 

Electrolyte:

Write the half-equation for the formation of gas bubbles at electrode 1.

[Ar] 3d¹⁰
OR
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ ✔

ΔH^θ = ΣΔH^θ_f (products) − ΣΔH^θ_f (reactants) ✔
ΔH^θ = 2(−241.8 «kJ mol⁻¹») − 4(−92.3 «kJ mol⁻¹») = −114.4 «kJ» ✔

NOTE: Award [2] for correct final answer.

E_{a (cat)} to the left of E_a ✔                        

peak lower AND E_{a (cat)} smaller ✔

«catalyst provides an» alternative pathway ✔

«with» lower E_a
OR
higher proportion of/more particles with «kinetic» E ≥ E_a(cat) «than E_a» ✔

mass of H₂O = «18.360 g – 17.917 g =» 0.443 «g» AND mass of CuCl₂ = «17.917 g – 16.221 g =» 1.696 «g» ✔

moles of H₂O = «$\frac{0.443 \text{g}}{18.02 \text{g mo}{\text{l}}^{-1}}$=» 0.0246 «mol»
OR
moles of CuCl₂ =«$\frac{1.696 \text{g}}{134.45 \text{g mo}{\text{l}}^{-1}}$= » 0.0126 «mol» ✔

«water : copper(II) chloride = 1.95 : 1»
«$x$ =» 2 ✔

NOTE: Accept «$x$ =» 1.95.

NOTE: Award [3] for correct final answer.

Wires:
«delocalized» electrons «flow» ✔

Electrolyte:
«mobile» ions «flow» ✔

2Cl⁻ → Cl₂ (g) + 2e⁻
OR
Cl⁻ → $\frac{1}{2}$Cl₂ (g) + e⁻ ✔

NOTE: Accept e for e⁻.

Outline one difference between the bonding of carbon atoms in C₆₀ and diamond.

State two features showing that propane and butane are members of the same homologous series.

Describe a test and the expected result to indicate the presence of carbon–carbon double bonds.

Draw the full structural formula of but-2-ene.

Write the equation for the reaction between but-2-ene and hydrogen bromide.

State the type of reaction.

Suggest two differences in the ¹H NMR of but-2-ene and the organic product from (d)(ii).

Calculate the enthalpy change of the reaction, ΔH, using section 11 of the data booklet.

Draw and label an enthalpy level diagram for this reaction.

C₆₀ fullerene: «each carbon is» bonded to 3 C AND diamond: bonded to 4 C
OR
C₆₀ fullerene: delocalized/resonance AND diamond: not delocalized/no resonance
OR
C₆₀ fullerene: single and double bonds AND diamond: single bonds ✔

Accept “C₆₀ fullerene: sp² AND diamond: sp³”.

Accept “C₆₀ fullerene: trigonal planar geometry / bond angles between 109.5°/109°/108°–120° AND diamond:  tetrahedral geometry / bond angle 109.5°/109°”.

Accept "bonds in fullerene are shorter/stronger/have higher bond order".

same general formula / C_nH_2n+2 ✔

differ by CH₂/common structural unit ✔

Accept "similar chemical properties".

Accept “gradation/gradual change in physical properties”.

ALTERNATIVE 1:

Test:

add bromine «water»/Br₂ (aq) ✔

Result:

«orange/brown/yellow» to colourless/decolourised ✔

Do not accept “clear” for M2.

ALTERNATIVE 2:

Test:

add «acidified» KMnO₄ ✔

Result:

«purple» to colourless/decolourised/brown ✔

Accept “colour change” for M2.

ALTERNATIVE 3:

Test:

add iodine /${\text{I}}_2$ ✔

Result:

«brown» to colourless/decolourised ✔

Accept

CH₃CH=CHCH₃ (g) + HBr (g) → CH₃CH₂CHBrCH₃ (l)

OR

C₄H₈ (g) + HBr (g) → C₄H₉Br (l) ✔

«electrophilic» addition/E_A ✔

Do not accept nucleophilic or free radical addition.

ALTERNATIVE 1: Any two of:

but-2-ene: 2 signals AND product: 4 signals ✔

but-2-ene: «area ratio» 3:1/6:2 AND product: «area ratio» 3:3:2:1 ✔

product: «has signal at» 3.5-4.4 ppm «and but-2-ene: does not» ✔

but-2-ene: «has signal at» 4.5-6.0 ppm «and product: does not» ✔

ALTERNATIVE 2:

but-2-ene: doublet AND quartet/multiplet/4 ✔

product: doublet AND triplet AND quintet/5/multiplet AND sextet/6/multiplet ✔

Accept “product «has signal at» 1.3–1.4 ppm «and but-2-ene: does not»”.

bond breaking: C–H + Cl–Cl / 414 «kJ mol^–1» + 242 «kJ mol^–1»/656 «kJ»
OR
bond breaking: 4C–H + Cl–Cl / 4 × 414 «kJ mol^–1» + 242 «kJ mol^–1» / 1898 «kJ» ✔

bond forming: «C–Cl + H–Cl / 324 kJ mol^–1 + 431 kJ mol^–1» / 755 «kJ»
OR
bond forming: «3C–H + C–Cl + H–Cl / 3 × 414 «kJ mol^–1» + 324 «kJ mol^–1» + 431 kJ mol^–1» / 1997 «kJ» ✔

«ΔH = bond breaking – bond forming = 656 kJ – 755 kJ» = –99 «kJ» ✔

Award [3] for correct final answer.

Award [2 max] for 99 «kJ».

reactants at higher enthalpy than products ✔

ΔH/-99 «kJ» labelled on arrow from reactants to products
OR
activation energy/E_a labelled on arrow from reactant to top of energy profile ✔

Accept a double headed arrow between reactants and products labelled as ΔH for M2.

This was a challenging question that asked about the difference between the bonding of carbon atoms in C₆₀ and diamond. 20% of the candidates gained the mark. The majority of the candidates did not have a specific enough answer for C₆₀ and mentioned the pentagons and hexagons but not the number of bonds or the geometry or the bond order or the electron delocalisation. Diamond was better known to candidates as expected.

About two-thirds of the candidates scored one of the two marks and stronger candidates scored both. The most common answers were the same general formula/C_nH_2n+2, the difference between the compounds was CH₂ and similar chemical properties. The same functional group was not accepted since alkanes do not have a functional group. Some candidates only stated that they are saturated hydrocarbons not gaining any marks.

About half of the candidates gave the bromine water test with the correct results. Iodine and KMnO4 were rarely seen in the scripts. There were candidates who used the term “clear” to mean “colourless” which was not accepted. Some candidates referred to the presence of UV light in a correct way and others in an incorrect way which was not penalized in this case. 10% of the candidates left the question blank. The most common incorrect answer was in terms of the IR absorptions. Other candidates referred to enthalpies of combustion and formation.

A well answered question. 70% of the candidates gave the correct structural formula for but-2-ene. Mistakes included too many hydrogens in the structure and an incorrect position of the C=C. Candidates should be reminded that the full structural formula requires all covalent bonds to be shown.

Half of the candidates wrote the correct equation for the reaction of but-2-ene with hydrogen bromide. Incorrect answers included hydrogen as a product. As expected, the question correlated well with highly achieving candidates.

Well answered. 60% of candidates identified the type of reaction between but-2-ene and HBr, some of them including the term “electrophilic”. ECF was generously awarded when substitution was stated based on the equation where H₂ was produced in part (ii). Candidates lost the mark if they only stated “hydrobromination” without mentioning addition. Some candidates lost the mark for stating “nucleophilic” or “free radical” addition.

The comparison of the ¹H NMR spectra of the two organic compounds was more challenging and 10% of the candidates left this question blank. The average mark was 0.7 out of 2 marks. Mistakes included non-specific answers that just stated “more signals” or “higher chemical shift”, and stating 3 signals in 2-bromobutane instead of 4 signals. Standard level candidates were expected to use the number of signals and the ratio of the areas under the signals to answer the question since they do not cover chemical shift, however, many of them did use the ¹H NMR section in the data booklet to obtain correct answers in terms of chemical shift.

This was the best answered question on the paper. Candidates identified the bonds and used bond enthalpies to calculate the enthalpy of reaction accurately. The most common mistakes were reversing the signs of bonds broken and bonds formed, assuming two Cl-Cl bonds were broken and using an incorrect value of bond enthalpy for one of the bonds.

The majority of candidates drew the enthalpy level diagram and labelled it correctly based on their answer to part (i). Some candidates reversed the products and reactants. A few candidates did not add any labels which prevented the awarding of the second mark. With 2 marks allocated to the question the second mark was awarded for correct labeling of either ΔH or E_a.

Several compounds can be synthesized from but-2-ene. Draw the structure of the final product for each of the following chemical reactions.

Determine the change in enthalpy, ΔH, for the combustion of but-2-ene, using section 11 of the data booklet. 

CH₃CH=CHCH₃(g) + 6O₂ (g) → 4CO₂(g) + 4H₂O (g)

Write the equation and name the organic product when ethanol reacts with methanoic acid.

Oxidation of ethanol with potassium dichromate, K₂Cr₂O₇, can form two different organic products. Determine the names of the organic products and the methods used to isolate them.

Deduce two features of this molecule that can be obtained from the mass spectrum. Use section 28 of the data booklet.

NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce
on behalf of the United States of America. All rights reserved.

Identify the bond responsible for the absorption at A in the infrared spectrum. Use section 26 of the data booklet.

NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce
on behalf of the United States of America. All rights reserved. 

Deduce the identity of the unknown compound using the previous information, the ¹H NMR spectrum and section 27 of the data booklet.

SDBS, National Institute of Advanced Industrial Science and Technology (AIST).

Penalize missing hydrogens in displayed structural formulas once only.

Accept condensed structural formulas: CH₃CH(OH)CH₂CH₃ / CH₃CH₂CH₂CH₃ or skeletal structures.

Bonds broken:
2(C–C) + 1(C=C) + 8(C–H) + 6O=O / 2(346) + 1(614) + 8(414) + 6(498) / 7606 «kJ» ✓

Bonds formed:
8(C=O) + 8(O–H) / 8(804) + 8(463) / 10 136 «kJ» ✓

Enthalpy change:
«Bonds broken – Bonds formed = 7606 kJ – 10 136 kJ =» –2530 «kJ» ✓

Award [2 max] for «+» 2530 «kJ».

Award [3] for correct final answer.

Equation:
CH₃CH₂OH + HCOOH $\rightleftharpoons$ HCOOCH₂CH₃ + H₂O ✓

Product name:
ethyl methanoate ✓

Accept equation without equilibrium arrows.

Accept equation with molecular formulas (C₂H₆O + CH₂O₂ $\rightleftharpoons$ C₃H₆O₂ + H₂O) only if product name is correct.

ethanal AND distillation ✓

ethanoic acid AND reflux «followed by distillation» ✓

Award [1 max] for both products OR both methods.

m/z 58:
molar/«relative» molecular mass/weight/Mr «is 58 g mol⁻¹/58» ✓

m/z 43:
«loses» methyl/CH₃ «fragment»
OR
COCH₃⁺ «fragment» ✓

Do not penalize missing charge on the fragments.

Accept molecular ion «peak»/ CH₃COCH₃⁺/C₃H₆O⁺.

Accept any C₂H₃O⁺ fragment/ CH₃CH₂CH₂⁺/C₃H₇⁺.

C=O ✓

Accept carbonyl/C=C.

Information deduced from ¹H NMR:

«one signal indicates» one hydrogen environment/symmetrical structure
OR
«chemical shift of 2.2 indicates» H on C next to carbonyl ✓

Compound:

propanone/CH₃COCH₃ ✓

Accept “one type of hydrogen”.

Accept .

Sketch the Lewis (electron dot) structure of the P₄ molecule, containing only single bonds.

Write an equation for the reaction of white phosphorus (P₄) with chlorine gas to form phosphorus trichloride (PCl₃).

Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl bond angle in PCl₃.

Explain the polarity of PCl₃.

Calculate the standard enthalpy change (ΔH^⦵) for the forward reaction in kJ mol⁻¹.

ΔH^⦵_f PCl₃(g) = −306.4 kJ mol⁻¹

ΔH^⦵_f PCl₅(g) = −398.9 kJ mol⁻¹

State the equilibrium constant expression, K_c, for this reaction.

State, with a reason, the effect of an increase in temperature on the position of this equilibrium.

Accept any diagram with each P joined to the other three.

Accept any combination of dots, crosses and lines.

P₄(s) + 6Cl₂ (g) → 4PCl₃ (l) ✔

Electron domain geometry: tetrahedral ✔

Molecular geometry: trigonal pyramidal ✔

Bond angle: 100«°» ✔

Accept any value or range within the range 91−108«°» for M3.

polar AND unsymmetrical distribution of charge
OR
polar AND dipoles do not cancel
OR
«polar as» dipoles «add to» give a «partial» positive «charge» at P and a «partial» negative «charge» at the opposite/Cl side of the molecule ✔

Accept “polar AND unsymmetrical molecule”.

«−398.9 kJ mol⁻¹ − (−306.4 kJ mol⁻¹) =» −92.5 «kJ mol⁻¹» ✔

«K_c =» $\frac{\left[{\mathrm{PCl}}_5\right]}{\left[{\mathrm{PCl}}_3\right]\left[{\mathrm{Cl}}_2\right]}$ ✔

«shifts» left/towards reactants AND «forward reaction is» exothermic/ΔH is negative ✔