Calculate the amount, in mol, of H₂SO₄.

Formulate the equation for the reaction of H₂SO₄ with Mg(OH)₂.

The excess sulfuric acid required 20.80 cm³ of 0.1133 mol dm⁻³ NaOH for neutralization.

Calculate the amount of excess acid present.

Calculate the amount of H₂SO₄ that reacted with Mg(OH)₂.

Determine the mass of Mg(OH)₂ in the antacid tablet.

Calculate the percentage by mass of magnesium hydroxide in the 1.24 g antacid tablet to three significant figures.

n(H₂SO₄) «= 0.0500 dm³ × 0.100 mol dm^–3» = 0.00500/5.00 × 10^–3«mol»

[1 mark]

H₂SO₄(aq) + Mg(OH)₂(s) → MgSO₄(aq) + 2H₂O(l)

Accept an ionic equation.

[1 mark]

«n(H₂SO₄) =$\frac{1}{2}$ × n(NaOH) = $\frac{1}{2}$ (0.02080 dm³ × 0.1133 mol dm^–3)»

0.001178/1.178 × 10^–3 «mol»

[1 mark]

n(H₂SO₄) reacted «= 0.00500 – 0.001178» = 0.00382/3.82 × 10^–3 «mol»

[1 mark]

n(Mg(OH)₂) «= n(H₂SO₄) =» = 0.00382/3.82 × 10^–3 «mol»

m(Mg(OH)₂) «= 0.00382 mol × 58.33 g mol^–1» = 0.223 «g»

Award [2] for correct final answer.

[2 marks]

% Mg(OH)₂ «= $\frac{0.223\text{ g}}{1.24\text{ g}}$ × 100» = 18.0 «%»

Answer must show three significant figures.

[1 mark]

Formulate an equation for the reaction of one mole of phosphoric acid with one mole of sodium hydroxide.

Formulate two equations to show the amphiprotic nature of H₂PO₄⁻.

Calculate the concentration of H₃PO₄ if 25.00 cm³ is completely neutralised by the addition of 28.40 cm³ of 0.5000 mol dm⁻³ NaOH.

Outline the reason that sodium hydroxide is considered a Brønsted–Lowry base.

H₃PO₄(aq) + NaOH (aq) → NaH₂PO₄(aq) + H₂O (l) ✔

Accept net ionic equation.

H₂PO₄⁻ (aq) + H⁺ (aq) → H₃PO₄ (aq) ✔

H₂PO₄⁻ (aq) + OH⁻ (aq) → HPO₄²⁻ (aq) + H₂O (l) ✔

Accept reactions of H₂PO₄⁻ with any acidic, basic or amphiprotic species, such as H₃O⁺, NH₃ or H₂O.

Accept H₂PO₄⁻ (aq) → HPO₄²⁻ (aq) + H⁺ (aq) for M2.

«NaOH $\frac{28.40 {\mathrm{cm}}^3}{1000}\times 0.5000 \mathrm{mol} {\mathrm{dm}}^{-3}=0.01420 \mathrm{mol}$»

«$\frac{0.01420 \mathrm{mol}}{3}=$» 0.004733 «mol» ✔

«$\frac{0.004733 \mathrm{mol}}{{\displaystyle \frac{25.00 {\mathrm{cm}}^3}{1000}}}=$» 0.1893 «mol dm⁻³» ✔

Award [2] for correct final answer.

«OH⁻ is a» proton acceptor ✔

Outline why metals, like iron, can conduct electricity.

Justify why sulfur is classified as a non-metal by giving two of its chemical properties.

Describe the bonding in this type of solid.

State the full electron configuration of the sulfide ion.

Outline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.

Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.

Write the equation for this reaction.

Deduce the change in the oxidation state of sulfur.

Suggest why this process might raise environmental concerns.

Explain why the addition of small amounts of carbon to iron makes the metal harder.

mobile/delocalized «sea of» electrons

Any two of:

forms acidic oxides «rather than basic oxides» ✔

forms covalent/bonds compounds «with other non-metals» ✔

forms anions «rather than cations» ✔

behaves as an oxidizing agent «rather than a reducing agent» ✔

Award [1 max] for 2 correct non-chemical properties such as non-conductor, high ionisation energy, high electronegativity, low electron affinity if no marks for chemical properties are awarded.

electrostatic attraction ✔

between oppositely charged ions/between Fe²⁺ and S²⁻ ✔

1s² 2s² 2p⁶ 3s² 3p⁶ ✔

Do not accept “[Ne] 3s² 3p⁶”.

«valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔

Accept 2,8 (for O^2–) and 2,8,8 (for S^2–)

allows them to explain the properties of different compounds/substances
OR
enables them to generalise about substances
OR
enables them to make predictions ✔

Accept other valid answers.

4FeS(s) + 7O₂(g) → 2Fe₂O₃(s) + 4SO₂(g) ✔

Accept any correct ratio.

+6
OR
−2 to +4 ✔

Accept “6/VI”.
Accept “−II, 4//IV”.
Do not accept 2− to 4+.

sulfur dioxide/SO₂ causes acid rain ✔

Accept sulfur dioxide/SO₂/dust causes respiratory problems
Do not accept just “causes respiratory problems” or “causes acid rain”.

disrupts the regular arrangement «of iron atoms/ions»
OR
carbon different size «to iron atoms/ions» ✔

prevents layers/atoms sliding over each other ✔

Calculate the percentage by mass of nitrogen in urea to two decimal places using section 6 of the data booklet.

Suggest how the percentage of nitrogen affects the cost of transport of fertilizers giving a reason.

The structural formula of urea is shown.

Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.

Urea can be made by reacting potassium cyanate, KNCO, with ammonium chloride, NH₄Cl.

                                      KNCO(aq) + NH₄Cl(aq) → (H₂N)₂CO(aq) + KCl(aq)

Determine the maximum mass of urea that could be formed from 50.0 cm³ of 0.100 mol dm⁻³ potassium cyanate solution.

Urea can also be made by the direct combination of ammonia and carbon dioxide gases.

                                   2NH₃(g) + CO₂(g) $\rightleftharpoons$ (H₂N)₂CO(g) + H₂O(g)     ΔH < 0

Predict, with a reason, the effect on the equilibrium constant, K_c, when the temperature is increased.

Suggest one reason why urea is a solid and ammonia a gas at room temperature.

Sketch two different hydrogen bonding interactions between ammonia and water.

The combustion of urea produces water, carbon dioxide and nitrogen.

Formulate a balanced equation for the reaction.

The mass spectrum of urea is shown below.

Identify the species responsible for the peaks at m/z = 60 and 44.

The IR spectrum of urea is shown below.

Identify the bonds causing the absorptions at 3450 cm⁻¹ and 1700 cm⁻¹ using section 26 of the data booklet.

Predict the number of signals in the ¹H NMR spectrum of urea.

molar mass of urea «= 4 × 1.01 + 2 × 14.01 + 12.01 + 16.00» = 60.07 «g mol^–1»

«% nitrogen = $\frac{2\times 14.01}{60.07}$ × 100 =» 46.65 «%»

Award [2] for correct final answer.

Award [1 max] for final answer not to two decimal places.

[2 marks]

«cost» increases AND lower N% «means higher cost of transportation per unit of nitrogen»

OR

«cost» increases AND inefficient/too much/about half mass not nitrogen

Accept other reasonable explanations.

Do not accept answers referring to safety/explosions.

[1 mark]

Note: Urea’s structure is more complex than that predicted from VSEPR theory.

[3 marks]

n(KNCO) «= 0.0500 dm³ × 0.100 mol dm^–3» = 5.00 × 10^–3 «mol»

«mass of urea = 5.00 × 10^–3 mol × 60.07 g mol^–1» = 0.300 «g»

Award [2] for correct final answer.

[2 marks]

«K_c» decreases AND reaction is exothermic

OR

«Kc» decreases AND ΔH is negative

OR

«K_c» decreases AND reverse/endothermic reaction is favoured

[1 mark]

Any one of:

urea has greater molar mass

urea has greater electron density/greater London/dispersion

urea has more hydrogen bonding

urea is more polar/has greater dipole moment

Accept “urea has larger size/greater van der Waals forces”.

Do not accept “urea has greater intermolecular forces/IMF”.

[1 mark]

Award [1] for each correct interaction.

If lone pairs are shown on N or O, then the lone pair on N or one of the lone pairs on O MUST be involved in the H-bond.

Penalize solid line to represent H-bonding only once.

[2 marks]

2(H₂N)₂CO(s) + 3O₂(g) → 4H₂O(l) + 2CO₂(g) + 2N₂(g)

correct coefficients on LHS

correct coefficients on RHS

Accept (H₂N)₂CO(s) + $\frac{3}{2}$O₂(g) → 2H₂O(l) + CO₂(g) + N₂(g).

Accept any correct ratio.

[2 marks]

60: CON₂H₄⁺

44: CONH₂⁺

Accept “molecular ion”.

[2 marks]

3450 cm^–1: N–H

1700 cm^–1: C=O

Do not accept “O–H” for 3450 cm^–1.

[2 marks]

1

[1 mark]

Complete combustion of 0.1595 g of menthol produces 0.4490 g of carbon dioxide and 0.1840 g of water. Determine the empirical formula of the compound showing your working.

0.150 g sample of menthol, when vaporized, had a volume of 0.0337 dm³ at 150 °C and 100.2 kPa. Calculate its molar mass showing your working.

carbon: «$\frac{0.4490\text{g}}{44.01\text{g}\text{mo}{\text{l}}^{-1}}$» = 0.01020 «mol» / 0.1225 «g»

OR

hydrogen: «$\frac{0.1840\times 2}{18.02}$» = 0.02042 «mol» / 0.0206 «g»

oxygen: «0.1595 – (0.1225 + 0.0206)» = 0.0164 «g» / 0.001025 «mol»

empirical formula: C₁₀H₂₀O

Award [3] for correct final answer.

temperature = 423 K

OR

«M $=\frac{mRT}{pV}$

«M $=\frac{0.150\text{g}\times 8.31\text{J}{\text{ K}}^{-1}\text{mol}{}^{-1}\times 423\text{K}}{100.2\text{kPa}\times 0.0337\text{d}{\text{m}}^3}=$» 156 «g mol^–1»

Award [1] for correct answer with no working shown.

Accept “pV = nRT AND n = $\frac{m}{M}$” for M1.

Deduce the equation for the decomposition of guanidinium nitrate.

Calculate the total number of moles of gas produced from the decomposition of 10.0 g of guanidinium nitrate.

Calculate the pressure, in kPa, of this gas in a 10.0 dm³ air bag at 127°C, assuming no gas escapes.

Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled, while nitrogen does not.

Another airbag reactant produces nitrogen gas and sodium.

Suggest, including an equation, why the products of this reactant present a safety hazard.

C(NH₂)₃NO₃(s) → 2N₂(g) + 3H₂O (g) + C (s) ✔

moles of gas = « $5\times \frac{10.0 \text{g}}{122.11 \text{g mo}{\text{l}}^{-1}}=$» 0.409 «mol» ✔

«$p=\frac{0.409 \text{mol}\times \text{8}{\text{.31\hspace{0.05em}J\hspace{0.05em}K}}^{-1}\text{mo}{\text{l}}^{-1}\times (127+273) \text{K}}{\text{10}\text{.0\hspace{0.05em}d}{\text{m}}^3}$» = 136 «kPa» ✔

Any two of:
nitrogen non-polar/London/dispersion forces AND water polar/H-bonding ✔
water has «much» stronger intermolecular forces ✔
water molecules attract/condense/occupy smaller volume «and therefore deviate from ideal behaviour» ✔

2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂(g) ✔

hydrogen explosive
OR
highly exothermic reaction
OR
sodium reacts violently with water
OR
forms strong alkali ✔

NOTE: Accept the equation of combustion of hydrogen.
Do not accept just “sodium is reactive/dangerous”.

Draw the structural formula of propan-2-ol.

Calculate the number of hydrogen atoms in 1.00 g of propan-2-ol.

Classify propan-2-ol as a primary, secondary or tertiary alcohol, giving a reason.

State a suitable oxidizing agent for the oxidation of propan-2-ol in an acidified aqueous solution.

Deduce the average oxidation state of carbon in propan-2-ol.

Deduce the product of the oxidation of propan-2-ol with the oxidizing agent in (d)(i).

CH₃CH(OH)CH₃

Accept the full or condensed structural formula.

«$\frac{1.00\text{g}}{\left(12.01\times 3+1.01\times 8+16.00\right)\text{g}\text{mo}{\text{l}}^{-1}}=$»  0.0166  «mol CH₃CH(OH)CH₃» ✔

«0.0166 mol × 6.02 × 10²³ molecules mol⁻¹ × 8 atoms molecule⁻¹ =» 8.01 × 10²² «atoms of hydrogen» ✔ 

Accept answers in the range 7.99 × 10²² to 8.19 × 10²².

Award [2] for correct final answer.

secondary AND OH/hydroxyl is attached to a carbon bonded to one hydrogen

OR

secondary AND OH/hydroxyl is attached to a carbon bonded to two C/R/alkyl/CH₃ «groups» ✔

Accept “secondary AND OH is attached to the second carbon in the chain”.

«potassium/sodium» manganate(VII)/permanganate/KMnO₄/NaMnO₄/MnO₄⁻

OR

«potassium/sodium» dichromate(VI)/K₂Cr₂O₇/Na₂Cr₂O₇/Cr₂O₇²⁻ ✔

−2 ✔

propanone/propan-2-one/CH₃COCH₃ ✔

Write a balanced equation for the reaction that occurs.

State the block of the periodic table in which magnesium is located.

Identify a metal, in the same period as magnesium, that does not form a basic oxide.

Calculate the amount of magnesium, in mol, that was used.

Determine the percentage uncertainty of the mass of product after heating.

Assume the reaction in (a)(i) is the only one occurring and it goes to completion, but some product has been lost from the crucible. Deduce the percentage yield of magnesium oxide in the crucible.

Evaluate whether this, rather than the loss of product, could explain the yield found in (b)(iii).

Suggest an explanation, other than product being lost from the crucible or reacting with nitrogen, that could explain the yield found in (b)(iii).

Calculate coefficients that balance the equation for the following reaction.

__ Mg₃N₂(s) + __ H₂O (l) → __ Mg(OH)₂(s) + __ NH₃(aq)

Determine the oxidation state of nitrogen in Mg₃N₂ and in NH₃.

Deduce, giving reasons, whether the reaction of magnesium nitride with water is an acid–base reaction, a redox reaction, neither or both.

State the number of subatomic particles in this ion.

Some nitride ions are ¹⁵N^3–. State the term that describes the relationship between ¹⁴N^3– and ¹⁵N^3–.

The nitride ion and the magnesium ion are isoelectronic (they have the same electron configuration). Determine, giving a reason, which has the greater ionic radius.

Suggest two reasons why atoms are no longer regarded as the indivisible units of matter.

State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.

2 Mg(s) + O₂(g) → 2 MgO(s) ✔

Do not accept equilibrium arrows. Ignore state symbols

s ✔

Do not allow group 2

aluminium/Al ✔

$⟨⟨\frac{53.726 \mathrm{g}-47.372 \mathrm{g}}{244.31 \mathrm{g} {\mathrm{mol}}^{-1}}=\frac{6.354 \mathrm{g}}{24.31 \mathrm{g} {\mathrm{mol}}^{-1}}⟩⟩=0.2614$ «mol» ✔

mass of product $«=56.941 \mathrm{g}-47.372 \mathrm{g}»=9.569 «\mathrm{g}»$ ✔

$\text{⟨⟨100 × }\frac{2\times 0.001 \mathrm{g}}{9.569 \mathrm{g}}\text{=0.0209⟩⟩ = 0.02 «\%»}$ ✔

Award [2] for correct final answer

Accept 0.021%

$⟨⟨ 0.2614 \mathrm{mol} \times (24.31 \mathrm{g} {\mathrm{mol}}^{-1}+16.00 \mathrm{g} {\mathrm{mol}}^{-1})=0.2614 \mathrm{mol}\times 40.31 \mathrm{g} {\mathrm{mol}}^{-1}⟩⟩=10.536 «\mathrm{g}»$ ✔

$⟨⟨100\times \frac{9.569 \mathrm{g}}{10.536 \mathrm{g}}= 90.822⟩⟩=91 «\%»$ ✔

Award «0.2614 mol x 40.31 g mol^–1»

Accept alternative methods to arrive at the correct answer.

Accept final answers in the range 91-92%

[2] for correct final answer.

yes
AND
«each Mg combines with $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ N, so» mass increase would be 14x$\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ which is less than expected increase of 16x
OR
3 mol Mg would form 101g of Mg₃N₂ but would form 3 x MgO = 121 g of MgO
OR
0.2614 mol forms 10.536 g of MgO, but would form 8.796 g of Mg₃N₂ ✔

Accept Yes AND “the mass of N/N₂ that combines with each g/mole of Mg is lower than that of O/O₂”

Accept YES AND “molar mass of nitrogen less than of oxygen”.

incomplete reaction
OR
Mg was partially oxidised already
OR
impurity present that evaporated/did not react ✔

Accept “crucible weighed before fully cooled”.

Accept answers relating to a higher atomic mass impurity consuming less O/O₂.

Accept “non-stoichiometric compounds formed”.

Do not accept "human error", "wrongly calibrated balance" or other non-chemical reasons.

If answer to (b)(iii) is >100%, accept appropriate reasons, such as product absorbed moisture before being weighed.

«1» Mg₃N₂(s) + 6 H₂O (l) → 3 Mg(OH)₂(s) + 2 NH₃(aq)

Mg₃N₂: -3
AND
NH₃: -3 ✔

Do not accept 3 or 3-

Acid–base:
yes AND N^3- accepts H⁺/donates electron pair«s»
OR
yes AND H₂O loses H⁺ «to form OH^-»/accepts electron pair«s» ✔

Redox:
no AND no oxidation states change ✔

Accept “yes AND proton transfer takes place”

Accept reference to the oxidation state of specific elements not changing.

Accept “not redox as no electrons gained/lost”.

Award [1 max] for Acid–base: yes AND Redox: no without correct reasons, if no other mark has been awarded

Protons: 7 AND Neutrons: 7 AND Electrons: 10 ✔

isotope«s» ✔

nitride AND smaller nuclear charge/number of protons/atomic number ✔

Any two of:

subatomic particles «discovered»
OR
particles smaller/with masses less than atoms «discovered»
OR
«existence of» isotopes «same number of protons, different number of neutrons» ✔

charged particles obtained from «neutral» atoms
OR
atoms can gain or lose electrons «and become charged» ✔

atom «discovered» to have structure ✔

fission
OR
atoms can be split ✔

Accept atoms can undergo fusion «to produce heavier atoms»

Accept specific examples of particles.

Award [2] for “atom shown to have a nucleus with electrons around it” as both M1 and M3.

Award [1] for all bonding types correct.

Award [1] for each correct description.

Apply ECF for M2 only once.

This was not as well done as one might have expected with the most common errors being O instead of O₂ oxygen and MgO rather than MgO₂.

Many students did not know what "block" meant, and often guessed group 2 etc.

Many students confused "period" and "group" and also many did not read metal, so aluminium was not chosen by the majority.

A number of students were not able to interpret the results and hence find the gain in mass and calculate the moles correctly.

Only a handful could work out the correct answer. Most had no real idea and quite a lot of blank responses. There also seems to be significant confusion between "percent uncertainty" and "percent error".

This was not well answered, but definitely better than the previous question with quite a few gaining some credit for correctly determining the theoretical yield.

This proved to be a very difficult question to answer in the quantitative manner required, with hardly any correct responses.

Quite a few students realised that incomplete reaction would lead to this, but only 30% of students gave a correct answer rather than a non-specific guess, such as "misread balance" or "impurities".

This was generally very well done with almost all candidates being able to determine the correct coefficients.

About 40% of students managed to correctly determine both the oxidation states, as -3, with errors being about equally divided between the two compounds.

Probably only about 10% could explain why this was an acid-base reaction. Rather more made valid deductions about redox, based on their answer to the previous question.

Most candidates could answer the question about subatomic particles correctly.

Identification of isotopes was answered correctly by most students.

In spite of being given the meaning of "isoelectronic", many candidates talked about the differing number of electrons and only about 30% could correctly analyse the situation in terms of nuclear charge.

The question was marked quite leniently so that the majority of candidates gained at least one of the marks by mentioning a subatomic particle. A significant number read "indivisible" as "invisible" however.

About a quarter of the students gained full marks and probably a similar number gained no marks. Metallic bonding was the type that seemed least easily recognised and least easily described. Another common error was to explain ionic bonding in terms of attraction of ions rather than describing electron transfer.

State the nuclear symbol notation, ${}_Z^{A}X$, for magnesium-26.

Mass spectroscopic analysis of a sample of magnesium gave the following results:

Calculate the relative atomic mass, A_r, of this sample of magnesium to two decimal places.

Magnesium burns in air to form a white compound, magnesium oxide. Formulate an equation for the reaction of magnesium oxide with water.

Describe the trend in acid-base properties of the oxides of period 3, sodium to chlorine.

In addition to magnesium oxide, magnesium forms another compound when burned in air. Suggest the formula of this compound

Describe the structure and bonding in solid magnesium oxide.

Magnesium chloride can be electrolysed.

Deduce the half-equations for the reactions at each electrode when molten magnesium chloride is electrolysed, showing the state symbols of the products. The melting points of magnesium and magnesium chloride are 922 K and 987 K respectively.

Anode (positive electrode):

Cathode (negative electrode):

${}_{12}^{26}\mathrm{M}\mathrm{g}$

«Ar =»$\frac{24\times 78.60+25\times 10.11+26\times 11.29}{100}$

«= 24.3269 =» 24.33

Award [2] for correct final answer.
Do not accept data booklet value (24.31).

MgO(s) + H₂O(l) → Mg(OH)₂(s)

OR

MgO(s) + H₂O(l) → Mg^₂+(aq) + 2OH^–(aq)

Accept $\rightleftharpoons$.

from basic to acidic

through amphoteric

Accept “alkali/alkaline” for “basic”.
Accept “oxides of Na and Mg: basic AND oxide of Al: amphoteric” for M1.
Accept “oxides of non-metals/Si to Cl acidic” for M2.
Do not accept just “become more acidic”

Mg₃N₂

Accept MgO₂, Mg(OH)₂, Mg(NOx)₂, MgCO₃.

«3-D/giant» regularly repeating arrangement «of ions»
OR
lattice «of ions»
Accept “giant” for M1, unless “giant covalent” stated.

electrostatic attraction between oppositely charged ions
OR
electrostatic attraction between Mg²⁺ and O^2– ions
Do not accept “ionic” without description.

Anode (positive electrode):
2Cl^– → Cl₂(g) + 2e^–

Cathode (negative electrode):
Mg²⁺ + 2e^– → Mg(l)

Penalize missing/incorrect state symbols at Cl₂ and Mg once only.
Award [1 max] if equations are at wrong electrodes.
Accept Mg (g).

Outline what is measured by BOD.

A student dissolved 0.1240 ± 0.0001 g of Na₂S₂O₃ to make 1000.0 ± 0.4 cm³ of solution to use in the Winkler Method.

Determine the percentage uncertainty in the molar concentration.

Calculate the amount, in moles of Na₂S₂O₃ used in the titration.

Deduce the mole ratio of O₂ consumed in step I to S₂O₃²⁻ used in step III.

Calculate the concentration of dissolved oxygen, in mol dm⁻³, in the sample.

The three steps of the Winkler Method are redox reactions.

Deduce the reduction half-equation for step II.

«amount of» oxygen used to decompose the organic matter in water ✔

«$\frac{0.0001 \mathrm{g}}{0.1240 \mathrm{g}}\times 100\%=$» 0.08 «%»
OR
«$\frac{0.4 {\mathrm{cm}}^3}{1000.0 {\mathrm{cm}}^3}\times 100\%=$» 0.04 «%» ✔

«0.08 % + 0.04 % =» 0.12/0.1 «%» ✔

Award [2] for correct final answer.

Accept fractional uncertainties for M1, i.e., 0.0008 OR 0.0004.

«$\frac{37.50 {\mathrm{cm}}^3}{1000}$× 5.000 × 10⁻⁴ mol dm⁻³ =» 1.875 × 10⁻⁵ «mol» ✔

1:4 ✔

Accept “4 mol S₂O₃^2– :1 mol O₂“, but not just 4:1.

«$1.875\times {10}^{-5} \mathrm{mol}\times \frac{1}{4}=$» 4.688 × 10⁻⁶«mol» ✔

«$\frac{4.688\times {10}^{-6} \mathrm{mol}}{{\displaystyle \frac{25.00 {\mathrm{cm}}^3}{1000}}}=$» 1.875 × 10⁻⁴«mol dm⁻³» ✔

Award [2] for correct final answer.

MnO₂(s) + 2e⁻ + 4H⁺(aq) → Mn²⁺(aq) + 2H₂O (l) ✔

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”.

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.

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”.

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]

Identify the formula and state symbol of X.

Suggest why the experiment should be carried out in a fume hood or in a well-ventilated laboratory.

The precipitate of sulfur makes the mixture cloudy, so a mark underneath the reaction mixture becomes invisible with time.

10.0 cm³ of 2.00 mol dm^-3 hydrochloric acid was added to a 50.0 cm³ solution of sodium thiosulfate at temperature, T1. Students measured the time taken for the mark to be no longer visible to the naked eye. The experiment was repeated at different concentrations of sodium thiosulfate.

Show that the hydrochloric acid added to the flask in experiment 1 is in excess.

Draw the best fit line of $\frac{1}{\mathrm{t}}$ against concentration of sodium thiosulfate on the axes provided.

A student decided to carry out another experiment using 0.075 mol dm^-3 solution of sodium thiosulfate under the same conditions. Determine the time taken for the mark to be no longer visible.

An additional experiment was carried out at a higher temperature, T₂.

(i) On the same axes, sketch Maxwell–Boltzmann energy distribution curves at the two temperatures T₁ and T₂, where T₂ > T₁.

(ii) Explain why a higher temperature causes the rate of reaction to increase.

Suggest one reason why the values of rates of reactions obtained at higher temperatures may be less accurate.

H₂O AND (l)
Do not accept H₂O (aq).

SO₂ (g) is an irritant/causes breathing problems
OR
SO₂ (g) is poisonous/toxic

Accept SO₂ (g) is acidic, but do not accept “causes acid rain”.
Accept SO₂ (g) is harmful.
Accept SO₂ (g) has a foul/pungent smell.

n(HCl) = «$\frac{10.0}{1000}$dm³ × 2.00 mol dm^-3 =» 0.0200 / 2.00 × 10^-2«mol»
AND
n(Na₂S₂O₃) = «$\frac{50}{1000}$dm³ × 0.150 mol × dm^-3 =» 0.00750 / 7.50 × 10^-3 «mol»

0.0200 «mol» > 0.0150 «mol»
OR
2.00 × 10^-2«mol» > 2 × 7.50 × 10^-3 «mol»
OR
$\frac{1}{2}$ × 2.00 × 10^-2 «mol» > 7.50 × 10^-3 «mol»

Accept answers based on volume of solutions required for complete reaction.
Award [2] for second marking point.
Do not award M2 unless factor of 2 (or half) is used.

five points plotted correctly
best fit line drawn with ruler, going through the origin

22.5 × 10^-3 «s^-1»

«Time = $\frac{1}{22.5\times {10}^{-3}}$ =» 44.4 «s»

Award [2] for correct final answer.
Accept value based on candidate’s graph.
Award M2 as ECF from M1.
Award [1 max] for methods involving taking mean of appropriate pairs of $\frac{1}{\mathrm{t}}$ values.
Award [0] for taking mean of pairs of time values.
Award [2] for answers between 42.4 and 46.4 «s».

(i)

correctly labelled axes
peak of T₂ curve lower AND to the right of T₁ curve

Accept “probability «density» / number of particles / N / fraction” on y-axis.

Accept “kinetic E/KE/E_K” but not just “Energy/E” on x-axis.

(ii)

greater proportion of molecules have E ≥ E_a or E > E_a
OR
greater area under curve to the right of the E_a

greater frequency of collisions «between molecules»
OR
more collisions per unit time/second

Accept more molecules have energy greater than E_a.
Do not accept just “particles have greater kinetic energy”.
Accept “rate/chance/probability/likelihood/” instead of “frequency”.
Accept suitably shaded/annotated diagram.
Do not accept just “more collisions”.

shorter reaction time so larger «%» error in timing/seeing when mark disappears

Accept cooling of reaction mixture during course of reaction.

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) ✔

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 ✔

Describe the bonding in metals.

Titanium exists as several isotopes. The mass spectrum of a sample of titanium gave the following data:

Calculate the relative atomic mass of titanium to two decimal places.

State the number of protons, neutrons and electrons in the ${}_{22}^{48}\text{Ti}$ atom.

State the full electron configuration of the ${}_{22}^{48}\text{Ti}$²⁺ ion.

Explain why an aluminium-titanium alloy is harder than pure aluminium.

State the type of bonding in potassium chloride which melts at 1043 K.

A chloride of titanium, TiCl₄, melts at 248 K. Suggest why the melting point is so much lower than that of KCl.

Formulate an equation for this reaction.

Suggest one disadvantage of using this smoke in an enclosed space.

electrostatic attraction

between «a lattice of» metal/positive ions/cations AND «a sea of» delocalized electrons

Accept mobile electrons.

Do not accept “metal atoms/nuclei”.

[2 marks]

$\frac{(46\times 7.98)+(47\times 7.32)+(48\times 73.99)+(49\times 5.46)+(50\times 5.25)}{100}$

= 47.93

Answer must have two decimal places with a value from 47.90 to 48.00.

Award [2] for correct final answer.

Award [0] for 47.87 (data booklet value).

[2 marks]

Protons: 22 AND Neutrons: 26 AND Electrons: 22

[1 mark]

1s²2s²2p⁶3s²3p⁶3d²

[1 mark]

titanium atoms/ions distort the regular arrangement of atoms/ions
OR
titanium atoms/ions are a different size to aluminium «atoms/ions» 

prevent layers sliding over each other

Accept diagram showing different sizes of atoms/ions.

[2 marks]

ionic
OR
«electrostatic» attraction between oppositely charged ions

[1 mark]

«simple» molecular structure
OR
weak«er» intermolecular bonds
OR
weak«er» bonds between molecules

Accept specific examples of weak bonds such as London/dispersion and van der Waals.

Do not accept “covalent”.

[1 mark]

TiCl₄(l) + 2H₂O(l) → TiO₂(s) + 4HCl(aq)

correct products

correct balancing

Accept ionic equation.

Award M2 if products are HCl and a compound of Ti and O.

[2 marks]

HCl causes breathing/respiratory problems
OR
HCl is an irritant
OR
HCl is toxic
OR
HCl has acidic vapour
OR
HCl is corrosive

Accept “TiO₂ causes breathing problems/is an irritant”.

Accept “harmful” for both HCl and TiO₂.

Accept “smoke is asphyxiant”.

[1 mark]

State the oxidation half-equation.

Deduce the overall redox equation for the reaction between acidic Sn²⁺(aq) and Cr₂O₇^2–(aq), using section 24 of the data booklet.

Calculate the percentage uncertainty for the mass of K₂Cr₂O₇(s) from the given data.

The sample of K₂Cr₂O₇(s) in (i) was dissolved in distilled water to form 0.100 dm³ solution. Calculate its molar concentration.

10.0 cm³ of the waste sample required 13.24 cm³ of the K₂Cr₂O₇ solution. Calculate the molar concentration of Sn²⁺(aq) in the waste sample.

Sn²⁺(aq) → Sn⁴⁺(aq) + 2e^–

Accept equilibrium sign.

Accept Sn²⁺(aq) – 2e^– → Sn⁴⁺(aq).

[1 mark]

Cr₂O₇^2–(aq) + 14H⁺(aq) + 3Sn²⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l) + 3Sn⁴⁺(aq)

Accept equilibrium sign.

[1 mark]

«13.239 g ± 0.002 g so percentage uncertainty» 0.02 «%»

Accept answers given to greater precision, such as 0.0151%.

[1 mark]

« [K₂Cr₂O₇] = $\frac{13.239\text{ g}}{294.20\text{ g}\text{mo}{\text{l}}^{-1}\times 0.100\text{ d}{\text{m}}^3}$ =» 0.450 «mol$$dm^–3»

[1 mark]

n(Sn²⁺) = «0.450 mol$$dm^–3 x 0.01324 dm³ x $\frac{3mol}{1mol}$ =» 0.0179 «mol»

«[Sn²⁺] = $\frac{0.0179mol}{0.0100mol}$ =» 1.79 «mol$$dm^–3»

Award [2] for correct final answer.

[2 marks]

State the full electron configuration of the chlorine atom.

State, giving a reason, whether the chlorine atom or the chloride ion has a larger radius.

Outline why the chlorine atom has a smaller atomic radius than the sulfur atom.

The mass spectrum of chlorine is shown.

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.

Outline the reason for the two peaks at $m/z=35$ and $37$.

Explain the presence and relative abundance of the peak at $m/z=74$.

Calculate the amount, in $\mathrm{mol}$, of manganese(IV) oxide added.

Determine the limiting reactant, showing your calculations.

Determine the excess amount, in $\mathrm{mol}$, of the other reactant.

Calculate the volume of chlorine, in ${\mathrm{dm}}^3$, produced if the reaction is conducted at standard temperature and pressure (STP). Use section 2 of the data booklet.

State the oxidation state of manganese in ${\mathrm{MnO}}_2$ and ${\mathrm{MnCl}}_2$.

Deduce, referring to oxidation states, whether ${\mathrm{MnO}}_2$ is an oxidizing or reducing agent.

Hypochlorous acid is considered a weak acid. Outline what is meant by the term weak acid.

State the formula of the conjugate base of hypochlorous acid.

Calculate the concentration of ${\mathrm{H}}^{+} \left(\mathrm{aq}\right)$ in a $\mathrm{HClO} \left(\mathrm{aq}\right)$ solution with a $\mathrm{pH}=3.61$.

State the type of reaction occurring when ethane reacts with chlorine to produce chloroethane.

Predict, giving a reason, whether ethane or chloroethane is more reactive.

Write the equation for the reaction of chloroethane with a dilute aqueous solution of sodium hydroxide.

Deduce the nucleophile for the reaction in d(iii).

Ethoxyethane (diethyl ether) can be used as a solvent for this conversion. Draw the structural formula of ethoxyethane

Deduce the number of signals and their chemical shifts in the ${}_1\mathrm{H} \mathrm{NMR}$ spectrum of ethoxyethane. Use section 27 of the data booklet.

Calculate the percentage by mass of chlorine in ${\mathrm{CCl}}_2{\mathrm{F}}_2$.

Comment on how international cooperation has contributed to the lowering of $\mathrm{CFC}$ emissions responsible for ozone depletion.

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^5$ ✔

Do not accept condensed electron configuration.

${\mathrm{Cl}}^{-}$ AND more «electron–electron» repulsion ✔

Accept $C{l}^{\mathit{-}}$ AND has an extra electron.

$\mathrm{Cl}$ has a greater nuclear charge/number of protons/${\mathrm{Z}}_{\mathrm{eff}}$ «causing a stronger pull on the outer electrons» ✔

same number of shells
OR
same «outer» energy level
OR
similar shielding ✔

«two major» isotopes «of atomic mass $35$ and $37$» ✔

«diatomic» molecule composed of «two» chlorine-37 atoms ✔

chlorine-37 is the least abundant «isotope»
OR
low probability of two ${}_{37}\mathrm{Cl}$ «isotopes» occurring in a molecule ✔

$«\frac{2.67 \mathrm{g}}{86.94 \mathrm{g} {\mathrm{mol}}^{-1}}=» 0.0307 «\mathrm{mol}»$ ✔

$«{\mathrm{n}}_{\mathrm{HCl}}=2.00 \mathrm{mol} {\mathrm{dm}}^{-3}\times 0.2000 {\mathrm{dm}}^3»=0.400 \mathrm{mol}$✔

$«\frac{0.400}{4}=» 0.100 \mathrm{mol}$ AND ${\mathrm{MnO}}_2$ is the limiting reactant ✔

Accept other valid methods of determining the limiting reactant in M2.

$«0.0307 \mathrm{mol}\times 4=0.123 \mathrm{mol}»$

$«0.400 \mathrm{mol}–0.123 \mathrm{mol}=» 0.277 «\mathrm{mol}»$ ✔

$«0.0307 \mathrm{mol}\times 22.7 {\mathrm{dm}}^3 {\mathrm{mol}}^{-1}=» 0.697 «{\mathrm{dm}}^3»$ ✔

Accept methods employing $pV=nRT$.

${\mathrm{MnO}}_2: +4$ ✔

${\mathrm{MnCl}}_2: +2$ ✔

oxidizing agent AND oxidation state of $\mathrm{Mn}$ changes from $+4$ to $+2$/decreases ✔

partially dissociates/ionizes «in water» ✔

${\mathrm{ClO}}^{-}$ ✔

$«\left[{\mathrm{H}}^{+}\right]={10}^{-3.61}=»2.5\times {10}^{-4} «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

«free radical» substitution/${\mathrm{S}}_{\mathrm{R}}$ ✔

Do not accept electrophilic or nucleophilic substitution.

chloroethane AND $\mathrm{C}–\mathrm{Cl}$ bond is weaker/$324 \mathrm{kJ} {\mathrm{mol}}^{-1}$ than $\mathrm{C}–\mathrm{H}$ bond/$414 \mathrm{kJ} {\mathrm{mol}}^{-1}$
OR
chloroethane AND contains a polar bond ✔

Accept “chloroethane AND polar”.

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\left(\mathrm{l}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$
OR
${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\left(\mathrm{l}\right)+\mathrm{NaOH}\left(\mathrm{aq}\right)\to {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\left(\mathrm{aq}\right)+\mathrm{NaCl}\left(\mathrm{aq}\right)$ ✔

Accept use of $C_{\mathit{2}}{H}_{\mathit{5}}Cl$ and $C_{\mathit{2}}{H}_{\mathit{5}}OH\mathrm{/}{C}_{\mathit{2}}{H}_{\mathit{6}}O$ in the equation.

hydroxide «ion»/${\mathrm{OH}}^{-}$ ✔

Do not accept $NaOH$.

 / ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{OCH}}_2{\mathrm{CH}}_3$ ✔

Accept $\mathit{(}C{H}_{\mathit{3}}C{H}_{\mathit{2}}{\mathit{)}}_{\mathit{2}}O$.

$2$ «signals» ✔

$0.9–1.0 «\mathrm{ppm}»$ AND $3.3–3.7 «\mathrm{ppm}»$✔

Accept any values in the ranges.

Award [1 max] for two incorrect chemical shifts.

$«M\left({\mathrm{CCl}}_2{\mathrm{F}}_2\right) =» 120.91 «\mathrm{g} {\mathrm{mol}}^{-1}»$ ✔

$\frac{2\times 35.45 \mathrm{g} {\mathrm{mol}}^{-1}}{120.91 \mathrm{g} {\mathrm{mol}}^{-1}}\times 100\%=» 58.64 «\%»$ ✔

Award [2] for correct final answer.

Any of:

research «collaboration» for alternative technologies «to replace $\mathrm{CFC}$s»
OR
technologies «developed»/data could be shared
OR
political pressure/Montreal Protocol/governments passing legislations ✔

Do not accept just “collaboration”.

Do not accept any reference to $CFC$ as greenhouse gas or product of fossil fuel combustion.

Accept reference to specific measures, such as agreement on banning use/manufacture of $CFC$s.

Most candidates wrote the electron configuration of chlorine correctly.

Only half of the candidates deduced that the chloride ion has a larger radius than the chlorine atom with a valid reason. Many candidates struggled with this question and decided that the extra electron in the chloride ion caused a greater attraction between the nucleus and the outer electrons.

Only about a third of the candidates identified the extra proton in the chlorine nucleus as the cause of the smaller atomic radius when compared to the sulfur atom, and only the stronger candidates also compared the shielding or the number of shells in the two atoms. Many candidates had a poor understanding of factors affecting atomic radius and could not explain the difference.

About 60% of the candidates recognized that the peaks at m/z 35 and 37 in the mass spectrum of chlorine are due to its isotopes. A few students wrote 'isomers' instead of 'isotopes'.

This was the lowest scoring question on the paper, that was also left blank by 10% of the candidates. About 20% of the candidates identified the peak at m/z = 74 to be due to a molecule made up of two 37Cl atoms. And only very few candidates commented that the low abundance of the peak was due to the low abundance of the 37Cl isotope. A common incorrect answer was that chlorine has an isotope of mass number 74.

Most candidates were able to determine the number of moles of MnO₂ using the mass.

It was pleasing that the majority of the candidates were able to determine the limiting reactant by using the stoichiometric ratio.

Half of the candidates were able to determine the amount of excess reactant. Some candidates who determined the limiting reactant in the previous part correctly, forgot to use the stoichiometric ratio in this part, and ended up with incorrect answers.

60% of the candidates determined the volume of chlorine produced correctly. Some candidates made mistakes in the units when using PV = nRT and had a power of 10 error.

The majority of candidates were able to determine the oxidation states of Mn in the two compounds correctly.

Less than half of the candidates were awarded the mark. Some did identify MnO2 as the oxidizing agent but did not give the explanation in terms of oxidation state as required in the question. Other candidates did not have an understanding of oxidizing and reducing agents. 

A very well answered question - 80% of candidates understood what is meant by the term weak acid. Incorrect answers included 'acids that have high pH'.

Half of the candidates deduced the formula of the conjugate base of hypochlorous acid. Incorrect answers included H₂O and HCl.

A well answered question. It was pleasing to see that 70% of the candidates were able to calculate [H⁺] from the given pH.

More than half of the candidates identified the type of reaction between ethane and chlorine as a substitution reaction. A few candidates lost the marks for writing 'electrophilic substitution' or 'nucleophilic substitutions'.

This was a challenging question that was answered correctly by only 30% of the candidates. A variety of incorrect answers were seen such as 'chlorine is a halogen and hence it is reactive', and 'ethane is more reactive because it is an alkane'. For students who answered correctly, the polarity was the most frequently given reason.

Half of the candidates wrote the correct equation for the hydrolysis of chloroethane. Incorrect answers often included carbon dioxide and water as the products.

This was a highly discriminating question. Only 30% of the candidates were able to identify the hydroxide ion as the nucleophile in the hydrolysis of chloroethane. Incorrect answers included NaOH where the ion was not specified. 14% of the candidates left this question blank.

Half of the candidates were able to give the structural formula of ethoxyethane. Incorrect answers included methoxymethane, ketones and esters.

Nearly half of the candidates were able to identify the number of signals obtained in the 1H NMR spectrum of ethoxyethane, obtaining the first mark of this question. Many candidates were awarded the mark as 'error carried forward' from an incorrect structure of ethoxyethane. The second mark for this question required candidates to look up values of chemical shift from the data booklet. Nearly a third of the candidates were able to match the chemical environments of the hydrogen atoms in ethoxyethane to those listed in the data booklet successfully. 

This was the highest scoring question in the paper. The majority of candidates were able to calculate the percentage by mass of chlorine in CCl₂F₂. Mistakes included incorrect rounding and arithmetic errors.

This nature of science question was well answered by half of the candidates. Some teachers commented that the wording was rather vague. Incorrect answers were mainly assuming that CFCs were related to the combustion of fuels and greenhouse gas emissions.

Determine the mole ratio of S₂O₃²⁻ to O₂, using the balanced equations.

Calculate the number of moles of oxygen in the day 0 sample.

The day 5 sample contained 5.03 × 10⁻⁵ moles of oxygen.

Determine the 5-day biochemical oxygen demand of the pond, in mg dm⁻³ (“parts per million”, ppm).

Calculate the percentage uncertainty of the day 5 titre.

Suggest a modification to the procedure that would make the results more reliable.

4 : 1 ✔

${\text{n}}_{{\text{s}}_2{{\text{o}}_3}^{2-}}=«0.0258\text{ d}{\text{m}}^3\times 0.010\text{ mol d}{\text{m}}^{-3}=» 2.58\times {10}^{-4} \text{«mol»}$ ✔

«$\frac{2.58\times {10}^{-4} \text{mol}}{4}=» 6.45\times {10}^{-5}\text{«mol»}$ ✔

NOTE: Award [2] for correct final answer.

«difference in moles per dm³ = (6.45 × 10⁻⁵ − 5.03 × 10⁻⁵) × $\frac{1000}{300.0}$ =»

4.73 × 10⁻⁵ «mol dm⁻³» ✔

«convert to mg per dm³: 4.73 × 10⁻⁵ mol dm⁻³ × 32.00 g mol⁻¹ × 1000 mg g^–1 = » 1.51 «ppm/mg dm⁻³» ✔

NOTE: Award [2] for correct final answer.

«$\frac{100\times 0.1 \text{c}{\text{m}}^3}{20.1 \text{c}{\text{m}}^3}=» 0.5$ «%»✔

repetition / take several samples «and average» ✔

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.

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 ✔

Explain why, as the reaction proceeds, the pressure increases by the amount shown.

Outline, in terms of collision theory, how a decrease in pressure would affect the rate of reaction.

The experiment is repeated using the same amount of dinitrogen monoxide in the same apparatus, but at a lower temperature.

Sketch, on the axes in question 2, the graph that you would expect.

The experiment gave an error in the rate because the pressure gauge was inaccurate. Outline whether repeating the experiment, using the same apparatus, and averaging the results would reduce the error.

The graph below shows the Maxwell–Boltzmann distribution of molecular energies at a particular temperature.

The rate at which dinitrogen monoxide decomposes is significantly increased by a metal oxide catalyst.

Annotate and use the graph to outline why a catalyst has this effect.

increase in the amount/number of moles/molecules «of gas»  [✔]

from 2 to 3/by 50 %  [✔]

«rate of reaction decreases»
concentration/number of molecules in a given volume decreases
OR
more space between molecules  [✔]

collision rate/frequency decreases
OR
fewer collisions per second/unit time  [✔]

Note: Do not accept just “larger space/volume” for M1.

smaller initial gradient  [✔]

initial pressure is lower AND final pressure of gas lower «by similar factor»  [✔]

no AND it is a systematic error/not a random error
OR
no AND «a similar magnitude» error would occur every time  [✔]

catalysed and uncatalysed Ea marked on graph AND with the catalysed being at lower energy  [✔]

«for catalysed reaction» greater proportion of/more molecules have E ≥ E_a / E > E_a
OR
«for catalysed reaction» greater area under curve to the right of the E_a  [✔]

Note: Accept “more molecules have the activation energy”.

About a quarter of the candidates gave the full answer. Some only gained the first marking point (M1) by recognizing the increase in the number of moles of gas. Some candidates wrote vague answers that did not receive credit such as “pressure increases as more gaseous products form” without explicitly recognizing that the reactants have fewer moles of gas than the products. Some candidates mistook it for a system at equilibrium when the pressure stops changing (although a straight arrow is shown in the equation). A teacher commented that the wording of the question was rather vague “not clear if question is asking about stoichiometry (i.e. how 200 & 300 connect to coefficients) or rates (i.e. explain graph shape)”. We did not see a discussion of the slope of the graph with time and most candidates understood the question as it was intended.

More than half of the candidates obtained the mark allocated for “less frequent collisions” at lower pressure, but only strong candidates explained that this was due to the lower concentration or increased spacing between molecules. Some candidates talked about a decrease in kinetic energy and they did not show a good understanding of collision theory. Some candidates lost M1 for stating “fewer collisions” without reference to time or probability.

This was a challenging question. Candidates usually obtained only one of the two marks allocated for the answer. Most of them scored the mark for a lower initial slope at low temperature, while others scored a mark for sketching their curve below the original curve as all pressures (initial and final) will be lower at the lower temperature. A teacher commented that the wording was unclear “sketch on the axes in question 2”, and it would have been better to label the graph instead.

This question was well answered by nearly 70 % of the candidates reflecting a good understanding of the impact of systematic errors. Some students did not gain the mark because of an incomplete answer. The question raised much debate among teachers. They worried if the error was clearly a systematic one. However, a high proportion of candidates had very clear and definite answers. In Spanish and French, the wording was a bit ambiguous which caused the markscheme in these languages to be more opened.

This question discriminated very well between high-scoring and low-scoring candidates. About half of the candidates annotated the Maxwell-Boltzmann distribution to show the effect of the catalyst. Some left it blank and some sketched a new distribution that would be obtained at a higher temperature instead. The majority of candidates knew that the catalyst provided an alternative route with lower E_a but only stronger candidates related it to the annotation of the graph and used the accurate language needed to score M2. A common mistake was stating that molecules have higher kinetic energy when a catalyst is added.

State why this equilibrium reaction is considered homogeneous.

0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm³ flask at 1000 K.

Predict the direction of the reaction showing your working.

all «species» are in same phase ✔

Accept “all species are in same state”.

Accept “all species are gases”.

«reaction quotient/Q =» $\frac{{\left[\text{S}{\text{O}}_3\right]}^2}{{\left[\text{S}{\text{O}}_2\right]}^2\left[{\text{O}}_2\right]}\text{/}\frac{{0.500}^2}{{0.200}^2\times 0.300}\text{/}20.8$ ✔

reaction quotient/Q/20.8/answer < K_c/280

OR

mixture needs more product for the number to equal Kc ✔

reaction proceeds to the right/products ✔

Do not award M3 without valid reasoning.

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.

Draw the Lewis (electron dot) structure of hydrogen sulfide.

Predict the shape of the hydrogen sulfide molecule.

State the formula of its conjugate base.

Saturated aqueous hydrogen sulfide has a concentration of 0.10 mol dm⁻³ and a pH of 4.0. Demonstrate whether it is a strong or weak acid.

Calculate the hydroxide ion concentration in saturated aqueous hydrogen sulfide.

A gaseous sample of nitrogen, contaminated only with hydrogen sulfide, was reacted with excess sodium hydroxide solution at constant temperature. The volume of the gas changed from 550 cm³ to 525 cm³.

Determine the mole percentage of hydrogen sulfide in the sample, stating one assumption you made.

 OR  ✔

Accept any combination of lines, dots or crosses to represent electrons.

bent/non-linear/angular/v-shaped✔

HS⁻ ✔

weak AND strong acid of this concentration/[H⁺] = 0.1 mol dm⁻³ would have pH = 1
OR
weak AND [H⁺] = 10⁻⁴ < 0.1 «therefore only fraction of acid dissociated» ✔

10⁻¹⁰ «mol dm⁻³» ✔

Mole percentage H₂S:
volume of H₂S = «550 − 525 = » 25 «cm³» ✔
mol % H₂S = «$\frac{25 {\mathrm{cm}}^3}{550 {\mathrm{cm}}^3}\times 100$ = » 4.5 «%» ✔

Award [2] for correct final answer of 4.5 «%»

Assumption:
«both» gases behave as ideal gases ✔

Accept “volume of gas $\mathrm{\alpha }$ mol of gas”.
Accept “reaction goes to completion”.
Accept “nitrogen is insoluble/does not react with NaOH/only H₂S reacts with NaOH”.

Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.

The pH of an aqueous solution of benzoic acid at 298 K is 2.95. Determine the concentration of hydroxide ions in the solution, using section 2 of the data booklet.

Formulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.

Suggest how benzoic acid, M_r = 122.13, forms an apparent dimer, M_r = 244.26, when dissolved in a non-polar solvent such as hexane.

 [✔]

Note: Accept Kekulé structures.

Negative sign must be shown in correct position- on the O or delocalised over the carboxylate.

ALTERNATIVE 1:
[H⁺] «= 10^−2.95» = 1.122 × 10⁻³ «mol dm⁻³»  [✔]

«[OH⁻] = $\frac{1.00\times {10}^{-14}\text{ mo}{\text{l}}^2\text{ d}{\text{m}}^{-6}}{1.22\times {10}^{-3}\text{ mol d}{\text{m}}^{-3}}$ =» 8.91 × 10⁻¹² «mol dm⁻³»  [✔]

ALTERNATIVE 2:
pOH = «14 − 2.95 =» 11.05  [✔]

«[OH⁻] = 10^−11.05 =» 8.91 × 10⁻¹² «moldm⁻³»  [✔]

Note: Award [2] for correct final answer.

Accept other methods.

2C₆H₅COOH(s) + 15O₂ (g) → 14CO₂ (g) + 6H₂O(l)

correct products  [✔]

correct balancing [✔]

«intermolecular» hydrogen bonding  [✔]

Note: Accept diagram showing hydrogen bonding.

Most failed to score a mark for the conjugate base of benzoic acid as either they didn’t show all bonds and atoms in the ring and/or they did not put the minus sign in the correct place. Some didn't read the question carefully so gave the structure of the acid form.

Many students could correctly calculate the hydroxide concentration, but some weaker students calculated hydrogen ion concentration only.

Most students earned at least one mark for writing the correct products of the combustion of benzoic acid but the balancing appeared to be difficult for some.

Very few students answered this question correctly, thinking benzoic would bond with the hexane even though it was a non-polar solvent. It was very rare for a student to realize there was intermolecular hydrogen bonding.

Determine the empirical formula of the compound using section 6 of the data booklet.

Determine the molecular formula of this compound if its molar mass is 88.12 g mol⁻¹. If you did not obtain an answer in (a) use CS, but this is not the correct answer.

Identify each compound from the spectra given, use absorptions from the range of 1700 cm⁻¹ to 3500 cm⁻¹. Explain the reason for your choice, referring to section 26 of the data booklet.

«$\frac{8.802 \mathrm{g}}{44.01 \mathrm{g} {\mathrm{mol}}^{-1}}=$» 0.2000 «mol of C/CO₂»

AND «$\frac{3.604 \mathrm{g}}{18.02 \mathrm{g} {\mathrm{mol}}^{-1}}=$» 0.2000 «mol of H₂O»/0.4000 «mol of H»

OR

 «$\frac{8.802 \mathrm{g}}{44.01 \mathrm{g} {\mathrm{mol}}^{-1}}\times 12.01 \mathrm{g} {\mathrm{mol}}^{-1}$» 2.402 «g of C»

OR

«$\frac{3.604 \mathrm{g}}{18.02 \mathrm{g} {\mathrm{mol}}^{-1}}\times 2\times 1.01 \mathrm{g} {\mathrm{mol}}^{-1}=$» 0.404 «g of H» ✔

«4.406 g − 2.806 g» = 1.600 «g of O» ✔

«$\frac{2.402 \mathrm{g}}{12.01 \mathrm{g} {\mathrm{mol}}^{-1}}=0.2000 \mathrm{mol} \mathrm{C}; \frac{0.404 \mathrm{g}}{1.01 \mathrm{g} {\mathrm{mol}}^{-1}}=0.400 \mathrm{mol} \mathrm{H}; \frac{1.600 \mathrm{g}}{16.00 \mathrm{g} {\mathrm{mol}}^{-1}}=0.1000 \mathrm{mol} \mathrm{O}$»

C₂H₄O ✔

Award [3] for correct final answer.

«$\frac{88.12 \mathrm{g} {\mathrm{mol}}^{-1}}{44.06 \mathrm{g} {\mathrm{mol}}^{-1}}=2$» C₄H₈O₂ ✔

C₂S₂ if CS used.

Award [1 max] for correctly identifying all 3 compounds without valid reasons given.

Accept specific values of wavenumbers within each range.