Identify the wavenumber of one peak in the IR spectrum of benzoic acid, using section 26 of the data booklet.

Identify the spectroscopic technique that is used to measure the bond lengths in solid benzoic acid.

Outline one piece of physical evidence for the structure of the benzene ring.

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

Outline why both C to O bonds in the conjugate base are the same length and suggest a value for them. Use section 10 of the data booklet.

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.

The combustion reaction in (f)(ii) can also be classed as redox. Identify the atom that is oxidized and the atom that is reduced.

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.

State the reagent used to convert benzoic acid to phenylmethanol (benzyl alcohol), C₆H₅CH₂OH.

Any wavenumber in the following ranges:
2500−3000 «cm⁻¹» [✔]
1700−1750 «cm⁻¹» [✔]
2850−3090 «cm⁻¹» [✔]

X-ray «crystallography/spectroscopy» [✔]

Any one of:

«regular» hexagon

OR

all «H–C–C/C-C-C» angles equal/120º [✔]
all C–C bond lengths equal/intermediate between double and single

OR

bond order 1.5 [✔]

     [✔]

Note: Accept Kekulé structures.
Negative sign must be shown in correct position.

electrons delocalized «across the O–C–O system»

OR

resonance occurs [✔]

122 «pm» < C–O < 143 «pm» [✔]

Note: Accept “delocalized π-bond”.
Accept “bond intermediate between single and double bond” or “bond order 1.5” for M1.
Accept any answer in range 123 to 142 pm.

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 «mol dm⁻³» [✔]

Note: Award [2] for correct final answer.
Accept other methods.

2C₆H₅COOH (s) + 15O₂ (g) → 14CO₂ (g) + 6H₂O (l)
correct products    [✔]
correct balancing    [✔]

Oxidized:

C/carbon «in C₆H₅COOH»

AND

Reduced:

O/oxygen «in O₂»      [✔]

«intermolecular» hydrogen bonding    [✔]

Note: Accept diagram showing hydrogen bonding.

lithium aluminium hydride/LiAlH₄    [✔]

Most candidates could identify a wavenumber or range of wavenumbers in the IR spectrum of benzoic acid.

Less than half the candidates identified x-ray crystallography as a technique used to measure bond lengths. There were many stating IR spectroscopy and quite a few random guesses.

Again less than half the candidates could accurately give a physical piece of evidence for the structure of benzene. Many missed the mark by not being specific, stating ‘all bonds in benzene with same length’ rather than ‘all C-C bonds in benzene have the same length’.

Very poorly answered with only 1 in 5 getting this question correct. Many did not show all the bonds and all the atoms or either forgot or misplaced the negative sign on the conjugate base.

This question was a challenge. Candidates were not able to explain the intermediate bond length and the majority suggested the value of either the bond length of C to O single bond or double bond.

Generally well done with a few calculating the pOH rather than the concentration of hydroxide ion asked for.

Most earned at least one mark by correctly stating the products of the reaction.

Another question where not reading correctly was a concern. Instead of identifying the atom that is oxidized and the atom that is reduced, answers included formulas of molecules or the atoms were reversed for the redox processes.

The other question where only 10 % of the candidates earned a mark. Few identified hydrogen bonding as the reason for carboxylic acids forming dimers. There were many G2 forms stating that the use of the word “dimer” is not in the syllabus, however the candidates were given that a dimer has double the molar mass and the majority seemed to understand that the two molecules joined together somehow but could not identify hydrogen bonding as the cause.

Very few candidates answered this part correctly and scored the mark. Common answers were H₂SO₄, HCl & Sn, H₂O₂. In general, strongest candidates gained the mark.

Write a balanced equation for the reaction.

Identify the major species, other than water and potassium ions, at these points.

State a suitable indicator for this titration. Use section 22 of the data booklet

Suggest, giving a reason, which point on the curve is considered a buffer region.

State the $K_{\mathrm{a}}$ expression for ethanoic acid.

Calculate the $K_{\mathrm{b}}$ of the conjugate base of ethanoic acid using sections 2 and 21 of the data booklet.

In a titration, $25.00 {\mathrm{cm}}^3$ of vinegar required $20.75 {\mathrm{cm}}^3$ of $1.00 \mathrm{mol} {\mathrm{dm}}^{-3}$ potassium hydroxide to reach the end-point.

Calculate the concentration of ethanoic acid in the vinegar.

Potassium hydroxide solutions can react with carbon dioxide from the air. The solution was made one day prior to using it in the titration.

State the type of error that would result from the student’s approach.

Potassium hydroxide solutions can react with carbon dioxide from the air. The solution was made one day prior to using it in the titration.

Predict, giving a reason, the effect of this error on the calculated concentration of ethanoic acid in 5(e).

${\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)+\mathrm{KOH}\left(\mathrm{aq}\right)\to {\mathrm{CH}}_3\mathrm{COOK} \left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ ✔

Accept the ionic equation.

B: ${\mathrm{CH}}_3\mathrm{COOH}$ AND ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ ✔

C: ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ ✔

Accept names.

Accept $C{H}_{\mathit{3}}COOK$ for $C{H}_{\mathit{3}}CO{O}^{\mathit{-}}$

phenolphthalein ✔

Accept “phenol red” or “bromothymol blue”.

B AND the region where small additions «of the base/$\mathrm{KOH}$ » result in little or no
change in $\mathrm{pH}$
OR
B AND the flattest region of the curve «at intermediate $\mathrm{pH}$/before equivalence
point »
OR
B AND half the volume needed to reach equivalence point
OR
B AND similar amounts of weak acid/${\mathrm{CH}}_3\mathrm{COOH}$/ethanoic acid AND conjugate base/${\mathrm{CH}}_3{\mathrm{COO}}^{-}$/ethanoate ✔

$K_{\mathrm{a}}=\frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

Accept $H^{+}$ instead of $H_3{O}^{+}$.

$«K_{\mathrm{a}}={10}^{-4.76}=1.7\times {10}^{-5}»$
$«K_{\mathrm{w}}=K_{\mathrm{a}}·K_{\mathrm{b}}=1.0\times {10}^{-14}=1.7\times {10}^{-5}\times K_{\mathrm{b}}»$
$«{\mathrm{K}}_{\mathrm{b}}=»5.8\times {10}^{-10}$ ✔

Accept answers between $\mathit{5}\mathit{.}\mathit{7}\mathit{-}\mathit{5}\mathit{.}\mathit{9}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{10}}$.

$«\mathrm{n}\left(\mathrm{KOH}\right)=0.02075 {\mathrm{dm}}^3\times 1.00 \mathrm{mol} {\mathrm{dm}}^{-3}=»0.0208 «\mathrm{mol}»$ ✔

$«\mathrm{n}\left(\mathrm{KOH}\right)=\mathrm{n}\left({\mathrm{CH}}_3\mathrm{COOH}\right)»$
$«\left[{\mathrm{CH}}_3\mathrm{COOH}\right]=\frac{0.0208 \mathrm{mol}}{0.02500 {\mathrm{dm}}^3}=»0.830 «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

Award [2] for correct final answer.

systematic «error» ✔

$\left[{\mathrm{CH}}_3\mathrm{COOH}\right]$ would be higher ✔

actual $\left[\mathrm{KOH}\right]$ is lower «than the value in calculation»
OR
larger volume of $\mathrm{KOH}$ «solution» needed to neutralize the acid ✔

Accept $KOH$ partially neutralised by $C{O}_2$ from air.

Most candidates could write a balanced neutralization equation.

Identifying species present at various points along a pH titration curve was one of the most poorly answered questions in the exam. Very few candidates realized there were two major species at point B even when they were able in general to realize that B was a buffer zone.

Almost all candidates could identify a suitable indicator to use in a titration of a weak acid with a strong base.

Most students could identify a buffer zone region in a titration but very few (50%) could coherently explain why.

Poorly answered with only 50% correctly writing a K_a expression. The major error was in candidates trying to calculate a K_a rather than write an expression for it.

Like with other calculations in this exam, the majority of candidates could correctly determine a concentration from titration data.

80% of candidates could identify the method used as a systematic error, with some stating human or random error.

Most candidates identified that the systematic error would result in the concentration of the alkali being lowered but then failed to propagate this through to the effect on the concentration of the acid.

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.

A voltaic cell is set up between the Fe²⁺(aq) | Fe (s) and Fe³⁺ (aq) | Fe²⁺ (aq) half-cells.

Deduce the equation and the cell potential of the spontaneous reaction. Use section 24 of the data booklet.

The figure shows an apparatus that could be used to electroplate iron with zinc. Label the figure with the required substances.

Outline why, unlike typical transition metals, zinc compounds are not coloured.

Transition metals like iron can form complex ions. Discuss the bonding between transition metals and their ligands in terms of acid-base theory.

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» ✔

Equation:
2Fe³⁺(aq) + Fe(s) → 3Fe²⁺(aq) ✔

Cell potential:
«+0.77 V − (−0.45 V) = +»1.22 «V» ✔

Do not accept reverse reaction or equilibrium arrow.

Do not accept negative value for M2.

left electrode/anode labelled zinc/Zn AND right electrode/cathode labelled iron/Fe ✔

electrolyte labelled as «aqueous» zinc salt/Zn²⁺ ✔

Accept an inert conductor for the anode.

Accept specific zinc salts such as ZnSO₄.

« Zn²⁺» has a full d-shell
OR
does not form « ions with» an incomplete d-shell ✔

Do not accept “Zn is not a transition metal”.

Do not accept zinc atoms for zinc ions.

ligands donate pairs of electrons to metal ions
OR
forms coordinate covalent/dative bond✔

ligands are Lewis bases
AND
metal «ions» are Lewis acids ✔

5.00 g of an impure sample of hydrated ethanedioic acid, (COOH)₂•2H₂O, was dissolved in water to make 1.00 dm³ of solution. 25.0 cm³ samples of this solution were titrated against a 0.100 mol dm^-3 solution of sodium hydroxide using a suitable indicator.

(COOH)₂ (aq) + 2NaOH (aq) → (COONa)₂ (aq) + 2H₂O (l)

The mean value of the titre was 14.0 cm³.

(i) Suggest a suitable indicator for this titration. Use section 22 of the data booklet.

(ii) Calculate the amount, in mol, of NaOH in 14.0 cm³ of 0.100 mol dm^-3 solution.

(iii) Calculate the amount, in mol, of ethanedioic acid in each 25.0 cm³ sample.

(iv) Determine the percentage purity of the hydrated ethanedioic acid sample.

Draw the Lewis (electron dot) structure of the ethanedioate ion, ^–OOCCOO^–.

Outline why all the C–O bond lengths in the ethanedioate ion are the same length and suggest a value for them. Use section 10 of the data booklet.

Explain how ethanedioate ions act as ligands.

i
phenolphthalein
OR
phenol red

ii
«n(NaOH) = $\left(\frac{14.0}{1000}\right)$ dm³ × 0.100 mol dm^-3 =» 1.40 × 10^-3 «mol»

iii
«$\frac{1}{2}$ × 1.40 × 10^-3 =» 7.00 × 10^-4 «mol»

iv
ALTERNATIVE 1:
«mass of pure hydrated ethanedioic acid in each titration = 7.00 × 10^-4 mol × 126.08 g mol^-1 =» 0.0883 / 8.83 × 10^-2 «g»

mass of sample in each titration = «$\frac{25}{1000}$ × 5.00 g =» 0.125 «g»

«% purity = $\frac{0.0883\mathrm{g}}{0.125\mathrm{g}}$ × 100 =» 70.6 «%»

ALTERNATIVE 2:
«mol of pure hydrated ethanedioic acid in 1 dm³ solution = 7.00 × 10^-4 × $\frac{1000}{25}$=» 2.80 × 10^-2 «mol»

«mass of pure hydrated ethanedioic acid in sample = 2.80 × 10^-2 mol × 126.08 g mol^-1 =» 3.53 «g»

«% purity = $\frac{3.53\mathrm{g}}{5.00\mathrm{g}}$ × 100 =» 70.6 «%»

ALTERNATIVE 3:
mol of hydrated ethanedioic acid (assuming sample to be pure) = $\frac{5.00\mathrm{g}}{126.08\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}}$ = 0.03966 «mol»

actual amount of hydrated ethanedioic acid = «7.00 × 10^-4 × $\frac{1000}{25}$ =» 2.80 × 10^-2 «mol»

«% purity = $\frac{2.80\times {10}^{-2}}{0.03966}$ × 100 =» 70.6 «%»

Award suitable part marks for alternative methods.
Award [3] for correct final answer.
Award [2 max] for 50.4 % if anhydrous ethanedioic acid assumed.

Accept single negative charges on two O atoms singly bonded to C.
Do not accept resonance structures.
Allow any combination of dots/crosses or lines to represent electron pairs.

electrons delocalized «across the O–C–O system»
OR
resonance occurs

Accept delocalized π-bond(s).
No ECF from (d).

122 «pm» < C–O < 143 «pm»

Accept any answer in range 123 «pm» to 142 «pm».
Accept “bond intermediate between single and double bond” or “bond order 1.5”.

coordinate/dative/covalent bond from O to «transition» metal «ion»
OR
acts as a Lewis base/nucleophile

can occupy two positions
OR
provide two electron pairs from different «O» atoms
OR
form two «coordinate/dative/covalent» bonds «with the metal ion»
OR
chelate «metal/ion»

Outline how this spectrum is related to the energy levels in the hydrogen atom.

A sample of magnesium has the following isotopic composition.

Calculate the relative atomic mass of magnesium based on this data, giving your answer to two decimal places.

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.

Determine the molecular formula of menthol using your answers from parts (d)(i) and (ii).

Deduce the order of reaction with respect to Cl₂ and NO.

State the rate expression for the reaction.

Calculate the value of the rate constant at 263 K.

electron transfer/transition between high«er» energy level to low«er» energy level

OR

electron transitions into first energy level causes UV series

OR

transition into second energy level causes visible series

OR

transition into third energy level causes infrared series

Accept any of the points shown on a diagram.

24 x 0.786 + 25 x 0.101 + 26 x 0.113

24.33

Award [2] for correct final answer.
Award [0] for 24.31 with no working (data booklet value).

carbon: «$\frac{0.4490\text{g}}{44.01\text{g}\text{mo}{\text{l}}^{-1}}$ =» 0.01020 «mol» / 0.1225 «g»
OR
hydrogen: «$\frac{0.1840\times 2}{18.02gmo{l}^{-1}}$ =» 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.

Do not award M3 for a hydrocarbon.

«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.

C₁₀H₂₀O

[1 Mark]

Cl₂: first
NO: second

rate = k [NO]² [Cl₂]

180 / 1.80 x 10² «dm⁶ mol^–2 min^–1»

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.

State the equilibrium constant expression, K_c.

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

Determine an approximate order of magnitude for K_c, using sections 1 and 2 of the data booklet. Assume ΔG^Θ for the forward reaction is approximately +50 kJ at 298 K.

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.

Calculate the maximum volume of CO₂, in cm³, produced at STP by the combustion of 0.600 g of urea, using sections 2 and 6 of the data booklet.

Describe the bond formation when urea acts as a ligand in a transition metal complex ion.

The C–N bonds in urea are shorter than might be expected for a single C–N bond. Suggest, in terms of electrons, how this could occur.

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.

Predict the splitting pattern of the ¹H NMR spectrum of urea.

Outline why TMS (tetramethylsilane) may be added to the sample to carry out ¹H NMR spectroscopy and why it is particularly suited to this role.

molar mass of urea «4 $\times$ 1.01 + 2 $\times$ 14.01 + 12.01 + 16.00» = 60.07 «g mol^_-1»

«% nitrogen = $\frac{\text{2}\times \text{14.01}}{\text{60.07}}$ $\times$ 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 $\times$ 0.100 mol dm^–3» = 5.00 $\times$ 10^–3 «mol»

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

Award [2] for correct final answer.

[2 marks]

$K_{\text{c}}=\frac{[{({\text{H}}_2\text{N})}_2\text{CO}]\times [{\text{H}}_2\text{O}]}{{[\text{N}{\text{H}}_3]}^2\times [\text{C}{\text{O}}_2]}$

[1 mark]

«K_c» decreases AND reaction is exothermic

OR

«K_c» decreases AND ΔH is negative

OR

«K_c» decreases AND reverse/endothermic reaction is favoured

[1 mark]

ln K « = $\frac{-\mathrm{\Delta }{G}^{\mathrm{\Theta }}}{RT}=\frac{-50\times {10}^3\text{ J}}{8.31\text{ J }{\text{K}}^{-1}\text{ mo}{\text{l}}^{-1}\times 298\text{ K}}$ » = –20

«K_c =» 2 $\times$ 10^–9

OR

1.69 $\times$ 10^–9

OR

10^–9

Accept range of 20-20.2 for M1.

Award [2] for correct final answer.

[2 marks]

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]

«V = $\frac{\text{0.600 g}}{\text{60.07 g mo}{\text{l}}^{-1}}$ $\times$ 22700 cm³ mol^–1 =» 227 «cm³»

[1 mark]

lone/non-bonding electron pairs «on nitrogen/oxygen/ligand» given to/shared with metal ion

co-ordinate/dative/covalent bonds

[2 marks]

lone pairs on nitrogen atoms can be donated to/shared with C–N bond

OR

C–N bond partial double bond character

OR

delocalization «of electrons occurs across molecule»

OR

slight positive charge on C due to C=O polarity reduces C–N bond length

[1 mark]

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

[2 marks]

singlet

Accept “no splitting”.

[1 mark]

acts as internal standard

OR

acts as reference point

one strong signal

OR

12 H atoms in same environment

OR

signal is well away from other absorptions

Accept “inert” or “readily removed” or “non-toxic” for M1.

[2 marks]

Plot the relative values of the first four ionization energies of sodium.

Outline why the alkali metals (group 1) have similar chemical properties.

Describe the structure and bonding in solid sodium oxide.

Calculate values for the following changes using section 8 of the data booklet.

ΔH_atomisation (Na) = 107 kJ mol⁻¹
ΔH_atomisation (O) = 249 kJ mol⁻¹

$\frac{1}{2}$O₂(g) → O^2- (g):

Na (s) → Na⁺ (g):

The standard enthalpy of formation of sodium oxide is −414 kJ mol⁻¹. Determine the lattice enthalpy of sodium oxide, in kJ mol⁻¹, using section 8 of the data booklet and your answers to (d)(i).

(If you did not get answers to (d)(i), use +850 kJ mol⁻¹ and +600 kJ mol⁻¹ respectively, but these are not the correct answers.)

Justify why K₂O has a lower lattice enthalpy (absolute value) than Na₂O.

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 5.00g of sodium oxide produces 5.50g 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 (g)(i).

An allotrope of molecular oxygen is ozone. Compare, giving a reason, the bond enthalpies of the O to O bonds in O₂ and O₃.

Outline why a real gas differs from ideal behaviour at low temperature and high pressure.

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.

     [✔]

Notes: Accept curve showing general trend.
Award mark only if the energy difference between the first two points is larger than that between points 2/3 and 3/4.

same number of electrons in outer shell

OR

all are s¹ [✔]

«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.

$\frac{1}{2}$O₂(g) → O^2- (g)

«ΔH_atomisation (O) + 1st EA + 2nd EA = 249 k Jmol⁻¹ − 141 kJmol⁻¹ + 753 kJmol⁻¹ =» «+»861 «kJmol⁻¹»    [✔]

Na (s) → Na⁺ (g)

«ΔH_atomisation (Na) + 1st IE = 107 kJmol⁻¹ + 496 kJmol⁻¹ =» «+»603 «kJmol⁻¹»     [✔]

lattice enthalpy = 861 «kJ mol⁻¹» + 2 × 603 «kJ mol⁻¹» −(−414 «kJ mol⁻¹»)     [✔]

«= +» 2481 «kJ mol⁻¹»    [✔]

Note: Award [2] for correct final answer.

If given values are used:
M1: lattice enthalpy = 850 «kJ mol⁻¹» +
2 × 600 «kJ mol⁻¹» −(−414 «kJ mol⁻¹»)
M2: «= +» 2464 «kJ mol⁻¹»

K⁺ ion is larger than Na⁺

OR

smaller attractive force because of greater distance between ion «centres»      [✔]

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 «= $\frac{5.00\text{g}}{61.98\text{g mo}{\text{l}}^{-1}}$» = 0.0807/8.07 × 10⁻² «mol»

OR

mass of Na₂O₂ theoretical yield «= $\frac{5.00\text{g}}{61.98\text{g mo}{\text{l}}^{-1}}$ × 77.98 gmol⁻¹» = 6.291 «g»    [✔]

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

Note: Award [2] for correct final answer.

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

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

ΔH = «∑ΔH_f products − ∑ΔH_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     [✔]

bond in O₃ has lower enthalpy AND bond order is 1.5 «not 2»    [✔]

Note: Accept “bond in ozone is longer”.

Any one of:

finite volume of particles «requires adjustment to volume of gas»     [✔]

short-range attractive forces «overcomes low kinetic energy»    [✔]

NaOH    [✔]

IV    [✔]

Generally well done with a correct plot of ionization energies.

The majority answered correctly stating same number of valence electrons as the reason. Some candidates stated same size or similar ionization energy but the majority scored well.

Many candidates lost one or two marks for missing “electrostatic forces” between “oppositely charged ions”, or “lattice”. Some candidates’ answers referred to covalent bonds and shapes of molecules.

Good performance with typical error being in the calculation for the first equation, ½O2 (g) → O₂⁻ (g), where the value for the first electron affinity of oxygen was left out.

Many candidates earned some credit for ECF based on (d)(i).

Average performance with answers using atomic size rather than ionic size or making reference to electronegativities of K and Na.

An average of 1.1 out of 3 earned here. Many candidates could write a balanced equation for the reaction of sodium oxide with water but not phosphorus(V) oxide. Mediocre performance in identifying the acid/base nature of the solutions formed.

The majority earned one or two marks in finding a % yield.

The average was 2.2 out 3 for this question on enthalpy of formation. Enthalpy calculations were generally well done.

The majority of candidates referred to “bond enthalpy values are average”, rather than not valid for solids or only used for gases.

Some candidates recognized that ozone had a resonance structure but then only compared bond length between ozone and oxygen rather than bond enthalpy.

Few candidates could distinguish the cause for difference in behaviour between real and ideal gases at low temperature or high pressure. Many answers were based on increase in number of collisions or faster rate or movement of gas particles.

Na₂O was a common formula in many candidates’ answers for the product of the reaction of sodium peroxide with water.

The vast majority of candidates could correctly state the oxidation number of carbon in sodium carbonate.

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.

Bubbles of gas were also observed at another electrode. Identify the electrode and the gas.

Electrode number (on diagram):

Name of gas: 

Deduce the half-equation for the formation of the gas identified in (c)(iii).

Determine the enthalpy of solution of copper(II) chloride, using data from sections 18 and 20 of the data booklet.

The enthalpy of hydration of the copper(II) ion is −2161 kJ mol⁻¹.

Calculate the cell potential at 298 K for the disproportionation reaction, in V, using section 24 of the data booklet.

Comment on the spontaneity of the disproportionation reaction at 298 K.

Calculate the standard Gibbs free energy change, ΔG^θ, to two significant figures, for the disproportionation at 298 K. Use your answer from (e)(i) and sections 1 and 2 of the data booklet.

Suggest, giving a reason, whether the entropy of the system increases or decreases during the disproportionation.

Deduce, giving a reason, the sign of the standard enthalpy change, ΔH^θ, for the disproportionation reaction at 298 K.

Predict, giving a reason, the effect of increasing temperature on the stability of copper(I) chloride solution.

Describe how the blue colour is produced in the Cu(II) solution. Refer to section 17 of the data booklet.

Deduce why the Cu(I) solution is colourless.

When excess ammonia is added to copper(II) chloride solution, the dark blue complex ion, [Cu(NH₃)₄(H₂O)₂]²⁺, forms.

State the molecular geometry of this complex ion, and the bond angles within it.

Molecular geometry:

Bond angles: 

Examine the relationship between the Brønsted–Lowry and Lewis definitions of a base, referring to the ligands in the complex ion [CuCl₄]²⁻.

[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 = «$\frac{0.443 \text{g}}{18.02 \text{g\hspace{0.05em}mo}{\text{l}}^{-1}}$=» 0.0246 «mol»
OR
moles of CuCl₂ =«$\frac{1.696 \text{g}}{134.45 \text{g\hspace{0.05em}mo}{\text{l}}^{-1}}$= » 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⁻.

«electrode» 3 AND oxygen/O₂ ✔

NOTE: Accept chlorine/Cl₂.

2H₂O (l) → 4H⁺ (aq) + O₂ (g) + 4e^– ✔

NOTE: Accept 2Cl^– (aq) → Cl₂ (g) + 2e^–.
Accept 4OH⁻ → 2H₂O + O₂ + 4e⁻

enthalpy of solution = lattice enthalpy + enthalpies of hydration «of Cu²⁺ and Cl⁻» ✔

«+2824 kJ mol^–1 − 2161 kJ mol^–1 − 2(359 kJ mol^–1) =» −55 «kJ mol^–1» ✔

NOTE: Accept enthalpy cycle.
Award [2] for correct final answer.

E^θ = «+0.52 – 0.15 = +» 0.37 «V» ✔

spontaneous AND E^θ positive ✔

ΔG^θ = «−nFE = −1 mol × 96 500 C Mol^–1 × 0.37 V=» −36 000 J/−36 kJ ✔

NOTE: Accept “−18 kJ mol^–1 «per mole of Cu⁺»”.

Do not accept values of n other than 1.

Apply SF in this question.

Accept J/kJ or J mol⁻¹/kJ mol⁻¹ for units.

2 mol (aq) → 1 mol (aq) AND decreases ✔

NOTE: Accept “solid formed from aqueous solution AND decreases”.
Do not accept 2 mol → 1 mol without (aq).

ΔG^θ < 0 AND ΔS^θ < 0 AND ΔH^θ < 0
OR
ΔG^θ + TΔS^θ < 0 AND ΔH^θ < 0 ✔

TΔS more negative «reducing spontaneity» AND stability increases ✔

NOTE: Accept calculation showing non-spontaneity at 433 K.

«ligands cause» d-orbitals «to» split ✔

light absorbed as electrons transit to higher energy level «in d–d transitions»
OR
light absorbed as electrons promoted ✔

energy gap corresponds to «orange» light in visible region of spectrum ✔

colour observed is complementary ✔

full «3»d sub-level/orbitals
OR
no d–d transition possible «and therefore no colour» ✔

octahedral AND 90° «180° for axial» ✔

NOTE: Accept square-based bi-pyramid.

Any two of:
ligand/chloride ion Lewis base AND donates e-pair ✔
not Brønsted–Lowry base AND does not accept proton/H⁺ ✔
Lewis definition extends/broader than Brønsted–Lowry definition ✔

Outline why repeating quantitative measurements is important.

to reduce random errors

OR

to increase precision

Accept “to ensure reliability”.

[1 mark]

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.

Suggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.

«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 × 10⁻⁴mol 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) ✔

rate of reaction of oxygen with impurities depends on temperature
OR
rate at which bacteria/organisms grow/respire depends on temperature ✔

Determine the oxidation state of vanadium in each of the following species.

Identify, from the table, a non-vanadium species that can reduce VO²⁺(aq) to V³⁺(aq) but no further.

Identify, from the table, a non-vanadium species that could convert ${\text{VO}}_2^{+}\text{(aq)}$ to V²⁺(aq).

Formulate an equation for the reaction between VO²⁺(aq) and V²⁺(aq) in acidic solution to form V³⁺(aq).

Comment on the spontaneity of this reaction by calculating a value for $\mathrm{\Delta }{G}^{\theta }$ using the data given in (b) and in section 1 of the data booklet.

$V_2{O}_5:+5$
$V{O}^{2+}:+4$

Do not penalize incorrect notation twice.

[2 marks]

H₂SO₃(aq)
OR
Pb(s)

[1 mark]

Zn(s)

[1 mark]

$\text{V}{\text{O}}^{2+}(\text{aq)}+{\text{V}}^{2+}(\text{aq)}+\text{2}{\text{H}}^{+}(\text{aq)}\to \text{2}{\text{V}}^{3+}(\text{aq)}+{\text{H}}_{\text{2}}\text{O(l)}$

Accept equilibrium sign.

[1 mark]

$E^{\theta }\ll =+0.34\text{ V}-(-0.26\text{ V})\gg =+0.60\ll \text{V}\gg$

$\mathrm{\Delta }{G}^{\theta }=\ll -nF{E}^{\theta }=-9.65\times {10}^4\text{ C}\text{mo}{\text{l}}^{-1}\times 0.60\text{ J}{\text{C}}^{-1}=\gg -57900\ll \text{J}\text{mo}{\text{l}}^{-1}\gg /-57.9\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg$

spontaneous as $\mathrm{\Delta }{G}^{\theta }$ is negative

Do not award M3 as a stand-alone answer.

Accept “spontaneous” for M3 if answer given for M2 is negative.

Accept “spontaneous as $E^{\theta }$ is positive” for M3.

[3 marks]

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.

Outline the reason why PCl₅ is a non-polar molecule, while PCl₄F is polar.

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

Calculate the entropy change, ΔS, in J K⁻¹mol⁻¹, for this reaction.

Chemistry 2e, Chpt. 21 Nuclear Chemistry, Appendix G: Standard Thermodynamic Properties for Selected Substances https://openstax.org/books/chemistry-2e/pages/g-standard-thermodynamic-properties-for- selectedsubstances# page_667adccf-f900-4d86-a13d-409c014086ea © 1999-2021, Rice University. Except where otherwise noted, textbooks on this site are licensed under a Creative Commons Attribution 4.0 International License. (CC BY 4.0) https://creativecommons.org/licenses/by/4.0/.

Calculate the Gibbs free energy change (ΔG), in kJ mol⁻¹, for this reaction at 25 °C. Use section 1 of the data booklet.

If you did not obtain an answer in c(i) or c(ii) use −87.6 kJ mol⁻¹ and −150.5 J mol⁻¹K⁻¹ respectively, but these are not the correct answers.

Determine the equilibrium constant, K, for this reaction at 25 °C, referring to section 1 of the data booklet.

If you did not obtain an answer in (c)(iii), use ΔG = –43.5 kJ mol⁻¹, but this is not the correct answer.

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.

PCl₅ is non-polar:

symmetrical
OR
dipoles cancel ✔

PCl₄F is polar:

P–Cl has a different bond polarity than P–F ✔

non-symmetrical «dipoles»
OR
dipoles do not cancel ✔

Accept F more electronegative than/different electronegativity to Cl for M2.

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

«ΔS = 364.5 J K^–1mol^–1 – (311.7 J K^–1mol^–1 + 223.0 J K^–1mol^–1)=» –170.2 «J K^–1mol^–1» ✔

«ΔS =» –0.1702 «kJ mol^–1K^–1»
OR
298 «K» ✔

«ΔG = –92.5 kJ mol^–1 – (298 K × –0.1702 kJ mol^–1K^–1) =» –41.8 «kJ mol^–1» ✔

Award [2] for correct final answer.

If –87.6 and -150.5 are used then –42.8.

«ΔG = –41.8 kJ mol^–1 = $-\frac{8.31 \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}}{1000}$ × 298 K × lnK»
OR
«ΔG = –41800 J mol^–1 = –8.31 J mol^–1K^–1 × 298 K × lnK»

«lnK = =» 16.9 ✔

«K = e^16.9 =» 2.19 × 10⁷ ✔

Award [2] for correct final answer.

Accept range of 1.80 × 10⁶–2.60 × 10⁷.

If –43.5 is used then 4.25 × 10⁷.

«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 ✔

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 reasons that sodium hydroxide is considered a Brønsted–Lowry and Lewis 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.

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

«$\frac{0.01420 \mathrm{mol}}{3}=$» 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.

Brønsted–Lowry base:
proton acceptor

AND

Lewis Base:
e^– pair donor/nucleophile ✔

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.

Predict the number of ¹H NMR signals, and splitting pattern of the –CH₃ seen for propanone (CH₃COCH₃) and propanal (CH₃CH₂CHO).

Predict the fragment that is responsible for a m/z of 31 in the mass spectrum of propan‑1‑ol. Use section 28 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.

CH₃O⁺ ✔

Accept any structure i.e. “CH₂OH⁺”.

State why this equilibrium reaction is considered homogeneous.

Predict, giving your reason, the sign of the standard entropy change of the forward reaction.

Calculate the standard Gibbs free energy change, ΔG^Θ, in kJ, for this reaction at 1000 K. Use sections 1 and 2 of the data booklet.

Predict, giving your reasons, whether the forward reaction is endothermic or exothermic. Use your answers to (b) and (c).

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”.

negative AND fewer moles/molecules «of gas» in the products ✔

ΔG^Θ =«–RT ln K_c =» –8.31 J K^–1 mol^–1 × 1000 K × ln 280

OR

ΔG^Θ = – 4.7 × 10⁴ «J» ✔

«ΔG^Θ =» – 47 «kJ» ✔

Award [2] for correct final answer.

ΔG^Θ < 0/spontaneous AND ΔS^Θ < 0/unfavourable ✔

exothermic AND ΔH^Θ  «must be» negative/favourable ✔

«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.

Write a balanced equation for the reaction that occurs.

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.

Ammonia is added to water that contains a few drops of an indicator. Identify an indicator that would change colour. Use sections 21 and 22 of the data booklet.

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, giving a reason, whether magnesium or nitrogen would have the greater sixth ionization energy.

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

aluminium/Al ✔

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

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 90.5-91.5%

[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) ✔

phenol red ✔

Accept bromothymol blue or phenolphthalein.

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 ✔

nitrogen AND electron lost from first «energy» level/s sub-level/s-orbital AND magnesium from p sub-level/p-orbital/second «energy» level
OR
nitrogen AND electron lost from lower level «than magnesium» ✔

Accept “nitrogen AND electron lost closer to the nucleus «than magnesium»”.

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.

Done very well. However, it was disappointing to see the formula of oxygen molecule as O and the oxide as Mg₂O and MgO₂ at HL level.

Average performance; the question asked to identify a metal; however, answers included S, Si, P and even noble gases besides Be and Na. The only choice of aluminium; however, since its oxide is amphoteric, it could not be the answer in the minds of some.

Very good performance; some calculated the mass of oxygen instead of magnesium for the calculation of the amount, in mol, of magnesium. Others calculated the mass, but not the amount in mol as required.

Mediocre performance; instead of calculating percentage uncertainty, some calculated percentage difference.

Satisfactory performance; however, a good number could not answer the question correctly on determining the percentage yield.

Poorly done. The question asked to evaluate and explain but instead many answers simply agreed with the information provided instead of assessing its strength and limitation.

Mediocre performance; explaining the yield found was often a challenge by not recognizing that incomplete reaction or Mg partially oxidized or impurities present that evaporated or did not react would explain the yield.

Calculating coefficients that balance the given equation was done very well.

Well done; some chose bromocresol green or methyl red as the indicator that would change colour, instead of phenol red, bromothymol blue or phenolphthalein.

Good performance; however, surprising number of candidates could not determine one or both oxidation states correctly or wrote it as 3 or 3−, instead of −3.

Average performance; choosing the given reaction as an acid-base or redox reaction was not done well. Often answers were contradictory and the reasoning incorrect.

Stating the number of subatomic particles in a ¹⁴N^3- was done very well. However, some answers showed a lack of understanding of how to calculate the number of relevant subatomic particles given formula of an ion with charge and mass number.

Exceptionally well done; A few candidates referred to isomers, rather than isotopes.

There was reference to nitrogen and magnesium, rather than nitride and magnesium ions. Also, instead identifying smaller nuclear charge in nitride ion, some referred to core electrons, Z_eff, increased electron-electron repulsion or shielding.

Common error in suggesting nitrogen would have the greater sixth ionization energy was that for nitrogen, electron is lost from first energy level without making reference to magnesium losing it from second energy level.

Good performance; some teachers were concerned about the expected answers. However, generally, students were able to suggest two reasons why matter is divisible.

One teacher commented that not asking to describe bonding in terms of electrostatic attractions as in earlier papers would have been confusing and some did answer in terms of electrostatic forces of attractions involved. However, the question was clear in its expectation that the answer had to be in terms of how the valence electrons produce the three types of bonds and the overall performance was good. Some had difficulty identifying the bond type for Mg, O₂ and MgO.

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.

Explain the mechanism of the reaction between chloroethane and aqueous sodium hydroxide, $\mathrm{NaOH} \left(\mathrm{aq}\right)$, using curly arrows to represent the movement of electron pairs.

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

Deduce the number of signals and chemical shifts with splitting patterns in the ¹H 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.

$\mathrm{CFC}$s produce chlorine radicals. Write two successive propagation steps to show how chlorine radicals catalyse the depletion of ozone.

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

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

Accept “chloroethane AND polar”.

curly arrow going from lone pair/negative charge on $\mathrm{O}$ in ⁻OH to $\mathrm{C}$ ✔

curly arrow showing $\mathrm{Cl}$ leaving ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

Accept $O{H}^{\mathit{-}}$ with or without the lone pair.

Do not accept curly arrows originating on $H$ in $O{H}^{-}$.

Accept curly arrows in the transition state.

Do not penalize if $HO$ and $Cl$ are not at 180°.

Do not award M3 if $OH-C$ bond is represented. 

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

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

2 «signals» ✔

0.9−1.0 AND triplet ✔

3.3−3.7 AND quartet ✔

Accept any values in the ranges.

Award [1] for two correct chemical shifts or two correct splitting patterns.

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

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

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.

${\mathrm{O}}_3+\mathrm{Cl}·\to \mathrm{O}2+\mathrm{ClO}·$ ✔

$\mathrm{ClO}·+\mathrm{O}·\to {\mathrm{O}}_2+\mathrm{Cl}·$
OR
$\mathrm{ClO}·+{\mathrm{O}}_3\to \mathrm{Cl}·+2{\mathrm{O}}_2$ ✔

Penalize missing/incorrect radical dot (∙) once only.

Well answered question with 90% of candidates correctly identifying the complete electron configuration for chlorine.

Most candidates could correctly explain the relative sizes of chlorine atom and chloride ion.

Fairly well answered though some candidates missed M2 for not recognizing the same number of shells affected.

More than 80% could identify that the two peaks in the MS of chlorine are due to different isotopes.

Not well answered. Some candidates were able to identify m/z 74 being due to the m/z of two Cl-37 atoms, however fewer candidates were able to explain the relative abundance of the isotope.

Stoichiometric calculations were generally well done and over 90% could calculate mol from a given mass.

90% of candidates earned full marks on this 2-mark question involving finding a limiting reactant.

Surprisingly, quite a number of candidates struggled with the quantity of excess reactant despite correctly identifying limiting reactant previously.

Most candidates could find the volume of gas produced in a reaction under standard conditions.

More than 90% could identify the oxidation number of manganese in both MnO₂ and MnCl₂.

Most candidates stated that MnO₂ is an oxidizing agent in the reaction but many did not get the mark because there was no reference to oxidation states.

Another well answered 1-mark question where candidates correctly identified a weak acid as an acid which partially dissociates in water. 

Roughly ⅓ of the candidates failed to identify the conjugate base, perhaps distracted by the fact it was not contained in the equation given.

Vast majority of candidates could calculate the concentration of H⁺ (aq) in a HClO (aq) solution with a pH =3.61.

Many identified the reaction of chlorine with ethane as free-radical substitution, or just substitution, with some erroneously stating nucleophilic or electrophilic substitution.

The underlying reasons for the relative reactivity of ethane and chloroethane were not very well known with a few giving erroneous reasons and some stating ethane more reactive.

Few earned full marks for the curly arrow mechanism of the reaction between sodium hydroxide and chloroethane. Mistakes being careless curly arrow drawing, inappropriate –OH notation, curly arrows from the hydrogen or from the carbon to the C–Cl bond, or a method that missed the transition state.

Approximately 60% could draw ethoxyethane however many demonstrated little knowledge of structure of an ether molecule.

A poorly answered question with some getting full marks on this 1HNMR spectrum of ethoxyethane question. Very few could identify all 3 of number of signals, chemical shift, and splitting pattern.

Another good example of candidates being well rehearsed in calculations with 90% earning 2/2 on this question of calculation percentage by mass composition. 

Somewhat disappointing answers on this question about how international cooperation has contributed to the lowering of CFC emissions. Many gave vague answers and some referred to carbon emissions and global warming.

Few could construct the propagation equations showing how CFCs affect ozone, and many lost marks by failing to identify ClO· as a radical.

Magnesium ions produce no emission or absorption lines in the visible region of the electromagnetic spectrum. Suggest why most magnesium compounds tested in a school laboratory show traces of yellow in the flame.

(i) Explain the convergence of lines in a hydrogen emission spectrum.

(ii) State what can be determined from the frequency of the convergence limit.

Magnesium chloride can be electrolysed.

(i) 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 922K and 987K respectively.

(ii) Identify the type of reaction occurring at the cathode (negative electrode).

(iii) State the products when a very dilute aqueous solution of magnesium chloride is electrolysed.

Standard electrode potentials are measured relative to the standard hydrogen electrode. Describe a standard hydrogen electrode.

A magnesium half-cell, Mg(s)/Mg²⁺(aq), can be connected to a copper half-cell, Cu(s)/Cu²⁺(aq).

(i) Formulate an equation for the spontaneous reaction that occurs when the circuit is completed.

(ii) Determine the standard cell potential, in V, for the cell. Refer to section 24 of the data booklet.

(iii) Predict, giving a reason, the change in cell potential when the concentration of copper ions increases.

contamination with sodium/other «compounds»

i
energy levels are closer together at high energy / high frequency / short wavelength

ii
ionisation energy

i)

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).

ii)

reduction

iii)

Anode (positive electrode):
oxygen/O₂
OR
hydogen ion/proton/H⁺ AND oxygen/O₂
Cathode (negative electrode):
hydrogen/H₂
OR
hydroxide «ion»/OH^– AND hydrogen/H₂

Award [1 max] if correct products given at wrong electrodes.

Any two of:

«inert» Pt electrode
OR
platinum black conductor

1 mol dm^–3 H⁺ (aq)

H₂ (g) at 100 kPa

Accept 1 atm H₂ (g).
Accept 1 bar H₂ (g)
Accept a labelled diagram.
Ignore temperature if it is specified.

i

Mg(s) + Cu²⁺ (aq) → Mg²⁺ (aq) + Cu(s)

ii

«+0.34V – (–2.37V) = +»2.71 «V»

iii

cell potential increases

reaction «in Q4(k)(i)» moves to the right
OR
potential of the copper half-cell increases/becomes more positive

Accept correct answers based on the Nernst equation

Suggest why many chemicals, including hydrogen peroxide, are kept in brown bottles instead of clear colourless bottles.

In a laboratory experiment solutions of potassium iodide and hydrogen peroxide were mixed and the volume of oxygen generated was recorded. The volume was adjusted to 0 at t = 0.

The data for the first trial is given below.

Plot a graph on the axes below and from it determine the average rate of
formation of oxygen gas in cm³ O₂ (g) s⁻¹.

Average rate of reaction:

Two more trials (2 and 3) were carried out. The results are given below.

Determine the rate equation for the reaction and its overall order, using your answer from (b)(i).

Rate equation: 

Overall order: 

Additional experiments were carried out at an elevated temperature. On the axes below, sketch Maxwell–Boltzmann energy distribution curves at two temperatures T₁ and T₂, where T₂ > T₁.

Apart from a greater frequency of collisions, explain, by annotating your graphs in (b)(iii), why an increased temperature causes the rate of reaction to increase.

MnO₂ is another possible catalyst for the reaction. State the IUPAC name for MnO₂.

Comment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.

H₂O₂ (aq) + CH₃COOH (aq) ⇌ CH₃COOOH (aq) + H₂O (l)

Sodium percarbonate, 2Na₂CO₃•3H₂O₂, is an adduct of sodium carbonate and hydrogen peroxide and is used as a cleaning agent.

M_r (2Na₂CO₃•3H₂O₂) = 314.04

Calculate the percentage by mass of hydrogen peroxide in sodium percarbonate, giving your answer to two decimal places.

decomposes in light    [✔]

Note: Accept “sensitive to light”.

points correctly plotted     [✔]

best fit line AND extended through (to) the origin   [✔]

Average rate of reaction:
«slope (gradient) of line =» 0.022 «cm³ O₂ (g) s⁻¹»   [✔]

Note: Accept range 0.020–0.024cm³ O₂ (g) s⁻¹.

Rate equation:
Rate = k[H₂O₂] × [KI]     [✔]

Overall order:
2     [✔]

Note: Rate constant must be included.

peak of T₂ to right of AND lower than T₁     [✔]

lines begin at origin AND T₂ must finish above T₁     [✔]

E_a marked on graph    [✔]

explanation in terms of more “particles” with E ≥ E_a

OR

greater area under curve to the right of E_a in T₂     [✔]

manganese(IV) oxide

OR

manganese dioxide     [✔]

Note: Accept “manganese(IV) dioxide”.

moves «position of» equilibrium to right/products    [✔]

Note: Accept “reactants are always present as the reaction is in equilibrium”.

M( H₂O₂) «= 2 × 1.01 + 2 × 16.00» = 34.02 «g»     [✔]

«% H₂O₂ = 3 × $\frac{34.02}{314.04}$ × 100 =» 32.50 «%»     [✔]

Note: Award [2] for correct final answer.

There were a couple of comments claiming that this NOS question on “why to store hydrogen peroxide in brown bottles” is not the syllabus. Most candidates were quite capable of reasoning this out.

Most candidates could plot a best fit line and find the slope to calculate an average rate of reaction.

Good performance but with answers that either typically included only [H₂O₂] with first or second order equation or even suggesting zero order rate equation.

Fair performance; errors including not starting the two curves at the origin, drawing peak for T2 above T1, T2 finishing below T1 or curves crossing the x-axis.

The majority of candidates earned at least one mark, many both marks. Errors included not annotating the graph with E_a and referring to increase of kinetic energy as reason for higher rate at T2.

A well answered question. Very few candidates had problem with nomenclature.

One teacher suggested that “stored” would have been better than “sold” for this question. There were a lot of irrelevant answers with many believing the back reaction was an acid dissociation.

It is recommended that candidates use the relative atomic masses given in the periodic table.

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 one reason why you would expect the value of ΔH calculated from the $∆H_f^{\mathrm{⦵}}$ values, given in section 12 of data booklet, to differ from your answer to (d)(i).

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.

The equilibrium constant, K_c, has a value of 1.01 at 298 K.

Calculate ΔG^⦵, in kJ mol^–1, for this reaction. Use sections 1 and 2 of the data booklet.

Calculate a value for the entropy change, ΔS^⦵, in J K^–1 mol^–1 at 298 K. Use your answers to (e)(i) and section 1 of the data booklet.

If you did not get answers to (e)(i) use –1 kJ, but this is not the correct answer.

Justify the sign of ΔS with reference to the equation.

Predict, giving a reason, how a change in temperature from 298 K to 273 K would affect the spontaneity of the reaction.

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»

bond energies are average values «not specific to the compound» ✔

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

K_c increases AND «forward» reaction endothermic ✔

«ΔG^⦵ = − RT lnK_c»
ΔG^⦵ = − 8.31 «J K⁻¹ mol⁻¹» × 298 «K» × ln (1.01) / −24.6 «J mol⁻¹» ✔

= −0.0246 «kJ mol^–1» ✔

Award [2] for correct final answer.

Award [1 max] for +0.0246 «kJ mol^–1».

«ΔG^⦵ = ΔH^⦵ − TΔS^⦵»

ΔG^⦵ = −129 «kJ mol^–1» − (298 «K» × ΔS) = −0.0246 «kJ mol^–1» ✔

ΔS^⦵ = « $\frac{\left(129 \mathrm{kJ} {\mathrm{mol}}^{-1}+0.0246 \mathrm{kJ} {\mathrm{mol}}^{-1}\right)\times {10}^3}{298 \mathrm{K}}$ = » −433 «J K^–1 mol^–1» ✔

Award [2] for correct final answer.

Award [1 max] for “−0.433 «kJ K^–1mol^–1»”.

Award [1 max] for “433” or “+433” «J K^–1 mol^–1».

Award [2] for −430 «J K^–1 mol^–1» (result from given values).

«negative as» product is liquid and reactants gases
OR
fewer moles of gas in product ✔

reaction «more» spontaneous/ΔG negative/less positive AND effect of negative entropy decreases/TΔS increases/is less negative/more positive
OR
reaction «more» spontaneous/ΔG negative/less positive AND reaction exothermic «so K_c increases » ✔

Award mark if correct calculation shown.

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.

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

Calculate the change in entropy, ΔS, in J K⁻¹, for the decomposition of calcium carbonate.

Determine the temperature, in K, at which the decomposition of calcium carbonate becomes spontaneous, using b(i), b(ii) and section 1 of the data booklet.

(If you do not have answers for b(i) and b(ii), use ΔH = 190 kJ and ΔS = 180 J K⁻¹, but these are not the correct answers.)

Sketch an energy profile for the decomposition of calcium carbonate based on your answer to b(i), labelling the axes and activation energy, E_a.

State how adding a catalyst to the reaction would impact the enthalpy change of reaction, ΔH, and the activation energy, E_a.

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 d(i). Calculate the percentage yield of CaCO₃.

(If you did not obtain an answer to d(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».

«ΔS = (40 J K⁻¹ + 214 J K⁻¹) − (93 J K⁻¹) =» 161 «J K⁻¹» ✓

Ignore an extra step to determine total entropy change in JK^–1: 161 J mol^–1K^–1 x 5.55 mol = 894 «J mol^–1K^–1»

Award [1] for 894 «J mol^–1K^–1».

«spontaneous» if ΔG = ΔH − TΔS < 0
OR
ΔH < TΔS ✓

«T >$\frac{179 \mathrm{kJ}}{0.16 \mathrm{kJ} {\mathrm{K}}^{-1}}$=» 1112 «K» ✓

Award [2] for correct final answer.

Accept “1056 K” if both of the incorrect values are used to solve the problem.

Do not award M2 for any negative T value.

endothermic sketch ✓

x-axis labelled “extent of reaction/progress of reaction/reaction coordinate/reaction pathway” AND y-axis labelled “potential energy/energy/enthalpy✓

activation energy/E_a ✓

Do not accept “time” for x-axis.

ΔH same AND lower E_a ✓

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 × 10⁻²mol dm⁻³ =» 0.0466 «mol dm⁻³» ✓