Draw the Lewis structures of oxygen, O₂, and ozone, O₃.

Outline why both bonds in the ozone molecule are the same length and predict the bond length in the ozone molecule. Refer to section 10 of the data booklet.

Reason: 

Length:

Predict the bond angle in the ozone molecule.

Discuss how the different bond strengths between the oxygen atoms in O₂ and O₃ in the ozone layer affect radiation reaching the Earth’s surface.

Identify the steps which absorb ultraviolet light.

Determine, showing your working, the wavelength, in m, of ultraviolet light absorbed by a single molecule in one of these steps. Use sections 1, 2 and 11 of the data booklet.

Ozone depletion is catalysed by nitrogen monoxide, NO, which is produced in aircraft and motor vehicle engines, and has the following Lewis structure.

Show how nitrogen monoxide catalyses the decomposition of ozone, including equations in your answer.

NOTES: Coordinate bond may be represented by an arrow.

Do not accept delocalized structure for ozone.

resonance «structures»
OR
delocalization of «the double/pi bond» electrons ✔
121 «pm» < length < 148 «pm» ✔

NOTE: Accept any length between these two values.

any value from 110°–119° ✔

«bond» in O₂ stronger than in O₃ ✔

ozone absorbs lower frequency/energy «radiation than oxygen»
OR
ozone absorbs longer wavelength «radiation than oxygen» ✔

NOTE: Accept ozone «layer» absorbs a range of frequencies.

steps 1 AND 3 ✔

ALTERNATIVE 1:
for oxygen:

$E=«\frac{498 000\text{\hspace{0.05em}J\hspace{0.05em}mo}{\text{l}}^{-1}}{6.02\times {10}^{23} \text{mo}{\text{l}}^{-1}}=» 8.27\times {10}^{-19} «\text{J}»$ ✔

$\mathrm{\lambda }\text{ = }«\frac{6.63\times {10}^{-34}\text{\hspace{0.05em}J\hspace{0.05em}s}\times \text{3}\text{.00}\times \text{1}{\text{0}}^8{\text{\hspace{0.05em}m\hspace{0.05em}s}}^{-1}}{8.27\times {10}^{-19} \text{J}}=» 2.40\times {10}^{-7} «\text{m}»$✔

ALTERNATIVE 2:
for ozone:

similar calculation using 200 < bond enthalpy < 400 for ozone, such as

$E=«\frac{300 000\text{\hspace{0.05em}J\hspace{0.05em}mo}{\text{l}}^{-1}}{6.02\times {10}^{23} \text{mo}{\text{l}}^{-1}}=» 4.98\times {10}^{-19}«\text{J}»$ ✔

$\mathrm{\lambda }\text{ = }«\frac{6.63\times {10}^{-34}\text{\hspace{0.05em}J\hspace{0.05em}s}\times \text{3}\text{.00}\times \text{1}{\text{0}}^8{\text{\hspace{0.05em}m\hspace{0.05em}s}}^{-1}}{4.98\times {10}^{-19} \text{J}}=» 3.99\times {10}^{-7} «\text{m}»$✔

NOTE: Award [2] for correct final answer.

•NO + O₃ → •NO₂ + O₂ ✔

•NO₂ + O₃ → •NO + 2O₂ ✔

NOTE: Accept •NO₂ → •NO + •O AND •O + O₃ → 2O₂ for M2.

Suggest an experiment that shows that magnesium is more reactive than zinc, giving the observation that would confirm this.

Calculate the standard potential, in V, of a cell formed by magnesium and steel half-cells. Use section 24 of the data booklet and assume steel has the standard electrode potential of iron.

Calculate the free energy change, ΔG^⦵, in kJ, of the cell reaction. Use sections 1 and 2 of the data booklet.

This cell causes the electrolytic reduction of water on the steel. State the half-equation for this reduction.

Use the graph to deduce the dependence of the reaction rate on the amount of Mg.

The reaction is first order with respect to HCl. Calculate the time taken, in seconds (s), for half of the Mg to dissolve when [HCl] = 0.5 mol dm^–3.

Carbonates also react with HCl and the rate can be determined by graphing the mass loss. Suggest why this method is less suitable for the reaction of Mg with HCl.

Alternative 1

put Mg in Zn²⁺(aq) ✔

Zn/«black» layer forms «on surface of Mg» ✔

Award [1 max] for “no reaction when Zn placed in Mg²⁺(aq)”.

Alternative 2

place both metals in acid ✔

bubbles evolve more rapidly from Mg
OR
Mg dissolves faster ✔

Alternative 3

construct a cell with Mg and Zn electrodes ✔

Accept “electrons flow from Mg to Zn”.

Accept Mg is negative electrode/anode
OR
Zn is positive electrode/cathode

bulb lights up
OR
shows (+) voltage
OR
size/mass of Mg(s) decreases <<over time>>
OR
size/mass of Zn increases <<over time>>

Accept other correct methods.

Cell potential: «(–0.45 V – (–2.37 V)» = «+»1.92 «V» ✔

«ΔGº = -nFEº»
n = 2
OR
ΔGº = «-»2×96500×1.92 / «-»370,560 «J» ✔

-371 «kJ» ✔

For n = 1, award [1] for –185 «kJ».

Award [1 max] for (+)371 «kJ»

2 H₂O + 2 e^- → H₂ + 2 OH^- ✔

Accept equation with equilibrium arrows.

independent / not dependent ✔

Accept “zero order in Mg”.

«2×170 s» = 340 «s» ✔

Accept 320 – 360 «s».

Accept 400 – 450 «s» based on no more gas being produced after 400 to 450s.

«relative/percentage» decrease in mass is «too» small/«much» less ✔

Accept “«relative/percentage» uncertainty in mass loss «too» great”. OR “density/molar mass of H₂ is «much» less than CO₂”.

Mediocre performance; some experiments would not have worked such as adding magnesium to zinc salt without reference to aqueous environment, adding Zn to magnesium ions, or Mg combustion reaction being more exothermic. In the last one, an inference wad made instead of identifying an observation or measuring temperature using a thermometer or a temperature probe.

Good performance; instead of E° = 1.92 V, answer such as −1.92 V + or −2.82 V showed a lack of understanding of how to calculate E° cell.

Satisfactory performance; two major challenges in applying the equation ΔG° = −nFE° from the data booklet included:

Using n = 1, not 2, the number of electrons transferred in the redox reaction.

ΔG° unit from the equation is in J; some did not convert J to kJ as asked for.

Mediocre performance; some candidates had difficulty writing the reduction half-equation for water, the typical error included O₂(g) gas in the reactant or product, rather than H₂(g) in the product or including an equation with Fe(s) and H₂O(l) as reactants.

Candidates found this to be a tough question (see comments for parts (ii) and (iii)).

Mediocre performance in calculating time from the graph for the data provided. Some wrote the rate expression, which only contains [HCl] and not mass or amount in mol Mg (as a solid, [Mg] is constant). This presented a challenge in arriving at a reasonable answer.

Poorly done; many candidates did not grasp the question and answer it appropriately. Candidates generally did not realize that decrease in mass (due to H₂(g) as a product for the reaction of Mg with HCl) is «too» small/«much» less compared to that of CO₂(g) from the reaction of carbonates with HCl.

Suggest how the change of iodine concentration could be followed.

A student produced these results with $[{\text{H}}^{+}]=0.15\text{ mol}\text{d}{\text{m}}^{-3}$. Propanone and acid were in excess and iodine was the limiting reagent. Determine the relative rate of reaction when $[{\text{H}}^{+}]=0.15\text{ mol}\text{d}{\text{m}}^{-3}$.

The student then carried out the experiment at other acid concentrations with all other conditions remaining unchanged.

Determine the relationship between the rate of reaction and the concentration of acid and the order of reaction with respect to hydrogen ions.

When the concentration of iodine is varied, while keeping the concentrations of acid and propanone constant, the following graphs are obtained.

Deduce, giving your reason, the order of reaction with respect to iodine.

When the reaction is carried out in the absence of acid the following graph is obtained.

Discuss the shape of the graph between A and B.

use a colorimeter/monitor the change in colour

OR

take samples AND quench AND titrate «with thiosulfate»

Accept change in pH.

Accept change in conductivity.

Accept other suitable methods.

Method must imply “change”.

[1 mark]

best fit line

relative rate of reaction $=\ll \frac{-\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{-(0.43-0.80)}{50}=\gg 0.0074/7.4\times {10}^{-3}$

Best fit line required for M1.

M2 is independent of M1.

Accept range from 0.0070 to 0.0080.

[2 marks]

Relationship:
rate of reaction is «directly» proportional to [H⁺]
OR
rate of reaction $\alpha$ [H⁺] 

Order of reaction with respect to [H⁺]:
first

Accept "doubling the concentration doubles the rate".

Do not accept “rate increases as concentration increases”.

[2 marks]

zero order

rate of reaction is the same for all concentrations of iodine

Accept “all graphs have same/similar gradient”.

[2 marks]

slow rate of reaction which gradually increases

as H⁺ ions are produced «to catalyse the reaction»
OR
reaction is autocatalytic

M1 should mention “rate of reaction”.

[2 marks]

(i) Using the graph, explain the order of reaction with respect to sodium thiosulfate.

(ii) In a different experiment, this reaction was found to be first order with respect to hydrochloric acid. Deduce the overall rate expression for the reaction.

i

first order
«because» [Na₂S₂O₃] is «directly» proportional to rate of reaction «$\frac{1}{\mathrm{t}}$»

Do not accept “linear” for M2.

ii

rate = k[Na₂S₂O₃][HCl]

Outline two ways in which the progress of the reaction can be monitored. No practical details are required.

Suggest why point D is so far out of line assuming human error is not the cause.

Draw the best fit line for the reaction excluding point D.

Suggest the relationship that points A, B and C show between the concentration of the acid and the rate of reaction.

Deduce the rate expression for the reaction.

Calculate the rate constant of the reaction, stating its units.

Predict from your line of best fit the rate of reaction when the concentration of HCl is 1.00 mol dm⁻³.

Describe how the activation energy of this reaction could be determined.

Any two of:

loss of mass «of reaction mixture/CO₂»

«increase in» volume of gas produced

change of conductivity

change of pH

change in temperature

Do not accept “disappearance of calcium carbonate”.

Do not accept “gas bubbles”.

Do not accept “colour change” or “indicator”.

[2 marks]

reaction is fast at high concentration AND may be difficult to measure accurately

OR

so many bubbles of CO₂ produced that inhibit contact of HCl(aq) with CaCO₃(s)

OR

insufficient change in conductivity/pH at high concentrations

OR

calcium carbonate has been used up/is limiting reagent/ there is not enough calcium carbonate «to react with the high concentration of HCl»

OR

HCl is in excess

OR

so many bubbles of CO₂ produced that inhibit contact of HCl(aq) with CaCO₃(s)

[1 mark]

straight line going through the origin AND as close to A, B, C as is reasonably possible

[1 mark]

«directly» proportional

Accept “first order” or “linear”.

Do not accept “rate increases as concentration increases” or “positive correlation”.

[1 mark]

rate = k [H⁺]

Accept “rate = k [HCl]”.

[1 mark]

0.02

s^–1

[2 marks]

20.5 $\times$ 10^–3 «mol dm^–3 s^–1»

Accept any answer in the range 19.5–21.5.

[1 mark]

ALTERNATIVE 1:

carry out reaction at several temperatures

plot $\frac{1}{\text{T}}$ against log rate constant

E_a = – gradient $\times$ R

ALTERNATIVE 2:

carry out reaction at two temperatures

determine two rate constants

OR

determine the temperature coefficient of the rate

use the formula  $\ln \frac{k_1}{k_2}=\frac{E_{\text{a}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Accept “gradient = $\frac{-E_{\text{a}}}{R}$” for M3.

Award both M2 and M3 for the formula  $\text{ln}\frac{rat{e}_1}{rat{e}_2}=\frac{E_{\text{a}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$.

Accept any variation of the formula, such as $\frac{rat{e}_1}{rat{e}_2}=e^{-\frac{E_{\text{a}}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$.

[3 marks]

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.

Deduce how the rate of reaction at t = 2 would compare to the initial rate.

It has been suggested that the reaction occurs as a two-step process:

Step 1: N₂O (g) → N₂ (g) + O (g)

Step 2: N₂O (g) + O (g) → N₂ (g) + O₂ (g)

Explain how this could support the observed rate expression.

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.

Determine the standard entropy change, in J K−1, for the decomposition of dinitrogen monoxide.

2N₂O (g) → 2N₂ (g) + O₂ (g)

Dinitrogen monoxide has a positive standard enthalpy of formation, ΔH_f^θ.

Deduce, giving reasons, whether altering the temperature would change the spontaneity of the decomposition reaction.

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 unit time     [✔]

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

half «of the initial rate»    [✔]

Note: Accept “lower/slower «than initial rate»”.

1 slower than 2
OR
1 rate determinant step/RDS    [✔]

1 is unimolecular/involves just one molecule so it must be first order
OR
if 1 faster/2 RDS, second order in N₂O
OR
if 1 faster/2 RDS, first order in O     [✔]

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 E_a 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”.

ΔS^θ = 2(S^θ(N₂)) + S^θ(O₂) – 2(S^θ(N₂O))
OR
ΔS^θ = 2 × 193 «J mol^-1 K^-1» + 205 «J mol^-1 K^-1» – 2 × 220 «J mol^-1 K^-1»     [✔]

«ΔS^θ = +»151 «J K^-1»     [✔]

Note: Award [2] for correct final answer.

exothermic decomposition
OR
ΔH_{(decomposition)} < 0    [✔]

TΔS^θ > ΔH^θ
OR
ΔG^θ «= ΔH^θ – TΔS^θ» < 0 «at all temperatures»     [✔]

reaction spontaneous at all temperatures    [✔]

Students were able in general to relate more moles of gas to increase in pressure.

Few students were able to relate the effect of reduced pressure at constant volume with a decrease in concentration of gas molecules and mostly did not even refer to this, but rather concentrated on lower rate of reaction and frequency of collisions. Many candidates lost a mark by failing to explain rate as collisions per unit time, frequency, etc.

Though the differential equation was considered to be misleading by teachers, most candidates attempted to answer this question, and more than half did so correctly, considering they had the graph to visualize the gradient.

Most students were able to identity step 1 as the RDS/slow but few mentioned unimolecularity or referred vaguely to NO₂ as the only reagent (which was obvious) and got only 1 mark.

Many students drew a lower initial gradient, but most did not reflect the effect of lower temperature on pressure at constant volume and started and finished the curve at the same pressure as the original one.

Almost all candidates identified the inaccurate pressure gauge as a systematic error, thus relating accuracy to this type of error.

The graph was generally well done, but in quite a few cases, candidates did not mention that increase of rate in the catalyzed reaction was due to E (particles) > E_a or did so too vaguely.

Candidates were able to calculate the ΔS of the reaction, though in some cases they failed to multiply by the number of moles.

Though the question asked for decomposition (in bold), most candidates ignored this and worked on the basis of a the ΔH of formation. However, many did write a sound explanation for that situation. On the other hand, in quite a number of cases, they did not state the sign of the ΔH (probably taking it for granted) nor explicitly relate ΔG and spontaneity, which left the examiner with no possibility of evaluating their reasoning.

Deduce the order of reaction with respect to hydrogen.

Deduce the rate expression for the reaction.

Calculate the value of the rate constant stating its units.

State two conditions necessary for a successful collision between reactants.

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

Calculate the entropy change of reaction, ΔS^⦵, in J K⁻¹ mol⁻¹.

Predict, giving a reason, how the value of the ΔS^⦵_reaction would be affected if ${\text{I}}_2$ (g) were used as a reactant.

Calculate the Gibbs free energy change, ΔG^⦵, in kJ mol⁻¹, for the reaction at 298 K. Use section 1 of the data booklet.

Calculate the equilibrium constant, K_c, for this reaction at 298 K. Use your answer to (d)(iii) and sections 1 and 2 of the data booklet.

(If you did not obtain an answer to (d)(iii) use a value of 2.0 kJ mol⁻¹, although this is not the correct answer).

first order ✔

Rate=k [H₂] [$\text{I}$₂]

$k=«\frac{1.2\times {10}^{-6} \mathrm{mol} {\mathrm{dm}}^{-3} {\mathrm{s}}^{-1}}{2.0\times {10}^{-3} \mathrm{mol} {\mathrm{dm}}^{-3}\times 3.0\times {10}^{-3} \mathrm{mol} {\mathrm{dm}}^{-3}}=»0.20$ ✔

mol^–1 dm³ s^–1 ✔

E ≥ E_a AND appropriate «collision» geometry/correct orientation ✔

$K_{\mathrm{c}}=\frac{{\left[\mathrm{HI}\right]}^2}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]}$ ✔

«Δ${S^{⦵}}_{\text{reaction}}$ = 2 × 206.6 – (130.6 + 116.1) =» 166.5 «J K^–1 mol^–1» ✔

Δ${S^{⦵}}_{\text{reaction}}$ lower/less positive AND same number of moles of gas

OR

Δ${S^{⦵}}_{\text{reaction}}$ lower/less positive AND a solid has less entropy than a gas ✔

«ΔG^⦵ = 53.0 kJ mol^–1 – (298K × 0.1665 kJ K^–1 mol^–1) =» 3.4 «kJ mol^–1» ✔

«ln K_c= – (3.4 × 10³ J mol^–1 /8.31 J K^–1 mol^–1 × 298 K)» = –1.37 ✔

«K_c =» 0.25 ✔

Award [2] for “0.45” for the use of 2.0 kJ mol^–1 for ΔG^⦵.

4(a)(i)-(iii): Deduction of rate orders and rate expression were very well done overall, with occasional errors in the units of the rate constant, but clearly among the best answered questions.

Generally well answered by all but very weak candidates. Some teachers thought this should be a 2-mark question but actually the marks were generally missed when students mentioned both required conditions but failed to refer the necessary energy to E_a.

One of the best answered questions.

ΔS was well calculated in general except for some inverted calculations or failure to consider the ratios of the reactants.

Some candidates confused the entropy change in this situation with absolute entropy of a solid and gas, or having realised that entropy would decrease lacked clarity in their explanations and lost the mark.

4(d)(ii)-(d)(iv): marks were lost due to inconsistency of units throughout, i.e., not because answers were given in different units to those required, but because candidates failed to convert all data to the same unit for calculations.

Classify substances B and D as reactant, product, catalyst, or intermediate, based on the proposed mechanism.

Deduce the rate expression.

Calculate the initial rate of reaction for experiment 2, if measured under the same conditions.

B: reactant ✔

D: intermediate ✔

rate = k[A][B] ✔

1.80 «mol dm^–3 s^–1» ✔

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.

State the type of reaction.

State the IUPAC name of the major product.

Outline why it is the major product.

Write an equation for the reaction of the major product with aqueous sodium hydroxide to produce a C₃H₈O compound, showing structural formulas.

Determine the rate expression from the results, explaining your method.

Deduce the type of mechanism for the reaction of this isomer of C₃H₇Cl with aqueous sodium hydroxide.

Sketch the mechanism using curly arrows to represent the movement of electrons.

Write an equation for the complete combustion of the compound C₃H₈O formed in (a)(iv).

Determine the enthalpy of combustion of this compound, in kJ mol⁻¹, using data from section 11 of the data booklet.

State the reagents for the conversion of the compound C₃H₈O formed in (a)(iv) into C₃H₆O.

Explain why the compound C₃H₈O, produced in (a)(iv), has a higher boiling point than compound C₃H₆O, produced in d(i).

Explain why the ¹H NMR spectrum of C₃H₆O, produced in (d)(i), shows only one signal.

Propene 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”.

2-chloropropane ✔

secondary carbocation/carbonium «ion» is more stable
OR
carbocation/carbonium «ion» stabilized by two/more alkyl groups ✔

CH₃CHClCH₃ (l) + OH⁻ (aq) → CH₃CH(OH)CH₃ (aq) + Cl⁻ (aq)
OR
CH₃CHClCH₃ (l) + NaOH (aq) → CH₃CH(OH)CH₃ (aq) + NaCl (aq) ✔

Rate = k [C₃H₇Cl] [OH⁻] ✔

«[OH⁻] held constant and» [C₃H₇Cl] triples AND rate triples «so first order wrt C₃H₇Cl» ✔

[C₃H₇Cl] doubles AND [OH⁻] doubles AND rate quadruples «so first order wrt OH⁻» ✔

SN2 ✔

NOTE: Accept ‘bimolecular nucleophilic substitution.’

curly arrow going from lone pair on O/negative charge on OH^– to C ✔

curly arrow showing C–Cl bond breaking ✔

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

formation of CH₃CH(OH)CH₃ AND Cl^– ✔

NOTE: Do not allow arrow originating on H in OH^–.

Allow curly arrow going from bond between C and Cl to Cl in either reactant or transition state.

Do not award M3 if OH–C bond is represented.

Accept formation of NaCl instead of Cl^–.

2C₃H₈O (l) + 9O₂ (g) → 6CO₂ (g) + 8H₂O (g)
OR
C₃H₈O (l) + 4.5O₂ (g) → 3CO₂ (g) + 4H₂O (g) ✔

bonds broken:
7(C–H) + C–O + O–H + 2(C–C) + 4.5(O=O)
OR
7(414 «kJ mol⁻¹») + 358 «kJ mol⁻¹» + 463 «kJ mol⁻¹» + 2(346 «kJ mol⁻¹») + 4.5(498 «kJ mol⁻¹») / 6652 «kJ» ✔

bonds formed:
6(C=O) + 8(O–H)
OR
6(804 «kJ mol⁻¹») + 8(463 «kJ mol⁻¹») / 8528 «kJ» ✔

«ΔH = bonds broken − bonds formed = 6652 – 8528 =» −1876 «kJ mol⁻¹» ✔

NOTE: Award [3] for correct final answer.

K₂Cr₂O₇/Cr₂O₇^2–/«potassium» dichromate «(VI)» AND acidified/H⁺
OR
«acidified potassium» manganate(VII) / «H⁺ and» KMnO₄ / «H⁺ and» MnO₄– ✔

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

C₃H₈O/propan-2-ol: hydrogen-bonding AND C₃H₆O/propanone: no hydrogen bonding/«only» dipole–dipole/dispersion forces ✔

hydrogen bonding stronger «than dipole–dipole» ✔

only one hydrogen environment
OR
methyl groups symmetrical «around carbonyl group» ✔

NOTE: Accept “all hydrogens belong to methyl groups «which are in identical positions»”.

✔

NOTE: Continuation bonds must be shown.

Methyl groups may be drawn on opposite sides of the chain or head to tail.

Ignore square brackets and “n”.

State the rate expression for the reaction.

The following mechanism is proposed for the reaction.

Identify the rate determining step giving your reason.

State one method that can be used to measure the rate for this reaction.

Sketch the relationship between the rate of reaction and the concentration of NO₂.

The Arrhenius equation, $k=A{e}^{-\frac{Ea}{RT}}$, gives the relationship between the rate constant and temperature.

State how temperature affects activation energy.

«rate =» k [NO₂]²

Accept rate = k [NO₂]²[CO]⁰.

[1 mark]

«step» I AND CO does not appear in the rate law expression
OR
«step» I AND only «2 molecules of» NO₂ appears in rate expression

Do not allow ECF from (i).

[1 mark]

«IR or UV-vis» spectroscopy
OR
colorimetry
OR
colour change «over time»

Accept GC/gas chromatography.

[1 mark]

Curve must go through origin.

[1 mark]

activation energy is independent of temperature

Accept “no relationship”.

[1 mark]

Determine the rate expression for the reaction.

Determine the value and unit of the rate constant using the rate expression in (a).

BrO₃^–: 1/first AND Br^–: 1/first AND H⁺: 2/second ✓

«Rate =» k[BrO₃⁻][Br⁻][H⁺]² ✓

M2: Square brackets required for the mark.

«k =$\frac{8.0\times {10}^{-4}}{0.10\times 0.10\times {\left(0.10\right)}^2}$=» 8.0 ✓

mol⁻³ dm⁹s⁻¹ ✓

Distinguish between a sigma and pi bond.

Identify the hybridization of carbon in ethane, ethene and ethyne.

State, giving a reason, if but-1-ene exhibits cis-trans isomerism.

State the type of reaction which occurs between but-1-ene and hydrogen iodide at room temperature.

Explain the mechanism of the reaction between but-1-ene with hydrogen iodide, using curly arrows to represent the movement of electron pairs.

State, giving a reason, if the product of this reaction exhibits stereoisomerism.

Deduce the rate expression for this reaction.

Deduce the units of the rate constant.

Determine the initial rate of reaction in experiment 4.

Deduce, with a reason, the mechanism of the reaction between 2-chloropentane and sodium hydroxide.

Discuss the reason benzene is more reactive with an electrophile than a nucleophile.

Sigma (σ) bond:

overlap «of atomic orbitals» along the axial / intermolecular axis / electron density is between nuclei
OR
head-on/end-to-end overlap «of atomic orbitals» ✔

Pi (π) bond:

overlap «of p-orbitals» above and below the internuclear axis/electron density above and below internuclear axis
OR
sideways overlap «of p-orbitals» ✔

Accept a suitable diagram.

All 3 required for mark.

no AND 2 groups on a carbon «in the double bond» are the same/hydrogen «atoms»

OR

no AND molecule produced by rearranging atoms bonded on a carbon «in the double bond» is the same as the original ✔

«electrophilic» addition ✔

Do not allow nucleophilic addition.

curly arrow going from C=C to H of HI AND curly arrow showing I leaving ✔

representation of carbocation ✔

curly arrow going from lone pair/negative charge on I^– to C⁺ ✔

2-iodobutane formed ✔

Penalize incorrect bond, e.g. –CH–H₃C or –CH₃C once only.

yes AND has a carbon attached to four different groups
OR
yes AND it contains a chiral carbon ✔

Accept yes AND mirror image of molecule different to original/non-superimposable on original.

«rate =» k[NaOH][C₅H₁₁Cl] ✔

mol^–1dm³s^–1 ✔

ALTERNATIVE 1:

«k = » 1.25 «mol^–1dm³s^–1» ✔

«rate = 1.25 mol^–1dm³s^–1 × 0.60 mol dm^–3 × 0.25 mol dm^–3»

1.9 x 10^–1 «mol dm^–3s^–1» ✔

ALTERNATIVE 2:

«[NaOH] exp. 4 is 3 × exp. 1»

«[C₅H₁₁Cl] exp. 4 is 2.5 × exp. 1»

«exp. 4 will be » 7.5× faster ✔

1.9 x 10^–1 «mol dm^–3s^–1» ✔

Award [2] for correct final answer.

S_N2 AND rate depends on both OH^– and 2-chloropentane ✔

Accept E2 AND rate depends on both OH^– and 2-chloropentane.

delocalized electrons/pi bonds «around the ring»
OR
molecule has a region of high electron density/negative charge ✔

electrophiles are attracted/positively charged AND nucleophiles repelled/negatively charged ✔

Do not accept just “nucleophiles less attracted” for M2.

Accept “benzene AND nucleophiles are both electron rich” for “repels nucleophiles”.

Nitrogen oxide is in equilibrium with dinitrogen dioxide.

2NO(g) $\rightleftharpoons$ N₂O₂(g)     ΔH^Θ < 0

Deduce, giving a reason, the effect of increasing the temperature on the concentration of N₂O₂.

A two-step mechanism is proposed for the formation of NO₂(g) from NO(g) that involves an exothermic equilibrium process.

First step: 2NO(g) $\rightleftharpoons$ N₂O₂(g)     fast

Second step: N₂O₂(g) + O₂ (g) → 2NO₂(g)     slow

Deduce the rate expression for the mechanism.

The rate constant for a reaction doubles when the temperature is increased from 25.0 °C to 35 °C.

Calculate the activation energy, E_a, in kJ mol⁻¹ for the reaction using section 1 and 2 of the data booklet.

[N₂O₂] decreases AND exothermic «thus reverse reaction favoured»

Accept “product” for [N₂O₂].

Do not accept just “reverse reaction favoured/shift to left” for “[N₂O₂] decreases”.

[1 mark]

ALTERNATIVE 1:

«from equilibrium, step 1»

$K_c=\frac{\text{[}{\text{N}}_2{\text{O}}_2\text{]}}{{\text{[NO]}}^2}$

OR

[N₂O₂] = K_c[NO]²

«from step 2, rate «= k₁[N₂O₂][O₂] = k₂K[NO]²[O₂]»

rate = k[NO]²[O₂]

ALTERNATIVE 2:

«from step 2» rate = k₂[N₂O₂][O₂]

«from step 1, rate₍₁₎ = k₁[NO]² = k_–1[N₂O₂], [N₂O₂] = $\frac{k_1}{k_{-1}}$ [NO]²»

«rate = $\frac{k_1}{k_{-1}}$ k₂[NO]²[O₂]»

rate = k[NO]²[O₂]

Award [2] for correct rate expression.

[2 marks]

«$\ln \frac{k_1}{k_2}=\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$»

T₂ = «273 + 35 =» 308 K AND T₁ = «273 + 25 =» 298 K

E_a = 52.9 «kJ mol^–1»

Award [2] for correct final answer.

[2 marks]

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»

Suggest how the extent of decomposition could be measured.

Plot the missing point on the graph and draw the best-fit line.

Outline why increasing the concentration of N₂O₅ increases the rate of reaction.

Write the rate expression for this reaction.

Calculate the value of the rate constant, k, giving its units.

use colorimeter
OR
change in colour
OR
change in volume
OR
change in pressure ✔

Accept suitable instruments, e.g. pressure probe/oxygen sensor.

point correct ✔

straight line passing close to all points AND through origin ✔

Accept free hand drawn line as long as attempt to be linear and meets criteria for M2.

greater frequency of collisions «as concentration increases»
OR
more collisions per unit time «as concentration increases» ✔

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

Do not accept just “more collisions”.

rate = k[N₂O₅] ✔

k = $\frac{∆\mathrm{rate}}{∆\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}$ ✔

«k = $\frac{0.75\times {10}^{-3}«\mathrm{mol} {\mathrm{dm}}^{-3} {\min }^{-1}»}{25\times {10}^{-3}«\mathrm{mol} {\mathrm{dm}}^{-3}»}$= » 0.030 «min^–1» ✔

min^–1 ✔

M1 can be awarded from correct M2 if not explicitly stated.

Accept k = gradient.

Accept values in the range 0.028–0.032.

Award [3] for correct final answer.

Draw a Lewis (electron dot) structure for ozone.

Discuss the relative length of the two O−O bonds in ozone.

Explain why there are frequencies of UV light that will dissociate O₃ but not O₂.

Explain, using equations, how the presence of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ results in a chain reaction that decreases the concentration of ozone in the stratosphere.

✔

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

Do not accept structures that represent 1.5 bonds.

both equal ✔

delocalization/resonance ✔

Accept bond length between 121 and 148 pm/ that of single O−O bond and double O=O bond for M1.

bond in O₃ is weaker
OR
O₃ bond order 1.5/< 2 ✔

Do not accept bond in O₃ is longer for M1.

lower frequency/longer wavelength «UV light» has enough energy to break the O–O bond in O₃ «but not that in O₂» ✔

Accept “lower frequency/longer wavelength «UV light» has lower energy”.

${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{(g) →∙CClF}}_{\text{2}}\text{(g) Cl}\bullet \text{(g)}$ ✔

${\text{Cl•(g)+O}}_{\text{3}}{\text{(g)→O}}_{\text{2}}\text{(g)+ClO•(g)}$
AND
${\text{ClO∙(g)+O}}_{\text{3}}{\text{(g)→2O}}_{\text{2}}\text{(g)+Cl}\bullet \text{(g)}$ ✔

Do not penalize missing radical.

Accept:for M2:
${\text{Cl∙(g) + O}}_{\text{3}}{\text{(g) → O}}_{\text{2}}\text{(g) + ClO}\bullet \text{(g)}$
AND
${\text{ClO∙(g) + O(g) → O}}_{\text{2}}\text{(g) + Cl}\bullet \text{(g)}$