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.

Predict the electron domain and molecular geometries around the oxygen atom of molecule A using VSEPR

State the type of hybridization shown by the central carbon atom in molecule B.

State the number of sigma ($\mathrm{\sigma }$) and pi ($\mathrm{\pi }$) bonds around the central carbon atom in molecule B.

The IR spectrum of one of the compounds is shown:

COBLENTZ SOCIETY. Collection © 2018 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.

Deduce, giving a reason, the compound producing this spectrum.

Compound A and B are isomers. Draw two other structural isomers with the formula ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$.

The equilibrium constant, $K_{\mathrm{c}}$, for the conversion of A to B is $1.0\times {10}^8$ in water at $298 \mathrm{K}$.

Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.

Calculate the standard Gibbs free energy change, $∆G^{⦵}$, in $\mathbf{kJ}\mathbf{ }{\mathbf{mol}}^{\mathbf{–}\mathbf{1}}$, for the reaction (A to B) at $298 \mathrm{K}$. Use sections 1 and 2 of the data booklet.

Propanone can be synthesized in two steps from propene. Suggest the synthetic route including all the necessary reactants and steps.

Propanone can be synthesized in two steps from propene.

Suggest why propanal is a minor product obtained from the synthetic route in (g)(i).

Electron domain geometry: tetrahedral ✔

Molecular geometry: bent/V-shaped ✔

${\mathrm{sp}}^2$ ✔

$\mathrm{\sigma }$-bonds: $3$
AND
$\mathrm{\pi }$-bonds: $1$ ✔

B AND $\mathrm{C}=\mathrm{O}$ absorption/$1750«{\mathrm{cm}}^{-1}»$ 
OR
B AND absence of $\mathrm{O}–\mathrm{H} /3200-3600«{\mathrm{cm}}^{-1} \mathrm{absorption}»$✔

Accept any value between $\mathit{1700}\mathit{-}\mathit{1750}\mathit{ }c{m}^{\mathit{-}\mathit{1}}$.

Accept any two $C_3{H}_{6}O$ isomers except for propanone and propen-2-ol:

✔✔

Penalize missing hydrogens in displayed structural formulas once only.

$\mathrm{B}$ AND $K_{\mathrm{c}}$ is greater than $1$/large ✔

$«\mathrm{\Delta }{G}^{\mathrm{\Theta }}=-RT\ln K=0.00831 \mathrm{kJ} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1} (298 \mathrm{K}) (\ln 1.0\times {10}^8)=»$
$-46«\mathrm{kJ} {\mathrm{mol}}^{-1}»$ ✔

${\mathrm{H}}_2\mathrm{O}$/water «and ${\mathrm{H}}^{+}$» ✔

${\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{OH}\right){\mathrm{CH}}_3$/propan-2-ol ✔

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$/«potassium» dichromate(VI) AND ${\mathrm{H}}^{+}$
OR
${\mathrm{KMnO}}_4$/«acidified potassium» manganate(VII) ✔

Accept $H_{\mathit{3}}{O}^{\mathit{+}}$.

primary carbocation «intermediate forms»
OR
minor product «of the water addition would be» propan-1-ol
OR
anti-Markovnikov addition of water ✔

primary alcohol/propan-1-ol oxidizes to an aldehyde/propanal ✔

The majority of students got at least one of electron domain geometry or molecular geometry correct.

The vast majority of students could identify the hybridization around a central carbon atom.

The vast majority of students could identify BOTH sigma and pi bonds in a molecule.

Many candidates identified B having C = O and a peak at 1750.

A surprising number of candidates drew propanone here as an option, either failing to read the question or perhaps finding the structural formulae provided difficult to understand.

Most candidates identified B, the product, as being in greater concentration at equilibrium however some lost the mark because they did not include a reason.

Most candidates could apply the formula for Gibbs free energy change, ΔG^Θ, correctly however some did not get the units correct.

The mean mark was ⅔ for the required synthetic route. Some candidates failed to identify water as a reagent in the hydration reaction, or note that dichromate ion oxidation requires acidic conditions. This was also the question with most No Response.

This question regarding the formation of a minor product was not well answered. Many candidates struggled to explain the formation of propan-1-ol and to then oxidize it to propanal.

Ethane, ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$, reacts with chlorine in sunlight. State the type of this reaction and the name of the mechanism by which it occurs.

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

Deduce the splitting patterns in the ¹H NMR spectrum of C₂H₅Cl.

Explain why tetramethylsilane (TMS) is often used as a reference standard in ¹H NMR.

One possible product, X, of the reaction of ethane with chlorine has the following composition by mass:

carbon: 24.27%, hydrogen: 4.08%, chlorine: 71.65%

Determine the empirical formula of the product.

The mass and ¹H NMR spectra of product X are shown below. Deduce, giving your reasons, its structural formula and hence the name of the compound.

When the product X is reacted with NaOH in a hot alcoholic solution, C₂H₃Cl is formed. State the role of the reactant NaOH other than as a nucleophile.

Chloroethene, ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$, can undergo polymerization. Draw a section of the polymer with three repeating units.

substitution AND «free-»radical

OR

substitution AND chain

Award [1] for “«free-»radical substitution” or “S_R” written anywhere in the answer.

[1 mark]

Two propagation steps:

${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}+\bullet \text{Cl}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet +\text{HCl}$

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet +\text{C}{\text{l}}_{\text{2}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl}+\bullet \text{Cl}$

One termination step:

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet +{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet \to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$

OR

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\bullet +\bullet \text{Cl}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl}$

OR

$\bullet \text{Cl}+\bullet \text{Cl}\to \text{C}{\text{l}}_{\text{2}}$

Accept radical without $\bullet$ if consistent throughout.

Allow ECF for incorrect radicals produced in propagation step for M3.

[3 marks]

triplet AND quartet

[1 mark]

chemical shift/signal outside range of common chemical shift/signal

strong signal/12/all H atoms in same environment
OR
singlet/no splitting of the signal

volatile/easily separated/easily removed
OR
inert/stabl

contains three common NMR nuclei/¹H and ¹³C and ²⁹Si

Do not accept chemical shift = 0.

[2 marks]

$\text{C}=\frac{24.27}{12.01}=2.021$ AND $\text{H}=\frac{4.08}{1.01}=4.04$ AND $\text{Cl}=\frac{71.65}{35.45}=2.021$

«hence» CH₂Cl

Accept $\frac{24.27}{12.01}$ : $\frac{4.08}{1.01}$ : $\frac{71.65}{35.45}.$

Do not accept C₂H₄Cl₂. 

Award [2] for correct final answer.

[2 marks]

molecular ion peak(s) «about» m/z 100 AND «so» C₂H₄Cl₂ «isotopes of Cl»

two signals «in ¹H NMR spectrum» AND «so» CH₃CHCl₂
OR
«signals in» 3:1 ratio «in ¹H NMR spectrum» AND «so» CH₃CHCl₂
OR
one doublet and one quartet «in ¹H NMR spectrum» AND «so» CH₃CHCl₂

1,1-dichloroethane

Accept “peaks” for “signals”.

Allow ECF for a correct name for M3 if an incorrect chlorohydrocarbon is identified.

[3 marks]

base
OR
proton acceptor

[1 mark]

Continuation bonds must be shown.

Ignore square brackets and “n”.

Accept  .

Accept other versions of the polymer, such as head to head and head to tail.

Accept condensed structure provided all C to C bonds are shown (as single).

[1 mark]

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.

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

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

Suggest why the enthalpy change of neutralization of CH₃COOH is less negative than that of HCl.

21.4 °C

Accept values in the range of 21.2 to 21.6 °C.
Accept two different values for the two solutions from within range.

HCl: 30.4 «°C»

Accept range 30.2 to 30.6 °C.

CH₃COOH: 29.0 «°C»

Accept range 28.8 to 29.2 °C.

CH₃COOH is weak acid/partially ionised

energy used to ionize weak acid «before reaction with NaOH can occur»

Deduce what information can be obtained from the ¹H NMR spectrum.

Identify the functional group that shows stretching at 1710 cm^–1 in the infrared spectrum of this compound using section 26 of the data booklet and the ¹H NMR.

Suggest the structural formula of this compound.

Bromine was added to hexane, hex-1-ene and benzene. Identify the compound(s) which will react with bromine in a well-lit laboratory.

Deduce the structural formula of the main organic product when hex-1-ene reacts with hydrogen bromide.

State the reagents and the name of the mechanism for the nitration of benzene.

Outline, in terms of the bonding present, why the reaction conditions of halogenation are different for alkanes and benzene.

Below are two isomers, A and B, with the molecular formula C₄H₉Br.

Explain the mechanism of the nucleophilic substitution reaction with NaOH(aq) for the isomer that reacts almost exclusively by an S_N2 mechanism using curly arrows to represent the movement of electron pairs.

Number of hydrogen environments: 3

Ratio of hydrogen environments: 2:3:9

Splitting patterns: «all» singlets

Accept any equivalent ratios such as 9:3:2.

Accept “no splitting”.

[3 marks]

carbonyl
OR
C=O

Accept “ketone” but not “aldehyde”.

[1 mark]

Accept (CH₃)₃CCH₂COCH₃.

Award [1] for any aldehyde or ketone with C₇H₁₄O structural formula.

[2 marks]

hexane AND hex-1-ene

Accept “benzene AND hexane AND hex-1-ene”.

[1 mark]

CH₃CH₂CH₂CH₂CHBrCH₃

Accept displayed formula but not molecular formula.

[1 mark]

Reagents: «concentrated» sulfuric acid AND «concentrated» nitric acid

Name of mechanism: electrophilic substitution

[2 marks]

benzene has «delocalized» $\pi$ bonds «that are susceptible to electrophile attack» AND alkanes do not

Do not accept “benzene has single and double bonds”.

[1 mark]

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

curly arrow showing Br leaving

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

Accept OH^– with or without the lone pair.

Do not allow curly arrows originating on H in OH^–.

Accept curly arrows in the transition state.

Do not penalize if HO and Br are not at 180°.

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

Award [2 max] if wrong isomer is used.

[3 marks]

State the class of compound to which ethene belongs.

State the molecular formula of the next member of the homologous series to which ethene belongs.

Justify why ethene has only a single signal in its ¹H NMR spectrum.

Deduce the chemical shift of this signal. Use section 27 of the data booklet.

Suggest two possible products of the incomplete combustion of ethene that would not be formed by complete combustion.

A white solid was formed when ethene was subjected to high pressure.

Deduce the type of reaction that occurred.

Sketch the mechanism for the reaction of propene with hydrogen bromide using curly arrows.

Explain why the major organic product is 2-bromopropane and not 1-bromopropane.

Explain why the major organic product is 2-bromopropane and not 1-bromopropane.

2-bromopropane can be converted directly to propan-2-ol. Identify the reagent required.

Propan-2-ol can also be formed in one step from a compound containing a carbonyl group.

State the name of this compound and the type of reaction that occurs.

alkene ✔

C₃H₆ ✔

Accept structural formula.

hydrogen atoms/protons in same chemical environment ✔

Accept “all H atoms/protons are equivalent”.
Accept “symmetrical”

4.5 to 6.0 «ppm» ✔

Accept a single value within this range.

carbon monoxide/CO AND carbon/C/soot ✔

«addition» polymerization ✔

curly arrow going from C=C to H of HBr AND curly arrow showing Br leaving ✔

representation of carbocation ✔

curly arrow going from lone pair/negative charge on Br⁻ to C⁺ ✔

Award [2 max] for mechanism producing 1-brompropane.

«2-bromopropane involves» formation of more stable «secondary» carbocation/carbonium ion/intermediate
OR
1-bromopropane involves formation of less stable «primary» carbocation/carbonium ion/intermediate ✔

«increased» positive inductive/electron-releasing effect of extra–R group/–CH₃/methyl «increases stability of secondary carbocation» ✔

Award [1] for “more stable due to positive inductive effect”.

Do not award marks for quoting Markovnikov’s rule without any explanation. 

«2-bromopropane involves» formation of more stable «secondary» carbocation/carbonium ion/intermediate
OR
1-bromopropane involves formation of less stable «primary» carbocation/carbonium ion/intermediate ✔

«increased» positive inductive/electron-releasing effect of extra–R group/–CH₃/methyl «increases stability of secondary carbocation» ✔

Award [1] for “more stable due to positive inductive effect”.
Do not award marks for quoting Markovnikov’s rule without any explanation.

sodium hydroxide/NaOH/potassium hydroxide/KOH ✔

Accept «aqueous» hydroxide ions/OH⁻

Name of carbonyl compound:
propanone ✔

Type of reaction:
reduction ✔

Accept other valid alternatives, such as “2-propyl ethanoate” for M1 and “hydrolysis” for M2.

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]

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]

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

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

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

State the hybridization of the carbon I and II atoms in but-2-ene.

Draw diagrams to show how sigma (σ) and pi (π) bonds are formed between atoms.

Sketch the mechanism for the reaction of 2-methylbut-2-ene with hydrogen bromide using curly arrows.

Explain why the major organic product is 2-bromo-2-methylbutane and not 2-bromo-3-methylbutane.

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

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

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

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

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

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

Draw the stereoisomers of butan-2-ol using wedge-dash type representations.

Outline how two enantiomers can be distinguished using a polarimeter.

Penalize missing hydrogens in displayed structural formulas once only.

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

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

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

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

Award [3] for correct final answer.

Award [2 max] for «+» 2530 «kJ».

Sigma (σ):

Accept any diagram showing end to end/direct overlap of atomic/hybridized orbitals and electron density concentrated between nuclei.

Pi (π):

Accept any diagram showing sideways overlap of unhybridized p/atomic orbitals and electron density above and below plane of bond axis.

Alternative 1

Penalize incorrect bond e.g., -CH-H₃C or –CH₃C only once in the paper.

Alternative 2

curly arrow going from C=C to H of HBr AND curly arrow showing Br leaving ✓

representation of carbocation ✓

curly arrow going from lone pair/negative charge on Br⁻ to C⁺ ✓

«2-bromo-2-methylbutane involves» formation of more stable «tertiary» carbocation/intermediate
OR
«2-bromo-3-methylbutane involves» formation of less stable «secondary» carbocation/intermediate ✓

«intermediate» more stable due to «increased positive» inductive/electron-releasing effect of extra –R/alkyl group/–CH₃/methyl ✓

Do not award marks for quoting Markovnikov’s rule without any explanation.

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

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

Do not penalize missing charge on the fragments.

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

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

C=O ✓

Accept carbonyl/C=C.

Information deduced from ¹H NMR:

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

Compound:

propanone/CH₃COCH₃ ✓

Accept “one type of hydrogen”.

Accept .

enantiomers rotate «plane of» plane-polarized light ✓

equal degrees/angles/amounts AND opposite directions/rotation ✓

Accept “optical isomers” for “enantiomers”.

Determine the limiting reactant showing your working.

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

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

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

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

The only significant uncertainty is in the temperature measurement.

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

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

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

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

A student electrolyzed aqueous iron(II) sulfate, FeSO₄ (aq), using platinum electrodes. State half-equations for the reactions at the electrodes, using section 24 of the data booklet.

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

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

CuSO₄ is the limiting reactant ✔

Do not award M2 if mole calculation is not shown.

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

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

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

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

Accept answers in the range 85–86 %.

Award [2] for correct final answer.

ALTERNATIVE 1:

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

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

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

ALTERNATIVE 2:

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

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

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

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

Award [2] for correct final answer.

density «of solution» is 1.00 g cm⁻³

OR

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

OR

reaction goes to completion

OR

iron/CuSO₄ does not react with other substances ✔

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

Do not accept “heat loss”.

ALTERNATIVE 1:

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

«0.03 × 160 kJ» = «±» 5 «kJ» ✔

ALTERNATIVE 2:

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

«0.03 × 180 kJ» = «±» 5 «kJ» ✔

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

Award [2] for correct final answer.

initial concentration is zero AND concentration increases with time ✔

decreasing gradient as reaction proceeds ✔

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

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

Accept suitable diagram.

piece has smaller surface area ✔

lower frequency of collisions

OR

fewer collisions per second/unit time ✔

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

Do not accept just “fewer collisions”.

Anode (positive electrode):

2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e^– ✔

Cathode (negative electrode):

2H₂O (l) + 2e^– → H₂ (g) + 2OH^– (aq)
OR
2H⁺ (aq) + 2e^– → H₂ (g) ✔

Accept “4OH^– (aq) → O₂ (g) + 2H₂O (l)  + 4e^–” OR “Fe²⁺ (aq) → Fe³⁺ (aq) + e^–” for M1.

Accept “Fe²⁺ (aq) + 2e^– → Fe (s)” OR “SO₄^2- (aq) 4H⁺ (aq) + 2e^– → 2H₂SO₃(aq) + H₂O (l)”
for M2.

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.

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 the equation for the reaction of each substance with water.

Draw a diagram showing the delocalization of electrons in the conjugate base of butanoic acid.

Deduce the average oxidation state of carbon in butanoic acid.

A 0.250 mol dm⁻³ aqueous solution of butanoic acid has a concentration of hydrogen ions, [H⁺], of 0.00192 mol dm⁻³. Calculate the concentration of hydroxide ions, [OH⁻], in the solution at 298 K.

Determine the pH of a 0.250 mol dm⁻³ aqueous solution of ethylamine at 298 K, using section 21 of the data booklet.

Sketch the pH curve for the titration of 25.0 cm³ of ethylamine aqueous solution with 50.0 cm³ of butanoic acid aqueous solution of equal concentration. No calculations are required.

Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room temperature.

State a suitable reagent for the reduction of butanoic acid.

Deduce the product of the complete reduction reaction in (e)(i).

Butanoic acid:
CH₃CH₂CH₂COOH (aq) + H₂O (l) $\rightleftharpoons$ CH₃CH₂CH₂COO⁻ (aq) + H₃O⁺ (aq) ✔

Ethylamine:
CH₃CH₂NH₂ (aq) + H₂O (l) $\rightleftharpoons$ CH₃CH₂NH₃⁺ (aq) + OH⁻ (aq) ✔

Diagram showing:
dotted line along O–C–O AND negative charge

Accept correct diagrams with pi clouds.

–1 ✔

«$\frac{1.00\times {10}^{-14}\text{mo}{\text{l}}^2\text{d}{\text{m}}^{-6}}{0.00192\text{mol}\text{d}{\text{m}}^{-3}}$» = 5.21 × 10^–12 «mol dm^–3» ✔

«pK_b = 3.35, K_b = 10^–3.35 = 4.5 × 10^–4»

«C₂H₅NH₂ + H₂O $\rightleftharpoons$ C₂H₅NH₃⁺ + OH^–»

K_b = $\frac{\left[\text{O}{\text{H}}^{-}-\right]\left[\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}\right]}{\left[\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{N}{\text{H}}_{\text{2}}\right]}$

OR

«K_b =» 4.5 × 10^–4 = $\frac{\left[\text{O}{\text{H}}^{-}\right]\left[\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}\right]}{0.250}$

OR

«K_b =» 4.5 × 10^–4 = $\frac{x^2}{0.250}$ ✔

« x = [OH^–] =» 0.011 «mol dm^–3» ✔

«pH = –log$\frac{1.00\times {10}^{-14}}{0.011}=$» 12.04

OR

«pH = 14.00 – (–log 0.011)=» 12.04 ✔

Award [3] for correct final answer.

decreasing pH curve ✔

pH close to 7 (6–8) at volume of 25 cm³ butanoic acid ✔

weak acid/base shape with no flat «strong acid/base» parts on the curve ✔

Any two of:
butanoic acid forms more/stronger hydrogen bonds ✔
butanoic acid forms stronger London/dispersion forces ✔
butanoic acid forms stronger dipole–dipole interaction/force ✔

Accept “butanoic acid forms dimers”

Accept “butanoic acid has larger M_r/hydrocarbon chain/number of electrons” for M2.

Accept “butanoic acid has larger «permanent» dipole/more polar” for M3.

lithium aluminium hydride/LiAlH₄ ✔

butan-1-ol/1-butanol/CH₃CH₂CH₂CH₂OH ✔

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.

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.

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”.

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 empirical formula of the compound, showing your working.

The infrared spectrum of the compound is shown. Deduce the functional group of the compound.

The mass spectrum of the compound is shown. Deduce the relative molecular mass of the compound.

The compound could not be oxidized using acidifi ed potassium dichromate(VI).

Deduce the structural formula of the compound.

«in 100 g sample» $\frac{62.02\text{g}}{12.01\text{g}\text{mo}{\text{l}}^{-1}}$ AND $\frac{10.43\text{g}}{1.01\text{g}\text{mo}{\text{l}}^{-1}}$

OR

«in 100 g sample» 5.164 mol C AND 10.33 mol H ✔

27.55 %

OR

1.722 mol O ✔

«empirical formula» C₃H₆O ✔

«absorption at wavenumber 1700−1750 cm^–1» C=O/carbonyl ✔

Do not accept “ketone” or “aldehyde”.

«m/z =» 58 ✔