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

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

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

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

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

Accept net ionic equation.

H₂PO₄⁻ (aq) + H⁺ (aq) → H₃PO₄ (aq) ✔

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

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

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

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

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

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

Award [2] for correct final answer.

«OH⁻ is a» proton acceptor ✔

After heating 3.760 g of a silver oxide 3.275 g of silver remained. Determine the empirical formula of Ag_xO_y.

Suggest why the final mass of solid obtained by heating 3.760 g of Ag_xO_y may be greater than 3.275 g giving one design improvement for your proposed suggestion. Ignore any possible errors in the weighing procedure.

Naturally occurring silver is composed of two stable isotopes, ¹⁰⁷Ag and ¹⁰⁹Ag.

The relative atomic mass of silver is 107.87. Show that isotope ¹⁰⁷Ag is more abundant.

Some oxides of period 3, such as Na₂O and P₄O₁₀, react with water. A spatula measure of each oxide was added to a separate 100 cm³ flask containing distilled water and a few drops of bromothymol blue indicator.

The indicator is listed in section 22 of the data booklet.

Deduce the colour of the resulting solution and the chemical formula of the product formed after reaction with water for each oxide.

Explain the electrical conductivity of molten Na₂O and P₄O₁₀.

Outline the model of electron configuration deduced from the hydrogen line emission spectrum (Bohr’s model).

n(Ag) = «$\frac{3.275\text{ g}}{107.87\text{ g}\text{mol}}=$» 0.03036 «mol»

AND

n(O) = «$\frac{3.760\text{ g}-3.275\text{ g}}{16.00\text{ g}\text{mo}{\text{l}}^{-1}}=\frac{0.485}{16.00}=$» 0.03031 «mol»

«$\frac{0.03036}{0.03031}\approx 1$ / ratio of Ag to O approximately 1 : 1, so»

AgO

Accept other valid methods for M1.

Award [1 max] for correct empirical formula if method not shown.

[2 marks]

temperature too low
OR
heating time too short
OR
oxide not decomposed completely

heat sample to constant mass «for three or more trials»

Accept “not heated strongly enough”.

If M1 as per markscheme, M2 can only be awarded for constant mass technique.

Accept "soot deposition" (M1) and any suitable way to reduce it (for M2).

Accept "absorbs moisture from atmosphere" (M1) and "cool in dessicator" (M2).

Award [1 max] for reference to impurity AND design improvement.

[2 marks]

A_r closer to 107/less than 108 «so more ¹⁰⁷Ag»
OR
A_r less than the average of (107 + 109) «so more ¹⁰⁷Ag»

Accept calculations that gives greater than 50% ¹⁰⁷Ag.

[1 mark]

Do not accept name for the products.

Accept “Na⁺ + OH^–” for NaOH.

Ignore coefficients in front of formula.

[3 marks]

«molten» Na₂O has mobile ions/charged particles AND conducts electricity

«molten» P₄O₁₀ does not have mobile ions/charged particles AND does not conduct electricity/is poor conductor of electricity

Do not award marks without concept of mobile charges being present.

Award [1 max] if type of bonding or electrical conductivity correctly identified in each compound.

Do not accept answers based on electrons.

Award [1 max] if reference made to solution.

[2 marks]

electrons in discrete/specific/certain/different shells/energy levels

energy levels converge/get closer together at higher energies
OR
energy levels converge with distance from the nucleus

Accept appropriate diagram for M1, M2 or both.

Do not give marks for answers that refer to the lines in the spectrum.

[2 marks] 

Identify the amphiprotic species.

Identify one conjugate acid-base pair in the reaction.

Calculate the amount, in mol, of NaOH(aq) used.

Calculate the molar mass of the acid.

Calculate [H⁺] in the NaOH solution.

water/H₂O

Accept “hydroxide ion/OH^–”.

[1 mark]

[1 mark]

«0.100 mol$$dm^–3 x 0.0250 dm³» = 0.00250 «mol»

[1 mark]

«M = $\frac{0.510\text{ g}}{0.00250\text{ mol}}$ =» 204 «g$$mol^–1»

[1 mark]

«1.00 x 10^–14 = [H⁺] x 0.100»

1.00 x 10^–13 «mol$$dm^–3»

[1 mark]

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

Ammonia reacts reversibly with water.

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

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

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

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

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

N₂H₄(l) → N₂(g) + 2H₂(g)

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

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

N₂H₄(g) → N₂(g) + 2H₂(g)

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

N₂H₄(l) → N₂H₄(g)

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

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

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

107^°

Accept 100° to < 109.5°.

Literature value = 105.8°

[1 mark]

removes/reacts with OH⁻

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

Accept ionic equation for M1.

[2 marks]

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

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

Equilibrium sign must be present.

[1 mark]

bubbles
OR
gas
OR
magnesium disappears

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

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

Accept "smell of ammonia".

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

Equation must be ionic.

[2 mark]

no oxygen required

[1 mark]

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

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

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

Award [3] for correct final answer.

Award [2 max] for +95 «kJ».

[3 marks]

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

«ΔH_vap =» +44 «kJ$$mol⁻¹»

Award [2] for correct final answer.

Award [1 max] for −44 «kJ$$mol⁻¹».

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

Award [1 max] for −34 «kJ$$mol⁻¹».

[2 marks]

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

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

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

Award [3] for correct final answer.

[3 marks]

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

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

Award [1] for correct final answer.

[1 mark]

The following reaction was allowed to reach equilibrium at 761 K.

H₂ (g) + I₂ (g) $\rightleftharpoons$ 2HI (g)               ΔH^θ < 0

Outline the effect, if any, of each of the following changes on the position of equilibrium, giving a reason in each case.

Identify two different amphiprotic species in the above reactions.

State what is meant by the term conjugate base.

State the conjugate base of the hydroxide ion, OH^–.

A student working in the laboratory classified HNO₃, H₂SO₄, H₃PO₄ and HClO₄ as acids based on their pH. He hypothesized that “all acids contain oxygen and hydrogen”.

Evaluate his hypothesis.

Award [1 max] if both effects are correct.

Reason for increasing volume:

Accept “concentration of all reagents reduced by an equal amount so cancels out in K_c expression”.

Accept “affects both forward and backward rates equally”.

HCO₃^– AND H₂O

species that has one less proton/H⁺ ion «than its conjugate acid»

OR

species that forms its conjugate acid by accepting a proton

OR

species that is formed when an acid donates a proton

Do not accept “differs by one proton/H⁺ from conjugate acid”.

oxide ion/O^2–

insufficient data to make generalization

OR

need to consider a «much» larger number of acids

OR

hypothesis will continue to be tested with new acids to see if it can stand the test of time

«hypothesis is false as» other acids/HCl/HBr/HCN/transition metal ion/BF₃ do not contain oxygen

OR

other acids/HCl/HBr/HCN/transition metal ion/BF₃ falsify hypothesis

correct inductive reasoning «based on limited sample»

«hypothesis not valid as» it contradicts current/accepted theories/Brønsted-Lowry/Lewis theory

[Max 2 Marks]

Outline, giving a reason, the effect of a catalyst on a reaction.

On the axes, sketch Maxwell–Boltzmann energy distribution curves for the reacting species at two temperatures T₁ and T₂, where T₂ > T₁.

Explain the effect of increasing temperature on the yield of SO₃.

State the product formed from the reaction of SO₃ with water.

State the meaning of a strong Brønsted–Lowry acid.

Draw the Lewis structure of NO₃⁻.

Explain the electron domain geometry of NO₃⁻.

increases rate AND lower E_a ✔

provides alternative pathway «with lower E_a»
OR
more/larger fraction of molecules have the «lower» E_a ✔

Accept description of how catalyst lowers E_a for M2 (e.g. “reactants adsorb on surface «of catalyst»”, “reactant bonds weaken «when adsorbed»”, “helps favorable orientation of molecules”).

both axes correctly labelled ✔

peak of T₂ curve lower AND to the right of T₁ curve ✔

lines begin at origin AND correct shape of curves AND T₂ must finish above T₁ ✔

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

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

decrease AND equilibrium shifts left / favours reverse reaction ✔

«forward reaction is» exothermic / ΔH is negative ✔

sulfuric acid/H₂SO₄ ✔

Accept “disulfuric acid/H₂S₂O₇”.

fully ionizes/dissociates ✔

proton/H⁺ «donor » ✔

Do not accept the delocalised structure.

Accept any combination of dots, crosses and lines.

Coordinate/dative bond may be represented by an arrow.

three electron domains repel

OR

three electron domains as far away as possible ✔

trigonal planar

OR

«all» angles are 120° ✔

A generally well-answered question. Most candidates explained the effect of a catalyst on a reaction correctly. A small proportion of candidates thought the catalyst increased the frequency of collisions. Some candidates focussed on the effect of the catalyst on an equilibrium since the equation above the question was that of a reversible reaction. These candidates usually still managed to gain at least the first marking point by stating that both forward and reverse reaction rates were increased due to the lower activation energy. Most candidates mentioned the alternative pathway for the second mark, and some gave a good discussion about the increase in the number of molecules or collisions with E≥E_a. A few candidates lost one of the marks for not explicitly stating the effect of a catalyst (that it increases the rate of the reaction).

The average mark scored for the Maxwell-Boltzmann distribution curves sketch was 1.5 out of 3 marks and the question had a strong correlation with the candidates who did well overall. The majority of candidates were familiar with the shapes of the curves. The most commonly lost mark was missing or incorrect labels on the axes. Sometimes candidates added the labels but did not specify “kinetic” energy for the x-axis. As for the curves, some candidates reversed the labels T₁ and T₂, some made the two curves meet at high energy or even cross, and some did not have the correct relationship between the peaks of T₁ and T₂.

Another question that showed a strong correlation with the candidates who did well overall. The average mark was 1 out of 2 marks. Many candidates explained the effect of an increase in temperature on the yield of SO₃ correctly and thoroughly. One of the common mistakes was to miss the fact that it was an equilibrium and reason that yield would not change due to an increase in the rate of reaction. Unfortunately, a number of candidates also deduced that yield would increase due to the increase in rate. Other candidates recognized that it was an exothermic reaction but deduced the equilibrium would shift to the right giving a higher yield of SO₃.

A very well answered question. 70% of the candidates stated H2SO4 as the product from the reaction of SO₃ with water.

While a straightforward question, many candidates only answered part of the question - either focussing on the “strong” or on the “Brønsted-Lowry acid”. The average mark on this question was 1.2 out of 2 marks.

Only 20% of the candidates scored the mark for the Lewis structure of NO₃^-. Mistakes included: missing charge, missing lone pairs, 3 single bonds, 2 double bonds.

The majority of candidates deduced the correct electron domain geometry scoring the first mark including cases of ECF. Only a small number of candidates satisfied the requirements of the markscheme for the explanation.

Distinguish between a weak acid and a strong acid.

Weak acid:

Strong acid:

Suggest why it is more convenient to express acidity using the pH scale instead of using the concentration of hydrogen ions.

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

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

The mean value of the titre was 14.0 cm³.

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

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

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

The Lewis (electron dot) structure of the ethanedioate ion is shown below.

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

Weak acid: partially dissociated/ionized «in solution/water»
AND
Strong acid: «assumed to be almost» completely/100% dissociated/ionized «in solution/water»

Accept answers relating to pH, conductivity, reactivity if solutions of equal concentrations stated.

«log scale» reduces a wide range of numbers to a small range
OR
simple/easy to use
OR
converts exponential expressions into linear scale/simple numbers

Do not accept “easy for calculations”

i

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

ii

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

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

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

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

ALTERNATIVE 2:
«mol of pure hydrated ethanedioic acid in 1 dm³ solution = 7.00 × 10^-4 × $\frac{1000}{25}$ =» 2.80×10^-2 «mol»
«mass of pure hydrated ethanedioic acid in sample = 2.80 × 10^-2 mol × 126.08 g mol^-1 =» 3.53 «g»
«% purity = $\frac{3.53\mathrm{g}}{5.00\mathrm{g}}$ × 100 =» 70.6 «%»

ALTERNATIVE 3:
mol of hydrated ethanedioic acid (assuming sample to be pure) = $\frac{5.00\mathrm{g}}{126.08\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}}$ = 0.03966 «mol»
actual amount of hydrated ethanedioic acid = «7.00 × 10^-4 × $\frac{1000}{25}$ =» 2.80 × 10^-2 «mol»

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

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

electrons delocalized «across the O–C–O system»
OR
resonance occurs
Accept delocalized π-bond(s).

122 «pm» < C–O < 143 «pm»

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

State the full electron configuration of the chlorine atom.

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

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

The mass spectrum of chlorine is shown.

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

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

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

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

Determine the limiting reactant, showing your calculations.

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

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

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

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

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

State the formula of the conjugate base of hypochlorous acid.

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

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

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

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

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

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

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

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

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

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

Do not accept condensed electron configuration.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Accept methods employing $pV=nRT$.

${\mathrm{MnO}}_2: +4$ ✔

${\mathrm{MnCl}}_2: +2$ ✔

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

partially dissociates/ionizes «in water» ✔

${\mathrm{ClO}}^{-}$ ✔

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

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

Do not accept electrophilic or nucleophilic substitution.

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

Accept “chloroethane AND polar”.

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

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

hydroxide «ion»/${\mathrm{OH}}^{-}$ ✔

Do not accept $NaOH$.

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

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

$2$ «signals» ✔

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

Accept any values in the ranges.

Award [1 max] for two incorrect chemical shifts.

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

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

Award [2] for correct final answer.

Any of:

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

Do not accept just “collaboration”.

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

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

Most candidates wrote the electron configuration of chlorine correctly.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Predict, giving a reason, a difference between the reactions of the same concentrations of hydrochloric acid and ethanoic acid with samples of calcium carbonate.

Dissolved carbon dioxide causes unpolluted rain to have a pH of approximately 5, but other dissolved gases can result in a much lower pH. State one environmental effect of acid rain.

slower rate with ethanoic acid

OR

smaller temperature rise with ethanoic acid

[H⁺] lower

OR

ethanoic acid is partially dissociated

OR

ethanoic acid is weak

Accept experimental observations such as “slower bubbling” or “feels less warm”.

[2 marks]

Any one of:

corrosion of materials/metals/carbonate materials

destruction of plant/aquatic life

«indirect» effect on human health

Accept “lowering pH of oceans/lakes/waterways”.

[1 mark]

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

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

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

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

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

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

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

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

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

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

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

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

Accept all 2p electrons pointing downwards.

Accept half arrows instead of full arrows.

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

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

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

«value of » K_c unchanged ✔

Accept “K_c only affected by changes in temperature”.

same/unaffected/unchanged ✔

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

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

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

Award [2 max] for (+)93 «kJ»

–92.4 «kJ» ✔

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

Accept it relies on average values not specific to NH₃

conjugate «acid and base» ✔

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

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

Award [2] for correct final answer.

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

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

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

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

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

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

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

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

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

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

Outline why the rate of the reaction decreases with time.

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

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

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

Calculate the activation energy for the reverse reaction.

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

concentration of acid decreases
OR
surface area of magnesium decreases

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

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

Do not accept “concentrations of reagents decrease”.

[1 mark]

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

[1 mark]

«E_a(rev) = 226 + 132 =» 358 «kJ»

Do not accept –358.

[1 mark]

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

Accept ionised forms of the acids.

[1 mark]

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

Predict the shape of the hydrogen sulfide molecule.

State the formula of its conjugate base.

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

Calculate the hydroxide ion concentration in saturated aqueous hydrogen sulfide.

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

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

 OR  ✔

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

bent/non-linear/angular/v-shaped✔

HS⁻ ✔

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

10⁻¹⁰ «mol dm⁻³» ✔

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

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

Assumption:
«both» gases behave as ideal gases ✔

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

Calcium carbonate is heated to produce calcium oxide, CaO.

CaCO₃(s) → CaO (s) + CO₂(g)

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

Thermodynamic data for the decomposition of calcium carbonate is given.

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

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

Outline why a catalyst has such an effect.

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

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

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

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

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

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

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

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

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

Award [2] for correct final answer.

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

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

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

«ΔH = + » 179 «kJ» ✓

Award [2] for correct final answer.

Award [1 max] for −179 kJ.

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

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

lower activation energy curve between same reactant and product levels ✓

Accept curve with or without an intermediate.

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

provides an alternative «reaction» pathway/mechanism ✓

Do not accept “lower activation energy” only.

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

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

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

Award [2] for correct final answer.

Award [1 max] for 0.058 «dm³».

Alternative 1:

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

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

Alternative 2:

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

«pOH = −log (0.0466) = 1.332»

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

Award [2] for correct final answer.

Award [1 max] for pH =12.367.

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

«CO₂ is the limiting reactant»

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

Only award ECF for M2 if limiting reagent is used.

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

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

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

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

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

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

Sketch a Maxwell–Boltzmann distribution curve for a chemical reaction showing the activation energies with and without a catalyst.

Sketch a curve on the graph to show the volume of gas produced over time if the same mass of crushed calcium carbonate is used instead of lumps. All other conditions remain constant.

State and explain the effect on the rate of reaction if ethanoic acid of the same concentration is used in place of hydrochloric acid.

Outline why pH is more widely used than [H⁺] for measuring relative acidity.

Outline why H₃PO₄/HPO₄²⁻ is not a conjugate acid-base pair.

both axes correctly labelled

correct shape of curve starting at origin

E_a(catalyst) < E_{a(without catalyst)} on x-axis

M1:

Accept “speed” for x-axis label.

Accept “number of particles”, “N”, “frequency” or “probability «density»” for y-axis label.

Do not accept “potential energy” for x-axis label.

M2:

Do not accept a curve that touches the x-axis at high energy.

Do not award M2 if two curves are drawn.

M3:

Ignore any shading under the curve.

[3 marks]

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

[1 mark]

rate decreases

OR

slower reaction

«ethanoic acid» partially dissociated/ionized «in solution/water»

OR

lower [H⁺]

Accept “weak acid” or “higher pH”.

[2 marks]

«pH» converts «wide range of [H⁺]» into simple «log» scale/numbers

OR

«pH» avoids need for exponential/scientific notation

OR

«pH» converts small numbers into values «typically» between 0/1 and 14

OR

«pH» allows easy comparison of values of [H⁺]

Accept “uses values between 0/1 and 14”.

Do not accept “easier to use”.

Do not accept “easier for calculations”.

[1 mark]

«species» do not differ by a «single» proton/H⁺

OR

conjugate base of H₃PO₄ is H₂PO₄^– «not HPO₄^2–»

OR

conjugate acid of HPO₄^2– is H₂PO₄^– «not H₃PO₄»

Do not accept “hydrogen/H” for “H⁺/proton”.

[1 mark]

Identify a conjugate acid–base pair in the equation.

The value of the equilibrium constant for the first dissociation at 298 K is 5.01 × 10⁻⁴.

State, giving a reason, the strength of citric acid.

The dissociation of citric acid is an endothermic process. State the effect on the hydrogen ion concentration, [H⁺], and on the equilibrium constant, of increasing the temperature.

Outline one laboratory methods of distinguishing between solutions of citric acid and hydrochloric acid of equal concentration, stating the expected observations.

C₆H₈O₇ AND C₆H₇O₇⁻
OR
H₂O AND H₃O⁺ ✔

weak acid AND partially dissociated
OR
weak acid AND equilibrium lies to left
OR
weak acid AND K_c/K_a<1 ✔

Any one of:
«electrical» conductivity AND HCl greater ✔
pH AND citric acid higher ✔
titrate with strong base AND pH at equivalence higher for citric acid ✔
add reactive metal/carbonate/hydrogen carbonate AND stronger effervescence/faster reaction with HCl ✔
titration AND volume of alkali for complete neutralisation greater for citric acid ✔
titrate with strong base AND more than one equivalence point for complete neutralisation of citric acid ✔
titrate with strong base AND buffer zone with citric acid ✔

NOTE: Accept “add universal indicator AND HCl more red/pink” for M2.

Accept any acid reaction AND HCl greater rise in temperature.

Accept specific examples throughout.

Do not accept “smell” or “taste”.

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

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

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

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

 [✔]

Note: Accept Kekulé structures.

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

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

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

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

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

Note: Award [2] for correct final answer.

Accept other methods.

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

correct products  [✔]

correct balancing [✔]

«intermolecular» hydrogen bonding  [✔]

Note: Accept diagram showing hydrogen bonding.

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

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

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

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

Write a balanced equation for the reaction that occurs.

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

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

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

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

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

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

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

Calculate coefficients that balance the equation for the following reaction.

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

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

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

State the number of subatomic particles in this ion.

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

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

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

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

2 Mg(s) + O₂(g) → 2 MgO(s) ✔

Do not accept equilibrium arrows. Ignore state symbols

s ✔

Do not allow group 2

aluminium/Al ✔

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

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

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

Award [2] for correct final answer

Accept 0.021%

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

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

Award «0.2614 mol x 40.31 g mol^–1»

Accept alternative methods to arrive at the correct answer.

Accept final answers in the range 91-92%

[2] for correct final answer.

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

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

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

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

Accept “crucible weighed before fully cooled”.

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

Accept “non-stoichiometric compounds formed”.

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

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

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

Mg₃N₂: -3
AND
NH₃: -3 ✔

Do not accept 3 or 3-

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

Redox:
no AND no oxidation states change ✔

Accept “yes AND proton transfer takes place”

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

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

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

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

isotope«s» ✔

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

Any two of:

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

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

atom «discovered» to have structure ✔

fission
OR
atoms can be split ✔

Accept atoms can undergo fusion «to produce heavier atoms»

Accept specific examples of particles.

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

Award [1] for all bonding types correct.

Award [1] for each correct description.

Apply ECF for M2 only once.

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

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

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

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

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

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

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

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

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

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

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

Most candidates could answer the question about subatomic particles correctly.

Identification of isotopes was answered correctly by most students.

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

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

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

Distinguish between a weak and strong acid.

Weak acid: 

Strong acid:

The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.

State the formula of its conjugate base.

When a bottle of carbonated water is opened, these equilibria are disturbed.

State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).

Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)

100.0 cm³ of soda water contains 3.0 × 10⁻² g NaHCO₃.

Calculate the concentration of NaHCO₃ in mol dm⁻³.

Identify the type of bonding in sodium hydrogencarbonate.

Between sodium and hydrogencarbonate:

Between hydrogen and oxygen in hydrogencarbonate:

Weak acid: partially dissociated/ionized «in solution/water»
AND
Strong acid: «assumed to be almost» completely/100 % dissociated/ionized «in solution/water»  [✔]

CO₃^2–  [✔]

shifts to left/reactants AND to increase amount/number of moles/molecules of gas/CO₂ (g)  [✔]

«additional HCO₃^–» shifts position of equilibrium to left  [✔]

pH increases  [✔]

Note:  Do not award M2 without any justification in terms of equilibrium shift in M1.

«molar mass of NaHCO₃ =» 84.01 «g mol^–1»  [✔]

«concentration = $\frac{3.0\times {10}^{-2}\text{ g}}{84.01\text{ g mo}{\text{l}}^{-1}}\times \frac{1}{0.100\text{ d}{\text{m}}^3}=$» 3.6 × 10^–3 «mol dm^–3»  [✔]

Note: Award [2] for correct final answer.

Between sodium and hydrogencarbonate:
ionic  [✔]

Between hydrogen and oxygen in hydrogencarbonate:
«polar» covalent  [✔]

It was rather disappointing that less than 70 % of the candidates could distinguish between weak and strong acids. Many candidates referred to pH differences.

A poorly answered question, though it discriminated very well between high-scoring and low-scoring candidates. Less than 40 % of the candidates were able to deduce the formula of the conjugate base of HCO₃^-. Wrong answers included water, the hydroxide ion and carbon dioxide.

This was a relatively challenging question. Only about a quarter of the candidates explained how a decrease in pressure affected the equilibrium. Some candidates stated there was no shift in the equilibrium as the number of moles is the same on both sides of the equation, not acknowledging that only gaseous substances need to be considered when deciding the direction of shift in equilibrium due to a change in pressure. Some candidates wrote that the equilibrium shifts right because the gas escapes.

This was one of the most challenging questions on the paper that required application of Le Chatelier’s Principle in an unfamiliar situation. Most candidates did not refer to equilibrium (2), as directed by the question, and hence could not gain any marks. Some candidates stated that NaHCO₃ was an acid and decreased pH. Some answers had contradictions that showed poor understanding of the pH concept.

Very well answered. Most candidates calculated the molar concentration correctly.

Many candidates identified the bonding between sodium and hydrogencarbonate as ionic. A much smaller proportion of candidates identified the bonding between hydrogen and oxygen in hydrogencarbonate as covalent. The most common mistake was “hydrogen bonding”.

Identify the formula and state symbol of X.

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

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

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

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

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

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

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

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

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

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

H₂O AND (l)
Do not accept H₂O (aq).

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

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

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

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

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

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

22.5 × 10^-3 «s^-1»

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

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

(i)

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

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

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

(ii)

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

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

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

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

Accept cooling of reaction mixture during course of reaction.

State the equation for the reaction of each substance with water.

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

State the formula of the salt formed when butanoic acid reacts with ethylamine.

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

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

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.

CH₃CH₂NH₃+ CH₃CH₂CH₂COO⁻
OR
CH₃CH₂CH₂COO⁻ CH₃CH₂NH₃⁺
OR
CH₃CH₂CH₂COO⁻ H₃N⁺CH₂CH₃ ✔

The charges are not necessary for the mark.

Describe the bonding in metals.

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

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

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

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

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

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

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

Formulate an equation for this reaction.

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

electrostatic attraction

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

Accept mobile electrons.

Do not accept “metal atoms/nuclei”.

[2 marks]

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

= 47.93

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

Award [2] for correct final answer.

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

[2 marks]

Protons: 22 AND Neutrons: 26 AND Electrons: 22

[1 mark]

1s²2s²2p⁶3s²3p⁶3d²

[1 mark]

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

prevent layers sliding over each other

Accept diagram showing different sizes of atoms/ions.

[2 marks]

ionic
OR
«electrostatic» attraction between oppositely charged ions

[1 mark]

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

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

Do not accept “covalent”.

[1 mark]

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

correct products

correct balancing

Accept ionic equation.

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

[2 marks]

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

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

Accept “harmful” for both HCl and TiO₂.

Accept “smoke is asphyxiant”.

[1 mark]

Outline why metals, like iron, can conduct electricity.

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

Describe the bonding in this type of solid.

State the full electron configuration of the sulfide ion.

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

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

Write the equation for this reaction.

Deduce the change in the oxidation state of sulfur.

Suggest why this process might raise environmental concerns.

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

mobile/delocalized «sea of» electrons

Any two of:

forms acidic oxides «rather than basic oxides» ✔

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

forms anions «rather than cations» ✔

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

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

electrostatic attraction ✔

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

1s² 2s² 2p⁶ 3s² 3p⁶ ✔

Do not accept “[Ne] 3s² 3p⁶”.

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

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

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

Accept other valid answers.

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

Accept any correct ratio.

+6
OR
−2 to +4 ✔

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

sulfur dioxide/SO₂ causes acid rain ✔

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

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

prevents layers/atoms sliding over each other ✔

Outline why ethanoic acid is classified as a weak acid.

A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.

Cl2 (g) + 2NaOH (aq) $\rightleftharpoons$ NaOCl (aq) + NaCl (aq) + H₂O (l)

Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.

Draw a Lewis (electron dot) structure of chloramine.

Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.

Molecular geometry:

H–N–H bond angle:

partial dissociation «in aqueous solution»  [✔]

ethanoic acid/vinegar reacts with NaOH  [✔]

moves equilibrium to left/reactant side [✔]

releases Cl₂ (g)/chlorine gas
OR
Cl₂ (g)/chlorine gas is toxic  [✔]

Note: Accept “ethanoic acid produces H⁺ ions”.

Accept “ethanoic acid/vinegar reacts with NaOCl”.

Do not accept “2CH₃COOH + NaOCl + NaCl → 2CH₃COONa + Cl₂ + H₂O” as it does not refer to equilibrium.

Accept suitable molecular or ionic equations for M1 and M3.

 [✔]

Note: Accept any combination of dots/crosses or lines to represent electron pairs.

Molecular geometry:
«trigonal» pyramidal  [✔]

H–N–H bond angle:
107°  [✔]

Note: Accept angles in the range of 100–109.

The definition of a weak acid was generally correct.

Explaining why it was dangerous to mix chlorine with vinegar was not well answered but most students gained at least one mark for stating that “chlorine gas will be produced”, but couldn’t link it to equilibrium ideas.

The Lewis structure of chloramine was correct for strong candidates, but many made the mistake of omitting electron pairs on N and Cl.

The molecular geometry and bond angles often did not correspond to each other with quite a few candidates stating trigonal planar and then 107 for the angle.