IB DP Chemistry Topic 1.3 Reacting masses and volumes SL Paper 2 part 2

IB DP PhysicsIB DP Maths IB DP ChemistryIB DP Biology

Marks available	1
Reference code	13M.2.sl.TZ2.1

Question

Iron tablets are often prescribed to patients. The iron in the tablets is commonly present as iron(II) sulfate, \({\text{FeS}}{{\text{O}}_{\text{4}}}\).

Two students carried out an experiment to determine the percentage by mass of iron in a brand of tablets marketed in Cyprus.

Experimental Procedure:

• The students took five iron tablets and found that the total mass was 1.65 g.

• The five tablets were ground and dissolved in \({\text{100 c}}{{\text{m}}^{\text{3}}}\) dilute sulfuric acid, \({{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\). The solution and washings were transferred to a \({\text{250 c}}{{\text{m}}^{\text{3}}}\) volumetric flask and made up to the mark with deionized (distilled) water.

• \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of this \({\text{F}}{{\text{e}}^{2 + }}{\text{(aq)}}\) solution was transferred using a pipette into a conical flask. Some dilute sulfuric acid was added.

• A titration was then carried out using a \(5.00 \times {10^{ – 3}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) standard solution of potassium permanganate, \({\text{KMn}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\). The end-point of the titration was indicated by a slight pink colour.

The following results were recorded.

This experiment involves the following redox reaction.

\[{\text{5F}}{{\text{e}}^{2 + }}{\text{(aq)}} + {\text{MnO}}_4^ – {\text{(aq)}} + {\text{8}}{{\text{H}}^ + }{\text{(aq)}} \to {\text{5F}}{{\text{e}}^{3 + }}{\text{(aq)}} + {\text{M}}{{\text{n}}^{2 + }}{\text{(aq)}} + {\text{4}}{{\text{H}}_2}{\text{O(l)}}\]

When the \({\text{F}}{{\text{e}}^{2 + }}{\text{(aq)}}\) solution was made up in the \({\text{250 c}}{{\text{m}}^{\text{3}}}\) volumetric flask, deionized (distilled) water was added until the bottom of its meniscus corresponded to the graduation mark on the flask. It was noticed that one of the two students measured the volume of the solution from the top of the meniscus instead of from the bottom. State the name of this type of error.

[1]

State what is meant by the term precision.

[1]

When the students recorded the burette readings, following the titration with KMnO₄(aq),the top of the meniscus was used and not the bottom. Suggest why the students read the top of the meniscus and not the bottom.

[1]

Define the term reduction in terms of electrons.

[1]

d.i.

Deduce the oxidation number of manganese in the \({\text{MnO}}_{\text{4}}^ – {\text{(aq)}}\) ion.

[1]

d.ii.

Determine the amount, in mol, of \({\text{MnO}}_{\text{4}}^ – {\text{(aq)}}\), used in each accurate titre.

[2]

e.i.

Calculate the amount, in mol, of \({\text{F}}{{\text{e}}^{2 + }}{\text{(aq)}}\) ions in \({\text{250 c}}{{\text{m}}^{\text{3}}}\) of the solution.

[1]

e.ii.

Determine the total mass of iron, in g, in the \({\text{250 c}}{{\text{m}}^{\text{3}}}\) solution.

[1]

e.iii.

Determine the percentage by mass of iron in the tablets.

[1]

e.iv.

One titration was abandoned because a brown precipitate, manganese(IV) oxide, formed. State the chemical formula of this compound.

[1]

f.i.

Markscheme

systematic (error);

Do not accept parallax.

closeness of agreement of a set of measurements to each other / OWTTE;

Allow reproducibility/consistency of measurement / measurements with small random errors/total amount of random errors/standard deviation / a more precise value contains more significant figures / OWTTE.

potassium permanganate has a very dark/deep (purple) colour so cannot read bottom of meniscus / OWTTE;

gain (of electrons);

d.i.

VII / +7;

Do not accept 7 or 7+.

d.ii.

volume \( = 16.80{\text{ }}({\text{c}}{{\text{m}}^3})/18.00 – 1.20{\text{ }}({\text{c}}{{\text{m}}^3})\);

amount \(\left( { = \frac{{16.80 \times 5.00 \times {{10}^{ – 3}}}}{{1000}}} \right) = 8.40 \times {10^{ – 5}}{\text{ (mol)}}\);

Award [2] for correct final answer.

e.i.

\((8.40 \times {10^{ – 5}} \times 5 \times 10) = 4.20 \times {10^{ – 3}}{\text{ (mol per 250 c}}{{\text{m}}^{\text{3}}}{\text{)}}\);

e.ii.

\((55.85 \times 4.20 \times {10^{ – 3}}) = 0.235{\text{ (g)}}\);

Do not penalize if 56 g mol^–1is used for atomic mass of iron.

e.iii.

\(\left( {\frac{{{\text{0.235}} \times {\text{100}}}}{{{\text{1.65}}}} = } \right){\text{ 14.2% }}\);

No ECF if answer >100 %.

e.iv.

MnO₂;

f.i.

Examiners report

Question 1 presented difficulties to many candidates. It is felt that the extended nature of the response distracted candidates from rather straight-forward quantitative chemistry calculations. Part (a) required candidates to determine whether an error was systematic or random and part (b) asked for the meaning of precision. Both of these questions are relevant to Topic 11.

Very few candidates related reading the top of the meniscus in the burette in part (c) to the colour of the KMnO₄solution. While it is acknowledged that few candidates would have performed this experiment themselves, it is reasonable that candidates should know the colour of KMnO₄.

Part d) (i) was answered very well with nearly all candidates correctly defining reduction.

d.i.

In d) (ii) many candidates correctly deduced the oxidation number of Mn in MnO₄􀀀. Several lost marks, however, for not using acceptable notation. 7 by itself is not correct.

d.ii.

Part (e) involved the calculations. Candidates were guided through the process of calculating number of moles from concentration and volume, finding mole ratios, and determining mass from moles and molar mass. Better candidates performed these calculations well. Weaker candidates often scored follow-through marks when working was shown.

e.i.

e.ii.

e.iii.

e.iv.

In f) (i) a common error was to write Mn₂O₄as the formula for manganese(IV) oxide. Also common was the use of the symbol Mg for manganese.

f.i.

Marks available	1
Reference code	13M.2.sl.TZ2.1

Question

Iron tablets are often prescribed to patients. The iron in the tablets is commonly present as iron(II) sulfate, \({\text{FeS}}{{\text{O}}_{\text{4}}}\).

Two students carried out an experiment to determine the percentage by mass of iron in a brand of tablets marketed in Cyprus.

Experimental Procedure:

• The students took five iron tablets and found that the total mass was 1.65 g.

The following results were recorded.

This experiment involves the following redox reaction.

[1]

State what is meant by the term precision.

[1]

Define the term reduction in terms of electrons.

[1]

d.i.

Deduce the oxidation number of manganese in the \({\text{MnO}}_{\text{4}}^ – {\text{(aq)}}\) ion.

[1]

d.ii.

Determine the amount, in mol, of \({\text{MnO}}_{\text{4}}^ – {\text{(aq)}}\), used in each accurate titre.

[2]

e.i.

Calculate the amount, in mol, of \({\text{F}}{{\text{e}}^{2 + }}{\text{(aq)}}\) ions in \({\text{250 c}}{{\text{m}}^{\text{3}}}\) of the solution.

[1]

e.ii.

Determine the total mass of iron, in g, in the \({\text{250 c}}{{\text{m}}^{\text{3}}}\) solution.

[1]

e.iii.

Determine the percentage by mass of iron in the tablets.

[1]

e.iv.

One titration was abandoned because a brown precipitate, manganese(IV) oxide, formed. State the chemical formula of this compound.

[1]

f.i.

Markscheme

systematic (error);

Do not accept parallax.

closeness of agreement of a set of measurements to each other / OWTTE;

potassium permanganate has a very dark/deep (purple) colour so cannot read bottom of meniscus / OWTTE;

gain (of electrons);

d.i.

VII / +7;

Do not accept 7 or 7+.

d.ii.

volume \( = 16.80{\text{ }}({\text{c}}{{\text{m}}^3})/18.00 – 1.20{\text{ }}({\text{c}}{{\text{m}}^3})\);

amount \(\left( { = \frac{{16.80 \times 5.00 \times {{10}^{ – 3}}}}{{1000}}} \right) = 8.40 \times {10^{ – 5}}{\text{ (mol)}}\);

Award [2] for correct final answer.

e.i.

\((8.40 \times {10^{ – 5}} \times 5 \times 10) = 4.20 \times {10^{ – 3}}{\text{ (mol per 250 c}}{{\text{m}}^{\text{3}}}{\text{)}}\);

e.ii.

\((55.85 \times 4.20 \times {10^{ – 3}}) = 0.235{\text{ (g)}}\);

Do not penalize if 56 g mol^–1is used for atomic mass of iron.

e.iii.

\(\left( {\frac{{{\text{0.235}} \times {\text{100}}}}{{{\text{1.65}}}} = } \right){\text{ 14.2% }}\);

No ECF if answer >100 %.

e.iv.

MnO₂;

f.i.

Examiners report

Part d) (i) was answered very well with nearly all candidates correctly defining reduction.

d.i.

In d) (ii) many candidates correctly deduced the oxidation number of Mn in MnO₄􀀀. Several lost marks, however, for not using acceptable notation. 7 by itself is not correct.

d.ii.

e.i.

e.ii.

e.iii.

e.iv.

In f) (i) a common error was to write Mn₂O₄as the formula for manganese(IV) oxide. Also common was the use of the symbol Mg for manganese.

f.i.

Marks available	1
Reference code	13N.2.sl.TZ0.1

Question

Reaction kinetics can be investigated using the iodine clock reaction. The equations for two reactions that occur are given below.

Reaction A: \({{\text{H}}_2}{{\text{O}}_2}{\text{(aq)}} + {\text{2}}{{\text{I}}^ – }{\text{(aq)}} + {\text{2}}{{\text{H}}^ + }{\text{(aq)}} \to {{\text{I}}_2}{\text{(aq)}} + {\text{2}}{{\text{H}}_2}{\text{O(l)}}\)

Reaction B: \({{\text{I}}_2}{\text{(aq)}} + {\text{2}}{{\text{S}}_2}{\text{O}}_3^{2 – }{\text{(aq)}} \to {\text{2}}{{\text{I}}^ – }{\text{(aq)}} + {{\text{S}}_4}{\text{O}}_6^{2 – }{\text{(aq)}}\)

Reaction B is much faster than reaction A, so the iodine, \({{\text{I}}_{\text{2}}}\), formed in reaction A immediately reacts with thiosulfate ions, \({{\text{S}}_{\text{2}}}{\text{O}}_3^{2 – }\), in reaction B, before it can react with starch to form the familiar blue-black, starch-iodine complex.

In one experiment the reaction mixture contained:

\(5.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of \({\text{2.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) hydrogen peroxide \({\text{(}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{)}}\)

\(5.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of 1% aqueous starch

\(20.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sulfuric acid (\({{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\))

\(20.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.0100 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sodium thiosulfate (\({\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\))

\(50.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of water with 0.0200 ± 0.0001 g of potassium iodide (KI) dissolved in it.

After 45 seconds this mixture suddenly changed from colourless to blue-black.

Calculate the amount, in mol, of KI in the reaction mixture.

[1]

Calculate the amount, in mol, of \({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\) in the reaction mixture.

[1]

The concentration of iodide ions, \({{\text{I}}^ – }\), is assumed to be constant. Outline why this is a valid assumption.

[1]

For this mixture the concentration of hydrogen peroxide, \({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\), can also be assumed to be constant. Explain why this is a valid assumption.

[2]

Explain why the solution suddenly changes colour.

[2]

Apart from the precision uncertainties given, state one source of error that could affect this investigation and identify whether this is a random error or a systematic error.

[2]

Calculate the total uncertainty, in \({\text{c}}{{\text{m}}^{\text{3}}}\), of the volume of the reaction mixture.

[1]

The colour change occurs when \(1.00 \times {10^{ – 4}}{\text{ mol}}\) of iodine has been formed. Use the total volume of the solution and the time taken, to calculate the rate of the reaction, including appropriate units.

[4]

In a second experiment, the concentration of the hydrogen peroxide was decreased to \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) while all other concentrations and volumes remained unchanged. The colour change now occurred after 100 seconds. Explain why the reaction in this experiment is slower than in the original experiment.

[2]

In a third experiment, 0.100 g of a black powder was also added while all other concentrations and volumes remained unchanged. The time taken for the solution to change colour was now 20 seconds. Outline why you think the colour change occurred more rapidly and how you could confirm your hypothesis.

[2]

Explain why increasing the temperature also decreases the time required for the colour to change.

[2]

Markscheme

\(\left( {\frac{{0.0200}}{{166.00}} = } \right){\text{ }}0.000120/1.20 \times {10^{ – 4}}{\text{ (mol)}}\);

Accept 1.21 \( \times \) 10^–4.

\((0.0050 \times 2.00 = ){\text{ }}0.010{\text{ (mol)}}/1.0 \times {10^{ – 2}}\);

KI/\({{\text{I}}^ – }\)/potassium iodide/iodide (ion) (rapidly) reformed (in second stage of reaction);

amount (in mol) of \({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\)/hydrogen peroxide \( \gg \) amount (in mol) \({\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{/}}{{\text{S}}_{\text{2}}}{\text{O}}_3^{2 – }\)/sodium thiosulfate/ thiosulfate (ion);

Accept amount (in mol) of H₂O₂/hydrogen peroxide \( \gg \) amount (in mol) KI/I^–/potassium iodide/iodide (ion).

Accept “[H₂O₂]/hydrogen peroxide is in (large) excess/high concentration”.

(at end of reaction) [\({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\)] is only slightly decreased/virtually unchanged;

all \({\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\)/sodium thiosulfate/\({{\text{S}}_{\text{2}}}{\text{O}}_3^{2 – }\)/thiosulfate consumed/used up;

Accept “iodine no longer converted to iodide”.

(free) iodine is formed / iodine reacts with starch / forms iodine-starch complex;

Random: synchronizing mixing and starting timing / (reaction) time / uncertainty of concentrations of solutions / temperature of solutions/room temperature;

Systematic: liquid remaining in measuring cylinders / not all solid KI transferred / precision uncertainty of stopwatch / ability of human eye to detect colour change / parallax error;

Accept concentration of stock solution and human reaction time as systematic error.

Award M1 for correctly identifying a source of error and M2 for classifying it.

Accept other valid sources of error.

Do not accept “student making mistakes” / OWTTE.

\((5 \times 0.1) = ( \pm )0.5{\text{ }}({\text{c}}{{\text{m}}^3})\);

total volume \( = 0.100{\text{ }}({\text{d}}{{\text{m}}^3})/100{\text{ }}({\text{c}}{{\text{m}}^3})\);

\(\left( {{\text{change in concentration}} = \frac{{1.00 \times {{10}^{ – 4}}}}{{{\text{0.100}}}} = } \right){\text{ }}1.00 \times {10^{ – 3}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{\text{)}}\);

\(\left( {{\text{rate}} = \frac{{1.00 \times {{10}^{ – 3}}}}{{45}} = } \right){\text{ }}2.2 \times {10^{ – 5}}\);

Award [3] for the correct final answer.

\({\text{mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{{\text{s}}^{ – 1}}\);

fewer particles (per unit volume);

lower collision rate/collision frequency / less frequent collisions;

Do not accept “less collisions”.

acting as a catalyst / black powder reacts with thiosulfate ions / solid dissolves to give blue-black solution;

Accept any other valid suggestion which will make colour change more rapid.

For catalyst: amount/mass of black powder remains constant / no new/different products formed / activation energy decreased;

For other suggestions: any appropriate way to test the hypothesis;

Award [1] for valid hypothesis, [1] for appropriate method of testing the stated hypothesis.

particles have greater (average) kinetic energy;

Do not accept energy instead of kinetic energy.

more frequent collisions/collision frequency/number of collisions in a given time increases;

Do not accept “more collisions” unless “less collisions” penalized in (i).

greater proportion of particles have energy \( \ge\) activation energy;

Accept “particles have sufficient energy for collisions to be successful”.

Examiners report

This was a data based question based on quantitative chemistry and proved difficult for many candidates. Majority of candidates were able to gain almost full marks in parts (a) and (b). In part (c), many candidates failing to recognise that KI is rapidly reformed in the second stage of the reaction. In part (d), majority of candidates could not interpret the information correctly and hence lost two marks. Similarly, only 1 mark was obtained in part (e) where candidates recognized that iodine forms the starch-iodine complex. Many candidates managed the systematic and random errors in part (f). Calculation of uncertainty in part (g) was relatively well done by many candidates. In part (h), calculation of rate of reaction occasionally saw the erroneous use of volume in \({\text{c}}{{\text{m}}^{\text{3}}}\). In part (i), the candidates just repeated the stem of the question but obtained credit for the second mark for stating less frequent collisions. Part (j) was quite open ended and elicited a number of interesting responses (instead of acting as catalyst) whereas the suggested tests would not in fact confirm the hypothesis suggested. In part (k), the effect of increasing temperature on the rate of reaction proved easy for majority of candidates.

Marks available	1
Reference code	13N.2.sl.TZ0.1

Question

Reaction kinetics can be investigated using the iodine clock reaction. The equations for two reactions that occur are given below.

In one experiment the reaction mixture contained:

\(5.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of \({\text{2.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) hydrogen peroxide \({\text{(}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{)}}\)

\(5.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of 1% aqueous starch

\(20.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sulfuric acid (\({{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\))

\(50.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of water with 0.0200 ± 0.0001 g of potassium iodide (KI) dissolved in it.

After 45 seconds this mixture suddenly changed from colourless to blue-black.

Calculate the amount, in mol, of KI in the reaction mixture.

[1]

Calculate the amount, in mol, of \({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\) in the reaction mixture.

[1]

The concentration of iodide ions, \({{\text{I}}^ – }\), is assumed to be constant. Outline why this is a valid assumption.

[1]

For this mixture the concentration of hydrogen peroxide, \({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\), can also be assumed to be constant. Explain why this is a valid assumption.

[2]

Explain why the solution suddenly changes colour.

[2]

Apart from the precision uncertainties given, state one source of error that could affect this investigation and identify whether this is a random error or a systematic error.

[2]

Calculate the total uncertainty, in \({\text{c}}{{\text{m}}^{\text{3}}}\), of the volume of the reaction mixture.

[1]

[4]

[2]

Explain why increasing the temperature also decreases the time required for the colour to change.

[2]

Markscheme

\(\left( {\frac{{0.0200}}{{166.00}} = } \right){\text{ }}0.000120/1.20 \times {10^{ – 4}}{\text{ (mol)}}\);

Accept 1.21 \( \times \) 10^–4.

\((0.0050 \times 2.00 = ){\text{ }}0.010{\text{ (mol)}}/1.0 \times {10^{ – 2}}\);

KI/\({{\text{I}}^ – }\)/potassium iodide/iodide (ion) (rapidly) reformed (in second stage of reaction);

Accept amount (in mol) of H₂O₂/hydrogen peroxide \( \gg \) amount (in mol) KI/I^–/potassium iodide/iodide (ion).

Accept “[H₂O₂]/hydrogen peroxide is in (large) excess/high concentration”.

(at end of reaction) [\({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\)] is only slightly decreased/virtually unchanged;

all \({\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\)/sodium thiosulfate/\({{\text{S}}_{\text{2}}}{\text{O}}_3^{2 – }\)/thiosulfate consumed/used up;

Accept “iodine no longer converted to iodide”.

(free) iodine is formed / iodine reacts with starch / forms iodine-starch complex;

Random: synchronizing mixing and starting timing / (reaction) time / uncertainty of concentrations of solutions / temperature of solutions/room temperature;

Systematic: liquid remaining in measuring cylinders / not all solid KI transferred / precision uncertainty of stopwatch / ability of human eye to detect colour change / parallax error;

Accept concentration of stock solution and human reaction time as systematic error.

Award M1 for correctly identifying a source of error and M2 for classifying it.

Accept other valid sources of error.

Do not accept “student making mistakes” / OWTTE.

\((5 \times 0.1) = ( \pm )0.5{\text{ }}({\text{c}}{{\text{m}}^3})\);

total volume \( = 0.100{\text{ }}({\text{d}}{{\text{m}}^3})/100{\text{ }}({\text{c}}{{\text{m}}^3})\);

\(\left( {{\text{rate}} = \frac{{1.00 \times {{10}^{ – 3}}}}{{45}} = } \right){\text{ }}2.2 \times {10^{ – 5}}\);

Award [3] for the correct final answer.

\({\text{mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{{\text{s}}^{ – 1}}\);

fewer particles (per unit volume);

lower collision rate/collision frequency / less frequent collisions;

Do not accept “less collisions”.

acting as a catalyst / black powder reacts with thiosulfate ions / solid dissolves to give blue-black solution;

Accept any other valid suggestion which will make colour change more rapid.

For catalyst: amount/mass of black powder remains constant / no new/different products formed / activation energy decreased;

For other suggestions: any appropriate way to test the hypothesis;

Award [1] for valid hypothesis, [1] for appropriate method of testing the stated hypothesis.

particles have greater (average) kinetic energy;

Do not accept energy instead of kinetic energy.

more frequent collisions/collision frequency/number of collisions in a given time increases;

Do not accept “more collisions” unless “less collisions” penalized in (i).

greater proportion of particles have energy \( \ge\) activation energy;

Accept “particles have sufficient energy for collisions to be successful”.

Examiners report

Marks available	2
Reference code	13N.2.sl.TZ0.1

Question

Reaction kinetics can be investigated using the iodine clock reaction. The equations for two reactions that occur are given below.

In one experiment the reaction mixture contained:

\(5.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of \({\text{2.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) hydrogen peroxide \({\text{(}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{)}}\)

\(5.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of 1% aqueous starch

\(20.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sulfuric acid (\({{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\))

\(50.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of water with 0.0200 ± 0.0001 g of potassium iodide (KI) dissolved in it.

After 45 seconds this mixture suddenly changed from colourless to blue-black.

Calculate the amount, in mol, of KI in the reaction mixture.

[1]

Calculate the amount, in mol, of \({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\) in the reaction mixture.

[1]

The concentration of iodide ions, \({{\text{I}}^ – }\), is assumed to be constant. Outline why this is a valid assumption.

[1]

For this mixture the concentration of hydrogen peroxide, \({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\), can also be assumed to be constant. Explain why this is a valid assumption.

[2]

Explain why the solution suddenly changes colour.

[2]

Apart from the precision uncertainties given, state one source of error that could affect this investigation and identify whether this is a random error or a systematic error.

[2]

Calculate the total uncertainty, in \({\text{c}}{{\text{m}}^{\text{3}}}\), of the volume of the reaction mixture.

[1]

[4]

[2]

Explain why increasing the temperature also decreases the time required for the colour to change.

[2]

Markscheme

\(\left( {\frac{{0.0200}}{{166.00}} = } \right){\text{ }}0.000120/1.20 \times {10^{ – 4}}{\text{ (mol)}}\);

Accept 1.21 \( \times \) 10^–4.

\((0.0050 \times 2.00 = ){\text{ }}0.010{\text{ (mol)}}/1.0 \times {10^{ – 2}}\);

KI/\({{\text{I}}^ – }\)/potassium iodide/iodide (ion) (rapidly) reformed (in second stage of reaction);

Accept amount (in mol) of H₂O₂/hydrogen peroxide \( \gg \) amount (in mol) KI/I^–/potassium iodide/iodide (ion).

Accept “[H₂O₂]/hydrogen peroxide is in (large) excess/high concentration”.

(at end of reaction) [\({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\)] is only slightly decreased/virtually unchanged;

all \({\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\)/sodium thiosulfate/\({{\text{S}}_{\text{2}}}{\text{O}}_3^{2 – }\)/thiosulfate consumed/used up;

Accept “iodine no longer converted to iodide”.

(free) iodine is formed / iodine reacts with starch / forms iodine-starch complex;

Random: synchronizing mixing and starting timing / (reaction) time / uncertainty of concentrations of solutions / temperature of solutions/room temperature;

Systematic: liquid remaining in measuring cylinders / not all solid KI transferred / precision uncertainty of stopwatch / ability of human eye to detect colour change / parallax error;

Accept concentration of stock solution and human reaction time as systematic error.

Award M1 for correctly identifying a source of error and M2 for classifying it.

Accept other valid sources of error.

Do not accept “student making mistakes” / OWTTE.

\((5 \times 0.1) = ( \pm )0.5{\text{ }}({\text{c}}{{\text{m}}^3})\);

total volume \( = 0.100{\text{ }}({\text{d}}{{\text{m}}^3})/100{\text{ }}({\text{c}}{{\text{m}}^3})\);

\(\left( {{\text{rate}} = \frac{{1.00 \times {{10}^{ – 3}}}}{{45}} = } \right){\text{ }}2.2 \times {10^{ – 5}}\);

Award [3] for the correct final answer.

\({\text{mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{{\text{s}}^{ – 1}}\);

fewer particles (per unit volume);

lower collision rate/collision frequency / less frequent collisions;

Do not accept “less collisions”.

acting as a catalyst / black powder reacts with thiosulfate ions / solid dissolves to give blue-black solution;

Accept any other valid suggestion which will make colour change more rapid.

For catalyst: amount/mass of black powder remains constant / no new/different products formed / activation energy decreased;

For other suggestions: any appropriate way to test the hypothesis;

Award [1] for valid hypothesis, [1] for appropriate method of testing the stated hypothesis.

particles have greater (average) kinetic energy;

Do not accept energy instead of kinetic energy.

more frequent collisions/collision frequency/number of collisions in a given time increases;

Do not accept “more collisions” unless “less collisions” penalized in (i).

greater proportion of particles have energy \( \ge\) activation energy;

Accept “particles have sufficient energy for collisions to be successful”.

Examiners report

Marks available	1
Reference code	13N.2.sl.TZ0.1

Question

Reaction kinetics can be investigated using the iodine clock reaction. The equations for two reactions that occur are given below.

In one experiment the reaction mixture contained:

\(5.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of \({\text{2.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) hydrogen peroxide \({\text{(}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{)}}\)

\(5.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of 1% aqueous starch

\(20.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sulfuric acid (\({{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\))

\(50.0 \pm 0.1{\text{ c}}{{\text{m}}^{\text{3}}}\) of water with 0.0200 ± 0.0001 g of potassium iodide (KI) dissolved in it.

After 45 seconds this mixture suddenly changed from colourless to blue-black.

Calculate the amount, in mol, of KI in the reaction mixture.

[1]

Calculate the amount, in mol, of \({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\) in the reaction mixture.

[1]

The concentration of iodide ions, \({{\text{I}}^ – }\), is assumed to be constant. Outline why this is a valid assumption.

[1]

For this mixture the concentration of hydrogen peroxide, \({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\), can also be assumed to be constant. Explain why this is a valid assumption.

[2]

Explain why the solution suddenly changes colour.

[2]

Apart from the precision uncertainties given, state one source of error that could affect this investigation and identify whether this is a random error or a systematic error.

[2]

Calculate the total uncertainty, in \({\text{c}}{{\text{m}}^{\text{3}}}\), of the volume of the reaction mixture.

[1]

[4]

[2]

Explain why increasing the temperature also decreases the time required for the colour to change.

[2]

Markscheme

\(\left( {\frac{{0.0200}}{{166.00}} = } \right){\text{ }}0.000120/1.20 \times {10^{ – 4}}{\text{ (mol)}}\);

Accept 1.21 \( \times \) 10^–4.

\((0.0050 \times 2.00 = ){\text{ }}0.010{\text{ (mol)}}/1.0 \times {10^{ – 2}}\);

KI/\({{\text{I}}^ – }\)/potassium iodide/iodide (ion) (rapidly) reformed (in second stage of reaction);

Accept amount (in mol) of H₂O₂/hydrogen peroxide \( \gg \) amount (in mol) KI/I^–/potassium iodide/iodide (ion).

Accept “[H₂O₂]/hydrogen peroxide is in (large) excess/high concentration”.

(at end of reaction) [\({{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\)] is only slightly decreased/virtually unchanged;

all \({\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\)/sodium thiosulfate/\({{\text{S}}_{\text{2}}}{\text{O}}_3^{2 – }\)/thiosulfate consumed/used up;

Accept “iodine no longer converted to iodide”.

(free) iodine is formed / iodine reacts with starch / forms iodine-starch complex;

Random: synchronizing mixing and starting timing / (reaction) time / uncertainty of concentrations of solutions / temperature of solutions/room temperature;

Systematic: liquid remaining in measuring cylinders / not all solid KI transferred / precision uncertainty of stopwatch / ability of human eye to detect colour change / parallax error;

Accept concentration of stock solution and human reaction time as systematic error.

Award M1 for correctly identifying a source of error and M2 for classifying it.

Accept other valid sources of error.

Do not accept “student making mistakes” / OWTTE.

\((5 \times 0.1) = ( \pm )0.5{\text{ }}({\text{c}}{{\text{m}}^3})\);

total volume \( = 0.100{\text{ }}({\text{d}}{{\text{m}}^3})/100{\text{ }}({\text{c}}{{\text{m}}^3})\);

\(\left( {{\text{rate}} = \frac{{1.00 \times {{10}^{ – 3}}}}{{45}} = } \right){\text{ }}2.2 \times {10^{ – 5}}\);

Award [3] for the correct final answer.

\({\text{mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{{\text{s}}^{ – 1}}\);

fewer particles (per unit volume);

lower collision rate/collision frequency / less frequent collisions;

Do not accept “less collisions”.

acting as a catalyst / black powder reacts with thiosulfate ions / solid dissolves to give blue-black solution;

Accept any other valid suggestion which will make colour change more rapid.

For catalyst: amount/mass of black powder remains constant / no new/different products formed / activation energy decreased;

For other suggestions: any appropriate way to test the hypothesis;

Award [1] for valid hypothesis, [1] for appropriate method of testing the stated hypothesis.

particles have greater (average) kinetic energy;

Do not accept energy instead of kinetic energy.

more frequent collisions/collision frequency/number of collisions in a given time increases;

Do not accept “more collisions” unless “less collisions” penalized in (i).

greater proportion of particles have energy \( \ge\) activation energy;

Accept “particles have sufficient energy for collisions to be successful”.

Examiners report

Marks available	3
Reference code	14M.2.sl.TZ1.1

Question

0.100 g of magnesium ribbon is added to \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sulfuric acid to produce hydrogen gas and magnesium sulfate.

\[{\text{Mg(s)}} + {{\text{H}}_2}{\text{S}}{{\text{O}}_4}{\text{(aq)}} \to {{\text{H}}_2}{\text{(g)}} + {\text{MgS}}{{\text{O}}_4}{\text{(aq)}}\]

Magnesium sulfate can exist in either the hydrated form or in the anhydrous form. Two students wished to determine the enthalpy of hydration of anhydrous magnesium sulfate. They measured the initial and the highest temperature reached when anhydrous magnesium sulfate, \({\text{MgS}}{{\text{O}}_{\text{4}}}{\text{(s)}}\), was dissolved in water. They presented their results in the following table.

The students repeated the experiment using 6.16 g of solid hydrated magnesium sulfate, \({\text{MgS}}{{\text{O}}_{\text{4}}} \bullet {\text{7}}{{\text{H}}_{\text{2}}}{\text{O(s)}}\), and \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\) of water. They found the enthalpy change, \(\Delta {H_2}\), to be \( + 18{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\).

The enthalpy of hydration of solid anhydrous magnesium sulfate is difficult to determine experimentally, but can be determined using the diagram below.

Magnesium sulfate is one of the products formed when acid rain reacts with dolomitic limestone. This limestone is a mixture of magnesium carbonate and calcium carbonate.

(i) The graph shows the volume of hydrogen produced against time under these experimental conditions.

Sketch two curves, labelled I and II, to show how the volume of hydrogen produced (under the same temperature and pressure) changes with time when:

I. using the same mass of magnesium powder instead of a piece of magnesium ribbon;

II. 0.100 g of magnesium ribbon is added to \({\text{50 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.500 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sulfuric acid.

(ii) Outline why it is better to measure the volume of hydrogen produced against time rather than the loss of mass of reactants against time.

[3]

(i) Calculate the amount, in mol, of anhydrous magnesium sulfate.

(ii) Calculate the enthalpy change, \(\Delta {H_1}\), for anhydrous magnesium sulfate dissolving in water, in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\). State your answer to the correct number of significant figures.

[3]

(i) Determine the enthalpy change, \(\Delta H\), in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\), for the hydration of solid anhydrous magnesium sulfate, \({\text{MgS}}{{\text{O}}_{\text{4}}}\).

(ii) The literature value for the enthalpy of hydration of anhydrous magnesium sulfate is \( – 103{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\). Calculate the percentage difference between the literature value and the value determined from experimental results, giving your answer to one decimal place. (If you did not obtain an answer for the experimental value in (c)(i) then use the value of \( – 100{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\), but this is not the correct value.)

[2]

Another group of students experimentally determined an enthalpy of hydration of \( – 95{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\). Outline two reasons which may explain the variation between the experimental and literature values.

[2]

(i) State the equation for the reaction of sulfuric acid with magnesium carbonate.

(ii) Deduce the Lewis (electron dot) structure of the carbonate ion, giving the shape and the oxygen-carbon-oxygen bond angle.

Lewis (electron dot) structure:

Shape:

Bond angle:

[[N/A]]

Markscheme

(i)

I: line which is steeper/increases faster and finishes at the same height;

II: line which is less steep/increases more slowly and finishes at the same height;

(ii) mass of hydrogen produced is very small (so not accurate) / decrease in mass is very small (so not accurate);

(i) \(n({\text{MgS}}{{\text{O}}_4}) = \left( {\frac{{3.01}}{{120.37}} = } \right){\text{ }}0.0250{\text{ (mol)}}\);

(ii) energy released \( = 50.0 \times 4.18 \times 9.7 \times 2027{\text{ (J)}}/2.027{\text{ (kJ)}}\);

\(\Delta {H_1} = – 81{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);

Award [2] for correct answer.

Award [2] if 53.01 is used giving an answer of –86 (kJ mol^–1).

Award [1 max] for +81/81/+86/86 (kJ mol⁻¹).

Award [1 max] for –81000/–86000 if units are stated as J mol⁻¹.

Allow answers to 3 significant figures.

(i) \(\Delta H{\text{ }}( = \Delta {H_1} – \Delta {H_2}) = – 99{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}})\);

Award [1] if –86 is used giving an answer of –104 (kJ mol⁻¹).

(ii) \(\frac{{(103 – 99)}}{{103}} \times 100 = 3.9\% \);

Accept answer of 2.9 % if –100 used but only if a value for (b)(i) is not
present.

Award [1] if –104 is used giving an answer of 1.0% .

Accept correct answers which are not to 1 decimal place.

\({\text{MgS}}{{\text{O}}_{\text{4}}}\) not completely anhydrous / OWTTE;

\({\text{MgS}}{{\text{O}}_{\text{4}}}\) is impure;

heat loss to the atmosphere/surroundings;

specific heat capacity of solution is taken as that of pure water;

experiment was done once only so it is not scientific;

density of solution is taken to be \(1{\text{ g}}\,{\text{c}}{{\text{m}}^{ – 3}}\);

mass of \(7{{\text{H}}_2}{\text{O}}\) ignored in calculation;

uncertainty of thermometer is high so temperature change is unreliable;

literature values determined under standard conditions but this experiment is not;

all solid not dissolved;

(i) \({{\text{H}}_2}{\text{S}}{{\text{O}}_4}{\text{(aq)}} + {\text{MgC}}{{\text{O}}_3}{\text{(s)}} \to {\text{MgS}}{{\text{O}}_4}{\text{(aq)}} + {\text{C}}{{\text{O}}_2}{\text{(g)}} + {{\text{H}}_2}{\text{O(l)}}\);

Ignore state symbols.

Do not accept H₂CO₃.

(ii) ;

Accept crosses, lines or dots as electron pairs.

Accept any correct resonance structure.

Award [0] if structure is drawn without brackets and charge.

Award [0] if lone pairs not shown on O atoms.

shape: trigonal/triangular planar;

bond angle: 120°;

Accept answers trigonal/triangular planar and 120° if M1 incorrect, but no other answer should be given credit.

Examiners report

Many candidates could sketch correct curves in (a)(i), though many did not realize that the same final volume of hydrogen is formed. Lines were generally poorly drawn with several lines for one curve, and curve I often did not join smoothly with the given curve, but dropped near the end or overshot the final volume and then fell back down. Candidates are advised to draw graphs in pencil first. In (a)(ii), very few students indicated that because the mass of hydrogen is very small it is better to measure reaction rate using gas volume; most indicated that it is not precise because the mass of a mixture is measured. It seems that very few candidates are aware that measuring loss of mass per unit time is a valid tool for determining the rate of a reaction when \({\text{C}}{{\text{O}}_{\text{2}}}\) is produced. The moles of magnesium sulfate were mostly calculated correctly in (b)(i), but in (b)(ii) most candidates had problems calculating the enthalpy change, working with the mass of magnesium sulfate instead of water or solution and not giving the enthalpy change a negative sign. Several candidates only found the temperature change and called this the enthalpy change, or found the energy change and ignored the number of moles. Few candidates correctly applied Hess’s law in (c)(i). Some respondents felt that this was not on the SL course, but it is clearly stated in 5.3.1. Some candidates had no idea how to calculate the percentage difference in (c)(ii) and several left this blank despite a value being given for the experimental results for candidates to use if they had not found a value themselves. Quite a few others determined the percentage difference correctly. In (d) most candidates stated heat loss to the surroundings as an error, mentioning further irrelevant errors. Only the better candidates also referred to the partial hydration of the anhydrous salt. The equation for the reaction between sulfuric acid and magnesium carbonate was generally done well in (e)(i) but \({{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\) was frequently (incorrectly) given as a product. A few candidates did not know the formulas for sulfuric acid and magnesium carbonate. Very few candidates could give a correct Lewis structure for the carbonate ion in (ii). Some almost scored but failed to include brackets and charge. Some decided that the carbonate ion was a synonym for carbon dioxide and drew that. The formula for the carbonate ion should be known (assessment statement 4.1.7) and only one Lewis structure was required so students did not need to know about resonance structures. Shape and bond angle were also done poorly but there were a few candidates who knew the shape and bond angle of the carbonate ion even though they couldn’t draw the Lewis structure.

Marks available	5
Reference code	14M.2.sl.TZ1.7

Question

\({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.200 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) ethanoic acid were added to \({\text{30.0 c}}{{\text{m}}^{\text{3}}}\) of a \({\text{0.150 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sodium hydrogencarbonate solution, \({\text{NaHC}}{{\text{O}}_{\text{3}}}{\text{(aq)}}\).

The molar mass of a volatile organic liquid, X, can be determined experimentally by allowing it to vaporize completely at a controlled temperature and pressure. 0.348 g of X was injected into a gas syringe maintained at a temperature of 90 °C and a pressure of \(1.01 \times {10^5}{\text{ Pa}}\). Once it had reached equilibrium, the gas volume was measured as \({\text{95.0 c}}{{\text{m}}^{\text{3}}}\).

Bromoethane, \({\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Br}}\), undergoes a substitution reaction to form ethanol, \({\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\).

Outline how electrical conductivity can be used to distinguish between a \({\text{0.200 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) solution of ethanoic acid, \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\), and a \({\text{0.200 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) solution of hydrochloric acid, HCl.

[1]

(i) State an equation for the reaction of ethanoic acid with a solution of sodium hydrogencarbonate.

(ii) Determine which is the limiting reagent. Show your working.

(iii) Calculate the mass, in g, of carbon dioxide produced.

[5]

(i) Determine the amount, in mol, of X in the gas syringe.

(ii) Calculate the molar mass of X.

[4]

(i) Identify the reagent necessary for this reaction to occur.

(ii) Deduce the mechanism for the reaction using equations and curly arrows to represent the movement of electron pairs.

[4]

Determine the enthalpy change, in kJ mol\(^{ – 1}\), for this reaction, using Table 10 of the Data Booklet.

[3]

e.ii.

Bromoethene, \({\text{C}}{{\text{H}}_{\text{2}}}{\text{CHBr}}\), can undergo polymerization. Draw a section of this polymer that contains six carbon atoms.

[1]

Markscheme

HCl is a strong acid and \({\text{C}}{{\text{H}}_3}{\text{COOH}}\) is a weak acid so HCl has higher conductivity / HCl dissociates completely in water and \({\text{C}}{{\text{H}}_3}{\text{COOH}}\) does not, so HCl has higher conductivity / HCl is stronger acid (than \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) so has higher \({\text{[}}{{\text{H}}^ + }{\text{]}}\) and higher conductivity;

(i) \({\text{C}}{{\text{H}}_3}{\text{COOH(aq)}} + {\text{HCO}}_3^ – {\text{(aq)}} \to {\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ – }{\text{(aq)}} + {{\text{H}}_2}{\text{O(l)}} + {\text{C}}{{\text{O}}_2}{\text{(g)}}\);

Accept NaHCO₃(aq) and CH₃COONa (aq) instead of ions.

Ignore state symbols.

(ii) \(n{\text{(C}}{{\text{H}}_3}{\text{COOH)}} = 0.00500{\text{ (mol)}}\) and \(n{\text{(NaHC}}{{\text{O}}_3}{\text{)}} = 0.00450{\text{ (mol)}}\);

\({\text{NaHC}}{{\text{O}}_3}\) is limiting;

(iii) \(n{\text{(C}}{{\text{O}}_2}{\text{)}} = n{\text{(NaHC}}{{\text{O}}_3}{\text{)}} = 0.00450{\text{ (mol)}}\);

\(m{\text{(C}}{{\text{O}}_2}{\text{)}} = 0.00450 \times 44.01 = 0.198{\text{ (g)}}\);

Award [2] for correct final answer.

(i) \(T = 363{\text{ K}}\) and \(V = 9.50 \times {10^{ – 5}}{\text{ }}{{\text{m}}^3}\);

Accept V = 9.5 \( \times \) 10^–2 dm³ if P is used as 101 kPa in calculation.

\(n = \frac{{PV}}{{RT}} = \frac{{1.01 \times {{10}^5} \times 9.50 \times {{10}^{ – 5}}}}{{8.31 \times 363}}\);

\( = 3.18 \times {10^{ – 3}}{\text{ (mol)}}\);

Award [3] for correct final answer.

(ii) \(M = \left( {\frac{m}{n} = \frac{{0.348}}{{3.18 \times {{10}^{ – 3}}}} = } \right)109{\text{ }}({\text{g}}\,{\text{mo}}{{\text{l}}^{ – 1}})\);

(i) (dilute aqueous) NaOH/sodium hydroxide / KOH/potassium hydroxide;

Do not accept hydroxide/OH^–.

(ii)

curly arrow going from lone pair/negative charge on O in HO– to C;

Do not allow curly arrow originating on H in HO^–.

curly arrow showing Br leaving;

Accept curly arrow either going from bond between C and Br to Br in bromoethane or in the transition state.

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

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

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

bonds broken:

1(C=C) \( + 1\) (H–Br) / \((612 + 366 = )978{\text{ (kJ)}}\);

Accept 2630 (kJ).

bonds formed:

1(C–C) \( + 1\) (C–H) \( + 1\) (C–Br) / \((1 \times 347 + 1 \times 413 + 1 \times 290 = )1050{\text{ (kJ)}}\);

Accept 2702 (kJ).

\(\Delta H = – 72{\text{ }}({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}})\);

Award [3] for correct final answer.

Award [2 max] for +72 (kJ mol⁻¹).

e.ii.

;

Extension bonds required.

Ignore brackets and n.

Examiners report

Question 7 was answered by relatively few candidates, and those who chose this question were usually not well-prepared. In (a) very few candidates indicated that HCl is a strong acid and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\) a weak one. Many candidates seemed unfamiliar with the distinction between state and outline and simply said that HCl would be a better conductor. In (b)(i) very few candidates could state a correct equation for the reaction between ethanoic acid and sodium hydrogencarbonate, even when the formulas were provided, but most could calculate the limiting reagent in (b)(ii) and the mass of \({\text{C}}{{\text{O}}_{\text{2}}}\) produced in (b)(iii). Part (c) gave details of a volatile organic liquid. Most candidates could calculate the moles of gas present in (c)(i), although the conversion to the correct units for pressure and volume gave many problems. The calculation of the molar mass of the gas, especially with ECF applied, was generally done well by the candidates. Part (d) referred to the substitution reaction of bromoethane to form ethanol. Identifying the reagent in (d)(i) for this reaction caused problems, with many stating \({\text{O}}{{\text{H}}^ – }\) as the reagent instead of NaOH or KOH. Only the best candidates could draw the mechanism for this substitution reaction in (d)(ii). Many candidates seemed to have very little idea of how to represent an \({{\text{S}}_{\text{N}}}{\text{2}}\) mechanism. Although most candidates identified HBr as the reagent which could produce bromoethane from ethene, they often gave UV as the required condition in (e)(i). Teachers should note that assessment statement 10.6.1 indicates that reagents, conditions and equations should be included for all reaction types listed in the syllabus. Calculation of the enthalpy change using bond enthalpies did not give problems to the good candidates in (e)(ii) but many of the weaker candidates failed to identify all the bonds broken and formed, and only scored the final mark through the application of ECF. Drawing a section of a polymer produced from bromoethene in (e)(iii) presented few problems for most candidates.

e.ii.

Marks available	4
Reference code	14M.2.sl.TZ1.7

Question

[1]

(i) State an equation for the reaction of ethanoic acid with a solution of sodium hydrogencarbonate.

(ii) Determine which is the limiting reagent. Show your working.

(iii) Calculate the mass, in g, of carbon dioxide produced.

[5]

(i) Determine the amount, in mol, of X in the gas syringe.

(ii) Calculate the molar mass of X.

[4]

(i) Identify the reagent necessary for this reaction to occur.

(ii) Deduce the mechanism for the reaction using equations and curly arrows to represent the movement of electron pairs.

[4]

Determine the enthalpy change, in kJ mol\(^{ – 1}\), for this reaction, using Table 10 of the Data Booklet.

[3]

e.ii.

Bromoethene, \({\text{C}}{{\text{H}}_{\text{2}}}{\text{CHBr}}\), can undergo polymerization. Draw a section of this polymer that contains six carbon atoms.

[1]

Markscheme

Accept NaHCO₃(aq) and CH₃COONa (aq) instead of ions.

Ignore state symbols.

(ii) \(n{\text{(C}}{{\text{H}}_3}{\text{COOH)}} = 0.00500{\text{ (mol)}}\) and \(n{\text{(NaHC}}{{\text{O}}_3}{\text{)}} = 0.00450{\text{ (mol)}}\);

\({\text{NaHC}}{{\text{O}}_3}\) is limiting;

(iii) \(n{\text{(C}}{{\text{O}}_2}{\text{)}} = n{\text{(NaHC}}{{\text{O}}_3}{\text{)}} = 0.00450{\text{ (mol)}}\);

\(m{\text{(C}}{{\text{O}}_2}{\text{)}} = 0.00450 \times 44.01 = 0.198{\text{ (g)}}\);

Award [2] for correct final answer.

(i) \(T = 363{\text{ K}}\) and \(V = 9.50 \times {10^{ – 5}}{\text{ }}{{\text{m}}^3}\);

Accept V = 9.5 \( \times \) 10^–2 dm³ if P is used as 101 kPa in calculation.

\(n = \frac{{PV}}{{RT}} = \frac{{1.01 \times {{10}^5} \times 9.50 \times {{10}^{ – 5}}}}{{8.31 \times 363}}\);

\( = 3.18 \times {10^{ – 3}}{\text{ (mol)}}\);

Award [3] for correct final answer.

(ii) \(M = \left( {\frac{m}{n} = \frac{{0.348}}{{3.18 \times {{10}^{ – 3}}}} = } \right)109{\text{ }}({\text{g}}\,{\text{mo}}{{\text{l}}^{ – 1}})\);

(i) (dilute aqueous) NaOH/sodium hydroxide / KOH/potassium hydroxide;

Do not accept hydroxide/OH^–.

(ii)

curly arrow going from lone pair/negative charge on O in HO– to C;

Do not allow curly arrow originating on H in HO^–.

curly arrow showing Br leaving;

Accept curly arrow either going from bond between C and Br to Br in bromoethane or in the transition state.

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

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

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

bonds broken:

1(C=C) \( + 1\) (H–Br) / \((612 + 366 = )978{\text{ (kJ)}}\);

Accept 2630 (kJ).

bonds formed:

1(C–C) \( + 1\) (C–H) \( + 1\) (C–Br) / \((1 \times 347 + 1 \times 413 + 1 \times 290 = )1050{\text{ (kJ)}}\);

Accept 2702 (kJ).

\(\Delta H = – 72{\text{ }}({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}})\);

Award [3] for correct final answer.

Award [2 max] for +72 (kJ mol⁻¹).

e.ii.

;

Extension bonds required.

Ignore brackets and n.

Examiners report

e.ii.

Marks available	4
Reference code	14M.2.sl.TZ2.1

Question

A class studied the equilibrium established when ethanoic acid and ethanol react together in the presence of a strong acid, using propanone as an inert solvent. The equation is given below.

\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} + {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} + {{\text{H}}_{\text{2}}}{\text{O}}\]

One group made the following initial mixture:

After one week, a \(5.00 \pm 0.05{\text{ c}}{{\text{m}}^{\text{3}}}\) sample of the final equilibrium mixture was pipetted out and titrated with \({\text{0.200 mol}}\,{\text{d}}{{\text{m}}^{ – 2}}\) aqueous sodium hydroxide to determine the amount of ethanoic acid remaining. The following titration results were obtained:

The density of ethanoic acid is \({\text{1.05 g}}\,{\text{c}}{{\text{m}}^{ – 3}}\). Determine the amount, in mol, of ethanoic acid present in the initial mixture.

[3]

The hydrochloric acid does not appear in the balanced equation for the reaction. State its function.

[1]

Identify the liquid whose volume has the greatest percentage uncertainty.

[1]

(i) Calculate the absolute uncertainty of the titre for Titration 1 (\({\text{27.60 c}}{{\text{m}}^{\text{3}}}\)).

(ii) Suggest the average volume of alkali, required to neutralize the \({\text{5.00 c}}{{\text{m}}^{\text{3}}}\) sample, that the student should use.

(iii) \({\text{23.00 c}}{{\text{m}}^{\text{3}}}\) of this \({\text{0.200 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) aqueous sodium hydroxide reacted with the ethanoic acid in the \({\text{5.00 c}}{{\text{m}}^{\text{3}}}\) sample. Determine the amount, in mol, of ethanoic acid present in the \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\) of final equilibrium mixture.

[4]

Referring back to your answer for part (a), calculate the percentage of ethanoic acid converted to ethyl ethanoate.

[1]

Deduce the equilibrium constant expression for the reaction.

[1]

Outline how you could establish that the system had reached equilibrium at the end of one week.

[1]

Outline why changing the temperature has only a very small effect on the value of the equilibrium constant for this equilibrium.

[1]

Outline how adding some ethyl ethanoate to the initial mixture would affect the amount of ethanoic acid converted to product.

[2]

Propanone is used as the solvent because one compound involved in the equilibrium is insoluble in water. Identify this compound and explain why it is insoluble in water.

[2]

Suggest one other reason why using water as a solvent would make the experiment less successful.

[1]

Markscheme

\({\text{M(C}}{{\text{H}}_3}{\text{COOH)}}\left( { = (4 \times 1.01) + (2 \times 12.01) + (2 \times 16.00)} \right) = 60.06{\text{ (g}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);

Accept 60 (g mol^–1).

\({\text{mass (C}}{{\text{H}}_3}{\text{COOH)}}( = 5.00 \times 1.05) = 5.25{\text{ (g)}}\);

\(\frac{{5.25}}{{{\text{60.06}}}} = 0.0874{\text{ (mol)}}\);

Award [3] for correct final answer.

Accept 0.0875 (comes from using Mr = 60 g mol^–1).

catalyst / OWTTE;

hydrochloric acid/HCl;

(i) \( \pm 0.1/0.10{\text{ (c}}{{\text{m}}^3}{\text{)}}\);

Do not accept without ±.

(ii) \({\text{26.00 (c}}{{\text{m}}^3}{\text{)}}\);

(iii) \(0.200 \times \frac{{23.00}}{{1000}} = 0.0046\);

\({\text{0.0046}} \times \frac{{{\text{50.0}}}}{{{\text{5.00}}}} = {\text{0.0460 (mol)}}\);

\(\frac{{0.0874 – 0.0460}}{{0.0874}} \times 100 = 47.4\% \);

\({\text{(}}{K_{\text{c}}} = {\text{)}}\frac{{{\text{[C}}{{\text{H}}_3}{\text{COO}}{{\text{C}}_2}{{\text{H}}_3}{\text{][}}{{\text{H}}_2}{\text{O]}}}}{{{\text{[}}{{\text{C}}_2}{{\text{H}}_5}{\text{OH][C}}{{\text{H}}_3}{\text{COOH]}}}}\);

Do not penalize minor errors in formulas.

Accept \({\text{(}}{K_{\text{c}}} = {\text{)}}\frac{{{\text{[}}ester{\text{][}}water{\text{]}}}}{{{\text{[}}ethanol / alcohol{\text{][(}}ethanoic{\text{)}} acid{\text{]}}}}\).

repeat the titration a day/week later (and result should be the same) / OWTTE;

Accept “concentrations/physical properties/macroscopic properties of the system do not change”.

enthalpy change/\(\Delta H\) for the reaction is (very) small / OWTTE;

decreases (the amount of ethanoic acid converted);

Accept “increases amount of ethanoic acid present at equilibrium” / OWTTE.

(adding product) shifts position of equilibrium towards reactants/LHS / increases the rate of the reverse reaction / OWTTE;

ethyl ethanoate/\({\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\);

forms only weak hydrogen bonds (to water);

Allow “does not hydrogen bond to water” / “hydrocarbon sections too long” / OWTTE.

M2 can only be given only if M1 correct.

(large excess of) water will shift the position of equilibrium (far to the left) / OWTTE;

Accept any other chemically sound response, such as “dissociation of ethanoic acid would affect equilibrium”.

Examiners report

Generally candidates found some elements of this question quite challenging but there were accessible marks of even the weakest candidates. The majority of students were able to determine the molar mass of ethanoic acid but some struggled to calculate the mass from the volume. Most candidates were able to identify the role of hydrochloric acid as a catalyst but some struggled to identify the liquid whose volume had the greatest uncertainty. Most candidates were able to calculate the absolute uncertainty of the titre but some lost a mark by omitting the \( + \)/\( – \) sign. Candidates did not identify the first titre as incongruent and simply averaged the three values which perhaps suggests limited experimental experience. Most students could determine an equilibrium constant expression, but many did not answer the question in (g) and did not suggest how the equilibrium could be established experimentally with many referring to the equal rate of the forward and backward reaction. Many candidates were aware of Le Chatelier effects on the position of equilibrium, but a significant number failed to use this information to answer the question asked and could not explain the small effect of temperature changes. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, many did not refer to its inability to form strong hydrogen bonds to water which was necessary for the mark. Quite a number of students came up with a valid reason why water would not be a suitable though some students appeared to have overlooked that the question asked for “one other reason” than that implied in (j).

Marks available	1
Reference code	14M.2.sl.TZ2.1

Question

A class studied the equilibrium established when ethanoic acid and ethanol react together in the presence of a strong acid, using propanone as an inert solvent. The equation is given below.

One group made the following initial mixture:

The density of ethanoic acid is \({\text{1.05 g}}\,{\text{c}}{{\text{m}}^{ – 3}}\). Determine the amount, in mol, of ethanoic acid present in the initial mixture.

[3]

The hydrochloric acid does not appear in the balanced equation for the reaction. State its function.

[1]

Identify the liquid whose volume has the greatest percentage uncertainty.

[1]

(i) Calculate the absolute uncertainty of the titre for Titration 1 (\({\text{27.60 c}}{{\text{m}}^{\text{3}}}\)).

(ii) Suggest the average volume of alkali, required to neutralize the \({\text{5.00 c}}{{\text{m}}^{\text{3}}}\) sample, that the student should use.

[4]

Referring back to your answer for part (a), calculate the percentage of ethanoic acid converted to ethyl ethanoate.

[1]

Deduce the equilibrium constant expression for the reaction.

[1]

Outline how you could establish that the system had reached equilibrium at the end of one week.

[1]

Outline why changing the temperature has only a very small effect on the value of the equilibrium constant for this equilibrium.

[1]

Outline how adding some ethyl ethanoate to the initial mixture would affect the amount of ethanoic acid converted to product.

[2]

Propanone is used as the solvent because one compound involved in the equilibrium is insoluble in water. Identify this compound and explain why it is insoluble in water.

[2]

Suggest one other reason why using water as a solvent would make the experiment less successful.

[1]

Markscheme

\({\text{M(C}}{{\text{H}}_3}{\text{COOH)}}\left( { = (4 \times 1.01) + (2 \times 12.01) + (2 \times 16.00)} \right) = 60.06{\text{ (g}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);

Accept 60 (g mol^–1).

\({\text{mass (C}}{{\text{H}}_3}{\text{COOH)}}( = 5.00 \times 1.05) = 5.25{\text{ (g)}}\);

\(\frac{{5.25}}{{{\text{60.06}}}} = 0.0874{\text{ (mol)}}\);

Award [3] for correct final answer.

Accept 0.0875 (comes from using Mr = 60 g mol^–1).

catalyst / OWTTE;

hydrochloric acid/HCl;

(i) \( \pm 0.1/0.10{\text{ (c}}{{\text{m}}^3}{\text{)}}\);

Do not accept without ±.

(ii) \({\text{26.00 (c}}{{\text{m}}^3}{\text{)}}\);

(iii) \(0.200 \times \frac{{23.00}}{{1000}} = 0.0046\);

\({\text{0.0046}} \times \frac{{{\text{50.0}}}}{{{\text{5.00}}}} = {\text{0.0460 (mol)}}\);

\(\frac{{0.0874 – 0.0460}}{{0.0874}} \times 100 = 47.4\% \);

Do not penalize minor errors in formulas.

Accept \({\text{(}}{K_{\text{c}}} = {\text{)}}\frac{{{\text{[}}ester{\text{][}}water{\text{]}}}}{{{\text{[}}ethanol / alcohol{\text{][(}}ethanoic{\text{)}} acid{\text{]}}}}\).

repeat the titration a day/week later (and result should be the same) / OWTTE;

Accept “concentrations/physical properties/macroscopic properties of the system do not change”.

enthalpy change/\(\Delta H\) for the reaction is (very) small / OWTTE;

decreases (the amount of ethanoic acid converted);

Accept “increases amount of ethanoic acid present at equilibrium” / OWTTE.

(adding product) shifts position of equilibrium towards reactants/LHS / increases the rate of the reverse reaction / OWTTE;

ethyl ethanoate/\({\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\);

forms only weak hydrogen bonds (to water);

Allow “does not hydrogen bond to water” / “hydrocarbon sections too long” / OWTTE.

M2 can only be given only if M1 correct.

(large excess of) water will shift the position of equilibrium (far to the left) / OWTTE;

Accept any other chemically sound response, such as “dissociation of ethanoic acid would affect equilibrium”.

Examiners report

Marks available	3
Reference code	14M.2.sl.TZ2.5

Question

A group of students investigated the rate of the reaction between aqueous sodium thiosulfate and hydrochloric acid according to the equation below.

\[{\text{N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3}{\text{(aq)}} + {\text{2HCl(aq)}} \to {\text{2NaCl(aq)}} + {\text{S}}{{\text{O}}_2}{\text{(g)}} + {\text{S(s)}} + {{\text{H}}_2}{\text{O(l)}}\]

The two reagents were rapidly mixed together in a beaker and placed over a mark on a piece of paper. The time taken for the precipitate of sulfur to obscure the mark when viewed through the reaction mixture was recorded.

Initially they measured out \({\text{10.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.500 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) hydrochloric acid and then added \({\text{40.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.0200 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) aqueous sodium thiosulfate. The mark on the paper was obscured 47 seconds after the solutions were mixed.

The teacher asked the students to measure the effect of halving the concentration of sodium thiosulfate on the rate of reaction.

The teacher asked the students to devise another technique to measure the rate of this reaction.

Another group suggested collecting the sulfur dioxide and drawing a graph of the volume of gas against time.

The teacher made up \({\text{2.50 d}}{{\text{m}}^{\text{3}}}\) of the sodium thiosulfate solution using sodium thiosulfate pentahydrate crystals, \({\text{N}}{{\text{a}}_2}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} \bullet {\text{5}}{{\text{H}}_{\text{2}}}{\text{O}}\). Calculate the required mass of these crystals.

[3]

(i) State the volumes of the liquids that should be mixed.

(ii) State why it is important that the students use a similar beaker for both reactions.

(iii) Explain, in terms of the collision theory, how decreasing the concentration of sodium thiosulfate would affect the time taken for the mark to be obscured.

[4]

(i) Sketch and label, indicating an approximate activation energy, the Maxwell–Boltzmann energy distribution curves for two temperatures, \({T_1}\) and \({T_2}{\text{ }}({T_2} > {T_1})\), at which the rate of reaction would be significantly different.

(ii) Explain why increasing the temperature of the reaction mixture would significantly increase the rate of the reaction.

[6]

(i) One group suggested recording how long it takes for the pH of the solution to change by one unit. Calculate the initial pH of the original reaction mixture.

(ii) Deduce the percentage of hydrochloric acid that would have to be used up for the pH to change by one unit.

[3]

(i) Calculate the volume of sulfur dioxide, in \({\text{c}}{{\text{m}}^{\text{3}}}\), that the original reaction mixture would produce if it were collected at \(1.00 \times {10^5}{\text{ Pa}}\) and 300 K.

(ii) Suggest why it is better to use a gas syringe rather than collecting the gas in a measuring cylinder over water.

[4]

Markscheme

\({\text{mol N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}( = 2.50 \times 0.0200) = 0.0500\);

\({{\text{M}}_{\text{r}}}{\text{N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} \bullet {\text{5}}{{\text{H}}_2}{\text{O}}\left( { = (2 \times 22.99) + (2 \times 32.06) + (3 \times 16.00) + (5 \times 18.02)} \right) = 248.20\);

Allow 248.

\({\text{mass N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} \bullet {\text{5}}{{\text{H}}_2}{\text{O}} = (0.0500 \times 248.20) = 12.4{\text{ g}}\);

Award [3] for correct final answer.

Award [2] for 7.91g (water of crystallization omitted in \({M_r}\) calculation).

(i) ;

Accept other volumes in a 1:2:2 ratio.

(ii) depth of liquid in the beaker must remain constant / OWTTE;

Accept “same thickness of glass” and any other valid point, such as answers framed around minimizing uncontrolled variables / making it a “fair test”.

(iii) increases the time;

decrease in collision frequency/number of collisions per unit time;

Do not award mark for decrease in number of collisions.

(i)

labelled y-axis: number of particles / probability of particles (with that kinetic energy) and labelled x-axis: (kinetic) energy;

Allow fraction/proportion/amount of particles (with kinetic energy) for y-axis label.

Allow speed/velocity for x-axis label.

T₂ curve broader and with maximum lower and to right of T₁ curve;

Do not award this mark if both curves not asymmetric.

Curves must pass through the origin and be asymptotic to x axis.

Do not award this mark if curves not labelled.

\({E_{\text{a}}}\) marked on graph;

(ii) kinetic energy of molecules increases;

This may be answered implicitly in the final marking point.

frequency of collision/number of collisions per unit time increases;

Only penalize use of “number of collisions” if not penalized in (b)(iii).

greater proportion of molecules have energy greater than/equal to activation energy / rate related to temperature by the Arrhenius equation;

Award [1 max] for statements such as “there will be more successful collisions” if neither of last two marking points awarded.

(i) \({\text{[}}{{\text{H}}^ + }{\text{]}} = 0.5 \times \frac{{10}}{{{\text{50}}}} = 0.1{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{\text{)}}\);

\({\text{pH }}\left( { = – \log {\text{[}}{{\text{H}}^ + }{\text{]}} = – \log (0.10)} \right) = 1\);

(ii) 90%;

(i) \({\text{mol N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} = {\text{mol S}}{{\text{O}}_2} = 0.0400 \times 0.0200 = 0.000800\);

\(V = \frac{{n \times R \times T}}{P}/\frac{{0.000800 \times 8.31 \times 300}}{{{{10}^5}}}\);

\((1.99 \times {10^{ – 5}}{\text{ }}{{\text{m}}^3}) = 19.9{\text{ }}({\text{c}}{{\text{m}}^3})\);

Award [3] for correct final answer.

Accept 20.0 cm³ if R = 8.314 is used.

Award [2] for 17.9 cm³or 19.2 cm³(result from using molar volume at standard temperature and pressure or at room temperature and pressure).

\({\text{mol N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} = {\text{mol S}}{{\text{O}}_2} = 0.0400 \times 0.0200 = 0.000800\);

\(V = 0.00080 \times 2.24 \times {10^{ – 2}} \times \left[ {\frac{{1.00 \times {{10}^5}}}{{1.01 \times {{10}^5}}}} \right] \times \frac{{300}}{{273}}\);

\((1.95 \times {10^{ – 5}}{\text{ }}{{\text{m}}^3}) = 19.5{\text{ }}({\text{c}}{{\text{m}}^3})\);

Award [3] for correct final answer.

Deduct [1] for answers based on amount of HCl, so correct calculation would score [2 max].

(ii) sulfur dioxide is soluble in water;

Accept other reasonable responses based on sound chemistry.

Accept “syringe more accurate/precise” or “less gas escapes”.

Examiners report

This was quite a popular question, but responses were mixed. As in question 1, students struggled to answer questions with a strong practical context, with very few able to devise a mixture that would halve the concentration of thiosulfate, whilst keeping other concentrations constant, and responses for the need for similar beakers to be used were often too vague. Explanations of changes of rates in terms of the collision theory were generally successful but a significant number referred to the “number” rather than “frequency” of collisions. Many candidates were able to sketch Maxwell–Boltzmann distribution curves for the two temperatures, \({T_1}\) and \({T_2}\), but marks were lost due to careless omissions; the graphs did not start at the origin, were not labelled or the activation energy was missing. Many struggled to calculate the pH and many teachers have commented that this question was beyond what is expected at Standard Level and it is acknowledged that the question would have been more accessible if candidates had been asked to calculate the concentration of \({{\text{H}}^ + }\) ions and state the pH. In part (e) many students could quote and substitute into the ideal gas equation, correctly converting the temperature to Kelvin, but converting from \({{\text{m}}^{\text{3}}}\) to \({\text{c}}{{\text{m}}^{\text{3}}}\) posed a problem for most candidates. Although not necessary for the mark, as answers which referred to improved accuracy and precision were accepted, most candidates did not refer to the solubility of sulfur dioxide as a problem when using measuring cylinders to measure its volume.

Marks available	4
Reference code	14M.2.sl.TZ2.5

Question

A group of students investigated the rate of the reaction between aqueous sodium thiosulfate and hydrochloric acid according to the equation below.

\[{\text{N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3}{\text{(aq)}} + {\text{2HCl(aq)}} \to {\text{2NaCl(aq)}} + {\text{S}}{{\text{O}}_2}{\text{(g)}} + {\text{S(s)}} + {{\text{H}}_2}{\text{O(l)}}\]

The teacher asked the students to measure the effect of halving the concentration of sodium thiosulfate on the rate of reaction.

The teacher asked the students to devise another technique to measure the rate of this reaction.

Another group suggested collecting the sulfur dioxide and drawing a graph of the volume of gas against time.

[3]

(i) State the volumes of the liquids that should be mixed.

(ii) State why it is important that the students use a similar beaker for both reactions.

(iii) Explain, in terms of the collision theory, how decreasing the concentration of sodium thiosulfate would affect the time taken for the mark to be obscured.

[4]

(ii) Explain why increasing the temperature of the reaction mixture would significantly increase the rate of the reaction.

[6]

(i) One group suggested recording how long it takes for the pH of the solution to change by one unit. Calculate the initial pH of the original reaction mixture.

(ii) Deduce the percentage of hydrochloric acid that would have to be used up for the pH to change by one unit.

[3]

(ii) Suggest why it is better to use a gas syringe rather than collecting the gas in a measuring cylinder over water.

[4]

Markscheme

\({\text{mol N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}( = 2.50 \times 0.0200) = 0.0500\);

Allow 248.

\({\text{mass N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} \bullet {\text{5}}{{\text{H}}_2}{\text{O}} = (0.0500 \times 248.20) = 12.4{\text{ g}}\);

Award [3] for correct final answer.

Award [2] for 7.91g (water of crystallization omitted in \({M_r}\) calculation).

(i) ;

Accept other volumes in a 1:2:2 ratio.

(ii) depth of liquid in the beaker must remain constant / OWTTE;

Accept “same thickness of glass” and any other valid point, such as answers framed around minimizing uncontrolled variables / making it a “fair test”.

(iii) increases the time;

decrease in collision frequency/number of collisions per unit time;

Do not award mark for decrease in number of collisions.

(i)

labelled y-axis: number of particles / probability of particles (with that kinetic energy) and labelled x-axis: (kinetic) energy;

Allow fraction/proportion/amount of particles (with kinetic energy) for y-axis label.

Allow speed/velocity for x-axis label.

T₂ curve broader and with maximum lower and to right of T₁ curve;

Do not award this mark if both curves not asymmetric.

Curves must pass through the origin and be asymptotic to x axis.

Do not award this mark if curves not labelled.

\({E_{\text{a}}}\) marked on graph;

(ii) kinetic energy of molecules increases;

This may be answered implicitly in the final marking point.

frequency of collision/number of collisions per unit time increases;

Only penalize use of “number of collisions” if not penalized in (b)(iii).

greater proportion of molecules have energy greater than/equal to activation energy / rate related to temperature by the Arrhenius equation;

Award [1 max] for statements such as “there will be more successful collisions” if neither of last two marking points awarded.

(i) \({\text{[}}{{\text{H}}^ + }{\text{]}} = 0.5 \times \frac{{10}}{{{\text{50}}}} = 0.1{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{\text{)}}\);

\({\text{pH }}\left( { = – \log {\text{[}}{{\text{H}}^ + }{\text{]}} = – \log (0.10)} \right) = 1\);

(ii) 90%;

(i) \({\text{mol N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} = {\text{mol S}}{{\text{O}}_2} = 0.0400 \times 0.0200 = 0.000800\);

\(V = \frac{{n \times R \times T}}{P}/\frac{{0.000800 \times 8.31 \times 300}}{{{{10}^5}}}\);

\((1.99 \times {10^{ – 5}}{\text{ }}{{\text{m}}^3}) = 19.9{\text{ }}({\text{c}}{{\text{m}}^3})\);

Award [3] for correct final answer.

Accept 20.0 cm³ if R = 8.314 is used.

Award [2] for 17.9 cm³or 19.2 cm³(result from using molar volume at standard temperature and pressure or at room temperature and pressure).

\({\text{mol N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} = {\text{mol S}}{{\text{O}}_2} = 0.0400 \times 0.0200 = 0.000800\);

\(V = 0.00080 \times 2.24 \times {10^{ – 2}} \times \left[ {\frac{{1.00 \times {{10}^5}}}{{1.01 \times {{10}^5}}}} \right] \times \frac{{300}}{{273}}\);

\((1.95 \times {10^{ – 5}}{\text{ }}{{\text{m}}^3}) = 19.5{\text{ }}({\text{c}}{{\text{m}}^3})\);

Award [3] for correct final answer.

Deduct [1] for answers based on amount of HCl, so correct calculation would score [2 max].

(ii) sulfur dioxide is soluble in water;

Accept other reasonable responses based on sound chemistry.

Accept “syringe more accurate/precise” or “less gas escapes”.

Examiners report

Marks available	1
Reference code	14N.2.sl.TZ0.2

Question

The graph below shows pressure and volume data collected for a sample of carbon dioxide gas at 330 K.

Draw a best-fit curve for the data on the graph.

[1]

Deduce the relationship between the pressure and volume of the sample of carbon dioxide gas.

[1]

Use the data point labelled X to determine the amount, in mol, of carbon dioxide gas in the sample.

[3]

Markscheme

smooth curve through the data;

Do not accept a curve that passes through all of the points or an answer that joins the points using lines.

inversely proportional / \({\text{V}}\alpha \frac{{\text{1}}}{{\text{p}}}\) / \({\text{P}}\alpha \frac{{\text{1}}}{{\text{V}}}\) ;

Accept inverse/negative correlation/relationship.

Do not accept \({\text{V}} = \frac{{\text{1}}}{{\text{p}}}\) / \({\text{P}} = \frac{{\text{1}}}{{\text{V}}}\) or descriptions like “one goes up as other goes down” / OWTTE.

\(p = 21 \times {10^5}/2.1 \times {10^6}{\text{ (Pa)}}/2.1 \times {10^3}{\text{ (kPa)}}\) and

\(V = 50 \times {10^{ – 6}}/5.0 \times {10^{ – 5}}{\text{ }}({{\text{m}}^3})/5.0 \times {10^{ – 2}}{\text{ }}({\text{d}}{{\text{m}}^3})\);

\(\left( {n = \frac{{pV}}{{RT}} = } \right){\text{ }}\frac{{2.1 \times {{10}^6} \times 5.0 \times {{10}^{ – 5}}}}{{8.31 \times 330}}\);

\(n = 0.038{\text{ (mol)}}\);

Award [3] for correct final answer.

For M3 apply ECF for correct computation of the equation the student has written, unless more than one mistake is made prior this point.

Examiners report

Almost all candidates gained the mark for drawing a best-fit curve through the data points on the graph, though some insisted in trying to put a straight line through obviously non-linear data. Many students identified the inverse proportionality of pressure and volume in Part (b), though the terminology often lacked precision. Most students could identify the correct equation to use in Part (c) in order to calculate the amount of gas from the specified data point, though quite often they had problems with units, either as a result of incorrectly reading the axis on the graph or as a result of conversion.

Marks available	3
Reference code	14N.2.sl.TZ0.2

Question

The graph below shows pressure and volume data collected for a sample of carbon dioxide gas at 330 K.

Draw a best-fit curve for the data on the graph.

[1]

Deduce the relationship between the pressure and volume of the sample of carbon dioxide gas.

[1]

Use the data point labelled X to determine the amount, in mol, of carbon dioxide gas in the sample.

[3]

Markscheme

smooth curve through the data;

Do not accept a curve that passes through all of the points or an answer that joins the points using lines.

inversely proportional / \({\text{V}}\alpha \frac{{\text{1}}}{{\text{p}}}\) / \({\text{P}}\alpha \frac{{\text{1}}}{{\text{V}}}\) ;

Accept inverse/negative correlation/relationship.

Do not accept \({\text{V}} = \frac{{\text{1}}}{{\text{p}}}\) / \({\text{P}} = \frac{{\text{1}}}{{\text{V}}}\) or descriptions like “one goes up as other goes down” / OWTTE.

\(p = 21 \times {10^5}/2.1 \times {10^6}{\text{ (Pa)}}/2.1 \times {10^3}{\text{ (kPa)}}\) and

\(V = 50 \times {10^{ – 6}}/5.0 \times {10^{ – 5}}{\text{ }}({{\text{m}}^3})/5.0 \times {10^{ – 2}}{\text{ }}({\text{d}}{{\text{m}}^3})\);

\(\left( {n = \frac{{pV}}{{RT}} = } \right){\text{ }}\frac{{2.1 \times {{10}^6} \times 5.0 \times {{10}^{ – 5}}}}{{8.31 \times 330}}\);

\(n = 0.038{\text{ (mol)}}\);

Award [3] for correct final answer.

For M3 apply ECF for correct computation of the equation the student has written, unless more than one mistake is made prior this point.

Examiners report

Marks available	2
Reference code	14N.2.sl.TZ0.6

Question

A sample of magnesium contains three isotopes: magnesium-24, magnesium-25 and magnesium-26, with abundances of 77.44%, 10.00% and 12.56% respectively.

Phosphorus(V) oxide, \({{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}{\text{ }}({M_{\text{r}}} = 283.88)\), reacts vigorously with water \(({M_{\text{r}}} = 18.02)\), according to the equation below.

\[{{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}{\text{(s)}} + {\text{6}}{{\text{H}}_{\text{2}}}{\text{O(l)}} \to {\text{4}}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\]

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

[2]

a.i.

Predict the relative atomic radii of the three magnesium isotopes, giving your reasons.

[2]

a.iii.

Describe the bonding in magnesium.

[2]

State an equation for the reaction of magnesium oxide with water.

[1]

A student added 5.00 g of \({{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}\) to 1.50 g of water. Determine the limiting reactant, showing your working.

[2]

d.i.

Calculate the mass of phosphoric(V) acid, \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\), formed in the reaction.

[2]

d.ii.

State a balanced equation for the reaction of aqueous \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\) with excess aqueous sodium hydroxide, including state symbols.

[2]

d.iii.

State the formula of the conjugate base of \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\).

[1]

d.iv.

(i) Deduce the Lewis structure of \({\text{PH}}_4^ + \).

(ii) Predict, giving a reason, the bond angle around the phosphorus atom in \({\text{PH}}_4^ + \).

(iii) Predict whether or not the P–H bond is polar, giving a reason for your choice.

[4]

Markscheme

\(\left( {\frac{{(77.44 \times 24) + (10.00 \times 25) + (12.56{\text{ }}26)}}{{100}}} \right)\);

24.35;

Award [2] for correct final answer.

Two decimal places are required for M2.

Do not award any marks for 24.31 without showing method (as the value can be copied from the Data Booklet).

a.i.

same atomic radii / 160 pm;

isotopes only differ by number of neutrons/size of nucleus / radius

determined by electron shells and number of protons / OWTTE;

Accept neutrons do not affect distance of electrons / OWTTE.

a.iii.

(lattice of) positive ions/cations and mobile/free/delocalized electrons;

Accept “sea of electrons” instead of “delocalized electrons”.

Award M1 for a suitable diagram.

electrostatic attraction (between ions and delocalized electrons);

\({\text{MgO}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {\text{Mg(OH}}{{\text{)}}_{\text{2}}}/{\text{M}}{{\text{g}}^{2 + }} + {\text{2O}}{{\text{H}}^ – }\);

Accept reversible arrow.

\({{\text{P}}_4}{{\text{O}}_{10}}{\text{: }}\left( {\frac{{{\text{5.00}}}}{{{\text{283.88}}}} = } \right){\text{ 0.0176 (mol)}}\) and \({{\text{H}}_2}{\text{O: }}\left( {\frac{{{\text{1.50}}}}{{{\text{18.02}}}} = } \right){\text{ 0.0832 (mol)}}\);

\({{\text{H}}_2}{\text{O}}\) is the limiting reactant and reason related to stoichiometry;

d.i.

\(\frac{{0.0832 \times 4}}{6}/0.0555{\text{ (mol)}}\);

\((0.0555 \times 98.00 = ){\text{ }}5.44{\text{ g}}\);

The unit is needed for M2.

Award [2] for correct final answer.

Do not penalize slight numerical variations due to premature rounding.

d.ii.

\({{\text{H}}_3}{\text{P}}{{\text{O}}_4}{\text{(aq)}} + {\text{3NaOH(aq)}} \to {\text{N}}{{\text{a}}_3}{\text{P}}{{\text{O}}_4}{\text{(aq)}} + {\text{3}}{{\text{H}}_2}{\text{O(l)}}\)

correct products and balancing;

correct state symbols;

Accept valid ionic equations.

d.iii.

\({{\text{H}}_2}{\text{PO}}_4^ – \);

d.iv.

(i) ;

Accept dots, crosses or lines for pairs of electrons.

No need to distinguish the dative covalent bond from the other bonds.

Charge is required for the mark.

Do not penalize missing square brackets.

(ii) \(109^\circ 27’/109.5^\circ /109^\circ \);

4 electron domains/pairs/(negative) charge centres (around central atom/P);

Accept ion is tetrahedral / electron pairs/domains repel each other.

(iii) non-polar and P and H have the same electronegativity / OWTTE;

Accept slightly polar as precise electronegativities of P and H are not identical / OWTTE.

Examiners report

In Part (a) most candidates gained full marks, with the most common error being a failure to quote the answer to the precision specified, but the explanations of deflection, and more particularly detection, in the mass spectrometer were weak. The prediction of relative atomic radii of the isotopes, something that required the application of reason rather than recall, also proved much more challenging. Part (b) revealed that many candidates have a very weak understanding of the metallic bond with many thinking the bonding was ionic.

Even when they knew about a cation lattice and delocalized electrons, a mark was frequently dropped by failing to specify that the attraction between them was electrostatic. Most candidates wrote the correct equation in Part (c), but it is still disturbing that some students at this level cannot write even the most straightforward chemical equation. In Part (d) many students proved capable of carrying out routine stoichiometric calculations to identify the limiting reactant and use the result to find the mass of the product.

Even if the final result was incorrect quite frequently students gained some credit through the application of ECF. Only the better candidates could write an equation for the neutralisation of phosphoric(V) acid and even the routine derivation of a conjugate base from the formula of the acid proved difficult for many. In Part (e) most students could manage the correct Lewis structure, though some lost the mark through omitting the charge. Many candidates also scored well on the shape of the ion and the polarity of the P-H bond.

a.i.

a.iii.

d.i.

d.ii.

d.iii.

d.iv.

Marks available	2
Reference code	14N.2.sl.TZ0.6

Question

A sample of magnesium contains three isotopes: magnesium-24, magnesium-25 and magnesium-26, with abundances of 77.44%, 10.00% and 12.56% respectively.

\[{{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}{\text{(s)}} + {\text{6}}{{\text{H}}_{\text{2}}}{\text{O(l)}} \to {\text{4}}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\]

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

[2]

a.i.

Predict the relative atomic radii of the three magnesium isotopes, giving your reasons.

[2]

a.iii.

Describe the bonding in magnesium.

[2]

State an equation for the reaction of magnesium oxide with water.

[1]

A student added 5.00 g of \({{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}\) to 1.50 g of water. Determine the limiting reactant, showing your working.

[2]

d.i.

Calculate the mass of phosphoric(V) acid, \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\), formed in the reaction.

[2]

d.ii.

State a balanced equation for the reaction of aqueous \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\) with excess aqueous sodium hydroxide, including state symbols.

[2]

d.iii.

State the formula of the conjugate base of \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\).

[1]

d.iv.

(i) Deduce the Lewis structure of \({\text{PH}}_4^ + \).

(ii) Predict, giving a reason, the bond angle around the phosphorus atom in \({\text{PH}}_4^ + \).

(iii) Predict whether or not the P–H bond is polar, giving a reason for your choice.

[4]

Markscheme

\(\left( {\frac{{(77.44 \times 24) + (10.00 \times 25) + (12.56{\text{ }}26)}}{{100}}} \right)\);

24.35;

Award [2] for correct final answer.

Two decimal places are required for M2.

Do not award any marks for 24.31 without showing method (as the value can be copied from the Data Booklet).

a.i.

same atomic radii / 160 pm;

isotopes only differ by number of neutrons/size of nucleus / radius

determined by electron shells and number of protons / OWTTE;

Accept neutrons do not affect distance of electrons / OWTTE.

a.iii.

(lattice of) positive ions/cations and mobile/free/delocalized electrons;

Accept “sea of electrons” instead of “delocalized electrons”.

Award M1 for a suitable diagram.

electrostatic attraction (between ions and delocalized electrons);

\({\text{MgO}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {\text{Mg(OH}}{{\text{)}}_{\text{2}}}/{\text{M}}{{\text{g}}^{2 + }} + {\text{2O}}{{\text{H}}^ – }\);

Accept reversible arrow.

\({{\text{H}}_2}{\text{O}}\) is the limiting reactant and reason related to stoichiometry;

d.i.

\(\frac{{0.0832 \times 4}}{6}/0.0555{\text{ (mol)}}\);

\((0.0555 \times 98.00 = ){\text{ }}5.44{\text{ g}}\);

The unit is needed for M2.

Award [2] for correct final answer.

Do not penalize slight numerical variations due to premature rounding.

d.ii.

\({{\text{H}}_3}{\text{P}}{{\text{O}}_4}{\text{(aq)}} + {\text{3NaOH(aq)}} \to {\text{N}}{{\text{a}}_3}{\text{P}}{{\text{O}}_4}{\text{(aq)}} + {\text{3}}{{\text{H}}_2}{\text{O(l)}}\)

correct products and balancing;

correct state symbols;

Accept valid ionic equations.

d.iii.

\({{\text{H}}_2}{\text{PO}}_4^ – \);

d.iv.

(i) ;

Accept dots, crosses or lines for pairs of electrons.

No need to distinguish the dative covalent bond from the other bonds.

Charge is required for the mark.

Do not penalize missing square brackets.

(ii) \(109^\circ 27’/109.5^\circ /109^\circ \);

4 electron domains/pairs/(negative) charge centres (around central atom/P);

Accept ion is tetrahedral / electron pairs/domains repel each other.

(iii) non-polar and P and H have the same electronegativity / OWTTE;

Accept slightly polar as precise electronegativities of P and H are not identical / OWTTE.

Examiners report

a.i.

a.iii.

d.i.

d.ii.

d.iii.

d.iv.

Marks available	2
Reference code	14N.2.sl.TZ0.6

Question

A sample of magnesium contains three isotopes: magnesium-24, magnesium-25 and magnesium-26, with abundances of 77.44%, 10.00% and 12.56% respectively.

\[{{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}{\text{(s)}} + {\text{6}}{{\text{H}}_{\text{2}}}{\text{O(l)}} \to {\text{4}}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\]

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

[2]

a.i.

Predict the relative atomic radii of the three magnesium isotopes, giving your reasons.

[2]

a.iii.

Describe the bonding in magnesium.

[2]

State an equation for the reaction of magnesium oxide with water.

[1]

A student added 5.00 g of \({{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}\) to 1.50 g of water. Determine the limiting reactant, showing your working.

[2]

d.i.

Calculate the mass of phosphoric(V) acid, \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\), formed in the reaction.

[2]

d.ii.

State a balanced equation for the reaction of aqueous \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\) with excess aqueous sodium hydroxide, including state symbols.

[2]

d.iii.

State the formula of the conjugate base of \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\).

[1]

d.iv.

(i) Deduce the Lewis structure of \({\text{PH}}_4^ + \).

(ii) Predict, giving a reason, the bond angle around the phosphorus atom in \({\text{PH}}_4^ + \).

(iii) Predict whether or not the P–H bond is polar, giving a reason for your choice.

[4]

Markscheme

\(\left( {\frac{{(77.44 \times 24) + (10.00 \times 25) + (12.56{\text{ }}26)}}{{100}}} \right)\);

24.35;

Award [2] for correct final answer.

Two decimal places are required for M2.

Do not award any marks for 24.31 without showing method (as the value can be copied from the Data Booklet).

a.i.

same atomic radii / 160 pm;

isotopes only differ by number of neutrons/size of nucleus / radius

determined by electron shells and number of protons / OWTTE;

Accept neutrons do not affect distance of electrons / OWTTE.

a.iii.

(lattice of) positive ions/cations and mobile/free/delocalized electrons;

Accept “sea of electrons” instead of “delocalized electrons”.

Award M1 for a suitable diagram.

electrostatic attraction (between ions and delocalized electrons);

\({\text{MgO}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {\text{Mg(OH}}{{\text{)}}_{\text{2}}}/{\text{M}}{{\text{g}}^{2 + }} + {\text{2O}}{{\text{H}}^ – }\);

Accept reversible arrow.

\({{\text{H}}_2}{\text{O}}\) is the limiting reactant and reason related to stoichiometry;

d.i.

\(\frac{{0.0832 \times 4}}{6}/0.0555{\text{ (mol)}}\);

\((0.0555 \times 98.00 = ){\text{ }}5.44{\text{ g}}\);

The unit is needed for M2.

Award [2] for correct final answer.

Do not penalize slight numerical variations due to premature rounding.

d.ii.

\({{\text{H}}_3}{\text{P}}{{\text{O}}_4}{\text{(aq)}} + {\text{3NaOH(aq)}} \to {\text{N}}{{\text{a}}_3}{\text{P}}{{\text{O}}_4}{\text{(aq)}} + {\text{3}}{{\text{H}}_2}{\text{O(l)}}\)

correct products and balancing;

correct state symbols;

Accept valid ionic equations.

d.iii.

\({{\text{H}}_2}{\text{PO}}_4^ – \);

d.iv.

(i) ;

Accept dots, crosses or lines for pairs of electrons.

No need to distinguish the dative covalent bond from the other bonds.

Charge is required for the mark.

Do not penalize missing square brackets.

(ii) \(109^\circ 27’/109.5^\circ /109^\circ \);

4 electron domains/pairs/(negative) charge centres (around central atom/P);

Accept ion is tetrahedral / electron pairs/domains repel each other.

(iii) non-polar and P and H have the same electronegativity / OWTTE;

Accept slightly polar as precise electronegativities of P and H are not identical / OWTTE.

Examiners report

a.i.

a.iii.

d.i.

d.ii.

d.iii.

d.iv.

Marks available	2
Reference code	15M.2.sl.TZ1.1

Question

Ethanedioic acid is a diprotic acid. A student determined the value of x in the formula of hydrated ethanedioic acid, \({\text{HOOC}}\)–\({\text{COOH}} \bullet {\text{x}}{{\text{H}}_{\text{2}}}{\text{O}}\), by titrating a known mass of the acid with a \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) solution of \({\text{NaOH(aq)}}\).

0.795 g of ethanedioic acid was dissolved in distilled water and made up to a total volume of \({\text{250 c}}{{\text{m}}^{\text{3}}}\) in a volumetric flask.

\({\text{25 c}}{{\text{m}}^{\text{3}}}\) of this ethanedioic acid solution was pipetted into a flask and titrated against aqueous sodium hydroxide using phenolphthalein as an indicator.

The titration was then repeated twice to obtain the results below.

State the uncertainty of the volume of NaOH added in \({\text{c}}{{\text{m}}^{\text{3}}}\).

[1]

Calculate the average volume of NaOH added, in \({\text{c}}{{\text{m}}^{\text{3}}}\), in titrations 2 and 3, and then calculate the amount, in mol, of NaOH added.

[2]

(i) The equation for the reaction taking place in the titration is:

\({\text{HOOC}}\)−\({\text{COOH(aq)}} + {\text{2NaOH(aq)}} \to {\text{NaOOC}}\)−\({\text{COONa(aq)}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{O(l)}}\)

Determine the amount, in mol, of ethanedioic acid that reacts with the average volume of \({\text{NaOH(aq)}}\).

(ii) Determine the amount, in mol, of ethanedioic acid present in \({\text{250 c}}{{\text{m}}^{\text{3}}}\) of the original solution.

(ii) Determine the molar mass of hydrated ethanedioic acid.

(iv) Determine the value of x in the formula \({\text{HOOC}}\)−\({\text{COOH}} \bullet {\text{x}}{{\text{H}}_{\text{2}}}{\text{O}}\).

[5]

Identify the strongest intermolecular force in solid ethanedioic acid.

[1]

Deduce the Lewis (electron dot) structure of ethanedioic acid, HOOC−COOH.

[1]

Markscheme

\(( \pm )0.10{\text{ (c}}{{\text{m}}^{\text{3}}}{\text{)}}\);

Accept ±0.1 (cm³).

Accept (±)0.09 (cm³) (based on more accurate method of calculating propagation of uncertainties).

\(\left( {\frac{{12.70 + 12.50}}{2} = } \right)12.60{\text{ (c}}{{\text{m}}^{\text{3}}}{\text{)}}\);

\((0.01260 \times 0.100 = )1.26 \times {10^{ – 3}}{\text{ (mol)}}\);

Award [2] for correct final answer.

(i) \(\left( {\frac{{1.26 \times {{10}^{ – 3}}}}{2} = } \right)6.30 \times {10^{ – 4}}{\text{ (mol)}}\);

(ii) \((6.30 \times {10^{ – 4}} \times 10 = )6.30 \times {10^{ – 3}}{\text{ (mol)}}\);

(iii) \(\left( {\frac{{0.795}}{{6.30 \times {{10}^{ – 3}}}} = } \right)126{\text{ (gmo}}{{\text{l}}^{ – 1}})\);

(iv) \({M_{\text{r}}}{\text{(}}{{\text{C}}_2}{{\text{H}}_2}{{\text{O}}_4}{\text{)}} = 90.04\) and \({M_{\text{r}}}{\text{(}}{{\text{H}}_2}{\text{O)}} = 18.02\);

\({\text{x}} = 2\);

Accept integer values for M_r’s of 90 and 18 and any reasonable calculation.

Award [1 max] if no working shown.

hydrogen bonding;

;

Mark cannot be scored if lone pairs are missing on oxygens.

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

Examiners report

This beginning of this question to state the uncertainty and to calculate the average volume added were well done and most students could also calculate the number of moles added. However, many candidates began to lose marks from this point onwards. Some could identify the ratio and correctly state the moles of ethanedioic acid, but fewer realized they needed to multiply 10 to get back to the original solution. The next step to calculate the \({M_{\text{r}}}\) was only correctly completed by a handful of students. Those that were correct with the molar mass always could calculate the moles of water, many students just guessed an answer though.

The intermolecular force was correctly described as hydrogen bonding, however there were some instances when it seemed unclear whether students realized this was between molecules and instead they seemed to suggest it was a bond between hydrogen and oxygen in the molecule. Some candidates could correctly draw the Lewis structure but a number of those lost marks for omitting the lone pairs on oxygen.

The intermolecular force was correctly described as hydrogen bonding, however there was some instances when it seemed unclear whether students realized this was between molecules and instead they seemed to suggest it was a bond between hydrogen and oxygen in the molecule. Some candidates could correctly draw the Lewis structure but a number of those lost marks for omitting the lone pairs on oxygen.

Marks available	5
Reference code	15M.2.sl.TZ1.1

Question

0.795 g of ethanedioic acid was dissolved in distilled water and made up to a total volume of \({\text{250 c}}{{\text{m}}^{\text{3}}}\) in a volumetric flask.

\({\text{25 c}}{{\text{m}}^{\text{3}}}\) of this ethanedioic acid solution was pipetted into a flask and titrated against aqueous sodium hydroxide using phenolphthalein as an indicator.

The titration was then repeated twice to obtain the results below.

State the uncertainty of the volume of NaOH added in \({\text{c}}{{\text{m}}^{\text{3}}}\).

[1]

Calculate the average volume of NaOH added, in \({\text{c}}{{\text{m}}^{\text{3}}}\), in titrations 2 and 3, and then calculate the amount, in mol, of NaOH added.

[2]

(i) The equation for the reaction taking place in the titration is:

\({\text{HOOC}}\)−\({\text{COOH(aq)}} + {\text{2NaOH(aq)}} \to {\text{NaOOC}}\)−\({\text{COONa(aq)}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{O(l)}}\)

Determine the amount, in mol, of ethanedioic acid that reacts with the average volume of \({\text{NaOH(aq)}}\).

(ii) Determine the amount, in mol, of ethanedioic acid present in \({\text{250 c}}{{\text{m}}^{\text{3}}}\) of the original solution.

(ii) Determine the molar mass of hydrated ethanedioic acid.

(iv) Determine the value of x in the formula \({\text{HOOC}}\)−\({\text{COOH}} \bullet {\text{x}}{{\text{H}}_{\text{2}}}{\text{O}}\).

[5]

Identify the strongest intermolecular force in solid ethanedioic acid.

[1]

Deduce the Lewis (electron dot) structure of ethanedioic acid, HOOC−COOH.

[1]

Markscheme

\(( \pm )0.10{\text{ (c}}{{\text{m}}^{\text{3}}}{\text{)}}\);

Accept ±0.1 (cm³).

Accept (±)0.09 (cm³) (based on more accurate method of calculating propagation of uncertainties).

\(\left( {\frac{{12.70 + 12.50}}{2} = } \right)12.60{\text{ (c}}{{\text{m}}^{\text{3}}}{\text{)}}\);

\((0.01260 \times 0.100 = )1.26 \times {10^{ – 3}}{\text{ (mol)}}\);

Award [2] for correct final answer.

(i) \(\left( {\frac{{1.26 \times {{10}^{ – 3}}}}{2} = } \right)6.30 \times {10^{ – 4}}{\text{ (mol)}}\);

(ii) \((6.30 \times {10^{ – 4}} \times 10 = )6.30 \times {10^{ – 3}}{\text{ (mol)}}\);

(iii) \(\left( {\frac{{0.795}}{{6.30 \times {{10}^{ – 3}}}} = } \right)126{\text{ (gmo}}{{\text{l}}^{ – 1}})\);

(iv) \({M_{\text{r}}}{\text{(}}{{\text{C}}_2}{{\text{H}}_2}{{\text{O}}_4}{\text{)}} = 90.04\) and \({M_{\text{r}}}{\text{(}}{{\text{H}}_2}{\text{O)}} = 18.02\);

\({\text{x}} = 2\);

Accept integer values for M_r’s of 90 and 18 and any reasonable calculation.

Award [1 max] if no working shown.

hydrogen bonding;

;

Mark cannot be scored if lone pairs are missing on oxygens.

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

Examiners report

Marks available	4
Reference code	15M.2.sl.TZ1.5

Question

Ethanol has many industrial uses.

(i) State an equation for the formation of ethanol from ethene and the necessary reaction conditions

Equation:

Conditions:

(ii) Deduce the volume of ethanol, in dm³, produced from \({\text{1.5 d}}{{\text{m}}^{\text{3}}}\) of ethene, assuming both are gaseous and at the same temperature and pressure.

[4]

Define the term average bond enthalpy.

[2]

b.i.

Ethanol can be used as a fuel. Determine the enthalpy of combustion of ethanol at 298 K, in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\), using the values in table 10 of the data booklet, assuming all reactants and products are gaseous.

[4]

b.ii.

Suggest why the value of the enthalpy of combustion of ethanol quoted in table 12 of the data booklet is different to that calculated using bond enthalpies.

[1]

b.iii.

Explain why the reaction is exothermic in terms of the bonds involved.

[1]

b.iv.

Identify the homologous series to which ethanol belongs and state two features of a homologous series.

[3]

Markscheme

(i) Equation:

\({\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}/{{\text{C}}_2}{{\text{H}}_4} + {{\text{H}}_2}{\text{O}} \to {{\text{C}}_2}{{\text{H}}_5}{\text{OH}}\);

Conditions:

(concentrated) sulfuric acid/\({{\text{H}}_2}{\text{S}}{{\text{O}}_{\text{4}}}\);

Do not accept dilute sulfuric acid.

Accept phosphoric acid/H₃PO₄ (on pellets of silicon dioxide) (for industrial preparation).

heat / high temperature;

Do not accept warm.

Accept high pressure (for industrial preparation) for M3 only if H₃PO₄ is given for M2.

(ii) \({\text{1.5 (d}}{{\text{m}}^{\text{3}}}{\text{)}}\);

energy needed to break (1 mol of) a bond in the gaseous state/phase;

(averaged over) similar compounds;

Do not accept “similar bonds” instead of “similar compounds”.

Concept of “similar” is important for M2.

b.i.

\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}} + {\text{3}}{{\text{O}}_2} \to {\text{2C}}{{\text{O}}_2} + {\text{3}}{{\text{H}}_2}{\text{O}}\);

Bonds broken:

\({\text{347}} + {\text{(5}} \times {\text{413)}} + {\text{358}} + {\text{464}} + {\text{(3}} \times {\text{498)}}/{\text{4728 (kJ)}}/\)

C–C + 5C–H + C–O + O–H + 3O=O;

Bonds made:

\((4 \times 746) + (6 \times 464)/5768{\text{ (kJ)}}/\) 4C=O + 6O—H;

\(\Delta H = (4728 – 5768 = ) – 1040{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\) / bonds broken − bonds formed;

Award [4] for correct final answer.

Award [3] for (+)1040 (kJ mol^–1).

b.ii.

ethanol and water are liquids / not all molecules are gaseous / in enthalpy of combustion molecules are in their standard states / bond enthalpies are average values;

Do not accept answer “ethanol/water is a liquid” alone.

b.iii.

less energy required to break bonds in reactants than is released when the bonds in products form / bonds stronger (overall) in products/weaker (overall) in reactants;

b.iv.

alcohols / alkanols;

Any two of the following for [2 max]:

differ by CH₂/methylene (unit);

similar chemical properties;

gradually changing physical properties;

same general formula;

same functional group;

Do not accept “same” instead of “similar”, or vice-versa.

Examiners report

This was not a popular question with few candidates choosing it. Some who chose it did very well but most scored poorly. Students needed to write an equation for the hydration of ethene which was generally answered well and then state the conditions, which were less well known. Applying Avogadro’s law to work out the volume of ethanol was only correctly answered by a few. The definition for bond enthalpy was not well known, however many candidates could calculate the energy change using bond enthalpies with some success although there were few completely correct answers as bonds were forgotten or incorrectly multiplied.

This was not a popular question with few candidates choosing it. Some who chose it did very well but most scored poorly. Students needed to write an equation for the hydration of ethene which was generally answered well and then state the conditions, which were less well known. Applying Avogadro’s law to work out the volume of ethanol was only correctly answered by a few. The definition for bond enthalpy was not well known, however many candidates could calculate the energy change using bond enthalpies with some success, although there were few completely correct answers as bonds were forgotten or incorrectly multiplied.

b.i.

This was not a popular question with few candidates choosing it. Some who chose it did very well but most scored poorly. Students needed to write an equation for the hydration of ethene which was generally answered well and then state the conditions, which were less well known. Applying Avogadro’s law to work out the volume of ethanol was only correctly answered by a few. The definition for bond enthalpy was not well known, however many candidates could calculate the energy change using bond enthalpies with some success, although there were few completely correct answers as bonds were forgotten or incorrectly multiplied.

b.ii.

This was not a popular question with few candidates choosing it. Some who chose it did very well but most scored poorly. Students needed to write an equation for the hydration of ethene which was generally answered well and then state the conditions, which were less well known. Applying Avogadro’s law to work out the volume of ethanol was only correctly answered by a few. The definition for bond enthalpy was not well known, however many candidates could calculate the energy change using bond enthalpies with some success, although there were few completely correct answers as bonds were forgotten or incorrectly multiplied.

b.iii.

This was not a popular question with few candidates choosing it. Some who chose it did very well but most scored poorly. Students needed to write an equation for the hydration of ethene which was generally answered well and then state the conditions, which were less well known. Applying Avogadro’s law to work out the volume of ethanol was only correctly answered by a few. The definition for bond enthalpy was not well known, however many candidates could calculate the energy change using bond enthalpies with some success, although there were few completely correct answers as bonds were forgotten or incorrectly multiplied.

b.iv.

This was not a popular question with few candidates choosing it. Some who chose it did very well but most scored poorly. Students needed to write an equation for the hydration of ethene which was generally answered well and then state the conditions, which were less well known. Applying Avogadro’s law to work out the volume of ethanol was only correctly answered by a few. The definition for bond enthalpy was not well known, however many candidates could calculate the energy change using bond enthalpies with some success, although there were few completely correct answers as bonds were forgotten or incorrectly multiplied.

Marks available	4
Reference code	15M.2.sl.TZ1.7

Question

A hydrocarbon has the empirical formula \({{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\). When 1.17 g of the compound is heated to 85 °C at a pressure of 101 kPa it occupies a volume of \({\text{400 c}}{{\text{m}}^{\text{3}}}\).

(i) Calculate the molar mass of the compound, showing your working.

(ii) Deduce the molecular formula of the compound.

[4]

\({{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}\) exists as three isomers. Identify the structure of the isomer with the lowest boiling point and explain your choice.

[2]

Ethanol is a primary alcohol that can be oxidized by acidified potassium dichromate(VI). Distinguish between the reaction conditions needed to produce ethanal and ethanoic acid.

Ethanal:

Ethanoic acid:

[2]

c.i.

Determine the oxidation number of carbon in ethanol and ethanal.

Ethanol:

Ethanal:

[2]

c.ii.

Deduce the half-equation for the oxidation of ethanol to ethanal.

[1]

c.iii.

Deduce the overall redox equation for the reaction of ethanol to ethanal with acidified potassium dichromate(VI) by combining your answer to part (c) (iii) with the following half-equation:

\[{\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_{\text{7}}^{2 – }{\text{(aq)}} + {\text{14}}{{\text{H}}^ + }{\text{(aq)}} + {\text{6}}{{\text{e}}^ – } \to {\text{2C}}{{\text{r}}^{3 + }}{\text{(aq)}} + {\text{7}}{{\text{H}}_{\text{2}}}{\text{O(l)}}\]

[2]

c.iv.

Describe two characteristics of a reaction at equilibrium.

[2]

d.i.

Describe how a catalyst increases the rate of a reaction.

[2]

d.ii.

State and explain the effect of a catalyst on the position of equilibrium.

[2]

d.iii.

Ethanoic acid reacts with ethanol to form the ester ethyl ethanoate.

\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH(l)}} + {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH(l)?????C}}{{\text{H}}_{\text{3}}}{\text{COOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{(l)}} + {{\text{H}}_{\text{2}}}{\text{O(l)}}\]

The esterification reaction is exothermic. State the effect of increasing temperature on the value of the equilibrium constant (\({K_{\text{c}}}\)) for this reaction.

[1]

Markscheme

(i) \({\text{temperature}} = 358{\text{ K}}\);

\(M = \frac{{mRT}}{{pV}}/1.17 \times 8.31 \times \frac{{358}}{{(0.40 \times 101)}}\);

\({\text{(}}M = {\text{) }}86.2{\text{ (gmo}}{{\text{l}}^{ – 1}}{\text{)}}\);

Award [1 max] for correct final answer without working.

(ii) \({{\text{C}}_6}{{\text{H}}_{14}}\);

\({\text{C(C}}{{\text{H}}_3}{{\text{)}}_4}\);

Accept correct name 2,2-dimethylpropane.

Do not penalize missing H atoms.

weakest London/dispersion/van der Waals’/vdW/instantaneous induced dipoleinduced dipole forces because of smallest surface area/contact

weakest London/dispersion/van der Waals’/vdW/ instantaneous induced dipoleinduced dipole forces because of least distortion of the electron cloud

weakest London/dispersion/van der Waals’/vdW/ instantaneous induced dipoleinduced dipole forces because polarizability of electrons (in electron cloud) is less;

Accept other words to that effect but student must mention a correct IMF and a correct reason.

Ethanal: distill off product as it forms;

Accept distillation.

Ethanoic acid: (heat under) reflux / use excess oxidizing agent;

c.i.

Ethanol: –2/–II;

Ethanal: –1/–I;

Do not accept 2– or 1–, but penalize only once.

c.ii.

\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}} \to {\text{C}}{{\text{H}}_3}{\text{CHO}} + {\text{2}}{{\text{H}}^ + } + {\text{2}}{{\text{e}}^ – }\);

Half-equation required. Do not accept \({C_2}{H_5}OH + 2[O] \to C{H_3}CHO + {H_2}O\).

Accept e for \({e^ – }\).

c.iii.

\({\text{3C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH(aq)}} + {\text{C}}{{\text{r}}_2}{\text{O}}_7^{2 – }{\text{(aq)}} + {\text{8}}{{\text{H}}^ + }{\text{(aq)}} \to {\text{2C}}{{\text{r}}^{{\text{3}} + }}{\text{(aq)}} + {\text{3C}}{{\text{H}}_3}{\text{CHO(l)}} + {\text{7}}{{\text{H}}_2}{\text{O(l)}}\)

correct reactants and products;

correct balancing;

M2 can only be scored if M1 correct.

Ignore state symbols.

c.iv.

rate of forward process/reaction = rate of backward/reverse process/reaction;

concentrations of reactants and products remain constant;

no change in macroscopic properties;

closed/isolated system / constant matter/energy;

d.i.

provides alternative pathway (of lower energy);

lowers activation energy (of the reaction) / more particles with \(E \geqslant {E_{\text{a}}}\);

d.ii.

no effect (on position of equilibrium);

increases rate of forward and reverse reactions (equally);

d.iii.

decreases;

Examiners report

This was the least popular question however many who chose it were successful in parts. Part (a) that required a calculation of \({M_{\text{r}}}\) was quite well done. However (b) that asked for the isomer of \({{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}\) with the lowest boiling point was not well answered. Identification of the methods to produce ethanal or ethanoic acid was done well by the strong candidates and others just guessed. Deduction of oxidation numbers and then writing of redox equations was not well answered. However (d) and (e) about equilibrium were answered well by many candidates although there were again some very poor answers.

c.i.

c.ii.

c.iii.

c.iv.

d.i.

d.ii.

d.iii.

Marks available	2
Reference code	15M.2.sl.TZ2.1

Question

A student carried out an experiment to determine the concentration of a hydrochloric acid solution and the enthalpy change of the reaction between aqueous sodium hydroxide and this acid by thermometric titration.

She added \({\text{5.0 c}}{{\text{m}}^{\text{3}}}\) portions of hydrochloric acid to \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sodium hydroxide solution in a glass beaker until the total volume of acid added was \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\), measuring the temperature of the mixture each time. Her results are plotted in the graph below.

The initial temperature of both solutions was the same.

By drawing appropriate lines, determine the volume of hydrochloric acid required to completely neutralize the \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of sodium hydroxide solution.

[2]

a.i.

Determine the concentration of the hydrochloric acid, including units.

[2]

a.ii.

Determine the change in temperature, \(\Delta T\).

[1]

b.i.

Calculate the enthalpy change, in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\), for the reaction of hydrochloric acid and sodium hydroxide solution.

[3]

b.ii.

The accepted theoretical value from the literature of this enthalpy change is \( – 58{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}\). Calculate the percentage error correct to two significant figures.

[1]

b.iii.

Suggest the major source of error in the experimental procedure and an improvement that could be made to reduce it.

[2]

b.iv.

Markscheme

drawing best-fit straight lines to show volume;

There should be approximately the same number of points above and below for both lines.

\({\text{27.0 (c}}{{\text{m}}^{\text{3}}}{\text{)}}\);

Accept any value in the range 26.0 to 28.0 (cm³) if consistent with student’s annotation on the graph.

Accept ECF for volumes in the range 27.0–30.0 cm³if it corresponds to maximum temperature of line drawn.

Volumes should be given to one decimal place.

a.i.

\({\text{[HCl]}} = \frac{{1.00 \times 0.0250}}{{0.0270}}\);

\( = 0.926{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\);

Volume of 26.0 gives [HCl] = 0.962 mol dm^–3. Volume of 28.0 gives [HCl] = 0.893 mol dm^–3

Award [2] for correct final answer with units.

Award [1 max] for correct concentration without units.

Accept M, mol L^–1, mol/dm³as units.

a.ii.

\((30.2 – 25.0 = )( + )5.2{\rm{ (^\circ C/K)}}\);

Any accepted value must be consistent with student’s annotation on the graph but do not accept \(\Delta T < 5.1\).

Accept \(( + )5.6{\rm{ (^\circ C/K) }}\) (ie, taking into account heat loss and using T when volume = 0.0 cm³).

b.i.

\({\text{Q}} = (m \times c \times \Delta T = (25.0 + 27.0) \times 4.18 \times 5.2 = 1130.272{\text{ J}} = )1.13{\text{ (kJ)}}\);

\(n = (1.00 \times 0.0250 = )0.0250{\text{ (mol)}}\);

\(\Delta H = \left( { – \frac{Q}{n} = – 45210.88{\text{ J}}\,{\text{mo}}{{\text{l}}^{ – 1}} = } \right) – 45{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);

Award [3] for correct final answer.

Award [2] for +45 (kJ mol ^–1).

Apply ECF for M3 even if both m and \(\Delta T\) are incorrect in M1.

Accept use of c = 4.2 Jg^–1K^–1.

b.ii.

\(\left( {\left| {\frac{{ – 45 – ( – 58)}}{{( – 58)}}} \right| \times 100 = } \right)22(\% )\);

Answer must be given to two significant figures.

Ignore sign.

b.iii.

heat losses;

better (thermal) insulation / using a polystyrene cup / putting a lid on the beaker;

Accept other suitable methods for better thermal insulation, but do not accept just “use a calorimeter” without reference to insulation.

b.iv.

Examiners report

Some teachers commented that thermometric titrations are not listed in the syllabus nor are they included as prescribed experiments for the new guide. A similar question was asked in a past examination and thermometric titrations are covered in Topic 5. The intention is that any data based questions should be accessible to all students, who have the appropriate practical experience. It is not intended that such questions will be constrained to experiments on this list. Most candidates were not able to access the first mark with by construction of lines of best fit. Some drew a ‘dot to dot’ curve, but with most just providing a construction line dropping down from the maximum point on the graph, which did allow them to access the second mark. There was some transferred error for 1aii), but many were not able to carry out the calculation. Scoring for the temperature difference was dependent upon on the candidate’s annotations, with a few extending the line of best fit back to the y axis. In the calculation of enthalpy change, the total mass of the solutions was often incorrect, but some salvaged the subsequent marks. The calculation of percentage error was generally done well, but a good third of the candidates failed to read the question stem and did not give the answer to two significant figures. The concept of heat loss in the experiment was well understood, but the solution was very often too vague.

a.i.

a.ii.

b.i.

b.ii.

b.iii.

b.iv.

Marks available	3
Reference code	15M.2.sl.TZ2.6

Question

Draw the Lewis (electron dot) structure of chloromethane.

[1]

a.i.

Predict the shape of the chloromethane molecule and the H–C–H bond angle.

Shape:

Bond angle:

[2]

a.ii.

Explain why chloromethane is a polar molecule.

[2]

a.iii.

Methanol has a lower molar mass than chloromethane. Explain why the boiling point of methanol is higher than that of chloromethane.

[2]

a.iv.

State the equation for the reaction between potassium and chlorine.

[1]

b.i.

Outline the nature of the metallic bonding present in potassium.

[1]

b.ii.

Describe the covalent bond present in the chlorine molecule and how it is formed.

[2]

b.iii.

Describe the ionic bonding present in potassium chloride and how the ions are formed.

[2]

b.iv.

Potassium also reacts with water to form hydrogen gas. Determine the volume, in \({\text{c}}{{\text{m}}^{\text{3}}}\), of hydrogen gas that could theoretically be produced at 273 K and \(1.01 \times {10^5}{\text{ Pa}}\) when 0.0587 g of potassium reacts with excess water.

[3]

b.v.

Identify the acid-base character of the oxides of each of the elements from sodium to chlorine in period 3.

[2]

c.i.

State the equations for the separate reactions of sodium oxide and phosphorus(V) oxide with water.

[2]

c.ii.

Markscheme

;

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

a.i.

Shape: tetrahedral;

Bond angle: accept any value in the range: 108° to 111°;

(Literature value is 108.2°).

a.ii.

Cl is more electronegative than C / C–Cl bond polar;

bond dipoles do not cancel / asymmetric distribution of electron cloud / (resultant) net dipole moment (from vectorial addition of bond dipoles) going in direction of C–Cl axis / OWTTE;

a.iii.

hydrogen bonding in methanol;

stronger than dipole-dipole/van der Waals’ attractions/forces in chloromethane;

Accept converse argument.

a.iv.

\({\text{2K(s)}} + {\text{C}}{{\text{l}}_2}{\text{(g)}} \to {\text{2KCl(s)}}\);

Ignore state symbols.

b.i.

(electrostatic) attraction between lattice of cations/positive ions and delocalized electrons;

b.ii.

(electrostatic) attraction between positively charged nuclei and a pair of electrons;

formed as a result of electron sharing;

b.iii.

(electrostatic) attraction between positive and negative ions/oppositely charged ions/cations and anions;

formed as a result of transfer of an electron from a K atom to a Cl atom / OWTTE;

b.iv.

amount of potassium \( = \left( {\frac{{0.0587}}{{39.10}} = } \right){\text{ }}1.5 \times {10^{ – 3}}{\text{ (mol)}}\);

\({\text{2K}} + {\text{2}}{{\text{H}}_2}{\text{O}} \to {\text{2KO}} + {{\text{H}}_2}\) / amount of hydrogen \( = 7.50 \times {10^{ – 4}}{\text{ (mol)}}\);

volume of hydrogen \( = (7.50 \times {10^{ – 4}} \times 22.4 \times 1000 = ){\text{ }}16.8{\text{ (c}}{{\text{m}}^3}{\text{)}}\);

Accept calculation of volume of hydrogen using PV = nRT (answer is 16.9 cm³).

Award [3] for correct final answer.

b.v.

Na, Mg (oxides): basic

Al (oxide): amphoteric

Do not accept amphiprotic.

Si to Cl (oxides): acidic

Award [2] for all three listed sets correct.

Award [1] for one or two listed sets correct.

Award [1] for stating oxides become more acidic towards right/Cl or more basic towards left/Na.

Do not penalize if reference is to Ar instead of Cl.

Do not penalize for incorrect formulas of oxides.

c.i.

\({\text{N}}{{\text{a}}_2}{\text{O(s)}} + {{\text{H}}_2}{\text{O(l)}} \to {\text{2NaOH(aq)}}\);

\({{\text{P}}_4}{{\text{O}}_{10}}{\text{(s)}} + {\text{6}}{{\text{H}}_2}{\text{O(l)}} \to {\text{4}}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}{\text{(aq)}}\);

Ignore state symbols.

Accept P₂O₅(s) + 3H₂O(l) \( \to \) 2H₃PO₄(aq).

Do not award marks if incorrect formulas of the oxides are used.

c.ii.

Examiners report

Probably the least popular option. The drawing of the diagram of chloromethane was generally excellent, as was the prediction/recall of the shape and bond angle. With the reasons for polarity, the concept of bond polarity was well understood, but the idea of asymmetry resulting in a dipole was less clearly appreciated. The construction of the chemical equation was disappointing, as was the description of the three types of bonding, very often missing the important point, in that they are attractions. With the calculation of volume of hydrogen, it was quite rare to get a fully correct answer. The biggest error was to use an incorrect value for the number of moles of hydrogen in the equation \(pV = nRT\), by failing to halve the moles of hydrogen. The use of \(pV = nRT\) also caused problems with units. The acid base nature of oxides of a period were generally well known. In contrast, the construction or recall of correct chemical equations for the reaction with water was a weakness.

a.i.

a.ii.

a.iii.

a.iv.

b.i.

b.ii.

b.iii.

b.iv.

b.v.

c.i.

c.ii.

Marks available	4
Reference code	16M.2.sl.TZ0.1

Question

Phosphine (IUPAC name phosphane) is a hydride of phosphorus, with the formula PH₃.

(i) Draw a Lewis (electron dot) structure of phosphine.

(ii) Outline whether you expect the bonds in phosphine to be polar or non-polar, giving a brief reason.

(iii) Explain why the phosphine molecule is not planar.

(iv) Phosphine has a much greater molar mass than ammonia. Explain why phosphine has a significantly lower boiling point than ammonia.

[6]

Phosphine is usually prepared by heating white phosphorus, one of the allotropes of phosphorus, with concentrated aqueous sodium hydroxide. The equation for the reaction is:

P4 (s) + 3OH⁻ (aq) + 3H₂O (l) → PH₃ (g) + 3H₂PO₂⁻ (aq)

(i) Identify one other element that has allotropes and list two of its allotropes.

Element:

Allotrope 1:

Allotrope 2:

(ii) The first reagent is written as P₄, not 4P. Describe the difference between P₄ and 4P.

(iii) The ion H₂PO₂⁻ is amphiprotic. Outline what is meant by amphiprotic, giving the formulas of both species it is converted to when it behaves in this manner.

(iv) State the oxidation state of phosphorus in P₄ and H₂PO₂⁻.

P₄:

H₂PO₂⁻:

(v) Oxidation is now defined in terms of change of oxidation number. Explore how earlier definitions of oxidation and reduction may have led to conflicting answers for the conversion of P₄ to H₂PO₂⁻ and the way in which the use of oxidation numbers has resolved this.

[10]

2.478 g of white phosphorus was used to make phosphine according to the equation:

P₄(s) +3OH⁻(aq)+3H₂O(l) → PH₃(g)+3H₂PO₂⁻(aq)

(i) Calculate the amount, in mol, of white phosphorus used.

(ii) This phosphorus was reacted with 100.0 cm³ of 5.00 mol dm⁻³ aqueous sodium hydroxide. Deduce, showing your working, which was the limiting reagent.

(iii) Determine the excess amount, in mol, of the other reagent.

(iv) Determine the volume of phosphine, measured in cm³ at standard temperature and pressure, that was produced.

[4]

Markscheme

(i)

Accept structures using dots and/or crosses to indicate bonds and/or lone pair.

(ii)
non-polar AND P and H have the same electronegativity

Accept “similar electronegativities”.
Accept “polar” if there is a reference to a small difference in electronegativity and apply ECF in 1 a (iv).

(iii)
4 electron domains/pairs/negative charge centres «around the central atom»
OR
a lone/non-bonding pair «and three bonding pairs around the central atom»
repulsion between electron domains/pairs/negative charge centres «produces non-planar shape»
OR
«repulsion causes» tetrahedral orientation/pyramidal shape

(iv)
PH₃ has London «dispersion» forces

NH₃ forms H-bonds

H-bonds are stronger
OR
London forces are weaker

Accept van der Waals’ forces, dispersion forces and instantaneous dipole – induced dipole forces.
Accept “dipole-dipole forces” as molecule is polar.

H-bonds in NH₃ (only) must be mentioned to score [2].
Do not award M2 or M3 if:
• implies covalent bond is the H-bond
• implies covalent bonds break.

Accept “dipole-dipole forces are weaker”.

(i)
Element
carbon/C
OR
oxygen/O/O₂

Allotropes
Award [1] for two of:
diamond
graphite
graphene
C₆₀ / buckminsterfullerene
OR
ozone/O₃ AND «diatomic/molecular» oxygen/O₂

Accept two correctly named allotropes of any other named element (S, Se, Sn, As, etc.).

Accept fullerene, “buckyballs” etc. instead of buckminsterfullerene.

(ii)
P₄ is a molecule «comprising 4P atoms» AND 4P is four/separate «P» atoms
OR
P₄ represents «4P» atoms bonded together AND 4P represents «4» separate/non-bonded «P» atoms

(iii)

can act as both a «Brønsted–Lowry» acid and a «Brønsted–Lowry» base
OR
can accept and/or donate a hydrogen ion/proton/H⁺

HPO2²⁻ AND H₃PO₂

(iv)

P₄ : 0
H₂PO₂⁻ : +1

Do not accept 1 or 1+ for H₂PO₂⁻.

(v)

oxygen gained, so could be oxidation

hydrogen gained, so could be reduction
OR
negative charge «on product/H₂PO₂⁻ » /gain of electrons so could be reduction

oxidation number increases so must be oxidation

Award [1 max] for M1 and M2 if candidate displays knowledge of at least two of these definitions but does not apply them to the reaction.
Do not award M3 for “oxidation number changes”.

(i)

«\(\left\langle {\frac{{2.478}}{{4 \times 30.97}}} \right\rangle \)»= 0.02000«mol»

(ii)
n(NaOH)=«0.1000×5.00=»0.500«mol» AND P₄/phosphorus is limiting reagent

Accept n(H₂O) =\(\frac{{100}}{{18}}\) = 5.50 AND P₄ is limiting reagent.

(iii)
amount in excess «= 0.500 – (3 × 0.02000)» = 0.440 «mol»

(iv)

«22.7 × 1000 × 0.02000» = 454 «cm³»

Accept methods employing pV = nRT, with p as either 100 (454 cm³) or 101.3 kPa (448 cm³).

Do not accept answers in dm³.

Examiners report

[N/A]

Marks available	5
Reference code	16N.2.sl.TZ0.2

Question

The concentration of a solution of a weak acid, such as ethanedioic acid, can be determined
by titration with a standard solution of sodium hydroxide, NaOH (aq).

Distinguish between a weak acid and a strong acid.

Weak acid:

Strong acid:

[1]

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

[1]

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.

[5]

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.

[2]

Markscheme

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”

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

«\(\frac{1}{2} \times 1.40 \times {10^{ – 3}} = \) \(7.00 \times {10^{ – 4}}\) «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 = \(\frac{{0.0883{\rm{g}}}}{{0.125{\rm{g}}}}\) × 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 = \(\frac{{3.53{\rm{g}}}}{{5.00{\rm{g}}}}\) × 100 =» 70.6 «%»

ALTERNATIVE 3:
mol of hydrated ethanedioic acid (assuming sample to be pure) = \(\frac{{5.00{\rm{g}}}}{{126.08{\rm{gmo}}{{\rm{l}}^{ – 1}}}}\) = 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”.

Examiners report

[N/A]

Marks available	2
Reference code	16N.2.sl.TZ0.3

Question

Sodium thiosulfate solution reacts with dilute hydrochloric acid to form a precipitate of sulfur at room temperature.

Na₂S₂O₃ (aq) + 2HCl (aq) → S (s) + SO₂(g) + 2NaCl (aq) + X

Identify the formula and state symbol of X.

[1]

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

[1]

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.

[2]

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

[2]

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.

[2]

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.

[4]

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

[1]

Markscheme

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 \(\frac{1}{{\rm{t}}}\) 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.

Examiners report

[N/A]