State the rate expression for the forward reaction.

Predict the effect on the rate of the forward reaction and on the rate constant if the concentration of NO is halved.

1.0 mol of \({\rm{C}}{{\rm{l}}_2}\) and 1.0 mol of NO are mixed in a closed container at constant temperature. Sketch a graph to show how the concentration of NO and NOCl change with time until after equilibrium has been reached. Identify the point on the graph where equilibrium is established.

Consider the following reaction.

\[{\text{N}}{{\text{O}}_2}{\text{(g)}} + {\text{CO(g)}} \to {\text{NO(g)}} + {\text{C}}{{\text{O}}_2}{\text{(g)}}\]

Possible reaction mechanisms are:

\(\begin{array}{*{20}{l}} {{\text{Above 775 K:}}}&{{\text{N}}{{\text{O}}_2} + {\text{CO}} \to {\text{NO}} + {\text{C}}{{\text{O}}_{\text{2}}}}&{{\text{slow}}} \\ {{\text{Below 775 K:}}}&{{\text{2N}}{{\text{O}}_2} \to {\text{NO}} + {\text{N}}{{\text{O}}_{\text{3}}}}&{{\text{slow}}} \\ {}&{{\text{N}}{{\text{O}}_3} + {\text{CO}} \to {\text{N}}{{\text{O}}_2} + {\text{C}}{{\text{O}}_2}}&{{\text{fast}}} \end{array}\)

Based on the mechanisms, deduce the rate expressions above and below 775 K.

State two situations when the rate of a chemical reaction is equal to the rate constant.

Consider the following graph of \(\ln k\) against \(\frac{1}{T}\) for the first order decomposition of \({{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\) into \({\text{N}}{{\text{O}}_{\text{2}}}\). Determine the activation energy in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) for this reaction.

Deduce the equilibrium constant expression, \({K_{\text{c}}}\), for the reaction.

Determine the value of the equilibrium constant, \({K_{\text{c}}}\).

If the temperature of the reaction is changed to 300 °C, predict, stating a reason in each case, whether the equilibrium concentration of \({\text{S}}{{\text{O}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\) and the value of \({K_{\text{c}}}\) will increase or decrease.

If the volume of the container is changed to \({\text{1.50 d}}{{\text{m}}^{\text{3}}}\), predict, stating a reason in each case, how this will affect the equilibrium concentration of \({\text{S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}\) and the value of \({K_{\text{c}}}\).

Suggest, stating a reason, how the addition of a catalyst at constant pressure and temperature will affect the equilibrium concentration of \({\text{S}}{{\text{O}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\).

\({\text{rate}} = k{{\text{[NO]}}^2}{\text{[C}}{{\text{l}}_{\text{2}}}{\text{]}}\);

rate of reaction will decrease by a factor of 4;

no effect on the rate constant;

y axis labelled concentration/\({\text{mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) and x axis is labelled time/s;

gradient for [NO];

gradient for [NOCl] will be equal and opposite;

equilibrium point identified / two curves level off at same time;

Above 775 K: \({\text{rate}} = k{\text{[N}}{{\text{O}}_2}{\text{][CO]}}\);

Below 775 K: \({\text{rate}} = k{{\text{[N}}{{\text{O}}_2}{\text{]}}^2}\);

zero order reaction;

all concentrations are \({\text{1.0 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\);

\({\text{slope}} = \frac{{9.2 - 8.4}}{{(3.53 - 3.65) \times {{10}^{ - 3}}}} =  - 6.67 \times {10^3}\);

\(({E_{\text{a}}} = 6.67 \times {10^3} \times 8.31)\)

\({\text{55.4 (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

Accept in range 55.0 – 56.0

Award [1] if 55454 (J) stated

Award [2] for the correct final answer

\(({K_{\text{c}}}) = \frac{{{\text{[S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}{\text{]}}}}{{{\text{[C}}{{\text{l}}_2}{\text{][S}}{{\text{O}}_2}{\text{]}}}}\);

Ignore state symbols.

Square brackets [ ] required for the equilibrium expression.

\({\text{7.84}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol of S}}{{\text{O}}_2}\) and \({\text{7.84}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol of C}}{{\text{l}}_2}\);

\({\text{7.84}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{ of S}}{{\text{O}}_2}\), \({\text{7.84}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{ of C}}{{\text{l}}_2}\) and

\({\text{7.65}} \times {\text{1}}{{\text{0}}^{ - 4}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{ of S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}\);

12.5;

Award [1] for 10.34

Award [3] for the correct final answer

value of \({K_{\text{c}}}\) increases;

\({\text{[S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}{\text{]}}\) increases;

decrease in temperature favours (forward) reaction which is exothermic;

Do not allow ECF.

no effect on the value of \({K_{\text{c}}}\) / depends only on temperature;

\({\text{[S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}{\text{]}}\) decreases;

increase in volume favours the reverse reaction which has more gaseous moles;

Do not allow ECF.

no effect;

catalyst increases the rate of forward and reverse reactions (equally) / catalyst decreases activation energies (equally);

In part (a) the rate expression was correctly stated although some confused this with an equilibrium constant expression.

Only the better candidates realized that the rate of reaction will decrease by a factor of four and there will be no effect on the rate constant.

Although most candidates were able to correctly sketch the concentration versus time graph many forgot to label the axes or include units.

Part (b) was well answered and candidates demonstrated a good understanding of rate expressions based on reaction mechanism.

The better candidates were able to figure out that the rate of a chemical reaction is equal to the rate constant when all concentrations are \({\text{1.0 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) or for a zero order reaction.

Most candidates had difficulty in calculating activation energy from the graph in part (d) and some gave the answer in \({\text{J}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) instead of \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) which showed that they missed this instruction in the question.

In part (e), the equilibrium constant expression was correctly stated by the majority but calculating the value of\({K_{\text{c}}}\) proved to be difficult.

A large number of candidates obtained the incorrect answer of 10.34 as a result of using the initial concentrations of the reactants instead of equilibrium concentrations.

The application of Le Chatelier’s principle was handled well by the majority with minor omissions such as not using the term gaseous particles in part (iv).

Some candidates stated that the addition of a catalyst does not affect the value of \({K_{\text{c}}}\) or the position of equilibrium, which did not answer the question and scored no marks because they had not commented on the concentration of \({\text{SOC}}{{\text{l}}_{\text{2}}}\). Some candidates correctly stated that a catalyst increases the rate of forward and reverse reactions equally.

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.

The concentration of ethanoic acid can be calculated as \({\text{1.748 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\). Determine the percentage uncertainty of this value. (Neglect any uncertainty in the density and the molar mass.)

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

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

\({\text{3.00 c}}{{\text{m}}^{\text{3}}}\) of the \({\text{0.200 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) aqueous sodium hydroxide reacted with the hydrochloric acid present in the \({\text{5.00 c}}{{\text{m}}^{\text{3}}}\) sample. Determine the concentration of ethanoic acid in the final equilibrium mixture.

Deduce the equilibrium constant expression for the reaction.

The other concentrations in the equilibrium mixture were calculated as follows:

Use these data, along with your answer to part (iii), to determine the value of the equilibrium constant. (If you did not obtain an answer to part (iii), assume the concentrations of ethanol and ethanoic acid are equal, although this is not the case.)

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

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

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

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.

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

\({\text{M(C}}{{\text{H}}_{\text{3}}}{\text{COOH)}}\left( { = (4 \times 1.01) + (2 \times 12.01) + (2 \times 16.00)} \right) = 60.06{\text{ (g}}\,{\text{mo}}{{\text{l}}^{ - 1}})\);

Accept 60 (g mol^–1).

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

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

Award [3] for correct final answer.

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

percentage uncertainty in volume of ethanoic acid \( = 100 \times \frac{{0.05}}{{5.00}}{\text{ }} = 1\% \);

percentage uncertainty in total volume \( = 100 \times \frac{{0.62}}{{50}} = 1.24\% \);

total percentage uncertainty \( = 1 + 1.24 = 2.24\% \);

Accept rounding down to 2.2/2%.

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

Do not accept without ±.

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

\(26.00 - 3.00 = 23.00{\text{ }}({\text{c}}{{\text{m}}^3})\);

If other methods used, award M1 for calculating amount of NaOH reacting with CH₃COOH.

\(0.200 \times \frac{{23.00}}{{5.00}} = 0.920{\text{ }}({\text{mol}}\,{\text{d}}{{\text{m}}^{ - 3}})\);

Award [2] for correct final answer.

If (ii) given as mean titre (26.5 cm³) then ECF answer comes to 0.94 (mol dm^–3).

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

Do not penalize minor errors in formulas.

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

\(({K_c} = )\frac{{0.828 \times 1.80}}{{0.884 \times 0.920}} = 1.83\);

If assumed [CH₃COOH] = 0.884 mol dm^-3, answer is 1.91 – allow this even if an answer was obtained for (iii).

If (ii) given as mean titre (26.5 cm³) then ECF answer comes to 1.79.

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

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”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Generally candidates found this question quite challenging and some left quite a number of parts unanswered. The tradition is that the first question on the paper is a data response question, which often addresses many aspects of the syllabus, and unfortunately candidates, especially those of average or below average ability, seem to have difficulty in tackling questions of this nature. One other issue with data response questions is that, of necessity, the data appears at the beginning of the question whilst, mainly because of the space left for candidates to answer, the later parts of the question referring to these data may not appear until a number of pages into the paper.

Part (a) concerning density, volume and amount of substance was generally reasonably well answered, but the following parts, concerning uncertainties, were rarely answered correctly and a number confused precision (uncertainty, either absolute or as a percentage) and accuracy (percentage error in the value obtained). Many candidates also seemed to lack experimental common sense, simply taking an average that included an initial titre that was much larger than the concordant second and third titres, rather than excluding it. This lack of experimental “know how” was also evident in responses to (c) (iii) where it was unusual for the approach to the question to indicate the candidate had realised that the alkali was neutralising two different acids (HCl and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\)) and again in part (d) where it was rare for the response to outline a practical solution to the problem, though quite a number of candidates suggested that the pH would become constant, presumably not realising that the pH would be dominated by the HCl catalyst. Most students could however carry out the more routine tasks of writing an equilibrium constant expression and determining its value from the data given. 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 actually asked and the unusual approach to the effect of temperature disconcerted many. Whilst most students managed to identify the ester as the component of the mixture that was insoluble in water, the reasons given were usually couched in terms of the polarity of the molecule (many quite polar molecules, halogenoalkanes for example, are insoluble in water) rather than its inability to form strong hydrogen bonds to water, which is the critical factor. Quite a number of students came up with a valid reason why water would not be a suitable solvent, though some students appeared to have overlooked the fact the question stated “other reason”.

Nitrogen oxide is in equilibrium with dinitrogen dioxide.

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

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

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

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

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

Deduce the rate expression for the mechanism.

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

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

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

Accept “product” for [N₂O₂].

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

[1 mark]

ALTERNATIVE 1:

«from equilibrium, step 1»

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

OR

[N₂O₂] = K_c[NO]²

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

rate = k[NO]²[O₂]

ALTERNATIVE 2:

«from step 2» rate = k₂[N₂O₂][O₂]

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

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

rate = k[NO]²[O₂]

Award [2] for correct rate expression.

[2 marks]

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

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

E_a = 52.9 «kJ mol^–1»

Award [2] for correct final answer.

[2 marks]

Two compounds, A and D, each have the formula \({{\text{C}}_{\text{4}}}{{\text{H}}_{\text{9}}}{\text{Cl}}\).

Compound A is reacted with dilute aqueous sodium hydroxide to produce compound B with a formula of \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\text{O}}\). Compound B is then oxidized with acidified potassium

manganate(VII) to produce compound C with a formula of \({{\text{C}}_{\text{4}}}{{\text{H}}_{\text{8}}}{\text{O}}\). Compound C resists further oxidation by acidified potassium manganate(VII).

Compound D is reacted with dilute aqueous sodium hydroxide to produce compound E with a formula of \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\text{O}}\). Compound E does not react with acidified potassium manganate(VII).

Deduce the structural formulas for compounds A, B, C, D and E.

A:

B:

C:

D:

E:

Deduce an equation for the reaction between propanoic acid and methanol. Identify the catalyst and state the name of the organic compound, X, formed.

Calculate the concentrations of the other three species present at equilibrium.

State the equilibrium constant expression, \({K_{\text{c}}}\), and calculate the equilibrium constant for this reaction at 25.0 °C.

2-chloro-3-methylbutane reacts with sodium hydroxide via an \({{\text{S}}_{\text{N}}}{\text{2}}\) mechanism. Explain the mechanism by using curly arrows to represent the movement of electron pairs.

Explain why the hydroxide ion is a better nucleophile than water.

1-chlorobutane can be converted to a pentylamine via a two stage process. Deduce equations for each step of this conversion including any catalyst required and name the organic product produced at each stage.

Accept condensed formulas.

Award [1 max] if A and D are other way round (and nothing else correct).

Award [2 max] if A and D are other way round but one substitution product B or E is correct based on initial choice of A and D.

Award [3 max] if A and D are other way round but both substitution products B and E are correct based on initial choice of A and D.

M2 (for B) and M5 (for E) may also be scored for substitution product if primary chloroalkane used.

Penalize missing hydrogens once only in Q.7.

\[{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOH}} + {\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOC}}{{\text{H}}_{\text{3}}} + {{\text{H}}_{\text{2}}}{\text{O}}\]

[1] for reactants and [1] for products.

(concentrated) sulfuric acid/\({{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\);

Do not accept just \({H^ + }\) or acid.

methyl propanoate;

[CH₃CH₂COOH]:

\((1.6 - 0.80 = ){\text{ }}0.8{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

[CH₃OH]:

\((2.0 - 0.80 = ){\text{ }}1.2{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

[H₂O]:

\({\text{0.80 (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\(({K_{\text{c}}} = )\frac{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOC}}{{\text{H}}_{\text{3}}}{\text{][}}{{\text{H}}_{\text{2}}}{\text{O]}}}}{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH][C}}{{\text{H}}_{\text{3}}}{\text{OH]}}}}\);

\(\left( {{K_{\text{c}}} = \frac{{[{{(0.80)}^2}]}}{{\left[ {(1.2 \times 0.8)} \right]}} = } \right){\text{ }}0.7\);

Allow 0.67.

Award [1 max] for 0.83.

curly arrow going from lone pair/negative charge on O in \({\text{H}}{{\text{O}}^ - }\) to C;

Do not allow curly arrow originating on H in \(H{O^ - }\).

curly arrow showing Cl leaving;

Accept curly arrow either going from bond between C and Cl to Cl in 2-chloro-3-methylbutane or in the transition state.

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

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

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

formation of organic product 3-methylbutan-2-ol and \({\text{C}}{{\text{l}}^ - }\);

\({\text{O}}{{\text{H}}^ - }\) has a negative charge/higher electron density;

greater attraction to the carbon atom (with the partial positive charge) / OWTTE;

Do not allow just greater attraction.

\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Cl}} + {\text{KCN}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CN}} + {\text{KCl}}\);

Accept \(C{N^ - }\) for KCN and \(C{l^ - }\) for KCl.

pentanenitrile;

Allow 1-cyanobutane.

\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{CN}} + {\text{2}}{{\text{H}}_2} \to {\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{N}}{{\text{H}}_2}\);

pentan-1-amine / 1-aminopentane / 1-pentylamine / 1-pentanamine;

Catalyst: nickel/Ni / palladium/Pd / platinum/Pt;

Penalise missing hydrogen once only in Q.7.

This was the least popular question in Section B. Most candidates either scored all five marks in (a) or just one.

(b) was usually well done, though it was disappointing that more candidates did not use the equilibrium sign.

In (c), a significant number of candidates omitted water from the equilibrium calculations.

In (c), a significant number of candidates omitted water from the equilibrium calculations.

The organic reaction mechanism in (d) (i) was very poorly presented. Many even tried drawing curly arrows from NaOH as an attacking species. The majority could identify the product of the reaction but a mechanism was far beyond them. Transition states were poor or missing completely.

In (ii) although many knew that \({\text{O}}{{\text{H}}^ - }\) has a negative charge, few linked this to the greater attraction to the carbon atom.

In (iii) very few candidates did well here and the name of pentan-1-amine was rarely given. Other mistakes included incorrect catalysts. Further common mistakes included some candidates not including all the hydrogens in the structural formulas. In general for this part there was very poor knowledge of organic synthesis amongst candidates. Very few had a good “stab” at this question. The fact that pentylamine was mentioned in the question initially meant that very few candidates accessed the last mark for the name of the product.

Discuss the bonding in the resonance structures of ozone.

Deduce one resonance structure of ozone and the corresponding formal charges on each oxygen atom.

The first six ionization energies, in kJ mol^–1, of an element are given below.

Explain the large increase in ionization energy from IE₃ to IE₄.

At a given time, the concentration of NO₂(g) and N₂O₄(g) were 0.52 and \(0.10{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) respectively.

Deduce, showing your reasoning, if the forward or the reverse reaction is favoured at this time.

Comment on the value of ΔG when the reaction quotient equals the equilibrium constant, Q = K.

lone pair on p orbital «of O atom» overlaps/delocalizes with pi electrons «from double bond»

both O–O bonds have equal bond length
OR
both O–O bonds have same/1.5 bond order
OR
both O–O are intermediate between O–O AND O=O 

both O–O bonds have equal bond energy

Accept “p/pi/\(\pi \) electrons are delocalized/not localized”.

[3 marks]

ALTERNATIVE 1:

FC: –1 AND +1 AND 0

ALTERNATIVE 2:

FC: 0 AND +1 AND –1

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

Do not accept structure that represents 1.5 bonds.

Do not penalize missing lone pairs if already penalized in 3(b).

If resonance structure is incorrect, no ECF.

Any one of the structures with correct formal charges for [2 max].

[2 marks]

Any two of:
IE₄: electron in lower/inner shell/energy level
OR
IE₄: more stable/full electron shell

IE₄: electron closer to nucleus
OR
IE₄: electron more tightly held by nucleus

IE₄: less shielding by complete inner shells

Accept “increase in effective nuclear charge” for M2.

[2 marks]

«Q_c = \(\frac{{0.10}}{{{{0.52}^2}}}\) =» 0.37
reaction proceeds to the left/NO₂(g) «until Q = K_c»
OR
reverse reaction «favoured»

Do not award M2 without a calculation for M1 but remember to apply ECF.

[2 marks]

ΔG = 0
reaction at equilibrium
OR
rate of forward and reverse reaction is the same
OR
constant macroscopic properties

[2 marks]

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

The following equilibrium concentrations in mol dm^–3 were obtained at 761 K.

Calculate the value of the equilibrium constant at 761 K.

Determine the value of ΔG^θ, in kJ, for the above reaction at 761 K using section 1 of the data booklet.

Calculate [H₃O⁺] in the solution and the dissociation constant, K_a , of the acid at 25 °C.

Calculate K_b for HCO₃^– acting as a base.

K_c = \(\frac{{{{{\text{[HI]}}}^{\text{2}}}}}{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{][}}{{\text{I}}_{\text{2}}}{\text{]}}}}\)

45.6

ΔG^θ = «– RT ln K = – (0.00831 kJ K⁻¹ mol⁻¹ x 761 K x ln 45.6) =» – 24.2 «kJ»

[H₃O⁺] = 6.76 x 10^–5 «mol dm^–3»

K_a = \(\frac{{{{\left( {6.76 \times {{10}^{ - 5}}} \right)}^2}}}{{\left( {0.010 - 6.76 \times {{10}^{ - 5}}} \right)}}/\frac{{{{\left( {6.76 \times {{10}^{ - 5}}} \right)}^2}}}{{0.010}}\)

4.6 x 10^–7

Accept 4.57 x 10^–7

Award [3] for correct final answer.

«\(\frac{{1.00 \times {{10}^{ - 14}}}}{{4.6 \times {{10}^{ - 7}}}}\) =» 2.17 x 10^–8

OR

«\(\frac{{1.00 \times {{10}^{ - 14}}}}{{4.57 \times {{10}^{ - 7}}}}\) =» 2.19 x 10^–8

Outline two characteristics of a reversible reaction in a state of dynamic equilibrium.

Predict, with a reason, how each of the following changes affects the position of equilibrium.

The volume of the container is increased.

Ammonia is removed from the equilibrium mixture.

Define the term activation energy, \({E_{\text{a}}}\).

Ammonia is manufactured by the Haber process in which iron is used as a catalyst.

Explain the effect of a catalyst on the rate of reaction.

Typical conditions used in the Haber process are 500 °C and 200 atm, resulting in approximately 15% yield of ammonia.

(i)     Explain why a temperature lower than 500 °C is not used.

(ii)     Outline why a pressure higher than 200 atm is not often used.

Deduce the equilibrium constant expression, \({K_{\text{c}}}\), for the reaction on page 10.

When 1.00 mol of nitrogen and 3.00 mol of hydrogen were allowed to reach equilibrium in a \({\text{1.00 d}}{{\text{m}}^{\text{3}}}\) container at a temperature of 500 °C and a pressure of 1000 atm, the equilibrium mixture contained 1.46 mol of ammonia.

Calculate the value of \({K_{\text{c}}}\) at 500 °C.

Define the term base according to the Lewis theory.

Define the term weak base according to the Brønsted–Lowry theory.

Deduce the formulas of conjugate acid-base pairs in the reaction below.

\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}{\text{(aq)}} + {{\text{H}}_{\text{2}}}{\text{O(l)}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{NH}}_{\text{3}}^ + {\text{(aq)}} + {\text{O}}{{\text{H}}^ - }{\text{(aq)}}\]

Determine the pH of a \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) solution of ammonia, \({\text{N}}{{\text{H}}_{\text{3}}}{\text{(aq)}}\), using tables 2 and 15 of the data booklet.

(i)     Sketch the pH titration curve obtained when \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{N}}{{\text{H}}_{\text{3}}}{\text{(aq)}}\) is added to \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{HCl (aq)}}\).

(ii)     Identify an indicator from table 16 of the data booklet that could be used for this titration.

rates of forward and reverse reactions are equal / opposing changes occur at equal rates;

the concentrations of all reactants and products remain constant / macroscopic properties remain constant;

closed/isolated system;

Accept “the same” for “equal” in M1 and for “constant” in M2.

The volume of the container is increased:

position of equilibrium shifts to the left/reactants and fewer moles of gas on the right hand side/pressure decreases / OWTTE;

Ammonia is removed from the equilibrium mixture:

position of equilibrium shifts to the right/products and \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) decreases so \({\text{[}}{{\text{N}}_{\text{2}}}{\text{]}}\) and \({\text{[}}{{\text{H}}_{\text{2}}}{\text{]}}\) must also decrease to keep K_c constant

OR

position of equilibrium shifts to the right/products and rate of reverse reaction decreases / OWTTE;

Award [1 max] if both predicted changes are correct.

Do not accept “to increase \([N{H_3}]\)” or reference to LCP without explanation.

minimum energy needed (by reactants/colliding particles) to react/start/initiate a reaction;

Accept “energy difference between reactants and transition state”.

more effective/successful collisions per unit time / greater proportion of collisions effective;

alternative pathway and a lower activation energy

OR

lowers activation energy so that more particles have enough energy to react;

Do not accept just “lowers/reduces the activation energy”.

Accept “provides a surface for reacting/reactants/reaction”.

(i)     slower rate / OWTTE;

uneconomic / OWTTE;

(ii)     high cost for building/maintaining plant / high energy cost of compressor / OWTTE;

Do not accept “high pressure is expensive” without justification.

Accept high pressure requires high energy.

\(({K_{\text{c}}} = )\frac{{{{{\text{[N}}{{\text{H}}_3}{\text{(g)]}}}^2}}}{{{\text{[}}{{\text{N}}_2}{\text{(g)]}} \times {{{\text{[}}{{\text{H}}_2}{\text{(g)]}}}^3}}}\);

Ignore state symbols.

Concentrations must be represented by square brackets.

moles at equilibrium: nitrogen 0.27, hydrogen 0.81 / concentrations at equilibrium: nitrogen \({\text{0.27 (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\), hydrogen \({\text{0.81 (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) (and ammonia \({\text{1.46 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\));

\({K_{\text{c}}} = 15\);

Actual calculation gives \({K_{\text{c}}}{\text{ = }}14{\text{.}}86\).

Award [2] for correct final answer.

Award [1 max] if \({K_{\text{c}}}\left( { = \frac{{{{1.46}^2}}}{{{3^3} \times 1}}} \right) = 0.079\)

electron pair donor;

Accept lone pair donor.

proton acceptor and partially/slightly ionized;

Accept “proton acceptor and partially/slightly dissociated”.

Award [1 max] for two correct acids OR two correct conjugate bases.

\({K_{\text{b}}} = \frac{{{\text{[NH}}_4^ + {\text{][O}}{{\text{H}}^ - }{\text{]}}}}{{{\text{[N}}{{\text{H}}_3}{\text{]}}}} = 1.8 \times {10^{ - 5}}/{10^{ - 4.75}}\);

\({\text{[NH}}_4^ + {\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\) and \({\text{[N}}{{\text{H}}_3}{\text{]}} \approx 1.00 \times {10^{ - 1}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[O}}{{\text{H}}^ - }{\text{]}} = (\sqrt {1.8 \times {{10}^{ - 6}}}  = )1.3 \times {10^{ - 3}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}/{\text{pOH}} = 2.89\);

\({\text{pH}} = (14.0 - 2.89 = )11.1\);

Award [4] for correct final answer.

(i)     

For volume \( = 0:{\text{ pH}} = 1\);

vertical jump should be positioned in volume range \({\text{24 c}}{{\text{m}}^{\text{3}}}\) to \({\text{26 c}}{{\text{m}}^{\text{3}}}\) and include pH range between 3 to 6;

For volume = 50: pH between 8 to 11;

(ii)     methyl orange / bromophenol blue / bromocresol green / methyl red;

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

Most candidates were able to give two characteristics of a dynamic equilibrium and explain the effect of changes in volume on the position of equilibrium but many had difficulty giving a complete explanation of the equilibrium shift resulting from the removal of ammonia. Candidates were expected to include a reference to the value of \({K_{\text{c}}}\) or the reduced rate of the reverse reaction when justifying their answer. The definition of activation energy was well known but some lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway in their explanation for the lower activation energy. The explanation of why lower temperatures were not used in the Haber process was also incomplete with many not considering the economic disadvantages of a slow reaction rate. Similarly many did not explain why high pressure was expensive in terms of energy or building costs. Most were able to deduce the equilibrium constant but many lost a mark in the calculation of \({K_{\text{c}}}\) as they used the initial concentrations of nitrogen and hydrogen. Some teachers identified an inconsistency in the question in that the total number of moles of gas under the conditions stated in the question was not consistent with the ideal gas equation however this did not appear to be a problem for the candidates. (However, the ideal gas law cannot be applied here as under these conditions ammonia would be in its supercritical state.) Most candidates were able to define Lewis bases but the definition of weak Brønsted-Lowry bases proved to be more problematic as many did not refer to partial ionisation in their response. Most students were able to identify the conjugate acid-base pairs. The calculation of the pH of an ammonia solution proved to be challenging with many confusing \({K_{\text{a}}}\) and \({K_{\text{b}}}\). Others did not recognize that since it is a weak base, \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}}\) at equilibrium is approximately equal to starting concentration \({\text{(0.100 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) or that \({\text{[NH}}{{\text{4}}^ + }{\text{]}} = {\text{[O}}{{\text{H}}^ - }{\text{]}}\). (The examination paper was rescaled for candidates sitting the examination in Spanish (due to the error in the question) and candidates close to a boundary given particular attention.) Only the strongest candidates were able to gain full marks for the pH curve although many recognised that the pH would be 1 before any ammonia was added given that HCl is a strong acid. A significant number had the final pH above 11 and did not allow for dilution of the \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution. Many correctly identified a possible indicator.

(i)     Use the graph to deduce whether the forward reaction is exothermic or endothermic and explain your choice.

(ii)     State and explain the effect of increasing the pressure on the yield of ammonia.

(iii)     Deduce the equilibrium constant expression, \({K_{\text{c}}}\), for the reaction.

(iv)     A mixture of 1.00 mol \({{\text{N}}_{\text{2}}}\) and 3.00 mol \({{\text{H}}_{\text{2}}}\) was placed in a \({\text{1.0 d}}{{\text{m}}^{\text{3}}}\) flask at 400 °C. When the system was allowed to reach equilibrium, the concentration of was found to be \({\text{0.062 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\). Determine the equilibrium constant, \({K_{\text{c}}}\), of the reaction at this temperature.

(v)     Iron is used as a catalyst in the Haber process. State the effect of a catalyst on the value of \({K_{\text{c}}}\).

(i)     Distinguish between the terms strong and weak acid and state the equations used to show the dissociation of each acid in aqueous solution.

(ii)     Deduce the expression for the ionization constant, \({K_{\text{a}}}\), of hydrocyanic acid and calculate its value from the \({\text{p}}{K_{\text{a}}}\) value given.

(iii)     Use your answer from part (b) (ii) to calculate the \({\text{[}}{{\text{H}}^ + }{\text{]}}\) and the pH of an aqueous solution of hydrocyanic acid of concentration \({\text{0.108 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\). State one assumption made in arriving at your answer.

A small piece of magnesium ribbon is added to solutions of nitric and hydrocyanic acid of the same concentration at the same temperature. Describe two observations that would allow you to distinguish between the two acids.

(i)     Calculate the volume of the sodium hydroxide solution required to react exactly with a \({\text{15.0 c}}{{\text{m}}^{\text{3}}}\) solution of \({\text{0.10 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) nitric acid.

(ii)     The following hypothesis was suggested by the student: “Since hydrocyanic acid is a weak acid it will react with a smaller volume of the \({\text{0.20 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) sodium hydroxide solution.” Comment on whether or not this is a valid hypothesis.

(iii)     Use Table 16 of the Data Booklet to identify a suitable indicator for the titration of sodium hydroxide and hydrocyanic acid.

(i)     exothermic;

Accept either of the following for the second mark.

increasing temperature favours endothermic/reverse reaction;

as yield decreases with increasing temperature;

(ii)     yield increases / equilibrium moves to the right / more ammonia;

increase in pressure favours the reaction which has fewer moles of gaseous products;

(iii)     \({K_{\text{c}}} = \frac{{{{{\text{[N}}{{\text{H}}_3}{\text{]}}}^2}}}{{{\text{[}}{{\text{N}}_2}{\text{][}}{{\text{H}}_2}{{\text{]}}^3}}}\);

(iv)     \({\text{[}}{{\text{N}}_2}{\text{]}}\): (at equilibrium \( = 1.00 - 0.031 = \)) \({\text{0.969 (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[}}{{\text{H}}_2}{\text{]}}\): (at equilibrium \( = 3.00 - 3(0.031) = \)) \({\text{2.91 (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({K_{\text{c}}}{\text{ }}\left( { = \frac{{{{{\text{(0.062)}}}^2}}}{{{\text{(0.969) (2.91}}{{\text{)}}^3}}}} \right) = {\text{1.6(1)}} \times {\text{1}}{{\text{0}}^{ - 4}}\);

Ignore units.

Award [1] for K_c = 1.4 \( \times \) 10^–4

(v)     no effect;

(i)     strong acid completely dissociated/ionized and weak acid partially dissociated/ionized;

\({\text{HN}}{{\text{O}}_3}{\text{(aq)}} \to {{\text{H}}^ + }{\text{(aq)}} + {\text{NO}}_3^ - {\text{(aq)}}\);

\({\text{HCN(aq)}} \rightleftharpoons {{\text{H}}^ + }{\text{(aq)}} + {\text{C}}{{\text{N}}^ - }{\text{(aq)}}\);

Insist on both arrows as shown.

State symbols not needed.

Accept H₂O and H₃O⁺.

(ii)     \({K_{\text{a}}} = \frac{{{\text{[}}{{\text{H}}^ + }{\text{][C}}{{\text{N}}^ - }{\text{]}}}}{{{\text{[HCN]}}}}\);

Allow H₃O⁺ instead of H⁺.

\({K_{\text{a}}} = {10^{ - 9.21}} = 6.17 \times {10^{ - 10}}\);

(iii)     \({[{{\text{H}}^ + }] = \sqrt {{K_{\text{a}}}[{\text{HCN}}]} /\sqrt {(6.17 \times {{10}^{ - 10}} \times 0.108)} }\);

\({ = 8.16 \times {{10}^{ - 6}}}\);

Allow in the range 8.13 \( \times \) 10^–6 to 8.16 \( \times \) 10^–6.

\({\text{pH}} = 5.09\);

OR

\({{\text{pH}} = \frac{1}{2}{\text{(p}}{K_{\text{a}}} - {\text{log}}[{\text{HCN}}])/\frac{1}{2}(9.21 - \log \,0.108)}\);

\({ = 5.09}\);

\({\text{[}}{{\text{H}}^ + }{\text{]}} = {10^{ - 5.09}} = 8.16 \times {10^{ - 6}}\);

Allow in the range 8.13 \( \times \) 10^–6 to 8.16 \( \times \) 10^–6.

If expression for [H⁺] missing but both answers correct, award [3], if one answer

correct, award [2].

assume \({\text{[}}{{\text{H}}^ + }{\text{]}} \ll 0.108\) / negligible dissociation;

With HNO₃:

faster rate of bubble/hydrogen/gas production;

faster rate of magnesium dissolving;

higher temperature change;

Accept opposite argument for HCN.

Reference to specific observations needed.

Award [1] if 2 observations given but acid is not identified.

(i)     (nitric acid) 7.5 cm³;

(ii)     not valid as hydrocyanic acid reacts with same volume/ 7.5 cm³;

(iii)     bromothymol blue / phenol red / phenolphthalein;

Equilibrium is a topic that has shown substantial improvement in recent sessions with some very well produced arguments. The reaction was correctly described as exothermic with a reason correctly given in most cases. Most candidates knew that yield would increase with increased pressure, but some failed to identify the change in the number of “gaseous” molecules as the reason. More candidates had difficulty with the equilibrium constant calculation often using the initial not equilibrium concentrations.

In (b) most correctly defined strong and weak acids and many also wrote correct equations. A few, however, missed the equilibrium sign for hydrocyanic acid. HA, CH₃COOH and HCl were commonly given instead of HCN and HNO₃, suggesting that students sometimes have difficulty applying general concepts to specific cases. It was encouraging to see many candidates determine the pH from the pK_a value including the assumption that there is negligible dissociation, as this has challenged students in previous sessions. A significant number of weaker candidates reported however that the acid solution would have pH values above 7.

Part (c) presented problems with many candidates unable to describe specific observations related to rate which would distinguish between a strong and weak acid and simply stated that the reaction would be faster.

The moles calculation was answered well in (d) with most candidates able to identify phenolphthalein as a suitable indicator.

(i)     Deduce the equilibrium constant expression for this reaction.

(ii)     Explain the effect on the position of equilibrium and the value of \({K_{\text{c}}}\) when pressure is decreased and temperature is kept constant.

(iii)     2.00 mol of NOCl was placed in a \({\text{1.00 d}}{{\text{m}}^{\text{3}}}\) container and allowed to reach equilibrium at 298 K. At equilibrium, 0.200 mol of NO was present. Determine the equilibrium concentrations of NOCl and \({\text{C}}{{\text{l}}_{\text{2}}}\), and hence calculate the value of \({K_{\text{c}}}\) at this temperature.

(iv)     The value of \({K_{\text{c}}}\) is \(1.60 \times {10^{ - 5}}\) at 318 K. State and explain whether the forward reaction is exothermic or endothermic.

(i)     Compare the two liquids in terms of their boiling points, enthalpies of vaporization and vapour pressures.

(ii)     Explain your answer given for part (b)(i).

Calculate the pH of a \({\text{1.50 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) solution of ammonia at 298 K to two decimal places, using Table 15 of the Data Booklet.

A buffer solution is made using \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.500 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) hydrochloric acid, HCl (aq), and \({\text{20.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.50 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ammonia solution, \({\text{N}}{{\text{H}}_{\text{3}}}{\text{(aq)}}\).

Describe the meaning of the term buffer solution.

Determine the pH of the buffer solution at 298 K.

A \({\text{1.50 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) solution of ammonia is added to \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of a \({\text{0.500 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) hydrochloric acid solution in a titration experiment.

Calculate the total volume of the solution at the equivalence point.

Calculate the pH of the solution at the equivalence point, using Table 15 of the Data Booklet.

Identify a suitable indicator for this titration, using Table 16 of the Data Booklet.

(i)     \(({K_{\text{c}}} = )\frac{{{\text{[C}}{{\text{l}}_2}{\text{(g)][NO(g)}}{{\text{]}}^2}}}{{{{{\text{[NOCl(g)]}}}^2}}}\);

Ignore state symbols.

(ii)     equilibrium shifts to right as there are more moles (of gas) on product side;

no change to \({K_{\text{c}}}\) as it is a constant at fixed temperature / OWTTE;

(iii)     \({\text{[NOCl(g)]}} = 1.80{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[C}}{{\text{l}}_2}{\text{(g)]}} = 0.100{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({K_{\text{c}}} = \left( {\frac{{0.100 \times {{(0.200)}^2}}}{{{{(1.80)}^2}}}} \right)1.23 \times {10^{ - 3}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

Award [3] for correct final answer.

(iv)     exothermic as \({K_{\text{c}}}\) is lower at higher temperature;

(i)     hexane has lower boiling point and enthalpy of vaporization than pentan-1-ol / OWTTE;

hexane has higher vapour pressure than pentan-1-ol / OWTTE;

(ii)     hexane is non-polar / has only van der Waals’/London/dispersion forces / has weaker intermolecular forces than pentan-1-ol;

pentan-1-ol has hydrogen bonding between molecules;

\({\text{[O}}{{\text{H}}^ - }{\text{]}} = \sqrt {1.50 \times 1.78 \times {{10}^{ - 5}}}  = 5.17 \times {10^{ - 3}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{pH}} = (14 - {\text{pOH}} = 14 - 2.29 = ){\text{ }}11.71\);

Award [2] for correct final answer.

Accept correct answer with more than 2 decimal places.

solution which resists change in pH / changes pH slightly / OWTTE;

when small amounts of acid or base are added;

\({\text{[N}}{{\text{H}}_3}{\text{] = }}\left( {\frac{{(1.50 \times 0.0200) - (0.500 \times 0.0250)}}{{0.0450}} = } \right){\text{ }}0.389{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[NH}}_4^ + {\text{]}} = \left( {\frac{{(0.500 \times 0.0250)}}{{0.0450}} = } \right){\text{ }}0.278{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[O}}{{\text{H}}^ - }{\text{]}} = \left( {\frac{{{K_b}{\text{[N}}{{\text{H}}_3}{\text{]}}}}{{{\text{[NH}}_4^ + {\text{]}}}} = } \right){\text{ }}\frac{{1.78 \times {{10}^{ - 5}} \times 0.389}}{{0.278}} = 2.49 \times {10^{ - 5}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{pH}} = (14.0 - {\text{pOH}} = 14.0 - 4.60 = ){\text{ }}9.40\);

OR

\({\text{pOH}} = {\text{p}}{K_b} + \log \frac{{[{\text{NH}}_4^ + ]}}{{{\text{[N}}{{\text{H}}_3}]}}{\text{  =  p}}{K_{\text{b}}} + \log \frac{{(12.5/1000)}}{{(17.5/1000)}}\);

\({\text{pOH}} = 4.75 + \log \left( {\frac{{12.5}}{{17.5}}} \right) = 4.75 - 0.146 = 4.604\);

\({\text{pH}} = 14.0 - 4.604 = 9.40\);

Award [4] for the correct final answer.

\(\left( {{\text{V(N}}{{\text{H}}_{\text{3}}}{\text{)}} = \frac{{25.0 \times 0.500}}{{1.50}} = 8.33{\text{ c}}{{\text{m}}^3}} \right)\)

\({\text{V}} = {\text{V(N}}{{\text{H}}_3}{\text{)}} + {\text{V(HCl)}} = 8.33 + 25.0 = 33.3{\text{ c}}{{\text{m}}^3}/0.0333{\text{ d}}{{\text{m}}^3}\);

(\({\text{NH}}_{\text{4}}^ + \) ions are present at equivalence point \({\text{N}}{{\text{H}}_3} + {\text{HCl}} \to {\text{NH}}_4^ +  + {\text{C}}{{\text{l}}^ - }\) at equivalence \({\text{n}}({\text{NH}}_4^ + {\text{ produced}}) = {\text{n}}({\text{N}}{{\text{H}}_3}{\text{ added}}) = {\text{n(HCl)}}\))

\([{\text{NH}}_4^ + ] = \frac{{0.500 \times 0.0250}}{{0.0333}} = 0.375{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}})\);

\({\text{(NH}}_4^ + {\text{(aq)}} \rightleftharpoons {\text{N}}{{\text{H}}_3}{\text{(aq)}} + {{\text{H}}^ + }{\text{(aq)}}/{\text{NH}}_4^ + {\text{(aq)}} + {{\text{H}}_2}{\text{O(l)}} \rightleftharpoons {\text{N}}{{\text{H}}_3}{\text{(aq)}} + {{\text{H}}_3}{{\text{O}}^ + }{\text{(aq)}}\)

\({\text{p}}{K_{\text{a}}}{\text{(NH}}_4^ + ) = 14 - {\text{p}}{K_{\text{b}}}{\text{(N}}{{\text{H}}_3}) = 14.00 - 4.75 = 9.25)\)

\({K_{\text{a}}} = \frac{{{\text{[N}}{{\text{H}}_3}{\text{(aq)][}}{{\text{H}}^ + }{\text{(aq)]}}}}{{{\text{[NH}}_4^ + {\text{(aq)]}}}} = 5.62 \times {10^{ - 10}}\);

\({\text{[}}{{\text{H}}^ + }{\text{(aq)]}} = \sqrt {5.62 \times {{10}^{ - 10}} \times 0.375}  = 1.45 \times {10^{ - 5}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{pH}} = 4.84\);

Award [4] for the correct final answer.

bromocresol green / methyl red;

ECF for answer in 7(c)(v) if pH given is below 7.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

The construction and use of equilibrium expressions for \({K_{\text{c}}}\) showed good understanding. The prediction of the effect of increasing pressure on the position of equilibria by applying Le Chatelier’s principle was good, but the fact that \({K_{\text{c}}}\) remains constant at fixed temperatures was less well known. 

pH calculations in c(i), c(ii) and c(v) tended to be very good or completely incorrect.

Define the term rate of reaction.

Describe the collision theory.

The Contact process involves this homogeneous equilibrium:

\[{\text{2S}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {{\text{O}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{2S}}{{\text{O}}_{\text{3}}}{\text{(g)}}\,\,\,\,\,\Delta H =  - 198{\text{ kJ}}\]

State and explain how increasing the pressure of the reaction mixture affects the yield of \({\text{S}}{{\text{O}}_{\text{3}}}\).

The Contact process involves this homogeneous equilibrium:

\[{\text{2S}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {{\text{O}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{2S}}{{\text{O}}_{\text{3}}}{\text{(g)}}\,\,\,\,\,\Delta H =  - 198{\text{ kJ}}\]

2.00 mol of \({\text{S}}{{\text{O}}_{\text{2}}}{\text{(g)}}\) are mixed with 3.00 mol of \({{\text{O}}_{\text{2}}}{\text{(g)}}\) in a \({\text{1.00 d}}{{\text{m}}^{\text{3}}}\) container until equilibrium is reached. At equilibrium there are 0.80 mol of \({\text{S}}{{\text{O}}_{\text{3}}}{\text{(g)}}\).

Determine the equilibrium constant (\({K_{\text{c}}}\)) assuming all gases are at the same temperature and pressure.

The Contact process involves this homogeneous equilibrium:

\[{\text{2S}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {{\text{O}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{2S}}{{\text{O}}_{\text{3}}}{\text{(g)}}\,\,\,\,\,\Delta H =  - 198{\text{ kJ}}\]

State the effect of increasing temperature on the value of \({K_{\text{c}}}\) for this reaction.

Outline the economic importance of using a catalyst in the Contact process.

change in concentration of reactant/product with time / rate of change of concentration;

Accept “increase” instead of “change” for product and “decrease” instead of “change” for reactant.

Accept “mass/amount/volume” instead of “concentration”.

Do not accept substance.

collision frequency;

two particles must collide;

particles must have sufficient energy to overcome the activation energy/\(E \geqslant {E_a}\);

Concept of activation energy must be mentioned.

appropriate collision geometry/orientation;

increases yield;

(equilibrium shifts to the right/products as) more gaseous moles in reactants/on left / fewer gaseous moles in products/on right;

\({\text{Eqm[}}{{\text{O}}_2}{\text{]}} = {\text{2.6 (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{Eqm[S}}{{\text{O}}_2}{\text{]}} = {\text{1.2 (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({K_{\text{c}}} = \frac{{{{{\text{[S}}{{\text{O}}_3}]}^2}}}{{{{{\text{[S}}{{\text{O}}_2}{\text{]}}}^2}{\text{[}}{{\text{O}}_2}{\text{]}}}}\);

\({K_{\text{c}}} = 0.17\);

Award [4] for correct final answer.

Ignore units.

\({\text{(}}{K_{\text{c}}}{\text{)}}\) decreases;

catalyst increases rate of reaction / equilibrium reached faster / increases yield of product per unit time;

reduces costs / reduces energy needed;

Do not accept just “increases the yield”.

The definitions of rate of reaction in (a) were poor with many referring to a measure of time rather than a change in concentration. The collision theory was described successfully for the most part with “frequency of collisions” less frequently mentioned. In (c) (i) most realized that the number of moles of gases is important and thus gave a correct answer. Whilst the \({K_{\text{c}}}\) expression was often given correctly in (ii), the calculation of equilibrium mole concentrations was more testing, particularly that for \({\text{[}}{{\text{O}}_{\text{2}}}{\text{]}}\). Many were able to answer (iii) correctly. In part (d) many suggested that it is good to make more of something rather than relating this to a reduction in costs.

The definitions of rate of reaction in (a) were poor with many referring to a measure of time rather than a change in concentration. The collision theory was described successfully for the most part with “frequency of collisions” less frequently mentioned. In (c) (i) most realized that the number of moles of gases is important and thus gave a correct answer. Whilst the \({K_{\text{c}}}\) expression was often given correctly in (ii), the calculation of equilibrium mole concentrations was more testing, particularly that for \({\text{[}}{{\text{O}}_{\text{2}}}{\text{]}}\). Many were able to answer (iii) correctly. In part (d) many suggested that it is good to make more of something rather than relating this to a reduction in costs.

The definitions of rate of reaction in (a) were poor with many referring to a measure of time rather than a change in concentration. The collision theory was described successfully for the most part with “frequency of collisions” less frequently mentioned. In (c) (i) most realized that the number of moles of gases is important and thus gave a correct answer. Whilst the \({K_{\text{c}}}\) expression was often given correctly in (ii), the calculation of equilibrium mole concentrations was more testing, particularly that for \({\text{[}}{{\text{O}}_{\text{2}}}{\text{]}}\). Many were able to answer (iii) correctly. In part (d) many suggested that it is good to make more of something rather than relating this to a reduction in costs.

The definitions of rate of reaction in (a) were poor with many referring to a measure of time rather than a change in concentration. The collision theory was described successfully for the most part with “frequency of collisions” less frequently mentioned. In (c) (i) most realized that the number of moles of gases is important and thus gave a correct answer. Whilst the \({K_{\text{c}}}\) expression was often given correctly in (ii), the calculation of equilibrium mole concentrations was more testing, particularly that for \({\text{[}}{{\text{O}}_{\text{2}}}{\text{]}}\). Many were able to answer (iii) correctly. In part (d) many suggested that it is good to make more of something rather than relating this to a reduction in costs.

The definitions of rate of reaction in (a) were poor with many referring to a measure of time rather than a change in concentration. The collision theory was described successfully for the most part with “frequency of collisions” less frequently mentioned. In (c) (i) most realized that the number of moles of gases is important and thus gave a correct answer. Whilst the \({K_{\text{c}}}\) expression was often given correctly in (ii), the calculation of equilibrium mole concentrations was more testing, particularly that for \({\text{[}}{{\text{O}}_{\text{2}}}{\text{]}}\). Many were able to answer (iii) correctly. In part (d) many suggested that it is good to make more of something rather than relating this to a reduction in costs.

The definitions of rate of reaction in (a) were poor with many referring to a measure of time rather than a change in concentration. The collision theory was described successfully for the most part with “frequency of collisions” less frequently mentioned. In (c) (i) most realized that the number of moles of gases is important and thus gave a correct answer. Whilst the \({K_{\text{c}}}\) expression was often given correctly in (ii), the calculation of equilibrium mole concentrations was more testing, particularly that for \({\text{[}}{{\text{O}}_{\text{2}}}{\text{]}}\). Many were able to answer (iii) correctly. In part (d) many suggested that it is good to make more of something rather than relating this to a reduction in costs.

At a temperature just above 700 K it is found that when 1.60 mol of hydrogen and 1.00 mol of iodine are allowed to reach equilibrium in a \({\text{4.00 d}}{{\text{m}}^{\text{3}}}\) flask, the amount of hydrogen iodide formed in the equilibrium mixture is 1.80 mol. Determine the value of the equilibrium constant at this temperature.

Outline why the value for the standard enthalpy change of formation of hydrogen is zero.

Calculate the standard enthalpy change for the hydrogenation of propene.

Calculate the standard entropy change for the hydrogenation of propene.

Determine the value of \(\Delta {G^\Theta }\) for the hydrogenation of propene at 298 K.

At 298 K the hydrogenation of propene is a spontaneous process. Determine the temperature above which propane will spontaneously decompose into propene and hydrogen.

amount of \({{\text{H}}_2}\) remaining at equilibrium \( = 1.60 - \frac{{1.80}}{2} = 0.70{\text{ mol}}\);

amount of \({{\text{I}}_2}\) remaining at equilibrium \( = 1.0 - \frac{{1.80}}{2} = 0.10{\text{ mol}}\);

\({K_{\text{c}}} = \frac{{{{(1.80/4.0)}^2}}}{{(0.70/4.00) \times (0.10/4.00)}}/\frac{{{{1.80}^2}}}{{0.70 \times 0.10}}\);

\({K_{\text{c}}} = \frac{{{{(1.80)}^2}}}{{0.70 \times 0.10}} = 46.3\);

Award [4] for correct final answer.

by definition \(\Delta H_{\text{f}}^\Theta \) of elements (in their standard states) is zero / no reaction involved / OWTTE;

\(\Delta H =  - 104 - ( + 20.4)\);

\( =  - 124.4{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

Award [1 max] for 124.4 (kJ\(\,\)mol⁻¹).

Award [2] for correct final answer.

\(\Delta S = 270 - (267 + 131)\);

\( =  - 128{\text{ (J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

Award [1 max] for +128 ( J\(\,\)K⁻¹mol⁻¹).

Award [2] for correct final answer.

\(\Delta G = \Delta H - {\text{T}}\Delta S =  - 124.4 - \frac{{( - 128 \times 298)}}{{1000}}\);

\( =  - 86.3{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\);

Units needed for the mark.

Award [2] for correct final answer.

Allow ECF if only one error in first marking point.

\(\Delta G = \Delta H - {\text{T}}\Delta S = 0/\Delta H = {\text{T}}\Delta S\);

\({\text{T}} = \frac{{ - 124.4}}{{ - 128/1000}} = 972{\text{ K}}/699{\text{ }}^\circ {\text{C}}\);

Only penalize incorrect units for T and inconsistent ΔS value once in (iv) and (v).

This was the most popularly answered question. Most candidates were able to give a good description of the characteristics of homogenous equilibrium, and apply Le Chatelier‟s Principle to explain the effect of catalysts and changes of temperature and pressure on the position of equilibrium and the equilibrium constant. A good majority were able to calculate the value of \({K_{\text{c}}}\) although a significant number of candidates incorrectly used the initial rather than the equilibrium concentrations.

Although most candidates clearly understood the concept of standard enthalpy change of formation many were unable to explain why the value for hydrogen is zero. Many responses neglected to mention that \({{\text{H}}_{\text{2}}}\) is an element in its standard state.

Most candidate were able to calculate \(\Delta H\) and \(\Delta S\) although some inverted the equation and gave a positive value instead of negative answer or confused the values for propane and propene.

There were some inconsistencies in the use of units and significant figures when calculating \(\Delta G\) from \(\Delta H\) and \(\Delta S\) values although there was a significant improvement in this area compared to previous.

There were some inconsistencies in the use of units and significant figures when calculating \(\Delta G\) from \(\Delta H\) and \(\Delta S\) values although there was a significant improvement in this area compared to previous.

There were some inconsistencies in the use of units and significant figures when calculating \(\Delta G\) from \(\Delta H\) and \(\Delta S\) values although there was a significant improvement in this area compared to previous. This error resulted in some very strange temperatures for the thermal decomposition of propane to propene.

Calculate the standard enthalpy change of this reaction, using the values of enthalpy of combustion in Table 12 of the Data Booklet.

Calculate the standard entropy change for this reaction, \(\Delta {S^\Theta }\), using Table 11 of the Data Booklet and given:

\({S^\Theta }{\text{(CO)}} = 198{\text{ J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}{\text{ and }}{S^\Theta }{\text{(}}{{\text{H}}_{\text{2}}}{\text{)}} = 131{\text{ J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}\).

Calculate, stating units, the standard free energy change for this reaction, \(\Delta {G^\Theta }\), at 298 K.

Predict, with a reason, the effect of an increase in temperature on the spontaneity of this reaction.

1.00 mol of \({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\) is placed in a closed container of volume \({\text{1.00 d}}{{\text{m}}^{\text{3}}}\) until equilibrium is reached with CO and \({{\text{H}}_{\text{2}}}\). At equilibrium 0.492 mol of \({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\) are present. Calculate \({K_{\text{c}}}\).

\({\text{C}}{{\text{H}}_3}{\text{OH}} + \frac{3}{2}{{\text{O}}_2} \to {\text{C}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}}\)     \(\Delta H_{\text{c}}^{^\Theta } =  - 726{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\)

\({\text{CO}} + \frac{1}{2}{{\text{O}}_2} \to {\text{C}}{{\text{O}}_2}\)     \(\Delta H_{\text{c}}^{^\Theta } =  - 283{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\)

\({{\text{H}}_2} + \frac{1}{2}{{\text{O}}_2} \to {{\text{H}}_2}{\text{O}}\)     \(\Delta H_{\text{c}}^{^\Theta } =  - 286{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\)

Award [1 max] for three correct values.

Mark can be implicit in calculations.

\((\Delta H_{\text{R}}^{^\Theta } = ){\text{ }}2( - 286) + ( - 283) - ( - 726)\);

\( - {\text{129 (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

Award [3] for correct final answer.

Award [2 max] for +129 (kJ\(\,\)mol^–1).

\((\Delta {S^\Theta } = 240 - 198 - 2 \times 131 = ){\text{ }} - 220{\text{ (J}}\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

\(\left( { - 129 - 298( - 0.220) = } \right){\text{ }} - 63.4{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\);

Award [1] for correct numerical answer and [1] for correct unit if the conversion has been made from J to kJ for \(\Delta {S^\Theta }\).

not spontaneous at high temperature;

\(T\Delta {S^\Theta } < \Delta {H^\Theta }\) and \(\Delta {G^\Theta }\) positive;

\(n{\text{(CO)}} = 0.508{\text{ (mol)}}\);

\(n({{\text{H}}_2}) = 2 \times 0.508{\text{ (mol)}}\);

\({K_{\text{c}}}{\text{ }}\left( { = \frac{{0.492}}{{0.508 \times {{(2 \times 0.508)}^2}}}} \right) = 0.938\);

Accept answer in range between 0.930 and 0.940.

Award [3] for correct final answer.

Award [2] for K_c = 1.066 if (c)(ii) is correct.

In (i), the most common error was \( + {\text{129 kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) but in (ii) the answer was often correct.

In (i), the most common error was \( + {\text{129 kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\) but in (ii) the answer was often correct.

Units tended to get muddled in (iii) and many marks were awarded as “error carried forward”.

Few were able to explain the \(\Delta H\) and \(T\Delta S\) relationship in detail in (iv).

Equilibrium was well understood in general with many candidates gaining one of the two available marks. “Equal rates” was more often given than the constancy of macroscopic properties for the second mark. The \({K_{\text{c}}}\) expression was given correctly by the vast majority of candidates (including the correct brackets and indices) but many had difficulty with the equilibrium concentrations in (iii).

The changes in equilibrium position were well understood for the most part although if a mark were to be lost it was for not mentioning the number of moles of gas.

State and explain the effect of increasing the pressure on the yield of sulfur trioxide.

State the effects of a catalyst on the forward and reverse reactions, on the position of equilibrium and on the value of \({K_{\text{c}}}\).

When a mixture of 0.100 mol NO, 0.051 mol \({{\text{H}}_{\text{2}}}\) and 0.100 mol \({{\text{H}}_{\text{2}}}{\text{O}}\) were placed in a \({\text{1.0 d}}{{\text{m}}^{\text{3}}}\) flask at 300 K, the following equilibrium was established.

\(2{\text{NO(g)}} + 2{{\text{H}}_2}{\text{(g)}} \rightleftharpoons {{\text{N}}_2}{\text{(g)}} + 2{{\text{H}}_2}{\text{O(g)}}\)

At equilibrium, the concentration of NO was found to be \({\text{0.062 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\). Determine the equilibrium constant, \({K_{\text{c}}}\), of the reaction at this temperature.

Outline two differences between an electrolytic cell and a voltaic cell.

Electroplating is an important application of electrolysis. State the composition of the electrodes and the electrolyte used in the silver electroplating process.

yield (of \({\text{S}}{{\text{O}}_{\text{3}}}\)) increases / equilibrium moves to right / more \({\text{S}}{{\text{O}}_{\text{3}}}\) formed;

3 gaseous molecules \( \to \) 2 gaseous molecules / decrease in volume of gaseous molecules / fewer gaseous molecules on right hand side;

Do not allow ECF.

rates of both forward and reverse reactions increase equally;

no effect on position of equilibrium;

no effect on value of [3] \({K_{\text{c}}}\);

\({\text{2NO(g)}} + {\text{2}}{{\text{H}}_2}{\text{(g)}} \rightleftharpoons {{\text{N}}_2}{\text{(g)}} + {\text{2}}{{\text{H}}_2}{\text{O(g)}}\)

\({\text{[}}{{\text{H}}_{\text{2}}}{\text{] at equilibrium}} = 0.013{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[}}{{\text{N}}_{\text{2}}}{\text{] at equilibrium}} = 0.019{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{[}}{{\text{H}}_{\text{2}}}{\text{O] at equilibrium}} = 0.138{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({K_{\text{c}}} = {\text{[}}{{\text{N}}_2}{\text{][}}{{\text{H}}_2}{\text{O}}{{\text{]}}^2}{\text{/[NO}}{{\text{]}}^2}{{\text{[}}{{\text{H}}_2}{\text{]}}^2} = {\text{(0.019)(0.138}}{{\text{)}}^2}{\text{/(0.062}}{{\text{)}}^2}{{\text{(0.013)}}^2} = 5.6 \times {10^2}\);

Award [4] for final correct answer.

Accept any value also in range 557–560.

Do not penalize significant figures.

electrolytic cell converts electrical energy to chemical energy and voltaic cell converts chemical energy to electrical energy / electrolytic cell uses electricity to carry out a (redox) chemical reaction and voltaic cell uses a (redox) chemical reaction to produce electricity / electrolytic cell requires a power supply and voltaic cell does not;

electrolytic cell involves a non-spontaneous (redox) reaction and voltaic cell involves a spontaneous (redox) reaction;

in an electrolytic cell, cathode is negative and anode is positive and vice-versa for a voltaic cell / electrolytic cell, anode is positive and voltaic cell, anode is negative / electrolytic cell, cathode is negative and voltaic cell, cathode is positive;

voltaic cell has two separate solutions and electrolytic cell has one solution / voltaic cell has salt bridge and electrolytic cell has no salt bridge;

electrolytic cell, oxidation occurs at the positive electrode/anode and voltaic cell, oxidation occurs at the negative electrode/anode and vice-versa;

Cathode/negative electrode:

object to be plated;

Allow a specific example here e.g. spoon.

Accept inert metal/graphite.

Do not accept silver halides or their formulae.

Anode/positive electrode:

Silver/Ag;

Electrolyte:

\({{\text{[Ag(CN}}{{\text{)}}_{\text{2}}}{\text{]}}^ - }\);

Allow silver nitrate/AgNO3 / silver cyanide/any other suitable silver salt/solution.

Do not accept AgCl.

In (ii) an overwhelming number of candidates were able to score the first mark but did not refer to the gaseous state and hence lost the second mark.

Part (iv) was another question where candidates easily scored the second and third mark. Although this has been asked a number of times in recent sessions, some candidates still do not state that the rates of both the forward and reverse reactions increase equally.

(b) was considered a very challenging question for candidates, and usually only the better candidates scored all four marks.

In (c) (i) most candidates scored two marks.

Electroplating was a topic only partially understood by candidates, and so only a few candidates obtained all three marks in (v). Often the nature of the electrode was mixed up or in many cases incorrect electrolytes were given.

Calculate the percentage by mass of nitrogen in urea to two decimal places using section 6 of the data booklet.

Suggest how the percentage of nitrogen affects the cost of transport of fertilizers giving a reason.

The structural formula of urea is shown.

Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.

Urea can be made by reacting potassium cyanate, KNCO, with ammonium chloride, NH₄Cl.

KNCO(aq) + NH₄Cl(aq) → (H₂N)₂CO(aq) + KCl(aq)

Determine the maximum mass of urea that could be formed from 50.0 cm³ of 0.100 mol dm⁻³ potassium cyanate solution.

State the equilibrium constant expression, K_c.

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

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

Suggest one reason why urea is a solid and ammonia a gas at room temperature.

Sketch two different hydrogen bonding interactions between ammonia and water.

The combustion of urea produces water, carbon dioxide and nitrogen.

Formulate a balanced equation for the reaction.

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

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

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

The mass spectrum of urea is shown below.

Identify the species responsible for the peaks at m/z = 60 and 44.

The IR spectrum of urea is shown below.

Identify the bonds causing the absorptions at 3450 cm⁻¹ and 1700 cm⁻¹ using section 26 of the data booklet.

Predict the number of signals in the ¹H NMR spectrum of urea.

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

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

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

«% nitrogen = \(\frac{{{\text{2}} \times {\text{14.01}}}}{{{\text{60.07}}}}\) \( \times \) 100 =» 46.65 «%»

Award [2] for correct final answer.

Award [1 max] for final answer not to two decimal places.

[2 marks]

«cost» increases AND lower N% «means higher cost of transportation per unit of nitrogen»

OR

«cost» increases AND inefficient/too much/about half mass not nitrogen

Accept other reasonable explanations.

Do not accept answers referring to safety/explosions.

[1 mark]

Note: Urea’s structure is more complex than that predicted from VSEPR theory.

[3 marks]

n(KNCO) «= 0.0500 dm³ \( \times \) 0.100 mol dm^–3» = 5.00 \( \times \) 10^–3 «mol»

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

Award [2] for correct final answer.

[2 marks]

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

[1 mark]

«K_c» decreases AND reaction is exothermic

OR

«K_c» decreases AND ΔH is negative

OR

«K_c» decreases AND reverse/endothermic reaction is favoured

[1 mark]

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

«K_c =» 2 \( \times \) 10^–9

OR

1.69 \( \times \) 10^–9

OR

10^–9

Accept range of 20-20.2 for M1.

Award [2] for correct final answer.

[2 marks]

Any one of:

urea has greater molar mass

urea has greater electron density/greater London/dispersion

urea has more hydrogen bonding

urea is more polar/has greater dipole moment

Accept “urea has larger size/greater van der Waals forces”.

Do not accept “urea has greater intermolecular forces/IMF”.

[1 mark]

Award [1] for each correct interaction.

If lone pairs are shown on N or O, then the lone pair on N or one of the lone pairs on O MUST be involved in the H-bond.

Penalize solid line to represent H-bonding only once.

[2 marks]

2(H₂N)₂CO(s) + 3O₂(g) → 4H₂O(l) + 2CO₂(g) + 2N₂(g)

correct coefficients on LHS

correct coefficients on RHS

Accept (H₂N)₂CO(s) + \(\frac{3}{2}\)O₂(g) → 2H₂O(l) + CO₂(g) + N₂(g).

Accept any correct ratio.

[2 marks]

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

[1 mark]

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

co-ordinate/dative/covalent bonds

[2 marks]

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

OR

C–N bond partial double bond character

OR

delocalization «of electrons occurs across molecule»

OR

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

[1 mark]

60: CON₂H₄⁺

44: CONH₂⁺

Accept “molecular ion”.

[2 marks]

3450 cm^–1: N–H

1700 cm^–1: C=O

Do not accept “O–H” for 3450 cm^–1.

[2 marks]

1

[2 marks]

singlet

Accept “no splitting”.

[1 mark]

acts as internal standard

OR

acts as reference point

one strong signal

OR

12 H atoms in same environment

OR

signal is well away from other absorptions

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

[2 marks]

(i) Deduce the equilibrium constant expression, K_c, for this reaction.

(ii) At exactly 600°C the value of the equilibrium constant is 0.200. Calculate the standard Gibbs free energy change, , for the reaction, in kJ, using sections 1 and 2 of the data booklet. State your answer to three significant figures.

(iii) The standard enthalpy change of formation of phosgene, \(\Delta H_f^\Theta \), is −220.1kJmol⁻¹. Determine the standard enthalpy change, \(\Delta H_{}^\Theta \), for the forward reaction of the equilibrium, in kJ, using section 12 of the data booklet.

(iv) Calculate the standard entropy change, \(\Delta S_{}^\Theta \), in JK⁻¹, for the forward reaction at 25°C, using your answers to (a) (ii) and (a) (iii).  (If you did not obtain an answer to (a) (ii) and/or (a) (iii) use values of +20.0 kJ and −120.0 kJ respectively, although these are not the correct answers.)

One important industrial use of phosgene is the production of polyurethanes. Phosgene is reacted with diamine X, derived from phenylamine.

(i) Classify diamine X as a primary, secondary or tertiary amine.

(ii) Phenylamine, C₆H₅NH₂, is produced by the reduction of nitrobenzene, C₆H₅NO₂. Suggest how this conversion can be carried out.

(iii) Nitrobenzene can be obtained by nitrating benzene using a mixture of concentrated nitric and sulfuric acids. Formulate the equation for the equilibrium established when these two acids are mixed.

(iv) Deduce the mechanism for the nitration of benzene, using curly arrows to indicate the movement of electron pairs.

The other monomer used in the production of polyurethane is compound Z shown below.

(i) State the name, applying IUPAC rules, of compound Z and the class of compounds to which it belongs.

Name:

Class:

(ii) Deduce the number of signals you would expect to find in the ¹H NMR spectrum of compound Z, giving your reasons.

The mass spectrum and infrared (IR) spectrum of compound Z are shown below:

Mass spectrum

IR spectrum

(iii) Identify the species causing the large peak at m/z=31 in the mass spectrum.

(iv) Identify the bond that produces the peak labelled Q on the IR spectrum, using section 26 of the data booklet.

Phenylamine can act as a weak base. Calculate the pH of a 0.0100 mol dm⁻³ solution of phenylamine at 298K using section 21 of the data booklet.

(i)
\( \ll {K_{\rm{C}}}{\rm{ = }} \gg \frac{{\left[ {{\rm{COC}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{CO}}} \right]\left[ {{\rm{C}}{{\rm{l}}_2}} \right]}}\)

(ii)
T«= 600 + 273» = 873K

ΔG^Θ = −8.31 × 873 × ln (0.200)
OR
ΔG^Θ = « + » 11676 «J»
ΔG^Θ = « + » 11.7 «kJ»

Accept 11.5 to 12.0.
Award final mark only if correct sig fig.
Award [3] for correct final answer.

(iii)
ΔH^Θ = −220.1 − (−110.5)
ΔH^Θ = −109.6 «kJ»

Award [2] for correct final answer.
Award [1] for −330.6, or +109.6 «kJ».

(iv)
ΔG^Θ= −109.6 − (298 × ΔS^Θ) = +11.7 «kJ»
ΔS^Θ«\(\frac{{\left( {11.7 + 109.6} \right) \times {{10}^3}}}{{298}}\)»= −407 «JK⁻¹»

Award [2] for correct final answer.
Award [2] for −470 «JK⁻¹» (result from given values).
Do not penalize wrong value for T if already done in (a)(ii).
Award [1 max] for −0.407 «kJ K⁻¹».
Award [1 max] for −138.9 «J K⁻¹».

(i)
primary

(ii)
ALTERNATIVE 1:
«heat with» tin/Sn AND hydrochloric acid/HCl
aqueous alkali/OH^–(aq)

(i)
Name: ethane-1,2-diol
Class: alcohol«s»

\({K_{\rm{b}}} \approx \frac{{{{\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}^2}}}{{\left[ {{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right]}} = {10^{ - 9.13}}/7.413 \times {10^{ - 10}}\)

\(\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = \sqrt {0.0100 \times {{10}^{ - 9.13}}}  = 2.72 \times {10^{ - 6}}\)

\(\left[ {{{\rm{H}}^ + }} \right] = \frac{{1 \times {{10}^{ - 14}}}}{{2.72 \times {{10}^{ - 6}}}} = 3.67 \times {10^{ - 9}}\)

OR

pOH = 5.57

pH = −log [H⁺] = 8.44

Accept other approaches to the calculation.
Award [4] for correct final answer.
Accept any answer from 8.4 to 8.5.