Chemical equilibrium and Kc Module 3

Answer

B

Mark scheme

Examiner's comment

Overall, candidates had little difficulty with this question and most selected B.

Answer

B

Mark scheme

Examiner's comment

The correct option of B was missed by a quarter of candidates. C was the main distractor, ignoring that a catalyst has no effect on the position of equilibrium.

Mark scheme

Examiner's comment

Most candidates identified at least one feature of a dynamic equilibrium. ‘In a closed system’ was given most often, and other mark-worthy features were ‘the same rate for forward and reverse reactions’ and ‘concentrations do not change’. Less successful responses that were not given marks included ‘forward reaction is the same as reverse’ and ‘concentrations are the same’.

Notes from the mark scheme and report

• DO NOT ALLOW concentration of reactants = concentration of products
• IGNORE 'reversible reaction'

Mark scheme

Examiner's comment

Following browsing of candidate responses, marks were given for correctly writing the Kc expression, substituting the terms with values, and finally, giving the calculated response to three significant figures and standard form. Most candidates scored either 2 or 3 marks. However, this question highlights the need for candidates to read the question carefully, with many responses not giving the Kc expression, or not giving the answer in standard form, or both.

Notes from the mark scheme and report

• square brackets required
• 3 sig figs AND standard form (DO NOT ALLOW 0.0261 × 10^0)
• ECF from inverted Kc only (3.83 × 10^1, 2 marks)
• IGNORE units

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This question was answered confidently with most candidates identified the conditions for a maximum equilibrium yield from the equation: a high pressure from fewer gaseous moles in the product and a low temperature from the exothermic reaction. This was all that the question asked but candidates often wrote far too much, considering operational conditions, rate, safely and expense. Unfortunately, candidates could end up contradicting themselves and wasting valuable time. Candidates are advised to read what the question is asking and to limit their response to this. The calculation was answered well, with most identifying that 2 significant figures is appropriate as the lowest number of significant figures present in the data provided in the question. Thus 0.48 mol dm–3 was commonly seen, rather than 0.476. The Kc expression was usually correct and most candidates rearranged this correctly to make [H2] the subject. The square root calculation showed that some candidates did not appear to have the required calculator skills. A significant number of candidates showed the Kc expression upside down and the mark scheme allowed ECF to be applied following this error.

Notes from the mark scheme and report

• high pressure because fewer gaseous moles on right (DO NOT ALLOW gaseous atoms)
• low temperature because forward reaction exothermic
• square brackets required in Kc
• square root needed
• 0.48 to 2 SF (0.476 loses a mark)
• ECF only from inverted Kc or no square root

Skills: M1.1 Use an appropriate number of significant figures; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Answer

C

Mark scheme

Examiner's comment

The correct answer was C. This was a well answered question with most candidates gaining the mark. The most common error was B, where reactants and product concentration became equal. It is important that candidates can apply definitions and theory to diagrams.

Notes from the mark scheme and report

• Dynamic equilibrium exists in a closed system when the rate of the forward reaction is equal to the rate of the reverse reaction and the concentrations of reactants and products do not change. It is a misconception that equilibrium is at the point where reactants and product concentration became equal.

Skills: M3.1 Translate information between graphical, numerical and algebraic forms

Answer

B

Mark scheme

Examiner's comment

Almost all candidates were aware of how to express the Kc expression for an equilibrium, choosing the correct option B.

Mark scheme

Examiner's comment

Most candidates identified at least one feature of a dynamic equilibrium. ‘In a closed system’ was given marks most often, and other mark-worthy features were ‘the same rate for forward and reverse reactions’ and ‘concentrations do not change’. Less successful responses that were not given marks included ‘reversible reaction’ and ‘concentrations are the same’.

Notes from the mark scheme and report

• rate of forward = rate of reverse
• concentrations constant
• closed system
• DO NOT ALLOW concentration of reactants = concentration of products

Mark scheme

Examiner's comment

This long-response question was approached very well and there were some excellent and concise responses. Only the less successful responses did not identify the main trends. Candidates are well-versed with tackling this type of question and most were able to write sensible explanations. However, marks could not be given for insufficiently specific responses. For example, some candidates stated that increased pressure would shift the equilibrium in the direction with fewer moles, without stating what that direction was for this equilibrium. It was also common to see explanations that contradicted earlier statements. Responses could be improved if candidates read through what they write to see if it makes cohesive sense. Nearly half of the scripts were given the full 3 marks, the question proving to be a good discriminator.

Notes from the mark scheme and report

• low temperature because forward reaction exothermic
• high pressure because fewer gaseous moles on right
• equilibrium shift stated correctly at least once
• DO NOT ALLOW gaseous atoms

Answer

B

Mark scheme

Examiner's comment

This question was for the most part answered correctly with B. Errors came from not recognising the reaction is endothermic and therefore its equilibrium would shift to the left when the temperature decreases, ruling out option 2. Section B overview The section contained questions from all aspects of the specification. Candidates found many of the questions, including those with an extended response nature, straightforward and the majority managed to cope with the mathematical content. Candidates should use calculator values in their intermediate calculations to avoid rounding errors in their final answer. Well set out calculations which clearly indicate what each numerical value represents allow error carried forward marks to be awarded in the event of an incorrect answer. Some candidates were less confident and knowledgeable with the less demanding aspects of some questions. These were specifically: ionic equations, equations involving group 2 metals and their compounds, equilibria shifts and enthalpy changes. There was evidence that candidates wrote far more than necessary in some responses. For example, Question16 (b) was allotted a very generous 9 lines for an answer but many candidates provided repetitive responses and requoted the question. Many more effective responses provided 4 or 5 line answers but gained full marks. With the exception of questions marked using a level of response mark scheme, candidates should look to the marks available to give insight to the amount of detail, and discrete points, that needed to be made.

Notes from the mark scheme and report

• Practice multiple choice questions can improve the skill in solving and identifying the distractors. Exposure to this type of question style will decrease the time taken over each question. These can often form the basis of end of topic tests.
• Multiple choice question quizzes can be found via the resource-finder on Teach Cambridge and there are instructions on how to use the online versions of the multiple choice quizzes.

Mark scheme

Examiner's comment

This question was answered for the most part correctly with many candidates scoring 2 marks for the explanations of the effect on the equilibrium position by the changing of the temperature and pressure. Most candidates were able to recognise the changes had opposite effects but could not score the final mark, as their response needed the concept of opposing factors, or ‘we don’t know which factor is dominant’. Some did not write anything about equilibrium and attempted answers based on rate, or loss of energy/chemicals to the surroundings.

Notes from the mark scheme and report

• higher T shifts left (forward reaction exothermic)
• higher P shifts right (fewer gaseous moles)
• cannot tell which opposing effect is dominant
• DO NOT ALLOW both shifts in the same direction

Mark scheme

Examiner's comment

Only about half the candidates obtained this mark suggesting that this term had not been recalled by many. A common error stated that the catalyst and reactants were in the same state. Responses from lower-attaining candidates sometimes had the appearance of guesses.

Notes from the mark scheme and report

• all species (reactants AND products) in the same state/phase
• IGNORE reactants and catalyst have same state

Mark scheme

Examiner's comment

This long-answer question was approached very well and there were some excellent and concise answers. Only the less successful responses did not identify the main trends. Marks were often lost for responses in general terms, rather than related to the scenario in the question. For example, some candidates stated that the equilibrium would shift in the direction with fewer moles, without stating what that direction was for this equilibrium. Some candidates contradicted themselves due to a lack of fully understanding how Le Chatelier’s Principle should be applied. It was also common to see lengthy responses in which candidates discussed the effect of rate and compromise (not asked for in this question) and then confused equilibrium yield with overall yield.

Notes from the mark scheme and report

• pressure: shifts right, fewer gaseous moles on right
• temperature: increased T shifts left as forward reaction exothermic
• catalyst: no shift in position
• IGNORE 'no increase in yield' for catalyst

Mark scheme

Examiner's comment

The Kc expression was written well by many inverted the expression or added rather than multiplying the values. Indices were usually included. Many candidates gave responses to more than 2 significant figures (2SF) and some did not give the response in standard form. Candidates should be made aware that the term in the question of ‘most appropriate’ means that the final response should be shown to the least number of significant figures in the supplied data. In this scenario, all values were provided to 2SF and so the final response should also be expressed to 2SF.

Notes from the mark scheme and report

• square brackets required
• 2 SF AND standard form (7.9 × 10^4)
• ECF only from inverted expression (1.3 × 10–5)
• DO NOT ALLOW + in denominator
• The Mathematical Skills Handbook provides guidance on the use of significant figures which can be shared with students.

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Few candidates successfully answered this difficult application question. Some candidates considered ensuring complete combustion, or not wasting reactants, or various environmental reasons. The question did supply a hint: ‘in terms of equilibrium’ but this was ignored by most candidates. The idea of equilibrium shift to the right was essential.

Notes from the mark scheme and report

• equilibrium shifts to the right/towards SO3

Mark scheme

Examiner's comment

As with the stock titration calculation in Question 5 (a), candidates were much more comfortable in tackling a stock Kc calculation. These have been assessed in several previous H432/01 exams and many excellent responses were seen, gaining all 5 marks. Error carried forwards allowed for an early slip in the calculation to be credited subsequently for a correct method. The hardest part of the calculation was the initial determination of the equilibrium moles and concentrations. The equilibrium moles of O2 was the main error, presumably from the 2:1:2 stoichiometry. Some candidates multiplied by 2 instead of dividing by 2 for the concentration. A common error was for candidates to use a partial pressure approach. These errors lost intermediate marks by credit could then be awarded for using the correct Kc expression to generate a Kc value with correct units. Very few candidates did not gain any marks.

Notes from the mark scheme and report

• equilibrium moles SO2 6.20 × 10–3 and O2 4.80 × 10–2
• ÷ 2.00 dm3 for concentrations
• Kc = 2930 dm3 mol–1
• common errors: no ÷ 2 (1465), original values used (21.6), SO3 not squared, inverted Kc
• units must match Kc expression used

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Answer

D

Mark scheme

Examiner's comment

Candidates produced a variety of responses with just over half choosing the correct option D. Option C was the main distractor.

Answer

A

Mark scheme

Examiner's comment

This question discriminated very well. Evidence from annotations showed that the successful candidates substituted the 0.1 mol dm–3 concentration into the Kc expression for each equilibrium and calculated Kc by calculator. This method should guarantee the correct answer.

Mark scheme, with the levels and indicative points

Examiner's comment

This Level of Response question was generally well answered with many candidates achieving maximum marks by simply considering what was required in the question. Responses were often split between a calculation on the main paper and the conditions explanation on extra pages. The calculation errors included no shift or incorrect shift in the equilibrium values. Not calculating the concentration or incorrectly multiplying by 8 rather than dividing by 8. Some candidates attempted a ‘hybrid’ calculation of Kp by trying to calculate a mole fraction and partial pressures. There was a number of candidates who confidently worked out the value of Kc. There were also some very good analyses of the operational conditions. Many of those who had done well on the calculation treated the explanation as an afterthought, not giving it enough attention to give them an answer that would access Level 3. An example of a complete answer, showing a good level of communication in the description and the calculation layout, achieving Level 3 (6 marks) is shown above. This candidate gave a clear method using an “ICE” table to calculate the number of moles at equilibrium. The calculation of the new concentrations can be seen by the use of [ ] and division by 8. The candidate then shows the Kc expression and substitutes the numerical values before successfully calculating the value and includes the units. This is followed by a concise explanation of the conditions used by industry.

Notes from the mark scheme and report

• equilibrium moles N2 1.12, H2 3.36 (0.08 and 0.24 reacted)
• divide by 8.00 dm3
• Kc = 0.0386 dm6 mol–2
• low temperature/high pressure for yield vs high temperature for rate, cost and safety of high pressure
• catalyst allows lower temperature

Skills: M0.0 Recognise and make use of appropriate units in calculations; M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

Many candidates gave the correct reason to agree with student 2. Those who agreed with student 1 did not see the equation as a heterogeneous equilibrium system. There were a small number of responses agreeing with student 2 but for the wrong reason – such as a confusion about how the position of equilibrium can change when the value of Kp stays constant. Candidates are advised to read through and address all parts of the question as a minority of students didn’t identify which student was correct but gave a correct explanation.

Notes from the mark scheme and report

• Student 2 correct AND same number of moles of GAS on each side (solids ignored)

Mark scheme, with the levels and indicative points

Examiner's comment

Only a small number of candidates were given all 6 marks for their response. Many responses made little or no reference to equilibrium, considering only the rate of reaction. A clear understanding of the difference between rate and equilibrium was essential to scoring highly on this question. Excellent descriptions were given regarding the role of the catalyst on the rate of reaction, possibly influenced by the previous question. Some gave detailed responses in terms of the impact of both pressure and temperature on the equilibrium position but then did not give the changes made by industry such as reducing temperature due to energy demand or increasing pressure to increase rate. Some candidates indicated that 15atm was a low pressure including reference to it being “lower than normal atmospheric pressure” which suggested that they did not understand either the measurement in atmospheres and/or scale of pressure. A common incorrect response was to suggest that a higher pressure gives increased yield as more moles on right/products. Some stated that the reaction was exothermic, clearly confusing the enthalpy change signs. Some also described the “endothermic side” demonstrating a lack of understanding that enthalpy changes require a difference between products and reactants. Many did not correctly consider the use of excess of steam to increase concentration of reactant and shift equilibrium to products/right but talked in terms of limiting reagents, e.g. “This is used so all of the methane is reacted” or increased collisions. Many were prone generalisations, e.g. “These conditions are used industrially to achieve the most hydrogen efficiently with the least amount of undesired (waste) products”. Candidates need to be encouraged to plan their answers, even using subheadings to organise information and follow the key approach: point, evidence, explain. Candidates should also be encouraged to re-read their answers to check communication is clear and accurate without any contradictory statements. Many included irrelevant or inaccurate information such as describing the toxicity of carbon monoxide, so reducing yield of that being beneficial, or “using a catalyst improves atom economy”. This exemplar shows a candidate who achieved L3 with all 6 marks given. They have considered both the effect of temperature and pressure on equilibrium and potential compromises used industrially. They have also considered the impact of using a catalyst. Their response is logically structured and has a clear line of reasoning.

Notes from the mark scheme and report

• endothermic so high temperature shifts right
• 2 mol gas to 4 mol so low pressure favours products but reduces rate
• compromise pressure
• high temperature costs energy
• catalyst increases rate without changing equilibrium position
• excess steam shifts right
• This question was good at highlighting candidates understanding of the impact of different conditions on equilibrium reactions. We have produced a delivery guide to help with teaching about equilibrium: https://www.ocr.org.uk/Images/261544-equilibrium.pdf

Mark scheme

Examiner's comment

This question was well answered with most candidates scoring both marks. Common errors seen included: inverse Kc expression, final answer not to 3 significant figures, adding rather than multiplying concentrations, calculator error ([H2]2 instead of [H2]3) or transcription errors (e.g. using 0.11 rather than 0.111).

Notes from the mark scheme and report

• 3 SF (24.1)
• ALLOW ECF to 3 SF only from inverted expression (0.0414)
• DO NOT ALLOW + signs in expression
• IGNORE units

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Examiner's comment

This question asked candidates to explain the different rates from a novel experiment carried out at 50ºC and 90ºC, and to predict the ∆H sign for the forward reaction. Candidate explanations for the rates were often superficial, solely in terms of greater energy at 90ºC. Many responses referred neither to the different frequency of collisions nor the greater number of particles exceeding the activation energy at 90ºC. Most candidates predicted that ∆H would have a negative sign. Candidates were expected to link the evidence from the absorbance data in the graph to less [Co(H2O)6]2+ being present at 90ºC. When experimental information has been presented, candidates are advised to look for the evidence responsible in their explanations.

Notes from the mark scheme and report

• at 90 °C faster rate AND more frequent collisions
• more particles have energy ≥ Ea
• equilibrium [Co(H2O)6]2+ is lower at 90 °C so forward reaction exothermic/∆H negative
• IGNORE more successful collisions

Skills: M3.1 Translate information between graphical, numerical and algebraic forms; PAG9 Rates of reaction – continuous monitoring

Mark scheme

Examiner's comment

In contrast with Question 4 (a), most candidates did interpret the graphical information provided and related this to the reduced concentration of CoCl42– ions and the increased concentration of [Co(H2O)6]2+ ions. Most candidates also referred to Equilibrium 4.1 to conclude that the equilibrium shifts to the right. Only the very best candidates recognised that the increase in Cl– concentration following the initial addition of AgNO3 was 4 times greater than the increase in the concentration of Co(H2O)6]2+, arising from the 4 : 1 ratio in the stoichiometry in the equation.

Notes from the mark scheme and report

• Cl– increase is 4 × the change in the complex ions
• equilibrium shifts to the right (to replace Cl–)

Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms

Answer

B

Mark scheme

Answer

C

Mark scheme

Mark scheme

Notes from the mark scheme and report

• rates of forward and reverse reactions equal OR concentrations constant
• DO NOT ALLOW concentrations 'are the same'

Mark scheme

Examiner's comment

Question 20(c): This question asked the candidate to calculate Kp.

Notes from the mark scheme and report

• marked by column: Kp smaller/same/same
• amount of NO2 smaller/greater/same
• initial rate greater for all three

Mark scheme

Notes from the mark scheme and report

• low pressure because more gaseous moles on right-hand side
• high temperature because forward reaction is endothermic
• ORA for reverse reaction

Mark scheme

Notes from the mark scheme and report

• square brackets essential – DO NOT ALLOW curved brackets

Mark scheme

Notes from the mark scheme and report

• rearrange Kc: 15.4 × 0.57 × 0.40^2 = 1.4
• ALLOW ECF from incorrect expression in (a)

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Notes from the mark scheme and report

• to the right/towards products

Skills: M0.3 Estimate results

Mark scheme

Examiner's comment

Q1(b): Candidates were required to apply their understanding of Kc to a novel context requiring some mathematical awareness.

Notes from the mark scheme and report

• square brackets required in Kc
• common errors: ÷2 instead of √ (2.00 × 10–4), inverse Kc expression (2.5)

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Answer

D

Mark scheme

Examiner's comment

D was the correct answer, with C being the main distractor. This straightforward question was answered correctly by only about half the candidates.

Answer

B

Mark scheme

Answer

B

Mark scheme

Mark scheme

Notes from the mark scheme and report

• high pressure AND low temperature
• fewer gaseous moles on right
• forward reaction exothermic
• marks independent

Mark scheme

Examiner's comment

4(a) This was a challenging example of predicting bond angles. Candidates often counted four pairs of electrons around the central atom, so based their bond angles on a tetrahedral arrangement (deducting 2.5 from 109.5o), losing this mark. Explanations were not always clear, but most candidates gained credit for describing electron pairs repelling. Many candidates state that they subtracted 2.5o rather than explaining why this is done in terms of electron pairs repelling. 4(b)(i) A wide range of responses, including addition signs, 2x, curved brackets, inverse. Candidates must take care to write the formulae correctly: Cl rather than Cl2 was seen a number of times. 4(b)(ii) Most recognised that the value for NO was 2 x Cl2 but some struggled to explain why. Answers needed to show clear evidence that they had taken into account the stoichiometry of the equation. Some gave responses such as using moles and volume to calculate or to put values into the Kc expression. 4(b)(iii) Those candidates that were able to give the correct expression in 4(b)(i) were often able to obtain the correct value here, demonstrating a good understanding of this topic. Some forgot to take the square root so only obtained one mark. 4(b)(iv) Candidates need to be well versed in how to tackle this type of question. Some, despite explaining what would happen, forgot to then state what would happen to the concentration of NO. Some said that the reaction would shift to the right but did not state that the reaction was endothermic. A small number of candidates thought that the reaction was exothermic, despite the + sign by the reaction, and some discussed other factors such as rate or changing pressure.

Notes from the mark scheme and report

• square brackets essential: DO NOT ALLOW curved brackets

Mark scheme

Examiner's comment

4(a) This was a challenging example of predicting bond angles. Candidates often counted four pairs of electrons around the central atom, so based their bond angles on a tetrahedral arrangement (deducting 2.5 from 109.5o), losing this mark. Explanations were not always clear, but most candidates gained credit for describing electron pairs repelling. Many candidates state that they subtracted 2.5o rather than explaining why this is done in terms of electron pairs repelling. 4(b)(i) A wide range of responses, including addition signs, 2x, curved brackets, inverse. Candidates must take care to write the formulae correctly: Cl rather than Cl2 was seen a number of times. 4(b)(ii) Most recognised that the value for NO was 2 x Cl2 but some struggled to explain why. Answers needed to show clear evidence that they had taken into account the stoichiometry of the equation. Some gave responses such as using moles and volume to calculate or to put values into the Kc expression. 4(b)(iii) Those candidates that were able to give the correct expression in 4(b)(i) were often able to obtain the correct value here, demonstrating a good understanding of this topic. Some forgot to take the square root so only obtained one mark. 4(b)(iv) Candidates need to be well versed in how to tackle this type of question. Some, despite explaining what would happen, forgot to then state what would happen to the concentration of NO. Some said that the reaction would shift to the right but did not state that the reaction was endothermic. A small number of candidates thought that the reaction was exothermic, despite the + sign by the reaction, and some discussed other factors such as rate or changing pressure.

Notes from the mark scheme and report

• rearrange for [NOCl]2 then take the square root (1.3 scores 1)
• ALLOW ECF from inverted Kc expression

Skills: M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M2.4 Solve algebraic equations

Mark scheme

Examiner's comment

4(a) This was a challenging example of predicting bond angles. Candidates often counted four pairs of electrons around the central atom, so based their bond angles on a tetrahedral arrangement (deducting 2.5 from 109.5o), losing this mark. Explanations were not always clear, but most candidates gained credit for describing electron pairs repelling. Many candidates state that they subtracted 2.5o rather than explaining why this is done in terms of electron pairs repelling. 4(b)(i) A wide range of responses, including addition signs, 2x, curved brackets, inverse. Candidates must take care to write the formulae correctly: Cl rather than Cl2 was seen a number of times. 4(b)(ii) Most recognised that the value for NO was 2 x Cl2 but some struggled to explain why. Answers needed to show clear evidence that they had taken into account the stoichiometry of the equation. Some gave responses such as using moles and volume to calculate or to put values into the Kc expression. 4(b)(iii) Those candidates that were able to give the correct expression in 4(b)(i) were often able to obtain the correct value here, demonstrating a good understanding of this topic. Some forgot to take the square root so only obtained one mark. 4(b)(iv) Candidates need to be well versed in how to tackle this type of question. Some, despite explaining what would happen, forgot to then state what would happen to the concentration of NO. Some said that the reaction would shift to the right but did not state that the reaction was endothermic. A small number of candidates thought that the reaction was exothermic, despite the + sign by the reaction, and some discussed other factors such as rate or changing pressure.

Notes from the mark scheme and report

• shifts right AND forward reaction endothermic

Mark scheme

Examiner's comment

Question 5 related an unfamiliar mathematical relationship with the y = mx + c equation for a straight line. Use of y = mx + c is a mathematical requirement of A level chemistry and candidates were expected to apply their understanding of the Arrhenius equation (part of the specification) which makes use of this straight-line equation in a novel context. Many candidates were unable to use this mathematical concept. In contrast, higher-attaining candidates were able to be credited with marks throughout question 5. 5(a) Candidates were expected to complete values for 1/T and ln Kp from supplied T and Kp values. Candidates were supplied with an example and this enabled most candidates to obtain both available marks. 5(b) Most candidates were aware that a decrease in Kp with increasing temperature signals that the forward reaction is exothermic. The question also asked for the effect on the equilibrium position. A significant number of candidates omitted this part of the question. Candidates are advised to check back to all the requirements in a question. 5(c) Candidates were required to plot a graph using their calculated values from 5(a). The axes for the graph had been provided. It was expected that the plotting of 5 points, with a best-fit straight line, would be straightforward. Many candidates plotted one or more points incorrectly, particularly the point at 1/T = 1.25 × 10–3. Candidates then needed to recognise that the gradient is equal to –∆H/R from Equation 5.1, to measure the gradient, and then to determine ∆H. Most candidates recognised that the gradient needed to be measured but its value was then not taken any further. The higher-attaining candidates correctly multiplied the gradient by R but did not always convert the calculated J mol–1 value into kJ mol–1, or to express their value to 3 significant figures, as required in the question. Significantly, nearly a third of candidates did not collect any of the four available marks. The question was an excellent discriminator. 5(d) Mathematically able candidates used the y = mx + c equation for a straight line with the supplied mathematical relationship (Equation 5.1) to identify the y intercept as ∆S/R. They then stated that ∆S could be determined by multiplying the value of the y intercept by R. Many candidates found the mathematical requirements of parts 5(c) and 5(d) difficult. Responses for 5(d) were often in terms of the gradient instead of ‘intercept’.

Notes from the mark scheme and report

• equilibrium shifts to the left AND forward reaction is exothermic (Kp decreases as T increases)

Skills: M0.3 Estimate results

Answer

B

Mark scheme

Mark scheme

Examiner's comment

Most candidates realised that the symbol indicated the equilibrium was (far) to the right hand side. The common error was to tell the examiner that Kp was bigger than 1.

Notes from the mark scheme and report

• equilibrium position (far) to the right
• ALLOW reaction nearly complete/high yield of products

Skills: M0.3 Estimate results; M2.1 Understand and use the symbols =, <, <<, >>, >, ∝, ~, ⇌

Mark scheme

Examiner's comment

This question was answered well with many candidates being given all 4 marks. Most candidates identified that there are fewer gaseous moles of products and that an increase the pressure will shift the equilibrium position to the right. Although the exothermic nature of the forward reaction was usually identified, candidates sometimes muddled the temperature conditions required, with ‘higher temperature’ being seen often instead of ‘low temperature’. Lower attaining candidates often seemed to confuse equilibrium (in this question) with rates.

Notes from the mark scheme and report

• fewer gaseous moles on right so high pressure
• forward reaction exothermic so low temperature
• ORA for reverse reaction

Mark scheme

Examiner's comment

This part discriminated well. Most candidates were able to write the correct expression for Kc as the starting point of the calculation. Candidates often got into a muddle in calculating Kc, perhaps due to issues inputting the calculation into their calculators. The question asked for ‘an appropriate number of significant figures and in standard form’. As the provided data was all to 3 significant figures, this also indicates the required number of significant figures in the answer. A calculated value to 2 significant figures was often seen (see the response); also 0.0286 rather than the standard form: 2.86 × 10–2. Some responses showed Kc inverted or added, rather than multiplying the two reactants in the denominator. Other candidates wrote the correct equilibrium expression but were then used 2.752, rather than 2.753, to obtain the standard form answer of 7.786 × 10–2 or 0.0786 with no standard form. Candidates are advised to check back through calculations to see if they have made any such errors.

Notes from the mark scheme and report

• 3 sig figs AND standard form required
• ECF only from inverted Kc (3.50 × 10^1)
• DO NOT ALLOW + in denominator
• IGNORE units

Skills: M0.1 Recognise and use expressions in decimal and ordinary form; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Answer

C

Mark scheme

Examiner's comment

This question was a good discriminator with well-prepared candidates usually selecting the correct option of C. Incorrect responses were reasonably evenly split across the other options, suggesting guesses and poor preparation.

Answer

B

Mark scheme

Examiner's comment

Most candidates responded with the correct response of B. The most common incorrect response was the inverse expression shown in A.

Mark scheme

Examiner's comment

The expression and the units were almost universally known by the candidates.

Notes from the mark scheme and report

• square brackets essential
• units dm3 mol–1
• ALLOW mol dm–3 as ECF from inverted expression

Skills: M0.0 Recognise and make use of appropriate units in calculations

Mark scheme

Examiner's comment

Generally, this question was well answered with only a small proportion of candidates adding the values together instead of multiplying.

Notes from the mark scheme and report

• square brackets required
• IGNORE state symbols

Mark scheme

Examiner's comment

Most candidates answered this question very well, with the most common mark being 4/5. Many candidates put a lot of effort into explaining, in depth, Le Chatelier’s principle, which was not required. The first three marking points were credited to most candidates. Responses were confident in their descriptions of equilibrium shifts and many candidates then went on to qualify their answers with operational factor considerations and/or rate. The explanation for pressure was described less commonly than temperature and many candidates did not appreciated that increased rate would lead to a decreased equilibrium yield. This candidate scored all five marks for this well-reasoned approach to the question.

Notes from the mark scheme and report

• low temperature (exothermic) and low pressure (fewer gaseous moles on left)
• compromise must relate temperature/pressure to rate
• IGNORE catalyst and stated values

Mark scheme

Examiner's comment

Candidates were expected to determine the type of energy change by linking their equilibrium equation in 2b(i) with the colour changes at different temperatures. Most candidates correctly concluded that the formation of a blue colour is endothermic. Many candidates did not explain this in terms of a shift in equilibrium position, considering bond breaking and bond making instead.

Notes from the mark scheme and report

• mark independently
• equilibrium shifts right at high temperature (temperature must be stated) or left at 0 °C - direction depends on equation written in (c)(i)
• endothermic
• IGNORE 'shifts to blue side'

Answer

B

Mark scheme

Mark scheme

Notes from the mark scheme and report

• low pressure AND high temperature
• more gaseous moles on right (DO NOT ALLOW atoms)
• forward reaction endothermic
• low pressure gives slow rate or high temperature uses a lot of energy
• IGNORE 'expensive' and catalyst

Mark scheme

Notes from the mark scheme and report

• units dm3 mol–1 (DO NOT ALLOW dm3/mol)
• ECF on units only if Kc upside down

Skills: M0.0 Recognise and make use of appropriate units in calculations

Mark scheme

Notes from the mark scheme and report

• 3 SF required (14.6)
• ECF to 3 SF only from inverted Kc (0.0687)
• DO NOT ALLOW [CO] + [H2]2 in expression

Skills: M0.2 Use ratios, fractions and percentages; M1.1 Use an appropriate number of significant figures; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities

Mark scheme

Notes from the mark scheme and report

• shift must be linked to exothermic forward reaction and to fewer gaseous moles on right
• cost/safety must be linked to pressure (IGNORE 'it's expensive')
• IGNORE activation energy

Mark scheme

Notes from the mark scheme and report

• square brackets required in Kc expression
• inverted Kc gives 5.48 (ECF, 2 marks)
• any other Kc expression scores no calculation marks

Skills: M0.3 Estimate results; M2.2 Change the subject of an equation; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities