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| Series | Paper | Q | What it asks |
|---|---|---|---|
| June 2022 | AS Paper 2 | 4(a)(i) | Explain why these conditions are used industrially. [6] |
| June 2017 | AS Paper 2 | 2 | A student was asked to carry out an experiment to determine the initial rate of reaction of zinc and Outline how the student could carry out the experiment and explain how the results could be |
![5 The graph shows the volume of hydrogen released over time in a chemical reaction.
100
90
80
70
60
Volume of
hydrogen 50
/ cm3
40
30
20
10
0
0 20 40 60 80 100
Time / s
What is the rate of reaction, in cm3 s–1, at 30 seconds?
A 0.8
B 1.2
C 2.1
D 2.4
Your answer [1]](/img/JUN25_H4321/JUN25_H4321_05.webp)
B
The correct answer was B. Candidates had to draw the tangent and calculate the gradient at 30 seconds. The answer should then be selected from the closest value to take the inaccuracy of the tangent into consideration. The most common incorrect answer was C, where candidates incorrectly calculated an average rate for the first 30 seconds.
![12 Hydrogen and iodine react in the reaction below.
H2(g) + I2(g) 2HI(g)
Which row is correct when pressure is increased?
Rate of reaction Equilibrium yield of HI
A Increases Increases
B Increases No effect
C Increases Decreases
D Decreases No effect
Your answer [1]](/img/JUN25_H0321/JUN25_H0321_12.webp)
B
Overall, candidates had little difficulty with this question and most selected B.
![13 Which statement is not a sustainability benefit of an industrial process using a catalyst?
A Lower temperatures can be used.
B The position of equilibrium moves towards the products.
C The reaction takes place in less time.
D There is reduced energy demand.
Your answer [1]](/img/JUN25_H0321/JUN25_H0321_13.webp)
B
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.
![18 Nitrogen monoxide, NO, reacts with oxygen to form nitrogen dioxide, NO2.
2NO(g) + O2(g) 2NO2(g) Reaction 18.1
(a) The rate of reaction increases in the presence of a heterogeneous platinum catalyst.
(i) Explain why platinum is an example of a heterogeneous catalyst for Reaction 18.1.
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..................................................................................................................................................... [1]](/img/JUN25_H4321/JUN25_H4321_18(a)(i).webp)
This was a well answered question. A few candidates, incorrectly, suggested that it was heterogeneous due to the reactants and products being in different states, and didn’t mention the catalyst or suggested the catalyst was in same state.
![(ii) Explain how the platinum catalyst increases the rate of reaction.
Include a labelled sketch of the Boltzmann distribution on the grid.
Number of
molecules
0
0
Energy
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..................................................................................................................................................... [2]](/img/JUN25_H4321/JUN25_H4321_18(a)(ii).webp)
Many candidates gave a good Boltzmann distribution graph, even though a few did not start at the origin, and some were horizontal on the far right-hand side, much above the 0 line. The quality of the Boltzmann distribution could be improved by the candidates, although marks were given. Occasionally candidates drew multiple lines, presumably addressing a change of temperature. This question proved difficult for some candidates to explain well. Collision theory linked to catalyst requires the link between number of particles having energy ≥ the activation energy. A lot of candidates wrote vague responses, often confusing or mixing temperature and concentration effects and did not gain credit. The candidate drew a curve clearly starting at 0,0 but then drew a horizontal line above 0 on the right- hand side. This line would never approach 0 and has a constant number of molecules with the increasing energy and so was not given credit.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(iii) In industry, catalysts have great economic importance and benefits for increased sustainability.
Explain how the use of catalysts is beneficial to the environment.
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..................................................................................................................................................... [1]](/img/JUN25_H4321/JUN25_H4321_18(a)(iii).webp)
Most responses were excellent, with many candidates mentioning more than one of the marking criteria. A typical response would be ‘lower temperatures can be used, decreasing use of (fossil) fuels, and reducing CO2 / greenhouse gas emissions.’ Some responses instead mentioned that catalysts can be reused. The most common incorrect response was to say fewer toxic products were released or similar.
![(ii) Explain the shape of the graph in Fig. 21.1, in terms of concentration, collisions and time.
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.................................................................................................................................................... [3]](/img/JUN25_H0321/JUN25_H0321_21(d)(ii).webp)
This question required candidates to appreciate that initially concentration of HCl is at its highest, leading to more frequent collisions. As the reaction proceeds, concentration and collision frequency decrease until HCl is used up and the reaction stops (shown by the graph plateauing). The first marking point ignored candidates incorrectly discussing concentration of solid magnesium, but candidates need to be clear that only the (limiting) reactant, HCl, was used up when the reaction stopped for the third marking point. This last point was often missed, and this would be a good question to use with high-attaining candidates. Apart from the last point, there was an even spread of marks given; the second point was often missed through lack of ‘frequency’ or other link to collision rate. Analysing data in graphs Lower scoring candidates often stated that ‘the line increases because rate/concentration increases’. Students should be encouraged to carefully read axes to make sure they are identifying the correct trend from a graph.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms; PAG9 Rates of reaction – continuous monitoring
![8 Which row in the table explains how a catalyst affects the activation energy (Ea) and the
proportion of molecules with energy > Ea?
How the activation Proportion of molecules
energy changes with energy > Ea
A decreases decreases
B decreases increases
C increases decreases
D increases increases
Your answer [1]](/img/JUN24_H4321/JUN24_H4321_08.webp)
B
The correct answer was B. Candidates had a good knowledge of the Boltzmann distribution and few incorrect responses were seen.
![12 A mixture of gases is heated in a closed container. The reaction rate increases.
Which statement explains why the rate increases?
A More molecules have an energy greater than the activation energy.
B The activation energy decreases.
C The activation energy increases.
D The concentration of the gases increases.
Your answer [1]](/img/JUN24_H0321/JUN24_H0321_12.webp)
A
Most candidates were aware that heat increases the number of molecules that have energy greater than the activation energy (A).
![(b) The rate of decomposition of hydrogen peroxide shown in Reaction 16.1 can be increased by
adding a small amount of powdered manganese(IV) oxide, MnO2.
The MnO2 acts as a catalyst.
(i) Complete the enthalpy profile diagram for Reaction 16.1 using formulae for the reactants and
products.
• Use Ea to label the activation energy without MnO2.
• Use Ec to label the activation energy with MnO2.
• Use ∆H to label the enthalpy change of reaction.
enthalpy
progress of reaction
[3]](/img/JUN24_H4321/JUN24_H4321_16(b)(i).webp)
This question proved more difficult for candidates with lots of inaccuracies. The profile was dependent on the calculation for ΔH in Question 16 (a) (i). The arrowhead for ΔH needs to be pointing from the reactants to the products. The activation energies, again, need to start at the reactant line and go to the maximum level of the curve. Those that needed to draw an endothermic profile were far more likely to make an error with the Ea and Ec arrows, often starting from the product line or even from the base line of the graph. A significant number of candidates did not add arrows and instead labelled the curves Ea and Ec. Some candidates drew a Boltzmann distribution curve scoring 0 marks. The candidate has the correct exothermic profile but has the incorrect starting point for the activation energy going from the product line.
![(ii) Explain why MnO2 is described as a heterogeneous catalyst for this reaction.
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..................................................................................................................................................... [1]](/img/JUN24_H4321/JUN24_H4321_16(b)(ii).webp)
This was a well answered question. A few candidates, incorrectly, suggested that it was heterogeneous due to the reactants and products being in different states, and did not mention the catalyst.
![17 Crystal violet (CV) is a purple dye. In the presence of an alkali, CV reacts to form a colourless
product.
A student uses a colorimeter to investigate the rate of the reaction between CV and sodium
hydroxide, NaOH.
• The student mixes 10.0 cm3 of 2.8 × 10–7 mol dm–3 CV with 10.0 cm3 of 0.016 mol dm–3
NaOH.
• A large excess of NaOH is used, so that the reaction is effectively zero-order with respect to
OH– ions.
• The student places a sample of the reaction mixture in a colorimeter and measures the
absorbance over time.
The student uses the absorbance readings to calculate the concentration of CV and plots a graph
of concentration of CV against time, as shown below.
1.6 × 10–7
1.4 × 10–7
1.2 × 10–7
1.0 × 10–7
concentration
of CV
0.8 × 10–7
/ mol dm–3
0.6 × 10–7
0.4 × 10–7
0.2 × 10–7
0
0 1 2 3 4 5 6 7 8 9 10
time / min
(a) Using collision theory, explain why the gradient decreases over time.
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..................................................................................................................................................... [1]](/img/JUN24_H4321/JUN24_H4321_17(a).webp)
This question proved difficult for candidates to explain well. Collision theory linked to rate requires a quantitative approach, e.g. less particles per unit volume and less collisions per unit time. A lot of candidates wrote vague responses about fewer particles so less collisions and did not gain credit.
![4 Aqueous hydrogen peroxide, H2O2(aq), gradually decomposes to produce water and oxygen.
2H2O2(aq) 2H2O(l) + O2(g) ΔH = –196 kJ mol–1 Equation 4.1
The rate of decomposition of H2O2 can be increased by adding a small amount of
manganese(IV) oxide, MnO2, which acts as a catalyst.
(a) Explain, using a Boltzmann distribution model, why the rate of a reaction increases in the
presence of a catalyst.
You are provided with the axes below, which you should label.
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..................................................................................................................................................... [4]](/img/JUN24_H0322/JUN24_H0322_04(a).webp)
This is a familiar question with around half of candidates scoring all 4 marks. Common errors included drawing 2 lines, as you would have with different temperatures, but labelling one with catalyst and other without. Some had incorrect or missing labels on the axes. The most frequently gained mark was for knowing that the activation energy was lowered by a catalyst; this could be given by correct lines and labelling shown on the distribution, although care needed to be taken as some contradicted their answers with their diagram. Some labelled Ea and Ec lines but didn’t have energy on x-axis scale. Some struggled with the final marking point, not recognising that more molecules have the required activation energy. For example, ‘more frequent successful collisions’ with no reference to the activation energy, i.e. why they are successful.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(ii) Use your graph to determine the rate of reaction, in cm3 s–1, at 50 s.
Show your working below and on the graph.
rate = ............................................. cm3 s–1 [2]](/img/JUN24_H0322/JUN24_H0322_04(b)(ii).webp)
More candidates were able to correctly draw a tangent than seen in previous years with similar questions. A generous range was given for both tangent and gradient so many scored both marks. The most common reasons for losing marks was for having a gap between the curve and the tangent or calculating the gradient incorrectly, e.g. misreading scales, dx/dy, or by using interpolation rather than a tangent.
Skills: M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring
![13 The Boltzmann distribution showing the activation energy, Ea, for an uncatalysed reaction is
shown below.
number of
molecules
Ea
energy
What is the difference for the catalysed reaction?
A The activation energy shifts to the left.
B The activation energy shifts to the right.
C The curve flattens.
D The curve shifts to the right.
Your answer [1]](/img/JUN23_H0321/JUN23_H0321_13.webp)
A
The behaviour of the Boltzmann distribution under different conditions is well known with most candidates making the correct choice of A.
![14 A student measures how the OH– concentration changes over time for a reaction.
The student plots the graph below.
0.20
0.15
[OH–(aq)]
0.10
/ mol dm–3
0.05
0.00
0 100 200 300 400 500 600
time / s
What is the rate of reaction, in mol dm–3 s–1, at 200 s?
A 2.2 × 10–4
B 2.8 × 10–4
C 1.8 × 10–3
D 4.4 × 10–2
Your answer [1]](/img/JUN23_H0321/JUN23_H0321_14.webp)
B
This question discriminated well, rewarding those candidates with a good understanding of the link between concentration, time and rate. Many candidates drew a tangent to the curve at 200 s, measuring its gradient to get the correct choice of B. Common errors focused on reading of the concentration at 200 s. This was then either matched to option D directly or divided by 200 to give option A.
![(ii) Explain why these two reactions of magnesium produce the same volume of gas but at
different rates.
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...................................................................................................................................... [3]](/img/JUN23_H4321/JUN23_H4321_20(d)(ii).webp)
This question proved challenging for the candidates to identify the three ideas: Those of comparing acids, comparing moles and comparing rates. Very few candidates were able to score the 3 marks. Most candidates recognised the different strength of the two acids, but some only used comparative language. Some linked the moles of acid used to the volume of gas produced but many simply restated the same volume and concentration which is given within the question. Only a few candidates linked the higher initial [H+] in HCl to the increased rate through more frequent collisions. A common issue was describing the rate of dissociation rather than the [H+] present in determining the rate of the reactions or mentioning that it dissociates more but not linking this to the H+ concentration.
![(iii) CO and NO are removed by use of a catalyst.
Explain the role of the catalyst.
Refer to your enthalpy profile diagram in Fig. 23.1 in your answer.
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Almost all candidates knew that a catalyst lowered activation energy and most were aware that an alternative pathway was made possible by a catalyst.
![SECTION B
Answer all the questions.
16 A catalytic converter in a car removes nitrogen monoxide, NO, and carbon monoxide, CO, from
the exhaust gases.
(a) One reaction that happens in a catalytic converter is shown below.
2CO(g) + 2NO(g) N2(g) + 2CO2(g) Reaction 16.1
(i) Explain how increasing the temperature increases the rate of Reaction 16.1.
Include a labelled sketch, using Boltzmann distributions, on the grid below.
Label the axes.
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Almost all candidates produced a graph which showed the correct shape curve, although they sometimes incorrectly identified the quantities on the axes, e.g. rate versus time. Those who produced an incorrect diagram were able to discuss ‘more’ particles reaching the activation energy threshold. A few limited their answer to discussing successful collisions. Candidates needed to identify which curve was the higher temperature and to make sure that the curves did intersect more than once.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(b) The Boltzmann distribution model can be used by chemists to explain how the rate of a
reaction is affected by temperature.
Fig. 25.1 shows the Boltzmann distribution for a gas at room temperature.
Fig. 25.1
Label the axes on Fig. 25.1 and add a second curve to show the Boltzmann distribution of
the gas at a higher temperature.
Explain why the Boltzmann distribution shows that the rate of a reaction is affected by
temperature.
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Most candidates answered this question well. Most graphs were drawn with care but some peaks were shown at the same height as the provided curve or meeting this curve at high energy. Some candidates labelled what should have been ‘energy’ on the x axis as ‘progress of reaction’, or used ‘atoms’ rather than molecules for the y axis label. Many candidates were able to explain that more molecules exceed the activation energy at a higher temperature.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![3 A student investigates some reactions of zinc compounds and zinc metal.
(a) The student investigates the rate of reaction between zinc carbonate, ZnCO3(s), and dilute
hydrochloric acid, HCl (aq).
ZnCO3(s) + 2HCl (aq) ZnCl 2(aq) + CO2(g) + H2O(l)
The student follows the method outlined below:
• Add 50 cm3 of dilute HCl (aq) into a conical flask at 20 °C.
• Place the flask on a top-pan balance.
• Add an excess of ZnCO3(s) to the flask.
• Record the mass of the flask and contents on the top-pan balance every 30 seconds.
The student plots a graph of mass against time, shown in Fig. 3.1 below.
92.6
92.5
92.4
92.3
92.2
92.1
92.0
91.9
mass / g 91.8
91.7
91.6
91.5
91.4
91.3
91.2
91.1
91.0
0 50 100 150 200 250 300 350 400
time / s
Fig. 3.1
(i) The graph shows that the reaction gets slower over time, and eventually stops.
Explain why, in terms of collision theory.
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Only a small number of candidates managed to score all 3 marks for this question. Lots began their response describing why the rate is high at the start. This often filled the space available with marks being gained only in the last sentence. Very few gave descriptions with this specific reaction in mind, such as recognising that the HCl was the limiting reagent so gets used up, leaving some MgCO3 remaining. Common errors included referring to “fewer collisions” but without any indication of time, or “less reactant” without considering the volume of the reaction. We have produced a delivery guide on rates with some useful resources to help consolidate ideas and avoid misconceptions such as these: https://www.ocr.org.uk/Images/231742-rates-delivery-guide.pdf
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(ii) Using the graph in Fig. 3.1, find the rate of reaction, in g s–1, at 50 seconds.
Show your working on the graph and in the space below.
rate of reaction = ................................................. g s–1 [2]](/img/JUN22_H0322/JUN22_H0322_03(a)(ii).webp)
Most candidates recognised that they needed to draw a tangent. Candidates need to be instructed to draw their line clearly and make sure that it covers several large squares on the graph – preferably extending to the axes. The range given for the calculated rate from the gradient of the tangent was generous, however some did not realise that the y-axis started at 91.0 so incorrectly calculated their gradient. Some candidates used interpolation so gained no marks.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring
![(iii) The student repeats the experiment but heats 50 cm3 of dilute hydrochloric acid up to
40 °C before adding the ZnCO3(s).
On Fig. 3.1, sketch the curve the student would obtain. [2]](/img/JUN22_H0322/JUN22_H0322_03(a)(iii).webp)
About a third of candidates didn’t score any marks here. Some did not follow the instructions to add their curve to the previous graph. The sketches produced in the blank space then usually lacked any scales on the axes so could not be given marks. Lines added were often inaccurately drawn with lines often extending below 91.55g before returning to this mass at the end. It was often difficult to interpret if a graph levelled off before the original with lots tracking the same curve with a very slightly steeper slope. Some gave a slower rate but did manage to gain marks for finishing at 91.55g.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms; PAG9 Rates of reaction – continuous monitoring
![(b) The student investigates the reaction between zinc and dilute sulfuric acid.
Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g) ΔH = –140 kJ mol–1
Copper(II) sulfate is a catalyst for this reaction.
• The student adds a piece of zinc to each of two test tubes.
• The student adds a few drops of aqueous copper(II) sulfate to one of the test tubes,
forming a pale blue solution.
• The student adds an excess of dilute sulfuric acid to each test tube.
(i) Describe two differences the student would observe between the test tubes.
1 ........................................................................................................................................
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2 ........................................................................................................................................
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[2]](/img/JUN22_H0322/JUN22_H0322_03(b)(i).webp)
Candidates found this question very challenging. Most recognised that the addition of the catalyst would speed up the reaction but could not translate this into observations for this specific practical. For example, “reacts faster”, “releases more hydrogen” or “stops reacting quicker”. Giving candidates the opportunity to observe a range of both familiar and unfamiliar reactions, noting qualitative observations can help to improve their skills here. Many just noted that one tube would be blue and the other colourless rather than considering the reaction taking place. Some indicated that only 1 tube would react and the other would not. A few candidates recognised that a competing displacement reaction would take place between Zn and CuSO4 so were given marks for this.
![(ii) Using the axes below, sketch an enthalpy profile diagram for the reaction with and
without the catalyst.
On your diagram, include the following labels:
• ΔH, the enthalpy change
• Ea, the activation energy without a catalyst
• Ec, the activation energy with a catalyst.
Enthalpy
Zn(s) + H2SO4(aq)
Progress of reaction
[3]](/img/JUN22_H0322/JUN22_H0322_03(b)(ii).webp)
Less than half of all candidates managed to gain all 3 marks giving the correct enthalpy profile diagram. Many demonstrated a lack of understanding about what an enthalpy profile diagram shows as seen by labelling of curves for Ea/Ec rather than the enthalpy change or arrows for ∆H extending to the x-axis. Other errors also seen included, -ΔH, double headed ΔH arrow, activation energy lines starting at products, endothermic profiles, Ea and Ec the wrong way round and missing or incorrect labels. Some candidates found it tricky to give activation lines for both catalysed and uncatalysed reactions on the same diagram, these often overlapped or did not extend to the top of the energy barrier.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![4 This question is about the manufacture of hydrogen, H2.
(a) In industry, hydrogen is manufactured from methane, as shown in Equilibrium 4.1.
CH4(g) + H2O(g) CO(g) + 3H2(g) ΔH = +206 kJ mol–1 Equilibrium 4.1
The industrial process is carried out at 15 atmospheres pressure and at a temperature of
800 °C using an excess of steam. A nickel catalyst is used.
(i)* Explain why these conditions are used industrially. [6]
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Additional answer space if required.
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...........................................................................................................................................](/img/JUN22_H0322/JUN22_H0322_04(a)(i).webp)
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.
![4 Two students plan to investigate Equilibrium 4.1, shown below.
CoCl 4 2−(aq) + 6H2O(l) [Co(H2O)6]2+(aq) + 4Cl −(aq) Equilibrium 4.1
blue pink
(a) The students are supplied with the equilibrium mixture in Equilibrium 4.1 at room
temperature.
• One student heats 20 cm3 of the mixture to 50 °C.
• The other student heats 20 cm3 of the mixture to 90 °C.
The students use colorimetry to observe how the colour of the equilibrium mixture changes
over time.
• The colorimeter is set up so that the greater the absorbance, the greater the
concentration of [Co(H2O)6]2+.
• The initial absorbance is set to zero.
• The absorbance is recorded every 30 seconds.
The students plot the graph below from the results of the experiment.
50 °C
90 °C
Absorbance
0,0 Time / s
Use the graph and relevant chemical theory to answer the following. Include all reasoning:
• Explain the different initial rates at 50 °C and 90 °C.
• Predict the sign of ∆H for the forward reaction in Equilibrium 4.1.
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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.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms; PAG9 Rates of reaction – continuous monitoring
![11 The Boltzmann distributions below show a gas at two different temperatures.
Which Boltzmann distribution shows the dotted curve at a higher temperature?
number of
A molecules
energy
number of
B molecules
energy
number of
C molecules
energy
number of
D molecules
energy
Your answer [1]](/img/NOV21_H0321/NOV21_H0321_11.webp)
A
![(ii) The chemist then repeats the experiment three times. In each experiment, the chemist
makes one change but uses the same initial amounts of NO and O2.
Complete the table to show the predicted effect of each change compared with the
original experiment.
Only use the words greater, smaller or same.
Equilibrium amount
Change Kp Initial rate
of NO2(g)
Temperature
increase
Pressure
increase
Catalyst
added
[3]](/img/NOV21_H4321/NOV21_H4321_20(c)(ii).webp)
Question 20(c): This question asked the candidate to calculate Kp.
![22 This question is about enthalpy changes.
Hydrogen, H2, can be manufactured by the reaction of methane and steam.
This is a reversible reaction, as shown in Equilibrium 22.1 below.
Equilibrium 22.1 CH4(g) + H2O(g) 3H2(g) + CO(g) ∆H = +206 kJ mol–1
(a) The rate of this reaction increases when a catalyst is present.
Complete the enthalpy profile diagram below.
On your diagram:
• label the activation energies, Ea (without catalyst) and Ec (with catalyst)
• label the enthalpy change of reaction, ∆H.
enthalpy CH4(g) + H2O(g)
progress of reaction
[3]](/img/NOV21_H0321/NOV21_H0321_22(a).webp)
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(c) The industrial manufacture of methanol has used a copper-based catalyst.
Chemists have recently developed a new method for making methanol that uses a nickel-
gallium catalyst. This allows methanol to be produced at a lower temperature than the old
method.
Suggest two reasons why using a lower temperature is beneficial to the environment.
1 ................................................................................................................................................
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2 ................................................................................................................................................
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[2]](/img/NOV21_H0322/NOV21_H0322_05(c).webp)
![6 A student investigates the rate of reaction between strontium and water.
Sr(s) + 2H2O(l) Sr(OH)2(aq) + H2(g)
The student’s method is shown below.
• Pour 100 cm3 of water into a conical flask.
• Add 0.26 g of strontium and quickly connect a 100 cm3 gas syringe.
• Measure the volume of gas produced every 10 seconds until all the strontium has reacted.
The student plots a graph of volume of gas produced against time as shown in Fig. 6.1.
80
70
60
50
volume
of gas 40
produced
/ cm3
30
20
10
0
0 10 20 30 40 50 60 70 80 90 100
time / s
Fig. 6.1
(a) Draw a best fit curve on Fig. 6.1.
Use the graph to determine the rate of reaction, in cm3 s–1, at 50 s.
Show your working below and on the graph.
rate at 50 s ............................................. cm3 s–1 [3]](/img/NOV21_H0322/NOV21_H0322_06(a).webp)
Skills: M3.2 Plot two variables from experimental or other data; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring
![(b) A second student suggests that the experiment could be improved by measuring the loss in
mass in the conical flask over time.
The student places a conical flask containing 100 cm3 of water on a 2 decimal place balance,
and then adds 0.26 g of strontium.
The mass is recorded every 10 seconds.
Suggest one advantage and one disadvantage of using this method compared to the gas
collection method.
Advantage: ...............................................................................................................................
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Disadvantage: ...........................................................................................................................
.............................................................................................................................................. [2]](/img/NOV21_H0322/NOV21_H0322_06(b).webp)
Skills: PAG9 Rates of reaction – continuous monitoring
![(ii) The student observes that the rate of reaction for barium is different from the rate of
reaction with strontium.
On Fig. 6.1 sketch the graph the student would obtain using barium instead of strontium.
[2]](/img/NOV21_H0322/NOV21_H0322_06(c)(ii).webp)
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![7 The diagram represents a Boltzmann distribution curve of molecules at a given temperature.
number of
molecules
energy
Which statement for this Boltzmann distribution curve is correct at a higher temperature?
A The peak increases in height and moves to the left.
B The peak increases in height and moves to the right.
C The peak decreases in height and moves to the left.
D The peak decreases in height and moves to the right.
Your answer [1]](/img/NOV20_H4321/NOV20_H4321_07.webp)
D
![(b) A catalyst is used in the production of methanol in equilibrium 1.
State two ways that the use of catalysts helps chemical companies to make their processes
more sustainable and less harmful to the environment.
1 ................................................................................................................................................
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2 ................................................................................................................................................
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[2]](/img/NOV20_H4321/NOV20_H4321_17(b).webp)
![(b) The industrial manufacture of NH3 from N2 and H2 is carried out at an increased temperature
and in the presence of a catalyst.
Explain, using Boltzmann distributions, why increasing the temperature and using a catalyst
both increase the reaction rate.
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24(a) Most candidates were able to make some progress with the enthalpy calculation. Most recognised that the provided ∆cH values had to be multiplied by the balancing numbers in the equation. Most candidates processed the data on either side of the equation correctly and subtracted the combined enthalpy change from one side of the equation from the other, although the subtraction was sometimes the wrong way round. This resulted in answers of +92, or –92 kJ mol–1, with the exothermic value being correct. An enthalpy change of –92 kJ mol–1 was the commonest final answer, with comparatively few candidates recognising that the enthalpy change of formation of NH3 is for the formation of 1 mol of NH3 and not 2 mol as in the equation. 24(b) This question was well-answered showing that most candidates were well-acquainted with the Boltzmann distribution. Some candidates used one diagram to show the effect of increasing the temperature and of using a catalyst. Other candidates showed two curves, one for temperature, the other for the presence of a catalyst. Either approach could produce a clear response. Most candidates knew that the activation energy was lower with a catalyst that without. A significant number of candidates limited their explanations to ‘successful collisions’ without referring to more molecules exceeding the lower activation energy in the presence of catalyst or at a higher temperature. The best responses secured all five marks from a well-drawn and annotated graph supported by a brief explanation. Candidate should make full use of annotated diagrams to communicate their answers.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![12 What is the main reason for the increase in reaction rate with increasing temperature?
A The activation energy decreases.
B The activation energy increases.
C More molecules have an energy greater than the activation energy.
D The molecules collide more frequently.
Your answer [1]](/img/JUN19_H0321/JUN19_H0321_12.webp)
C
The role of activation energy in the rate of a reaction with increasing temperature was well-known and most candidates chose the correct option C. From the annotations on candidate scripts, many had ruled out options A and B entirely. D was anticipated as being the main distractor and this proved to be the case. Activation energy has a much greater effect than increasing collision frequency.
![(c) Vanadium(V) oxide, V2O5(s), is used as a catalyst in equilibrium 1.
2SO2(g) + O2(g) 2SO3(g) ∆H = −197 kJ mol−1 Equilibrium 1
(i) Explain how the presence of V2O5(s) increases the rate of reaction.
Include a labelled sketch of the Boltzmann distribution, on the grid below.
Label the axes.
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...................................................................................................................................... [4]](/img/JUN19_H4321/JUN19_H4321_19(c)(i).webp)
Most candidates were able to achieve all four marks. Occasionally candidates didn’t start their graph from zero or mislabelled the x-axis as ‘progress of reaction’.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(ii) Explain whether vanadium(V) oxide is acting as a homogeneous or heterogeneous
catalyst.
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...................................................................................................................................... [1]](/img/JUN19_H4321/JUN19_H4321_19(c)(ii).webp)
The number of candidates who considered V2O5 to be a homogeneous catalyst was small.
![(b) The student uses the apparatus in the diagram.
100 cm3 gas syringe
The student’s method is outlined below:
• Pour 50.0 cm3 of 0.100 mol dm–3 HCl into the conical flask.
• Add 0.200 g of zinc (an excess), and quickly attach the delivery tube and gas syringe.
• Measure the volume of gas collected every 20 seconds until the reaction stops.
The student obtains the results shown in Table 4.1.
Time / s 0 20 40 60 80 100 120 160 200
Volume of gas / cm3 0 16 27 37 39 50 53 58 58
Table 4.1
(i) On the graph paper in Fig. 4.1, label the x axis and plot the results in Table 4.1. [1]](/img/JUN19_H0322/JUN19_H0322_04(b)(i).webp)
Most candidates obtained this mark, some lost the mark because they did not use a linear scale or provide units.
Skills: M3.2 Plot two variables from experimental or other data; PAG9 Rates of reaction – continuous monitoring
![(iii) Draw a best-fit smooth curve on the graph you have drawn in Fig. 4.1. [1]](/img/JUN19_H0322/JUN19_H0322_04(b)(iii).webp)
Nearly all the candidates obtained this mark
Skills: M3.2 Plot two variables from experimental or other data; PAG9 Rates of reaction – continuous monitoring
![80
70
60
50
volume
40
/ cm3
30
20
10
0
Fig. 4.1
(c) The student repeats the experiment using:
• zinc with the same mass (0.200 g) and same surface area
• the same temperature and pressure
• 40.0 cm3 of 0.125 mol dm–3 HCl, instead of 50.0 cm3 of 0.100 mol dm–3 HCl.
On your graph in Fig. 4.1 sketch the curve you would expect in this experiment. [2]](/img/JUN19_H0322/JUN19_H0322_04(c).webp)
Many students did not sketch this curve or sketched a curved that was less steep and did not finish at 58cm3.
Skills: M0.2 Use ratios, fractions and percentages; M3.1 Translate information between graphical, numerical and algebraic forms
![(d) The graph shows that rate of reaction decreases over time.
Explain why, in terms of collision theory.
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.............................................................................................................................................. [2]](/img/JUN19_H0322/JUN19_H0322_04(d).webp)
Many responses detailed why the graph was steep at the beginning, rather than answering the question. Those that did explain the decrease often omitted the words concentration and frequency so the majority did not gain 2 marks. A large number of candidates discussed particles “losing energy” and “less successful collisions” so were not given any marks.
![(e) (i) The rate of the reaction between zinc and hydrochloric acid can be increased using a
solution of copper(II) sulfate as a catalyst.
Explain how a catalyst increases the rate of reaction.
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...................................................................................................................................... [2]](/img/JUN19_H0322/JUN19_H0322_04(e)(i).webp)
Most candidates scored the first marking point but many did not achieve the second marking point as their explanations were too vague.
![(ii) Why is it difficult to classify the solution of copper(II) sulfate as a homogeneous or
heterogeneous catalyst in this reaction?
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...................................................................................................................................... [1]](/img/JUN19_H0322/JUN19_H0322_04(e)(ii).webp)
Almost half of the candidates answered this question correctly, the remainder did not realise that the question gave them the answer to the state that the copper sulphate solution was in. Many answers stated that it could be solid or aqueous, so difficult to classify.
![14 Which statement(s) explain(s) why reaction rates increase as temperature increases?
1 The activation energy is less.
2 Collisions between molecules are more frequent.
3 A greater proportion of molecules have energy greater than the activation energy.
A 1, 2 and 3
B Only 1 and 2
C Only 2 and 3
D Only 1
Your answer [1]](/img/JUN18_H4321/JUN18_H4321_14.webp)
C
This was almost universally correct.
![(b) Hydrogen iodide, HI, is decomposed by heat into its elements:
2HI(g) H2(g) + I2(g) ∆H = +9.5 kJ mol–1
The decomposition is much faster in the presence of a platinum catalyst.
Complete the enthalpy profile diagram for this reaction using formulae for the reactants and
products.
• Use Ea to label the activation energy without a catalyst.
• Use Ec to label the activation energy with a catalyst.
• Use ∆H to label the enthalpy change of reaction.
enthalpy
progress of reaction
[3]](/img/JUN18_H4321/JUN18_H4321_20(b).webp)
Apart from lower ability candidates, who opted to draw Boltzmann distributions, most candidates scored well here. Common errors included: omission of formulae of reactants and/or products; using a double-headed arrow for ∆H; and masking the lower part of the Ea line with the Ec line so it appeared that Ea was an additional energy increase above the Ec value.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(b) Using the Boltzmann distribution model, explain how the rate of a reaction is affected by
temperature.
You are provided with the axes below, which should be labelled.
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.............................................................................................................................................. [4]](/img/JUN18_H0321/JUN18_H0321_23(b).webp)
Overall, this question was answered well. Most candidates showed two Boltzmann distribution curves at different temperatures. Labelling of the axes was usually correct, although the labels were sometimes seen the wrong way around. Most candidates were aware that more molecules possessed the required activation energy at a higher temperature, although lower ability candidates discussed frequency of collisions instead. Strangely, many good responses were spoilt by not labelling which of the two curves was at higher temperature. This is shown in the otherwise excellent response in the exemplar.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(c) Predict the conditions of temperature and pressure for a maximum equilibrium yield of
nitrogen monoxide in equilibrium 4.1.
• Explain your prediction in terms of le Chatelier’s principle.
• State and explain how these conditions could be changed to achieve a compromise
between equilibrium yield, rate and other operational factors.
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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.
![(b) A student investigates the rate of reaction of 1-bromobutane with aqueous hydroxide ions.
The graph shows how the hydroxide ion concentration, [OH–(aq)], changes during the
reaction.
0.45
0.40
0.35
0.30
[OH–(aq)] 0.25
/ mol dm–3
0.20
0.15
0.10
0.05
0.00
0 50 100 150 200
time / min
Using the graph, calculate the rate of reaction, in mol dm–3 min–1, at 30 minutes.
Show your working on the graph and in the space below.
rate of reaction = ................................. mol dm–3 min–1 [2]](/img/JUN18_H0322/JUN18_H0322_05(b).webp)
Most candidates knew they had to draw a tangent to the curve but calculating the gradient led to careless errors. Many drew tiny triangles and mis-read readings of coordinates. It is far easier, and more accurate, to use large triangles using the axes for coordinates. Lower ability candidates just interpolated the graph at 30s and were not credited marks for this.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change
![2 This question looks at reactions of hydrogen peroxide and of cobalt(II) ions.
(a) Aqueous hydrogen peroxide decomposes as shown in equation 2.1.
2H2O2(aq) → 2H2O(l) + O2(g) Equation 2.1
The reaction is catalysed by manganese(IV) oxide, MnO2.
A student investigates the decomposition of a hydrogen peroxide solution as outlined below.
• The student adds 50.00 cm3 of H2O2(aq) to a conical flask.
• The student adds a small spatula measure of MnO2 and quickly connects the flask
to a gas syringe.
• The student measures the volume of oxygen every 200 seconds.
Results
Time/s Volume of O2/cm3
0 0
200 15
400 28
600 36
800 41
1000 46
1200 48
1400 50
(i) Process the results as outlined below.
• On page 5, plot a graph of volume of O2 against time.
• Use your graph to find the rate of the reaction, in cm3 s–1, at t = 500 s.
Show your working on the graph and in the space below.
rate = ..............................................cm3 s–1 [5]](/img/JUN18_H4323/JUN18_H4323_02(a)(i).webp)
The graph was generally well drawn although many didn’t realise that a tangent needed to be drawn. A few tried to draw a straight line of best fit, rather than a curve. Most candidates were able to construct a correct graph, assigning the axes correctly and making good use of the graph paper. Points were usually plotted correctly and most candidates made a good attempt to draw a curved line of best fit. Candidates then needed to draw a tangent at 500 seconds and to measure its gradient to calculate the rate. Lower ability candidates did not draw a tangent to their curve and just calculated the volume formed at 500 seconds.
Skills: M3.1 Translate information between graphical, numerical and algebraic forms; M3.2 Plot two variables from experimental or other data; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG9 Rates of reaction – continuous monitoring
![(b) Suggest a different experimental method that would allow the rate of this reaction to be
followed over time.
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.............................................................................................................................................. [1]](/img/JUN17_H4321/JUN17_H4321_19(b).webp)
Skills: PAG9 Rates of reaction – continuous monitoring
![23 Methanol can be prepared industrially by reacting carbon monoxide with hydrogen in the presence
of a copper catalyst. This is a reversible reaction.
CO(g) + 2H2(g) CH3OH(g)
(a) Using the Boltzmann distribution model, explain why the rate of a reaction increases in the
presence of a catalyst.
You are provided with the axes below, which should be labelled.
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.............................................................................................................................................. [4]](/img/JUN17_H0321/JUN17_H0321_23(a).webp)
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(b) The reaction for the production of methanol in the presence of the copper catalyst is carried
out at 200–300 °C.
Explain why use of the catalyst reduces energy demand and benefits the environment.
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.............................................................................................................................................. [2]](/img/JUN17_H0321/JUN17_H0321_23(b).webp)
![2* A student was asked to carry out an experiment to determine the initial rate of reaction of zinc and
hydrochloric acid.
The student plans to collect a total of about 72 cm3 of hydrogen at RTP and to use an excess of
zinc.
The student selects the following apparatus:
• the apparatus shown in the diagram
• 100 cm3 measuring cylinder
• stop clock
• 2 decimal place balance 100 cm3 gas syringe
Outline how the student could carry out the experiment and explain how the results could be
processed graphically.
Show all working in your calculations.
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Skills: M0.2 Use ratios, fractions and percentages; M2.3 Substitute numerical values into algebraic equations using appropriate units for physical quantities; M3.2 Plot two variables from experimental or other data; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change; PAG1 Moles determination; PAG9 Rates of reaction – continuous monitoring
![12
5 A student investigates the reaction between strontium carbonate and dilute nitric acid.
SrCO3 + 2HNO3 Sr(NO3)2 + CO2 + H2O
The rate of reaction is determined from the loss in mass over a period of time.
(a) (i) Explain why there is a loss in mass during the reaction.
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...................................................................................................................................... [1]](/img/JUN16_H0322/JUN16_H0322_05(a)(i).webp)
Skills: PAG9 Rates of reaction – continuous monitoring
![(b) The student plots a graph of total mass (reagents + container) against time.
96.40
96.30
96.20
mass / g
96.10
96.00
95.90
0 50 100 150 200 250 300 350 400
time / s
13
(i) Describe and explain the change in the rate of the reaction during the first 200 seconds
of the experiment.
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...................................................................................................................................... [2]](/img/JUN16_H0322/JUN16_H0322_05(b)(i).webp)
Skills: M3.1 Translate information between graphical, numerical and algebraic forms
![(ii) Using the graph, calculate the rate of reaction, in g s–1, at 200 seconds.
Show your working on the graph.
rate of reaction = ................................................. g s–1 [2]](/img/JUN16_H0322/JUN16_H0322_05(b)(ii).webp)
Skills: M3.1 Translate information between graphical, numerical and algebraic forms; M3.5 Draw and use the slope of a tangent to a curve as a measure of rate of change
![(c) Outline a method that could be used to obtain the results that are plotted on the graph.
Your answer should include the apparatus required and the procedure for the experiment.
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.............................................................................................................................................. [3]](/img/JUN16_H0322/JUN16_H0322_05(c).webp)
Skills: PAG9 Rates of reaction – continuous monitoring
Dropping marks on calculations, mechanisms or the six-markers? Book a trial session with a specialist OCR chemistry tutor.
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