Edexcel A-Level Chemistry Paper 2, June 2023: Question 5
10 marks · Medium difficulty · Short Open Response
Investigate the kinetics, rate equations, reaction profiles, and economics of catalysis for nitrogen oxide reactions.
Practise this questionQuestion
![The question concerns the reaction 2NO(g) + O2(g) -> 2NO2(g), where the rate is proportional to [O2] and [NO]^2. Part (a) asks why total gas pressure can be used to measure rate and two factors to keep constant. Part (b) features a graph with four lines (P, Q, R, S) plotting quantity Y against X, asking to identify specific relationships. Part (c) is a multiple-choice calculation on how the rate changes when reactant concentrations are altered. Part (d) provides a reaction profile for the exothermic reaction 2NO(g) + 2CO(g) -> 2CO2(g) + N2(g) and asks to complete it for a catalysed reaction and explain the economic benefits of using a catalyst.](/.netlify/images?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fexamquestions-bf477.firebasestorage.app%2Fo%2Fquestions%252FEdexcel.Q23.L2.05.SRC.edexcel-alevel-2%25257C2023%25257CJUN%25257Cuntiered-1791657398237.png%3Falt%3Dmedia&w=1000)
Question text
5 Nitrogen monoxide reacts with oxygen to form nitrogen dioxide.
2NO(g) + O2(g) → 2NO2(g)
The rate is proportional to the concentration of oxygen and to the square of the
concentration of nitrogen monoxide.
(a) The rate of this reaction can be determined by measuring the change in the total
gas pressure.
(i) Give a reason why this method can be used in this reaction.
(1)
(ii) State two factors, other than initial amounts of reactants, that must be kept
constant for this method to work.
(1)
(b) The graph shows four lines of a quantity Y plotted against a quantity X.
P
Q
Y R
S
X
(i) Which line shows the relationship between the concentration of
nitrogen monoxide (Y) and time (X)?
(1)
A line P
B line Q
C line R
10 D line S
*P71913A01028*
(ii) Which line shows the relationship between rate (Y) and concentration of
oxygen (X)?
(1)
A line P
B line Q
C line R
D line S
(c) The rate of this reaction is z mol dm–3 s–1 under certain conditions.
The concentration of nitrogen monoxide is doubled and the concentration of
oxygen is halved. All other conditions remain the same.
What will be the new rate of reaction in mol dm–3 s–1?
(1)
A z/2
B z
C 2z
D 4z
*P71913A01128*
(d) Nitrogen monoxide is formed in car engines. It is removed by the catalytic
converter in the car exhaust.
2NO(g) + 2CO(g) → 2CO2(g) + N2(g)
The reaction is exothermic and the most active catalyst is platinum.
(i) Complete the labelled reaction profile for the catalysed reaction.
(3)
2NO(g) + 2CO(g)
Enthalpy
Progress of reaction
(ii) Catalysts, such as platinum, are very expensive.
Explain an economic benefit of using a catalyst in an industrial process.
(2)
(Total for Question 5 = 10 marks)
Mark scheme
Show the mark scheme
Question
Answer Additional Guidance Mark
Number
5(a)(i) An answer that makes reference to the following point: (1)
• there is a change / decrease in the number of (gas) molecules / moles (so If numbers are given, they must be correct (3 to 2)
there will be a change in the total gas pressure) Ignore there is a change in the total volume of gas
Ignore references to partial pressure
Question
Answer Additional Guidance Mark
Number
5(a)(ii) An answer that includes: Do not award ‘pressure’. (1)
Do not award ‘heat’ for temperature.
• temperature and volume Allow ‘volume of container’
Do not award ‘volume of reactants’.
Ignore catalyst
Do not award if more than two factors given
Question Answer Mark
number
5(b)(i) (1)
The only correct answer is D (line S)
A is incorrect because [NO(g)] is decreasing during the reaction
B is incorrect because [NO(g)] is decreasing during the reaction
C is incorrect because [NO(g)] is decreasing during the reaction
number
5(b)(ii) (1)
The only correct answer is B (line Q)
A is incorrect because rate is directly proportional to [O2(g)]
C is incorrect because rate is directly proportional to [O2(g)]
D is incorrect because rate is directly proportional to [O2(g)]
number
5(c) (1)
The only correct answer is C (2z)
A is incorrect because this is the rate if only the concentration of oxygen is halved
B is incorrect because this is the rate if the concentration of nitrogen monoxide is doubled, the concentration of oxygen is
halved and the reaction is first order with respect to nitrogen monoxide
D is incorrect because this is the rate if only the concentration of nitrogen monoxide is doubled
Question
Answer Additional Guidance Mark
Number
5(d)(i) Example of reaction profile (3)
intermediate
enthalpy
2NO(g) + 2CO(g)
∆H
N2(g) + 2CO2(g)
progress of reaction
• products line to the right and lower than reactants line Allow (unbalanced) formula of product or just ‘product’
and labelled (1) Ignore missing state symbols
• single - headed arrow downwards and labelled (1) Allow other labels, e.g. enthalpy change / energy change
Do not award − H but H negative is awarded
Do not award double-headed arrow
Arrow should be close to the reactant line and finish close to
the product line
Ignore activation energy arrows
• two curves to show enthalpy with a catalyst (1) Both curves must be above the reactants line
Ignore missing line and / or label for intermediate
Question
Answer Additional Guidance Mark
Number
5(d)(ii) An explanation that makes reference to the following points: (2)
• a lower temperature can be used / no need for high(er) temperature Ignore reference to decreased pressure
(as the forward reaction is exothermic) there will be a higher Ignore environmental issues (e.g. greenhouse gases)
equilibrium yield at lower temperature (1) Ignore less heat.
• this reduces the cost of the energy / fuel needed (1) Ignore just ‘to reduce cost’
If no other mark is awarded, allow (1) for
‘even though they are expensive, they are not used up
so last a long time’ / ‘catalysts reduce the activation
energy and speed up the reaction’
(Total for Question 5 = 10 marks)
How to answer it
Kinetics, Rate Equations, and Energetics Study Guide
This question assesses core physical chemistry concepts including reaction kinetics (rate equations, order of reaction, graphical interpretation, and experimental tracking via pressure changes), alongside energetics (interpreting exothermic reaction profiles with catalysts) and the industrial economic factors of using catalysts.
Part (a): Measuring Reaction Rates via Pressure
Investigating 2NO(g) + O₂(g) → 2NO₂(g)
✅ Correct Answers
- (i) Reason: There is a change/decrease in the number of gas molecules (3 moles on the reactant side turning into 2 moles on the product side), leading to a measurable change in total gas pressure.
- (ii) Two factors to keep constant: Temperature and volume (of the container).
❌ Common Errors
- Saying "pressure" must be kept constant in part (ii) — pressure is what you are measuring/changing!
- Using vague terms like "heat" instead of "temperature", or writing "volume of reactants".
Part (b): Interpreting Graphical Relationships
Analyzing Concentration, Time, and Rate Curves
✅ Correct Answers
- (i) Line for [NO] against time (X): D (line S) — reactant concentration decreases as time increases, leveling off towards zero or equilibrium.
- (ii) Line for Rate against [O₂]: B (line Q) — since the rate equation is Rate = k[NO]²[O₂] , rate is directly proportional to [O₂], giving a straight line through the origin.
🧠 Exam Technique
- Always look closely at the axis labels before choosing a multiple-choice graph letter.
- Relate orders directly to mathematical proportionality: first-order dependence yields a linear positive gradient passing through (0,0).
Part (c): Rate Equation Calculations
Determining the Effect of Concentration Changes
📐 Step-by-Step Calculation
- Step 1: Write out the rate equation provided in the stem:
Rate = k[NO]²[O₂] - Step 2: Substitute the changes into the rate expression:
[NO] is doubled ( ×2 ), which gets squared ( 2² = 4 ).
[O₂] is halved ( ×0.5 or ÷2 ). - Step 3: Calculate the overall multiplier:
New Rate = k × (2[NO])² × (0.5[O₂]) = k × 4[NO]² × 0.5[O₂] = 4 × 0.5 = 2 times the original rate. - Correct Option: C (2z)
Part (d): Energetics and Catalysts
Reaction Profiles and Industrial Economics
💡 Key Knowledge: Catalysed Reaction Profile
- Products position: Drawn to the right and lower than the reactants line, clearly labeled as N₂(g) + 2CO₂(g) (since the reaction is exothermic).
- Enthalpy change arrow: A single-headed arrow pointing downwards between the reactant and product levels, labeled with ΔH or enthalpy change.
- Catalyst curves: Two distinct curves showing lower activation energies (humps) than the uncatalysed route, remaining above the reactant level and featuring an intermediate.
✅ Economic Benefit of Catalysts (5d ii)
- Allows the reaction to run at a lower temperature because the catalyst lowers the activation energy.
- Lower temperatures reduce energy/fuel costs required to heat the system, saving money industrially.
❌ Common Errors in Profiles
- Drawing double-headed arrows for enthalpy change ( ΔH ).
- Placing the product line higher than reactants on an exothermic profile.
- Failing to draw a lower alternative pathway for the catalysed route.
Topics
Physical Chemistry · Topic 9: Kinetics I · Topic 16: Kinetics II · Topic 8: Energetics I
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.