Edexcel A-Level Chemistry AS Paper 2, June 2017: Question 6

13 marks · Hard difficulty · Extended Writing

Evaluate the feasibility of changing industrial conditions for the oxidation of sulfur dioxide, sketch reaction profiles for catalysed and uncatalysed reactions, write the Kc expression and determine its units.

Practise this question

Question

Question text

6 One of the stages in the production of sulfuric acid from sulfide ores involves the

oxidation of sulfur dioxide to sulfur trioxide. The equation for the reaction is

2SO (g) + O (g) `_ 2SO (g) ǻ H = –197 kJ mol–1

22 3 r

The conditions used in one industrial process are: 420°C and a pressure of 1.7 atm

together with a vanadium(V) oxide catalyst.

It is proposed to change the conditions to 600°C and 10 atm pressure, while still using

the same catalyst.

*(a) Evaluate the feasibility of each of these changes in terms of their effect on the

rate, yield and economics of the reaction.

(6)

(b) (i) On the axes provided, sketch the reaction profiles for the uncatalysed and*P49857A01324*

catalysed reaction.

2SO (g) + O (g) `_ 2SO (g) ǻ H = –197 kJ mol–1

22 3 r

Label the uncatalysed reaction, A, and the reaction catalysed by

vanadium(V) oxide, B.

(3)

(ii) On your reaction profile, identify and label both the enthalpy change and the

activation energy for the catalysed reaction.

(2)

2SO2(g) + O2(g) `_ 2SO3(g)

(1)

(ii) What are the units, if any, of the equilibrium constant,*P49857A01424*Kc?

(1)

A mol dm–3

B dm3 mol–1

C no units

D mol2 dm–6

(Total for Question 6 = 13 marks)

Mark scheme

Show the mark scheme

Question Acceptable Answer Additional Guidance Mark

Number

*6(a) This question assesses a student’s ability to show a Guidance on how the mark scheme (6)

coherent and logically structured answer with linkages should be applied:

and fully-sustained reasoning.

Marks are awarded for indicative content and for how The mark for indicative content should

be added to the mark for lines of

the answer is structured and shows lines of reasoning.

reasoning. For example, an answer

The following table shows how the marks should be with five indicative marking points that

awarded for indicative content. is partially structured with some

Number of indicative Number of marks awarded linkages and lines of reasoning, scores

marking points seen in for indicative marking 4 marks (3 marks for indicative content

answer points and 1 mark for partial structure and

64 some linkages and lines of reasoning).

5-4 3

3-2 2 If there are no linkages between points,

11 the same five indicative marking points

00 would yield an overall score of 3 marks

(3 marks for indicative content and no

The following table shows how the marks should be marks for linkages).

awarded for structure and lines of reasoning.

Number of marks awarded In general it would be expected that 5

for structure and sustained or 6 indicative points would get 2

lines of reasoning reasoning marks, and 3 or 4 indicative

Answer shows a coherent points would get 1 mark for reasoning,

and logical structure with 2 and 0, 1 or 2 indicative points would

linkages and fully sustained score zero marks for reasoning.

lines of reasoning

demonstrated throughout. If there is any incorrect chemistry,

Answer is partially structured deduct mark(s) from the reasoning. If

no reasoning mark(s) awarded do not

with some linkages and lines 1

deduct mark(s).

of reasoning.

Comment: Look for the indicative

Answer has no linkages

marking points first, then consider the

between points and is 0

mark for the structure of the answer

unstructured.

and sustained line of reasoning.

*6(a) Indicative content:

IP1 increase in temperature will increase rate

Decreased yield with no reference to

IP2 (but) increase in temperature will decrease

exothermic reaction does not get IP2.

yield/move the equilibrium to the LHS/ produce

less SO3 because it is an exothermic reaction Allow increases yield of reactants/SO2

(in the forward direction) and O2 (with reference to exothermic

reaction)

IP3 increase in temperature increases energy

costs

IP4 increase in pressure has no effect on rate

(because all the active sites are already

occupied on a heterogeneous catalyst).

increase in pressure will increase rate (of

reaction)

IP5 increase in pressure will move position of Increased yield with no reference to

eqm to RHS/increase yield because there are number of moles does not get IP5.

less moles/molecules (of gas) on the RHS

Award one mark for IP2 and IP5 if

correct references to yield in both but

reasons not given

IP6 but increased pressure increases Allow IP3 and IP6 if increased costs of

(construction and running) costs/reduces higher temperature and pressure are

economic viability mentioned together provided that the

temperature costs are linked to energy

costs. Otherwise only IP6 can be

awarded.

Ignore any reference to catalyst

Question Acceptable Answer Additional Guidance Mark

Number

6(b)(i)

(3)

vertical axis labelled:

H/enthalpy/energy/E (1) Do not award H

Ignore horizontal axis label

Ignore units if given

level of reactants / 2SO2 + O2 above ignore state symbols even if incorrect

level of products / 2SO3 (1)

correct profile for uncatalysed reaction allow vertical lines for catalysed and uncatalysed reactions

labelled A to run together

and allow double hump profile

peak lower for catalysed reaction

labelled B (1)

Question Acceptable Answer Additional Guidance Mark

Number

6(b)(ii) enthalpy change, H/ H/(-)197(kJ mol-1), shown Ignore presence/absence of arrowheads (2)

correctly (1) Allow a degree of imprecision in the start/finish

points of the lines for H and Ea

Ea shown on double hump profile - shown in this

diagram as Ea1

activation energy, Ea, shown correctly (upper Ignore Ea2 if also shown

diagram) (1)

Question Acceptable Answer Additional Guidance Mark

Number

6(c)(i) [SO ]2 Do not award just K or K . (1)

3 p

(Kc = ) must be square brackets

[O ][SO ]2 do not accept partial pressures

ignore units or lack of units

ignore state symbols

Allow x sign in the denominator but

not +

Question Answer Mark

Number

6(c)(ii) 6(c)(ii). The only correct answer is B (1)

A is not correct because it refers to the inverted expression for Kc

C is not correct because units do not cancel for concentration2/concentration3

D is not correct because it refers to concentration3/concentration or similar ratio of powers

(Total for Question 6 = 13 marks)

How to answer it

Industrial Equilibria and Reaction Kinetics

What this question tests

This question assesses your ability to apply Le Chatelier's principle and collision theory to industrial conditions (The Contact Process), evaluate economic trade-offs (rate vs. yield vs. cost), sketch and label reaction profiles for catalyzed vs. uncatalyzed pathways, write expressions for equilibrium constants ( K_c ), and derive units for equilibrium constants.

Question 6(a)

Evaluating Changes in Industrial Conditions

💡 Key Knowledge

Temperature Effect: Increasing temperature increases rate (more particles have E >= E_a), but decreases yield because the forward reaction is exothermic ( Δ_r H = -197 kJ mol⁻¹ ).
Pressure Effect: Increasing pressure shifts the position of equilibrium to the right (fewer moles of gas on products side: 2 mol vs 3 mol total on left). It also increases rate by increasing collision frequency.
Economic Balance: Higher temperatures and pressures increase energy and plant/capital equipment costs, requiring a compromise between reaction speed, percentage conversion, and safe running costs.

🧠 Exam Technique (Quality of Written Communication)

This is a 6-mark extended-response question assessed for both indicative chemistry content (up to 4 marks) and structure/logical flow (up to 2 marks).

State the direct effects of changing from 420°C to 600°C and 1.7 atm to 10 atm on rate, yield, and economics.
Explicitly link temperature changes to the exothermicity of the reaction.
Maintain a structured, coherent line of reasoning to capture both quality marks.

❌ Common Errors

Stating that "increasing temperature decreases yield" without mentioning that the forward reaction is exothermic.
Failing to mention economic running costs when evaluating high temperature and pressure together.
Discussing catalysts (which do not change in this scenario, as the question states "while still using the same catalyst").

✅ Mark Scheme Breakdown

IP1: Increase in temperature increases rate.
IP2: Increase in temperature decreases yield (linked to exothermic forward reaction).
IP3: Increase in temperature increases energy costs.
IP4: Increase in pressure increases rate.
IP5: Increase in pressure shifts equilibrium to RHS / increases yield (fewer moles of gas on RHS).
IP6: Increased pressure increases construction and running costs.

Question 6(b)(i)

Reaction Profiles (Uncatalyzed vs. Catalyzed)

💡 Key Knowledge

A catalyst provides an alternative reaction pathway with a lower activation energy ( E_a ).
For an exothermic reaction ( Δ_r H is negative ), the energy level of the products must be drawn lower than the energy level of the reactants.

🧠 Exam Technique & Sketching Rules

Axes: Label the vertical axis as Energy , Enthalpy , or E (Do NOT label as ΔH ).
Levels: Ensure the reactant line ( 2SO₂(g) + O₂(g) ) is positioned horizontally above the product line ( 2SO₃(g) ).
Curves: Draw curve A (uncatalyzed) with a high activation energy hump, and curve B (catalyzed) with a lower activation energy hump. Label both clearly.

Question 6(b)(ii)

Identifying Enthalpy Change and Activation Energy

✅ Correct Answers & Mark Allocation (2 Marks)

Mark 1: Enthalpy change ( ΔH or -197 kJ mol⁻¹ ) shown correctly as the vertical distance between the reactant and product energy levels.
Mark 2: Activation energy ( E_a ) shown correctly as the vertical energy gap from the reactants up to the peak of the catalyzed curve (B).

Question 6(c)(i)

Equilibrium Constant Expression

✅ Correct Answer

K_c = [SO₃]² / ([O₂][SO₂]²)

Examiner Note: You must use square brackets [ ] . Do not use round brackets or curly braces. State symbols are ignored, but formula expressions must be exact based on stoichiometric balancing coefficients.

Question 6(c)(ii)

Units of the Equilibrium Constant

✅ Correct Answer: B (dm³ mol⁻¹)

Substitute concentration units into the K_c expression:

K_c = (mol dm⁻³)² / ((mol dm⁻³)(mol dm⁻³)²)

Cancel out common terms:

K_c = 1 / (mol dm⁻³) = dm³ mol⁻¹

❌ Why other options are incorrect

A: Refers to the inverted/reciprocal expression.
C: Incorrect because units do not fully cancel out (powers of concentration remain).
D: Results from incorrect power ratios or inverted fractions.

Topics

Physical Chemistry · Inorganic Chemistry · Topic 9: Kinetics I · Topic 10: Equilibrium I · Topic 11: Equilibrium II

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.