AQA A-Level Chemistry Paper 1, 2017: Question 9

14 marks · Medium difficulty · State/Explain/Numerical

Use graphs to deduce optimum temperature and pressure for the industrial methane + steam equilibrium, then calculate equilibrium amounts of CO and H2 given formation of methanol and determine Kp (with units) for CO + 2H2 ⇌ CH3OH at 600 K.

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Question

Question text

09 There are several stages in the industrial production of methanol

from methane.

09.1 The first stage involves a gaseous equilibrium between the reactants

(methane and steam), and some gaseous products. Figures 1 and 2 show

the percentage conversion of methane into the gaseous products under

different conditions at equilibrium.

Figure 1 Figure 2

Deduce the optimum conditions for the industrial conversion of methane and

steam into the gaseous products.

Explain your deductions.

[6 marks]

18 ou

09.2 The equation shows the final stage in the production of methanol.

CO(g) + 2H2(g) ⇌ CH3OH(g)

20.1 mol of carbon monoxide and 24.2 mol of hydrogen were placed in a

*18* sealed container. An equilibrium was established at 600 K. The equilibrium

mixture contained 2.16 mol of methanol.

Calculate the amount, in moles, of carbon monoxide and of hydrogen in the

equilibrium mixture.

[2 marks]

Amount of carbon monoxide mol

Amount of hydrogen20 mol ou

09.3 A different mixture of carbon monoxide and hydrogen was allowed to reach

equilibrium at 600 K

At equilibrium, the mixture contained 2.76 mol of carbon monoxide, 4.51 mol of

hydrogen and 0.360 mol of methanol. The total pressure was 630 kPa

Calculate a value for the equilibrium constant, Kp, for this reaction at 600 K and

state its units.

[6 marks]

Value of Kp Units

Mark scheme

Show the mark scheme

Question Answers Mark Additional Comments/Guidance

This question is marked using levels of response. Refer to the

Indicative Chemistry content

Mark Scheme Instructions for Examiners for guidance on how

to mark this question. Stage 1: Deductions from graph

Level 3 All stages are covered and the explanation of each 1a Yield increases as temperature increases (or converse)

stage is generally correct and virtually complete.

5–6 1b After a certain temperature yield no longer increases

marks To access Level 3, statement 3a must be

considered. 1c Yield decreases as pressure increases (or converse)

Answer is communicated coherently and shows a

logical progression from stage 1 (including 1b) to Stage 2: Optimum temperature and explanation

stage 2 and stage 3

2a High temperature results in high energy costs/expensive

Level 2 All stages are covered but the explanation of each

stage may be incomplete or may contain 2b (After a certain temperature) yield no longer increases

3–4

09.1 marks inaccuracies OR two stages are covered and the therefore there is no gain in using a higher temperature

explanations are generally correct and virtually 6 o

complete. 2c Optimum temperature is between 780-880 C

Answer is mainly coherent and shows progression

from stage 1 to stage 2 and/or stage 3. Stage 3: Optimum pressure and explanation

Level 1 Two stages are covered but the explanation of 3a Low pressure may be too slow

each stage may be incomplete or may contain

1–2 3b So compromise pressure required

marks inaccuracies, OR only one stage is covered but the

explanation is generally correct and virtually 3c Optimum pressure is 1000-2000kPa or moderate pressure

complete. used

– – –

Answer includes isolated statements but these are

presented in a logical order, with sensible

reasoning.

Level 0

Insufficient correct chemistry to gain a mark.

30 of 35

0 marks

Moles of carbon monoxide 17.9 1 Allow 17.94

09.2

Moles of hydrogen 19.9 1 Allow 19.88

𝑝𝑝(CH3OH)

𝐾p = ignore brackets 1 If Kp expression incorrect can only score M2 & M3 & M4

𝑝𝑝(CO)×𝑝𝑝(H2)2

1 If CE in M2 allow ecf for M3, M4 and M6

Total moles of gas = (2.76 + 4.51 + 0.36) = 7.63

If no total moles calculated then can only score M1 and M6

2.76

pp(CO) = x 630 (kPa) (= 228 (kPa))

7.63

4.51

pp(H2) = x 630 (kPa) (= 372 (kPa)) All 3 pp of CO, H2 and CH3OH = 2 marks

7.63 2

2 pp correct = 1 mark

0.36

09.3 pp(CH3OH) = x 630 (kPa) (= 29.7 (kPa))

7.63

29.7 -7 -13 1 -7 -2

Kp = = 9.4(1) x 10 or 9.4(1) x 10 if pp in Pa Allow 9.39 to 9.50 x 10 (kPa )

228x(372)2

can also score M1 from this expression

If no marks awarded allow M6 only for kPa−2 or Pa-2

kPa−2 or Pa-2 (if converted to 630 000) 1

Total 14

How to answer it

Methanol from methane: equilibria + Kp calculations

AQA A-Level Chemistry

What this question tests

Equilibrium from graphs (industrial thinking)

Reading trends: conversion vs temperature and pressure.
Choosing optimum conditions using yield + cost/rate compromise.
Explaining plateaus on graphs (no further gain past a certain point).

Stoichiometry + equilibrium composition

Using the balanced equation to link moles reacted to moles formed.
ICE-style thinking (initial → change → equilibrium).

Kp from partial pressures

Writing correct Kp expression for: CO(g) + 2H₂(g) ⇌ CH₃OH(g).
Finding partial pressures from mole fraction × total pressure.
Units of Kp from powers in the expression (kPa⁻² or Pa⁻²).

Total marks: 14. Q09.1 is levels of response (6 marks) so coherence + justified industrial compromise matters.

Q09.1 (6 marks)

Part (a): Optimum conditions for conversion of methane + steam to gaseous products

Use Figure 1 and Figure 2 to justify temperature and pressure choices

✅ Correct answer (what to state)

Optimum temperature: about 780–880°C (high temperature region where conversion is high but then levels off).
Optimum pressure: about 1000–2000 kPa (a moderate pressure compromise).

Mark scheme wording expects: temperature range 780–880°C and pressure range 1000–2000 kPa, plus clear reasoning.

💡 Key knowledge (what the graphs are telling you)

Figure 1 (constant pressure): % conversion increases as temperature increases.
Figure 1 plateau: after a certain temperature, conversion no longer increases much (little benefit going higher).
Figure 2 (constant temperature): % conversion decreases as pressure increases.

🧠 Exam technique (how to hit Level 3)

Write in a logical chain: (1) trend from graph → (2) industrial issue (cost/rate) → (3) chosen optimum range.
For full marks, you must include the idea that after a certain temperature there is no gain (plateau) and link it to cost.
For pressure, don’t just say “low pressure”: explain why not extremely low (can be too slow) and therefore a compromise.

❌ Common errors (examiner-style)

Stating “use high pressure for higher yield” even though Figure 2 shows conversion falls with pressure.
Choosing the highest temperature possible without mentioning the plateau (no further yield increase) and high energy costs.
Not giving a range (the mark scheme credits approx 780–880°C and 1000–2000 kPa).
Listing statements without linking them into a coherent “industrial compromise” argument (drops you down levels).

🧠 Model Level 3 explanation (how to write it)

From Figure 1, the percentage conversion increases as temperature increases. However, above roughly 780–880°C the curve levels off, so increasing temperature further gives little extra conversion. A very high temperature would also increase energy costs, so an optimum is a high temperature in the region 780–880°C.

From Figure 2, the percentage conversion decreases as pressure increases, so lower pressure favours conversion. But very low pressure can make the process too slow, so an industrial compromise is to use a moderate pressure, about 1000–2000 kPa.

This hits the mark scheme “stages”: (1) deduce trends (including plateau), (2) explain temperature compromise (cost + no gain), (3) explain pressure compromise (rate/feasibility).

Q09.2 (2 marks)

Part (b): Equilibrium moles from CO + H₂ ⇌ CH₃OH

CO(g) + 2H₂(g) ⇌ CH₃OH(g) at 600 K

✅ Correct answers

Amount of carbon monoxide at equilibrium = 17.9 mol (allow 17.94)
Amount of hydrogen at equilibrium = 19.9 mol (allow 19.88)

Marks: 1 mark for CO, 1 mark for H₂ (correct to appropriate rounding).

📐 Calculation (step-by-step)

Use stoichiometry: 1 mol CH₃OH forms from 1 mol CO and 2 mol H₂.
Moles CH₃OH at equilibrium = 2.16 mol ⇒ moles CO reacted = 2.16 mol .
Moles H₂ reacted = 2 × 2.16 = 4.32 mol .
CO remaining = 20.1 − 2.16 = 17.94 mol ≈ 17.9 mol.
H₂ remaining = 24.2 − 4.32 = 19.88 mol ≈ 19.9 mol.

❌ Common errors

Forgetting the 2 in 2H₂, so subtracting 2.16 instead of 4.32 from H₂.
Rounding too early (keep 17.94 and 19.88 until the end).
Assuming CH₃OH formed equals limiting reagent amount without using the equation ratio.

🧠 Exam technique

Write a mini ICE table in words: initial moles → change using coefficients → equilibrium moles.
Always link “moles formed” to “moles reacted” using the balanced equation coefficients.

Q09.3 (6 marks)

Part (c): Calculate Kp and its units

Given equilibrium moles and total pressure at 600 K

💡 Key knowledge

Kp expression uses partial pressures raised to stoichiometric powers.
Partial pressure: pp(gas) = (moles of gas / total moles) × total pressure
Units come from powers: numerator kPa¹ / denominator kPa³ ⇒ overall kPa⁻².

✅ Correct answer (from mark scheme)

Kp = 9.41 × 10⁻⁷ kPa⁻² (acceptable range ≈ 9.39 to 9.50 × 10⁻⁷ kPa⁻²)
Units: kPa⁻² (or Pa⁻² if pressure used in Pa consistently)

Mark points include: correct Kp expression, total moles, partial pressures, Kp value, and correct units.

🧠 Examiner insight: where marks are won/lost

If your Kp expression is wrong, the scheme says you can only score later calculation marks in limited circumstances—so get it correct first.
You must calculate total moles to get mole fractions; skipping this prevents access to most partial-pressure marks.
Units are often forgotten: the scheme awards a separate mark for kPa⁻² or Pa⁻².

❌ Common errors (calculation traps)

Using moles directly in Kp instead of partial pressures.
Not squaring pp(H₂) (must be (pp(H₂))² ).
Mixing units: using total pressure in kPa but converting only one partial pressure to Pa.
Rounding pp values too aggressively before calculating Kp (carry at least 3 s.f. through).

📐 Full calculation (step-by-step, mark-scheme method)

Reaction: CO(g) + 2H₂(g) ⇌ CH₃OH(g)

Write Kp:
Kp = pp(CH₃OH) / (pp(CO) × (pp(H₂))²)
This is M1 in the mark scheme.
Total moles at equilibrium:
n(total) = 2.76 + 4.51 + 0.360 = 7.63 mol
This is a required step to access the partial pressure marks.
Calculate partial pressures (pp = mole fraction × total pressure):
Total pressure = 630 kPa

pp(CO) = (2.76 / 7.63) × 630 = 228 kPa
pp(H₂) = (4.51 / 7.63) × 630 = 372 kPa
pp(CH₃OH) = (0.360 / 7.63) × 630 = 29.7 kPa
Mark scheme: all 3 correct = 2 marks; 2 correct = 1 mark.
Substitute into Kp:
Kp = 29.7 / (228 × (372)²)
First evaluate denominator:
(372)² = 138384
228 × 138384 = 31551552
So:
Kp = 29.7 / 31551552 = 9.41 × 10⁻⁷
Acceptable: 9.39 to 9.50 × 10⁻⁷ (kPa⁻²).
Units:
Numerator has kPa¹; denominator has kPa¹ × (kPa¹)² = kPa³
So Kp units = kPa¹ / kPa³ = kPa⁻²
If you used Pa throughout, units become Pa⁻² (and Kp value changes accordingly).

Final checklist

🧠 30-second exam checklist

Q09.1: Quote trends from graphs + plateau + industrial compromise + give the ranges.
Q09.2: Use coefficients: CH₃OH formed = CO reacted; H₂ reacted = 2×.
Q09.3: Kp in partial pressures, calculate total moles, then pp values, then units.

❌ Fast “don’t do this” list

Don’t say “increase pressure increases yield” when the graph shows the opposite.
Don’t forget to square pp(H₂).
Don’t omit Kp units.

✅ Key final numbers to remember

Optimum T: 780–880°C
Optimum P: 1000–2000 kPa
Q09.2 equilibrium moles: CO 17.9, H₂ 19.9
Q09.3: Kp 9.41 × 10⁻⁷ kPa⁻²

Topics

Physical Chemistry · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc · 3.1.10 Equilibrium Constant Kp · 3.1.2 Amount of Substance · 3.1.5 Kinetics

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.