OCR A-Level Chemistry AS Depth in chemistry (02), June 2022: Question 4

15 marks · Hard difficulty · Structured Questions

Explain the industrial conditions for hydrogen manufacture from methane, calculate equilibrium constant Kc, enthalpy change from bond enthalpies, and explain shapes and bond angles of CO2 and H2O.

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Question

Exam question about the manufacture of hydrogen featuring three main sections: (a)(i) asks to explain industrial conditions for methane steam reforming (equilibrium 4.1), (a)(ii) provides a table of equilibrium concentrations to calculate Kc, (b) gives a reaction equation with structural formulas and average bond enthalpies to calculate delta H, and (c) asks to state and explain bond angles in CO2 and H2O molecules using electron pair repulsion theory.
Question text

4 This question is about the manufacture of hydrogen, H2.

(a) In industry, hydrogen is manufactured from methane, as shown in Equilibrium 4.1.

CH (g) + H O(g) CO(g) + 3H (g) ΔH = +206 kJ mol–1 Equilibrium 4.1

42 2

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]

Additional answer space if required.

(ii) A chemist mixes CH4(g) and H2O(g) and leaves the mixture to reach equilibrium.

CH (g) + H O(g) CO(g) + 3H (g) ΔH = +206 kJ mol–1 Equilibrium 4.1

42 2

The equilibrium mixture contains the following concentrations.

Substance Concentration / mol dm–3

CH4(g) 0.111

H2O(g) 0.682

CO(g) 0.510

H2(g) 1.530

Write an expression for the equilibrium constant, Kc, for Equilibrium 4.1 and calculate

the numerical value of Kc.

Give your answer to 3 significant figures.

K = … mol2 dm–6 [2]

c

(b) Hydrogen can also be manufactured by reacting ethanol with steam, as shown in

Equilibrium 4.2.

H H

H C C O H + 3H O H 6H H + 2O C O Equilibrium 4.2

H H

Average bond enthalpies are shown in the table below.

Bond C–H C–C C–O O–H H–H C=O

Average bond

–1 +415 +347 +358 +464 +435 +805

enthalpy / kJ mol

Calculate ΔH, in kJ mol–1, for the forward reaction in Equilibrium 4.2.

ΔH = … kJ mol–1 [3]

(c) CO2 and H2O molecules have different shapes.

State the bond angles in CO2 and H2O molecules and explain, in terms of electron pair

repulsion, why the bond angles are different.

… [4]

Mark scheme

Show the mark scheme Mark scheme providing detailed marking points for question 4: Level of response rubric for the 6-mark industrial conditions question, Kc calculation steps leading to 24.1 with units, bond enthalpy calculation steps leading to +198 kJ mol-1, and VSEPR explanation points for CO2 (180 degrees, linear/2 bond pairs) and H2O (104.5 degrees, 2 bond pairs and 2 lone pairs).

AO

Question Answer Marks Guidance

element

4 (a) (i)* Please refer to the marking instructions on page 5 of the 6 Indicative scientific points may include:

mark scheme for guidance on how to mark this question. AO1.2 ALLOW reverse arguments throughout

× 3

Level 3 (5-6 marks) Effect of Temperature on equilibrium position

A comprehensive explanation of effect of temperature AO2.5 • (Forward) reaction is endothermic/ΔH is +ve

AND pressure on equilibrium is given with some details × 3

about rate AND operating conditions • High temperature shifts equilibrium to right

There is a well-developed line of reasoning which is clear Effect of Pressure on equilibrium position

and logically structured. The information presented is • Left-hand side has fewer (gaseous) moles

relevant and substantiated. • OR 2 (gaseous) moles form 4 (gaseous) moles

• Low pressure shifts equilibrium to right

Level 2 (3–4 marks)

The candidate attempts three scientific points, but Effect on rate of reaction

explanations are incomplete. • High temp increases rate

There is a line of reasoning presented with some • Low pressure reduces rate

structure. The information presented is relevant and • Catalyst increases rate

supported by some evidence. • Catalyst lowers activation energy

Level 1 (1–2 marks) • Discussion using collision theory to support

A simple description based on at least two of the main arguments

scientific points.

Operating conditions (not inclusive)

There is an attempt at a logical structure with a line of • Compromise conditions needed

reasoning. The information is in the most part relevant.

• High temperatures increase energy

0 marks No response or no response worthy of credit. demand/costs

• Slightly higher pressure used than optimum

• Higher pressures unsafe

• Catalyst reduces need for higher temperatures

• Catalyst doesn’t effect the position of equilibrium

• Excess steam shifts equilibrium to right

17 AO

element

(ii) FIRST CHECK THE ANSWER ON ANSWER LINE 2 IF there is an alternative answer, check for any ECF

IF answer = 24.1, award 2 marks credit possible using working below.

-------------------------------------------------------------------- -------------------------------------------------------------

Kc expression

ALLOW calculated value 24.12887731 correctly

[CO] [H ]3 (0.510) (1.53)3 rounded to 3 or more SF for 1st marking point

(Kc =) OR

[CH4][H2O] (0.111) (0.682)

OR 24.12 … AO2.5

ALLOW ECF to 3 SF

Answer to 3 SF ONLY from inverted Kc expression → 0.0414

Kc = 24.1 AO2.6

[CO] + [H ]3

DO NOT ALLOW (no marks)

[CH4] + [H2O]

IGNORE attempts at units

AO

18 element

(b) FIRST CHECK THE ANSWER ON ANSWER LINE 3 FULL ANNOTATIONS MUST BE USED

IF answer = (+)198 award 3 marks --------------------------------------------------------------------

------------------------------------------------------------------------

Energy for bonds broken

(1 × C–C + 5 × C–H + 1 × C–O + 7 × O–H)

347 + 5(415) + 358 + 7(464)

OR 6028 (kJ) AO2.2 IGNORE sign

×2

Energy for bonds made ( 6 × H–H + 4 × C=O )

6 × 435 + 4 × 805

OR 2610 + 3220

OR 5830 (kJ) IGNORE sign

----------------------------------------

∆H correctly calculated from above ALLOW ECF

∆H = 6028 – 5830 DO NOT ALLOW – sign

= (+)198 (kJ mol–1) AO2.6

Common errors for 2 marks

–198 (incorrect cycle)

–149 (missed C-C from bonds broken)

–2586 (missing 6 x O-H from H2O)

(c) CO2 bond angle = 180º 4

AND ALLOW 104–105

H2O bond angle = 104.5º AO1.1 IGNORE Names of shapes even if incorrect

CO2 has 2 double bonds / 2 bonding regions AO2.1

×3

H2O has 2 bonded pairs AND 2 lone pairs

ALLOW alternative phrases/words for repel

Lone pairs repel more than bonding pairs e.g. ‘push apart’

DO NOT ALLOW atoms repel

Total 15

How to answer it

Manufacture of Hydrogen Study Guide

OCR AS Level Chemistry • Equilibria, Enthalpy Changes & Shapes of Molecules

What this question tests

This question assesses your ability to apply Le Chatelier's principle and collision theory to industrial equilibria, write equilibrium constant expressions and calculate numerical values of Kc, use average bond enthalpies to calculate reaction enthalpy changes (ΔH), and predict bond angles using Electron Pair Repulsion Theory (VSEPR).

Part (a)(i): Industrial Conditions (6 Marks)

Explain why these conditions (15 atm, 800 °C, nickel catalyst, excess steam) are used industrially for: CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂ (ΔH = +206 kJ mol⁻¹)

💡 Key Knowledge

  • Temperature (800 °C): The forward reaction is endothermic (ΔH is positive). High temperatures shift the equilibrium position to the right (products) to absorb heat. It also increases the rate of reaction by increasing collision frequency and energy.
  • Pressure (15 atm): There are 2 moles of gas on the left and 4 moles on the right. High pressure shifts equilibrium to the left (fewer moles), reducing yield. However, too low a pressure slows down the rate. 15 atm is a compromise between acceptable yield and fast rate, while maintaining safe operating limits.
  • Catalyst (Nickel): Lowers activation energy, increasing rate. It does not affect the position of equilibrium or yield, but saves energy by allowing a lower temperature to be used.
  • Excess Steam: Shifts equilibrium to the right, increasing the conversion of methane and ensuring complete reaction of the limiting reagent.

🧠 Exam Technique (Level of Response)

This is a 6-mark extended-response question marked using levels (Level 1: 1-2 marks, Level 2: 3-4 marks, Level 3: 5-6 marks).

  • To hit Level 3, you must link temperature AND pressure effects on equilibrium position alongside rate and operating/economic factors.
  • Use a well-developed, logical line of reasoning rather than isolated bullet points.

Part (a)(ii): Equilibrium Constant Calculation (2 Marks)

Write an expression for Kc and calculate its numerical value to 3 significant figures.

✅ Correct Answer

  • Kc Expression: Kc = ([CO][H₂]³) / ([CH₄][H₂O])
  • Calculated Value: 24.1 (from unrounded 24.1288... )

📐 Step-by-Step Calculation

  1. Substitute values: ((0.510) × (1.530)³) / ((0.111) × (0.682))
  2. Calculate numerator: 0.510 × 3.581577 = 1.8266...
  3. Calculate denominator: 0.111 × 0.682 = 0.075702
  4. Divide and round to 3 SF: 24.1

❌ Common Errors & Examiner Pitfalls

  • Power Errors: Forgetting to cube the concentration of H₂ in the expression or calculation.
  • Inverted Expressions: Putting reactants on the numerator and products on the denominator yields 0.0414 , which scores 0 marks.
  • Significant Figures: Failing to round to the requested 3 significant figures.

Part (b): Enthalpy Change Calculation (3 Marks)

Calculate ΔH, in kJ mol⁻¹, for the forward reaction in Equilibrium 4.2 using average bond enthalpies.

✅ Correct Answer

ΔH = (+)198 kJ mol⁻¹

📐 Step-by-Step Calculation

  1. Bonds Broken (Reactants):
    1 × C–C (347) + 5 × C–H (5 × 415 = 2075) + 1 × C–O (358) + 1 × O–H (464) + 3 × O–H (3 × 464 = 1392)
    Total bonds broken = 347 + 2075 + 358 + 464 + 1392 = 6028 kJ
  2. Bonds Made (Products):
    6 × H–H (6 × 435 = 2610) + 4 × C=O (4 × 805 = 3220)
    Total bonds made = 2610 + 3220 = 5830 kJ
  3. Calculate ΔH:
    ΔH = (Energy of bonds broken) − (Energy of bonds made)
    ΔH = 6028 − 5830 = +198 kJ mol⁻¹

❌ Common Errors & Examiner Pitfalls

  • Sign Convention: Examiners allow the value without a sign or with a '+' sign, but missing the correct sign entirely or reversing the calculation (making −198) loses marks.
  • Counting Bonds: Missing bonds from the expanded display formula (e.g. forgetting the C–C bond or undercounting O–H bonds from steam).

Part (c): Molecule Shapes and Bond Angles (4 Marks)

State the bond angles in CO₂ and H₂O molecules and explain, in terms of electron pair repulsion, why the bond angles are different.

✅ Correct Answers & Marking Points

  • CO₂ Bond Angle: 180°
  • H₂O Bond Angle: 104.5° (Accept 104°–105°)
  • CO₂ Electron Pairs: Has 2 double bonds / 2 bonding regions around the central atom with no lone pairs.
  • H₂O Electron Pairs: Has 2 bonded pairs AND 2 lone pairs around the central oxygen atom.
  • Repulsion Rule: Lone pairs repel more than bonding pairs , compressing the bond angle in water.

❌ Common Errors & Examiner Pitfalls

  • Attribution Error: Never state that "atoms repel each other." Marks are strictly awarded for stating that electron pairs repel.
  • Vague Explanations: Stating water has "lone pairs" without quantifying them (stating explicitly that there are 2 lone pairs and 2 bonding pairs) fails to gain the structural marks.

Topics

Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Module 2: Foundations in chemistry · 3.2 Physical chemistry · 5.1 Rates, equilibrium and pH · 5.2 Energy · 2.2 Electrons, bonding and structure

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.