Edexcel A-Level Chemistry AS Paper 2, November 2020: Question 4

14 marks · Medium difficulty · Calculations

Calculate the enthalpy of combustion of methanol using calorimetry and Hess's law, and explain the effects of temperature and pressure on equilibrium and reaction rate.

Practise this question

Question

A multi-part exam question about the combustion and synthesis of methanol. Part (a) provides experimental data for burning methanol to heat water and asks for the enthalpy of combustion. Part (b) gives two chemical equilibria for methanol synthesis and asks to explain the effects of pressure and temperature. Part (c) provides a table of standard enthalpies of combustion for CO and H2 and asks to calculate the standard enthalpy of combustion of gaseous methanol using Hess's law.
Question text

4 Methanol, CH3OH, is a liquid fuel.

An experiment was carried out to determine the enthalpy change of combustion of

liquid methanol.

thermometer

beaker

water

spirit burner

methanol

The energy obtained from burning 2.08g of methanol was used to heat 75.0g of water.

The temperature of the water rose from 25.0°C to 91.0°C.

[Specific heat capacity of water = 4.18 J g−1 °C−1]

(a) Use the data to calculate a value for the enthalpy change of combustion of

one mole of methanol.

Give your answer to an appropriate number of significant figures and include

a sign and units.

(4)

(b) Methanol can be synthesised from methane and steam by a process that occurs

in two steps.

Step 1 CH (g) + H O(g) 3H (g) + CO(g) ΔH = +206 kJ mol−1

42 2

Step 2 CO(g) + 2H (g) CH OH(g) ΔH = −91 kJ mol−1

(i) Explain the effects of increasing the pressure on the yield of the products and

on the rate of the reaction in Step 1.

(4)

… 11

… *P62307A01128*

(ii) Step 2 is carried out at a compromise temperature of 500K.

Explain why 500K is considered to be a compromise for Step 2 by considering

what would happen at higher and lower temperatures.

(3)

(c) Calculate a value for the standard enthalpy change of combustion of gaseous

12 methanol using the enthalpy change for Step 2 and the standard enthalpy

change of combustion of gaseous carbon monoxide and of hydrogen.*P62307A01228*

Standard enthalpy change

Substance −1

of combustion/kJmol

CO −283

H2 −286

(3)

(Total for Question 4 = 14 marks)

Mark scheme

Show the mark scheme The mark scheme providing detailed answers and guidance for all parts of question 4, including calculated values for enthalpy changes, moles, and point-by-point explanations for equilibrium and rate effects.

How to answer it

Enthalpy Changes, Equilibria & Hess's Law

What this question tests

This comprehensive AS Chemistry question assesses your ability to calculate enthalpy changes from calorimetry experiments, apply Le Chatelier's Principle to predict the effects of pressure and temperature changes on chemical equilibria and reaction rates, and construct Hess's Law cycles using standard enthalpies of combustion.

Question Part (a) - Calorimetry Calculation

Calculating Enthalpy Change of Combustion

📐 Step-by-Step Calculation

  1. Calculate heat energy absorbed by water (Q):
    Q = m × c × ΔT
    Q = 75.0 × 4.18 × (91.0 - 25.0) = 75.0 × 4.18 × 66.0 = 20691 J
  2. Calculate moles of methanol burned:
    Moles = mass / molar mass
    Molar mass of CH₃OH = 12.0 + (4 × 1.0) + 16.0 = 32.0 g mol⁻¹
    Moles = 2.08 / 32.0 = 0.0650 mol
  3. Calculate enthalpy change per mole:
    ΔH = Q / moles = 20691 / 0.0650 = 318323 J mol⁻¹ = 318.3 kJ mol⁻¹
  4. Apply sign and significant figures:
    Combustion is exothermic, so add a negative sign: -320 kJ mol⁻¹ (to 2 SF) or -318 kJ mol⁻¹ (to 3 SF).

❌ Common Errors & Traps

  • Missing negative sign: Forgetting that combustion reactions are exothermic (-). Examiners will penalize a positive final value.
  • Incorrect significant figures: Giving too many SF (e.g. 4 or 5) when data in the stem is given to 2 or 3 significant figures.
  • Molar mass errors: Incorrectly calculating the relative molecular mass of CH₃OH.
Marks available: 4 marks (1 for calculating Q, 1 for moles of methanol, 1 for scaling to 1 mole, 1 for correct sign, units, and appropriate 2 or 3 SF).
Question Part (b)(i) - Pressure & Equilibria

Effect of Pressure on Yield and Rate (Step 1)

CH₄(g) + H₂O(g) ⇌ 3H₂(g) + CO(g) (ΔH = +206 kJ mol⁻¹)

✅ Correct Answer Breakdown

  • Yield: Increasing pressure decreases the yield of products.
  • Reasoning: There are fewer moles of gas on the reactant side (2 moles) compared to the product side (4 moles). The equilibrium shifts to the side with fewer moles to oppose the increase in pressure.
  • Rate: Increasing pressure increases the rate of reaction.
  • Reasoning: Particles are closer together in a smaller volume, leading to an increased frequency of successful collisions.

💡 Key Knowledge

Always separate your discussion into position of equilibrium (yield) and collision theory (rate). Never mix up the two concepts! A higher rate does not mean a higher yield in reversible reactions.

Marks available: 4 marks (1 for yield decreasing, 1 for reference to more moles of gas on the right, 1 for rate increasing, 1 for increased collision frequency).
Question Part (b)(ii) - Temperature & Compromise

Evaluating Compromise Temperature (Step 2)

CO(g) + 2H₂(g) ⇌ CH₃OH(g) (ΔH = -91 kJ mol⁻¹)

🧠 Exam Technique & Answer

  • Higher temperatures: Would decrease the yield of methanol because the forward reaction is exothermic. However, it would give a faster rate.
  • Lower temperatures: Would increase the yield of methanol, but the reaction would become too slow to be economically viable.
  • Conclusion (500 K): Acts as a compromise, balancing a reasonable/acceptable yield with an economically viable reaction rate.

❌ Common Errors

  • Vague statements like "lower temperatures give a better yield" without explaining why (mentioning exothermic/endothermic) or failing to link the low temperature to an unacceptably slow rate.
Marks available: 3 marks (1 for effect of higher temperature on yield due to exothermicity, 1 for effect of lower temperature on rate, 1 for explaining the compromise between yield and rate).
Question Part (c) - Hess's Law Calculation

Enthalpy Change of Combustion of Gaseous Methanol

📐 Step-by-Step Hess's Law Cycle

  1. Set up the relationship or cycle:
    Combustion of gaseous methanol produces CO₂(g) + 2H₂O(l) (or corresponding combustion products).
  2. Identify standard enthalpy values from data table:
    Δc H(CO) = -283 kJ mol⁻¹
    Δc H(H₂) = -286 kJ mol⁻¹ (note: must multiply by 2 for the equation!)
  3. Apply Hess's Law equation:
    Δc H(CH₃OH) = ΔH (Step 2) + Δc H(CO) + 2Δc H(H₂)
    Alternative route using cycle:
    -91 + (-283) + (2 × -286)
  4. Calculate final value:
    = -91 - 283 - 572 = -764 kJ mol⁻¹

❌ Common Calculation Traps

  • Forgetting stoichiometry: Forgetting to multiply the enthalpy of combustion of hydrogen (-286) by 2 to account for the 2H₂ in Step 2. This is the single most common mark loss in this question!
  • Sign errors: Reversing signs when rearranging algebraic loops. Always double-check your arrows in a Hess cycle.
Marks available: 3 marks (1 for correct equation or cycle linking the three processes, 1 for correct numerical substitution including multiplying hydrogen by 2, 1 for final correct value with negative sign).

Topics

Physical Chemistry · Core Practicals · Topic 5: Formulae, Equations and Amounts of Substance · Topic 8: Energetics I · Topic 9: Kinetics I · Topic 10: Equilibrium I · Core Practical 8: Determine the enthalpy change of a reaction using Hess’s law

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.