AQA A-Level Chemistry AS Paper 1, June 2023: Question 15
1 mark · Easy difficulty · Multiple Choice
Identify which reaction equation represents the standard enthalpy of formation of potassium oxide.
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Question text
15 Which reaction has an enthalpy change equal to the standard enthalpy of formation of
potassium oxide?
[1 mark]
A 4K(s) + O2(g) → 2K2O(s)
B 2K(s) + O(g) → K2O(s)
C 2 K+(g) + O2–(g) → K O(s)
D 2K(s) + O2(g) → K2O(s)
Mark scheme
Show the mark scheme
15 D 1 (AO1) 2K(s) + O2(g) → K2O(s)
How to answer it
Standard Enthalpy of Formation: Potassium Oxide
This question evaluates your core knowledge of thermodynamic definitions, specifically whether you can translate the formal definition of standard enthalpy of formation (ΔfH°) into a correctly balanced thermochemical equation with correct stoichiometric coefficients and standard states.
- Definition of standard enthalpy of formation (ΔfH°).
- Standard physical states of elements under standard conditions (298 K, 100 kPa).
- Recognising the distinction between enthalpy of formation, lattice enthalpy, and enthalpy of combustion.
Identifying the Formation Equation for K₂O(s)
Which reaction has an enthalpy change equal to the standard enthalpy of formation of potassium oxide?
✅ Correct Answer
D: 2 K(s) + ½ O₂(g) → K₂O(s)
This matches the definition precisely: exactly one mole of compound is formed from its constituent elements in their standard states.
💡 Key Knowledge
The standard enthalpy of formation (ΔfH°) is defined as:
"The enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (298 K and 100 kPa)."
- Potassium standard state: Solid metal, K(s)
- Oxygen standard state: Diatomic gas, O₂(g)
- Product required: Exactly 1 K₂O(s)
📐 Option-by-Option Breakdown
- Option A: 4 K(s) + O₂(g) → 2 K₂O(s)
Forms 2 moles of K₂O(s). The enthalpy change for this reaction is 2 × ΔfH° . Therefore, this does not equal ΔfH°. - Option B: 2 K(s) + O(g) → K₂O(s)
Uses monoatomic oxygen gas, O(g) . Oxygen does not exist as isolated gaseous atoms in its standard state; standard elemental oxygen is O₂(g) . - Option C: 2 K⁺(g) + O²⁻(g) → K₂O(s)
This represents the lattice enthalpy of formation (formation of 1 mole of an ionic lattice from its constituent gaseous ions), not the standard enthalpy of formation from elements. - Option D: 2 K(s) + ½ O₂(g) → K₂O(s)
Forms exactly 1 mole of solid compound from elements in their standard states. A fractional coefficient ( ½ ) for O₂ is required and fully valid.
🧠 Exam Technique: 3-Check Rule
Whenever you need to write or identify an equation for ΔfH°, run these three quick checks:
- Look at the product side: Is there strictly 1 mole of the compound? If the coefficient is 2 or more, eliminate it immediately!
- Look at the reactant side: Are all reactants uncombined elements (never ions, never radicals)?
- Check standard states: Are state symbols correct at 298 K and 100 kPa? (e.g., metals as solids, except Hg; oxygen as O₂(g), not O(g)).
❌ Common Student Traps
- Avoiding fractions: Many students mistakenly pick A because they have been taught in GCSE to avoid fractional coefficients like ½ . In energetics/thermodynamics, fractions are mandatory when fixing product amounts to 1 mole.
- Confusing with Lattice Enthalpy: Selecting C occurs when students mix up standard enthalpy of formation with lattice enthalpy.
- Overlooking Diatomic Gases: Selecting B happens when students balance atoms without considering how the element naturally exists under standard conditions.
Topics
Physical Chemistry · 3.1.4 Energetics
Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.