Edexcel A-Level Chemistry AS Paper 2, June 2019: Question 3

✅ Correct Answer

The enthalpy change when one mole of a substance burns completely in excess oxygen (or fully combusts) under standard conditions (100 kPa and a stated temperature, usually 298 K).

💡 Key Knowledge

• Key phrase 1: "one mole of a substance". Mentioning atoms instead of a substance loses the mark.
• Key phrase 2: "burns completely in excess oxygen".
• Key phrase 3: "standard conditions" (100 kPa / 1 bar and specified temperature).

❌ Common Errors

Students frequently lose the second mark by writing vague terms like "under rtp" or just "under standard conditions" without defining what those conditions (pressure/temperature) actually are.

✅ Correct Answer

C₈H₁₈(l) + 12.5O₂(g) → 8CO₂(g) + 9H₂O(l)

(Accept fractions like 25/2 O₂ to keep 1 mole of octane reacting).

🧠 Exam Technique

Because standard enthalpy of combustion requires 1 mole of the reactant, never scale up the equation to get whole numbers for oxygen if it means having 2 moles of C₈H₁₈. Fractions are completely acceptable and required here.

❌ Common Errors

Forgetting state symbols or misidentifying water as a gas ( g ) instead of liquid ( l ) under standard conditions will forfeit the state symbol mark.

✅ Correct Answer

A reaction profile curve starting at reactants, rising smoothly to a peak (transition state), and falling down to products positioned at a significantly lower energy level than the reactants (exothermic).

💡 Key Knowledge

Label requirements:

• An arrow for activation energy ( Eₐ ) pointing upwards from the reactants line to the absolute peak of the curve.
• An arrow for enthalpy change ( ΔcH or -5470 kJ mol⁻¹ ) spanning vertically between the level of the reactants and the level of the products.

❌ Common Errors

Using double-headed arrows (arrows must have a single clear point), or drawing arrows that float in mid-air instead of clearly touching/starting directly on the reactant and product energy levels.

📐 Step-by-Step Calculation

1. Complete the Hess cycle: Place the combustion products ( CO₂(g) + 2H₂O(l) ) in the bottom box with downwards arrows pointing from both reactants and products towards them.
2. Calculate Reactants Combustion ( ΔcH(reactants) ):
   (-394) + (2 × -286) = -966 kJ mol⁻¹
3. Identify Products Combustion ( ΔcH(products) ):
   CH₄(g) combustion value given in table = -890 kJ mol⁻¹
4. Apply Hess's Law:
   Δr_H = ΣΔcH(reactants) - ΣΔcH(products)
   = -966 - (-890) = -76 kJ mol⁻¹

🧠 Exam Technique & Signs

With combustion cycles, arrows go downwards towards combustion products. Follow the path: go with the reactant arrow (keep sign) and against the product arrow (reverse sign), or use the formula: Reactants minus Products .

📐 Part (i): C—H Bond Enthalpy

Equation given: C(g) + 4H(g) → CH₄(g) has ΔH = -1652 kJ mol⁻¹

Methane contains 4 moles of C—H bonds formed.

Mean bond enthalpy = -1652 / 4 = +413 kJ mol⁻¹

📐 Part (ii): C—C Bond Enthalpy

Equation given: 3C(g) + 8H(g) → C₃H₈(g) has ΔH = -3998 kJ mol⁻¹

Propane (C₃H₈) contains 8 C—H bonds and 2 C—C bonds.

1. Energy released by forming 8 C—H bonds:
   8 × (-413) = -3304 kJ mol⁻¹
2. Subtract from total enthalpy to find remaining C—C energy:
   -3998 - (-3304) = -694 kJ mol⁻¹
3. Divide by 2 for the two C—C bonds:
   -694 / 2 = -347 kJ mol⁻¹
4. Express as a positive bond enthalpy value:
   +347 kJ mol⁻¹

❌ Common Errors & Signs

• Sign convention: Bond enthalpies are always quoted as positive values because energy is required to break bonds (even though energy is released when they form). Leaving a negative sign loses the final mark.
• Consequential error (TE): If you got part (i) wrong, examiners will apply error carried forward if your incorrect value is used consistently in part (ii).