Edexcel A-Level Chemistry AS Paper 2, June 2016: Question 6

Enthalpy Changes of Solution and Hess Cycles

🔍 What this question tests

This question assesses your ability to calculate enthalpy changes from experimental calorimetric data ( q = mcΔT ), construct and apply Hess's Law enthalpy cycles, and evaluate how experimental flaws and impure/degraded reagents impact thermochemical data.

Part (a) — Calorimetry & Enthalpy of Solution

Calculate the enthalpy change of solution for anhydrous sodium carbonate (4 marks)

📐 Step-by-Step Calculation

Calculate heat energy transferred (q):
ΔT = 32.4 - 27.0 = 5.4 °C
q = m × c × ΔT = 50.0 × 4.18 × 5.4 = 1128.6 J (or 1.1286 kJ )
Calculate moles of anhydrous Na₂CO₃:
Mr of Na₂CO₃ = (23.0 × 2) + 12.0 + (16.0 × 3) = 106.0 g mol⁻¹
Moles = 5.09 / 106.0 = 0.04802 mol
Calculate enthalpy change per mole:
ΔH = - q / moles = - 1.1286 kJ / 0.04802 mol = - 23.504 kJ mol⁻¹
Format final answer:
- 23.5 kJ mol⁻¹ (to 3 significant figures, with the negative sign).

❌ Common Errors & Traps

Forgetting the negative sign: Enthalpy of solution is exothermic here, so a minus sign is mandatory for the final mark.
Mass confusion: Using 55.09 g (adding solid mass to water mass) for m in mcΔT . The standard Edexcel convention for dilute solutions is to use just the mass of the water (50.0 g).
Significant figures: Giving 1 or 4+ SF when data is given to 3 SF. Stick to 2 or 3 SF.

Mark Breakdown (4 marks): (1) Calculation of heat energy (q); (2) Calculation of moles of Na₂CO₃; (3) Calculation of enthalpy change value; (4) Correct negative sign and 2 or 3 SF.

Part (b) — Hess's Law Cycles

Complete the Hess cycle and calculate enthalpy change of hydration (2 marks)

💡 Hess Cycle Construction

The cycle links anhydrous sodium carbonate and water directly to hydrated sodium carbonate via aqueous ions/solutions:

Top line: Na₂CO₃(s) + 10H₂O(l) reacting to form Na₂CO₃·10H₂O(s)
Both pathways lead to the common intermediate state: Na₂CO₃(aq) (+ 10H₂O(l))
Arrows pointing downwards from both solid forms to the aqueous solution state representing the respective enthalpies of solution.

✅ Correct Calculation & Answer

Using Hess's Law (vector routing):

ΔH = ΔH₁ - ΔH₂ (or routing around the cycle)

Answer from (a): - 23.5 kJ mol⁻¹
Given in (b): + 53.7 kJ mol⁻¹

Calculation: - 23.5 - 53.7 = - 77.2 kJ mol⁻¹

Mark Breakdown (2 marks): (1) Both arrows pointing in correct directions with correct intermediate species in the cycle box; (2) Correct arithmetic resulting in - 77.2 kJ mol⁻¹ (allowing consequential error from part a).

Part (c) — Evaluation of Experimental Conditions

Explain how using an old (weathered) sample affects the second experiment (2 marks)

🧠 Exam Technique & Key Insight

When hydrated crystals lose water of crystallisation to the air (efflorescence), the sample becomes partially anhydrous. Think about what happens when this mixture dissolves:

Anhydrous sodium carbonate dissolving in water is exothermic (releases energy, as calculated in part a).
Therefore, having less water of crystallisation already bound means more exothermic contributions/less endothermic heat absorption.

✅ Acceptable Answers for Full Marks

Point 1: The enthalpy change of solution will be lower / less endothermic / more exothermic (smaller positive value or negative).
Point 2: Because anhydrous sodium carbonate reacts exothermically with water / less energy is required to break bonds in the partial crystal.

Mark Breakdown (2 marks): (1) Stating the enthalpy of solution is lower/less endothermic; (2) Valid chemical reason regarding anhydrous/partially hydrated nature releasing energy or needing less bond-breaking energy.