Edexcel A-Level Chemistry Paper 1, June 2019: Question 8

✅ Correct Answers

• Error 1: The arrow for the enthalpy change of formation must point downwards (or be reversed/labelled correctly as exothermic).
• Error 2: The word "half" must be deleted from the enthalpy change of atomisation of hydrogen, because the equation shows a full mole of H atoms being produced from H₂.
• Calculation (8aii): First electron affinity = -73 kJ mol⁻¹

💡 Key Knowledge

• Enthalpy of formation (delta f H) is formation of 1 mole of a compound from its elements in standard states, which is exothermic for stable ionic hydrides (arrow points down).
• Atomisation of hydrogen produces 1 mole of gaseous hydrogen atoms: ½H₂(g) → H(g) . Therefore, using "half" in front of atomisation double-counts the fraction.

🧠 Exam Technique

• When spotting cycle errors, check arrow directions relative to energy levels and check stoichiometry matching the definitions of enthalpy changes.
• For calculation cycles, set up an algebraic sum of steps using Hess's Law: sum of clockwise arrows = sum of anticlockwise arrows.

📐 Calculation Steps (8aii)

1. Step 1: Identify the path around the cycle: delta_f H + Lattice Energy = Atomisation(Na) + 1st IE(Na) + Atomisation(H) + 1st EA(H)
2. Step 2: Substitute values: -56 + (-804) = +107 + +496 + +218 + (1st EA)
3. Step 3: Rearrange for 1st EA: -860 = 821 + (1st EA)
4. Step 4: Solve: 1st EA = -860 - 821 = -73 kJ mol⁻¹ (Watch signs carefully!)

✅ Correct Answers

• 8bi: delta_G = -33.2 kJ mol⁻¹ (or -33203 J mol⁻¹ ). State that the reaction is feasible because delta G is negative (< 0).
• 8bii: T = 732 K (or 459 °C ).

❌ Common Errors

• The Unit Trap: Enthalpy change is typically given in kJ mol⁻¹ while entropy change is in J K⁻¹ mol⁻¹ . Failing to divide delta S by 1000 when mixing units leads to massive calculation errors!
• Omitting the conclusion ("reaction is feasible because delta G is negative") loses the second mark in 8(b)(i).

📐 Step-by-Step Calculations

For 8(b)(i) [delta G = delta H - T delta S]:

1. Convert delta S to kJ: -76.5 / 1000 = -0.0765 kJ K⁻¹ mol⁻¹
2. Substitute into equation: delta_G = -56 - (298 × -0.0765)
3. Calculate: -56 - (-22.797) = -33.2 kJ mol⁻¹

For 8(b)(ii) [At equilibrium, delta G = 0, so T = delta H / delta S]:

1. Use full joules to keep numbers clean: T = 56000 / 76.5
2. Calculate: 732 K

✅ Correct Answers

• Reduction half-equation: ½O₂ + H₂O + 2e⁻ → 2OH⁻ (Accept valid multiples like O₂ + 2H₂O + 4e⁻ → 4OH⁻ )

🧠 Exam Technique & Derivation

• Method 1 (Subtraction): Subtract the given oxidation half-equation from the overall fuel cell equation ( Overall - Oxidation = Reduction ).
• Method 2 (Ion-Electron Method): Balance oxygen atoms using H₂O and hydrogen/charge using OH⁻ ions since the condition is alkaline.
• State symbols are not required as per the question stem, but adding them correctly never penalises you.

✅ Correct Answers

• Option B is correct: NaH (-804), KH (-711), MgH₂ (-2718).

💡 Key Knowledge

• Lattice energy becomes more exothermic (larger negative value) when ionic radius decreases and ionic charge increases (higher charge density, stronger electrostatic attraction).
• Potassium (K⁺) is larger than sodium (Na⁺), so KH has a less exothermic lattice energy than NaH (-711 vs -804). Eliminate options C and D.
• Magnesium (Mg²⁺) has a 2+ charge and a small ionic radius, resulting in a massively more exothermic lattice energy (-2718 kJ mol⁻¹) compared to 1+ ions. Eliminate option A.