Energetics — A/L Chemistry

How do I practise Energetics?

1 question with a worked solution

Check before you answer: Hess's Law & Enthalpy Cycles: Constructing accurate thermochemical cycles to calculate enthalpy of reaction from enthalpies of formation (ΔH = ΣΔH_f(products) - ΣΔH_f(reactants)) or combustion.

1. Given the standard enthalpies of formation (ΔH_f°): C₂H₆(g) = -84.7 kJ mol⁻¹, CO₂(g) = -393.5 kJ mol⁻¹, and H₂O(l) = -285.8 kJ mol⁻¹, calculate the standard enthalpy of combustion (ΔH_c°) of ethane gas.
   1. A) -3119.4 kJ mol⁻¹
   2. B) -1559.7 kJ mol⁻¹
   3. C) -1475.0 kJ mol⁻¹
   4. D) -1644.4 kJ mol⁻¹
   Check my answer

   Correct answer: B) -1559.7 kJ mol⁻¹

   The combustion equation is C₂H₆(g) + 3.5O₂(g) → 2CO₂(g) + 3H₂O(l). Using Hess's law: ΔH_c° = [2 × ΔH_f°(CO₂) + 3 × ΔH_f°(H₂O)] - [ΔH_f°(C₂H₆) + 3.5 × ΔH_f°(O₂)]. Since ΔH_f°(O₂) = 0: ΔH_c° = [2(-393.5) + 3(-285.8)] - [-84.7] = [-787.0 - 857.4] + 84.7 = -1644.4 + 84.7 = -1559.7 kJ mol⁻¹.

   Show worked solution

   Step 1: Write balanced equation for 1 mole of fuel: C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l). Step 2: Apply Hess's Law formula: ΔH_rxn° = Σ n ΔH_f°(products) - Σ m ΔH_f°(reactants). Step 3: Substitute values: ΔH_c° = [2(-393.5) + 3(-285.8)] - [1(-84.7) + 0] = (-787.0 - 857.4) - (-84.7) = -1644.4 + 84.7 = -1559.7 kJ mol⁻¹. Answer: -1559.7 kJ mol⁻¹.
   Modelled on: the A/L Chemistry examination pattern

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Where does Energetics fit in the syllabus?

Grade 12 · Term 2