Calculate the enthalpy change for the combustion of 2 mol of methane (CH4) at standard conditions. Use the following standard enthalpies of formation: ΔHf°(CH4) = –74.8 kJ mol⁻¹, ΔHf°(CO₂) = –393.5 kJ mol⁻¹, ΔHf°(H₂O(l)) = –285.8 kJ mol⁻¹. The combustion reaction is CH₄ + 2 O₂ → CO₂ + 2 H₂O.
Answer
–1 780.6 kJ
The correct answer is –1 780.6 kJ — State the final answer.
Step-by-step solution
- Determine the enthalpy change of the reaction using ΔHrxn = ΣΔHf(products) – ΣΔHf(reactants).
- Write the products: 2 mol CO₂ and 4 mol H₂O.
- Calculate their contribution: 2(–393.5) + 4(–285.8) = –787 + (–1 143.2) = –1 930.2 kJ.
- Write the reactants: 2 mol CH₄ and 4 mol O₂ (ΔHf of O₂ = 0).
- Calculate their contribution: 2(–74.8) + 0 = –149.6 kJ.
- Compute ΔHrxn: –1 930.2 – (–149.6) = –1 780.6 kJ.
- State the final answer.
Common mistake: Using ΔHf of O₂ as non‑zero or mis‑counting moles
Concept tested: Enthalpy change calculation
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