Calculate the standard enthalpy of combustion of propane (C₃H₈) given standard enthalpies of formation: ΔHf°[C₃H₈(g)] = -104 kJ/mol, ΔHf°[CO₂(g)] = -394 kJ/mol, ΔHf°[H₂O(l)] = -286 kJ/mol.

Answer and explanation

Correct answer: C. -2,222 kJ/mol

  1. Step 1: Combustion equation: C₃H₈(g) + 5O₂(g) -> 3CO₂(g) + 4H₂O(l). Step 2. ΔH° = Σ ΔHf°(products) - Σ ΔHf°(reactants). Step 3. Products: 3(-394) + 4(-286) = -1182 - 1144 = -2326 kJ/mol. Step 4. Reactants: -104 kJ/mol (O₂ is element in standard state, ΔHf° = 0). Step 5. ΔH° = -2326 - (-104) = -2326 + 104 = -2,222 kJ/mol. Answer: -2,222 kJ/mol. Why not the others: -2,326 kJ/mol forgets to subtract ΔHf° of propane; -2,118 kJ/mol and -1,984 kJ/mol are calculation errors.

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