Given the standard enthalpies of formation: ΔHf°[CO2(g)] = -393.5 kJ/mol, ΔHf°[H2O(l)] = -285.8 kJ/mol, and ΔHf°[CH4(g)] = -74.8 kJ/mol, what is the standard enthalpy of combustion for methane: CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(l)?
Answer and explanation
Correct answer: D. -890.3 kJ/mol
The enthalpy of reaction is calculated as ΔH°_rxn = Σ ΔHf°(products) - Σ ΔHf°(reactants). Products: (-393.5) + 2 (-285.8) = -393.5 - 571.6 = -965.1 kJ/mol. Reactants: -74.8 + 2(0) = -74.8 kJ/mol. ΔH°_rxn = -965.1 - (-74.8) = -890.3 kJ/mol.
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