Given standard enthalpies of formation CH₄(g) = −74.8, CO₂(g) = −393.5 and H₂O(l) = −285.8 kJ mol⁻¹, what is ΔH for CH₄ + 2O₂ → CO₂ + 2H₂O(l)?
Correct answer B. −890.3 kJ mol⁻¹
Explanation
ΔH = ΣΔHf(products) − ΣΔHf(reactants) = [−393.5 + 2(−285.8)] − (−74.8) = −965.1 + 74.8 = −890.3 kJ mol⁻¹. O₂ has ΔHf = 0.
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