Use bond enthalpies (C–H 413, O=O 498, C=O 805, O–H 464 kJ/mol) to find ΔH for CH₄ + 2O₂ → CO₂ + 2H₂O.

Answer and explanation

Correct answer: C. −818 kJ/mol

  1. Step 1: Bonds broken: 4(413) + 2(498) = 1652 + 996 = 2648 kJ.
  2. Step 2: Bonds formed: 2(805) + 4(464) = 1610 + 1856 = 3466 kJ.
  3. Step 3: ΔH = broken − formed = 2648 − 3466 = −818 kJ/mol.
Answer
−818 kJ/mol.
Why not the others: +818 kJ/mol: the sign was reversed. −2648 kJ/mol: only bonds broken were used. −3466 kJ/mol: only bonds formed were used.

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