Spontaneous combustion indicates oxidation state stability

Steps:

• Reaction: 2CO(g) + O₂(g) → 2CO₂(g), where CO has C in +2 state and CO₂ has C in +4 state.
• "Burns readily" means the reaction is spontaneous and exothermic, favoring products.
• This implies CO₂ is more stable than CO.
• Therefore, +4 oxidation state of C is more stable than +2.

Why A is correct:

• Thermodynamics dictate that spontaneous reactions form more stable products; here, CO₂'s +4 state for C is thermodynamically preferred over CO's +2 state per Hess's law.

Why the others are wrong:

• B: ΔH_f° of CO₂ (-393 kJ/mol) is more negative than CO (-110 kJ/mol), contradicting the claim.
• C: High K follows from spontaneity but does not directly address oxidation state stability.
• D: Option incomplete; insufficient information.

Final answer: A