Which of the following molecules will have different values of standard molar enthalpy of formation, one calculated using bond energy concept and other calculated calorimetrically?

Question

Prashant Kotian · Accepted Answer

Correct answer: B,D. The standard molar enthalpy of formation calculated from bond energy assumes localized bonds. If a molecule exhibits significant resonance stabilization or electron delocalization, the actual calorimetric enthalpy of formation (which accounts for this extra stability) will differ from the value calculated using localized bond energies.
(a) (ethane) is a saturated alkane with no resonance or delocalization. Thus, both values are nearly identical.
(b) 1,3-Butadiene has conjugated double bonds (CH2=CH-CH=CH2) and is stabilized by resonance. Therefore, the calorimetric value differs from the simple bond energy calculation.
(c) 1,4-Cyclohexadiene has isolated double bonds (not conjugated), so its resonance energy is negligible and the two calculations match closely.
(d) (nitrous oxide) is a resonance hybrid (N=N=O ⇌ N≡N→O) with substantial resonance stabilization. Its calorimetric value differs significantly from the bond-energy-derived value.
Therefore, the correct options are B and D.

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