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CGP EDU Academic Team
Published on: August 14, 2026
Standard enthalpy of combustion of cyclopropane is – 2091 kJ/mole at 25ºC then calculate the enthalpy of formation of cyclopropane. If Δ Hº f (CO 2 ) = – 393.5 kJ/mole and Δ Hº f (H 2 O) = – 285.8 kJ/mole.
Text Solution
Verified by ExpertsThe correct answer is:
53
(53)
(CH 2 ) 3 (g) + 
O 2 (g) ⎯→ 3CO 2 (g) + 3H 2 O ( λ )
Δ Hº f (CH 2 ) 3 = – Δ Hº C + 3 Δ Hº f (CO 2 ) (g) + 3 Δ Hº f (H 2 O) (g)
= 2091 + 3 × (– 393.5) + 3 (– 285.8)
= 53.1 kJ/mole
≈ 53 kJ/mole
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