Given the bond dissociation enthalpy of the C 2 H 5 -H bond as 100 kcal/mol and the enthalpy of formation of C 2 H 6 (g) as -20 kcal/mol, find the enthalpy of formation of the ethyl radical. The dissociation energy of H 2 (g) into H (atoms) is 104 kcal/mol.

C 2 H 6 (g) → C 2 H 5 (g) + H(g), ΔH = +100 kcal. Δ f H(H, g) = 104/2 = 52 kcal/mol. So Δ f H(C 2 H 5 , g) = 100 + Δ f H(C 2 H 6 ) - Δ f H(H) = 100 - 20 - 52 = 28 kcal/mol. This is option (c).

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Chemistry - Thermochemistry Question with Solution | TestHub

ChemistryThermochemistryBond EnthalpyHard2 minQB

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Given the bond dissociation enthalpy of the C₂H₅-H bond as 100 kcal/mol and the enthalpy of formation of C₂H₆(g) as -20 kcal/mol, find the enthalpy of formation of the ethyl radical. The dissociation energy of H₂(g) into H (atoms) is 104 kcal/mol.

Options:

Answer:

Solution:

C₂H₆(g) → C₂H₅(g) + H(g), ΔH = +100 kcal. Δ_f H(H, g) = 104/2 = 52 kcal/mol. So Δ_f H(C₂H₅, g) = 100 + Δ_f H(C₂H₆) - Δ_f H(H) = 100 - 20 - 52 = 28 kcal/mol. This is option (c).

Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy

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