From the following bond energies: H–H bond energy: 431.37 kJ mol⁻¹ C=C bond energy: 606.10 kJ mol⁻¹ C–C bond energy: 336.49 kJ mol⁻¹ C–H bond energy: 410.50 kJ mol⁻¹ Enthalpy for the reaction H₂C=CH₂ + H–H → H₃C–CH₃ will be:

Question

Dr. Abubacker · Accepted Answer

[Solution from source] ΔH = bonds broken − bonds formed = [BE(C=C) + BE(H–H)] − [BE(C–C) + 2BE(C–H)] = (606.10 + 431.37) − (336.49 + 821.00) = −120.0 kJ mol⁻¹. The four C–H bonds common to both sides cancel out. Answer (2).

Q. From the following bond energies: H–H bond energy: 431.37 kJ mol⁻¹ C=C bond energy: 606.10 kJ mol⁻¹ C–C bond energy: 336.49 kJ mol⁻¹ C–H bond energy: 410.50 kJ mol⁻¹ Enthalpy for the reaction H₂C=CH₂ + H–H → H₃C–CH₃ will be:

Explanation

Reviewed by Dr. Abubacker

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