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Question
Calculate the enthalpy of hydrogenation of ethylene from the following data.
Bond energies of C − H, C − C, C = C and H − H are 414, 347, 618 and 435 kJ mol−1.
Solution
Given, EC−H = 414 kJ mol−1
EC−C = 347 kJ mol−1
EC=C = 618 kJ mol−1
EH−H = 435 kJ mol−1
ΔHr = Σ (Bond energy)r − Σ (Bond energy)p
ΔHr = (EC=C + 4EC−H + EH−H) − (EC−C + 6EC−H)
ΔHr = (618 + (4 × 414) + 435) − (347 + (6 × 414))
ΔHr = 2709 − 2831
ΔHr = −122 kJ mol−1
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