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Calculate enthalpy change of the followi...

Calculate enthalpy change of the following reaction `:`
`H_(2)C=CH_(2(g))+H_(2(g)) rarr H_(3)C-CH_(3(g))`
The bond energy of `C-H,C-C,C=C,H-H` are `414,347,615` and `435k J mol^(-1)` respectively.

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The correct Answer is:
125 kj
`Delta H_("reaction") =` Bond energy for the formation of bond `+` Bond energy data for the dissociation of bond `= - [1 (C - C) + 6 (C - H) ] + [1 (C = C) + 4 (C - H) + 1 (H - H)] = - 347 - 2 xx 414 + 615 + 435 = - 125 kJ`
Enthalpy change for the the reaction `= -125 kJ`
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