Calculate the enthalpy of the following reaction:
`H_(2)C =CH_(2)(g) +H_(2)(g) rarr CH_(3)-CH_(3)(g)`
The bond enegries of `C-H, C-C,C=C`, and `H-H` are `99,83,147`,and `104kcal` respectively.

Explanation

The reaction is
`underset(H)underset(|)overset(H)overset(|)(C)=underset(H)underset(|)overset(H)overset(|)(C)(g)+H-H(g)toH-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(H)overset(|)(C)-H(g),DeltaH=?`
`DeltaH=`Sum of bond energies of reactants
`-`Sum of bond energies of products.
`=[DeltaH_(C=C)+4xxDeltaH_(C-H)+Delta_(H-H)]`
`-[Delta_(C-C)+6xxDeltaH_(C-H)]`
`=(147+4xx99+104)-(83+6xx99)=-30`kcal