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 `410.5 kJ mol^(-1)`
Enthalpy for the reaction
`overset(H)overset(|)underset(H)underset(|)(C )=overset(H)overset(|)underset(H)underset(|)(C )+H-HtoH-overset(H)overset(|)underset(H)underset(|)(C )-overset(H)overset(|)underset(H)underset(|)(C )-H`
will be

Options

A. `553.6 kJ mol^(-1)`
B. `1573.6 kJ mol^(-1)`
C. `-243.6 kJ mol^(-1)`
D. `-120.0 kJ mol^(-1)`

Answer

D

Explanation

`DeltaH-"reaction"=sum"Bond energies of reactants"-sum "Bond energies of products"`
`=[BE(C=C+4xxBE(C-H)+BE(H-H)]-[BE(C-C)+6xxBE(C-H)]`
`:. [606.10+1642.00+431.37]-[336.49+2463.49]`
`=2679.47-2799.40=-120 kJ mol^(-1)`