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

From the following bond energies `H-H` bond energy `431.37 kJ mol^(-1)

1.	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 beA. `553.6 kJ mol^(-1)`B. `1573.6 kJ mol^(-1)`C. `-243.6 kJ mol^(-1)`D. `-120.0 kJ mol^(-1)`
Answer» Correct Answer - D `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)`

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