From the following bond energies: `H--H` bond energy: `431.37KJmol^(-1)`
`C=C` bond energy: `606.10KJmol^(-1)`
`C--C` bond energy: `336.49KJmol^(-1)`
`C--H` bond energy: `410.50KJmol^(-1)`
Enthalpy for the reaction will be:
`overset(H)overset(|)underset(H)underset(|)(C)=overset(H)overset(|)underset(H)underset(|)(C)+H-H to Hoverset(H)overset(|)underset(H)underset(|)(C)-overset(H)overset(|)underset(H)underset(|)(C)-H`

Options

A. `553.0 kJ mol^(-1)`
B. `1523.6 kJ mol^(-1)`
C. `-243. kJ mol^(-1)`
D. `-120.0 kJ mol^(-1)`

Answer

D

Explanation

`Delta H= Sigma (B.E.)_(R )- Sigma (B.E.)_(P)`
`Delta H=[4xx(B.E.)_(H-H)+1xx(B.E.)_(C=C)+1xx(B.E.)_(H-H)]-[6xx(B.E.)_(H-H)+1xx(B.E.)_(C-C)]`