If a `298K` the bond energies of `C-H,C-C,C=C` and `H-H` bonds are respectivly `414, 347, 615KJmol^(-1)` , the vlaue of enthalpy change for the reaction
`H_(2)C=CH_(2)(g)+H_(2)(g)+H_(2)(g)rarrH_(3)C-CH_(3)(g)` at `298K` will be

Options

A. `+250 kJ`
B. `-250 kJ`
C. `+125 kJ`
D. `-125 kJ`

Answer

D

Explanation

`DeltaH=sumB.E."of reactants"-sumB.E."of products"`
`=B.E.(C=C)+4xxB.E.(C-H)+B.E.(H-H)-B.E.(C-C)-6xxB.E.(C-H)`
`=B.E.(C=C)+B.E.(H-H)-B.E.(C-C)-2xxB.E.(C-H)`
`=615+435-347-2xx414=1050-1175`
`=-125 kJ`