Calculate the enthalpy of hydrogenation of ethylene from the following data. Bond energies of C-H , C-C , C=C and H-H are 414,347,618 and 435 kJ `mol^(-1)` .

Explanation

`E_(C-H)=414 "kJ mol"^(-1)`
`E_(C-C)=347 "kJ mol"^(-1)`
`E_(C=C)=618 "kJ mol"^(-1)`
`E_(H-H)=435 "kJ mol"^(-1)`

`DeltaH_(r)=sum ("Bond energy")_(r)-sum("Bond energy")p`
`DeltaH_(r)=(E_(C=C)+4E_(C-H)+E_(H-H))-(E_(C-C)+6E_(C-H))`
`DeltaH_(r)=(618+(4xx414)+435)-(347+(6xx414))`
`DeltaH_(r)=2709-2831`
`DeltaH_(r)=-122 "kJ mol"^(-1)`