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

Given : `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 + ( 4 x 414 ) + 435 ) -(347 +(6 x 414))
`DeltaH_r`=2709-2831
`DeltaH_r=-122 "kJ mol"^(-1)`