Calculate enthalpy change of the following reaction `:`
`H_(2)C=CH_(2(g))+H_(2(g)) rarr H_(3)C-CH_(3(g))`
The bond energy of `C-H,C-C,C=C,H-H` are `414,347,615` and `435k J mol^(-1)` respectively.
A. `+125kJ`
B. `-125kJ`
C. `+250kJ`
D. `-250kJ`
`H_(2)C=CH_(2(g))+H_(2(g)) rarr H_(3)C-CH_(3(g))`
The bond energy of `C-H,C-C,C=C,H-H` are `414,347,615` and `435k J mol^(-1)` respectively.
A. `+125kJ`
B. `-125kJ`
C. `+250kJ`
D. `-250kJ`
Correct Answer – B
`DeltaH=-[e_(C-C)+6xxe_(C-H)]+[e_(C=C)+4xxe_(C-H)+e_(H-H)]`
`=-[347+6xx414]+[615+4xx414+435]`
`=-125kJ`