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

Answer

C

Explanation

`CH_(2) =CH_(2(g))+H_(2(g))rarr CH_(3)-CH_(3)`
`Delta H =1 (C =C) +4 (C-H)+ 1 (H-H)-1(C-C)-6(C-H)`
`=1 (C=C)+1(H-H)-1 (C-C)-2(C-H)`
`=615 +435-347-2xx414=1050-1175 =-125 kJ`