`CH_(2)=CH_(2)(g)+H_(2)(g) to CH_(3) to CH_(3)(g)`
The heat of reaction is [bond energy of C=C=80 kcal, C=C=145 kcal, C-H=98 kcal, H-H=103 kcal]

Options

A. `-14` kcal
B. `-28` kcal
C. `-42` kcal
D. `-56` kcal

Answer

B

Explanation

`DeltaH=sumH_(R)-sumH_(P)`
Given, bond energy of C-C=80 kcal, C=C=145 kcal
`C-H=98kcal,H-H=103` kcal
`H-overset(H)overset(|)(C)=overset(H)overset(H)overset(|)(C)-H+H-H to H-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(H)overset(|)(C)-H`
`DeltaH=["B E of "4C-H" bond"]+["B E of "C=C]+["B E of "H-H]`
`-["B E of "6C-H " bond+B E of "C-C]`
`=[98xx4)+(145)+(103)]-[(6xx98)+80]`
`=640-668=-28kcal`