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From the following bond energies: H-H bo...

From the following bond energies: `H-H` bond energy: `431.37KJmol^(-1)`
`C=C` bond energy: `606.10KJmol^(-1)`
`C-C` bond energy: `336.49KJmol^(-1)`
`C-H` bond energy: `410.50KJmol^(-1)`
Enthalpy for the reaction will be:
`overset(H)overset(|)underset(H)underset(|)(C)=overset(H)overset(|)underset(H)underset(|)(C)+H-H to Hoverset(H)overset(|)underset(H)underset(|)(C)-overset(H)overset(|)underset(H)underset(|)(C)-H`

A

553.0kJ `mol^(-1)`

B

1523.6 kJ `mol^(-1) `

C

`-243.6 " kJ mol"^(-1)`

D

`-120.0 "kJ mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`DeltaH` = dissociation energy of reactant -Bond dissociation of energy of product .
`DeltaH=(606.10+4xx410.5+431.37)-(6xx410.50+336.49)`
` =-120.0 kJ//"mol"`
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