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Sameedha Modi
Asked: 2022-11-05T19:13:37+05:30 In:

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. `553.0 kJ mol^(-1)`
B. `1523.6 kJ mol^(-1)`
C. `-243. kJ mol^(-1)`
D. `-120.0 kJ mol^(-1)`

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. `553.0 kJ mol^(-1)`
B. `1523.6 kJ mol^(-1)`
C. `-243. kJ mol^(-1)`
D. `-120.0 kJ mol^(-1)`
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  1. Correct Answer – D
    `Delta H= Sigma (B.E.)_(R )- Sigma (B.E.)_(P)`
    `Delta H=[4xx(B.E.)_(H-H)+1xx(B.E.)_(C=C)+1xx(B.E.)_(H-H)]-[6xx(B.E.)_(H-H)+1xx(B.E.)_(C-C)]`

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