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. `553.0KJmol^(-1)`B. `1523.6KJmol^(-1)`C. `-243.6KJmol^(-1)`D. `-120.0KJmol^(-1)`

From the following bond energies: `H--H` bond energy: `431.37KJmol^(-1

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. `553.0KJmol^(-1)`B. `1523.6KJmol^(-1)`C. `-243.6KJmol^(-1)`D. `-120.0KJmol^(-1)`
Answer» Correct Answer - B For reaction `DeltaH_(r)=-[4xxBE_(C-H)+1BE_(H=-H)]` `-[6xxBE_(C-H)+1xxBE_(C-C)]` `=(4xx410.50+1xx606.10+1xx431.37)` `-[(6xx410.50)+(1xx336.49)]` `=-120.0KJmol^(-1)`

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