InterviewSolution
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InterviewSolution
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From the following bond energies `H-H` bond energy `431.37 kJ mol^(-1)` `C=C` bond energy `606.10 kJ mol^(-1)` `C-C` bond energy `336.49 kJ mol^(-1)` `C-H` bond energy `410.5 kJ mol^(-1)` Enthalpy for the reaction `overset(H)overset(|)underset(H)underset(|)(C )=overset(H)overset(|)underset(H)underset(|)(C )+H-HtoH-overset(H)overset(|)underset(H)underset(|)(C )-overset(H)overset(|)underset(H)underset(|)(C )-H` will beA. 553.0kJ `mol^(-1)`B. 1523.6 kJ `mol^(-1) `C. `-243.6 " kJ mol"^(-1)`D. `-120.0 "kJ mol"^(-1)` |
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Answer» Correct Answer - 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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