Question

From the following bond energies:
$H-H$ bond energy $:431.37kJmo{l}^{-1}$
$C=C$ bond energy $:606.10kJmo{l}^{-1}$
$C-C$ bond energy $:336.49kJmo{l}^{-1}$
$C-H$ bond energy $:410.50kJmo{l}^{-1}$
Enthalpy for the reaction,

will be

• A. $-243.6kJmo{l}^{-1}$
• B. $-120.0kJmo{l}^{-1}$
• C. $553.0kJmo{l}^{-1}$
• D. $1523.6kJmo{l}^{-1}$

Answer 1

Answer: (B)

For the given reaction, enthalpy of reaction can be calculated as $=B.E.($ reactant $)-B.E.$ (product) $=\left[B.E{.}_{(C=C)}+B.E_{\cdot (H-H)}+4\times B.E{.}_{(C-H)}\right]$

$-[B.E.(C-C)+6\times B.E.(C-H)$

$=[606.10+431.37+4\times 410.50]-[336.49+6\times 410.50]$

$=2679.47-2799.49=-120.02kJmo{l}^{-1}$