From the following bond energies : $\mathrm{H}-\mathrm{H}$ bond energy : $431.37 \mathrm{~kJ}…
From the following bond energies :
$\mathrm{H}-\mathrm{H}$ bond energy : $431.37 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{C}=\mathrm{C}$ bond energy : $606.10 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{C}-\mathrm{C}$ bond energy : $336.49 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{C}-\mathrm{H}$ bond energy : $410.50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Enthalpy for the reaction,
will be :
$553.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$1523.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$-243.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$-120.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\Delta \mathrm{H}=$ dissociation energy of reactant - Bond dissociation of energy of product.
$\begin{aligned}
\Delta \mathrm{H}= & (606.10+4 \times 410.5+431.37) \\
& -(6 \times 410.50+336.49) \\
& =-120.0 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}$
will be :