From the following bond energies:H-H bond ene | vedclass

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(B) The enthalpy of reaction $(\Delta H)$ is calculated using the formula:$\Delta H = \sum 	ext{Bond Energy (Reactants)} - \sum 	ext{Bond Energy (Pr

Vedclass · Accepted Answer

$-120$

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(B) The enthalpy of reaction $(\Delta H)$ is calculated using the formula:$\Delta H = \sum 	ext{Bond Energy (Reactants)} - \sum 	ext{Bond Energy (Products)}$For the reaction: $CH_2=CH_2 + H-H 	o CH_3-CH_3$Reactants have: $1 	imes (C=C)$,$4 	imes (C-H)$,and $1 	imes (H-H)$ bonds.Products have: $1 	imes (C-C)$ and $6 	imes (C-H)$ bonds.$\Delta H = [BE(C=C) + 4 	imes BE(C-H) + BE(H-H)] - [BE(C-C) + 6 	imes BE(C-H)]$$\Delta H = BE(C=C) + BE(H-H) - BE(C-C) - 2 	imes BE(C-H)$$\Delta H = 606.10 + 431.37 - 336.49 - 2(410.50)$$\Delta H = 1037.47 - 336.49 - 821.00$$\Delta H = -120.02 \approx -120 	ext{ kJ mol}^{-1}$.

$H-H$ bond energy	$431.37 \text{ kJ mol}^{-1}$
$C=C$ bond energy	$606.10 \text{ kJ mol}^{-1}$
$C-C$ bond energy	$336.49 \text{ kJ mol}^{-1}$
$C-H$ bond energy	$410.50 \text{ kJ mol}^{-1}$

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