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

Vedclass · Accepted Answer

$-488$

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(A) The heat of reaction $\Delta H$ is calculated using the formula: $\Delta H = \sum 	ext{Bond Energy of Reactants} - \sum 	ext{Bond Energy of Products}$.For the reaction $2H_2 + O_2 \longrightarrow 2H_2O$,the bonds broken are $2 	imes (H-H)$ and $1 	imes (O=O)$.Total energy required to break bonds = $2 	imes 436 + 496 = 872 + 496 = 1368 \ kJ/mol$.The bonds formed are $4 	imes (O-H)$ (since there are two $H_2O$ molecules,each with two $O-H$ bonds).Total energy released during bond formation = $4 	imes 464 = 1856 \ kJ/mol$.$\Delta H = 1368 - 1856 = -488 \ kJ/mol$.

Given that the bond energy of hydrogen-hydrogen bond is $436 \ kJ/mol$,that of hydrogen-oxygen bond is $464 \ kJ/mol$,and those in oxygen molecules $496 \ kJ/mol$,what is the approximate heat of reaction for $2H_2 + O_2 \longrightarrow 2H_2O$ ? .....$kJ/mol$

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