For the reaction 2H_{2(g)} + O_{2(g)} to 2H_2 | vedclass

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(A) The enthalpy of reaction is given by the sum of bond energies of reactants minus the sum of bond energies of products: $\Delta H = \sum BE_{	ext{

Vedclass · Accepted Answer

$484$

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(A) The enthalpy of reaction is given by the sum of bond energies of reactants minus the sum of bond energies of products: $\Delta H = \sum BE_{	ext{reactants}} - \sum BE_{	ext{products}}$.For the reaction $2H_2 + O_2 	o 2H_2O$,the bonds broken are $2(H-H)$ and $1(O=O)$,and the bonds formed are $4(O-H)$.$\Delta H = [2 	imes BE(H-H) + 1 	imes BE(O=O)] - [4 	imes BE(O-H)]$.Substituting the given values: $-571 = [2 	imes 435 + 498] - 4 	imes BE(O-H)$.$-571 = [870 + 498] - 4 	imes BE(O-H)$.$-571 = 1368 - 4 	imes BE(O-H)$.$4 	imes BE(O-H) = 1368 + 571 = 1939$.$BE(O-H) = \frac{1939}{4} = 484.75 \ kJ \ mol^{-1} \approx 484 \ kJ \ mol^{-1}$.

For the reaction $2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}$,the enthalpy change is $\Delta H = -571 \ kJ$. If the $H-H$ bond energy is $435 \ kJ \ mol^{-1}$ and the $O=O$ bond energy is $498 \ kJ \ mol^{-1}$,calculate the average bond energy of the $O-H$ bond in $kJ \ mol^{-1}$.

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