If the bond energies of H-H,Br-Br and H-Br ar | vedclass

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(D) The enthalpy change for the reaction is calculated using the formula: $\Delta H^o = \sum (	ext{Bond Energy})_{Reactants} - \sum (	ext{Bond Energ

Vedclass · Accepted Answer

$-103$

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(D) The enthalpy change for the reaction is calculated using the formula: $\Delta H^o = \sum (	ext{Bond Energy})_{Reactants} - \sum (	ext{Bond Energy})_{Products}$.For the reaction $H_{2(g)} + Br_{2(g)} 	o 2HBr_{(g)}$,the reactants are $1$ mole of $H-H$ and $1$ mole of $Br-Br$,and the products are $2$ moles of $H-Br$.$\Delta H^o = [BE(H-H) + BE(Br-Br)] - [2 	imes BE(H-Br)]$.Substituting the given values: $\Delta H^o = [433 + 192] - [2 	imes 364]$.$\Delta H^o = 625 - 728$.$\Delta H^o = -103 \ kJ$.

If the bond energies of $H-H$,$Br-Br$ and $H-Br$ are $433$,$192$ and $364 \ kJ \ mol^{-1}$ respectively,$\Delta H^o$ for the reaction $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$ is.....$kJ$

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