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(B) The enthalpy change of a reaction is given by the difference between the sum of bond energies of reactants and the sum of bond energies of product

Vedclass · Accepted Answer

$368$

Vedclass · Answer

(B) The enthalpy change of a reaction is given by the difference between the sum of bond energies of reactants and the sum of bond energies of products:$\Delta H = \sum(BE)_{reactants} - \sum(BE)_{products}$For the reaction $H_{2(g)} + Br_{2(g)} ightarrow 2HBr_{(g)}$,the equation is:$\Delta H = [BE(H-H) + BE(Br-Br)] - [2 	imes BE(H-Br)]$Substituting the given values:$-109 = [435 + 192] - 2 	imes BE(H-Br)$$-109 = 627 - 2 	imes BE(H-Br)$$2 	imes BE(H-Br) = 627 + 109$$2 	imes BE(H-Br) = 736$$BE(H-Br) = 368 \; kJ \ mol^{-1}$

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