Question

Calculate ΔH∘ for the reaction H2(g)+Br2(g)→2HBr(g) using the bond energy values.

The ΔH∘f of HBr(g) is not equal to one-half of the value calculated. Account for the difference. Explain in 3-4 sentences.

Answer 1

H2(g) + Br2(g)-------------------------> 2HBr(g)

ΔH∘rxn = Bond energy of reactants - bond energy of products

             = (BE of H-H   + BE of Br-Br) - ( 2 x BE of H-Br)

             = (436 + 194) - ( 2 x 366)

             = -102 kJ

ΔH∘ = -102 kJ

here we are calculation ΔH∘of the reaction . the value is for entire reaction . we need to do half this value.

if it is asks calculate the ΔH∘f of HBr(g) then we have to do 1/2 of the ΔH∘ value . because standard enthalphy of formation is for only one mole of substance formed .