Given reaction represents dissociation of bromine gas to form bromine atoms
Br2(g) ↔ 2Br(g)
The enthalpy of the above reaction is given as:
ΔH = ∑n(products)Δ[tex]H^{0}f(products)[/tex] - ∑n(reactants)Δ[tex]H^{0}f(reactants)[/tex]
where n = number of moles
Δ[tex]H^{0}f[/tex]= enthalpy of formation
ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol
Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol
