Here's what I got.
In simple terms, the bond dissociation energy tells you much energy is needed in order to break a covalent bond so that both atoms retain the electrons they shared to form the bond, i.e. the bond is cleaved by homolysis.
Now, the problem provides you with the enthalpy of formation,
As you know, bromine exists as a diatomic molecule in its standard state,
#"Br"_ ((g)) + "Br"_ ((g)) -> "Br"_ (2(g))#
This means that when you break a bromine molecule via homolysis, i.e. to reform two bromine atoms, you have
#"Br"_ (2(g)) -> "Br"_ ((g)) + "Br"_ ((g))#
#"Br" - "Br" -> "Br" * + * "Br"#
This can be rewritten as
#"Br"_ (2(g)) -> color(red)(2)"Br"_ ((g))#
Now, you know that when one mole of bromine atoms is formed, the enthalpy change is equal to
This means that you need to invest twice as much energy into breaking the bond as you would when forming one mole of bromine atoms.
Therefore, the energy needed to break a
#color(red)(2)color(red)(cancel(color(black)("moles of Br atoms"))) * overbrace((+"193 kJ")/(1color(red)(cancel(color(black)("mole of Br atoms")))))^(color(blue)(DeltaH_fcolor(white)(a) "for Br"_ ((g)))) = "386 kJ"#
You can now say that the bond dissociation energy of the
#color(green)(bar(ul(|color(white)(a/a)color(black)("BDE" = +"386 kJ mol"^(-1))color(white)(a/a)|)))#
SIDE NOTE I'm not sure about the value of the enthalpy of formation of a gaseous bromine atom given to you here because the actual bond dissociation energy for the
The idea remains the same, though. Focus on how much energy is needed to produce one atom of bromine and keep in mind that you produce two atoms of bromine when you cleavage the