# Question ea30d

Sep 13, 2016

Here's what I got.

#### Explanation:

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, $\Delta {H}_{f}$, of a bromine atom in the gaseous state, ${\text{Br}}_{\left(g\right)}$.

The value $+ {\text{193 kJ mol}}^{- 1}$ tells you that when one mole of bromine atoms is formed, the enthalpy change is equal to $+ \text{193 kJ}$. Keep this in mind.

As you know, bromine exists as a diatomic molecule in its standard state, ${\text{Br}}_{2 \left(g\right)}$. A bromine molecule is formed when two bromine atoms share one electron to form a nonpolar covalent bond.

${\text{Br"_ ((g)) + "Br"_ ((g)) -> "Br}}_{2 \left(g\right)}$

This means that when you break a bromine molecule via homolysis, i.e. to reform two bromine atoms, you have

${\text{Br"_ (2(g)) -> "Br"_ ((g)) + "Br}}_{\left(g\right)}$

Here the $\text{Br"-"Br}$ bond is being broken and two atoms of bromine are being formed

$\text{Br" - "Br" -> "Br" * + * "Br}$

This can be rewritten as

${\text{Br"_ (2(g)) -> color(red)(2)"Br}}_{\left(g\right)}$

Now, you know that when one mole of bromine atoms is formed, the enthalpy change is equal to $+ \text{193 kJ}$. When a mole of bromine is cleaved, $\textcolor{red}{2}$ moles of of bromine atoms are formed.

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 $\text{Br"- "Br}$ bond via homolysis will be

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 $\text{Br"-"Br}$ bond is

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{BDE" = +"386 kJ mol}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 $\text{Br"-"Br}$ bond is $+ {\text{193 kJ mol}}^{- 1}$.

https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf

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 $\text{Br"-"Br}$ bond via homolysis.