Question

Calculate the bond energy of a $BrF$ bond using the following reaction equation and with the help of given data. ($\mathrm{\Delta }{H}_f$ of $Br{F}_5(g)=-429kJ/mol,\mathrm{\Delta }{H}_f$ of $Br(g)=112kJ/mol,\mathrm{\Delta }{H}_f$ of $F(g)=79kJ/mol$)
$$Br(g)+5F(g)\to Br{F}_5(g)$$
A. $936kJ/mol$
B. $187kJ/mol$
C. $86kJ/mol$
D. $47kJ/mol$

Answer 1

Hint: Bond energy, or bond dissociation enthalpy, is the energy that needs to be supplied for all the bonds of a particular molecule to be broken down into individual elements. It gives a measure of the strength of the bond.

Formulae used: $\mathrm{\Delta }{H}_{total}=\mathrm{\Delta }{H}_{products}-\mathrm{\Delta }{H}_{reactants}$

Complete step by step answer:
The values given to us are actually nothing but the enthalpy of formation of each constituent in the given reaction. The difference between formation enthalpies of the products and the reactants will give us the total enthalpy change for the reaction. The total enthalpy change is also thus, the energy that must be supplied to break all the bonds of the molecule at the given temperature. In other words, bond energy of the total molecule is equivalent to the enthalpy change involved while forming the product.

Firstly, we have to calculate the total enthalpy change:
$\mathrm{\Delta }{H}_{total}=\mathrm{\Delta }{H}_{products}-\mathrm{\Delta }{H}_{reactants}$
Where $\mathrm{\Delta }{H}_{total}$ is the total enthalpy change of the reaction, $\mathrm{\Delta }{H}_{products}$ and $\mathrm{\Delta }{H}_{reactants}$ are the enthalpies of formation of products and reactants respectively. As the product here is $Br{F}_5$, its enthalpy of formation is $-429kJ/mol$. On the reactant side, we have $Br$ atom, whose formation enthalpy is $-112kJ/mol$ and five fluorine atoms. So, we have to account for the formation of all five of these atoms. Hence, we have to multiply the formation enthalpy of fluorine ( $79kJ/mol$) with $5$ to account for all five atoms.

Substituting the given values, we get:
$\mathrm{\Delta }{H}_{total}=(-429)-(-112+(5\times 79))$
On solving, we get:
$\mathrm{\Delta }{H}_{total}=-936kJ/mol$
This is equal to the bond energy of the entire $Br{F}_5$ molecule. As there are five $Br-F$ bonds in it, bond energy of a single $Br-F$ bond is one-fifth of the total bond energy.
$\Rightarrow B.E={\displaystyle \frac{936}{5}}=187kJ/mol$

Hence, the correct option is option B.

Additional Information: The negative sign in the enthalpy change shows that it is an exothermic reaction, that is, it gives off heat. Had the enthalpy change been a positive quantity, it would signify an endothermic reaction.

Note: While enthalpy change was negative, bond energy is always a positive quantity as it signifies the amount of energy we need to supply. Whenever we are supplying energy, the convention is to mention it as positive and vice versa. So, in the final answer, always write bond enthalpy as a positive quantity. Note that in the question, only the bond energy of the $Br-F$ is asked. That is why in the end,we divided the total bond energy by five.