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Question: Calculate the bond energy of a \(BrF\) bond using the following reaction equation and with the help ...

Calculate the bond energy of a BrFBrF bond using the following reaction equation and with the help of given data. (ΔHf\Delta {H_f} of   BrF5(g)=429kJ/mol,  ΔHf\;Br{F_5}(g) = - 429kJ/mol,\;\Delta {H_f} of Br(g)=112kJ/mol,  ΔHfBr(g) = 112kJ/mol,\;\Delta {H_f} of F(g)=79kJ/molF(g) = 79kJ/mol)
Br(g)+5F(g)BrF5(g)Br(g) + 5F(g) \to Br{F_5}(g)
A. 936kJ/mol936kJ/mol
B. 187kJ/mol187kJ/mol
C. 86kJ/mol86kJ/mol
D. 47kJ/mol47kJ/mol

Explanation

Solution

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: ΔHtotal=ΔHproductsΔHreactants\Delta {H_{{{total}}}} = \Delta {H_{{{products}}}} - \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:
ΔHtotal=ΔHproductsΔHreactants\Delta {H_{{{total}}}} = \Delta {H_{{{products}}}} - \Delta {H_{{{reactants}}}}
Where ΔHtotal\Delta {H_{{{total}}}} is the total enthalpy change of the reaction, ΔHproducts\Delta {H_{{{products}}}} and ΔHreactants\Delta {H_{{{reactants}}}} are the enthalpies of formation of products and reactants respectively. As the product here is BrF5Br{F_5}, its enthalpy of formation is 429kJ/mol - 429kJ/mol. On the reactant side, we have BrBr atom, whose formation enthalpy is 112kJ/mol - 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/mol79kJ/mol) with 55 to account for all five atoms.

Substituting the given values, we get:
ΔHtotal=(429)(112+(5×79))\Delta {H_{{{total}}}} = ( - 429) - ( - 112 + (5 \times 79))
On solving, we get:
ΔHtotal=936kJ/mol\Delta {H_{{{total}}}} = - 936kJ/mol
This is equal to the bond energy of the entire BrF5Br{F_5} molecule. As there are five BrFBr - F bonds in it, bond energy of a single BrFBr - F bond is one-fifth of the total bond energy.
B.E=9365=187kJ/mol\Rightarrow B.E = \dfrac{{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 BrFBr - F is asked. That is why in the end,we divided the total bond energy by five.