Question

Consider the following reaction: Br2(g) + 3 F2(g) LaTeX: \rightarrow→ 2 BrF3(g) LaTeX: \Delta H_{rxn}Δ H r x n= ‒836 kJ/mol Bond Bond Energy (kJ/mol) Br–Br 193 F–F 155 Using the above bond dissociation energies, calculate the energy, in kJ/mol, of a Br–F bond.

Answer 1

Answer: The energy of a Br–F bond is 110 kJ/mol

Step-by-step explanation:

The balanced chemical reaction is,

`Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)`

The expression for enthalpy change is,

`\Delta H=\sum [n* B.E(reactant)]-\sum [n* B.E(product)]`

`\Delta H=[(n_(Br_2)* B.E_(Br_2))+(n_(F_2)* B.E_(F_2)) ]-[(n_(BrF_3)* B.E_(BrF_3))]`

`\Delta H=[(n_(Br_2)* B.E_(Br-Br))+(n_(F_2)* B.E_(F_F)) ]-[(n_(BrF_3)* 3* B.E_(Br-F))]`

where,

n = number of moles

Now put all the given values in this expression, we get

`\Delta H=[(1* 193)+(3* 155)]-[(2* 3* B.E_(Br-F))]`

`B.E_(Br-F)=110kJ/mol`

Thus the energy, in kJ/mol, of a Br–F bond is 110