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

Explanation:

The balanced chemical reaction is,

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

The expression for enthalpy change is,

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

$\Delta H=[(n_{Br_2}\times B.E_{Br_2})+(n_{F_2}\times B.E_{F_2}) ]-[(n_{BrF_3}\times B.E_{BrF_3})]$

$\Delta H=[(n_{Br_2}\times B.E_{Br-Br})+(n_{F_2}\times B.E_{F_F}) ]-[(n_{BrF_3}\times 3\times B.E_{Br-F})]$

where,

n = number of moles

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

$\Delta H=[(1\times 193)+(3\times 155)]-[(2\times 3\times B.E_{Br-F})]$

$B.E_{Br-F}=110kJ/mol$

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