The reaction of H2 with F2 produces HF with ΔH = - 269 kJ/mol of HF. If the H-H and H-F bond energies are 432 and 565 kJ/mol, respectively, what is the F-F bond energy?

The bond energy of F-F = 429 kJ/mol

Explanation:

Given:

The bond energy of H-H = 432 kJ/mol

The bond energy of H-F = 565 kJ/mol

The bond energy of F-F = ?

Given that the standard enthalpy of the reaction:

H₂ (g) + F₂ (g) ⇒ 2HF (g)

ΔH = - 269 kJ/mol

So,

ΔH = Bond energy of reactants - Bond energy of products.

-269 kJ/mol = [1. (H-H) + 1. (F-F) ] - [2. (H-F) ]

Applying the values as:

-269 kJ/mol = [1. (432 kJ/mol) + 1. (F-F) ] - [2. (565 kJ/mol) ]

Solving for, The bond energy of F-F, we get:

The bond energy of F-F = 429 kJ/mol