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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?

Respuesta :

Answer:

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

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