Consider the following reaction: $$A_{2}+B_{2} \longrightarrow 2 A B \quad \Delta H=-285 \mathrm{kJ}$$ The bond energy for \(\mathrm{A}_{2}\) is one-half the amount of the AB bond energy. The bond energy of \(\mathrm{B}_{2}=432 \mathrm{kJ} / \mathrm{mol}\) . What is the bond energy of \(\mathrm{A}_{2}\) ?

The change in enthalpy (ΔH) for the reaction can be represented by the formula: ΔH = Bond energy of reactants - Bond energy of products For our given reaction, A₂ + B₂ → 2AB, we have: ΔH = (Bond energy of A₂ + Bond energy of B₂) - 2 × (Bond energy of AB)

We are given that the bond energy of A₂ is one-half of the bond energy of AB. Therefore, we can write: Bond energy of A₂ = (1/2) × (Bond energy of AB)

Now, we can substitute the information given in the problem into the enthalpy change equation, and solve for the bond energy of A₂: -285 kJ = (Bond energy of A₂ + 432 kJ) - 2 × (Bond energy of AB) We can also substitute the relationship between bond energies of A₂ and AB: -285 kJ = ((1/2) × (Bond energy of AB) + 432 kJ) - 2 × (Bond energy of AB) Now, let's solve for the Bond energy of AB: -285 kJ = - (3/2) × (Bond energy of AB) + 432 kJ -147 kJ = -(3/2) × (Bond energy of AB) Bond energy of AB = 294 kJ/mol Finally, we can find the bond energy of A₂ using the given relationship: Bond energy of A₂ = (1/2) × (Bond energy of AB) Bond energy of A₂ = (1/2) × 294 kJ/mol Bond energy of A₂ = 147 kJ/mol The bond energy of A₂ is 147 kJ/mol.