A _ 2 + B _ 2 → 2AB . Δ H _ f = - 200kJ mol ^ - 1 AB, A _ 2 and B _...

First write balanced reaction for the formation of one mole of $AB$ with the help of diatomic molecule:

$\dfrac{1}{2}{A}_{2} + \dfrac{1}{2}{B}_{2} \to AB{\Delta H}_{f} = - 200kJ / mol$

Now,

$∆ {H}_{reaction} = \dfrac{1}{2} ∆ {H}_{A - A} + \dfrac{1}{2} ∆ {H}_{B - B} - ∆ {H}_{AB}\left(let, ∆ {H}_{AB} = x\right)$

$\dfrac{1}{2}x + \dfrac{1}{2} ( 0 . 5x ) - x = - 200 ⇒ \dfrac{x}{2} + 0 . 25x - x = - 200$

$⇒ - 0 . 25x = - 200 ⇒ x = 800kJ / mol$

Bond enthalpy $= 800kJ / mol .$