A 2 + B 2 → 2 AB . Δ H f = - 200 kJ mol - 1 AB , A 2 and B 2 are diatomic molecules. If the bond enthalpies of A 2 , B 2 and AB are in the ratio 1 : 0 . 5 : 1 , then the bond enthalpy of A 2 is _________kJ mol – 1 (Nearest integer)

Step-by-step solution

First write balanced reaction for the formation of one mole of AB with the help of diatomic molecule: 1 2 A 2 + 1 2 B 2 → AB           ΔH f = - 200   kJ / mol Now, ∆ H reaction = 1 2 ∆ H A - A + 1 2 ∆ H B - B - ∆ H AB           let ,   ∆ H AB = x   1 2 x + 1 2 ( 0 . 5 x ) - x = - 200    ⇒    x 2 + 0 . 25 x - x = - 200 ⇒    - 0 . 25 x = - 200    ⇒  &