Bond enthalpies of A 2 ,   B 2 and AB are in the ratio 2 :   1 :   2 . If bond enthalpy of formation of AB is - 100   KJ   mol - 1 . The bond enthalpy of B 2 is

Step-by-step solution

Assume bond strength of A 2 = 2 X , then, B 2 = X ,   AB = 2 X Δ r H Θ = ∑ Δ diss H Θ - ( R ) ∑ Δ diss H Θ ( P ) 1 2   A 2 + 1 2   B 2 → AB X + X 2   -   2 X   =   - 100   kJ   mol - 1 X   =   200   kJ   mol - 1