KCET 2021 Chemistry Solved Paper

Given,
(BE)A2:(BE)B2:(BE)AB=2:1:2
where, BE= Bond enthalpy
(ΔHf)AB=−100‌kJ
Let the BE of A2,B2 and AB be 2x,x and 2x respeclively.
As we know, ΔfH°=ΣΔdissH°R−ΣΔdissH°P
where,
ΔdissH*= Enthalpy of dissociation

1

2

A2+

1

2

B2→AB
∴ −100=(

1

2

2x+

1

2

x)−x
x=200
∴ Bond enthalpy of B2 is 200‌kJ‌mol−1.