\(\Rightarrow \Delta H _{ p }^0( AB )=-200\,kJ\,mol\,m ^{-1}\)
\(\therefore \Delta H _{ R }^0\) for reaction \(A _2+ B _2 \rightarrow 2 AB\) is \(-400\,kJ\,mol ^{-1}\)
Given: Bond Enthalpy of \(A _2, B _2\) and \(AB\) is \(1: 0.5: 1\)
Assuming bond enthalpy of \(A _2\) be \(x\, kJ\, mol { }^{-1}\)
\(\therefore\) Bond enthalpy \(B _2=0.5 \times kJ\, mol ^{-1}\)
\(\therefore\) Bond enthalpy \(AB =( x ) kJ\, mol\, m ^{-1}\)
\(-400= x +0.5 x -2 x\)
\(-400=-0.5 x\)
\(\therefore x =800\,kJ / mol\)