A-B bonds present in `AB_(3)(g)` molecule undergo stewise dissociation by the following sequence of steps.
(i) `AB_(3)(g)toAB_(2)(g)+B(g)`
(ii) `AB_(2)(g) to AB(g)+B(g)`
(iii) `AB(g)toA(g)+B(g)`
At `25^(@)C`, if the enthalpy changes in steps (i) and (iii) are x and z `kJ*mol^(-1)`, respectively, and the bond dissociation energy of A-B bond is y `kJ*mol^(-1)`, then what would be the enthalpy change in step (ii)?

Explanation

Bond energy of A-B bond in `AB_(3)(g)` molecule=the average bond dissocation energy of three A-B bonds in `AB_(3)(g)` molecule.
If the standard enthalpy change in step (ii) be `DeltaH^(0)kJ*mol^(-1)`, then the bond dissociation energy of A-B bond in `AB_(3)(g)` molecule `=(x+DeltaH^(0)+z)/(3)kJ*mol^(-1)`.
Now, the bond dissociation energy of A-B bond=y `kJ*mol^(-1)`
`therefore y=(x+DeltaH^(0)+z)/(3) or, DeltaH^(0)=3y-(x+z)kJ*mol^(-1)`
Therefore, the standard enthalpy change in step (ii) is `[3y-(x+z)]kJ*mol^(-1)`.