The bond dissociation energy of a diatomic molecule is also called bond energy. However, the bond dissociation energy depends upon the nature of bond and also the molecule in which the bond is present. The bond energy N-H bind in `NH_(3)` is equal to one-third of the energy of dissociation of `NH_(3)` because there are three N-H bonds and those of C-H bond in `CH_(4)` is equal to one-fourth of the energy of dissociation of `CH_(4)` Heat of a reaction = Bond energy of reactants - Bond energy of products
The enthalpy changes for the reaction `H_(2)O(g) rarr H(g)+OH(g)` and `OH (g) rarr O(g)+H(g)` are `501.87 kJ mol^(-1)` and `423.3 kJ mol^(-1)`. The bond enthalpy of O-H bond is

Options

A. `-462.625 kJ mol^(-1)`
B. `462.625 kJ mol^(-1)`
C. `-713.54 kJ mol^(-1)`
D. `713.54 kJ mol^(-1)`

Answer

B

Explanation

The enthalpy of dissociation of the O-H bond depends on the molecular species from which H-atom is being separated.
`H_(2)O(g) rarr H(g) + OH(g), Delta H^(@) = 501.87 kJ mol^(-1)`. However to break O-H bond in hydroxyl a different quantity of heat `(423.38 Kj mol^(-1))` is BE (O-H) `= (510.87 kJ mol^(-1)+423.38 kJ mol^(-1))/(2) = 462.625 kJ mol^(-1)`