Bond enthalpy of hydroxide
The Masterton Hurley (6th ed.) textbook has a bond dissociation energy of +499 kJ/mol for breaking water into a hydrogen atom and the hydroxyl radical OH. The bond dissociation energy for breaking the hydroxyl radical into another hydrogen atom and an oxygen atom is given as +428 kJ/mol.

You could make the argument that you are starting out with a situation where all electrons are paired and end up with two radicals in the first step, whereas in the second step you already start with unpaired electrons (the hydroxyl radical has seven valence electrons).

In the title of your question, you mention the hydroxide ion. This is not a relevant species because bond energies usually refer to homolytic bond cleavage, with one electron each remaining with the two fragments. It would require heterolytic bond cleavage to go from water to the hydroxide ion (water reacting as a Bronsted acid). This reaction merely requires +55 kJ/mol in aqueous solution.

Some sources contribute to the confusion between hydroxide ion and hydroxyl radical in this context. For example, Boundless states: 


  For instance, the HO-H bond in a water molecule requires 493 kJ/mol to break and generate the hydroxide ion (OH–). Breaking the O-H bond in the hydroxide ion requires an additional 424 kJ/mol. Therefore, the bond energy of the covalent O-H bonds in water is reported to be the average of the two values, or 458.9 kJ/mol. These energy values (493 and 424 kJ/mol) required to break successive O-H bonds in the water molecule are called ‘ bond dissociation energies,’ and they are different from the bond energy. The bond energy is the average of the bond dissociation energies in a molecule.


They incorrectly refer to the hydroxide ion (OH-) and should be referring to the hydroxyl radical (OH).