The enthalpy changes at 298 K in successive breaking of `O-H` bonds of water, are
`H_(2)O(g) rarr H(g)+OH(g),DeltaH=498kJ mol^(-1)`
`OH(g) rarr H(g)+O(g), DeltaH =428 kJ mol^(-1)`
The bond energy of the O-H bond is

Explanation

To find the bond energy of the O-H bond in water, we will analyze the enthalpy changes given for the breaking of O-H bonds in two reactions.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Enthalpy Changes**:
- The first reaction is:
\[ H_2O(g) \rightarrow H(g) + OH(g) \quad \Delta H_1 = 498 \, \text{kJ/mol} \]
- The second reaction is:
\[ OH(g) \rightarrow H(g) + O(g) \quad \Delta H_2 = 428 \, \text{kJ/mol} \]

2. **Understand the Bond Breaking Process**:
- In the first reaction, one O-H bond in water is broken to form H and OH.
- In the second reaction, the O-H bond in the hydroxyl radical (OH) is broken to form H and O.

3. **Calculate the Total Energy for Breaking Both Bonds**:
- The total energy required to break both O-H bonds (one from water and one from hydroxyl) is the sum of the enthalpy changes:
\[ \text{Total Energy} = \Delta H_1 + \Delta H_2 = 498 \, \text{kJ/mol} + 428 \, \text{kJ/mol} = 926 \, \text{kJ/mol} \]

4. **Determine the Bond Energy of One O-H Bond**:
- Since two O-H bonds are broken in the process, the bond energy of one O-H bond can be calculated by dividing the total energy by 2:
\[ \text{Bond Energy of O-H} = \frac{\text{Total Energy}}{2} = \frac{926 \, \text{kJ/mol}}{2} = 463 \, \text{kJ/mol} \]

5. **Final Result**:
- The bond energy of the O-H bond is:
\[ \text{Bond Energy of O-H} = 463 \, \text{kJ/mol} \]