The enthalpy changes at 298 kJ 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

Options

A. `498 kJ mol^(-1)`
B. `463 kJ mol^(-1)`
C. `428 kJ mol^(-1)`
D. `70 kJ mol^(-1)`

Explanation

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

### Step-by-Step Solution:

1. **Identify the 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. **Calculate the Total Enthalpy Change**:
- To find the total energy required to break both O-H bonds, we sum the enthalpy changes:
\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 = 498 \, \text{kJ/mol} + 428 \, \text{kJ/mol} = 926 \, \text{kJ/mol} \]

3. **Determine the Average Bond Energy**:
- Since we are breaking two O-H bonds, the average bond energy (which represents the energy required to break one O-H bond) can be calculated by dividing the total enthalpy change by 2:
\[ \text{Bond Energy} = \frac{\Delta H_{\text{total}}}{2} = \frac{926 \, \text{kJ/mol}}{2} = 463 \, \text{kJ/mol} \]

4. **Conclusion**:
- The bond energy of the O-H bond in water is:
\[ \text{Bond Energy of O-H} = 463 \, \text{kJ/mol} \]