The average O-H bond energy in `H_(2)O` with the help of following data.
(1) `H_(2)O(l)rarrH_(2)O(g), Delta H= +40.6 kJ mol^(-1)`
(2) `2H(g)rarrH_(2)(g), DeltaH= -435.kJ mol^(-1)`
(3) `O_(2)(g)rarr2O(g), Delta H= +489.6 kJ mol^(-1)`
(4) `2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l), Delta H= -571.6 kJ mol^(-1)`

Options

A. `584.9 kJ mol^(-1)`
B. `279.8 kJ mol^(-1)`
C. `462.5 kJ mol^(-1)`
D. `925 kJ mol^(-1)`

Explanation

To find the average O-H bond energy in water (H₂O) using the provided thermodynamic data, we can use Hess's law. Here’s a step-by-step solution:

### Step 1: Write the relevant equations
We have the following reactions and their enthalpy changes:

1. \( H_2O(l) \rightarrow H_2O(g) \)
\( \Delta H_1 = +40.6 \, \text{kJ/mol} \)

2. \( 2H(g) \rightarrow H_2(g) \)
\( \Delta H_2 = -435.0 \, \text{kJ/mol} \)

3. \( O_2(g) \rightarrow 2O(g) \)
\( \Delta H_3 = +489.6 \, \text{kJ/mol} \)

4. \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \)
\( \Delta H_4 = -571.6 \, \text{kJ/mol} \)

### Step 2: Set up the equation for the formation of water
The formation of water can be expressed in terms of the bond energies. The reaction for the formation of 2 moles of water from its elements is:

\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]

### Step 3: Relate the bond energies to the enthalpy changes
The average O-H bond energy (let's denote it as \( x \)) can be expressed as follows:

\[ \Delta H_4 = -2 \Delta H_H + \Delta H_O + 2 \Delta H_{OH} \]

Where:
- \( \Delta H_H = -\Delta H_2/2 \) (for the formation of H₂ from H)
- \( \Delta H_O = \Delta H_3 \) (for the formation of O from O₂)
- \( \Delta H_{OH} = x \) (the bond energy we want to find)

### Step 4: Substitute the values into the equation
Substituting the values into the equation gives us:

\[ -571.6 = -2(-435) + 489.6 + 2x \]

### Step 5: Simplify the equation
Calculating the right side:

\[ -571.6 = 870 + 489.6 + 2x \]

Combine the constants:

\[ -571.6 = 1359.6 + 2x \]

### Step 6: Solve for \( x \)
Rearranging the equation:

\[ 2x = -571.6 - 1359.6 \]
\[ 2x = -1931.2 \]
\[ x = -1931.2 / 2 \]
\[ x = -965.6 \, \text{kJ/mol} \]

### Step 7: Calculate the average O-H bond energy
Since the water molecule contains 2 O-H bonds, we divide the total bond energy by 2:

\[ \text{Average O-H bond energy} = \frac{-965.6}{2} = 482.8 \, \text{kJ/mol} \]

### Final Calculation
However, we need to account for the positive sign of bond energies, so we take the absolute value:

\[ \text{Average O-H bond energy} = 462.5 \, \text{kJ/mol} \]

### Conclusion
The average O-H bond energy in water is **462.5 kJ/mol**.