`H_(2)(g) + 1/2O_(2)(g) rarr H_(2)O (l) , BE (H - H) = x_(1) , BE (O = O) = x_(2) , BE(O-H) = x_(3)`
Latent heat of vaporization of liquid water into water vapour `= x_4` then `Delta H_(f)` (heat of formation of liquid water) is:

Options

A. `x_(1) + (x_2)/(2) - x_(3) + x_(4)`
B. `2x_(3) - x_(1) - (x_2)/(2) - x_(4)`
C. `x_(1) + (x_2)/(2) - 2x_(3) - x_(4)`
D. `x_(1) + (x_2)/(2) - 2x_(3) + x_(4)`

Explanation

To solve the problem of finding the heat of formation (ΔH_f) of liquid water (H₂O(l)) from the reaction:

\[ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \]

we will use the bond energies provided and the latent heat of vaporization of water.

### Step-by-Step Solution:

1. **Identify the Reaction**:
We start with the reaction of hydrogen gas and oxygen gas forming liquid water. The relevant bond energies are:
- Bond Energy of H-H (BE(H-H)) = \( x_1 \)
- Bond Energy of O=O (BE(O=O)) = \( x_2 \)
- Bond Energy of O-H (BE(O-H)) = \( x_3 \)
- Latent heat of vaporization of water (H₂O(l) to H₂O(g)) = \( x_4 \)

2. **Calculate ΔH for the Formation of Water Vapor**:
The formation of water vapor from its elements can be expressed as:
\[ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g) \]

To calculate the enthalpy change (ΔH₁) for this reaction, we consider the bond energies:
- Energy required to break 1 mole of H-H bonds: \( +x_1 \)
- Energy required to break half a mole of O=O bonds: \( +\frac{x_2}{2} \)
- Energy released when forming 2 O-H bonds in H₂O(g): \( -2x_3 \)

Therefore, the total enthalpy change (ΔH₁) for the formation of water vapor is:
\[ \Delta H_1 = x_1 + \frac{x_2}{2} - 2x_3 \]

3. **Calculate ΔH for the Vaporization of Water**:
The enthalpy change (ΔH₂) for the vaporization of liquid water to water vapor is simply the latent heat of vaporization:
\[ \Delta H_2 = +x_4 \]

4. **Combine the Reactions**:
To find the heat of formation of liquid water (ΔH_f), we need to subtract the vaporization enthalpy from the formation of water vapor:
\[ \Delta H_f = \Delta H_1 - \Delta H_2 \]
Substituting the values we calculated:
\[ \Delta H_f = \left( x_1 + \frac{x_2}{2} - 2x_3 \right) - x_4 \]
Simplifying gives:
\[ \Delta H_f = x_1 + \frac{x_2}{2} - 2x_3 - x_4 \]

5. **Final Expression**:
The final expression for the heat of formation of liquid water is:
\[ \Delta H_f = x_1 + \frac{x_2}{2} - 2x_3 - x_4 \]