`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)`

Answer

To calculate the heat of formation of liquid water (ΔH_f) from the given reaction and bond energies, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \]

2. **Write the Bond Energies**:
- Bond energy of H-H (x1)
- Bond energy of O=O (x2)
- Bond energy of O-H (x3)
- Latent heat of vaporization of water (x4)

3. **Calculate the Enthalpy Change for the Formation of Gaseous Water (H2O(g))**:
The enthalpy change (ΔH) for the formation of gaseous water can be calculated using the bond energies:
\[ ΔH_{f(g)} = \text{Bond energy of reactants} - \text{Bond energy of products} \]

For the reactants:
- 1 H-H bond: x1
- 1 O=O bond (half): \(\frac{x2}{2}\)

For the products (2 O-H bonds in H2O):
- 2 O-H bonds: \(2 \times x3 = 2x3\)

Therefore, we can write:
\[ ΔH_{f(g)} = \left(x1 + \frac{x2}{2}\right) - 2x3 \]

4. **Account for the Latent Heat of Vaporization**:
The formation of liquid water from gaseous water involves the latent heat of vaporization (x4):
\[ ΔH_{f(l)} = ΔH_{f(g)} - x4 \]

5. **Combine the Equations**:
Substituting the expression for ΔH_{f(g)} into the equation for ΔH_{f(l)}:
\[ ΔH_{f(l)} = \left(x1 + \frac{x2}{2} - 2x3\right) - x4 \]

6. **Final Expression**:
Therefore, the heat of formation of liquid water is:
\[ ΔH_{f(l)} = x1 + \frac{x2}{2} - 2x3 - x4 \]

### Final Answer:
\[ ΔH_{f(l)} = x1 + \frac{x2}{2} - 2x3 - x4 \]