The heat of formation of `H_(2)O(l)` is `-68.0` kcal, the heat of formation of `H_(2)O(g)` can logically be

Options

A. `-68.0` kcal
B. `-67.4` kcal
C. 80.0 kcal
D. `-58.3` kcal

Answer

To determine the heat of formation of \( H_2O(g) \) based on the given heat of formation of \( H_2O(l) \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- The heat of formation of \( H_2O(l) \) is given as \( -68.0 \) kcal.

2. **Recognize the Relationship**:
- The heat of formation of \( H_2O(g) \) can be derived from the heat of formation of \( H_2O(l) \) by considering the phase change from liquid to gas. This process is endothermic, meaning it requires heat.

3. **Set Up the Equation**:
- The heat of formation of \( H_2O(g) \) can be expressed as:
\[ \Delta H_f(H_2O(g)) = \Delta H_f(H_2O(l)) + \Delta H_{vap} \]
- Here, \( \Delta H_{vap} \) is the heat of vaporization, which is the amount of heat required to convert \( H_2O(l) \) to \( H_2O(g) \).

4. **Estimate the Heat of Vaporization**:
- The heat of vaporization for water is approximately \( 9.7 \) kcal (this value can vary slightly with temperature).

5. **Calculate the Heat of Formation of \( H_2O(g) \)**:
- Substitute the known values into the equation:
\[ \Delta H_f(H_2O(g)) = -68.0 \, \text{kcal} + 9.7 \, \text{kcal} \]
- Performing the calculation:
\[ \Delta H_f(H_2O(g)) = -68.0 + 9.7 = -58.3 \, \text{kcal} \]

6. **Conclusion**:
- The heat of formation of \( H_2O(g) \) is approximately \( -58.3 \) kcal.

### Final Answer:
The heat of formation of \( H_2O(g) \) can logically be estimated to be approximately \( -58.3 \) kcal.