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 1: Understand the Concept of Heat of Formation
The heat of formation (\( \Delta H_f \)) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. For water, we have two states: liquid (\( H_2O(l) \)) and gas (\( H_2O(g) \)).

**Hint:** Remember that the heat of formation is always defined for the formation of one mole of a substance from its elements.

### Step 2: Analyze the Given Data
We know that the heat of formation of \( H_2O(l) \) is \( -68.0 \) kcal. This value indicates that the formation of liquid water from its elements (hydrogen and oxygen) releases \( 68.0 \) kcal of heat.

**Hint:** A negative value indicates an exothermic reaction, meaning heat is released.

### Step 3: Consider the Phase Change from Liquid to Gas
When \( H_2O(l) \) is converted to \( H_2O(g) \), heat is absorbed because energy is required to overcome intermolecular forces in the liquid state. This process is endothermic.

**Hint:** The phase change from liquid to gas requires energy input, which means the enthalpy change for this process will be positive.

### Step 4: Relate the Heats of Formation
Since converting \( H_2O(l) \) to \( H_2O(g) \) absorbs heat, we can say:
\[ \Delta H_f(H_2O(g)) = \Delta H_f(H_2O(l)) + \Delta H_{vap} \]
Where \( \Delta H_{vap} \) is the heat of vaporization of water. The heat of vaporization is a positive value.

**Hint:** The heat of vaporization is the amount of energy required to convert a unit mass of a liquid into vapor without a change in temperature.

### Step 5: Estimate the Value of \( \Delta H_f(H_2O(g)) \)
The heat of vaporization of water is approximately \( 540 \) kcal/mol. Therefore, we can estimate:
\[ \Delta H_f(H_2O(g)) = -68.0 \text{ kcal} + 540 \text{ kcal} = 472 \text{ kcal} \]
However, since we are looking for a logical estimate, we can consider that the heat of formation of \( H_2O(g) \) should be less negative than that of \( H_2O(l) \) but still negative.

**Hint:** Since the heat of formation of \( H_2O(g) \) must be less negative than \( -68.0 \) kcal, we can logically deduce that it could be around \( -80 \) kcal.

### Conclusion
Thus, the heat of formation of \( H_2O(g) \) can logically be estimated to be around \( -80.0 \) kcal.

**Final Answer:** The heat of formation of \( H_2O(g) \) can logically be \( -80.0 \) kcal.