`H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l) , Delta H` at `298 K = - 285.8 kJ`
The molar enthalpy of vapourization of water at 1 atm and `25^(@)C` is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at is

Options

A. `-241.8 kJ`
B. 241.8 kJ
C. 329.8 kJ
D. `-329.8 kJ`

Answer

To find the standard enthalpy of formation of 1 mole of water vapor, we can use the information provided about the enthalpy of formation of liquid water and the enthalpy of vaporization of water.

### Step-by-Step Solution:

1. **Write the reaction for the formation of water vapor:**
The formation of water vapor from its elements can be represented as:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \]
We will denote the standard enthalpy change for this reaction as \( \Delta H_f^\circ \) for water vapor.

2. **Write the reaction for the vaporization of water:**
The vaporization of liquid water can be represented as:
\[ H_2O(l) \rightarrow H_2O(g) \]
The enthalpy change for this process is given as the molar enthalpy of vaporization, which is \( 44 \, \text{kJ} \).

3. **Use the given enthalpy of formation of liquid water:**
The enthalpy change for the formation of liquid water from its elements is given as:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta H = -285.8 \, \text{kJ} \]

4. **Combine the reactions:**
To find the enthalpy of formation of water vapor, we can combine the two reactions:
- First, we write the formation of liquid water:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta H = -285.8 \, \text{kJ} \]
- Then, we add the vaporization reaction:
\[ H_2O(l) \rightarrow H_2O(g) \quad \Delta H = +44 \, \text{kJ} \]

5. **Calculate the overall enthalpy change:**
By adding the enthalpy changes of the two reactions:
\[ \Delta H_f^\circ (H_2O(g)) = \Delta H_f^\circ (H_2O(l)) + \Delta H_{vap} \]
\[ \Delta H_f^\circ (H_2O(g)) = -285.8 \, \text{kJ} + 44 \, \text{kJ} \]
\[ \Delta H_f^\circ (H_2O(g)) = -241.8 \, \text{kJ} \]

### Final Answer:
The standard enthalpy of formation of 1 mole of water vapor is:
\[ \Delta H_f^\circ (H_2O(g)) = -241.8 \, \text{kJ} \]