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

Options

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

Explanation

To find the standard enthalpy of formation of 1 mole of water vapor at 25°C, we can use the given information about the reaction and the enthalpy changes involved.

### Step-by-Step Solution:

1. **Write the reaction for the formation of liquid water:**
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \]
The enthalpy change for this reaction at 298 K is given as:
\[ \Delta H = -285.8 \, \text{kJ} \]

2. **Write the reaction for the vaporization of liquid water:**
\[ H_2O(l) \rightarrow H_2O(g) \]
The molar enthalpy of vaporization of water at 25°C is given as:
\[ \Delta H_{vap} = +44 \, \text{kJ} \]

3. **Combine the two reactions to find the enthalpy of formation of water vapor:**
To find the standard enthalpy of formation of water vapor, we need to add the enthalpy change for the formation of liquid water and the enthalpy change for the vaporization of water:
\[ \Delta H_{f, H_2O(g)} = \Delta H_{f, H_2O(l)} + \Delta H_{vap} \]
Substituting the values:
\[ \Delta H_{f, H_2O(g)} = -285.8 \, \text{kJ} + 44 \, \text{kJ} \]

4. **Calculate the final value:**
\[ \Delta H_{f, H_2O(g)} = -285.8 + 44 = -241.8 \, \text{kJ} \]

Thus, the standard enthalpy of formation of 1 mole of water vapor at 25°C is:
\[ \Delta H_{f, H_2O(g)} = -241.8 \, \text{kJ} \]

### Final Answer:
The standard enthalpy of formation of 1 mole of water vapor at 25°C is **-241.8 kJ**.