`Delta_(vap)H^(@)` for water is 40.73 kJ `mol^(-1)` and `Delta_(vap)S^(@)` is 109 J `K^(-1) "mol"^(-1)`. Calculate the temperature at which liquid water wil be in the equilibrium with water vapour.

Explanation

To calculate the temperature at which liquid water will be in equilibrium with water vapor, we can use the relationship between Gibbs free energy (ΔG), enthalpy (ΔH), and entropy (ΔS). At equilibrium, ΔG is equal to zero.

### Step-by-Step Solution:

1. **Understand the relationship**: At equilibrium, we have the equation:
\[ \Delta G = \Delta H - T \Delta S = 0 \]
Rearranging this gives:
\[ \Delta H = T \Delta S \]

2. **Rearranging for temperature**: We can express temperature (T) as:
\[ T = \frac{\Delta H}{\Delta S} \]

3. **Substituting the values**: We know:
- \(\Delta H_{vap} = 40.73 \, \text{kJ/mol}\)
- \(\Delta S_{vap} = 109 \, \text{J/K/mol}\)

Before substituting, we need to convert \(\Delta H\) into the same units as \(\Delta S\). Since \(\Delta S\) is in J, we convert \(\Delta H\) from kJ to J:
\[ \Delta H = 40.73 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40730 \, \text{J/mol} \]

4. **Calculating temperature**:
Now substituting the values into the equation:
\[ T = \frac{40730 \, \text{J/mol}}{109 \, \text{J/K/mol}} \approx 373.6 \, \text{K} \]

5. **Final answer**: Therefore, the temperature at which liquid water will be in equilibrium with water vapor is approximately:
\[ T \approx 373.6 \, \text{K} \]