At what temperature liquid water will be in equilibrium with water vapour?
`DeltaH_("vap")="40.73 kJ mol"^(-1),DeltaS_("vap")="0.109 kJ K"^(-1)"mol"^(-1)`

Options

A. 282.4 K
B. 373.6 K
C. 100 K
D. 400 K

Explanation

To determine the temperature at which liquid water is in equilibrium with water vapor, we can use the relationship between enthalpy of vaporization (ΔH_vap) and entropy of vaporization (ΔS_vap). The equation that relates these quantities is:

\[ \Delta S_{vap} = \frac{\Delta H_{vap}}{T} \]

Where:
- ΔS_vap is the entropy of vaporization,
- ΔH_vap is the enthalpy of vaporization,
- T is the temperature in Kelvin.

### Step-by-step Solution:

1. **Identify the given values**:
- ΔH_vap = 40.73 kJ/mol
- ΔS_vap = 0.109 kJ/K·mol

2. **Rearrange the equation to solve for temperature (T)**:
\[ T = \frac{\Delta H_{vap}}{\Delta S_{vap}} \]

3. **Substitute the given values into the equation**:
\[ T = \frac{40.73 \text{ kJ/mol}}{0.109 \text{ kJ/K·mol}} \]

4. **Perform the calculation**:
\[ T = \frac{40.73}{0.109} \approx 373.6 \text{ K} \]

5. **Conclusion**:
The temperature at which liquid water is in equilibrium with water vapor is approximately **373.6 K**.