Calculate the entropy change for the conversion of2 moles of liquid water at 373K to vapours , if `Delta_(vap)` H is `37.3kJ mol^(-1)`.

Answer

To calculate the entropy change (ΔS) for the conversion of 2 moles of liquid water at 373 K to vapor, we can use the formula:

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

where:
- ΔH_vap is the enthalpy of vaporization,
- T is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the given values**:
- ΔH_vap = 37.3 kJ/mol
- T = 373 K
- Number of moles = 2

2. **Convert ΔH_vap to J/mol**:
Since 1 kJ = 1000 J, we convert ΔH_vap:
\[ \Delta H_{vap} = 37.3 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 37300 \, \text{J/mol} \]

3. **Calculate the entropy change for 1 mole**:
Using the formula for entropy change:
\[ \Delta S_{1 \, \text{mole}} = \frac{\Delta H_{vap}}{T} = \frac{37300 \, \text{J/mol}}{373 \, \text{K}} = 100 \, \text{J/(K·mol)} \]

4. **Calculate the total entropy change for 2 moles**:
Since we have 2 moles of water:
\[ \Delta S_{2 \, \text{moles}} = 2 \times \Delta S_{1 \, \text{mole}} = 2 \times 100 \, \text{J/(K·mol)} = 200 \, \text{J/K} \]

### Final Answer:
The entropy change for the conversion of 2 moles of liquid water at 373 K to vapor is **200 J/K**.

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