Enthalpy change `( Delta_(vap)H )` for the transition of liquid water to steam at `100^(@)C` is`40.8 kJ mol^(-1)`. Calculate the entropy change for the process.

Answer

To calculate the entropy change (ΔS) for the transition of liquid water to steam at 100°C, we can use the relationship between enthalpy change (ΔH) and entropy change (ΔS) at a constant temperature. The formula is:

\[ \Delta S = \frac{\Delta H}{T} \]

where:
- ΔS is the entropy change,
- ΔH is the enthalpy change,
- T is the absolute temperature in Kelvin.

### Step-by-step Solution:

1. **Convert the temperature from Celsius to Kelvin:**
\[ T = 100°C + 273.15 = 373.15 \, K \]

2. **Convert the enthalpy change from kJ/mol to J/mol:**
Given ΔH = 40.8 kJ/mol, we convert it to joules:
\[ \Delta H = 40.8 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40800 \, \text{J/mol} \]

3. **Substitute the values into the entropy change formula:**
\[ \Delta S = \frac{\Delta H}{T} = \frac{40800 \, \text{J/mol}}{373.15 \, K} \]

4. **Calculate ΔS:**
\[ \Delta S \approx \frac{40800}{373.15} \approx 109.3 \, \text{J/K/mol} \]

Thus, the entropy change for the process is approximately **109.3 J/K/mol**.