If the enthalpy change for the transition of liquid water to steam is `30 kJ mol^(-1)` at `27 ^(@)C`, the entropy change for the process would be

Options

A. `100 J mol^(-1) K^(-1)`
B. `10 J mol^(-1) K^(-1)`
C. `1.0 J mol^(-1) K^(-1)`
D. `0.1 J mol^(-1) K^(-1)`

Answer

To find the entropy change (ΔS) for the transition of liquid water to steam, we can use the relationship between enthalpy change (ΔH) and entropy change at a given temperature (T). The formula is given by:

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

### Step-by-Step Solution:

1. **Identify the given values**:
- Enthalpy change (ΔH) = 30 kJ/mol
- Temperature (T) = 27°C

2. **Convert the enthalpy change from kJ to J**:
\[ \Delta H = 30 \text{ kJ/mol} = 30 \times 10^3 \text{ J/mol} = 30,000 \text{ J/mol} \]

3. **Convert the temperature from Celsius to Kelvin**:
\[ T = 27°C + 273 = 300 \text{ K} \]

4. **Substitute the values into the entropy change formula**:
\[ \Delta S = \frac{\Delta H}{T} = \frac{30,000 \text{ J/mol}}{300 \text{ K}} \]

5. **Calculate the entropy change**:
\[ \Delta S = \frac{30,000}{300} = 100 \text{ J/K/mol} \]

### Final Answer:
The entropy change (ΔS) for the transition of liquid water to steam at 27°C is **100 J/K/mol**.