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

Explanation

To find the entropy change for the transition of liquid water to steam, we can use the following steps:

### Step 1: Identify the given data
- Enthalpy change (ΔH) = 30 kJ/mol
- Temperature (T) = 27°C

### Step 2: Convert the enthalpy change from kJ to J
Since we need the entropy change in J/(mol·K), we convert kJ to J:
\[ \Delta H = 30 \text{ kJ/mol} = 30 \times 10^3 \text{ J/mol} = 30000 \text{ J/mol} \]

### Step 3: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we add 273.15:
\[ T = 27°C + 273.15 = 300.15 \text{ K} \approx 300 \text{ K} \]

### Step 4: Use the formula for entropy change
The formula for calculating the change in entropy (ΔS) during a phase transition is:
\[ \Delta S = \frac{\Delta H}{T} \]

### Step 5: Substitute the values into the formula
Now we substitute the values of ΔH and T into the formula:
\[ \Delta S = \frac{30000 \text{ J/mol}}{300 \text{ K}} = 100 \text{ J/(mol·K)} \]

### Step 6: Conclusion
The entropy change for the transition of liquid water to steam at 27°C is:
\[ \Delta S = 100 \text{ J/(mol·K)} \]

### Final Answer:
The entropy change for the process is **100 J/(mol·K)**.

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