The enthalpy change for the transition of liquid water to steam, `Delta_(vap) " H=37.3 kJ " mol^-1` at 373 K. The entropy change for the process is __________ .

Options

A. `119.9 J mol^(-1) K^(-1)`
B. `37.3 J mol^(-1) K^(-1)`
C. `100 J mol^(-1) K^(-1)`
D. `74.6 J mol^(-1) K^(-1)`

Answer

To find the entropy change (\( \Delta S \)) for the transition of liquid water to steam, we can use the relationship between enthalpy change and entropy change at a constant temperature. The formula is:

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

Where:
- \( \Delta S \) = Entropy change
- \( \Delta H \) = Enthalpy change
- \( T \) = Temperature in Kelvin

### Step 1: Convert Enthalpy Change to Joules
The enthalpy change given is \( \Delta_{vap} H = 37.3 \, \text{kJ/mol} \). We need to convert this value to joules because the standard unit for entropy is Joules per mole per Kelvin.

\[ \Delta H = 37.3 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 37300 \, \text{J/mol} \]

### Step 2: Use the Temperature
The temperature is given as \( T = 373 \, \text{K} \).

### Step 3: Calculate the Entropy Change
Now we can substitute the values into the entropy change formula:

\[ \Delta S = \frac{\Delta H}{T} = \frac{37300 \, \text{J/mol}}{373 \, \text{K}} \]

### Step 4: Perform the Calculation
Now, we perform the division:

\[ \Delta S = 100 \, \text{J/mol K} \]

### Final Answer
The entropy change for the transition of liquid water to steam is:

\[ \Delta S = 100 \, \text{J/mol K} \]