What is the entropy change `(in JK^(-1) mol^(-1))` when one mole of ice is converted into water at `0^(@)C`? (The enthalpy change for the conversion of ice to liquid water is `6.0 kJ mol^(-1) at 0^(@)C`)

Options

A. `20.13`
B. `2.013`
C. `2.198`
D. `21.98`

Answer

To calculate the entropy change when one mole of ice is converted into water at \(0^\circ C\), we can use the formula:

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

where:
- \(\Delta S\) is the change in entropy,
- \(\Delta H\) is the change in enthalpy,
- \(T\) is the temperature in Kelvin.

### Step 1: Convert the enthalpy change from kJ to J
The enthalpy change for the conversion of ice to liquid water is given as \(6.0 \, \text{kJ mol}^{-1}\). We need to convert this to joules:

\[ \Delta H = 6.0 \, \text{kJ mol}^{-1} \times 1000 \, \text{J/kJ} = 6000 \, \text{J mol}^{-1} \]

### Step 2: Convert the temperature from Celsius to Kelvin
The temperature is given as \(0^\circ C\). To convert this to Kelvin, we use the formula:

\[ T = 0 + 273.15 = 273.15 \, \text{K} \]

### Step 3: Calculate the change in entropy
Now we can substitute the values of \(\Delta H\) and \(T\) into the entropy change formula:

\[ \Delta S = \frac{\Delta H}{T} = \frac{6000 \, \text{J mol}^{-1}}{273.15 \, \text{K}} \]

Calculating this gives:

\[ \Delta S \approx \frac{6000}{273.15} \approx 21.98 \, \text{J K}^{-1} \text{ mol}^{-1} \]

### Final Answer
The entropy change when one mole of ice is converted into water at \(0^\circ C\) is approximately:

\[ \Delta S \approx 21.98 \, \text{J K}^{-1} \text{ mol}^{-1} \]