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 follow these steps:

### Step 1: Identify the given data
- Enthalpy change (\(\Delta H\)) for the conversion of ice to liquid water = \(6.0 \, \text{kJ mol}^{-1}\)
- Temperature (\(T\)) = \(0^\circ C\)

### Step 2: Convert temperature to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
Thus,
\[ T = 0 + 273.15 = 273.15 \, K \]

### Step 3: Convert enthalpy change from kJ to J
Since we need the entropy change in \(J \, K^{-1} \, mol^{-1}\), we convert \(\Delta H\) from kJ to J:
\[ \Delta H = 6.0 \, \text{kJ mol}^{-1} \times 1000 \, \frac{J}{kJ} = 6000 \, J \, mol^{-1} \]

### Step 4: Use the formula for entropy change
The formula for the change in entropy (\(\Delta S\)) is given by:
\[ \Delta S = \frac{\Delta H}{T} \]
Substituting the values we have:
\[ \Delta S = \frac{6000 \, J \, mol^{-1}}{273.15 \, K} \]

### Step 5: Calculate the entropy change
Now we perform the calculation:
\[ \Delta S \approx \frac{6000}{273.15} \approx 21.96 \, J \, K^{-1} \, mol^{-1} \]

### Step 6: Round off the answer
Rounding off to two decimal places, we get:
\[ \Delta S \approx 21.98 \, J \, K^{-1} \, mol^{-1} \]

### Final Answer
The entropy change when one mole of ice is converted into water at \(0^\circ C\) is approximately:
\[ \Delta S \approx 21.98 \, J \, K^{-1} \, mol^{-1} \]

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