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 `6.0KJmol^(-1)` at `0^(@)C`)

Options

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

Explanation

To find 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 \(6.0 \, \text{kJ/mol}\). To convert this to joules:

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

### Step 2: Convert the temperature from Celsius to Kelvin

The temperature is given as \(0^\circ C\). To convert this to Kelvin:

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

### Step 3: Calculate the entropy change

Now we can substitute the values of \(\Delta H\) and \(T\) into the entropy change formula:

\[ \Delta S = \frac{6000 \, \text{J/mol}}{273.15 \, \text{K}} \approx 21.96 \, \text{J/K/mol} \]

### Step 4: Round to appropriate significant figures

Rounding to two decimal places, we get:

\[ \Delta S \approx 21.98 \, \text{J/K/mol} \]

### Final Answer

The entropy change when one mole of ice is converted into water at \(0^\circ C\) is approximately \(21.98 \, \text{J/K/mol}\).

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