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

Answer

To find the entropy change when one mole of ice is converted into water at \(0^\circ C\), we can use the formula for entropy change (\(\Delta S\)) in a reversible process:

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

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

### Step 1: Convert the enthalpy change from kilojoules to joules.
Given that the enthalpy change for the conversion of ice to liquid water is \(6.0 \, \text{kJ/mol}\), we need 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 given is \(0^\circ C\). To convert this to Kelvin:

\[ T = 0 + 273 = 273 \, \text{K} \]

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

\[ \Delta S = \frac{\Delta H}{T} = \frac{6000 \, \text{J/mol}}{273 \, \text{K}} \]

Calculating this gives:

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

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

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

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