At `0^(@)C,DeltaH_(fus)^(@)=6kJ //mol`, change of entropy for freezing of one mole of ice will be `:`

Options

A. `oo`
B. 0
C. `21.98J//mol-K`
D. `13.6J//mol-K`

Explanation

To find the change in entropy for the freezing of one mole of ice at 0°C, we can use the relationship between enthalpy change and entropy change. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
We are provided with:
- The enthalpy change of fusion (ΔH_fus) = 6 kJ/mol
- The temperature at which this change occurs = 0°C

### Step 2: Convert Units
Since we need to calculate the change in entropy in Joules, we should convert the enthalpy change from kilojoules to joules:
\[ \Delta H_{fus} = 6 \text{ kJ/mol} = 6000 \text{ J/mol} \]

### Step 3: Convert Temperature to Kelvin
The temperature must be in Kelvin for the calculation:
\[ T = 0°C + 273.15 = 273.15 \text{ K} \]

### Step 4: Use the Formula for Change in Entropy
The change in entropy (ΔS) during the freezing process can be calculated using the formula:
\[ \Delta S = \frac{\Delta H_{fus}}{T} \]
Substituting the values we have:
\[ \Delta S = \frac{6000 \text{ J/mol}}{273.15 \text{ K}} \]

### Step 5: Calculate the Change in Entropy
Now, performing the calculation:
\[ \Delta S \approx \frac{6000}{273.15} \approx 21.95 \text{ J/mol K} \]

### Step 6: Conclusion
The change in entropy for the freezing of one mole of ice at 0°C is approximately:
\[ \Delta S \approx 21.95 \text{ J/mol K} \]

### Final Answer
The change of entropy for freezing of one mole of ice will be approximately **21.95 J/mol K**.

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