At `127^oC`, heat of fusion of a compound is 4.8kJ/mol. Entropy change for the process is

Options

A. `37.8 JK^-1mol^-1`
B. `15 JK^-1mol^-1`
C. `18.6 JK^-1mol^-1`
D. `12 JK^-1mol^-1`

Answer

To find the entropy change (ΔS) for the process, we can use the formula:

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

where:
- \( Q \) is the heat absorbed or released during the process,
- \( T \) is the absolute temperature in Kelvin.

### Step 1: Convert the heat of fusion from kJ to J
The heat of fusion is given as 4.8 kJ/mol. We need to convert this to joules:

\[ Q = 4.8 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 4800 \, \text{J/mol} \]

### Step 2: Convert the temperature from Celsius to Kelvin
The temperature is given as 127°C. To convert this to Kelvin, we use the formula:

\[ T = 127 + 273 = 400 \, \text{K} \]

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

\[ \Delta S = \frac{Q}{T} = \frac{4800 \, \text{J/mol}}{400 \, \text{K}} = 12 \, \text{J/K/mol} \]

### Final Answer
The entropy change for the process is:

\[ \Delta S = 12 \, \text{J/K/mol} \]

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