At `27^(@)C` latent heat of fusion of a compount is 2930 j/mol. Entropy change is:

Options

A. 9.77 J/mol-K
B. 10.77 J/mol-K
C. 9.07 J/mol-K
D. 0.977 J/mol-K

Answer

To calculate the entropy change (ΔS) during the phase transition of a compound at a given temperature, we can use the formula derived from the second law of thermodynamics:

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

Where:
- ΔS = change in entropy
- ΔQ = heat absorbed or released during the phase transition (latent heat of fusion in this case)
- T = absolute temperature in Kelvin

### Step-by-Step Solution:

1. **Identify the given values**:
- Latent heat of fusion (ΔQ) = 2930 J/mol
- Temperature (T) = 27°C

2. **Convert the temperature to Kelvin**:
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- Therefore:
\[ T = 27 + 273 = 300 \, K \]

3. **Substitute the values into the entropy change formula**:
- Now that we have ΔQ and T, we can substitute these values into the formula:
\[ \Delta S = \frac{2930 \, J/mol}{300 \, K} \]

4. **Calculate ΔS**:
- Performing the division:
\[ \Delta S = \frac{2930}{300} \approx 9.77 \, J/mol \cdot K \]

5. **Final Result**:
- The entropy change (ΔS) during the fusion process is approximately:
\[ \Delta S \approx 9.77 \, J/mol \cdot K \]