At `37 ^(@)C` latent heat of fusion of a compound is 3930 J/mol. Entropy change is

Options

A. `13.10 J/mol K`
B. `10.77 J/mol K`
C. `9.07 J/mol K`
D. `0.977 J/mol K`

Answer

To find the entropy change (ΔS) during the fusion of a compound at a given temperature, we can use the formula:

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

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

### Step-by-Step Solution:

1. **Convert the Temperature to Kelvin:**
The given temperature is \(37^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
So,
\[ T = 37 + 273.15 = 310.15 \, K \approx 310 \, K \]

2. **Identify the Latent Heat of Fusion:**
The latent heat of fusion given in the problem is:
\[ Q = 3930 \, J/mol \]

3. **Substitute the Values into the Entropy Change Formula:**
Now we can substitute the values of \(Q\) and \(T\) into the entropy change formula:
\[ \Delta S = \frac{Q}{T} = \frac{3930 \, J/mol}{310 \, K} \]

4. **Calculate the Entropy Change:**
Performing the division:
\[ \Delta S = \frac{3930}{310} \approx 12.68 \, J/mol \cdot K \]

5. **Final Result:**
Therefore, the entropy change during the fusion of the compound at \(37^\circ C\) is approximately:
\[ \Delta S \approx 12.68 \, J/mol \cdot K \]