What is the melting point of benzene if `DeltaH_("fusion")= 9.95kJ//mol` and `DeltaS_("fusion")= 35.7J//K-"mol"`

Options

A. `278.7^(@)C`
B. `278.7^(@)K`
C. 300 K
D. 298 K

Explanation

To find the melting point of benzene using the given values for the enthalpy of fusion (ΔH_fusion) and the entropy of fusion (ΔS_fusion), we can use the relationship between these thermodynamic quantities at the melting point.

### Step-by-Step Solution:

1. **Understand the Relationship**: At the melting point (T_m), the change in Gibbs free energy (ΔG) for the phase transition (solid to liquid) is zero. Therefore, we can use the equation:
\[ \Delta G = \Delta H - T_m \Delta S = 0 \]
Rearranging this gives us:
\[ T_m = \frac{\Delta H}{\Delta S} \]

2. **Convert ΔH_fusion to Joules**: The value of ΔH_fusion is given as 9.95 kJ/mol. We need to convert this to Joules:
\[ \Delta H_fusion = 9.95 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 9950 \, \text{J/mol} \]

3. **Use the Given ΔS_fusion**: The value of ΔS_fusion is already provided in Joules:
\[ \Delta S_fusion = 35.7 \, \text{J/K·mol} \]

4. **Calculate the Melting Point (T_m)**: Now we can substitute the values into the equation for T_m:
\[ T_m = \frac{9950 \, \text{J/mol}}{35.7 \, \text{J/K·mol}} \]

5. **Perform the Calculation**:
\[ T_m = \frac{9950}{35.7} \approx 278.7 \, \text{K} \]

6. **Final Result**: The melting point of benzene is approximately **278.7 K**.