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

Options

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

Answer

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 follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between ΔG, ΔH, and ΔS**:
When a substance melts, the process occurs at a constant temperature and pressure. At this point, the change in Gibbs free energy (ΔG) is zero. The relationship is given by the equation:
\[ ΔG = ΔH - TΔS \]
Since ΔG = 0 at the melting point, we can rearrange this equation to:
\[ ΔH = TΔS \]

2. **Rearranging the equation to find temperature (T)**:
From the equation \(ΔH = TΔS\), we can solve for T:
\[ T = \frac{ΔH}{ΔS} \]

3. **Substituting the values**:
We need to ensure that the units are consistent. The given values are:
- \(ΔH_{fusion} = 9.95 \, kJ/mol = 9.95 \times 10^3 \, J/mol\) (since 1 kJ = 1000 J)
- \(ΔS_{fusion} = 35.7 \, J/(K \cdot mol)\)

Now, substituting these values into the equation:
\[ T = \frac{9.95 \times 10^3 \, J/mol}{35.7 \, J/(K \cdot mol)} \]

4. **Calculating T**:
Performing the division:
\[ T = \frac{9.95 \times 10^3}{35.7} \approx 278.7 \, K \]

5. **Conclusion**:
Therefore, the melting point of benzene is approximately:
\[ T \approx 278.7 \, K \]

### Final Answer:
The melting point of benzene is **278.7 K**.

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