The enthalpy of vaporization of liquid is 40 kJ `mol^(−1)` and entropy of vaporization is 64 J `mol^(−1)k^(−1)`. The boiling point of the liquid is

Options

A. 625 K
B. 254 K
C. 456 K
D. 725 K

Answer

To find the boiling point of the liquid using the given enthalpy of vaporization and entropy of vaporization, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Enthalpy of vaporization (ΔH) = 40 kJ/mol
- Entropy of vaporization (ΔS) = 64 J/mol·K

2. **Convert Enthalpy from kJ to J**:
- Since ΔH is given in kJ, we need to convert it to joules for consistency with ΔS.
- ΔH = 40 kJ/mol × 1000 J/kJ = 40,000 J/mol

3. **Use the Formula for Boiling Point**:
- At the boiling point, the Gibbs free energy change (ΔG) is zero. Therefore, we can use the equation:
\[ ΔG = ΔH - TΔS = 0 \]
- Rearranging this gives:
\[ T = \frac{ΔH}{ΔS} \]

4. **Substitute the Values into the Equation**:
- Now, substitute the values of ΔH and ΔS into the equation:
\[ T = \frac{40,000 \text{ J/mol}}{64 \text{ J/mol·K}} \]

5. **Calculate the Temperature**:
- Performing the division:
\[ T = \frac{40,000}{64} = 625 \text{ K} \]

6. **Conclusion**:
- The boiling point of the liquid is **625 K**.