Molar heat of vaporization of a liquid is 6 `KJ mol^(-1)`. If its entropy change is `16 JK^(-1)mol^(-1)`, then boiling point of the liquid is

Options

A. `375^(@)C`
B. 375 K
C. 273 K
D. 102 K

Answer

To find the boiling point of the liquid given the molar heat of vaporization and the entropy change, we can use the relationship between Gibbs free energy, enthalpy, and entropy.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Molar heat of vaporization (ΔH) = 6 kJ/mol = 6000 J/mol (since 1 kJ = 1000 J)
- Change in entropy (ΔS) = 16 J/K·mol

2. **Set Up the Gibbs Free Energy Equation:**
At equilibrium (boiling point), the Gibbs free energy change (ΔG) is zero. The equation is:
\[ \Delta G = \Delta H - T \Delta S \]
Setting ΔG to zero gives:
\[ 0 = \Delta H - T_b \Delta S \]

3. **Rearranging the Equation:**
Rearranging the equation to solve for the boiling point (T_b):
\[ T_b = \frac{\Delta H}{\Delta S} \]

4. **Substituting the Values:**
Substitute the values of ΔH and ΔS into the equation:
\[ T_b = \frac{6000 \text{ J/mol}}{16 \text{ J/K·mol}} \]

5. **Calculating the Boiling Point:**
Performing the division:
\[ T_b = \frac{6000}{16} = 375 \text{ K} \]

6. **Final Answer:**
The boiling point of the liquid is **375 K**.