A liquid is in equilibrium with its vapour at a certain temperature. If `DeltaH_(vap) = 60.24 kJ mol^-1 and DeltaS_(vap) = 150.6 J mol^-1` then the boiling point of the liquid will be

Options

A. 275 K
B. 375 K
C. 350 K
D. 400 K

Answer

To find the boiling point of the liquid given the enthalpy of vaporization (ΔH_vap) and the entropy of vaporization (ΔS_vap), we can use the following relationship derived from the Gibbs free energy equation:

\[ \Delta G = \Delta H - T\Delta S \]

At the boiling point, the liquid is in equilibrium with its vapor, which means that the Gibbs free energy change (ΔG) is zero. Therefore, we can set up the equation as follows:

\[ 0 = \Delta H_{vap} - T_b \Delta S_{vap} \]

Rearranging this equation gives us:

\[ T_b = \frac{\Delta H_{vap}}{\Delta S_{vap}} \]

### Step 1: Convert ΔH_vap to the same units as ΔS_vap
Given:
- ΔH_vap = 60.24 kJ/mol
- ΔS_vap = 150.6 J/mol·K

We need to convert ΔH_vap from kJ to J:

\[ \Delta H_{vap} = 60.24 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 60240 \, \text{J/mol} \]

### Step 2: Substitute the values into the equation
Now we can substitute the values into the equation for T_b:

\[ T_b = \frac{60240 \, \text{J/mol}}{150.6 \, \text{J/mol·K}} \]

### Step 3: Calculate T_b
Now perform the calculation:

\[ T_b = \frac{60240}{150.6} \approx 400 \, \text{K} \]

### Conclusion
The boiling point of the liquid is approximately 400 K.