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 in equilibrium with its vapor, we can use the relationship between the change in enthalpy (ΔH), change in entropy (ΔS), and temperature (T) at equilibrium.

### Step-by-step Solution:

1. **Understand the Equilibrium Condition**:
At equilibrium, the Gibbs free energy change (ΔG) is zero. The relationship is given by:
\[ \Delta G = \Delta H - T \Delta S \]
Since ΔG = 0 at equilibrium, we can rearrange the equation:
\[ \Delta H = T \Delta S \]

2. **Rearranging for Temperature (T)**:
We can express the temperature (T) in terms of ΔH and ΔS:
\[ T = \frac{\Delta H}{\Delta S} \]

3. **Substituting the Given Values**:
We have:
- ΔH (enthalpy of vaporization) = 60.24 kJ/mol
- ΔS (entropy of vaporization) = 150.6 J/mol·K

First, we need to convert ΔH from kJ to J:
\[ \Delta H = 60.24 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 60240 \, \text{J/mol} \]

4. **Calculating Temperature (T)**:
Now substitute the values into the equation:
\[ T = \frac{60240 \, \text{J/mol}}{150.6 \, \text{J/mol·K}} \]
\[ T = 399.2 \, \text{K} \]

5. **Rounding the Result**:
Rounding 399.2 K gives us approximately:
\[ T \approx 400 \, \text{K} \]

6. **Conclusion**:
The boiling point of the liquid is approximately **400 K**.

### Final Answer:
The boiling point of the liquid is **400 K**.

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