The value for `DeltaH_(vap) and Delat S_(vap)` for ethanol are respectively `38.594 kJ mol^(-1)` and `109.8 JK^(-1)`. The boiling point of ethanol will be

Options

A. `78.5 K`
B. `431.5 K`
C. `351.5 K`
D. `331.5 K`

Answer

To find the boiling point of ethanol using the given values for the enthalpy of vaporization (ΔH_vap) and the entropy of vaporization (ΔS_vap), we can use the Gibbs free energy equation. At the boiling point, the system is in equilibrium, and thus the change in Gibbs free energy (ΔG) is equal to zero.

### Step-by-Step Solution:

1. **Convert ΔH_vap to Joules:**
Given:
\[ \Delta H_{vap} = 38.594 \, \text{kJ mol}^{-1} \]
To convert to Joules:
\[ \Delta H_{vap} = 38.594 \times 10^3 \, \text{J mol}^{-1} = 38594 \, \text{J mol}^{-1} \]

**Hint:** Remember that 1 kJ = 1000 J.

2. **Identify ΔS_vap:**
Given:
\[ \Delta S_{vap} = 109.8 \, \text{J K}^{-1} \, \text{mol}^{-1} \]

**Hint:** Ensure the units for ΔS are consistent with the units for ΔH.

3. **Set up the Gibbs free energy equation:**
At the boiling point, we have:
\[ \Delta G = \Delta H - T \Delta S \]
Since at boiling point, ΔG = 0:
\[ 0 = \Delta H - T \Delta S \]

4. **Rearranging the equation to solve for T:**
\[ T = \frac{\Delta H}{\Delta S} \]

5. **Substituting the values:**
\[ T = \frac{38594 \, \text{J mol}^{-1}}{109.8 \, \text{J K}^{-1} \, \text{mol}^{-1}} \]

6. **Calculating T:**
\[ T = \frac{38594}{109.8} \approx 351.493 \, \text{K} \]

7. **Rounding the result:**
\[ T \approx 351.5 \, \text{K} \]

### Final Answer:
The boiling point of ethanol is approximately **351.5 K**.

---