The value for `DeltaH_(vap) and Delta S_(vap)` for ethanol are respectively `38.594 kJ mol^(-1)` and `109.8 JK^(-1) mol^-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 of enthalpy of vaporization (ΔH_vap) and entropy of vaporization (ΔS_vap), we can use the following relationship:

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

Where:
- \( \Delta S_{vap} \) is the change in entropy of vaporization,
- \( \Delta H_{vap} \) is the change in enthalpy of vaporization,
- \( T_b \) is the boiling point in Kelvin.

### Step-by-Step Solution:

1. **Identify the given values:**
- \( \Delta H_{vap} = 38.594 \, \text{kJ mol}^{-1} \)
- \( \Delta S_{vap} = 109.8 \, \text{J K}^{-1} \, \text{mol}^{-1} \)

2. **Convert ΔH_vap from kJ to J:**
Since ΔS_vap is given in J, we need to convert ΔH_vap from kJ to J:
\[ \Delta H_{vap} = 38.594 \, \text{kJ mol}^{-1} \times 1000 \, \text{J/kJ} = 38594 \, \text{J mol}^{-1} \]

3. **Rearrange the formula to solve for \( T_b \):**
\[ T_b = \frac{\Delta H_{vap}}{\Delta S_{vap}} \]

4. **Substitute the values into the equation:**
\[ T_b = \frac{38594 \, \text{J mol}^{-1}}{109.8 \, \text{J K}^{-1} \, \text{mol}^{-1}} \]

5. **Calculate \( T_b \):**
\[ T_b = \frac{38594}{109.8} \approx 351.50 \, \text{K} \]

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