Chloroform has `DeltaH_("vaporization")=29.2 kJ//"mol"` and boils at `61.2^(@)C`. What is the value of `Delta``S`_`("vaporization")` for chloroform ?

Options

A. `87.3J//"mol"-K`
B. `477.1 J//"mol"-K`
C. `-87.3 J//"mol"-K`
D. `-477.1 J//"mol"-K`

Explanation

To find the value of ΔS_vaporization for chloroform, we can use the relationship between enthalpy of vaporization (ΔH_vaporization) and entropy of vaporization (ΔS_vaporization) at the boiling point. The formula is given by:

\[ \Delta S_{vaporization} = \frac{\Delta H_{vaporization}}{T_{boiling}} \]

Where:
- ΔH_vaporization is the enthalpy of vaporization in joules per mole.
- T_boiling is the boiling point in Kelvin.

### Step 1: Convert ΔH_vaporization to Joules
Given:
\[ \Delta H_{vaporization} = 29.2 \, \text{kJ/mol} \]

Convert kJ to J:
\[ \Delta H_{vaporization} = 29.2 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 29200 \, \text{J/mol} \]

### Step 2: Convert the boiling point from Celsius to Kelvin
Given:
\[ T_{boiling} = 61.2^\circ C \]

Convert to Kelvin:
\[ T_{boiling} = 61.2 + 273.15 = 334.35 \, \text{K} \]

### Step 3: Calculate ΔS_vaporization
Now, substitute the values into the formula:
\[ \Delta S_{vaporization} = \frac{29200 \, \text{J/mol}}{334.35 \, \text{K}} \]

Calculating this gives:
\[ \Delta S_{vaporization} \approx 87.3 \, \text{J/mol·K} \]

### Final Answer
The value of ΔS_vaporization for chloroform is approximately:
\[ \Delta S_{vaporization} \approx 87.3 \, \text{J/mol·K} \]

---