Enthalpy of vaporization of benzene is +353 kJ `mol^(-1)` at its boiling point, `80^(@)C`. The entropy change in the transition of the vapour to liquid at its boiling point [in `JK^(-1) mol^(-1)`] is .....

Options

A. `-441`
B. `-100`
C. `+441`
D. `+100`

Explanation

To find the entropy change in the transition of vapor to liquid at the boiling point of benzene, we can use the relationship between enthalpy change, temperature, and entropy change. Here’s the step-by-step solution:

### Step 1: Understand the given data
- Enthalpy of vaporization of benzene (ΔH_vap) = +353 kJ/mol
- Boiling point of benzene = 80°C

### Step 2: Convert the enthalpy of vaporization to joules
Since the enthalpy is given in kJ, we need to convert it to joules for consistency in units:
\[ \Delta H_{vap} = 353 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 353000 \, \text{J/mol} \]

### Step 3: Convert the boiling point to Kelvin
To use the temperature in calculations, we convert the boiling point from Celsius to Kelvin:
\[ T = 80°C + 273 = 353 \, \text{K} \]

### Step 4: Use the formula for entropy change
The relationship between enthalpy change (ΔH), temperature (T), and entropy change (ΔS) is given by:
\[ \Delta G = \Delta H - T \Delta S \]
At equilibrium (phase change), the Gibbs free energy change (ΔG) is zero:
\[ 0 = \Delta H - T \Delta S \]
Rearranging gives:
\[ \Delta S = \frac{\Delta H}{T} \]

### Step 5: Substitute the values into the formula
Since we are looking for the transition from vapor to liquid, we will take ΔH as negative:
\[ \Delta S = \frac{-353000 \, \text{J/mol}}{353 \, \text{K}} \]

### Step 6: Calculate the entropy change
Now, we perform the calculation:
\[ \Delta S = \frac{-353000}{353} \approx -1000 \, \text{J/K·mol} \]

### Conclusion
The entropy change in the transition of the vapor to liquid at its boiling point is approximately:
\[ \Delta S \approx -1000 \, \text{J/K·mol} \]