Enthalpy of vapourization of benzene is `+35.3 kJ mol^(-1)` at its boiling point of `80^(@)C`. The entropy change in the transition of the vapour to liquid at its boiling points [in `K^(-1) mol^(-1)`] is

Options

A. `-100`
B. `+100`
C. `+342`
D. `-342`

Answer

To find the entropy change in the transition of benzene from vapor to liquid at its boiling point, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data**:
- Enthalpy of vaporization of benzene, ΔH_vap = +35.3 kJ/mol
- Boiling point of benzene = 80°C

2. **Convert Enthalpy of Vaporization to Joules**:
- Since 1 kJ = 1000 J, we convert ΔH_vap:
\[ ΔH_vap = 35.3 \, \text{kJ/mol} = 35.3 \times 1000 \, \text{J/mol} = 35300 \, \text{J/mol} \]

3. **Determine the Heat of Condensation**:
- The heat of condensation (ΔH_cond) is the negative of the enthalpy of vaporization:
\[ ΔH_{cond} = -ΔH_{vap} = -35300 \, \text{J/mol} \]

4. **Convert Boiling Point to Kelvin**:
- To convert the boiling point from Celsius to Kelvin:
\[ T(K) = 80°C + 273 = 353 \, K \]

5. **Calculate the Entropy Change (ΔS)**:
- The relationship between the entropy change and the heat of condensation at constant temperature is given by:
\[ ΔS = \frac{ΔH_{cond}}{T} \]
- Substituting the values:
\[ ΔS = \frac{-35300 \, \text{J/mol}}{353 \, K} \]
- Performing the calculation:
\[ ΔS = -100 \, \text{J/K/mol} \]

### Final Answer:
The entropy change in the transition of the vapor to liquid at its boiling point is:
\[ ΔS = -100 \, \text{J/K/mol} \]