The enthalpy of vaporization of a certain liquid at its boiling point of `35^(@)C` is `24.64 kJ "mol"^(-1)`.The value of change in entropy for the process is

Options

A. `704 J K^(-1) "mol"^(-1)`
B. `80 J K^(-1) "mol"^(-1)`
C. `24.64 J K^(-1) "mol"^(-1)`
D. `7.04 J K^(-1) "mol"^(-1)`

Explanation

To find the change in entropy (ΔS) for the vaporization process of the liquid at its boiling point, we can follow these steps:

### Step 1: Understand the relationship between ΔG, ΔH, and ΔS
At equilibrium (which is the case at the boiling point), the change in Gibbs free energy (ΔG) is zero. According to the Gibbs-Helmholtz equation:
\[ ΔG = ΔH - TΔS \]
Since ΔG = 0, we can rearrange the equation:
\[ ΔH = TΔS \]

### Step 2: Rearranging to find ΔS
From the rearranged equation, we can express ΔS as:
\[ ΔS = \frac{ΔH}{T} \]

### Step 3: Convert the enthalpy of vaporization to Joules
The enthalpy of vaporization is given as 24.64 kJ/mol. To convert this to Joules:
\[ ΔH = 24.64 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 24640 \, \text{J/mol} \]

### Step 4: Convert the temperature to Kelvin
The boiling point of the liquid is given as 35°C. To convert this to Kelvin:
\[ T = 35 + 273 = 308 \, \text{K} \]

### Step 5: Calculate ΔS
Now, we can substitute the values of ΔH and T into the equation for ΔS:
\[ ΔS = \frac{24640 \, \text{J/mol}}{308 \, \text{K}} \]

### Step 6: Perform the calculation
Calculating the above expression:
\[ ΔS = \frac{24640}{308} \approx 80 \, \text{J/K/mol} \]

### Final Answer
Thus, the change in entropy (ΔS) for the vaporization process is approximately:
\[ ΔS \approx 80 \, \text{J/K/mol} \]