The entropy change when `36g` of water evaporates at `373 K` is `:-` `(DeltaH=40.63(KJ)/(mol))`

Options

A. `0`
B. `218JK^(-1)`
C. `118JK^(-1)`
D. `200JK^(-1)`

Explanation

To find the entropy change when 36 g of water evaporates at 373 K, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Enthalpy Change to Joules**:
Given that \(\Delta H = 40.63 \, \text{kJ/mol}\), we need to convert this to Joules:
\[ \Delta H = 40.63 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40630 \, \text{J/mol} \]

2. **Calculate the Entropy of Vaporization**:
The entropy change for vaporization (\(\Delta S_{vaporization}\)) can be calculated using the formula:
\[ \Delta S_{vaporization} = \frac{\Delta H}{T_B} \]
where \(T_B = 373 \, \text{K}\). Substituting the values:
\[ \Delta S_{vaporization} = \frac{40630 \, \text{J/mol}}{373 \, \text{K}} \approx 108.9 \, \text{J/K/mol} \]

3. **Calculate the Number of Moles of Water**:
The number of moles of water in 36 g can be calculated using the molar mass of water (H₂O), which is approximately 18 g/mol:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{36 \, \text{g}}{18 \, \text{g/mol}} = 2 \, \text{mol} \]

4. **Calculate the Total Entropy Change for 36 g of Water**:
Now, we can calculate the total entropy change (\(\Delta S\)) for the evaporation of 36 g of water:
\[ \Delta S = \Delta S_{vaporization} \times \text{Number of moles} = 108.9 \, \text{J/K/mol} \times 2 \, \text{mol} = 217.8 \, \text{J/K} \]
Rounding this, we can say:
\[ \Delta S \approx 218 \, \text{J/K} \]

### Final Answer:
The entropy change when 36 g of water evaporates at 373 K is approximately **218 J/K**.