The enthalpy of vaporization for water is 186.5 KJ `mol^(−1)` ,the entropy of its vaporization will be:

Options

A. 4.07 J `mol^(−1)k^(−1)`
B. 1.02 KJ `mol^(−1)k^(−1)`
C. 0.7 J `mol^(−1)k^(−1)`
D. 0.5 KJ `mol^(−1)k^(−1)`

Answer

To find the entropy of vaporization for water, we can use the relationship between enthalpy change (ΔH) and entropy change (ΔS) at a given temperature (T). The formula we will use is:

\[ \Delta S = \frac{\Delta H}{T} \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Enthalpy of vaporization (ΔH) = 186.5 kJ/mol
- Boiling point of water = 100 °C

2. **Convert the boiling point to Kelvin:**
- To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
- So, for water:
\[ T = 100 + 273.15 = 373.15 \, K \approx 373 \, K \]

3. **Substitute the values into the entropy formula:**
- Now we can substitute ΔH and T into the formula:
\[ \Delta S = \frac{186.5 \, \text{kJ/mol}}{373 \, K} \]

4. **Convert ΔH from kJ to J for consistency:**
- Since 1 kJ = 1000 J, we convert:
\[ \Delta H = 186.5 \, \text{kJ/mol} = 186500 \, \text{J/mol} \]

5. **Calculate ΔS:**
- Now we can calculate:
\[ \Delta S = \frac{186500 \, \text{J/mol}}{373 \, K} \approx 500.67 \, \text{J/mol·K} \]
- Converting this back to kJ for final answer:
\[ \Delta S \approx 0.50067 \, \text{kJ/mol·K} \approx 0.5 \, \text{kJ/mol·K} \]

6. **Final answer:**
- The entropy of vaporization for water is approximately **0.5 kJ/mol·K**.