If the enthalpy change for the transition of liquid water to steam is 300kJ `mol^(-1)" at "27^(@)C`, the entropy change for the proces would be

Options

A. `1000 JK^(-1)mol^(-1)`
B. `10 JK^(-1)mol^(-1)`
C. `1.0 JK^(-1)mol^(-1)`
D. `0.1JK^(-1)mol^(-1)`

Answer

To find the entropy change for the transition of liquid water to steam, we can use the formula:

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

where:
- \(\Delta S\) is the entropy change,
- \(\Delta H\) is the enthalpy change,
- \(T\) is the temperature in Kelvin.

### Step 1: Convert the enthalpy change from kJ to J
The enthalpy change given is 300 kJ/mol. To convert this to Joules, we multiply by \(10^3\):

\[ \Delta H = 300 \, \text{kJ/mol} \times 10^3 = 300,000 \, \text{J/mol} \]

### Step 2: Convert the temperature from Celsius to Kelvin
The temperature is given as 27°C. To convert this to Kelvin, we add 273:

\[ T = 27 + 273 = 300 \, \text{K} \]

### Step 3: Calculate the entropy change
Now we can substitute the values of \(\Delta H\) and \(T\) into the entropy change formula:

\[ \Delta S = \frac{300,000 \, \text{J/mol}}{300 \, \text{K}} = 1000 \, \text{J/(K·mol)} \]

### Conclusion
The entropy change for the process of transitioning from liquid water to steam at 27°C is:

\[ \Delta S = 1000 \, \text{J/(K·mol)} \]