At 373 K, steam and water are in equilibrium and `DeltaH=40.98" kJ mol"^(-1)`. What will be `DeltaS` for conversion of water into system ?
`H_(2)O_((l))rarrH_(2)O_((g))`

Options

A. `"109.8 J K"^(-1)"mol"^(-1)`
B. `"31 J K"^(-1)"mol"^(-1)`
C. `"21.98 J K"^(-1)"mol"^(-1)`
D. `"326 J K"^(-1)"mol"^(-1)`

Explanation

To solve the problem, we need to calculate the change in entropy (ΔS) for the conversion of water (H₂O) from liquid to gas (steam) at a temperature of 373 K, given that the enthalpy change (ΔH) for this process is 40.98 kJ/mol.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction we are considering is:
\[ H_2O_{(l)} \rightarrow H_2O_{(g)} \]
This process is known as vaporization.

2. **Convert ΔH from kJ to J**:
The given ΔH is in kilojoules per mole. We need to convert it to joules per mole for our calculations:
\[ \Delta H = 40.98 \, \text{kJ/mol} = 40.98 \times 10^3 \, \text{J/mol} = 40980 \, \text{J/mol} \]

3. **Use the Gibbs Free Energy Equation**:
At equilibrium, the change in Gibbs free energy (ΔG) is zero. The relationship between ΔG, ΔH, and ΔS is given by:
\[ \Delta G = \Delta H - T \Delta S \]
Setting ΔG to zero for equilibrium:
\[ 0 = \Delta H - T \Delta S \]

4. **Rearrange to Solve for ΔS**:
From the equation above, we can rearrange it to find ΔS:
\[ T \Delta S = \Delta H \implies \Delta S = \frac{\Delta H}{T} \]

5. **Substitute the Values**:
Now, substitute the values of ΔH and T into the equation:
\[ \Delta S = \frac{40980 \, \text{J/mol}}{373 \, \text{K}} \]

6. **Calculate ΔS**:
Performing the calculation:
\[ \Delta S = \frac{40980}{373} \approx 109.8 \, \text{J/K/mol} \]

### Final Answer:
The change in entropy (ΔS) for the conversion of water into steam at 373 K is approximately:
\[ \Delta S \approx 109.8 \, \text{J/K/mol} \]