For the process :
`H_(2)O(l) (1 bar, 373 K) to H_(2)O(g) (1 bar, 373 K)`, the correct set of thermodynamic parameters is :

Options

A. `Delta G = 0, DeltaS = + ve`
B. `DeltaG = 0, DeltaS = -ve`
C. `DeltaG = + ve, DeltaS =0`
D. `DeltaG = -ve, DeltaS = + ve`

Explanation

To analyze the thermodynamic parameters for the process of converting liquid water (H₂O(l)) at 1 bar and 373 K to water vapor (H₂O(g)) at the same pressure and temperature, we can follow these steps:

### Step 1: Identify the Process
The process is a phase change from liquid water to water vapor, which is also known as vaporization or evaporation.

### Step 2: Determine the Gibbs Free Energy Change (ΔG)
For a phase change at equilibrium (which is the case here since both the liquid and vapor phases are at the same temperature and pressure), the change in Gibbs free energy (ΔG) is zero. This is because the system is at equilibrium, and there is no net change in the system.

**ΔG = 0**

### Step 3: Determine the Enthalpy Change (ΔH)
The enthalpy change for the process of vaporization at the boiling point (373 K for water) is known as the enthalpy of vaporization (ΔH_vap). For water, this value is approximately +40.79 kJ/mol. This value is positive because energy is absorbed when liquid water is converted to vapor.

**ΔH = +40.79 kJ/mol**

### Step 4: Determine the Entropy Change (ΔS)
The entropy change for the process can be calculated using the relationship between ΔH and ΔG at constant temperature:

\[ \Delta G = \Delta H - T\Delta S \]

Since ΔG = 0 at equilibrium, we can rearrange this to find ΔS:

\[ 0 = \Delta H - T\Delta S \implies T\Delta S = \Delta H \implies \Delta S = \frac{\Delta H}{T} \]

Substituting the values:

\[ \Delta S = \frac{40.79 \, \text{kJ/mol}}{373 \, \text{K}} \approx 0.109 \, \text{kJ/(mol·K)} \approx 109 \, \text{J/(mol·K)} \]

### Summary of Thermodynamic Parameters
- **ΔG = 0** (at equilibrium)
- **ΔH = +40.79 kJ/mol** (enthalpy of vaporization)
- **ΔS ≈ 109 J/(mol·K)** (entropy change)