A contributions of both heat (enthalpy) and randomness(entropy) shall be considered to the overall spontaneity of process.When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change `(DeltaG)`.
`DeltaG=DeltaH-TDeltaS`
where,`DeltaH`=Enthalpy change, `DeltaS`=Entropy change , T=Temperature in kelvin.
If `DeltaGlt0`, Process is spontaneous , `DeltaG=0`, Process is at equilibrium , `DeltaGgt0`, Process is non-spontaneous.
For the change `H_2O(s),(273 K`,2 atm)`to H_2O(l)`,(273 K, 2 atm), choose the correct option.

Options

A. `DeltaG=0`
B. `DeltaGlt0`
C. `DeltaGgt0`
D. None

Explanation

To determine the spontaneity of the phase change from solid water (ice) to liquid water at 273 K and 2 atm, we will analyze the Gibbs free energy change (ΔG) using the formula:

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

### Step 1: Identify the Phase Change
We are considering the phase change from solid (H₂O(s)) to liquid (H₂O(l)) at a temperature of 273 K and a pressure of 2 atm.

**Hint:** Remember that the phase change from solid to liquid is known as melting.

### Step 2: Understand the Conditions
At 273 K, ice and water are in equilibrium at 1 atm. However, in this case, the pressure is increased to 2 atm.

**Hint:** Think about how increasing pressure affects the equilibrium of a phase change involving liquids and solids.

### Step 3: Apply Le Chatelier's Principle
According to Le Chatelier's principle, if the pressure is increased, the equilibrium will shift towards the side with fewer moles of gas. In this case, since both ice and water are in the liquid phase, the reaction will favor the formation of liquid water.

**Hint:** Consider how the equilibrium shifts when pressure changes and how that relates to spontaneity.

### Step 4: Analyze ΔH and ΔS
- **ΔH (Enthalpy Change):** The melting of ice is an endothermic process, meaning ΔH is positive.
- **ΔS (Entropy Change):** The transition from solid to liquid increases randomness, so ΔS is positive.

**Hint:** Recall that for melting, ΔH is positive and ΔS is also positive.

### Step 5: Evaluate ΔG
Substituting the values into the Gibbs free energy equation:

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

Since ΔH is positive and ΔS is positive, we need to consider the temperature (T = 273 K). The term \(T \Delta S\) will also be positive.

The key point here is that at higher pressures, the process is more favorable, and the increase in pressure shifts the equilibrium towards the formation of liquid water.

**Hint:** Think about how the signs of ΔH and ΔS affect the overall sign of ΔG.

### Step 6: Conclusion
Since the process is favored under the increased pressure, we can conclude that:

\[ \Delta G < 0 \]

Thus, the process of converting solid water to liquid water at 273 K and 2 atm is spontaneous.

**Final Answer:** The correct option is **ΔG < 0** (Option B).