The sign of `DeltaG` for the process of melting of ice at `273K` and `1` atm pressure is

Options

A. positive
B. negative
C. neither positive nor negative
D. either positive nor negative

Explanation

To determine the sign of ΔG for the process of melting of ice at 273 K and 1 atm pressure, we can follow these steps:

### Step 1: Understand the process
The melting of ice is a phase transition where solid ice turns into liquid water. This process occurs at a specific temperature and pressure, which in this case are 273 K and 1 atm.

**Hint:** Recall that phase transitions occur at specific temperatures and pressures where the two phases can coexist.

### Step 2: Identify the equilibrium condition
At 273 K and 1 atm, ice and water are in equilibrium. This means that both solid ice and liquid water can exist simultaneously without favoring one phase over the other.

**Hint:** Equilibrium indicates that the forward and reverse processes occur at the same rate.

### Step 3: Apply the Gibbs Free Energy concept
The Gibbs Free Energy change (ΔG) is a thermodynamic potential that can predict the spontaneity of a process. The sign of ΔG is determined as follows:
- ΔG < 0: The process is spontaneous.
- ΔG > 0: The process is non-spontaneous.
- ΔG = 0: The system is at equilibrium.

**Hint:** Remember that at equilibrium, the system does not favor either direction of the process.

### Step 4: Determine the sign of ΔG
Since the melting of ice at 273 K and 1 atm is an equilibrium condition, the value of ΔG for this process is zero. This means that the melting of ice is neither spontaneous nor non-spontaneous; it is at a state of balance.

**Hint:** Equilibrium conditions always result in ΔG being zero.

### Conclusion
Therefore, the sign of ΔG for the process of melting of ice at 273 K and 1 atm pressure is **zero**.

### Final Answer
ΔG = 0 (neither positive nor negative).