the value of `DeltaG` for the process
`H_(2)O(s) rarr H_(2)O` at `1` atm and `260` K is

Options

A. `lt 0`
B. `= 0`
C. ` gt 0`
D. Unpredictable

Answer

To determine the value of ΔG for the process \( H_2O(s) \rightarrow H_2O(l) \) at 1 atm and 260 K, we can follow these steps:

### Step 1: Identify the phase change
The reaction describes the melting of ice (solid water) into liquid water.

### Step 2: Determine the temperature in Celsius
Convert the temperature from Kelvin to Celsius:
\[ T = 260 \, K - 273.15 = -13.15 \, °C \]

### Step 3: Compare with the melting point of water
The melting point of ice is 0 °C (273 K). Since -13.15 °C is below the melting point, ice will be the predominant phase at this temperature.

### Step 4: Determine spontaneity of the process
For a process to be spontaneous, ΔG must be less than 0. Since the temperature is below the melting point, the conversion of ice to liquid water is non-spontaneous at this temperature.

### Step 5: Conclude the value of ΔG
Since the process is non-spontaneous, we conclude that:
\[ \Delta G > 0 \]

Thus, the value of ΔG for the process \( H_2O(s) \rightarrow H_2O(l) \) at 1 atm and 260 K is greater than 0.

### Final Answer
\(\Delta G > 0\)

---