At `0^(@)C` , ice and water are in equilibrium and `DeltaG` and `DeltaS` for the conversion of ice to liquid water is,if `Delta H=6J mol^(-1)`

Options

A. 0 and 21.98 `JK^(-1) "mole"^(-1)`
B. `-ve` and zero
C. `+ve` and `21.98 JK^(-1) "mole"^(-1)`
D. zero and zero

Explanation

To solve the problem, we need to find the values of ΔG (Gibbs free energy change) and ΔS (entropy change) for the conversion of ice to liquid water at 0°C, given that ΔH (enthalpy change) is 6 J/mol.

### Step-by-Step Solution:

**Step 1: Understanding the conditions**
- At 0°C (273 K), ice and water are in equilibrium.
- The process of converting ice to liquid water is known as fusion.

**Step 2: Using the relationship between ΔG, ΔH, and ΔS**
- The relationship is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
- At equilibrium, ΔG = 0.

**Step 3: Rearranging the equation for equilibrium**
- Since ΔG = 0 at equilibrium, we can rearrange the equation:
\[ 0 = \Delta H - T \Delta S \]
\[ \Delta H = T \Delta S \]

**Step 4: Calculating ΔS (entropy change)**
- We know ΔH = 6 J/mol and T = 273 K.
- Rearranging gives us:
\[ \Delta S = \frac{\Delta H}{T} \]
- Substituting the values:
\[ \Delta S = \frac{6 \, \text{J/mol}}{273 \, \text{K}} \approx 0.02198 \, \text{J/K/mol} \]
- Converting to a more standard unit:
\[ \Delta S \approx 21.98 \, \text{J/K/mol} \]

**Step 5: Finding ΔG**
- Since we established that at equilibrium ΔG = 0:
\[ \Delta G = 0 \]

### Final Answers:
- ΔG = 0 J/mol
- ΔS = 21.98 J/K/mol

### Conclusion:
The correct answer is:
- ΔG = 0 and ΔS = 21.98 J/K/mol.