At `0^(@)C` , ice and water are in equilibrium and `DeltaS` and `DeltaG` for the conversion of ice to liquid water is,if `Delta H`=`7kJmol^-1`

Options

A. `DeltaS=25.64` `DeltaG=0.54`
B. `DeltaS=2.198` `DeltaG=0`
C. `DeltaS=2.198` `DeltaG=-0.54`
D. `DeltaS=25.64` `DeltaG=0`

Explanation

To solve the problem, we need to find the values of ΔS (change in entropy) and ΔG (change in Gibbs free energy) for the conversion of ice to liquid water at 0°C, given that ΔH (enthalpy change) is 7 kJ/mol.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The conversion of ice (solid water) to liquid water can be represented as:
\[ \text{H}_2\text{O (s)} \rightarrow \text{H}_2\text{O (l)} \]
This process is known as fusion.

2. **Convert ΔH to Joules**:
Given ΔH = 7 kJ/mol, we need to convert this to Joules:
\[ \Delta H = 7 \, \text{kJ/mol} = 7 \times 10^3 \, \text{J/mol} = 7000 \, \text{J/mol} \]

3. **Determine the Temperature in Kelvin**:
The temperature at which the equilibrium occurs is given as 0°C. To convert this to Kelvin:
\[ T = 0 + 273 = 273 \, \text{K} \]

4. **Calculate ΔS (Change in Entropy)**:
Using the formula for the change in entropy during fusion:
\[ \Delta S = \frac{\Delta H}{T} \]
Substituting the values we have:
\[ \Delta S = \frac{7000 \, \text{J/mol}}{273 \, \text{K}} \approx 25.64 \, \text{J/K/mol} \]

5. **Calculate ΔG (Change in Gibbs Free Energy)**:
At equilibrium, ΔG is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
Since we are at equilibrium, ΔG = 0. Thus, we can set up the equation:
\[ 0 = \Delta H - T \Delta S \]
Rearranging gives:
\[ \Delta H = T \Delta S \]
Substituting the known values:
\[ 7000 \, \text{J/mol} = 273 \, \text{K} \times 25.64 \, \text{J/K/mol} \]
Calculating the right side:
\[ 273 \times 25.64 \approx 7000 \, \text{J/mol} \]
This confirms that ΔG = 0 at equilibrium.

### Final Answers:
- ΔS = 25.64 J/K/mol
- ΔG = 0 J/mol