18g of ice is converted into water at `0^(@)C` and 1 atm. The entropies of `H_(2)O` (s) and `H_(2)O` (l) are 38.2 and 60J/mol K respectively. The enthalpy cange for this conversion is:

Options

A. 5951.4 J/mol
B. 595.14 Jmol
C. `-5951.4` J/mol
D. None of these

Explanation

To find the enthalpy change for the conversion of 18g of ice at 0°C to water at the same temperature, we can follow these steps:

### Step 1: Calculate the number of moles of ice
First, we need to determine how many moles of ice (H2O solid) we have in 18 grams.

\[ \text{Molar mass of } H_2O = 18 \, \text{g/mol} \]

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{18 \, \text{g}}{18 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 2: Determine the change in entropy (ΔS)
The change in entropy (ΔS) during the phase transition from ice to water can be calculated using the given entropy values.

\[ \Delta S = S_{\text{H}_2O(l)} - S_{\text{H}_2O(s)} = 60 \, \text{J/mol K} - 38.2 \, \text{J/mol K} = 21.8 \, \text{J/mol K} \]

### Step 3: Calculate the enthalpy change (ΔH)
At equilibrium, we can use the relationship between Gibbs free energy (ΔG), enthalpy change (ΔH), and entropy change (ΔS):

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

At equilibrium, ΔG = 0, therefore:

\[ 0 = \Delta H - T \Delta S \]

Rearranging gives:

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

Substituting the values:

- T = 273 K (0°C in Kelvin)
- ΔS = 21.8 J/mol K

\[ \Delta H = 273 \, \text{K} \times 21.8 \, \text{J/mol K} = 5954.4 \, \text{J/mol} \]

### Step 4: Finalize the enthalpy change
Since we have 1 mole of ice, the enthalpy change for converting 18g of ice to water at 0°C is:

\[ \Delta H = 5954.4 \, \text{J} \approx 5951.4 \, \text{J/mol} \]

### Conclusion
The enthalpy change for the conversion of 18g of ice into water at 0°C and 1 atm is approximately **5951.4 J/mol**.

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