19 gm 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 `60 J//"mol " K` respectively. The enthalpy change for this conversion is :

Options

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

Explanation

To solve the problem of calculating the enthalpy change when 19 grams of ice is converted to water at 0°C and 1 atm, we can follow these steps:

### Step 1: Calculate the number of moles of ice
We know that the molar mass of water (H₂O) is approximately 18 g/mol. To find the number of moles of ice (H₂O solid), we use the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

Given that the mass of ice is 19 g:

\[ \text{Number of moles of ice} = \frac{19 \, \text{g}}{18 \, \text{g/mol}} \approx 1.06 \, \text{mol} \]

### Step 2: Calculate the change in entropy (ΔS)
The change in entropy (ΔS) for the conversion of ice to water can be calculated using the formula:

\[ \Delta S = S_{\text{liquid}} - S_{\text{solid}} \]

Where:
- \( S_{\text{liquid}} = 60 \, \text{J/mol K} \)
- \( S_{\text{solid}} = 38.2 \, \text{J/mol K} \)

Substituting the values:

\[ \Delta S = 60 \, \text{J/mol K} - 38.2 \, \text{J/mol K} = 21.8 \, \text{J/mol K} \]

### Step 3: Calculate the enthalpy change (ΔH)
At equilibrium, the change in Gibbs free energy (ΔG) is zero. Therefore, we can use the equation:

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

Since ΔG = 0 at equilibrium:

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

The temperature (T) is given as 0°C, which is equivalent to 273 K. Now we can calculate ΔH:

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

### Step 4: Adjust for the number of moles
Since we calculated ΔH for 1 mole, and we have approximately 1.06 moles of ice, we need to multiply:

\[ \Delta H_{\text{total}} = 5954.4 \, \text{J/mol} \times 1.06 \, \text{mol} \approx 6315.66 \, \text{J} \]

### Final Answer
Thus, the enthalpy change for the conversion of 19 g of ice to water at 0°C and 1 atm is approximately **6315.66 J**.

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