One mole of ice is converted into water at `273K`. The entropies of `H_2O(s)` and `H_2O(l)` are `38.20` and `60.01Jmol^(-1)K^(-1)` respectively. The enthalpy change for the conversion is:

Options

A. `59.54 J mol^(-1)`
B. `5954 J mol^(-1)`
C. `595.4 J mol^(-1)`
D. `320.6 J mol^(-1)`

Explanation

To find the enthalpy change (ΔH) for the conversion of 1 mole of ice (H₂O(s)) to water (H₂O(l)) at 273 K, we can follow these steps:

### Step 1: Write the Reaction
The conversion can be represented as:
\[ \text{H}_2\text{O(s)} \rightarrow \text{H}_2\text{O(l)} \]

### Step 2: Identify Given Data
We are given:
- Entropy of ice (H₂O(s)): \( S_{\text{ice}} = 38.20 \, \text{J/mol·K} \)
- Entropy of water (H₂O(l)): \( S_{\text{water}} = 60.01 \, \text{J/mol·K} \)

### Step 3: Calculate the Change in Entropy (ΔS)
The change in entropy (ΔS) for the reaction can be calculated as:
\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]
Substituting the values:
\[ \Delta S = S_{\text{water}} - S_{\text{ice}} = 60.01 \, \text{J/mol·K} - 38.20 \, \text{J/mol·K} \]
\[ \Delta S = 21.81 \, \text{J/mol·K} \]

### Step 4: Use the Gibbs Free Energy Equation
At equilibrium (which is the case here at 273 K), the Gibbs free energy change (ΔG) is zero. The relationship between ΔG, ΔH, and ΔS is given by:
\[ \Delta G = \Delta H - T \Delta S \]
Since ΔG = 0 at equilibrium:
\[ 0 = \Delta H - T \Delta S \]

### Step 5: Rearranging the Equation
Rearranging the equation to solve for ΔH:
\[ \Delta H = T \Delta S \]

### Step 6: Substitute the Values
Substituting the temperature (T = 273 K) and the calculated ΔS:
\[ \Delta H = 273 \, \text{K} \times 21.81 \, \text{J/mol·K} \]
\[ \Delta H = 5954.13 \, \text{J/mol} \]

### Step 7: Final Answer
Thus, the enthalpy change for the conversion of 1 mole of ice to water at 273 K is:
\[ \Delta H = 5954.13 \, \text{J/mol} \]