1 mol of ice is converted to liquid at 273 K, `H_2O(s) and H_2O` (l) have entropies 38.20 and 60.03 `"Jmol"^(-1)``k^(-1)`. Enthalpy changes in the conversion will be

Options

A. 59.59 J/mol
B. 5959.59 J/mol
C. 595.959 J/mol
D. 5.959 J/mol

Answer

To solve the problem of calculating the enthalpy change during the conversion of 1 mole of ice to liquid water at 273 K, we will use the relationship between Gibbs free energy, enthalpy, and entropy.

### Step-by-Step Solution:

1. **Understanding the Given Data:**
- We have the entropies of ice (H2O(s)) and liquid water (H2O(l)):
- \( S_{H2O(s)} = 38.20 \, \text{J/mol·K} \)
- \( S_{H2O(l)} = 60.03 \, \text{J/mol·K} \)
- The temperature at which the conversion occurs is \( T = 273 \, \text{K} \).

2. **Calculate the Change in Entropy (ΔS):**
- The change in entropy (ΔS) during the phase change from ice to liquid water can be calculated as:
\[ \Delta S = S_{H2O(l)} - S_{H2O(s)} = 60.03 \, \text{J/mol·K} - 38.20 \, \text{J/mol·K} \]
- Performing the calculation:
\[ \Delta S = 21.83 \, \text{J/mol·K} \]

3. **Using the Gibbs Free Energy Equation:**
- At equilibrium, the Gibbs free energy change (ΔG) is zero:
\[ \Delta G = \Delta H - T \Delta S = 0 \]
- Rearranging the equation to solve for ΔH:
\[ \Delta H = T \Delta S \]

4. **Substituting the Values:**
- Now, substitute the values of T and ΔS into the equation:
\[ \Delta H = 273 \, \text{K} \times 21.83 \, \text{J/mol·K} \]
- Performing the multiplication:
\[ \Delta H = 5960.19 \, \text{J/mol} \]

5. **Converting to Kilojoules:**
- Since the answer is often required in kilojoules, we convert:
\[ \Delta H = 5.96019 \, \text{kJ/mol} \approx 5.96 \, \text{kJ/mol} \]

6. **Final Answer:**
- The enthalpy change for the conversion of 1 mole of ice to liquid water at 273 K is approximately:
\[ \Delta H \approx 5.96 \, \text{kJ/mol} \]