Enthalpy change for the process,
`H_(2)O"(ice")hArr H_(2)O"(water)"`
is `"6.01 kJ mol"^(-1)`. The entropy change of 1 mole of ice at its melting point will be

Options

A. `"12 J K"^(-1)"mol"^(-1)`
B. `"22 J K"^(-1)"mol"^(-1)`
C. `"100 J K"^(-1)"mol"^(-1)`
D. `"30 J K"^(-1)"mol"^(-1)`

Explanation

To find the entropy change (ΔS) for the process of melting ice (H₂O solid) to water (H₂O liquid), we can use the formula:

\[ \Delta S = \frac{\Delta H}{T} \]

where:
- ΔH is the enthalpy change,
- T is the temperature in Kelvin.

### Step 1: Identify the given values
- The enthalpy change (ΔH) for the melting of ice is given as 6.01 kJ/mol.
- The melting point of ice is 0°C.

### Step 2: Convert the enthalpy change from kJ to J
Since we need to express ΔH in joules for our calculations, we convert it:

\[ \Delta H = 6.01 \, \text{kJ/mol} = 6.01 \times 1000 \, \text{J/mol} = 6010 \, \text{J/mol} \]

### Step 3: Convert the temperature from Celsius to Kelvin
The melting point of ice is 0°C. To convert this to Kelvin:

\[ T = 0°C + 273.15 = 273.15 \, \text{K} \approx 273 \, \text{K} \]

### Step 4: Substitute the values into the entropy change formula
Now we can substitute the values of ΔH and T into the entropy change formula:

\[ \Delta S = \frac{6010 \, \text{J/mol}}{273 \, \text{K}} \]

### Step 5: Calculate the entropy change
Now we perform the calculation:

\[ \Delta S \approx \frac{6010}{273} \approx 22.0 \, \text{J/K/mol} \]

### Final Answer
The entropy change (ΔS) for the melting of 1 mole of ice at its melting point is approximately:

\[ \Delta S \approx 22.0 \, \text{J/K/mol} \]

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