At `0^(@)C` ice and water are in equilibrium and `Delta H= 6.0KJ " then " Delta S` will be

Options

A. `21.98JK^(-10 mol^(-1)`
B. `35 JK^(-1) mol^(-1)`
C. `48 JK^(-1) mol^(-1)`
D. `100 JK^(-1) mol^(-1)`

Answer

To solve the problem of finding the change in entropy (ΔS) when ice and water are in equilibrium at 0°C, we can follow these steps:

### Step 1: Understand the relationship between ΔH and ΔS
The change in entropy (ΔS) during a phase change can be calculated using the formula:
\[ \Delta S = \frac{\Delta H}{T} \]
where:
- ΔH is the change in enthalpy (in joules),
- T is the absolute temperature in Kelvin.

### Step 2: Convert ΔH to Joules
Given that ΔH = 6.0 kJ, we need to convert this value to joules:
\[ \Delta H = 6.0 \, \text{kJ} = 6.0 \times 1000 \, \text{J} = 6000 \, \text{J} \]

### Step 3: Convert the temperature from Celsius to Kelvin
The temperature at which ice and water are in equilibrium is given as 0°C. We convert this to Kelvin:
\[ T = 0°C + 273.15 = 273.15 \, \text{K} \]

### Step 4: Calculate ΔS
Now we can substitute the values of ΔH and T into the formula for ΔS:
\[ \Delta S = \frac{6000 \, \text{J}}{273.15 \, \text{K}} \approx 21.96 \, \text{J/K} \]

### Step 5: Round off the answer
Rounding off to two decimal places, we get:
\[ \Delta S \approx 21.96 \, \text{J/K} \]

### Final Answer
Thus, the change in entropy (ΔS) at 0°C when ice and water are in equilibrium is approximately:
\[ \Delta S \approx 21.96 \, \text{J/K} \]

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