Calculate `DeltaS` for following process :
`underset("at" 100K)(X(s)) underset("at" 200 K)(rarrX(l))`
Given : Melting point of `X_(s)=100K,DeltaH_("Fusion")=20kJ//"mol",C_(p.m)(X,l)=10J//"mol".K`

Options

A. `26.93 J//K`
B. `206.93 J//K`
C. `203 J//K`
D. `206.93 kJ//K`

Explanation

To calculate the change in entropy (ΔS) for the process of a solid (X) melting into a liquid (X) and then heating from 100 K to 200 K, we will break the problem into two parts: the melting process and the heating process.

### Step 1: Calculate ΔS during melting (ΔS1)

The change in entropy during the melting process can be calculated using the formula:

\[ \Delta S_1 = \frac{\Delta H_{\text{fusion}}}{T} \]

Where:
- ΔH_fusion = 20 kJ/mol = 20,000 J/mol (since we need to convert kJ to J)
- T = 100 K (the melting point)

Substituting the values:

\[ \Delta S_1 = \frac{20,000 \, \text{J/mol}}{100 \, \text{K}} = 200 \, \text{J/mol·K} \]

### Step 2: Calculate ΔS during heating (ΔS2)

For the heating process from 100 K to 200 K, we use the formula:

\[ \Delta S_2 = \int_{T_1}^{T_2} \frac{C_{p,m}}{T} \, dT \]

Where:
- \(C_{p,m} = 10 \, \text{J/mol·K}\) (given)
- \(T_1 = 100 \, \text{K}\)
- \(T_2 = 200 \, \text{K}\)

This integral can be simplified as:

\[ \Delta S_2 = C_{p,m} \cdot \ln\left(\frac{T_2}{T_1}\right) \]

Substituting the values:

\[ \Delta S_2 = 10 \, \text{J/mol·K} \cdot \ln\left(\frac{200}{100}\right) \]

Calculating the logarithm:

\[ \ln\left(\frac{200}{100}\right) = \ln(2) \approx 0.693 \]

Thus,

\[ \Delta S_2 = 10 \cdot 0.693 = 6.93 \, \text{J/mol·K} \]

### Step 3: Calculate total ΔS

Now, we can find the total change in entropy (ΔS) by adding ΔS1 and ΔS2:

\[ \Delta S = \Delta S_1 + \Delta S_2 \]

Substituting the values:

\[ \Delta S = 200 \, \text{J/mol·K} + 6.93 \, \text{J/mol·K} = 206.93 \, \text{J/mol·K} \]

### Final Answer

Thus, the total change in entropy for the process is:

\[ \Delta S = 206.93 \, \text{J/mol·K} \]

---