When 20g of `CaCO_3` is put into a 11.45L flask and heated to `800^@C` , 35% of `CaCO_3` remains undissociated at equilibrium. Calculate the value of `K_p`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of CaCO₃
Given:
- Mass of CaCO₃ (W) = 20 g
- Molar mass of CaCO₃ (M) = 100 g/mol

Using the formula for moles:
\[ N = \frac{W}{M} \]
Substituting the values:
\[ N_{\text{CaCO}_3} = \frac{20 \, \text{g}}{100 \, \text{g/mol}} = 0.2 \, \text{mol} \]

### Step 2: Calculate the initial concentration of CaCO₃
Given:
- Volume of the flask (V) = 11.45 L

Using the formula for concentration:
\[ C_0 = \frac{N_{\text{CaCO}_3}}{V} \]
Substituting the values:
\[ C_0 = \frac{0.2 \, \text{mol}}{11.45 \, \text{L}} \approx 0.0175 \, \text{mol/L} \]

### Step 3: Determine the amount of CaCO₃ that dissociates
Given:
- 35% of CaCO₃ remains undissociated at equilibrium.

Thus, the percentage that dissociates is:
\[ 100\% - 35\% = 65\% \]
This means:
\[ \alpha = 0.65 \]

### Step 4: Calculate the equilibrium concentrations
At equilibrium:
- The concentration of CaCO₃ that remains = \(C_0 \times (1 - \alpha)\)
- The concentration of CO₂ produced = \(C_0 \times \alpha\)

Calculating these:
\[ [\text{CaCO}_3]_{\text{eq}} = C_0 \times (1 - 0.65) = 0.0175 \times 0.35 \approx 0.006125 \, \text{mol/L} \]
\[ [\text{CO}_2]_{\text{eq}} = C_0 \times \alpha = 0.0175 \times 0.65 \approx 0.011375 \, \text{mol/L} \]

### Step 5: Write the expression for Kc
The reaction is:
\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]
The equilibrium constant \(K_c\) is given by:
\[ K_c = \frac{[\text{CO}_2]}{1} = [\text{CO}_2]_{\text{eq}} \]
Substituting the value:
\[ K_c = 0.011375 \, \text{mol/L} \]

### Step 6: Convert Kc to Kp
Using the relation:
\[ K_p = K_c \times R \times T^{\Delta N_g} \]
Where:
- \(R = 0.0821 \, \text{L atm/(K mol)}\)
- \(T = 800 + 273 = 1073 \, \text{K}\)
- \(\Delta N_g = 1\) (1 mole of gas product - 0 moles of gas reactants)

Calculating \(K_p\):
\[ K_p = K_c \times R \times T \]
Substituting the values:
\[ K_p = 0.011375 \times 0.0821 \times 1073 \]
Calculating this gives:
\[ K_p \approx 0.967 \, \text{atm} \]

### Final Answer
\[ K_p \approx 0.967 \, \text{atm} \]