20 gm of `CaCO_(3)` is allowed to dissociate in a 5.6 litres container at `819^(@)` C. If 50% of `CaCO_(3)` is dissocitated at equilibrium the `'K_(p)'` value is

Options

A. 5 atm
B. 1.6 atm
C. 4.8 atm
D. 10 atm

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of CaCO₃
Given mass of CaCO₃ = 20 g
Molar mass of CaCO₃ = 100 g/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \text{ g}}{100 \text{ g/mol}} = 0.2 \text{ mol} \]

### Step 2: Determine the amount dissociated
Since 50% of CaCO₃ is dissociated at equilibrium:
\[ \text{Moles dissociated} = 0.5 \times 0.2 \text{ mol} = 0.1 \text{ mol} \]

### Step 3: Calculate the remaining moles of CaCO₃ at equilibrium
\[ \text{Moles of CaCO₃ remaining} = 0.2 \text{ mol} - 0.1 \text{ mol} = 0.1 \text{ mol} \]

### Step 4: Determine the moles of products formed
From the reaction:
\[ \text{CaCO₃ (s)} \rightleftharpoons \text{CaO (s)} + \text{CO₂ (g)} \]
For every mole of CaCO₃ that dissociates, 1 mole of CaO and 1 mole of CO₂ is formed. Thus, at equilibrium:
\[ \text{Moles of CaO} = 0.1 \text{ mol} \text{Moles of CO₂} = 0.1 \text{ mol} \]

### Step 5: Calculate the molar concentrations at equilibrium
Volume of the container = 5.6 L
\[ \text{Concentration of CaCO₃} = \frac{0.1 \text{ mol}}{5.6 \text{ L}} = 0.01786 \text{ mol/L} \text{Concentration of CaO} = \frac{0.1 \text{ mol}}{5.6 \text{ L}} = 0.01786 \text{ mol/L} \text{Concentration of CO₂} = \frac{0.1 \text{ mol}}{5.6 \text{ L}} = 0.01786 \text{ mol/L} \]

### Step 6: Write the expression for Kc
The equilibrium constant Kc is given by:
\[ K_c = \frac{[\text{CaO}][\text{CO₂}]}{[\text{CaCO₃}]} \]
Substituting the concentrations:
\[ K_c = \frac{(0.01786)(0.01786)}{0.01786} = 0.01786 \text{ mol/L} \]

### Step 7: Convert Kc to Kp
Using the relation:
\[ K_p = K_c \cdot R^{\Delta n} \cdot T \]
Where:
- R = 0.0821 L·atm/(K·mol)
- T = 819 + 273 = 1092 K
- \(\Delta n\) = moles of gaseous products - moles of gaseous reactants = 1 - 0 = 1

Calculating Kp:
\[ K_p = K_c \cdot R \cdot T = 0.01786 \cdot 0.0821 \cdot 1092 \]
Calculating this gives:
\[ K_p \approx 1.6 \text{ atm} \]

### Final Answer
The value of Kp is approximately **1.6 atm**.

---