When 20 g of `CaCO_3` were put into 10 litre flask and heated to `800^@C, 30%` of `CaCO_3` remained unreacted at equilibrium `K_p` for decomposition of `CaCO_3` will be

Options

A. 1.145 atm
B. 1.231 atm
C. 2.146 atm
D. 3.145 atm

Answer

To find the equilibrium constant \( K_p \) for the decomposition of calcium carbonate (\( CaCO_3 \)), we can follow these steps:

### Step 1: Calculate the number of moles of \( CaCO_3 \)
Given:
- Mass of \( CaCO_3 = 20 \, g \)
- Molar mass of \( CaCO_3 = 100 \, g/mol \) (calculated as follows: \( Ca = 40 \, g/mol + C = 12 \, g/mol + 3 \times O = 3 \times 16 \, g/mol = 48 \, g/mol \))

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

### Step 2: Determine the moles of \( CaCO_3 \) that remain unreacted
Given that 30% of \( CaCO_3 \) remains unreacted at equilibrium:
\[ \text{Unreacted moles of } CaCO_3 = 0.3 \times 0.2 = 0.06 \, moles \]

### Step 3: Calculate the moles of \( CaCO_3 \) that reacted
The initial moles of \( CaCO_3 \) were 0.2 moles. Therefore, the moles that reacted are:
\[ \text{Moles reacted} = \text{Initial moles} - \text{Unreacted moles} = 0.2 - 0.06 = 0.14 \, moles \]

### Step 4: Write the decomposition reaction
The decomposition of \( CaCO_3 \) can be represented as:
\[ CaCO_3 (s) \rightleftharpoons CaO (s) + CO_2 (g) \]
Here, \( CaCO_3 \) and \( CaO \) are solids, and \( CO_2 \) is a gas.

### Step 5: Determine the moles of \( CO_2 \) produced
From the stoichiometry of the reaction, for every mole of \( CaCO_3 \) that decomposes, one mole of \( CO_2 \) is produced. Therefore, the moles of \( CO_2 \) produced is equal to the moles of \( CaCO_3 \) that reacted:
\[ \text{Moles of } CO_2 = 0.14 \, moles \]

### Step 6: Calculate the partial pressure of \( CO_2 \)
Using the ideal gas law, \( PV = nRT \), we can express the partial pressure \( P \):
\[ P = \frac{nRT}{V} \]
Where:
- \( n = 0.14 \, moles \) (moles of \( CO_2 \))
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- \( T = 800 + 273 = 1073 \, K \)
- \( V = 10 \, L \)

Substituting the values:
\[ P_{CO_2} = \frac{0.14 \times 0.0821 \times 1073}{10} \]
Calculating:
\[ P_{CO_2} = \frac{0.14 \times 0.0821 \times 1073}{10} \approx 1.23 \, atm \]

### Step 7: Calculate \( K_p \)
Since \( K_p \) for the reaction is defined as:
\[ K_p = \frac{P_{CO_2}}{1} = P_{CO_2} \]
Thus:
\[ K_p = 1.23 \, atm \]

### Final Answer
The equilibrium constant \( K_p \) for the decomposition of \( CaCO_3 \) at \( 800^\circ C \) is approximately \( 1.23 \, atm \).

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