Equilibrium constant `K_(p)` for the reaction `CaCO_(3)(s) hArr CaO(s) + CO_2(g)` is 0.82 atm at `727^@C`.
If 1 mole of `CaCO_(3)` is placed in a closed container of 20 L and heated to this temperature, what amount of `CaCO_(3)` would dissociate at equilibrium?

Options

A. 0.2 g
B. 80 g
C. 20 g
D. 50 g

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]

### Step 2: Understand the equilibrium constant
The equilibrium constant \( K_p \) for this reaction is given as:
\[ K_p = P_{CO_2} = 0.82 \, \text{atm} \]
Since solids do not appear in the expression for the equilibrium constant, we only consider the gaseous product.

### Step 3: Set up the initial conditions
Initially, we have 1 mole of \( \text{CaCO}_3 \) in a closed container of volume 20 L. The initial moles of \( \text{CO}_2 \) are 0.

### Step 4: Define the change in moles at equilibrium
Let \( x \) be the number of moles of \( \text{CaCO}_3 \) that dissociate at equilibrium. Therefore, at equilibrium:
- Moles of \( \text{CaCO}_3 \) remaining = \( 1 - x \)
- Moles of \( \text{CO}_2 \) formed = \( x \)

### Step 5: Calculate the partial pressure of \( \text{CO}_2 \)
Using the ideal gas law, we can find the partial pressure of \( \text{CO}_2 \):
\[ P_{CO_2} = \frac{nRT}{V} \]
Where:
- \( n = x \) (moles of \( \text{CO}_2 \))
- \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 727 + 273 = 1000 \, \text{K} \)
- \( V = 20 \, \text{L} \)

Substituting these values into the equation gives:
\[ P_{CO_2} = \frac{x \cdot 0.082 \cdot 1000}{20} \]

### Step 6: Set the equation equal to the given \( K_p \)
Since we know that \( P_{CO_2} = 0.82 \, \text{atm} \), we can set up the equation:
\[ 0.82 = \frac{x \cdot 0.082 \cdot 1000}{20} \]

### Step 7: Solve for \( x \)
Rearranging the equation to solve for \( x \):
\[ x = \frac{0.82 \cdot 20}{0.082 \cdot 1000} \]

Calculating this gives:
\[ x = \frac{16.4}{82} = 0.2 \, \text{moles} \]

### Step 8: Calculate the mass of \( \text{CaCO}_3 \) that dissociated
The molar mass of \( \text{CaCO}_3 \) is approximately 100 g/mol. Therefore, the mass of \( \text{CaCO}_3 \) that dissociated is:
\[ \text{mass} = x \cdot \text{molar mass} = 0.2 \cdot 100 = 20 \, \text{grams} \]

### Final Answer
Thus, the amount of \( \text{CaCO}_3 \) that would dissociate at equilibrium is **20 grams**.

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