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 need to determine how much calcium carbonate (CaCO₃) dissociates at equilibrium when placed in a closed container. We will follow these steps:

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

### Step 2: Define the initial conditions
Initially, we have 1 mole of CaCO₃ in a 20 L container. The initial concentrations are:
- \([CaCO_3] = 1 \text{ mole}\) (solid, does not affect equilibrium)
- \([CaO] = 0 \text{ moles}\)
- \([CO_2] = 0 \text{ moles}\)

### Step 3: Define the change in concentration at equilibrium
Let \(x\) be the amount of CaCO₃ that dissociates at equilibrium. Then, at equilibrium:
- \([CaCO_3] = 1 - x\)
- \([CaO] = x\)
- \([CO_2] = x\)

### Step 4: Calculate the partial pressure of CO₂
Using the ideal gas law, we can find the partial pressure of CO₂:
\[ P = \frac{nRT}{V} \]
Where:
- \(n = x\) (moles of CO₂)
- \(R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}\)
- \(T = 727 + 273 = 1000 \text{ K}\) (temperature in Kelvin)
- \(V = 20 \text{ L}\)

Thus, the partial pressure of CO₂ is:
\[ P_{CO_2} = \frac{x \cdot 0.082 \cdot 1000}{20} \]

### Step 5: Substitute into the equilibrium expression
The equilibrium constant \(K_p\) is given as 0.82 atm. Since only CO₂ is a gas, we can write:
\[ K_p = P_{CO_2} = 0.82 \]
So we have:
\[ 0.82 = \frac{x \cdot 0.082 \cdot 1000}{20} \]

### Step 6: Solve for \(x\)
Rearranging the equation to solve for \(x\):
\[ 0.82 = \frac{x \cdot 82}{20} \]
\[ 0.82 \cdot 20 = 82x \]
\[ 16.4 = 82x \]
\[ x = \frac{16.4}{82} \]
\[ x \approx 0.2 \text{ moles} \]

### Step 7: Calculate the mass of CaCO₃ that dissociates
The molar mass of CaCO₃ is approximately 100 g/mol. Therefore, the mass of CaCO₃ that dissociates is:
\[ \text{mass} = x \cdot \text{molar mass} = 0.2 \text{ moles} \cdot 100 \text{ g/mol} = 20 \text{ grams} \]

### Final Answer
Thus, the amount of CaCO₃ that would dissociate at equilibrium is **20 grams**.

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