When 20.0 g of `CaCO_3` in a 10.0-litre flask is heated to `800^@C,` 35% of it did not dissociate, calculate `K_p` for the equilibrium `CaCO_(3) (s) Leftrightarrow CaO(s)+CO_(2)(g)`.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the number of moles of CaCO3
First, we need to calculate the molar mass of calcium carbonate (CaCO3).

- Molar mass of Ca = 40.08 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol × 3 = 48.00 g/mol

So, the molar mass of CaCO3 = 40.08 + 12.01 + 48.00 = 100.09 g/mol.

Now, we can calculate the number of moles of CaCO3 in 20.0 g:

\[ \text{Number of moles of CaCO3} = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.1998 \, \text{mol} \]

### Step 2: Determine the amount that dissociates
According to the problem, 35% of CaCO3 did not dissociate. Therefore, the percentage that did dissociate is:

\[ 100\% - 35\% = 65\% \]

Now, we can calculate the moles that dissociated:

\[ \text{Moles that dissociated} = 0.1998 \, \text{mol} \times 0.65 \approx 0.1299 \, \text{mol} \]

### Step 3: Write the equilibrium expression
The equilibrium reaction is:

\[ \text{CaCO}_3 (s) \leftrightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

At equilibrium, the solid calcium carbonate does not appear in the equilibrium expression, and we only consider the gaseous product CO2.

### Step 4: Calculate the partial pressure of CO2
To find the partial pressure of CO2, we need to know the number of moles of CO2 produced. Since 1 mole of CaCO3 produces 1 mole of CO2, the moles of CO2 produced is equal to the moles of CaCO3 that dissociated:

\[ \text{Moles of CO2} = 0.1299 \, \text{mol} \]

Next, we can calculate the concentration of CO2 in the 10.0-litre flask:

\[ \text{Concentration of CO2} = \frac{\text{moles of CO2}}{\text{volume}} = \frac{0.1299 \, \text{mol}}{10.0 \, \text{L}} = 0.01299 \, \text{mol/L} \]

Now, we can calculate the partial pressure of CO2 using the ideal gas law:

\[ P = C \cdot R \cdot T \]

Where:
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 800 \, \text{C} = 800 + 273.15 = 1073.15 \, \text{K} \)

Substituting the values:

\[ P_{CO2} = 0.01299 \, \text{mol/L} \times 0.0821 \, \text{L atm/(K mol)} \times 1073.15 \, \text{K} \approx 1.005 \, \text{atm} \]

### Step 5: Calculate Kp
The equilibrium constant \( K_p \) for the reaction is given by the expression:

\[ K_p = P_{CO2} \]

Thus, we find:

\[ K_p \approx 1.005 \, \text{atm} \]

### Final Answer:
The value of \( K_p \) for the equilibrium \( \text{CaCO}_3 (s) \leftrightarrow \text{CaO} (s) + \text{CO}_2 (g) \) is approximately **1.005 atm**.

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