Consider the following heterogeneous equilibrium, `CaCO_(3)(s)CaO(s)+CO_(2)(g)`. At `800^(@)C`, the pressure of `CO_(2)` is 0.236 atm. The value of `K_(c)` for the abvoe given equilibrium reaction at this temperature is

Options

A. `4xx10^(-5)`
B. `2.68xx10^(-3)`
C. `2.64xx10^(-5)`
D. `1.5xx10^(-3)`

Answer

To find the equilibrium constant \( K_c \) for the reaction

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

at \( 800^\circ C \) with a given pressure of \( \text{CO}_2 \) as \( 0.236 \) atm, we can follow these steps:

### Step 1: Identify the components of the equilibrium expression

In this equilibrium, we have solid calcium carbonate (\( \text{CaCO}_3 \)) and solid calcium oxide (\( \text{CaO} \)), along with gaseous carbon dioxide (\( \text{CO}_2 \)). Since the equilibrium constant \( K_c \) only includes the concentrations of gaseous and aqueous species, we can ignore the solids in our expression.

### Step 2: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) for the reaction can be expressed as:

\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} \]

Since the reactants are solids, they do not appear in the expression. Thus, we have:

\[ K_c = [\text{CO}_2] \]

### Step 3: Convert pressure to concentration

To find the concentration of \( \text{CO}_2 \), we can use the ideal gas law:

\[ PV = nRT \]

Where:
- \( P \) is the pressure of the gas
- \( V \) is the volume of the gas
- \( n \) is the number of moles of the gas
- \( R \) is the ideal gas constant
- \( T \) is the temperature in Kelvin

From the ideal gas law, we can express concentration \( C \) as:

\[ C = \frac{n}{V} = \frac{P}{RT} \]

### Step 4: Substitute the known values

Given:
- \( P = 0.236 \) atm
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 800^\circ C = 800 + 273 = 1073 \, K \)

Substituting these values into the concentration formula:

\[ C = \frac{0.236}{0.0821 \times 1073} \]

### Step 5: Calculate the concentration

Calculating the denominator:

\[ 0.0821 \times 1073 \approx 88.0733 \]

Now substituting back:

\[ C = \frac{0.236}{88.0733} \approx 0.00268 \, \text{mol/L} \]

### Step 6: Conclude the value of \( K_c \)

Since \( K_c = [\text{CO}_2] \), we have:

\[ K_c \approx 0.00268 \, \text{mol/L} \]

### Final Answer

Thus, the value of \( K_c \) for the reaction at \( 800^\circ C \) is approximately:

\[ K_c \approx 2.68 \times 10^{-3} \]

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