For `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g), K_(c)` is equal to …………..

Options

A. `K_(c)=(1)/([CO_(2)])`
B. `K_(c)=[CO_(2)]`
C. `K_(c)=([CaO][CO_(2)])/([CaCO_(3)])`
D. `K_(c)=([CaCO_(3)])/([CaO][CO_(2)])`

Explanation

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

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

we need to consider the definition of the equilibrium constant in terms of the concentrations (or active masses) of the reactants and products.

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
The reaction involves solid calcium carbonate (\( \text{CaCO}_3 \)), solid calcium oxide (\( \text{CaO} \)), and gaseous carbon dioxide (\( \text{CO}_2 \)).

2. **Understand Active Mass**:
The active mass (or concentration) of a substance in equilibrium expressions is represented by square brackets \([ ]\). For solids and pure liquids, the active mass is considered to be constant and is taken as 1.

3. **Write the Expression for \( K_c \)**:
The general expression for the equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{Products}]}{[\text{Reactants}]} \]

In this case, the products are \( \text{CaO} \) and \( \text{CO}_2 \), and the reactant is \( \text{CaCO}_3 \).

4. **Substituting the Active Mass Values**:
Since both \( \text{CaCO}_3 \) and \( \text{CaO} \) are solids, their active masses are 1. Therefore, we can write:
\[ K_c = \frac{[\text{CaO}] \cdot [\text{CO}_2]}{[\text{CaCO}_3]} \]
This simplifies to:
\[ K_c = \frac{[\text{CO}_2]}{1} = [\text{CO}_2] \]

5. **Final Expression for \( K_c \)**:
Thus, the equilibrium constant \( K_c \) for the reaction is simply the active mass of \( \text{CO}_2 \):
\[ K_c = [\text{CO}_2] \]

### Conclusion:
The correct answer is:
\[ K_c = [\text{CO}_2] \]