The reaction `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)` reaches the state of equilibrium when

Options

A. Partial pressure of `CaCO_(3)` become constant
B. Partial pressure of CaO become constant
C. Pressure of `CO_(2)` become constant
D. All partial pressure keep changing

Answer

To determine when the reaction \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \) reaches a state of equilibrium, we can analyze the components involved in the reaction.

### Step-by-Step Solution:

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

2. **Understand the Concept of Equilibrium**:
Equilibrium in a chemical reaction occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products.

3. **Focus on the Gaseous Component**:
In this reaction, the only component that can change its concentration and affect the equilibrium is the gaseous product, \( \text{CO}_2 \). The solids (\( \text{CaCO}_3 \) and \( \text{CaO} \)) do not contribute to the equilibrium expression because their activities are considered constant.

4. **Establish the Condition for Equilibrium**:
The reaction reaches equilibrium when the partial pressure of \( \text{CO}_2 \) becomes constant. This means that the amount of \( \text{CO}_2 \) produced from the decomposition of \( \text{CaCO}_3 \) is equal to the amount of \( \text{CO}_2 \) that might react back to form \( \text{CaCO}_3 \).

5. **Conclusion**:
Therefore, the equilibrium state for the reaction is reached when the partial pressure of \( \text{CO}_2 \) is constant.

### Final Answer:
The reaction \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \) reaches the state of equilibrium when the partial pressure of \( \text{CO}_2 \) becomes constant.

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