For the equilibrium `CaCO_3 (s) Leftrightarrow CaO(s)+CO_2(g)` which of the following expression is correct?

Options

A. `K_p=([CaO].[Co_2])/([CaCO_3])`
B. `K_(p)=(p_(CaO) xx p_(CO_(2)))/(p_(CaCO_(3))`
C. `K_(p)=p_(CO_(2))`
D. `K_(p)=(p_(CaO)+p_(CO_(2)))/(p_(CaCO_(3))`

Answer

To solve the question regarding the equilibrium expression for the reaction:

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

we will follow these steps:

### Step 1: Identify the states of the reactants and products
In the given equilibrium reaction, we have:
- Calcium carbonate \(\text{CaCO}_3\) in solid state (s)
- Calcium oxide \(\text{CaO}\) in solid state (s)
- Carbon dioxide \(\text{CO}_2\) in gaseous state (g)

### Step 2: Understand the equilibrium constant expression
The equilibrium constant expression for a reaction is derived from the concentrations (or partial pressures) of the gaseous and aqueous species involved in the reaction. Solids and liquids do not appear in the equilibrium constant expression because their activities are considered to be equal to 1.

### Step 3: Write the equilibrium constant expression
For the reaction:

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

The equilibrium constant \(K_p\) will only include the gaseous component, which is \(\text{CO}_2\). Therefore, the expression for \(K_p\) is:

\[ K_p = P_{\text{CO}_2} \]

where \(P_{\text{CO}_2}\) is the partial pressure of carbon dioxide.

### Step 4: Calculate the change in the number of moles of gas (\(\Delta N_g\))
To find \(\Delta N_g\), we look at the number of moles of gaseous products and reactants:
- Moles of gaseous products = 1 (from \(\text{CO}_2\))
- Moles of gaseous reactants = 0 (since there are no gaseous reactants)

Thus,

\[ \Delta N_g = \text{Moles of products} - \text{Moles of reactants} = 1 - 0 = 1 \]

### Step 5: Conclusion
The correct expression for the equilibrium constant \(K_p\) for the given reaction is:

\[ K_p = P_{\text{CO}_2} \]