According to law of mass action, for `CaCO_(3(s)) Leftrightarrow CaO+CO_(2)) (R_(r)"=Rate of forward and "R_b= "Rate of backward reaction").` Which of the following is true at equilibrium?

Options

A. `R_(b)=K_(b) [CaCO_(3)]^(2)`
B. `R_(f)=K_(f)[CaO_(3)]^(2)`
C. `R_(f)=K_(b) [CO_(2)]`
D. `R_(f)/R_(b)=[CO_(2)]^(1)`

Answer

To solve the problem regarding the equilibrium of the reaction \( \text{CaCO}_3 (s) \leftrightarrow \text{CaO} (s) + \text{CO}_2 (g) \), we will analyze the situation step by step.

### Step 1: Understand the Reaction
The reaction involves solid calcium carbonate (\( \text{CaCO}_3 \)) decomposing into solid calcium oxide (\( \text{CaO} \)) and gaseous carbon dioxide (\( \text{CO}_2 \)).

### Step 2: Identify the Rates of Reaction
Let:
- \( R_f \) = Rate of the forward reaction (decomposition of \( \text{CaCO}_3 \))
- \( R_b \) = Rate of the backward reaction (formation of \( \text{CaCO}_3 \) from \( \text{CaO} \) and \( \text{CO}_2 \))

### Step 3: Apply the Law of Mass Action
According to the law of mass action, at equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction:
\[ R_f = R_b \]

### Step 4: Consider the Role of Solids
In equilibrium expressions, the concentrations of pure solids and liquids are not included. Therefore, the concentration of \( \text{CaCO}_3 \) and \( \text{CaO} \) (both solids) will be considered constant and equal to 1. This means:
\[ R_f = k_f \cdot [\text{CaCO}_3] = k_f \cdot 1 = k_f \]
\[ R_b = k_b \cdot [\text{CaO}] \cdot [\text{CO}_2] = k_b \cdot 1 \cdot [\text{CO}_2] = k_b \cdot [\text{CO}_2] \]

### Step 5: Set the Rates Equal
At equilibrium:
\[ k_f = k_b \cdot [\text{CO}_2] \]

### Step 6: Conclusion
From the above relationship, we can conclude that the concentration of \( \text{CO}_2 \) is directly related to the rate constants of the forward and backward reactions. This indicates that the equilibrium condition is dependent on the concentration of the gaseous product \( \text{CO}_2 \).

### Final Answer
Thus, the correct statement at equilibrium is that the rate of the forward reaction equals the rate of the backward reaction, and it can be expressed as:
\[ R_f = R_b \]

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