Given
`i) A harr B+C K_(eq)(1)
ii) B+C harr P K_(eq)(2)`
then `K_(eq)` for reaction `Aharrp`is

Options

A. `K_(eq)(1).K_(eq)(2)`
B. `K_(eq)(1)//K_(eq)(2)`
C. `K_(eq)(1)+K_(eq)(2)`
D. `K_(eq)(1)-K_(eq)(2)`

Answer

To find the equilibrium constant \( K_{eq} \) for the reaction \( A \rightleftharpoons P \), we will use the given reactions and their equilibrium constants.

### Step-by-Step Solution:

1. **Write the Given Reactions:**
- Reaction 1: \( A \rightleftharpoons B + C \) with \( K_{eq} = K_{eq}(1) \)
- Reaction 2: \( B + C \rightleftharpoons P \) with \( K_{eq} = K_{eq}(2) \)

2. **Reverse Reaction 2:**
- To find the equilibrium constant for \( A \rightleftharpoons P \), we need to manipulate the given reactions. First, we reverse Reaction 2:
\[ P \rightleftharpoons B + C \]
- The equilibrium constant for the reversed reaction is the reciprocal of the original:
\[ K_{eq} = \frac{1}{K_{eq}(2)} \]

3. **Add the Reactions:**
- Now, we can add the modified Reaction 2 to Reaction 1:
\[ A \rightleftharpoons B + C \quad (K_{eq} = K_{eq}(1)) \]
\[ P \rightleftharpoons B + C \quad (K_{eq} = \frac{1}{K_{eq}(2)}) \]
- When we add these two reactions, the \( B + C \) cancels out:
\[ A \rightleftharpoons P \]

4. **Calculate the Overall Equilibrium Constant:**
- According to the principle of equilibrium constants, when reactions are added, their equilibrium constants multiply:
\[ K_{eq} = K_{eq}(1) \times \frac{1}{K_{eq}(2)} \]
- Therefore, we can express the equilibrium constant for the reaction \( A \rightleftharpoons P \) as:
\[ K_{eq} = \frac{K_{eq}(1)}{K_{eq}(2)} \]

### Final Answer:
The equilibrium constant \( K_{eq} \) for the reaction \( A \rightleftharpoons P \) is:
\[ K_{eq} = \frac{K_{eq}(1)}{K_{eq}(2)} \]