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 equilibrium constants for the two reactions provided.

### Step 1: Write down the given reactions and their equilibrium constants.

1. For the reaction \( A \rightleftharpoons B + C \), the equilibrium constant is given as:
\[ K_{eq1} = \frac{[B][C]}{[A]} \]

2. For the reaction \( B + C \rightleftharpoons P \), the equilibrium constant is given as:
\[ K_{eq2} = \frac{[P]}{[B][C]} \]

### Step 2: Rearrange the equations to express concentrations.

From the first equilibrium constant \( K_{eq1} \):
\[ [B][C] = K_{eq1} [A] \]

From the second equilibrium constant \( K_{eq2} \):
\[ [B][C] = \frac{[P]}{K_{eq2}} \]

### Step 3: Equate the expressions for \( [B][C] \).

Since both expressions equal \( [B][C] \), we can set them equal to each other:
\[ K_{eq1} [A] = \frac{[P]}{K_{eq2}} \]

### Step 4: Solve for \( K_{eq} \) for the reaction \( A \rightleftharpoons P \).

Rearranging the equation gives:
\[ [P] = K_{eq1} [A] K_{eq2} \]

Now, we can express the equilibrium constant \( K_{eq} \) for the reaction \( A \rightleftharpoons P \):
\[ K_{eq} = \frac{[P]}{[A]} = \frac{K_{eq1} [A] K_{eq2}}{[A]} \]

This simplifies to:
\[ K_{eq} = K_{eq1} \cdot K_{eq2} \]

### Final Answer:
Thus, the equilibrium constant \( K_{eq} \) for the reaction \( A \rightleftharpoons P \) is:
\[ K_{eq} = K_{eq1} \cdot K_{eq2} \]

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