`A+BhArrC+D" constant "=K_(1)`
`E+FhArrG+H" Constent "=K_(2)`
then C + D + E + F `implies` product. The constant of reaction will be :

Options

A. `K_(1)/K_(2)`
B. `K_(2)/K_(1)`
C. `K_(1)K_(2)`
D. None of these

Answer

To solve the problem, we need to determine the equilibrium constant for the reaction involving the species C, D, E, and F leading to products. We are given two equilibrium reactions with their respective equilibrium constants.

### Step-by-Step Solution:

1. **Identify the Given Reactions and Their Constants:**
- Reaction 1: \( A + B \rightleftharpoons C + D \) with equilibrium constant \( K_1 \).
- Reaction 2: \( E + F \rightleftharpoons G + H \) with equilibrium constant \( K_2 \).

2. **Reverse the First Reaction:**
- To find the equilibrium constant for the reaction involving \( C \) and \( D \), we need to reverse the first reaction:
\[ C + D \rightleftharpoons A + B \]
- The equilibrium constant for the reversed reaction becomes:
\[ K' = \frac{1}{K_1} \]

3. **Write the Second Reaction:**
- The second reaction remains unchanged:
\[ E + F \rightleftharpoons G + H \]
- Its equilibrium constant is \( K_2 \).

4. **Combine the Reactions:**
- Now, we can add the reversed first reaction and the second reaction:
\[ (C + D \rightleftharpoons A + B) + (E + F \rightleftharpoons G + H) \]
- This gives us:
\[ C + D + E + F \rightleftharpoons A + B + G + H \]

5. **Determine the New Equilibrium Constant:**
- When two equilibrium reactions are added, their equilibrium constants multiply:
\[ K_{\text{new}} = K' \cdot K_2 = \left(\frac{1}{K_1}\right) \cdot K_2 = \frac{K_2}{K_1} \]

6. **Final Answer:**
- Therefore, the equilibrium constant for the reaction \( C + D + E + F \) leading to products is:
\[ K = \frac{K_2}{K_1} \]