For `A+BhArrC+D` , the equilibrium constant is `K_(1)` and for `C+DhArrA+B`, the equilibrium constant is `K_(2)`. The correct relation between `K_(1)` and `K_(2)` is

Options

A. `K_(1)xxK_(2)=1`
B. `K_(1)xx(K_(2)-1)=0`
C. `K_(1)//K_(2)=1`
D. All of these

Explanation

To solve the problem, we need to analyze the two given chemical reactions and their equilibrium constants.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Equilibrium Constants**:
- The first reaction is:
\[ A + B \rightleftharpoons C + D \]
The equilibrium constant for this reaction is denoted as \( K_1 \).
- The second reaction is:
\[ C + D \rightleftharpoons A + B \]
The equilibrium constant for this reaction is denoted as \( K_2 \).

2. **Write the Expression for \( K_1 \)**:
- The equilibrium constant \( K_1 \) for the first reaction can be expressed as:
\[ K_1 = \frac{[C][D]}{[A][B]} \]
where \([C]\), \([D]\), \([A]\), and \([B]\) are the molar concentrations of the respective substances at equilibrium.

3. **Write the Expression for \( K_2 \)**:
- The equilibrium constant \( K_2 \) for the second reaction can be expressed as:
\[ K_2 = \frac{[A][B]}{[C][D]} \]

4. **Relate \( K_1 \) and \( K_2 \)**:
- Notice that the expression for \( K_2 \) is the inverse of the expression for \( K_1 \):
\[ K_2 = \frac{1}{K_1} \]
- This implies:
\[ K_1 K_2 = 1 \]

5. **Conclusion**:
- The correct relation between \( K_1 \) and \( K_2 \) is:
\[ K_1 K_2 = 1 \quad \text{or} \quad K_1 = \frac{1}{K_2} \]

### Final Answer:
The correct relation between \( K_1 \) and \( K_2 \) is \( K_1 K_2 = 1 \).