For the following three reactions I, II and III, equilibrium constants are given :
I. `H_(2)(g) + 1/2O_(2)(g) rarr H_(2)O ....K_(1)`
II. `C_(2)H_(4)(g) + 3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(g) ....K_(2)`
III.`C_(2)H_(6)(g) + 7/2O_(2)(g) rarr 2CO_(2)(g) + 3H_(2)O(g) ...K_(3)` IV. `C_(2)H_(4) + H_(2) rarr C_(2)H_(6) ...K_(4)`
Which of the following relations is correct?

Options

A. `K_(4) = K_(1)K_(2)K_(3)`
B. `K_(4) = K_(2)K_(3)/K_(1)`
C. `K_(4) = K_(1)K_(2)/K_(3)`
D. `K_(4) = K_(1)^(2)K_(2)/K_(3)`

Explanation

To solve the problem, we need to analyze the given reactions and their equilibrium constants. We will derive the relationship between the equilibrium constants based on the transformations of the reactions.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Equilibrium Constants:**
- Reaction I: \( H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g) \) with equilibrium constant \( K_1 \)
- Reaction II: \( C_2H_4(g) + 3O_2(g) \rightleftharpoons 2CO_2(g) + 2H_2O(g) \) with equilibrium constant \( K_2 \)
- Reaction III: \( C_2H_6(g) + \frac{7}{2}O_2(g) \rightleftharpoons 2CO_2(g) + 3H_2O(g) \) with equilibrium constant \( K_3 \)
- Reaction IV: \( C_2H_4(g) + H_2(g) \rightleftharpoons C_2H_6(g) \) with equilibrium constant \( K_4 \)

2. **Combine Reactions I and II:**
- We will add Reaction I and Reaction II to create a new reaction.
- The combined reaction will be:
\[ H_2(g) + \frac{1}{2}O_2(g) + C_2H_4(g) + 3O_2(g) \rightleftharpoons H_2O(g) + 2CO_2(g) + 2H_2O(g) \]
- Simplifying the products gives:
\[ C_2H_4(g) + H_2(g) + \frac{7}{2}O_2(g) \rightleftharpoons 2CO_2(g) + 3H_2O(g) \]
- The equilibrium constant for this combined reaction is:
\[ K = K_1 \cdot K_2 \]

3. **Subtract Reaction III from the Combined Reaction:**
- Now, we need to subtract Reaction III from the combined reaction to isolate Reaction IV.
- Reaction III is:
\[ C_2H_6(g) + \frac{7}{2}O_2(g) \rightleftharpoons 2CO_2(g) + 3H_2O(g) \]
- When we subtract Reaction III from the combined reaction, we get:
\[ C_2H_4(g) + H_2(g) \rightleftharpoons C_2H_6(g) \]
- The equilibrium constant for this reaction (Reaction IV) is:
\[ K_4 = \frac{K_1 \cdot K_2}{K_3} \]

4. **Final Relationship:**
- Therefore, the relationship between the equilibrium constants is:
\[ K_4 = \frac{K_1 \cdot K_2}{K_3} \]

### Conclusion:
The correct relation among the equilibrium constants is:
\[ K_4 = \frac{K_1 \cdot K_2}{K_3} \]