Two gaseous equilibria `SO_(2)(g)+(1)/(2)O_(2)(g)hArr SO_(3)(g) and 2SO_(3)(g) hArr 2SO_(2)(g)+O_(2)(g)` have equilibrium constants `K_(1) and K_(2)` respectively at 298 K. Which of the following relationships is correct?

Options

A. `K_(1)=K_(2)`
B. `K_(1)= sqrt(K_(2))`
C. `K_(1)= sqrt((1)/(K_(2)))`
D. `K_(2)=(1)/(K_(1))`

Answer

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

### Step-by-Step Solution:

1. **Write the Given Reactions**:
- Reaction 1: \( SO_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons SO_3(g) \) with equilibrium constant \( K_1 \).
- Reaction 2: \( 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \) with equilibrium constant \( K_2 \).

2. **Relate the Equilibrium Constants**:
- For the first reaction, the equilibrium constant \( K_1 \) is given by:
\[ K_1 = \frac{[SO_3]}{[SO_2][O_2]^{1/2}} \]
- For the second reaction, the equilibrium constant \( K_2 \) is given by:
\[ K_2 = \frac{[SO_2]^2[O_2]}{[SO_3]^2} \]

3. **Manipulate the First Reaction**:
- If we multiply the first reaction by 2 to match the stoichiometry of the second reaction, we get:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]
- The equilibrium constant for this modified reaction becomes:
\[ K' = K_1^2 \]

4. **Reverse the Modified Reaction**:
- Reversing this reaction gives:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
- The equilibrium constant for the reversed reaction is:
\[ K_{reverse} = \frac{1}{K'} = \frac{1}{K_1^2} \]

5. **Equate the Two Equilibrium Constants**:
- Now, we can equate this to \( K_2 \):
\[ K_2 = \frac{1}{K_1^2} \]

6. **Final Relationship**:
- Rearranging gives us:
\[ K_1^2 = \frac{1}{K_2} \quad \text{or} \quad K_1 = \frac{1}{\sqrt{K_2}} \]

### Conclusion:
The correct relationship between the equilibrium constants \( K_1 \) and \( K_2 \) is:
\[ K_1 = \frac{1}{\sqrt{K_2}} \]