If `K_(1)` and `K_(2)` are the equilibrium constants of the equilibria (a) and (b) respectivel, what is the relationship between the two constants ?
(a) `SO_(2)+(1)/(2) O_(2)(g)hArr SO_(3)(g),K_(1)`
(b) `2SO_(3)(g) hArr 2SO_(2)(g)+O_(2)(g),K_(2)`

Options

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

Explanation

To find the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) for the given reactions, we need to analyze the reactions step by step.

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

1. Reaction (a):
\[ SO_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons SO_3(g) \quad K_1 \]

2. Reaction (b):
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \quad K_2 \]

### Step 2: Relate the two reactions.

Notice that reaction (b) can be derived from reaction (a) by reversing the reaction and multiplying the entire equation by 2.

- Reversing reaction (a):
\[ SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \]
The equilibrium constant for the reverse reaction is given by:
\[ K' = \frac{1}{K_1} \]

- Now, if we multiply the reversed reaction by 2:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
The equilibrium constant for this reaction (which is \( K_2 \)) is related to \( K' \) by squaring it:
\[ K_2 = (K')^2 = \left(\frac{1}{K_1}\right)^2 = \frac{1}{K_1^2} \]

### Step 3: Rearranging the equation.

From the relationship derived, we can express \( K_1 \) in terms of \( K_2 \):
\[ K_1^2 = \frac{1}{K_2} \]
Taking the square root of both sides gives:
\[ K_1 = \sqrt{\frac{1}{K_2}} = \frac{1}{\sqrt{K_2}} \]

### Conclusion:

The relationship between the two equilibrium constants is:
\[ K_1^2 = \frac{1}{K_2} \]